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Nonequilibrium Thermodynamics of Cell Signaling
Journal of Thermodynamics
Volume 2012 (2012), Article ID 432143, 10 pages
Research Article
Nonequilibrium Thermodynamics of Cell Signaling
Computational Genomics Department, National Institute of Genomic Medicine, Periférico Sur 4809, Col. Arenal Tepepan, Delegación Tlalpan, 14610 Mexico City, DF, Mexico
Received 12 March 2012; Revised 7 June 2012; Accepted 9 June 2012
Academic Editor: Ali-Akbar Saboury
Copyright © 2012 Enrique Hernández-Lemus. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
Signal transduction inside and across the cells, also called cellular signaling, is key to most biological functions and is ultimately related with both life and death of the organisms. The processes
giving rise to the propagation of biosignals are complex and extremely cooperative and occur in a far-from thermodynamic equilibrium regime. They are also driven by activation kinetics strongly
dependent on local energetics. For these reasons, a nonequilibrium thermodynamical description, taking into account not just the activation of second messengers, but also transport processes and
dissipation is desirable. Here we present a proposal for such a formalism, that considers cells as small thermodynamical systems and incorporates the role of fluctuations as intrinsic to the dynamics
in a spirit guided by mesoscopic nonequilibrium thermodynamics. We present also a minimal model for cellular signaling that includes contributions from activation, transport, and intrinsic
fluctuations. We finally illustrate its feasibility by considering the case of FAS signaling which is a vital signal transduction pathway that determines either cell survival or death by apoptosis.
1. Introduction
Survival of living organisms is intimately linked to their ability to react quite efficiently to even extremely weak external signals. Common examples are the reaction of the human eye to single
light photons [1, 2], the reaction of a male butterfly to a single pheromone molecule coming from a female at a distance that sometimes is in the order of kilometers [3], and so forth. Cellular
receptors react to hormones, cytokines, or antigens at very low concentrations. This strong reaction to a weak impulse is attained by an amplification process which is performed by means of special
pathways of free energy transduction. Mechanisms such as immune system response, thermal-shock inhibitions, and cardiovascular rearrangement in response to environmental changes are all mediated by
signaling processes. Signal transduction (information flow) is, thus, equally important, if not more important, for the functioning of a living organisms than metabolism and energy flow.
Signal transduction or cell signaling is the generic name of the set of concatenated processes or stages in which a cell transforms a certain signal or stimulus—either intercellular or
intracellular—into another signal or a specific response. Cell signaling affects the complex arrangement of biochemical reactions inside the cell that takes place by means of enzymes that are bounded
to other molecules called second messengers. Each process takes place in fast times, with dynamic ranges between a few milliseconds in most cases, to a few seconds in the case of more complex
signaling cascades. Intricate and very sensible molecular biology experiments have shown cell signaling to be rate processes, that is, kinetic-guided phenomena determined by previous systems settings
The wide variety of physicochemical signals to which cells may respond may seem to imply on a wide range of signal transduction mechanisms. However, only a handful of event chains is able to generate
proper response to every stimulus in different cell subtypes which points to generalistic strategies commonly beginning with the action of cell receptors. Many signal transduction processes are then
usually started by the adhesion of a ligand protein to a membrane receptor that then activates either itself or other receptor (or series of receptors) thus converting the initial stimulus into a
response that once inside the cell provokes a chain of biochemical events known as a signaling cascade or second messenger pathway which results in the amplification of the signal.
The archetypal example here is that of the epinephrine cascade. It is known that epinephrine (adrenaline) stimulates the liver to convert glycogen to glucose in liver cells, but epinephrine alone
would not convert glycogen to glucose. In an outstanding experimental tour de force that granted him the 1971 Nobel Prize in Physiology or Medicine, Earl Sutherland found that epinephrine had to
trigger a second messenger, cyclic AMP, for the liver to convert glycogen to glucose [5]. Secondary messenger systems can be synthesized and activated by the action of enzymes, for instance, cyclases
that synthesize cyclic nucleotides. Second messengers also form by opening of ion channels to allow influx of metal ions (e.g., calcium signaling). These second messengers then bind and activate
protein kinases, ion channels, and other proteins continuing the signaling cascade.
The role that activation kinetics and other energy-driven dynamic processes play in cell signaling makes evident the need for a thermodynamic description. Most studies to date are based in
equilibrium thermodynamics assumptions [6, 7] or, in any case, coarse-grained approaches [8]. Specific applications of nonequilibrium thermodynamics have been studied in the past [9–13] focusing on
single features such as switching, sensitivity, and controllability. Thus, a nonequilibrium thermodynamics analysis of cell signaling, describing transport processes, activation kinetics and nonlocal
effects is desirable.
Some particular cases of signaling dynamics have been studied, even at the nonequilibrium statistical physics level of description, for instance, by means of information theoretical approaches [14].
In such studies, a positive correlation between the channel capacity (i.e., the information-carrying capacity of the signaling networks) and free-energy expenditure has been observed. For
phosphorylation-dephosphorylation switches, hydrolysis-free energy is in the sustained high concentration of ATP and low concentrations of ADP, that is, away from thermochemical equilibrium. This
deviation from equilibrium implies, among other things, that useful hydrolysis-free energy does not come from the phosphate bond of the ATP molecule alone but from more complex-systemic mechanisms.
Nonequilibrium thermodynamic entropy and entropy production have been studied, to gain dynamic signaling information transfer, in insulin transduction [11]. In that case, entropy production rates
show a broad secondary peak in time that represents a possible evidence of the decrease of the concentration of membrane GLUT4 (a so called backflow), thus to the reduction of insulin efficiency.
Interestingly, at least in that case entropic contributions take a leading role in controlling signaling efficiency. Pathway selectivity driven by receptor-receptor interaction has also been studied
by means of thermodynamic models [8]. Ligand-induced oligomerization of cell-surface receptors is driven by cooperative behavior. Oligomerization occurs due to interaction between nearest-neighbor
receptors. This type of cooperativity can exhibit a first-order phase transition, corresponding to a jump in the surface density of ligand-receptor complexes. Clustering could be described by the
statistical mechanics of a simple lattice Hamiltonian. Receptor-receptor interaction may lead to a first-order phase transition with a discontinuous jump in the receptor density as a function of the
receptor chemical potential and/or the ligand concentration [8].
Thermodynamical studies of biomolecular switches could be quantitatively described by a simple 3-state population-shift model, in which the equilibrium between a nonbinding, nonsignaling state and
the binding-competent, signaling state is shifted toward the latter upon target binding. Performance of biomolecular switches can be sensibly tuned via mutations that alter their switching
thermodynamics [7]. Thermodynamic conditions in the intracellular medium hence alter sensitivity, control, and effective information transfer in signaling networks [14].
Moreover, as we may see later, typical settings in signal transduction correspond with complex nonequilibrium stages [15]. In fact, even relatively simple signaling models such as the
phosphorylation-dephosphorylation switches exhibit bistability due to feedback, and the related nonequilibrium steady state even presents a phase transition [16]. Such complex behavior led some
researchers to propose that non-linear deterministic biochemical behavior is dynamically trapped between stochastic dynamics, both at the molecular signaling level and at the cellular evolution level
[16]. This proposition raised from an analysis of the so-called chemical master equation which is founded in the tenets of non-linear nonequilibrium thermodynamics [10, 17].
Thermodynamic models of cell signaling aim to model and describe these phenomena at the basic level, and applications of thermodynamic modeling in search of therapeutic action have been recently
developed [18]. Claims have been made that fast binding kinetics was advantageous for most targets with a couple of exceptions, that targeting some protein kinases could enhance rather than attenuate
the pathway, and that therapeutic doses could be sensitive to the kinetic parameters of drug binding. Thermo-kinetic rates have been shown to play an important role in the dynamics of signaling and
immune response. Plasmon resonance-based thermodynamics points out to slow-signaling modified second-messenger variants have similar affinities but distinctly faster dissociation rates that compared
with the original messengers and that this may be behind their lower activity. Signaling deregulation could be starting not at the biochemical (recognition) but at the thermodynamical (dissociation
rates) levels [19]. In fact, thermodynamic studies are now part of the drug-design tools of pharmaceutical chemistry. In fact, thermodynamic and kinetic analyses are sources of deeper insight into
specificity of molecular recognition processes and signaling [20]. Such advances had led to research efforts combining statistical thermodynamic models in combination with experimental data [21].
Preliminary results of these studies are very promising.
A proposal for a nonequilibrium thermodynamics formalism including the role of fluctuations as intrinsic to the dynamics, and the role of transport processes is made in this work. We detail a minimal
model for cellular signaling that considers activation, transport, and intrinsic fluctuations.
The rest of the paper is organized as follows: in Section 2 we discuss the role that stochastic fluctuations and transport processes play in biological signal transduction, in Section 3 we develop a
nonequilibrium thermodynamics formalism of cell signaling and from it we derive a minimalist model, Section 4 deals with the biology of direct FAS signaling in apoptosis that in its simplest version
(here presented) is akin to our minimalist model, finally in Section 5 we discuss some potential applications, the scope and limitations of our formalism as well as some perspectives.
2. Cell Signaling and Stochasticity and Nonlocality
One source of complexity in the nonequilibrium thermodynamical characterization of cell signaling is the fact that a cell is a small system; that is, cellular dimensions do not permit the immediate
application of the thermodynamic limit. Specifically, the role that fluctuations and stochasticity may play within such scenarios is not completely clear. A formalism to study small systems
thermodynamics in equilibrium has been developed [22, 23], and some results were even expected to extend to local equilibrium settings within cellular sized biosystems [24]. However, one important
drawback in completing such theoretical frameworks at that time was the lack of proper experimental settings to test their hypotheses. Nevertheless, with the development of modern techniques, such as
microscopic manipulation by means of atomic force microscopes, optical tweezers, and cold traps, this situation has been overcome at least partially. Theories have been developed including mesoscopic
thermodynamical approaches [25–29] and also studies made by means of fluctuation theorems [30–33]. Some of these theoretical results have been even experimentally tested.
Due to the low copy number of many reactants in cells, and the nonequilibrium nature of the many intracellular reactions, signal transduction may result from stochastic intracellular events.
Distribution analyses of cell responses provide a means to probe the stochastic character of intracellular signaling. A goal is to determine the class of stochasticity that affects intracellular
pathways [34]. Stochasticity has been measured experimentally, it has been also incorporated in molecular simulations, and it has been discovered that locality and Gaussian behavior are not always
present. In fact, transient multipeak distributions have been observed in computer simulations of cell-signaling dynamics. The emergence of these complex distributions cannot be explained using
either deterministic chemical kinetics or simple Gaussian noise approximation [35].
Multipeak distributions are typically transient and eventually evolve into single-peak distributions in certain cases these distributions may be stable in the limit of long times. It has also been
shown that introducing positive feedback loops results in diminution of the probability distribution complexity. This effect is so strong that even stochastic resonance has been reported in signaling
cascades [36] where certain optimal reaction rates minimize the average threshold-crossing time. A noisy signal reaches the threshold more easily when the upstream and downstream reaction time scales
are related in a specific way, indicating the existence of internal resonances embedded in cellular signaling cascades [36]; that is, nonequilibrium thermodynamic couplings exist between different
modes (as characterized by their corresponding relaxation times) a feature that can be accounted by certain nonequilibrium thermodynamics formalisms (see next section). This may seem to point out on
how the rates of various nodes could be collectively tuned in protein-signaling networks in such a way that signals are optimally picked up and biological information is transmitted through the
signaling cascade.
3. Irreversible Thermodynamics of Cell Signaling
As we have just pointed out, systems outside the thermodynamic limit are characterized by large fluctuations and hence stochastic effects. The classic thermodynamic theory of irreversible process
(also called linear irreversible thermodynamics, LIT) [37] provides us with a coarse-grained description of the systems, thus ignoring the molecular nature of matter studying it as a continuum media
by means of a phenomenological field theory. As such LIT is not suitable for the description of small systems because it ignores fluctuations that could become the dominant factor in the system's
response. Nevertheless, in many instances, it would be desirable to have a thermodynamic theoretical framework to study small systems, most noticeably in cellular and subcellular processes like
signal transduction. One way to do so is by considering the stochastic nature of the time evolution of small nonequilibrium systems. This is the approach of Mesoscopic Nonequilibrium Thermodynamics
(MNETs) [26]. MNET for small systems can be understood as the extension of the equilibrium thermodynamics of small systems developed by Hill and Chamberlin [22] and Hill [23, 24].
MNET, for instance, was developed to analyze nonequilibrium small systems. Any reduction of the spatio-temporal scale description of a system implies an increase in the number of noncoarse-grained
degrees of freedom. These degrees of freedom could be related with the extended variables in extended irreversible thermodynamics (EITs) [38]. In order to characterize such variables, let us say that
there exist a set of such non-equilibrated degrees of freedom. is the probability that the system is at a state given by at time . If one assumes [27] that the evolution of the degrees of freedom
could be described as a diffusion process in -space, then the corresponding Gibbs equation could be written as is a generalized chemical potential related to the probability density, whose
time-dependent expression could be explicitly be written as or in terms of a nonequilibrium work term as follows:
The time-evolution of the system could be described as a generalized diffusion process over a potential landscape in the space of mesoscopic variables . This process is driven by a generalized
mesoscopic-thermodynamic force whose explicit stochastic origin could be tracked back by means of a Fokker-Planck-like analysis [26, 27]. MNET seems to be a good candidate theory for describing
nonequilibrium thermodynamics for small systems, provided that one has a suitable model or microscopic means to infer the probability distribution .
MNET and similar approaches are appropriate to deal with activated processes, like a system crossing a potential barrier. Biochemical reactions like the ones involved in signal transduction are
clearly in this case. According to [27] the diffusion current in this space could be written in terms of a local fugacity defined as and the expression for the associated flux it will be
is an Onsager-like coefficient. After defining a diffusion coefficient and the associated affinity , the integrated rate is given as with .
MNET then gives rise to nonlinear kinetic laws like (6). MNET has been applied successfully to biomolecular processes at the cellular level or description [28]. Non-linear kinetics are used to
express, for example, RNA unfolding rates as diffusion currents, modeled via transition state theory, giving rise to Arrhenius-type non-linear equations. In that case the current was proportional to
the chemical potential difference, so the entropy production was quadratic in that chemical potential gradient.
Signal transduction consists of a series of biochemical reactions, and many of these have unexplored chemical kinetics, due to this fact a detailed MNET analysis such as the one described above is
unattainable at the present moment. On what follows, we will explore a phenomenologically based approach that takes into account similar considerations as the MNET framework already sketched but does
so in a more informal, modeling-like manner. This phenomenological approach is based on the EIT assumption of enlargement of the thermodynamical variables space [39, 40].
Assuming that a generalized entropy-like function exists, we can write down a Gibbs equation which may be written in the following form [38, 41]: or as a differential form
Quantities are defined as usual, is the internal energy per mol, T the absolute temperature, p the pressure, the molar volume, is the chemical potential for the species, its concentration (mole
fraction), the molar chemical affinity for the reaction producing species, (i.e., being the stoichiometric coefficient for the th species in the th reaction and the corresponding chemical potential),
the reaction coordinate for the production of species, , , and are extended fluxes and forces for diverse processes, and is the appropriate scalar product.
Here we are considering the presence of thermal processes, but also the energetics of three different contributions due to signal transduction: the effect of bulk chemical potentials related to
concentration changes of the signaling molecules in the cellular environment (identified with the subscript ), activation kinetics (considered as generalized chemical reactions) related to the
chemical affinities between ligand proteins, membrane receptors and effector proteins in the signaling cascade (identified with the subscript ), and generalized transport processes (including the
effects of nonlocal dynamics and delays) considered as extended variables or generalized fluxes and forces (identified with the the subscript ).
3.1. Activation Kinetics
We will introduce a simple—although general—model for signal transduction including the action of ligand proteins (LPs), membrane receptors (MRs), effector proteins (EPs), and finally response
proteins (RPs). In this idealized model, LPs and MRs play the role of pulls and triggers, then a series of EP steps (not necessarily, but possibly sequential) constitute the core of the signaling
cascade and finally, and the RPs when activated constitute the cell’s response to the initial stimulus. The pseudo-chemical reactions could be written as follows:
Here the superscript refers to the activated form of the molecule, that is, the form which presents the corresponding biological signaling activity. For a pictorial representation, please refer to
Figure 1.
3.2. Generalized Transport Processes
Let us recall (8). If we write down explicit expression for second and third terms in the r.h.s. of (8) in the context of signal transduction, we have the following generalized Gibbs form:
Since signal transduction occurs within the cell, it is possible to relate an internal work term with the regulation process itself, being this a far-from equilibrium contribution. This nonlocal
contribution is given by the generalized force-flux term (last term in the r.h.s. of (14)). This is so as cell signaling often does not occur in situ and also since is the only way to take into
account (albeit indirectly) the changes in the local chemical potentials that cause the long tails in the fluctuations distributions characteristic of nonequilibrium small systems (e.g., cells). The
term relating second messenger flows due to transduction could be written as a product of extended fluxes and forces . Here refers to the different second messenger species involved.
3.3. A Minimalist Model of Signal Transduction
In order to present a full detail of the different energetic contributions in (14), let us consider a minimalist model consisting of just four molecules under isothermal conditions: one triggering
ligand protein (LP), one membrane receptor (MR), one effector protein (EP), and one response protein (RP). In such case we have the following 3 pseudoreactions:
That will give rise to the following form of (14):
Equation (18) considers the energies of formation for four molecules (as given by the ’s), the energies of activation of three species (as given by the chemical affinities ’s) as well as the energies
related with transport of the active species, given by their respective thermodynamic forces (’s). If we now refer to (6) for the definition of signaling fluxes [25, 28], we can write down
expressions for the ’s, namely,
Hence their temporal derivatives are given by:
If we consider, as it is often done in irreversible thermodynamics, that the thermodynamic forces ’s are proportional to the fluxes ’s, with proportionality constant , we have
We now define the following generalized transport coefficients:
By substitution of (22) to (28) in (18) we have
Equation (29) gives a complete irreversible thermodynamical description of the minimal model given by (15) to (17). The model is then to be supplemented with the appropriate constitutive relations;
in this case, the time evolution for the concentrations, chemical potentials, and chemical affinities as given by biochemical kinetics.
The free-energy coupling given by the corresponding generalized Maxwell relations (since is an exact differential form, integrability conditions imply the existence of Maxwell-like relations [41]),
as well as Gibbs-Duhem constrains (not all the concentrations and chemical potentials are independent) once explicit kinetics are given, constitutes the energetic core behind the complex processes of
signal transduction. This is possibly the key contribution of this work, the explicit derivation from a nonequilibrium thermodynamics formalism showing that cell signaling control is, indeed, an
energy-driven process. Of course, free-energy transduction has been known to be responsible for the initiation of signaling cascades. However, our model has shown that every step of the process is
controlled and locked via the local chemical potentials even in the presence of stochasticity and fluctuations, provided that the assumptions of MNET hold.
4. Case Study: FAS Signaling in Apoptosis
An important family of signal transduction pathways is related with the onset and regulation of programmed cellular death or, apoptosis. Any functional disruption in the balance that apoptotic cells
encounter may affect death signaling thus leading to diseases ranging from cancer in the case of subnormal apoptosis to degenerative disorders in supernormal apoptosis. Hence the control of the
process as given by signal transduction pathways is of foremost relevance. One of the simplest example of such pathways is apoptosis regulation by FAS signaling. FAS is a cell-surface receptor
protein that when triggered by an stimulus induces apoptosis in FAS-expressing cells. This process is highly linked with immune response, as the ligand for FAS, FAS-L, is mostly present on cytotoxic
T cells and TH1 cells central players in innate immunity. FAS is composed of an extracellular region, one transmembrane domain, and an intracellular region. FAS activity is governed by interaction
with its ligand (FAS-L). Activation of FAS through binding to its ligand or FAS antibody induces apoptosis, which has been confirmed by many experiments [42].
In order for signal transduction to occur, cross-linking of FAS with its ligand must occur. FAS trimerizes to properly bind to its ligand, which exists as a trimer. This creates a clustering of FAS
that is necessary for signaling. In its intracellular region, FAS contains a conserved sequence deemed as death domain. An adaptor protein, FADD, interacts with the death domain on the FAS receptor.
Subsequent binding to another region of FADD by procaspase 8 promotes grouping of pro-caspase 8 molecules bound to each of the clustered FADD proteins. This entire cluster is sometimes called a
death-inducing signaling complex, or DISC [43]. Pro-caspase 8 transactivates itself once grouped, cleaving and releasing active caspase 8 molecules intracellularly.
As is clear from Figure 2, there is a correspondence between the model given by (15) to (17) and direct FAS signaling (Figure 2). In this case the ligand protein (LP) is the FAS-L molecule, the
membrane receptor (MR) is FAS-R that when activated () becomes FAS and then interacts with the effector protein (EP), in this case FADD, that carries the biosignal activating procaspase-8 (a response
protein) that when activated () becomes caspase-8, the molecule responsible for the no-return apoptotic response leading to cellular death.
FAS signaling is a well-characterized process [45], some thermodynamic parameters may be thus obtained by experiments [46, 47] or by means of molecular simulations [48]. This is the case of
activation energies—especially when activation occurs by means of ATP produced by oxidative phosphorylation—and free energies of formation. However, transport processes have not been measured
accurately (and in most cases have not been even measured at all). Being signaling pathways so important for the understanding of cell function, and in many instances for their biomedical importance
as pharmacological targets; we hope that this situation soon will change. At the present moment, some insight on particular signaling pathways may be obtained by molecular dynamics simulations [49–54
Experimental techniques have been refined that allow thermomolecular characterization of signaling processes. The technical challenges are, however, gigantic. Cell-signaling thermodynamic parameters
must be experimentally measured by combining many different methodologies involving different scales of description: protein-protein electrostatic interactions, the electrohydrodynamic effect of the
medium, cleavage and protein structure, free energies of folding/unfolding, transport processes, and so forth. Nonetheless, progress is being made in the actual realization of such experimental
challenges, by using a clever combination of surface plasmon resonance (SPR), isothermal titration calorimetry (ITC), and ultracentrifugation (UC) of the thermodynamics of T-cell signaling in the MHC
pathway have been unveiled [55].
SPR is extensively used to study receptor/ligand binding both qualitatively and quantitatively. However, results are commonly ambiguous, and every conclusion needs to be independently verified. SPR
provides both kinetic and equilibrium data; data acquisition is fast, comparative studies are easily performed, and low affinities can be detected with relatively low amounts of protein. The accuracy
of the SPR-derived kinetic constants depends crucially on other various parameters such as mass transport, sensor chip capacity, and flow rate. In conclusion, SPR experiments are useful but partial
and sometimes even dubious. In contrast, equilibrium methods, such as ITC, often result in more reliable, specially when used in conjunction with SPR and analytical techniques as ultracentrifugation
in which sedimentation velocity, and equilibrium experiments provide insight into the hydrodynamic and thermodynamic properties of the sample, thus enabling the inference of transport parameters and
free energies via association constants; on the other hand, SPR experiments could shed some light on biochemical kinetics and their associated relaxation times [55]. ITC has also used in the
experimental analysis of the interaction between TRAF and tumor necrosis factor receptors [56].
FAS signaling proceeds by typical physicochemical mechanisms. Being this the case, common ranges for the parameters in cell signaling may be used as proxy values instead. For instance, the
characteristic concentration of signaling protein molecules ranges from 0.01 to 1-molar, with molecule counts between 120,000 and 20,000,000 depending on molecular weights and type of cell [57]. RAS
concentration in HeLa cells has been measured to be 0.4-molar [58]. Thresholding signal duration times for whole processes range between 2 minutes and 24 hours. NF-B signaling (which is related with
FAS signaling) takes about 320 minutes in epithelial cells [59, 60].
In order to figure out the order of magnitude of kinetic parameters, let us consider the case of the values of fluxes and dissociation constants in the Wnt-signaling pathway [61]. Dissociation
constants for several second messenger molecules range around 10–1200nM, with protein concentrations in the 15 to 100nM regime. The degradation flux of -catenin via the proteasome is 25nM/h. The
characteristic time of the associated phosphorylation-dephosphorylation switch is 2.5 minutes for APC. Relaxation times for GSk3- association/dissociation is 1 minute, and that of Axin degradation is
6 minutes [61]. Decay rates (half-life times) , for signaling molecules in the MAPK pathway and the STAT pathway, are valued between 0.75 and 24h [57]. More closely related with FAS signaling,
duration times on switch of apoptosis and duration of apoptotic death in HeLa cells exposed to different levels of TRAIL are between 19 and 27 minutes for switching and between 140 and 660 minutes
for cell death [62]. Rate constants for diffusion-limited enzymes may vary around and [63] although there are other kinetic mechanisms for second messengers that in some cases seem to be cell-type
specific [64], these figures may serve as reference to infer chemical kinetic behavior since the general behavior seems to be quite common [65]. Once the rate constants are measured, one can infer
activation energies from them, following a kinetic model [66].
In relation to energetics, it is known that ATP hydrolysis releases between 28 and 33.5kJ/mol [67] depending on cell type and condition whereas for other energy-rich compounds involved in substrate
level phosphorylation range around 23.44 and 88kJ/mol [68]. Free energy profiles may also help us to understand the role that protein-coupling plays in cell signaling. In Ras signaling, for
instance, binding free energy determines the fate of Ras/Raf dynamics [69]. Diffusion coefficients measured inside the cell differ according to cell medium and molecule size. Inside the cell nucleus
typical diffusion constants vary between 10 and 100 [70]. In reference with signaling proteins, this is usually also the case even in cytoplasm and/or aqueous solution. The diffusion rate of
phosphoglycerate kinase (around 45kDa) has been measured as 63.8 [71], while heavier molecules as 62kDa Dextran move slowly, around 39 [72]. Smaller second messenger molecules like insulin
(5.808kDa) can diffuse much faster, [73]. The combination of different experimental/computational modeling techniques and estimated parameters just sketched may allow to construct quantitative
thermodynamic models for cell signaling, following the lines of the present work, in the near future.
5. Discussion
Signaling transduction is a quite complex yet extremely important physicochemical phenomenon in cell biology. As we have seen cellular signaling is characterized by a combination of stochastic
effects, activation biochemical kinetics, and multiple transport processes all setup in a far-from thermodynamic equilibrium setting. Is is known that free-energy transduction plays a key role in the
process of signaling cascades. For this reason a nonequilibrium thermodynamics description at the mesoscopic level is desirable. In the present work we have presented such a formalism in the context
of MNET [26].
The role of stochasticity is taken into account (albeit in an indirect way) by means of incorporating the probability distribution for the nonequilibrated degrees of freedom into a generalized
chemical potential as is described in (2) to (6). In this scheme, the thermodynamic forces (equations (25) to (27))—that reflect in a coarse grained way the effect of stochasticity—are identified as
the gradients in the space of mesoscopic variables of the logarithm of the ratio of the probability density to its equilibrium value. The main idea is to generalize the definition of the chemical
potential to account for these additional mesoscopic variables. Thus it is possible to assume that the evolution of these degrees of freedom is described by a diffusion process and formulate the
corresponding Gibbs equation.
The effect of generalized transport processes related with the distribution of relaxation times of the kinetics (it takes some (in general different) time for every biochemical reaction to activate
the corresponding signaling molecule) and nonlocalities in the molecular processes (i.e., a second messenger has to travel some distance, say by diffusion, until it reaches its target molecule) is
given by the last term at the r.h.s. of (14). In particular, relaxation times for the coarse-grained processes are given in terms of generalized transport coefficients (28). Our formalism is written
in such a way that we can distinguish between local equilibrium effects (corresponding to the energetics of non-activated molecules at the top of the signaling cascade, as given by the first 4 terms
at the r.h.s. of (29)), deterministic activation kinetics depending exclusively on the rate equations for the chemical reactions (corresponding to the 5th to 7th terms at the r.h.s. of (29)), and
far-from equilibrium effects, involving both stochasticity and transport processes (terms 8th–13th at the r.h.s. of (29)) involving the dynamics of the evolution of nonconserved variables. We could
think of this structure as a hierarchy in which trains of signals are coupled with each other via their relative relaxation times (as given by the corresponding generalized transport coefficients, (
The potential application of such a formalism is wide, in particular with respect to the detailed study of the dynamics for important biological pathways. Consider, for instance, the extremely
important scenario of calcium signaling. Phenomena like calcium waves [74, 75] and calcium-induced calcium release [76] could be understood more clearly (and even modeled and simulated) in the light
of a nonequilibrium thermodynamic description like the one presented above. For instance, the role of energy releasing pathways in the dynamics and control of cell signaling under a system
biology-like philosophy becomes almost crystal clear. In turn these free-energy triggers may be appropriate candidates for pharmacological targets for drug-therapy in cases of diseases associated
with abnormal signaling. This may be the case of cardiac arrhythmias, neurological disorders [75, 77], and metabolic diseases [76].
Of course, such general modeling strategy has some drawbacks. On the one hand being a fully thermodynamical description, this framework depends entirely on experimental data for the activation
kinetics and other constitutive relations or in any case in a good set of molecular simulations. On the other hand, ultrafast kinetics may be accompanied by noncompensated stochasticity that could
not be handled entirely under the MNET paradigm. In conclusion, much work has still to be done in order to establish the validity and feasibility of these physical models into biological and
biomedical research.
The author gratefully acknowledges support by Grant PIUTE10-92 (ICyT-DF) (Contract 281-2010) as well as federal funding from the National Institute of Genomic Medicine (México). He also acknowledges
suggestions of the anonymous reviewers that, no doubt, contributed greatly to improve this work.
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How to Recognize Recursive Arithmetic Sequences
A recursive sequence is an arithmetic sequence in which each term depends on the term(s) before it; the Fibonacci sequence is a well-known example. When your pre-calculus teacher asks you to find any
term in a recursive sequence, you use the given term (at least one term, usually the first, is given) and the given formula that allows you to find the other terms in the sequence.
You can recognize recursive sequences because the given formula typically has a[n] (the nth term of the sequence) as well as a[n][ – 1] (the term before the nth term of the sequence). In these
sequences, you're given a formula (a different one for each sequence), and the directions ask you to find the terms of the sequence.
For example, the most famous recursive sequence is the Fibonacci sequence, in which each term after the second term is defined as the sum of the two terms before it. The first term of this sequence
is 1, and the second term is 1 also. The formula for the Fibonacci sequence is
So if you were asked to find the next three terms of the sequence, you'd have to use the formula as follows:
a[3] = a[3 – 2] + a[3 – 1] = a[1] + a[2] = 1 + 1 = 2
a[4] = a[4 – 2] + a[4 – 1][ ]= a[2] + a[3] = 1 + 2 = 3
a[5] = a[5 – 2] + a[5 – 1] = a[3] + a[4] = 2 + 3 = 5
The first ten terms of this sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55. It is very famous because many things in the natural world follow the pattern of the Fibonacci sequence. For examples, the
florets in the head of a sunflower form two oppositely directed spirals, 55 of them clockwise and 34 counterclockwise; lilies and irises both have 3 petals; buttercups have 5 petals; and corn
marigolds have 13 petals. Seeds of coneflowers and sunflowers have also been observed to follow the same pattern as the Fibonacci sequence. Pine cones and cauliflower also follow this pattern.
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Find the value of each expression. 9P3 and10C4
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nPr = n!/ (n-r)!
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nCr = n!/(r! * (n-r)! )
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i am new to this please explain
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Rainbow attack
Elcomsoft.com » Password Recovery Software » Proactive Password Auditor » Help
Rainbow attack Previous Top Next
Rainbow attack is an implementation of the Faster Cryptanalytic Time-Memory Trade-Off method developed by Dr Philippe Oechslin. The idea is to generate the password hash tables in advance (only
once), and during the audit/recovery process, simply look up the hash in these pre-computed tables. Such process dramatically reduces the auditing time (especially for complex passwords). Due to the
nature of this attack, not all passwords can be found (although with a probability which can be as high as needed).
To access Rainbow attack settings, switch the Attack type to Rainbow, and click on the Rainbow attack tab (second tab, next to the Hashes tab). If you already have the tables, click on the Rainbow
tables list button, and you will be able to browse for the tables for further attack (you can add several tables at once), remove the tables from the list, and move them up and down; when completed,
press Close, and proceed with the attack itself.
The program also supports indexed rainbow tables that are available at http://www.freerainbowtables.com.
To create your own tables, press the Generate tables button. There are a few settings there:
Hash type
LM and NTLM hash tables can be generated; see About Windows passwords for details on hash types.
Password length
Minimum and Maximum; typically, from 1 to 7 (to cover all password space for LM hashes). However, if you want to audit just 6-character passwords (and second halves of passwords that are from 8 to 15
characters long), you can create more effective and still relatively small tables for length from 1 to 6.
Available choices:
•alpha: capital letters only (26)
•alpha-space: capital letters plus space character (27)
•alpha-numeric: capital letters plus digits (36)
•alpha-numeric-space: capital letters plus digits and space character (37)
•alpha-numeric-symbol14: capital letters, digits, and 14 most-common symbols: !@#$%^&*()-_+= (50)
•alpha-numeric-symbol14: capital letters, digits, space and 14 most-common symbols: !@#$%^&*()-_+= (51)
•all: capital letters, digits and 32 printable symbols including space (69)
Chain length
Typical values are from 1000 to 10000. When this value is increased, you get better probability, but worse generation and cryptanalysis times.
Chain count
Chain count affects the table size (and so disk space), table size, probability and generation time (but not cryptanalysis time).
Number of tables and Indexes
Number of tables to generate, or indexes of tables if you distribute the table generation process across a few computers. More tables you have, the better success rate is achieved. For example, if
one table gives a probability of 60% (0,6), two tables will give 1 - (1 - 0,6) * (1 - 0,6) = 0,84 (84%). With three such tables, the probability is already 1 - (1 - 0,6) ^ 3 = 0,936 (93,6%). But of
course, the total space also increases dramatically.
Output folder
Press Browse to select the folder to save generated tables to (before starting the generation process, please verify that there is enough free space there).
Once all parameters are selected, PPA immediately calculates the key space (the total number of passwords in the given range; actually, it depends only on the character set and password length), disk
space (size of each table multiplied by number of tables), and success probability. You can also run the benchmark: press Start, and PPA calculates the speed of your computer on these operations, and
so the table precomputation time, total precomputation time, and maximum cryptanalysis time.
There are some typical configurations (for LM hash type, length from 1 to 7; the time is calculated for Pentium 4 3.0GHz CPU) you can use, for example:
#1 #2 #3 #4
Charset alpha alpha-numeric alpha-num-sym14 all
Chain length 2,100 2,400 12,000 20,000
Chain count 8,000,000 40,000,000 40,000,000 100,000,000
Tables 5 7 13 20
Success rate 99.9% 99.9% 99.9% 99,3%
Total space 640 Mb 4,480 Mb 8,320 Mb 32,000 Mb
Max gen. time 17h 5d 14h 52d 332d
Max analysis time 7 s 14 s 11 m 48 m
The tables for first three configurations can fit into one CD, DVD (Single Layer) and DVD (Double Layer), respectively. For the last configuration (with a complete character set), they take about 32
gigabytes and need 369 days to generate (so you have to use multiple computers), but with such tables, any password can be recovered in just about an hour with 99,3% probability. Normally, it would
take up to 3 weeks to recover such password using a brute-force attack.
Get more information about Proactive Password Auditor
Get full version of Proactive Password Auditor
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Z-Score Example #1
This video contains a detailed example of a statistics problem involving the use of the standardized normal curve and z-scores.
Language: English
Audience: College
Duration: 4 minutes 52 seconds
Released on: Sep 23, 2009 11:25pm
Groups: Boston College Statistics & Probability
Students: 10
Tags: statistics normal distribution z-score z score z-score problem normal distribution problem
Hey, I'm Zach Hagopian, and I'm a sophmore majoring in Finance in the Carroll School of Management at Boston College.
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Sunpower EG-1000 Stirling
Problem 3.2 - The Sunpower EG-1000 Stirling Engine/Generator
This exercise concerns the ideal performance of the EG-1000 Stirling engine (developed by Sunpower, Inc) which is gas fired and has been designed to generate electricity as well as to provide hot
water for a private home. This engine is shown in the figure below together with a simplified schematic diagram. Notice that there are two pistons - a power piston which allows compression and
expansion of the working gas (helium) and a displacer piston which shuttles the working gas between the hot expansion space V[E] and the cold compression space V[C], through the series connected
heater, regenerator and cooler. Conceptually the Stirling engine is the simplest of all heat engines. The working gas is sealed within its cylinder by the power piston. The displacer piston shuttles
the gas such that the gas will compress while it is mainly in the cool compression space and expand while in the hot expansion space. Since the gas is at a higher temperature, and therefore pressure,
during its expansion than during its compression, more power is produced during expansion than is reabsorbed during compression, and this net excess power is the useful output of the engine. Note
that there are no valves or intermittent combustion, which is the major source of noise in an internal combustion engine. The same working gas is used over and over again, making the Stirling engine
a sealed, closed cycle system. All that is added to the system is steady high temperature heat, and all that is removed from the system is low temperature heat and mechanical power.
Figure 1 - The Sunpower EG-1000 free-piston Stirling engine/generator
││The linear electrical generator (not shown in the above schematic) is comprised of powerful rare-earth magnets in the piston cutting a magnetic circuit and coils in the cylinder. This produces 240│
││Volts at 50 Herz - designed for operation in Europe, and is capable of producing more than one kilowatt of electrical power output at an efficiency of around 90%. │
││ │
││ │
││The hot water is provided by operating the cooling water at a temperature of 50°C. │
Unfortunately the analysis of actual Stirling cycle machines is extremely complex and requires sophisticated computer analysis. We consider an idealized model of this engine defined in terms of the
P-V diagram shown below, and will attempt to quantify the performance characteristics from this ideal model. These are the mechanical output power, the thermal efficiency, thermal power for home
water heating and the effect of the regenerator on the thermal efficiency.
The working gas used is helium, which has the advantage of having a low molecular weight and high thermal conductivity compared to air, allowing a high efficiency system. Process (1)-(2) is the
isothermal compression of the helium at temperature T[C] = 50°C, during which heat Q[C] is rejected to the cooling water. Process (2)-(3) is the constant volume displacement process during which heat
Q[R] is absorbed from the regenerator matrix. Process (3)-(4) is the power producing isothermal expansion process at temperature T[E] = 500°C, during which heat Q[E] is absorbed from the gas burner,
and finally process (4)-(1) is the constant volume displacement process during which heat Q[R] is lost to the regenerator matrix. Thus the ideal Stirling cycle consists of four distinct processes,
each one of which can be separately analysed in accordance with the methods that are described in Chapter 3b. State (1) is defined at a maximum volume of 650 cu.cm and a pressure of 10 bar, and State
(2) is defined at a minimum volume of 550 cu.cm. This large minimum volume is the dead space due to the unswept volumes including the heater, regenerator and cooler spaces. (Note that the values
presented here are not actual values of the EG-1000, however were devised by your instructor for purposes of this exercise only).
Figure 2. The ideal Stirling cycle engine P-V diagram
Since the Stirling cycle is a closed cycle, we can consider each process separately. Thus the work done for each process can be determined by integration. This is equivalent to evaluating the area
under the P-v curve, as follows:
The working fluid is helium which is an ideal gas, we use the ideal gas equation of state throughout. Thus P V = m R T, where R = 2.077 kJ/kg K, and [V] , where C[V] = 3.116 kJ/kg K. (refer: Ideal
Gas Properties)
1. From the given conditions at state 1 (P = 10 bar = 1000 kPa, V = 650 cc, T = 50°C) determine the mass of working gas (helium) used in the cycle. [m = 0.00097 kg (close to 1 gm)]
2. Determine the net work done per cycle (kiloJoules): W[E] + W[C] (Note that the compression work WC is always negative). At a frequency of 50 Herz (cycles per second) determine the power output
produced by the engine. [Wnet = 0.151 kJ/cycle, Power = 7.55 kW]
3. Determine the heat absorbed in the expansion space Q[E] during the expansion process (3)-(4). [Q[E] = 0.260 kJ]
4. Evaluate the Thermal Efficiency [th], defined as: [th] = (W[E] + W[C]) / Q[E]. (Net mechanical work done divided by the heat supplied externally by the gas burner). [58 %]
5. Determine the amount of thermal power rejected to the cooling water. Note that at a temperature of 50°C this is suitable for providing hot water for the home, as well as providing home space
heating capability. [Q[C] = -0.109 kJ/cycle, Thermal power to cooling water = 5.45 kW]
6. Determine the amount of heat transferred to the working fluid Q[R] as it passes through the regenerator during process (2)-(3). [1.36 kJ] If this heat were to be supplied externally by the gas
burner, (i.e. no regenerator) what would be the new value of thermal efficiency [th]? [9.3%]
││In this photograph we see the Sunpower EG-1000 being demonstrated using sawdust pellets as the fuel, and generating more than 1000W of electricity to a light panel. This was done at the │
││Sustainability Fair in the Fairgrounds of Athens Ohio, 2001. A closeup photograph of the basic system is shown. Notice the closed cycle radiator and vibration pump used in the water cooling │
││system. │
Engineering Thermodynamics by Israel Urieli is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License
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CCSS.Math.Content.HSG-GPE.B.5 - Wolfram Demonstrations Project
US Common Core State Standard Math HSG-GPE.B.5
Demonstrations 1 - 1 of 1
Description of Standard: Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given
line that passes through a given point).
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View high resolution
This is the Google trend for the search query “Quadratic formula”
It repeats in the same pattern every year. Down in summer, up in September, down again in December and up again in spring time before going down again in the summer.
And so it goes on forever.
The Quadratic Formula Formula.
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omplex number
Complex number
complex numbers
are an extension of the
real numbers
, in which all non-constant
. The complex numbers contain a number
, the
imaginary unit
, with
= −1, i.e.,
is a
square root
of −1. Every complex number can be represented in the form
, where
are real numbers called the
real part
and the
imaginary part
of the complex number respectively.
The sum and product of two complex numbers are:
Complex numbers were first introduced in connection with explicit formulas for the roots of
polynomials. In mathematics, the term "complex" when used as an
means that the field of complex numbers is the underlying number field considered. For example
complex matrix
complex polynomial
complex Lie algebra
The earliest fleeting reference to square roots of negative numbers occurred in the work of the Greek mathematician and inventor Heron of Alexandria in the 1st century AD, when he considered the
volume of an impossible frustum of a pyramid. They became more prominent when in the 16th century closed formulas for the roots of third and fourth degree polynomials were discovered by Italian
mathematicians (see Tartaglia, Cardano). It was soon realized that these formulas, even if one was only interested in real solutions, sometimes required the manipulation of square roots of negative
numbers. This was doubly unsettling since not even negative numbers were considered to be on firm ground at the time. The term "imaginary" for these quantities was coined by René Descartes in the
17th century and was meant to be derogatory. (See imaginary number for a discussion of the "reality" of complex numbers.) The 18th century saw the labors of Abraham de Moivre and Leonhard Euler. To
De Moivre is due (1730) the well-known formula which bears his name, de Moivre's formula:
and to Euler (1748) Euler's formula of complex analysis:
The existence of complex numbers was not completely accepted until the geometrical interpretation (see below) had been described by
Caspar Wessel
; it was rediscovered several years later and popularized by
Carl Friedrich Gauss
, and as a result the theory of complex numbers received a notable expansion. The idea of the graphic representation of complex numbers had appeared, however, as early as 1685, in
De Algebra tractatus
Wessel's memoir appeared in the Proceedings of the Copenhagen Academy for 1799, and is exceedingly clear and complete, even in comparison with modern works. He also considers the sphere, and gives a
quaternion theory from which he develops a complete spherical trigonometry. In 1804 the Abbé Buée independently came upon the same idea which Wallis had suggested, that should represent a unit line,
and its negative, perpendicular to the real axis. Buée's paper was not published until 1806, in which year Jean-Robert Argand also issued a pamphlet on the same subject. It is to Argand's essay that
the scientific foundation for the graphic representation of complex numbers is now generally referred. Nevertheless, in 1831 Gauss found the theory quite unknown, and in 1832 published his chief
memoir on the subject, thus bringing it prominently before the mathematical world. Mention should also be made of an excellent little treatise by Mourey (1828), in which the foundations for the
theory of directional numbers are scientifically laid. The general acceptance of the theory is not a little due to the labors of Cauchy and Abel, and especially the latter, who was the first to
boldly use complex numbers with a success that is well known.
The common terms used in the theory are chiefly due to the founders. Argand called the direction factor, and the modulus; Cauchy (1828) called the reduced form (l'expression réduite); Gauss used for
, introduced the term complex number for , and called the norm.
The expression direction coefficient, often used for , is due to Hankel (1867), and absolute value, for modulus, is due to Weierstrass.
Following Cauchy and Gauss have come a number of contributors of high rank, of whom the following may be especially mentioned: Kummer (1844), Kronecker (1845), Scheffler (1845, 1851, 1880),
Bellavitis (1835, 1852), Peacock (1845), and De Morgan (1849). Möbius must also be mentioned for his numerous memoirs on the geometric applications of complex numbers, and Dirichlet for the expansion
of the theory to include primes, congruences, reciprocity, etc., as in the case of real numbers.
Other types have been studied, besides the familiar , in which is the root of . Thus Eisenstein has studied the type , being a complex root of . Similarly, complex types have been derived from (
prime). This generalization is largely due to Kummer, to whom is also due the theory of Ideal numbers, which has recently been simplified by Klein (1893) from the point of view of geometry. A further
complex theory is due to Galois, the basis being the imaginary roots of an irreducible congruence,
(mod , a prime). The late writers (from 1884) on the general theory include
, Schwarz,
, Berloty,
, and Macfarlane.
The formally correct definition using pairs of real numbers was given in the 19th century.
Formally we may define complex numbers as ordered pairs of real numbers (a, b) together with the operations:
So defined, the complex numbers form a field, the complex number field, denoted by C (or in blackboard bold).
We identify the real number a with the complex number (a, 0), and in this way the field of real numbers R becomes a subfield of C. The imaginary unit i is the complex number (0,1).
In C, we have:
• additive identity ("zero"): (0,0)
• multiplicative identity ("one"): (1,0)
• additive inverse of (a,b): (−a,−b)
• multiplicative inverse of non-zero (a,b):
could also be defined as the topological closure of
algebraic numbers
and the
algebraic closure
A complex number can also be viewed as a point or a position vector on the two dimensional Cartesian coordinate system. This representation is sometimes called an Argand diagram. In the figure, we
The latter expression is sometimes shorthanded as r cis φ, where r = |z| is called the absolute value of z and φ = arg(z) is called the complex argument of z. However, Euler's formula states that e^i
φ = cisφ. The exponential form gives us a better insight than the shorthand rcisφ, which is almost never used in serious mathematical articles. By simple trigonometric identities, we see that
and that
Now the addition of two complex numbers is just the
vector addition
of two vectors, and the multiplication with a fixed complex number can be seen as a simultaneous rotation and stretching.
Multiplication with i corresponds to a counter clockwise rotation by 90 degrees. The geometric content of the equation i^2 = -1 is that a sequence of two 90 degree rotation results in a 180 degree
rotation. Even the fact (-1) · (-1) = +1 from arithmetic can be understood geometrically as the combination of two 180 degree turns.
Absolute value, conjugation and distance
Recall that the absolute value (or modulus or magnitude) of a complex number z = r e^iφ is defined as |z| = r. Algebraically, if z = a + ib, then |z| = &radic(a² + b² ).
One can check readily that the absolute value has three important properties:
for all complex numbers z and w. By defining the distance function d(z, w) = |z - w| we turn the complex numbers into a metric space and we can therefore talk about limits and continuity. The
addition, subtraction, multiplication and division of complex numbers are then continuous operations. Unless anything else is said, this is always the metric being used on the complex numbers.
The complex conjugate of the complex number z = a + ib is defined to be a - ib, written as or z^*. As seen in the figure, is the "reflection" of z about the real axis. The following can be checked:
if and only if z is real
if z is non-zero
The latter formula is the method of choice to compute the inverse of a complex number if it is given in rectangular coordinates.
That conjugate commutes with all the algebraic operations (and many functions; e.g. i (-1 has two square roots); note, however, that conjugate is not differentiable (see holomorphic).
The complex argument of z=re^iφ is φ. Note that the complex argument is unique modulo 2π.
Matrix representation of complex numbers
While usually not useful, alternative representations of complex field can give some insight into their nature. One particularly elegant representation interprets every complex number as 2×2 matrix
with real entries which stretches and rotates the points of the plane. Every such matrix has the form
with real numbers a and b. The sum and product of two such matrices is again of this form. Every non-zero such matrix is invertible, and its inverse is again of this form. Therefore, the matrices of
this form are a field. In fact, this is exactly the field of complex numbers. Every such matrix can be written as
which suggests that we should identify the real number 1 with the matrix
and the imaginary unit i with
a counter-clockwise rotation by 90 degrees. Note that the square of this latter matrix is indeed equal to -1.
The absolute value of a complex number expressed as a matrix is equal to the square root of the determinant of that matrix. If the matrix is viewed as a transformation of a plane, then the
transformation rotates points through an angle equal to the argument of the complex number and scales by a factor equal to the complex number's absolute value. The conjugate of the complex number z
corresponds to the transformation which rotates through the same angle as z but in the opposite direction, and scales in the same manner as z; this can be described by the transpose of the matrix
corresponding to z.
Some properties
Real vector space
C is a two-dimensional real vector space. Unlike the reals, complex numbers cannot be ordered in any way that is compatible with its arithmetic operations: C cannot be turned into an ordered field.
Solutions of polynomial equations
A root of the polynomial p is a complex number z such that p(z) = 0. A most striking result is that all polynomials of degree n with real or complex coefficients have exactly n complex roots
(counting multiple roots according to their multiplicity). This is known as the Fundamental Theorem of Algebra, and shows that the complex numbers are an algebraically closed field.
Indeed, the complex number field is the algebraic closure of the real number field. It can be identified as the quotient ring of the polynomial ring R[X] by the ideal generated by the polynomial X^2
+ 1:
This is indeed a field because X^2 + 1 is irreducible. The image of X in this quotient ring becomes the imaginary unit i.
Algebraic characterization
The field C is (up to field isomorphism) characterized by the following three facts:
Consequently, C contains many proper subfields which are isomorphic to C.
Complex analysis
The study of functions of a complex variable is known as complex analysis and has enormous practical use in applied mathematics as well as in other branches of mathematics. Often, the most natural
proofs for statements in real analysis or even number theory employ techniques from complex analysis (see prime number theorem for an example). Unlike real functions which are commonly represented as
two dimensional graphs, complex functions have four dimensional graphs and may usefully be illustrated by color coding a three dimensional graph to suggest four dimensions, or by animating the
complex function's dynamic transformation of the complex plane.
Control theory
In control theory, systems are often transformed from the time domain to the frequency domain using the Laplace transform. The system's poles and zeros are then analyzed in the complex plane. The
root locus, nyquist plot, and nichols plot techniques all make use of the complex plane.
In the root locus method, it is especially important whether the poles and zeros are in the left or right half planes, i.e. have real part greater than or less than zero. If a system has poles that
• in the right half plane, it will be unstable,
• all in the left half plane, it will be stable,
• on the imaginary axis, it will be marginally stable.
If a system has zeros in the right half plane, it is a nonminimum phase system.
Signal analysis
Complex numbers are used in signal analysis and other fields as a convenient description for periodically varying signals. The absolute value |z| is interpreted as the amplitude and the argument arg(
z) as the phase of a sine wave of given frequency.
If Fourier analysis is employed to write a given real-valued signal as a sum of periodic functions, these periodic functions are often written as the real part of complex valued functions of the form
where ω represents the angular frequency and the complex number z encodes the phase and amplitude as explained above.
In electrical engineering, this is done for varying voltages and currentss. The treatment of resistors, capacitors and inductors can then be unified by introducing imaginary frequency-dependent
resistances for the latter two and combining all three in a single complex number called the impedance. (Electrical engineers and some physicists use the letter j for the imaginary unit since i is
typically reserved for varying currents.)
Improper integrals
The residue theorem of complex analysis is often used in applied fields to compute certain improper integrals. See examples of contour integration.
Quantum mechanics
The complex number field is also of utmost importance in quantum mechanics since the underlying theory is built on (infinite dimensional) Hilbert spaces over C.
In Special and general relativity, some formulas for the metric on spacetime become simpler if one takes the time variable to be imaginary.
Applied mathematics
In differential equations, it is common to first find all complex roots r of the characteristic equation of a linear differential equation and then attempt to solve the system in terms of base
functions of the form f(t) = e^rt.
Fluid dynamics
In fluid dynamics, complex functions are used to describe potential flow in 2d.
Certain fractals employ complex numbers in the plotting of their function, e.g. Mandelbrot set and Lyapunov fractal.
See also
quaternions, complex geometry, local fields, phasors, Leonhard Euler, Euler's identity, Hypercomplex number, De Moivre's formula,
Further reading
• An Imaginary Tale, by Paul J. Nahin; Princeton University Press; ISBN 0691027951 (hardcover, 1998). A gentle introduction to the history of complex numbers and the beginnings of complex analysis.
Topics in mathematics related to quantity Edit
Numbers | Natural numbers | Integers | Rational numbers | Real numbers | Complex numbers | Hypercomplex numbers | Quaternions | Octonions | Sedenions | Hyperreal numbers | Surreal numbers |
Ordinal numbers | Cardinal numbers | p-adic numberss | Integer sequences | Mathematical constants | Infinity
Topics in mathematics related to spaces Edit
Topology | Geometry | Trigonometry | Algebraic geometry | Differential geometry and topology | Algebraic topology | Linear algebra | Fractal geometry | Compact space
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TI Math Nspired
• Students will recognize that the correlation coefficient describes the strength and direction of the linear association between two variables.
• Students will recognize that when two variables are highly linearly correlated, their correlation coefficient will be close to , and when they have little correlation, the correlation coefficient
will be close to 0.
• Students will recognize that two variables with a high correlation coefficient might have a scatterplot that displays a nonlinear pattern.
• Students will recognize that correlation is not affected by the choice of x or y, that is, by the choice of which variable is explanatory and which is response.
• Students will make sense of problems and persevere in solving them (CCSS Mathematical Practices).
• Students will reason abstractly and quantitatively (CCSS Mathematical Practices).
• correlation coefficient
• explanatory variable
• linear
• outlier
• response variable
• scatterplot
About the Lesson
This lesson involves investigating the connection between the scatterplot of bivariate data and the numerical value of the correlation coefficient.
As a result, students will:
• Consider a scatterplot of points that lie in a straight line and one whose points do not line in a straight line and interpret the correlation coefficient for each plot.
• Look at pairs of scatterplots to estimate which plot has the higher correlation coefficient.
• Move points to try to match a given correlation coefficient.
• Investigate a plot of ordered pairs and a plot of the inverse relation by inspecting the coordinates of the points and, by dragging the points in either plot, observing that the correlation
coefficients are the same for both plots.
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High Frequency and Microwave Engineering
This book is dedicated to my wife,
for her help and encouragement in the writing of this book
High Frequency and
Microwave Engineering
E. da Silva
The Open University
OXFORD AUCKLAND BOSTON JOHANNESBURG MELBOURNE NEW DELHI
Linacre House, Jordan Hill, Oxford OX2 8DP
225 Wildwood Avenue, Woburn, MA 01801-2041
A division of Reed Educational and Professional Publishing Ltd
A member of the Reed Elsevier plc group
First published 2001
© E. da Silva 2001
All rights reserved. No part of this publication may be reproduced or
transmitted in any form or by any means, electronically or mechanically,
including photocopying, recording or any information storage or retrieval
system, without either prior permission in writing from the publisher or a
licence permitting restricted copying. In the United Kingdom such licences
are issued by the Copyright Licensing Agency: 90 Tottenham Court Road,
London W1P 0LP.
Whilst the advice and information in this book are believed to be true and
accurate at the date of going to press, neither the author[s] nor the publisher
can accept any legal responsibility or liability for any errors or omissions
that may be made.
British Library Cataloguing in Publication Data
A catalogue record for this book is available from the British Library
ISBN 0 7506 5646 X
Typeset in 10/12 pt Times by Cambrian Typesetters, Frimley, Surrey
Printed and bound by MPG Books Ltd, Bodmin, Cornwall
Preface ix
1 BASIC FEATURES OF RADIO COMMUNICATION SYSTEMS 1
1.1 Introduction 1
1.2 Radio communication systems 2
1.3 Modulation and demodulation 3
1.4 Radio wave propagation techniques 9
1.5 Antennas and Aerials 14
1.6 Antenna arrays 23
1.7 Antenna distribution systems 25
1.8 Radio receivers 32
1.9 Radio receiver properties 33
1.10 Types of receivers 37
1.11 Summary 41
2 TRANSMISSION LINES 43
2.1 Introduction 43
2.2 Transmission line basics 45
2.3 Types of electrical transmission lines 47
2.4 Line characteristic impedances and physical parameters 50
2.5 Characteristic impedance (Z0) from primary electrical parameters 54
2.6 Characteristic impedance (Z0) by measurement 58
2.7 Typical commercial cable impedances 60
2.8 Signal propagation on transmission lines 61
2.9 Waveform distortion and frequency dispersion 63
2.10 Transmission lines of finite length 64
2.11 Reflection and transmission coefficients 64
2.12 Propagation constant (γ) of transmission lines 72
2.13 Transmission lines as electrical components 77
2.14 Transmission line couplers 82
2.15 Summary 98
3 SMITH CHARTS AND SCATTERING PARAMETERS 88
3.1 Introduction 88
3.2 Smith charts 89
vi Contents
3.3 The immittance Smith chart 97
3.4 Admittance manipulation on the chart 98
3.5 Smith chart theory and applications 98
3.6 Reflection coefficients and impedance networks 102
3.7 Impedance of distributed circuits 106
3.8 Impedance matching 110
3.9 Summary of Smith charts 125
3.10 Scattering parameters (s-parameters) 125
3.11 Applied examples of s-parameters in two port networks 131
3.12 Summary of scattering parameters 140
4 PUFF SOFTWARE 142
4.1 Introduction 142
4.2 CalTech’s PUFF Version 2.1 143
4.3 Installation of PUFF 143
4.4 Running PUFF 144
4.5 Examples 147
4.6 Bandpass filter 152
4.7 PUFF commands 156
4.8 Templates 157
4.9 Modification of transistor templates 163
4.10 Verification of some examples given in Chapters 2 and 3 165
4.11 Using PUFF to evaluate couplers 170
4.12 Verification of Smith chart applications 172
4.13 Verification of stub matching 176
4.14 Scattering parameters 186
4.15 Discontinuities: physical and electrical line lengths 189
4.16 Summary 192
5 AMPLIFIER BASICS 195
5.1 Introduction 195
5.2 Tuned circuits 196
5.3 Filter design 202
5.4 Butterworth filter 208
5.5 Tchebyscheff filter 224
5.6 Summary on filters 231
5.7 Impedance matching 232
5.8 Three element matching networks 252
5.9 Broadband matching networks 259
5.10 Summary of matching networks 261
6 HIGH FREQUENCY TRANSISTOR AMPLIFIERS 262
6.1 Introduction 262
6.2 Bi-polar transistors 262
6.3 Review of field effect transistors 277
6.4 A.C. equivalent circuits of transistors 291
6.5 General r.f. design considerations 304
Contents vii
6.6 Transistor operating configurations 316
6.7 Summary 319
7 MICROWAVE AMPLIFIERS 320
7.1 Introduction 320
7.2 Transistors and s-parameters 321
7.3 Design of amplifiers with conjugately matched impedances 322
7.4 Design of amplifiers for a specific gain 332
7.5 Design of amplifiers for optimum noise figure 345
7.6 Design of broadband amplifiers 349
7.7 Feedback amplifiers 352
7.8 R.F. power transistors 355
7.9 Summary 35S
8 OSCILLATORS AND FREQUENCY SYNTHESISERS 357
8.1 Introduction 357
8.2 Sine wave type oscillators 358
8.3 Low frequency sine wave oscillators 361
8.4 Wien bridge oscillator 361
8.5 Phase shift oscillators 364
8.6 Radio frequency (LC) oscillators 368
8.7 Colpitts oscillator 368
8.8 Hartley oscillator 371
8.9 Clapp oscillator 372
8.10 Voltage-controlled oscillator 375
8.11 Comparison of the Hartley, Colpitts, Clapp and voltage-controlled
oscillators 376
8.12 Crystal control oscillators 376
8.13 Phase lock loops 380
8.14 Frequency synthesisers 393
8.15 Summary 399
9 FURTHER TOPICS 400
9.1 Aims 400
9.2 Signal flow graph analysis 400
9.3 Small effective microwave CAD packages 410
9.4 Summary of software 420
References 421
Index 000
This book was started while the author was Professor and Head of Department at Etisalat
College which was set up with the technical expertise of the University of Bradford,
England. It was continued when the author returned to the Open University, England.
Many thanks are due to my colleagues Dr David Crecraft and Dr Mike Meade of the Open
University, Dr L. Auchterlonie of Newcastle University and Dr N. McEwan and Dr D.
Dernikas of Bradford University. I would also like to thank my students for their many
helpful comments.
High Frequency and Microwave Engineering has been written with a view to ease of
understanding and to provide knowledge for any engineer who is interested in high
frequency and microwave engineering. The book has been set at the third level standard of
an electrical engineering degree but it is eminently suitable for self-study. The book
comprises standard text which is emphasised with over 325 illustrations. A further 120
examples are given to emphasise clarity in understanding and application of important
A software computer-aided-design package, PUFF 2.1 produced by California Institute
of Technology (CalTech) U.S.A., is supplied free with the book. PUFF can be used to
provide scaled layouts and artwork for designs. PUFF can also be used to calculate the
scattering parameters of circuits. Up to four scattering parameters can be plotted simulta-
neously and automatically on a Smith chart as well as in graphical form. In addition to the
PUFF software I have also included 42 software application examples. These examples
have been chosen to calculate and verify some of the examples given in the text, but many
are proven designs suitable for use in practical circuits. The confirmation of manual design
and CAD design is highly gratifying to the reader and it helps to promote greater confi-
dence in the use of other types of software. An article ‘Practical Circuit Design’ explain-
ing how PUFF can be used for producing layout and artwork for circuits is explained in
detail. There is also a detailed microwave amplifier design which uses PUFF to verify
circuit calculations, match line impedances, and produce the artwork for amplifier fabri-
cation. There is also a copy of CalTech’s manual on disk. This will prove useful for more
advanced work.
The book commences with an explanation of the many terms used in radio, wireless,
high frequency and microwave engineering. These are explained in Chapter 1. Chapter 2
provides a gentle introduction to the subject of transmission lines. It starts with a gradual
introduction of transmission lines by using an everyday example. Diagrams have been
x Preface
used to illustrate some of the characteristics of transmission lines. Mathematics has been
kept to a minimum. The chapter ends with some applications of transmission lines espe-
cially in their use as inductors, capacitors, transformers and couplers.
Chapter 3 provides an introduction to Smith charts and scattering parameters. Smith
charts are essential in understanding and reading manufacturers’ data because they also
provide a ‘picture’ of circuit behaviour. Use of the Smith chart is encouraged and many
examples are provided for the evaluation and manipulation of reflection coefficients,
impedance, admittance and matching circuits. For those who want it, Smith chart theory is
presented, but it is stressed that knowledge of the theory is not essential to its use.
The installation of PUFF software is introduced in Chapter 4. The chapter goes on to
deal with the printing and fabrication of artwork and the use and modification of templates.
Particular attention is paid to circuit configurations including couplers, transformers and
matching of circuits. Scattering parameters are re-introduced and used for solving scatter-
ing problems. Many of the examples in this chapter are used to confirm the results of the
examples given in Chapters 2 and 3.
Amplifier circuitry components are dealt with in Chapter 5. Particular attention is paid
to the design of Butterworth and Tchebyscheff filters and their uses as low pass, bandpass,
high pass and bandstop filters. Impedance matching is discussed in detail and many meth-
ods of matching are shown in examples.
Chapter 6 deals with the design of amplifiers including transistor biasing which is
vitally important for it ensures the constancy of transistor parameters with temperature.
Examples are given of amplifier circuits using unconditionally stable transistors and condi-
tionally stable amplifiers. The use of the indefinite matrix in transistor configurations is
shown by examples.
The design of microwave amplifiers is shown in Chapter 7. Design examples include
conjugately matched amplifiers, constant gain amplifiers, low noise amplifiers, broadband
amplifiers, feedback amplifiers and r.f. power amplifiers.
Oscillators and frequency synthesizers are discussed in Chapter 8. Conditions for oscil-
lation are discussed and the Barkhausen criteria for oscillation is detailed in the early part
of the chapter. Oscillator designs include the Wien bridge, phase shift, Hartley, Colpitts,
Clapp, crystal and the phase lock loop system. Frequency synthesizers are discussed with
reference to direct and indirect methods of frequency synthesis.
Chapter 9 is a discussion of topics which will prove useful in future studies. These
include signal flow diagrams and the use of software particularly the quasi-free types.
Comments are made regarding the usefulness of Hewlett Packard’s AppCAD and
Motorola’s impedance matching program, MIMP.
Finally, I wish you well in your progress towards the fascinating subject of high
frequency and microwave engineering.
Ed da Silva
Basic features of radio
communication systems
1.1 Introduction
This chapter describes communication systems which use radio waves and signals. Radio
signals are useful for two main reasons. They provide a relatively cheap way of commu-
nicating over vast distances and they are extremely useful for mobile communications
where the use of cables is impractical.
Radio signals are generally considered to be electromagnetic signals which are broad-
cast or radiated through space. They vary in frequency from several kilohertz1 to well over
100 GHz (1011 Hz). They include some well known public broadcasting bands: long-wave
(155–280 kHz), medium-wave (522–1622 kHz), short-wave (3–30 MHz), very high
frequency FM band (88–108 MHz), ultra high frequency television band (470–890 MHz)
and the satellite television band (11.6 to 12.4 GHz). The frequencies2 quoted above are
approximate figures and are only provided to give an indication of some of the frequency
bands used in Radio and TV broadcasting.
1.1.1 Aims
The aims of this chapter are to introduce you to some basic radio communications princi-
ples and methods. These include modulation (impressing signal information on to radio
carrier waves), propagation (transmission of radio carrier waves) and demodulation
(detection of radio carrier waves) to recover the original signal information.
The method we use here is to start with an overview of a communication system. The
system is then divided to show its sub-systems and the sub-systems are then expanded to
show individual circuits and items.
1.1.2 Objectives
The general objectives of this chapter are:
• to help you understand why certain methods and techniques are used for radio frequency
and high frequency communication circuits;
1 One hertz (Hz) means 1 cyclic vibration per second: 1 kHz = 1000 cyclic vibrations per second, 1 MHz =
1 000 000 cyclic vibrations per second, and 1 GHz = 1 000 000 000 cyclic vibrations per second. The word Hertz
is named after Heinrich Hertz, one of the early pioneers of physics.
2 The frequencies quoted are for Europe. Other countries do not necessarily follow the exact same frequen-
cies but they do have similar frequency bands.
2 Basic features of radio communication systems
• to appreciate the need for modulation;
• to understand the basic principles of modulation and demodulation;
• to understand the basic principles of signal propagation using antennas;
• to introduce radio receivers;
• to introduce you to the requirements of selectivity and bandwidth in radio communica-
tion circuits.
1.2 Radio communication systems
1.2.1 Stages in communication
Let’s commence with a simple communications example and analyse the important stages
necessary for communication. This is shown diagramatically in Figure 1.1. We start by
writing a letter-message, putting it in an envelope, and sending it through a post-carrier
(postal carrier system) to our destination. At the other end, our recipient receives the letter
from the post office, opens the envelope and reads our message. Why do we carry out these
We write a letter because it contains the information we want to send to our recipient.
In radio communications, we do the same thing; we use a message signal, which is an elec-
trical signal derived from analogue sound or digitally encoded sound and/or video/data
signals, as the information we want to convey. The process of putting this information into
an ‘envelope’ for transmission through the carrier is called modulation and circuits
designed for this purpose are known as modulation circuits or modulators.
We use the post office as the carrier for our letters because the post office has the abil-
ity to transmit messages over long distances. In radio communications, we use a radio
frequency carrier because a radio carrier has the ability to carry messages over long
distances. A radio frequency carrier with an enveloped message impressed on it is often
called an enveloped carrier wave or a modulated carrier wave.
When the post office delivers a letter to a destination, the envelope must be opened to
enable the message to be read. In radio communications when the enveloped carrier wave
Fig. 1.1 Analogy between the postal system and a radio system
Modulation and demodulation 3
arrives at its destination, the enveloped carrier must be ‘opened’ or demodulated to
recover the original message from the carrier. Circuits which perform this function are
known as demodulation circuits or demodulators.
The post office uses a system of postal codes and addresses to ensure that a letter is
selected and delivered to the correct address. In radio communications, selective or tuned
circuits are used to select the correct messages for a particular receiver. Amplifiers are also
used to ensure that the signals sent and received have sufficient amplitudes to operate the
message reading devices such as a loudspeaker and/or a video screen.
In addition to the main functions mentioned above, we need a post box to send our
letter. The electrical equivalent of this is the transmitting antenna. We require a letter box
at home to receive letters. The electrical equivalent of this is the receiving antenna.
1.2.2 Summary of radio communications systems
A pictorial summary of the above actions is shown in Figure 1.1. There are three main
functions in a radio communications system. These are: modulation, transmission and
demodulation. There are also supplementary functions in a radio communications
system. These include transmitting antennas,3 receiving antennas, selective circuits,
and amplifiers. We will now describe these methods in the same order but with more
1.3 Modulation and demodulation
Before discussing modulation and demodulation, it is necessary to clarify two points: the
modulation information and the modulation method.
In the case of a letter in the postal system, we are free to write our messages (modula-
tion information) in any language, such as English, German, French, pictures, data, etc.
However, our recipient must be able to read the language we use. For example it is useless
to write our message in Japanese if our recipient can only read German. Hence the modu-
lation information system we use at the transmitter must be compatible with the demod-
ulation information system at the receiver.
Secondly, the method of putting information (modulation method) on the letter is
important. For example, we can type, use a pencil, ultra violet ink, etc. However, the reader
must be able to decipher (demodulate) the information provided. For example, if we use
ultra violet ink, the reader must also use ultra violet light to decipher (demodulate) the
message. Hence the modulation and demodulation methods must also be compatible.
In the discussions that follow we are only discussing modulation and demodulation
methods; not the modulation information. We also tend to use sinusoidal waves for our
explanation. This is because a great mathematician, Joseph Fourier,4 has shown that peri-
odic waveforms of any shape consist of one or more d.c. levels, sine waves and cosine
waves. This is similar to the case in the English language, where we have thousands of
words but, when analysed, all come from the 26 letters of the alphabet. Hence, the sinu-
soidal wave is a useful tool for understanding modulation methods.
3 Antennas are also known as aerials.
4 Fourier analysis will be explained fully in a later section.
4 Basic features of radio communication systems
Fig. 1.2 A sinusoidal radio carrier wave
We now return to our simple radio carrier wave which is the sinusoidal wave5 shown in
Figure 1.2.
A sinusoidal wave can be described by the expression
vc = Vc cos (ωct + fc) (1.1)
vc = instantaneous carrier amplitude (volts)
Vc = carrier amplitude (peak volts)
ωc = angular frequency in radians and ωc = 2πf c where
f c = carrier frequency (hertz)
φc = carrier phase delay (radians)
If you look at Figure 1.2, you can see that a sinusoidal wave on its own provides little
information other than its presence or its absence. So we must find some method of modu-
lating our information on to the radio carrier wave. We can change:
• its amplitude (Vc) according to our information – this is called amplitude modulation
and will be described in Section 1.3.1;
• its frequency (ωc) according to our information – this is called frequency modulation
and will be described in Section 1.3.2;
• its phase (φc) according to our information – this is known as phase modulation and
will be described in Section 1.3.3;
• or we can use a combination of one or more of the methods described above – this
method is favoured by digital modulation.
1.3.1 Amplitude modulation (AM)
This is the method used in medium-wave and short-wave radio broadcasting. Figure 1.3
shows what happens when we apply amplitude modulation to a sinusoidal carrier wave.
5 A sinusoidal wave is a generic name for a sine or cosine wave. In many cases, cosine waves are used
because of ease in mathematical manipulation.
Modulation and demodulation 5
Fig. 1.3 Amplitude modulation waveforms: (a) modulating wave; (b) carrier wave; (c) modulated wave
Figure 1.3(a) shows the modulating wave on its own.6 Figure 1.3(b) shows the carrier wave
on its own. Figure 1.3(c) shows the resultant wave. The resultant wave shape is due to the
fact that at times the modulating wave and the carrier wave are adding (in phase) and at
other times, the two waves are opposing each other (out of phase).
Amplitude modulation can also be easily analysed mathematically. Let the sinusoidal
modulating wave be described as
6 I have used a cosine wave here because you will see later when we use Fourier analysis that waveforms, no
matter how complicated, can be resolved into a series of d.c., sine and cosine terms and their harmonics.
6 Basic features of radio communication systems
vm = Vm cos (ωmt) (1.2)
vm = instantaneous modulating amplitude (volts)
Vm = modulating amplitude (peak volts)
ωm = angular frequency in radians and ωm = 2πf m where
fm = modulating frequency (hertz)
When the amplitude of the carrier is made to vary about Vc by the message signal vm, the
modulated signal amplitude becomes
[Vc + Vm cos (ωmt)] (1.3)
The resulting envelope AM signal is then described by substituting Equation 1.3 into
Equation 1.1 which yields
[Vc + Vm cos (ωmt)] cos (ωct + φc) (1.4)
It can be shown that when this equation is expanded, there are three frequencies, namely
(f c – f m), f c and (f c + f m). Frequencies (f c – f m) and (f c + f m) are called sideband frequen-
cies. These are shown pictorially in Figure 1.4.
Fig. 1.4 Frequency spectrum of an AM wave
The modulating information is contained in one of the sideband frequencies which must
be present to extract the original message. The bandwidth (bw) is defined as the highest
frequency minus the lowest frequency. In this case, it is (f c + f m) – (f c – f m) = 2f m where
f m is the highest modulation frequency. Hence, a radio receiver must be able to accommo-
date the bandwidth of a signal.7
1.3.2 Frequency modulation (FM)
Frequency modulation is the modulation method used in VHF radio broadcasting. Figure
1.5 shows what happens when we apply frequency modulation to a sinusoidal carrier wave.
Figure 1.5(a) shows the modulating wave on its own. Figure 1.5(b) shows the carrier wave
on its own. Figure 1.5(c) shows the resultant wave. The resultant wave shape is due to the
7 This is not unusual because speech or music also have low notes and high notes and to hear them our own
ears (receivers) must be able to accommodate their bandwidth. Older people tend to lose this bandwidth and often
are unable to hear the high notes.
Modulation and demodulation 7
Fig. 1.5 Frequency modulation waveforms: (a) modulating wave; (b) carrier wave; (c) FM wave
fact that the carrier wave frequency increases when the modulating signal is positive and
decreases when the modulating signal is negative. Note that in pure FM, the amplitude of
the carrier wave is not altered.
The frequency deviation (∆f c) of the carrier is defined as [f c (max) – f c (min)] or
∆f c = f c (max) – f c (min) (1.5)
According to Carson’s rule, the frequency bandwidth required for wideband FM is approx-
imately 2 × (maximum frequency deviation + highest frequency present in the message
signal) or
bw = 2 [∆f c + f m (max)] (1.6)
In FM radio broadcasting, the allocated channel bandwidth is about 200 kHz.
8 Basic features of radio communication systems
Fig. 1.6 Phase modulation waveforms: (a) modulating wave; (b) carrier wave; (c) modulated wave
1.3.3 Phase modulation (PM)
Phase modulation is particularly useful for digital waveforms. Figure 1.6 shows what
happens when we apply phase modulation to a sinusoidal carrier wave. Figure 1.6(a)
shows a digital modulating wave on its own. We have used a pulse waveform as opposed
to a sine wave in this instance because it demonstrates phase modulation more clearly.
Figure 1.6(b) shows the carrier wave on its own. Figure 1.6(c) shows the resultant wave.
Note particularly how the phase of the carrier waveform changes when a positive modu-
lating voltage is applied. In this particular case, we have shown you a phase change of
180°, but smaller phase changes are also possible.
Phase modulation is popularly used for digital signals. Phase modulation is synony-
mous with frequency modulation in many ways because an instantaneous change in phase8
is also an instantaneous change in frequency and vice-versa. Hence, much of what is said
about FM also applies to PM.
8 Phase (φ) = angular velocity (ω) multiplied by time (t). Hence φ = ωt. Note this equation is similar to that
of distance = velocity × time. This is because φ = amount of angle travelled = velocity (ω) × time (t).
Radio wave propagation techniques 9
Fig. 1.7 An eight level coded signal modulated on to a radio carrier
1.3.4 Combined modulation methods
Digital signals are often modulated on to a radio carrier using both phase and amplitude
modulation. For example, an eight level coded digital signal can be modulated on to a
carrier by using distinct 90° phase changes and two amplitude levels. This is shown
diagrammatically in Figure 1.7 where eight different signals, points A to H, are encoded on
to a radio carrier. This method is also known as quadrature amplitude modulation (QAM).
1.3.5 Summary of modulation systems
In this section, we have shown you four methods by which information signals can be
modulated on to a radio carrier.
1.4 Radio wave propagation techniques
1.4.1 Properties of electromagnetic waves
In Figure 1.8 we show the case of a radio generator feeding energy into a load via a two
wire transmission line. The radio generator causes voltage and current waves to flow
towards the load. A voltage wave produces a voltage or electric field. A current wave
produces a cuurent or magnetic field. Taken together these two fields produce an electro-
magnetic field which at any instant varies in intensity along the length of the line.
The electromagnetic field pattern is, however, far from stationary. Like the voltage
on the line, it propagates from end to end with finite velocity which – for an air spaced
line – is close to the velocity of light in free space.9 The flow of power from source to
9 Strictly speaking ‘free space’ is a vacuum. However, the velocity of propagation of electro-magnetic waves
in the atmosphere is practically the same as that in a vacuum and is approximately 3 × 108 metres per second.
Wavelength (λ) is defined as the ratio, velocity/frequency.
10 Basic features of radio communication systems
Fig. 1.8 Energy propagation in a transmission line
load is then regarded as that of an electromagnetic wave propagating between the
The equivalence between the circuit and field descriptions of waves on transmission
lines is demonstrated by the fact that at any point in the electromagnetic field the instan-
taneous values of the electric field (E) (volts/metre) and the magnetic field (H)
(amperes/metre) are related by
———— = Z0 (ohms) (1.7)
where Z0 is the characteristic impedance of the transmission line.10 It can also be
shown that both approaches give identical results for the power flow along a matched
In the two wire transmission line shown in Figure 1.8, the parallel conductors produce
electromagnetic fields which overlap and cancel in the space beyond the conductors. The
radio frequency energy is thus confined and guided by the conductors from the source to
its destination. If, however, the conductor spacing is increased so that it becomes com-
parable with the wavelength of operation the line will begin to radiate r.f. energy to its
surroundings. The energy is lost in the form of free-space electromagnetic waves which
radiate away from the line with the velocity of light.
The 19th century mathematician James Clerk Maxwell was the first to recognise that
electromagnetic waves can exist and transport energy quite independently of any system
of conductors. We know now that radio waves, heat waves, visible light, X-rays are all
electromagnetic waves differing only in frequency. Figure 1.9 shows the range of frequen-
cies and the regions occupied by the different types of radiation. This is known as the elec-
tromagnetic spectrum.
Fig. 1.9 The electromagnetic frequency spectrum
10 Transmission lines have impedances because they are constructed from physical components which have
resistance, self inductance, conductance and capacitance.
Radio wave propagation techniques 11
1.4.2 Free-space radiation
At operational frequencies, where the operational wavelengths are comparable in size to
circuit components,11 any circuit consisting of components connected by conductors will
tend to act as an imperfect transmission line. As a result, there will always be some loss of
r.f. energy by way of radiation. In other words, the circuit will tend to behave like a crude
radio transmitter antenna.
It follows that for minimal radiation, components should be small with respect to their
operational wavelengths. Conversely, if radiation is desired, then the physical components
should be large, approximately 1/4 wavelength for optimum radiation. This is why anten-
nas are physically large in comparison with their operational wavelength.
Energy radiates from an r.f. source or transmitter in all directions. If you imagine a
spherical surface surrounding the transmitter, then the interior of the surface would be
‘illuminated’ with radiated energy, just like the inside of a globular lamp-shade. The illu-
mination is not necessarily uniform, however, since all transmitters are, to some extent,
If the r.f. source is sinusoidal, then the electric and magnetic fields will also be varying
sinusoidally at any point in the radiation field. Now it is difficult to depict a propagating elec-
tromagnetic field but some of its important properties can be identified. To do this we
consider propagation in a particular direction on a straight line connecting a transmitter to a
distant receiver as shown in Figure 1.10. You will see that this line coincides with the z-direc-
tion in Figure 1.10. Measurements at the radio receiver would then indicate that the oscillat-
ing electric field is acting all in one direction, the x-direction in Figure 1.10. The magnetic
field is in-phase with the electric field but acts at right-angles to the electric field, in the y-
direction. The two fields are thus at right-angles to each other and to the direction of propa-
gation. An electromagnetic wave with these characteristics is known as a plane wave.
Fig. 1.10 Electric and magnetic field directions for an electromagnetic wave propagating in the z-direction
11 Generally taken to be the case when the operational wavelength is about 1/20 of the physical size of compo-
12 Basic features of radio communication systems
Provided there is no disturbance in the propagation path, the electric and magnetic field
orientations with respect to the earth’s surface will remain unchanged. By convention, the
orientation of the electric field with respect to the earth’s surface is called the polarisation
of the electromagnetic wave. If the electric field is vertical, the wave is said to be verti-
cally polarised; if horizontal, the wave is horizontally polarised. A wave is circularly
polarised if its electric field rotates as the wave travels. Circular polarisation can be either
clockwise or anti-clockwise.
Polarisation is important because antennas must be mounted in the correct plane for
optimum signal reception.12 Terrestrial broadcasting stations tend to use either vertical or
horizontal polarisation. Satellite broadcasting stations use circular polarisation. The polar-
isation of a wave is sometimes ‘twisted’ as it propagates through space. This twisting is
caused by interfering electric or magnetic fields. It is particularly noticeable near steel-
structured buildings where aerials are mounted at odd angles to the vertical and horizontal
planes to compensate for these effects.
Field strength
The strength of a radio wave can be expressed in terms of the strength of its electric
field or by the strength of its magnetic field. You should recall that these are measured
in units of volts per metre and amperes per metre respectively. For a sinusoidally vary-
ing field it is customary to quote r.m.s. values E rms and H rms. What is the physical
significance of E rms? This is numerically equal to the r.m.s. voltage induced in a
conductor of length 1 m when a perpendicular electromagnetic wave sweeps over the
conductor with the velocity of light.
As stated earlier, the electric and magnetic fields in a plane wave are everywhere in
phase. The ratio of the field strengths is always the same and is given by
electric field strength Erms ( V m )
= 377Ω (1.8)
magnetic field strength Hrms ( A m )
This ratio is called the free-space wave impedance. It is analogous to the characteristic
impedance of a transmission line.
Example 1.1
The electric field strength at a receiving station is measured and found to have an r.m.s
value of 10 microvolts/m. Calculate (a) the magnetic field strength; (b) the amount of
power incident on a receiving aerial with an effective area of 5 m2.
Given: Electric field strength = 10 microvolts/m.
Required: (a) Magnetic field strength, (b) incident power on a receiving aerial with effec-
tive area of 5 m2.
12 You can see this effect by looking at TV aerials mounted on houses. In some districts, you will see aerials
mounted horizontally whilst in other areas you will find aerials mounted vertically. As a general rule, TV broad-
casting authorities favour horizontal polarisation for main stations and vertical polarisation for sub or relay
Radio wave propagation techniques 13
Solution. Using equation 1.8
(a) Hrms = 10 µV/m/377 Ω = 2.65 × 10–8 A/m
(b) Power density is given by
Erms × Hrms = 10 × 10–6 × 2.65 × 10–8 W/m2 = 2.65 × 10–13 W/m2
This is the amount of power incident on a surface of area 1 m2. For an aerial with area
5 m2, the total incident power will be
P = 2.65 × 10–13 W/m2 × 5 m2 = 1.33 pW
Power density
The product Erms × Hrms has the dimensions of ‘volts per metre’ times ‘amps per metre’,
giving watts per square metre. This is equivalent to the amount of r.f. power flowing
through one square metre of area perpendicular to the direction of propagation and is
known as the power density of the wave. The power density measures the intensity of the
‘illumination’ falling on a receiving aerial.
A plane wave expands outwards as it travels through space from a point source. As a
result, the power density falls off with increasing distance from the source. If you have
studied any optics then you will be familiar with the idea that the power density falls off
as the square of the distance from the source, i.e.
PD2 ⎡ D1 ⎤
=⎢ ⎥ (1.9)
PD1 ⎣ D2 ⎦
where P D1, P D2 = power densities at distances D 1 and D 2 respectively.
Example 1.2
If the data in Example 1.1 applies to a receiver located 10 km from the transmitter, what
will be the values of E rms and H rms at a distance of 100 km?
Given: Data of Example 1.1 applied to a receiver at 10 km from transmitter.
Required: (a) E rms at 100 km, (b) Hrms at 100 km.
Solution. Using Equation 1.9 at a distance of 100 km, the power density will be reduced
by a factor (10/100)2 = 0.01, so power density = 2.65 × 10–15 W/m2. Now, power density
= E rms × H rms and since H rms = E rms/377 (Equation 1.7)
= 2.65 × 10 −15 W m 2
Erms = 2.65 × 10 −15 × 377 = 1 µV m
Hrms = 1 µV/m/377 Ω = 2.65 × 10–9 A/m
14 Basic features of radio communication systems
Summary of propagation principles
Several important points have been established in Section 1.4.
• R.F. energy is radiated by way of travelling electric and magnetic fields which together
constitute an electromagnetic wave propagating in free space with the velocity of light.
• In a plane wave, the electric and magnetic fields vary in phase and act at right-angles to
each other. Both fields are at right-angles to the direction of propagation.
• The direction of the electric field determines the polarisation of a plane wave.
• At any point, the ratio of the electric and magnetic fields is the same and equal to the
wave impedance. This impedance is 377 W approximately.
• The product Erms × Hrms gives the power density of the wave.
• The power density falls off as the square of the distance from the r.f. source.
• To obtain optimum signal reception from free space a receiving aerial should be set for
the correct polarisation and be suitably located with regard to height and direction.
1.5 Antennas and aerials
1.5.1 Introduction
An antenna or aerial is a structure, usually made from good conducting material, that has
been designed to have a shape and size so that it will provide an efficient means of trans-
mitting or receiving electromagnetic signals through free space. Many of the principles
used in the construction of antennas can be easily understood by analogy to the headlamp
of your car (see Figure 1.11).
An isotropic light source is a light source which radiates light equally in all directions.
The radiation pattern from an isotropic light source can be altered by placing a reflecting
mirror on one side of the light source. This is carried out in car headlamps where a quasi-
parabolic reflecting mirror (reflector) is placed behind a bulb to increase the light intensity
of the lamp in the forward direction. The reflector has therefore produced a change in the
directivity of the light source. The increase or ‘gain’ of light intensity in the forward direc-
tion has been gained at the expense of losing light at the back of the lamp. This gain is not
a ‘true gain’ because total light energy from the lamp has not been increased; light energy
has only been re-directed to produce an intensity gain in the forward direction.
Fig. 1.11 Radiation patterns from a car headlamp: (a) top view; (b) side view
Antennas and aerials 15
The forward light intensity of a car lamp can be further improved by using one or more
lenses to concentrate its forward light into a main beam or main lobe. Again, this ‘gain’
in light intensity has been achieved by confining the available light into a narrower beam
of illumination; there has been no overall gain in light output from the bulb.
There are also optimum sizes and distances for the placement of reflectors and lenses.
These are dictated by the physical size of the bulb, the desired gain intensity of the main
beam or main lobe, the required width of the main beam and the requirement to suppress
minor or spurious light lobes which consume energy and cause unnecessary glare to on-
coming motorists.
A car headlamp (Figure 1.11) has two main light-emitting patterns; a horizontal pattern
and a vertical pattern. The horizontal pattern (Figure 1.11(a)) is a bird’s eye view of the
illumination pattern. A plot of the horizontal pattern is called a polar diagram. The verti-
cal or azimuth pattern (Figure 1.11(b)) is the pattern seen by an observer standing to one
side of the lamp. The vertical pattern is sometimes called the end-fire pattern. Both light
patterns must be considered because modern headlamp reflectors tend to be elliptical and
affect emitted light in the horizontal and vertical planes differently.
In the above description, light has been assumed to travel from bulb to free space but
the effect is equally true for light travelling in the opposite direction, i.e. the system is bi-
directional. It can be used either for transmitting light from the bulb or for receiving exter-
nal light at the point source usually occupied by the bulb filament. This can be easily
verified by shining an external light source through the lens and the reflector in the oppo-
site direction from which light had emerged, and seeing it converge on the bulb source.13
Many of the principles introduced above apply to antennas as well. Because of its bi-
directional properties, a radio antenna can be used for transmitting or receiving signals.
1.5.2 Radiating resistance
The relationship, power (watts) = (volts2/ohms), is used for calculating power loss in a circuit.
It is not always possible to apply this law directly to a radiating circuit because a physical
resistor does not always exist. Yet we cannot deny that there is a radiated power loss when a
voltage is applied across a radiating circuit. To overcome this problem, engineers postulate an
‘equivalent’ resistor to represent a physical resistor which would absorb the same radiated
power loss. This equivalent resistor is called the radiating resistance of the circuit.
The radiating resistance of an antenna should not be confused with its input impedance.
The input impedance is the value used when considering the connection of an antenna to
a transmission line with a specified characteristic impedance. Antennas are bi-directional
and it is not uncommon to use the same antenna for transmitting and receiving signals.
Example 1.3
A transmitter with an output resistance of 72 W and an r.m.s. output of 100 V is connected
via a matched line to an antenna whose input resistance is 72 W. Its radiation resistance is
also 72 W. Assuming that the antenna is 100% efficient at the operating frequency, how
much power will be transmitted into free space?
13 If you have any doubts about the system being bi-directional, you should visit a lighthouse which uses a
similar reflector and lens system. Curtains must be drawn around the system during daylight hours because
sunlight acting on the system has been known to produce such high light and heat intensities that insulation melt-
down and fires have been caused.
16 Basic features of radio communication systems
Given: Transmitter output = 100 V, transmitter output impedance = 72 Ω, antenna input
impedance = 72 Ω, radiation resistance = 72 Ω, antenna efficiency = 100%.
Required: Power radiated into free space.
Solution. The antenna has an input impedance Z in = 72 Ω and provides a matched termi-
nation to the 72 Ω line. The r.f. generator then ‘sees’ an impedance of 72 Ω, so the r.m.s.
voltage applied to the line will be 100/2 = 50 V. The amount of power radiated is calcu-
lated using
radiated power =
where R = 72 Ω is the radiation resistance. The radiated power is therefore 34.7 W. Notice
that, because in this case R = Z in, maximum power is radiated into free space.
1.5.3 The half-wave dipole antenna
Most antennas can be analysed by considering them to be transmission lines whose config-
urations and physical dimensions have been altered to present easy energy transfer from
transmission line to free space. In order to do this effectively, most antennas have physical
sizes comparable to their operational wavelengths.
Figure 1.12(a) shows a two wire transmission line, open-circuited at one end and
driven by a sinusoidal r.f. generator. Electromagnetic waves will propagate along the
line until it reaches the open-circuit end of the line. At the open-circuit end of the line,
the wave will be reflected and travel back towards the sending end. The forward wave
and the reflected wave then combine to form a voltage standing wave pattern on the line.
The voltage is a maximum at the open end. At a distance of one quarter wavelength from
Fig. 1.12 (a) Voltage standing-wave pattern on an open-circuited transmission line; (b) open-circuited line forming a
Antennas and aerials 17
Fig. 1.13 Polar pattern of a half-wave dipole
the end, the voltage standing wave is at a minimum because the sending wave and the
reflected wave oppose each other. Suppose now that the wires are folded out from the λ/4
points, as in Figure 1.12(b). The resulting arrangement is called a half-wave dipole antenna.
Earlier we said that the electromagnetic fields around the parallel conductors overlap
and cancel outside the line. However, the electromagnetic fields along the two (λ/4) arms
of the dipole are now no longer parallel. Hence there is no cancellation of the fields. In
fact, the two arms of the dipole now act in series and are additive. They therefore reinforce
each other. Near to the dipole the distribution of fields is complicated but at a distance of
more than a few wavelengths electric and magnetic fields emerge in phase and at right-
angles to each other which propagate as an electromagnetic wave.
Besides being an effective radiator, the dipole antenna is widely used as a VHF and TV
receiving antenna. It has a polar diagram which resembles a figure of eight (see Figure
1.13). Maximum sensitivity occurs for a signal arriving broadside on to the antenna. In this
direction the ‘gain’ of a dipole is 1.5 times that of an isotropic antenna. An isotropic
antenna is a theoretical antenna that radiates or receives signals uniformly in all directions.
The gain is a minimum for signals arriving in the ‘end-fire’ direction. Gain decreases
by 3 dB from its maximum value when the received signal is ±39° off the broadside direc-
tion. The maximum gain is therefore 1.5 and the half-power beam-width is 78°. The input
impedance of a half-wave dipole antenna is about 72 Ω. It turns out that the input imped-
ance and the radiation resistance of a dipole antenna are about the same.
1.5.4 Folded dipole antenna
The folded dipole (Figure 1.14) is a modified form of the dipole antenna. The antenna is
often used for VHF FM receivers. The impedance of a folded λ/2 dipole is approximately
292 W. This higher input impedance is advantageous for two main reasons:
18 Basic features of radio communication systems
Fig. 1.14 Folded dipole antenna
• it allows easy connection to 300 W balanced lines.
• its higher impedance makes it more compatible for use in directive aerials (particularly
Yagi arrays) which will be described in Section 1.6.
1.5.5 The monopole or vertical rod antenna
The monopole or vertical rod antenna (Figure 1.15) is basically a coaxial cable14 whose
outer conductor has been removed and connected to earth. It is usually about λ/4 long
except in cases where space restrictions or other electrical factors restrict its length. At
high frequencies, the required λ/4 length is short and the antenna can be made self-
supporting by the use of hollow metal tubing. At low frequencies where a greater length is
required, the antenna is often supported by poles.
Fig. 1.15 Rod or monopole antenna
This antenna is favoured for use in low frequency transmitting stations, in portable radio
receivers, in mobile radio-telephones, and for use on motor vehicles because it has a circu-
lar polar receiving pattern, i.e. it transmits and receives signals equally well in all directions
around its circumference. This is particularly important in mobile radio-phones and in
motor vehicles because a motor vehicle may be moving in any direction with respect to a
transmitting station. To minimise interference from the engine of the vehicle and for
14 A typical example of a coaxial cable is the TV lead which connects your television set to the antenna.
Antennas and aerials 19
maximum receiving height, rod aerials are frequently mounted on the roofs of vehicles.
These aerials are also often mounted at an angle of about 45° to the horizon to enable them
to be receptive to both horizontal and vertical polarisation transmissions.
1.5.6 Single loop antennas
Another type of antenna which is frequently used for TV reception is the single loop
antenna shown in Figure 1.16. This loop antenna usually has an electrical length equal to
approximately λ/2 at its operating frequency. It is popular with TV manufacturers because
it is comparatively cheap and easy to manufacture. The antenna’s input impedance is
approximately 292 Ω and it is easily coupled to 300 Ω balanced transmission lines. The
antenna is directive and has to be positioned for maximum signal pick-up.
Fig. 1.16 Single loop antenna
1.5.7 Multi-loop antennas
At low frequencies, particularly at frequencies in the medium wave band where wave-
lengths are long, single loop λ/2 length antennas are not practical; multi-loop antennas
(Figure 1.17) are used instead. The multi-loop antenna can be reduced even further in size
if a ferrite rod is inserted within the loop.
The open-circuit voltage induced in multiple loop antennas can be calculated by making
use of Faraday’s Law of Electromagnetic Induction which states that the voltage induced
in a coil of n turns is proportional to the rate of change of magnetic flux linkage. For
simplicity in derivation, it will be assumed that the incident radiation is propagating along
the axis of the coil (see Figure 1.18).
Fig. 1.17 Multi-loop antenna
20 Basic features of radio communication systems
Fig. 1.18 Multi-looped antenna aligned for maximum flux linkage
Expressing Faraday’s Law mathematically,
e=n (1.10)
e = open-circuit voltage in volts
n = number of turns on coil
dφ/dt = rate of change of magnetic flux linkage (φ = webers and t = seconds)
Some fundamental magnetic relations are also required. These include:
total flux φ = flux density (B) per unit area × area (A)
φ webers = Btesla × Asquare metres (1.11)
By definition flux density in air cored coil ( Btesla ) is given by
free-space permeability (µ0) × magnetic field strength (H)
B(tesla) = µ 0 (henry metre) × H(ampere metre) (1.12)
Suppose that the incident wave has a magnetic field strength
H = H max sin ωt (1.13)
where ω is the angular frequency of the r.f. signal. Then substituting Equations 1.12 and
1.13 in Equation 1.11 yields
φ = BA = µ 0 Hmax sin ωt × A (1.14)
Taking the rate-of-change15 in Equation 1.14, then the induced voltage is
= nωµ 0 AHmax cos ωt (1.15)
15 If you do not know how to differentiate to get the rate of change of a value, then please refer to a maths
Antennas and aerials 21
For a coil with a ferrite core, the flux density is increased by the relative effective perme-
ability (µr), giving
e = nωµ 0 µ r AHmax cos ωt (1.16)
You will see that the ferrite core has increased the effective area of the coil by a factor µr.
Ferrite cores with effective relative permeabilities of 100–300 are readily available but
even with these values, the effective area of the aerial is relatively small when compared
with a λ/2 aerial length. The ferrite rod aerial is therefore very inefficient when compared
to an outdoor aerial but it is popular because of its convenient size and portability. At
medium wave frequencies, the inherent poor signal pick-up is acceptable because broad-
cast stations radiate large signals.
In the foregoing derivation, it has been assumed that the magnetic field has been cutting
the coil along its axis. Occasions arise when the incident magnetic field arrives at an angle
a with respect to the axis of the coil. This is shown in Figure 1.19. In this case the effec-
tive core area is reduced by cos a, and the induced voltage becomes
e = nωµ 0 AHmax cos ωt cos α (1.17)
This expression shows that the induced open-circuit voltage, e, is dependent on the axial
direction of the aerial coil with respect to the direction of the propagation. It is maximum
when cos a = 1, i.e. a = 0°, and minimum when cos a = 0, i.e. a = 90°. This explains why
it is necessary to position a loop aerial to receive maximum signal from a particular broad-
casting station and this is done in a portable radio receiver by orienting its direction.
The above reasons apply equally well to ferrite rod aerials and for these cases we have
an induced voltage
e = nωµ r µ 0 AHmax cos ωt cos α (1.18)
If the magnetic field strength is given as an r.m.s. value (Hrms), then the r.m.s. value of the
induced voltage is
erms = nωµ r µ 0 AHrms cos α (1.19)
Finally, ferrite aerials are seldom used at the higher frequencies because ferrite can be
extremely lossy above 10 MHz.
Fig. 1.19 H field arriving at an angle α
22 Basic features of radio communication systems
Example 1.4
A coil of 105 turns is wound on a ferrite rod with an effective cross-sectional area of 8 ×
10–5 m2. The relative permeability of the ferrite is 230 and the permeability of air is 4π ×
10–7 henry/m. The r.m.s. field strength is 10 µA/m. If the magnetic field is incident along
the axis of the coil and the frequency of operation is 1 MHz, what is the r.m.s. open-circuit
voltage induced in the coil?
Given: No. of coil turns = 105, effective cross-sectional area of ferrite rod = 8 × 10–5 m2,
relative permeability (µr) = 230, permeability of air (µ0) = 4π × 10–7 Henry/metre, r.m.s.
field strength = 10 µA/m, frequency = 1 MHz.
Required: r.m.s. open-circuit voltage induced in coil.
Solution. Using Equation 1.19
erms = nωµ r µ 0 AHrms cos α
= 105 × 2π × 1 × 10 6 × 230 × 4π × 10 −7 × 10 × 10 −6 × 8 × 10 −5 × cos 0°
= 152.5 µV
Broadcasting authorities tend to quote electric field strengths rather than magnetic field
strengths for their radiated signals. This creates no problems because the two are
related by the wave impedance formula given earlier as Equation 1.8. This is repeated
electric field strength ( E )
= 377 Ω
magnetic field strength ( H )
Example 1.5
A coil of 100 turns is wound on a ferrite rod with an effective cross-sectional area of 8 ×
10–5 m2 . The relative permeability of the ferrite is 200 and the permeability of air is 4π ×
10–7 henry/m. The magnetic field is incident at an angle of 60° to the axis of the coil and
the frequency of operation is 1 MHz. If the electric field strength is 100 µV/m, what is the
r.m.s. open-circuit voltage induced in the coil?
Given: No. of coil turns = 105, effective cross-sectional area of ferrite rod = 8 × 10–5
m2, relative permeability (µr) = 200, permeability, of air (µ0) = 4π × 10−7 henry/metre,
incidence of magnetic field = 60°, frequency = 1 MHz, electric field strength = 100
Required: Open-circuit voltage (erms).
Solution. Substituting Equation 1.8 in Equation 1.19 yields
erms = nωµ r µ 0 A cos α
100 × 10 −6
= 100 × 2π × 1 × 10 6 × 200 × 4π × 10 −7 × 8 × 10 −5 × × cos 60°
= 1.68 µV
Antenna arrays 23
1.6 Antenna arrays
1.6.1 Introduction
Antenna arrays are used to shape and concentrate energy in required patterns. One of the
more common domestic arrays is the Yagi-Uda array used for the reception of television
1.6.2 Yagi-Uda array
The Yagi-Uda aerial array shown in Figure 1.20 is one of the most commonly used antenna
arrays. It is used extensively for the reception of TV signals and can be seen on the roofs of
most houses. The Yagi array is an antenna system designed with very similar principles to
the car headlamp system described in Section 1.5.1. Its main elements are a folded dipole, a
reflector, and directivity elements which serve as ‘electrical lenses’ to concentrate the signal
into a more clearly defined beam. The number of directors per array varies according to the
gain required from the aerial. The length of directors and the spacing between them are also
dependent on the number of elements used in the array. In general, gain increases with the
number of directors, but greater gain needs more careful alignment with the transmitting
station and requires that the antenna be more sturdily mounted otherwise its pointing direc-
tion will waver in high winds which can cause fluctuations in the received signal strength.
The Yagi array is usually designed to be connected to a 75 W transmission line.16 Yagi
Fig. 1.20 Yagi-Uda array: (a) physical arrangement;(b) radiation pattern
16 Earlier on, we said that the impedance of a folded dipole aerial was 292 W, yet now we say that this antenna
is designed to operate with a 75 W system. This apparent discrepancy arises because the use of reflector and direc-
tors loads the folded dipole and causes its impedance to fall. Judicious director spacing is then used to set the
array to the required impedance.
24 Basic features of radio communication systems
arrays suitable for operation over the entire TV band can be obtained commercially, but
these broadband arrays are usually designed to ‘trade off’ bandwidth against aerial gain.
Broadband Yagi arrays are extremely useful for mobile reception where minimum space
and convenience are of importance. (You often see them on top of mobile caravans.)
Domestic Yagi arrays are usually designed to provide greater gain but with a more
restricted operational frequency band. The latter is not a disadvantage because TV stations
operating from a common transmitting site confine their broadcasts to well defined
frequency bands. The common practice for domestic Yagi arrays17 is to use three or more
designs (scaled in size) to provide reception for the complete TV band.
Typical values for Yagi arrays operating in the TV band are shown in Table 1.1. These
figures have been taken from a well known catalogue but some of the terms need expla-
Table 1.1 Typical values for Yagi arrays operating in the TV band
No. of elements Forward gain Front/back ratio Acceptance angle
(±0.5 dB) (± 2 dB) (±3°)
• ‘Number of elements’ means the total number of directors, folded dipoles and reflectors
used in the array. For example, if the number of elements in an array is 10, the array
includes eight directors, one folded dipole and one reflector.
• ‘Forward gain’ is the maximum ‘gain’ which the antenna can provide with respect to an
isotropic aerial. A maximum aerial gain of 10 dB means that the antenna will provide 10
times the ‘gain’ you would get from an isotropic aerial when the array is pointed in its
maximum gain direction.
• ‘Front to back ratio’ is the difference in gain between the direction of maximum antenna
gain and the minimum direction of gain which is usually in the opposite direction. This
ratio is important because it provides a measure of how the array behaves towards inter-
fering signals arriving from different directions. It is particularly useful in confined areas
such as cities where interfering signals ‘bounce’ off high buildings and interfere with a
strong desired signal. In such cases, it is often better to select an antenna with a large
front to back ratio to provide rejection to the interfering signal than trying to get maxi-
mum antenna gain.
• ‘Acceptance angle’ is the beamwidth angle in degrees where antenna gain remains
within 3 dB of its stated maximum gain. An acceptance angle of 20° and a maximum
array gain of 10 dB means that for any signal arriving within ±10° of the maximum
gain direction the antenna will provide at least (10 – 3) dB, i.e. 7 dB of gain. However,
you should be aware that the acceptance angle itself is not accurate and that it can vary
by ±3° as well.
17 There is a class of Yagi arrays known as Log Periodic Yagis. These have greater bandwidths because the
directors are spaced differently. They do cover the entire TV bands but their gain is a compromise between
frequency bandwidth and gain.
Antenna distribution systems 25
The values given in the table are representative of the middle range of commercially
available Yagi arrays. The figures quoted above have been measured by manufacturers
under ideal laboratory conditions and proper installation is essential if the specification is
to be achieved in practice.
1.7 Antenna distribution systems
Occasions often arise where it is desired to have one antenna supply signal to several tele-
vision and radio receivers. A typical example is that of an apartment block, where a single
aerial on the roof supplies signals to all the apartments. Another possible use for such a
system is in your own home where you would like to distribute signals to all rooms from
a single external aerial. In such cases, and for maximum efficiency, an aerial distribution
system is used. There are many ways of designing such a system but before discussing
them, it is best to understand some of the terms used.
1.7.1 Balanced and unbalanced systems
Examples of balanced and unbalanced aerials and distribution lines are shown in Figures
1.21 and 1.22. You should refer to these figures while you are reading the descriptions
given below.
A balanced antenna (Figure 1.21(a) and (b)) is an aerial which has neither conductor
connected directly to earth; it is balanced because the impedance between earth and each
conductor is the same. A folded dipole is a typical example of a balanced antenna because
the impedance from each end of the antenna to earth is equal and balanced. An unbalanced
Fig. 1.21 Balanced antenna system and (a) balanced distribution system; (b) unbalanced distribution system
26 Basic features of radio communication systems
Fig. 1.22 Unbalanced antenna system and (a) unbalanced distribution system; (b) balanced distribution system
antenna (Figure 1.22(a) and (b)) is an aerial which has one of its conductors connected
directly to earth. The impedance between earth and each conductor is not the same. A
monopole aerial is a typical example of an unbalanced aerial because its other end (see
Figure 1.15) is connected to earth.
A balanced line (Figure 1.21(a) and (b)) is a transmission line where the impedance
between earth and each conductor is identical. A twin pair cable is an example of a
balanced line because the impedance between earth and each conductor is the same. An
unbalanced line (Figures 1.22(a), 1.22(b) is a transmission line where the impedance
between earth and each conductor is not equal. A coaxial cable is an example of an unbal-
anced line because the impedance between earth and the outer shield is different to the
impedance between earth and the inner conductor.
The key to the connections in Figures 1.21 and 1.22 is the balanced/unbalanced trans-
former. These transformers are carefully wound to produce maximum energy transfer by
magnetic coupling. Coil windings are designed to have minimum self-capacitance, mini-
mum inter-winding capacitance and minimum capacity coupling between each winding
and earth. No direct connection is used between input and output circuits. The above
conditions are necessary, otherwise balanced circuits will become unbalanced when parts
of the circuit are connected together. The balanced/unbalanced transformer is bi-direc-
tional; it can be used to pass energy in either direction.
As the operational frequencies become higher and higher (above 2 GHz), it becomes
increasingly difficult to make such a good transformer and a transformer is simply not used
and antennas and transmission lines are connected directly. In such cases, the systems
resolve to either an unbalanced antenna and distribution system or a balanced antenna and
distribution system. The unbalanced system is almost always used because of convenience
and costs.
Antenna distribution systems 27
1.7.2 Multi-point antenna distribution systems
In the design of antenna distribution systems, transmission lines connecting signal distribution
points must function efficiently; they must carry signal with minimum loss, minimum inter-
ference and minimum reflections. Minimum loss cables are made by using good conductivity
materials such as copper conductors and low loss insulation materials. Minimum interference
is obtained by using coaxial cables whose outer conductor shields out interference signals.
Reflections in the system are minimised by proper termination of the cables. For proper termi-
nation and no reflections in the system, two conditions must be fulfilled:
• the antenna and cable must be terminated in its characteristic impedance Z 0;
• the source impedance (Z s) feeding each receiver must be matched to the input imped-
ance of the receiver (Z in), i.e. Z s = Z in, otherwise there will be signal reflections and
minimum cable transmission loss will not be obtained.
In Figure 1.23, an aerial of characteristic impedance (Z 0) is used to feed a transmission
(TX) line with a characteristic impedance Z 0. The output of the line is fed to a number (n)
of receivers, each of which is assumed to have an input impedance (Z in) equal to Z 0. Resis-
tors R represent the matching network resistors which must be evaluated to ensure prop-
erly terminated conditions.
For the system to be properly terminated, it is essential that the aerial and cable system
be terminated with Z 0, i.e. the impedance to the right of the plane ‘AE’ must present an
impedance Z 0 to the antenna and cable system. It is also essential that each receiver be
energised from a source impedance (Z s) matched to its own input impedance (Z in), i.e. Z s
= Z in. For ease of analysis we will assume the practical case, Z s = Z in = Z 0.
Now for the transmission line in Figure 1.23 to be properly terminated:
R + [R + Z 0]/n = Z 0
Multiplying both sides by n:
nZ 0 = nR + R + Z 0
Collecting and transposing terms gives:
(n – 1)
R = ——— Z 0 (1.20)
(n + 1)
Fig. 1.23 Aerial distribution system for n receivers, each with an input impedance of Zin
28 Basic features of radio communication systems
This equation is all we need to calculate the value of the matching resistors in Figure
Example 1.6
A 75 W aerial system is used to supply signals to two receivers. Each receiver has an input
impedance of 75 W. What is the required value of the matching resistor?
Given: 75 W aerial system, input impedance of each receiver = 75 Ω, no. of receivers = 2.
Required: Value of matching resistor.
Solution. Using Equation 1.20 with n = 2, we obtain
(n – 1) (2 – 1)
R = ——— Z 0 = ——— 75 = 25 W
(n + 1) (2 + 1)
Example 1.7
A 50 W aerial receiving system is to be used under matched conditions to supply signal to
four receivers, each of input impedance 50 W. If the configuration shown in Figure 1.23 is
used, calculate the value of the resistor, R, which must be used to provide matching condi-
Given: 50 Ω aerial system, input impedance of each receiver = 50 Ω, no. of receivers = 4.
Required: Value of matching resistor.
Solution. From Equation 1.20
(n – 1) (4 – 1)
R = ——— Z 0 = ——— 50 = 30 W
(n + 1) (4 + 1)
From the answers above, it would appear that an aerial system can be matched to any
number of receivers. This is true only within limits because the signal level supplied to
individual receivers decreases with the number of distribution points. With large numbers
of receivers, network losses become prohibitive.
Transmission losses associated with the matching network of Figure 1.23 can be calculated
by reference to Figure 1.24. The network has been re-drawn for easier derivation of circuit
losses but Z 0, R and n still retain their original definitions.
Fig. 1.24 Calculating the signal loss in an antenna distribution system
Antenna distribution systems 29
In Figure 1.24
Voc = open-circuit source voltage from the aerial
Vce = terminated voltage at an intermediate point in the network
Vout = terminated voltage at the input to a receiver
By inspection
⎧ R + Z0 ⎫
⎨ ⎬
Vout =
Vce and Vce = ⎩ n ⎭ Voc
R + Z0 ⎧ R + Z0 ⎫ + +
⎨ ⎬ R Z0
⎩ n ⎭
Z0 R + Z0 Z0
Vout = × Voc = Voc
R + Z0 (n + 1)( R + Z0 ) (n + 1)( R + Z0 )
Using Equation 1.20 and substituting R = [(n – 1)/(n + 1)]Z 0 in the above equation
Z0 V
Vout = Voc = oc
⎡ (n − 1) ⎤ 2n
(n + 1)⎢ Z0 + Z0 ⎥
⎣ (n + 1) ⎦
Transposing, we find that
Vout 1
voltage transmission loss = = (1.21)
Voc 2n
voltage transmission loss = 20 log ⎡ ⎤dB18
⎢ 2n ⎥
⎣ ⎦
Example 1.8
A broadcast signal induces an open-circuit voltage of 100 µV into a rod aerial. The aerial
system has a characteristic impedance of 50 Ω and it is used to supply signal to three iden-
tical receivers each of which has an input impedance of 50 Ω. If the matching network type
shown in Figure 1.23 is used, calculate (a) the value of the resistance (R) required for the
matching network and (b) the terminated voltage appearing across the input terminals of
the receiver.
18 dB is short for decibel. The Bel is a unit named after Graham Bell, the inventor of the telephone. 1 Bel =
log10 [power 1(P1)/power 2(P2)]. In practice the unit Bel is inconveniently large and another unit called the deci-
bel is used. This unit is 1/10 of a Bel. Hence 1 Bel = 10 dB or dB = 10 log10 [P1 /P2] = 10 log10 [(V12/R)/(V22/R)]
= 20 log10 [V1/V2]
30 Basic features of radio communication systems
Given: 50 Ω aerial system, input impedance of each receiver = 50 Ω, no. of receivers = 3,
open-circuit voltage in aerial = 100 µV.
Required: (a) Value of matching resistor, (b) terminated voltage at receiver input termi-
(a) For the matching network of Figure 1.23
(n − 1) (3 − 1)
R= Z0 = × 50 = 25Ω
(n + 1) (3 + 1)
(b) Using Equation 1.20
Vreceiver = Vantenna = × 100 µV = 16.67 µV
2n 6
1.7.3 Other aerial distribution systems
The matching network shown in Figure 1.23 is only one type of matching network. Figure
1.25 shows a commercially available matching network for two outlets. This network is
sometimes called a two way splitter because it splits the signal from a single input port
into two output ports. The circuit has been designed for low insertion loss and it does this
by trading off proper matching against insertion loss.
Fig. 1.25 Two way splitter
Example 1.9
Figure 1.25 shows a commercially available 75 Ω matching network. Calculate: (a) the
ratio Vout/Voc when all ports are each terminated with 75 Ω, (b) the input impedance to the
matching network when the output ports are each terminated with 75 Ω and (c) the source
impedance to either receiver when the remaining ports are each terminated in 75 Ω.
Given: 75 Ω network splitter of Figure 1.25 with 75 Ω terminations.
Required: (a) Ratio Vout/Voc, (b) input impedance of matching network, (c) source
impedance to either receiver.
Solution. By inspection:
75 (43 + 75)/2
(a) Vout = ———— × ——————— × Voc = 0.28
43 + 75 (43 + 75)/2 + 75
Antenna distribution systems 31
(b) input impedance to the network = (43 + 75)/2 = 59 W
(43 + 75)(75)
(c) receiver source impedance = 43 + ——————— = 89 W
(43 + 75) + (75)
From the answers to Example 1.9, it can be seen that the insertion loss is slightly reduced
but this has been carried out at the expense of system match. The manufacturer is fully
aware of this but relies on the fact that the reflected signal will be weak and that it will not
seriously affect signal quality. The manufacturer also hopes that the cable system will be
correctly matched by the antenna and that any reflections set up at the receiver end will be
absorbed by the hopefully matched antenna termination to the cable system. This design
is popular because the installation cost of an additional resistor is saved by the manufac-
1.7.4 Amplified antenna distribution systems
Amplified aerial distribution systems are aerial distribution systems which incorporate
amplifiers to compensate for signal transmission, distribution and matching losses. Two
systems will be discussed here. The first concerns a relatively simple distribution system
where indoor amplifiers are used. The second system deals with a more elaborate system
using amplifiers mounted on the aerial (masthead amplifiers) to compensate for distribu-
tion and matching losses.
1.7.5 Amplified aerial distribution systems using amplifiers
The block diagram of an amplified aerial distribution system using an amplifier is shown
in Figure 1.26. This system is often used in domestic environments. Outdoor aerials
provide the incoming signals, UHF for TV and/or VHF for FM-radio to the input of an
amplifier. The gain of this amplifier is nominally greater than 10 dB but this varies accord-
ing to the particular amplifier used. The amplifier is usually placed on the antenna mast-
head or near the aerial down-lead cables and a power supply point in the attic.
Fig. 1.26 Antenna amplifier distribution network
32 Basic features of radio communication systems
Output signals from the amplifier are fed into matching networks for distribution to
individual terminals. To save cabling costs, both UHF and VHF signals are often carried
on the same cables. Filters (a high pass filter for UHF and a low pass filter for VHF) are
installed at individual terminals to feed the signals to their designated terminals.
The main advantage of such a system is that it is relatively easy to install especially if
wiring for the distribution already exists. It also compensates for signal loss in the distrib-
ution network. The amplifier casing is relatively cheap when the amplifier is used indoors
as it does not have to be protected from extreme weather conditions. The main disadvan-
tage of indoor mounting is that signals are attenuated by the aerial down-lead cables before
amplification. This signal loss decreases the available signal before amplification and
therefore a poorer signal-to-noise ratio is available to the distribution points than if the
amplifier was to be mounted on the masthead.
1.8 Radio receivers
1.8.1 Aims
The aims of this section are to introduce you to:
• the tuned radio frequency receiver
• the superhet receiver
• the double superhet receiver
• selectivity requirements in receivers
• sensitivity requirements in receivers
• concepts of signal-to-noise and sinad ratios
• noise figures of receivers
1.8.2 Objectives
After reading this section you should be able to understand:
• the basic principles of tuned radio frequency receivers
• the basic principles of superhet receivers
• the basic principles of satellite receivers
• the concepts of selectivity
• the concepts of sensitivity
• the concepts of signal-to-noise and sinad ratios
• the concepts of noise figures
1.8.3 Introduction
Radio receivers are important because they provide a valuable link in communications and
entertainment. Early receivers were insensitive, inefficient, cumbersome, and required
large power supplies. Modern designs using as little as one integrated circuit have
overcome most of these disadvantages and relatively inexpensive receivers are readily
Radio receiver properties 33
Fig. 1.27 Spacing of broadcast stations in the medium wave band
1.8.4 Fundamental radio receiver requirements
In the AM medium wave band, broadcasting stations transmit their signals centred on
assigned carrier frequencies. These carrier frequencies are spaced 9 kHz apart from each
other as in Figure 1.27 and range from 522 kHz to 1620 kHz. The information bandwidth
allocated for each AM transmission is 9 kHz. This means that modulation frequencies
greater than 4.5 kHz are not normally used. To receive information from a broadcast
signal, an AM broadcast receiver must be tuned to the correct carrier frequency, have a
bandwidth that will pass the required modulated signal, and be capable of extracting infor-
mation from the required radio signal to operate desired output devices such as loud-
speakers and earphones.
Discussions that follow pertain mainly to receivers operating in this band. This is not a
limitation because many of the principles involved apply equally well to other frequency
bands. When the need arises, specific principles applying to a particular frequency band
will be mentioned but these occasions will be clearly indicated.
1.9 Radio receiver properties
A radio receiver has three main sections (see Figure 1.28).
• A radio frequency section to select and if necessary to amplify a desired radio
frequency signal to an output level sufficient to operate a demodulator.
• A demodulator section to demodulate the required radio signal and extract its modu-
lated information.
• A post-demodulation section to amplify demodulated signals to the required level to
operate output devices such as loudspeakers, earphones and/or TV screens.
Fig. 1.28 Three main sections of a radio receiver
34 Basic features of radio communication systems
1.9.1 Radio frequency section
A radio frequency section is designed to have the following properties.
Receiver selectivity is a measure of the ability of a radio receiver to select the desired
transmitted signal from other broadcast signals. An ideal selectivity response curve for an
AM broadcast receiver centred on a desired carrier frequency ( f 0) is shown in Figure 1.29.
Two main points should be noted about the ideal selectivity response curve. First, it should
have a wide enough passband (9 kHz approx.) to pass the entire frequency spectrum of the
desired broadcast signal. Second, the passband should present equal transmission charac-
teristics to all frequencies of the desired broadcast signal. In addition, the bandwidth
should be no wider than that required for the desired signal because any additional band-
width will allow extraneous signals and noise from adjacent channels to impinge on the
receiver. Notice that the skirts of the ideal selectivity curve are vertical, so that the attenu-
ation of any signal outside the passband is infinitely high.
Fig. 1.29 Ideal selectivity curve for an AM medium wave broadcast receiver
In practice, costs and stability constraints prevent the ideal selectivity response curve
from ever being attained and it is more rewarding to examine what is achieved by commer-
cial receivers.
An overall receiver selectivity response curve for a typical domestic transistor receiver
for the reception of AM broadcast signals is shown in Figure 1.30. In the table supplied
with Figure 1.30, you should note that the selectivity curve is not symmetrical about its
centre frequency. This is true of most tuned circuits because the effective working quality
factor (Qw) of components, particularly inductors, varies with frequency. Note also that the
3 dB bandwidth is only 3.28 kHz and that the 6 dB bandwidth points are approximately
4.82 kHz apart. The 60 dB points are 63.1 kHz apart.
Consider the case of a carrier signal (f0) modulated with two inner sideband frequen-
cies f1L, f1U (±1.64 kHz) and two outer sideband frequencies f2L, f2U (±2.4 kHz) away
from the carrier. The frequency spectrum of this signal is shown in Figure 1.31(a). When
Radio receiver properties 35
Fig. 1.30 Typical selectivity curve of a commercial AM six transistor receiver
this frequency spectrum is passed through a receiver with the selectivity response shown
in Figure 1.30, the inner sidebands f1L, f1U (±1.64 kHz) and the outer sidebands f2L, f2U
(±2.4 kHz) will suffer attenuations of 3 dB and 6 dB approximately with respect to the
carrier. (See the table in Figure 1.30.) The new spectrum of the signal is shown in Figure
Comparison of Figures 1.31(a) and (b) shows clearly that amplitude distortion of the
sidebands has occurred but what does this mean in practice? If the transmitted signal had
been music, there would have been an amplitude reduction in high notes. If the transmit-
ted signal had been speech, the speaker’s voice would sound less natural.
Fig. 1.31(a) Transmitted spectrum Fig. 1.31(b) Distorted spectrum
From the above discussion, it should be noted that for good quality reproduction the
selectivity curve of a receiver should be wide enough to pass all modulation frequencies
without discriminatory frequency attenuation.
Adjacent channel selectivity
A graphical comparison of the selectivity curves of Figures 1.29 and 1.30 is shown in
Figure 1.32. From these curves, it can be seen that the practical selectivity curve does not
provide complete rejection of signals to stations broadcasting on either side of the desired
36 Basic features of radio communication systems
Fig. 1.32 A comparison between the ideal and practical selectivity curve
response channel. The breakthrough of signal from adjacent channels into the desired
channel is known as adjacent channel interference.
Adjacent channel interference causes signals from adjacent channels to be heard in the
desired channel. It is particularly bad when strong adjacent channel signals are present
relative to the desired station. What does this mean in practice? It means that you will
obtain interference from the unwanted station. In a broadcast receiver, you often hear
signals from both channels simultaneously.
Broadcasting authorities minimise adjacent channel interference by forbidding other
transmitters situated near the desired station to broadcast on an adjacent channel. Stations
geographically distant from the desired station are allowed to operate on adjacent channels
because it is likely that their signals will have suffered considerable transmission loss by
the time they impinge on the desired channel.
The sensitivity of a radio receiver is a measure of the modulated signal input level which
is required to produce a given output level. A receiver with good sensitivity requires a
smaller input signal than a receiver with poor sensitivity to produce a given output level.
The sensitivity of a small portable receiver (audio output rated at 250 mW) may be quoted
as 200 µV/m. What this means is that a modulated AM carrier (modulated with a 400 Hz
tone and with an AM modulation depth of 30%) will produce an audio output of 50 mW
under its maximum gain conditions when the input signal is 200 mV/m.
1.9.2 Signal-to-noise ratios
Any signal transmitted through a communications system suffers attenuation in the
passive (non-amplifying) parts of the system. This is particularly true for radio signals
propagating between transmitting and receiving aerials. Attenuation is compensated for
by subsequent amplification, but amplifiers add their own inherent internally generated
random noise to the signal. Noise levels must always be less than the required signal,
otherwise the required signal will be lost in noise. Some means must be provided to
Types of receivers 37
specify the level of the signal above noise. This means is called the Signal-to-Noise
ratio. It is defined as:
S signal power
signal-to-noise ratio = — = —————— (1.23)
N noise power
Notice that S/N is specified as a ratio of power levels.
An alternative way of specifying signal-to-noise ratios is to quote the ratio in decibels.
This is defined by Equation 1.24:
S ⎡ signal power ⎤
signal-to-noise ratio = (dB) = 10 log10 ⎢ ⎥ dB (1.24)
N ⎣ noise power ⎦
A strong signal relative to the noise at the receiver input is essential for good reception. In
practice, we require an S/N of 10–20 dB to distinguish speech, an S/N of 30 dB to hear
speech clearly, and an S/N of 40 dB or better for good television pictures.
Noise figure
Certain amplifiers have more inherent electrical noise than others. Manufacturers usually
produce a batch of transistors, then classify and name the transistors according to their
inherent electrical noise levels. The inherent noise produced by a transistor is dependent
on its general operating conditions, particularly frequency, temperature, voltage and oper-
ating current, and these conditions must be specified when its noise level is measured.
Engineers use the ratio term noise figure to specify noise levels in transistors.
Noise figure is defined as
[ ]
noise figure (N.F.) = ——— at 290 K (1.25)
If a transistor introduces no noise, then its S/N at both the input and output is the same,
therefore from Equation 1.25, N.F. = 1 or in dB N.F. = 10 log 1 = 0 dB. Hence a ‘perfect’
or ‘noiseless’ amplifier has a noise figure of 0 dB. An imperfect amplifier has a noise
figure greater than 0 dB. For example an amplifier with a noise figure of 3 dB (= 2 ratio)
means that it is twice as bad as a perfect amplifier.
1.10 Types of receivers
There are many types of radio receivers. These include:
• tuned radio frequency receivers (TRF)
• superheterodyne receivers (superhets)
• double superheterodyne receivers (double superhets)
1.10.1 Tuned radio frequency receiver
A tuned radio frequency receiver (Figure 1.33) has three main sections, a radio frequency
amplifier section, a detector section, and an audio amplifier section.
38 Basic features of radio communication systems
Fig. 1.33 Main sections of a tuned radio frequency receiver
The radio frequency section consists of one or more r.f. amplifiers connected in
cascade.19 For efficient operation, all tuned circuit amplifiers must be tuned to exactly the
same broadcast frequency and to ensure that this is the case, all tuning adjusters are fixed
on to a common tuning shaft. Tuning capacitors which are connected in this manner are
said to be ‘ganged’ and two and three stage ganged tuning capacitors are common.
The detector is usually a conventional AM diode type detector. This type of detector is
usually a diode which detects the positive peaks of the modulated carrier and filters the r.f.
out, so that the remaining signal is the inital low frequency modulation frequency.
The audio section uses audio amplifiers which serve to amplify the signals to operate a
loudspeaker. This section is similar to the amplifier in your home which is used for play-
ing compact disks (CD) and cassettes.
The main advantages of TRF receivers are that they are relatively simple, easy to construct,
and require a minimum of components. A complete TRF receiver can be constructed using
a single integrated circuit such as a ZN414 type chip.
TRF receivers suffer from two main disadvantages, gain/bandwidth variations and poor
selectivity. The inevitable change of gain and bandwidth as the receiver is tuned through
its frequency range is due to changes in the selectivity circuits.
Circuit instability can be a problem because it is relatively easy for any stray or leaked
signal to be picked up by one of the many r.f. amplifiers in the receiver. R.F. signal can also
be easily coupled from one r.f. stage to another through the common power supply. To
minimise these risks, r.f. amplifiers are usually shielded and de-coupled from the common
power supply.
1.10.2 Superheterodyne receiver
Block diagram
A block diagram of a superheterodyne (commonly called superhet) receiver is shown in
Figure 1.34.
This receiver features an r.f. section which selects the desired signal frequency ( frf).
This signal is then mixed with a local carrier at frequency (fo) in a frequency changer to
produce an intermediate frequency ( fif) which retains the modulated information initially
19 Here, cascade is meant to imply one amplifier following another amplifer and so on.
Types of receivers 39
Fig. 1.34 Block diagram of a superhet radio receiver
carried by frf. The intermediate frequency (fif) then undergoes intensive amplification
(60–80 dB) in the intermediate frequency amplifiers to bring the signal up to a suitable
level for detection and subsequent application to the post-detection (audio) amplifiers.
Radio frequency amplifiers are sometimes included in the r.f. section in order to make
the noise figure of the receiver as small as possible. Frequency changers have compara-
tively larger noise figures (6–12 dB) than r.f. amplifiers.
The frequency of the local oscillator ( fo) is always set so that its frequency differs from
the desired frequency ( frf) by an amount equal to the intermediate frequency ( fif), i.e.
fo – frf = fif (1.26)
frf – fo = fif (1.27)
Equation 1.26 is more usual for medium-wave receivers. Typical tuning ranges for a
medium-wave receiver with fif = 465 kHz are 522–1620 kHz for frf and 987–2085 kHz for
The main advantages of the superhet receiver are as follows.
• Better selectivity because fixed bandpass filters with well defined cut-off frequency points
can be used in the i.f. stages of a superhet. Filters and tuned circuits are also less complex
because they need only operate at one frequency, namely the intermediate frequency.
• In a superhet, tuning is relatively simple. A two ganged capacitor can be used to tune the
r.f. and oscillator sections simultaneously to produce the intermediate frequency for the
i.f. amplifiers.
• R.F. circuit bandwidths are not critical because receiver selectivity is mainly determined
by the i.f. amplifiers.
The main disadvantages of superhets are as follows.
• Image channel interference is caused by the local oscillator (fo) combining with an
undesired frequency (fim) which is separated from the desired frequency (frf) by twice the
i.f. frequency (fif). Expressed mathematically
40 Basic features of radio communication systems
fim = frf ± 2fif (1.28)
The term 2nd channel interference is another name for image channel interference.
Image channel interference is more easily understood by substituting some arbi-
trary values into Equations 1.26 and 1.27. For example, assume that the local oscil-
lator of a superhet is set to 996 kHz and that its intermediate frequency amplifiers
operate at 465 kHz. Then, either of two input frequencies, 996 – 465 = 531 kHz
(Equation 1.26) or 996 + 465 = 1461 kHz (Equation 1.27), will mix with the local
oscillator to produce a signal in the i.f. amplifiers. If the desired frequency is
531 kHz, then the undesired frequency of 1461 kHz is 2( fif) or 930 kHz away, i.e. it
forms an image on the other side of the oscillator frequency. This condition is shown
graphically in Figure 1.35.
• There is the possibility that any strong signal or sub-harmonics of 465 kHz (fif) might
impinge directly on the i.f. amplifiers and cause interference.
• Any harmonic of the oscillator (fo) could mix with an unwanted signal to produce
unwanted responses. For example
2 × 996(fo) kHz – 1527 kHz = 465 kHz
The spurious responses stated above are minimised in superhets by using tuned circuits in
the r.f. section of the receiver to select the desired signal and to reject the undesired ones.
The local oscillator is also designed to be ‘harmonic free’.
Fig. 1.35 Image response in superhet receivers
1.10.3 Double superheterodyne receivers
A block diagram of a double conversion superhet used for receiving direct broadcast
signals (DBS) from satellites is shown in Figure 1.36. Direct broadcasting satellites for the
United Kingdom region transmit in the 11.6–12.4 GHz band. Each TV channel uses a
26 MHz bandwidth.
The double superhet is basically a superhet receiver with two i.f. sections. The first i.f.
section operates at a much higher i.f. frequency than the second i.f. section. This choice is
deliberate because a higher 1st i.f. frequency gives better image channel rejection. You
have already seen this in the calculations of p. 40. The 2nd i.f. section is made to operate
at a lower frequency because it gives better adjacent channel selectivity.
In this receiver, the input signal, f1, is selected, mixed with a local oscillator carrier, fx,
and frequency translated to form the first i.f. frequency ( fif1). This signal is applied to the
Summary 41
Fig. 1.36 Block diagram of a double conversion superhet receiver
1st i.f. amplifier section, then mixed with fo to produce a second intermediate frequency
(fif2) and amplified prior to detection and low frequency amplification.
In a typical direct broadcast satellite receiver, the first r.f. amplifier section operates in
the band 11.6–12.4 GHz. The first local oscillator ( fx) is operated at a fixed frequency of
10.650 GHz. The resultant first i.f. frequency bandwidth range is 950 to 1750 MHz and
is really the r.f. band translated to a lower frequency band. This i.f. is then amplified by
the first set of 1st i.f. amplifiers. All the foregoing action takes place in a masthead unit
which is mounted directly on the antenna. The total gain including r.f. amplification,
frequency conversion and i.f. amplification is about 55 dB. This high order of gain is
necessary to compensate for the losses which occur in the down-lead coaxial cable to the
satellite receiver which is situated within the domestic environment. The satellite receiver
treats the 1st i.f. frequency band (950–1750 MHz) as a tuning band and fo is varied to
select the required TV channel which is amplified by the 2nd i.f. section before signal
1.11 Summary
The main purpose of Chapter 1 has been to introduce you to the radio environment in your
home. The knowledge you have gained will assist you in understanding basic radio prop-
agation and reception principles. It will also help you to remedy some of the simpler radio
and TV problems which you are likely to encounter in your home.
In Sections 1.1–1.3, we started with the necessity for modulation and demodulation and
you were introduced to the basic principles of modulation, demodulation and radio prop-
agation. You should now understand the meaning of terms such as amplitude modulation
(AM), frequency modulation (FM), phase modulation (PM) and digital modulation.
In Section 1.4, you were introduced to radio propagation, wave polarisation, field
strength and power density of radio waves.
42 Basic features of radio communication systems
In Sections 1.5 and 1.6, you learned about the properties of several antennas. These
included l/2 dipole, folded dipole, monopole, loop antennas and the Yagi-Uda array.
Section 1.7 dealt with various antenna distributions and matching systems.
In Section 1.8, you encountered some basic concepts concerning the reception of radio
signals. You should now be able to carry out simple calculations with regard to selectivity,
adjacent channel selectivity, sensitivity, S/N ratio and noise figure ratio as applied to radio
Sections 1.9 and 1.10 described the main functions required in a radio receiver, and also
the main advantages and disadvantages of three basic radio receiver types, namely the
TRF, superhet and double superhet receivers. The first type is used in very simple
receivers, the second type is used extensively in domestic receivers and the last type is used
for direct broadcast reception from satellites.
You have now been provided with an overview of a basic radio communication system.
Having established this overview, we will now be in a position to deal with individual sub-
systems and circuits in the next chapters.
Transmission lines
2.1 Introduction
At this stage, I would like to prepare you for the use of the software program called PUFF
which accompanies this book. PUFF (Version 2.1) is very useful for matching circuits, and
the design of couplers, filters, line transformers, amplifiers and oscillators. Figure 2.1
shows what you see when you first open the PUFF program. Figure 2.2 shows you how
the program can be used in the design of a filter. In Figure 2.1 you can see for yourself that
Fig. 2.1 PUFF 2.1 – blank screen (words in italics have been added for explanation)
44 Transmission lines
Fig. 2.2 Bandpass filter design using PUFF
to understand and use the program, you must be familiar with Smith charts (top right hand
corner) and scattering or ‘s-parameters’ (top left hand corner), transmission lines and the
methods of entering data (F3 box) into the program. Within limits, the layout window (F1
box) helps to layout your circuit for etching.
2.1.1 Aims
We shall cover the basic principles of transmission lines in this part, and Smith charts and
s-parameters in Part 3. We will then be in a position to save ourselves much work and avoid
most of the tedious mathematical calculations involved with radio and microwave engi-
The main aims of this chapter are:
• to introduce you to various types of transmission lines;
• to explain their characteristic impedances from physical parameters;
• to provide and also to derive expressions for their characteristic impedances;
• to explain their effects on signal transmission from physical and electrical parameters;
• to explain and derive expressions for reflection coefficients;
• to explain and derive expressions for standing wave ratios;
• to explain and derive the propagation characteristics of transmission lines;
• to provide an understanding of signal distortion, phase velocity and group delay;
Transmission line basics 45
• to show how transmission lines can be used as inductors;
• to show how transmission lines can be used as capacitors;
• to show how transmission lines can be used as transformers.
2.1.2 Objectives
This part is mainly devoted to transmission lines. Knowledge of transmission lines is
necessary in order to understand how high frequency engineering signals can be efficiently
moved from one location to another. For example, the antenna for your domestic TV
receiver is usually mounted on the roof and it is therefore necessary to find some means of
efficiently transferring the received signals into your house. In the commercial world, it is
not unusual for a radio transmitter to be situated several hundred metres from a mast-
mounted transmitting antenna. Here again, we must ensure that minimal loss occurs when
the signal is transferred to the antenna for propagation.
2.2 Transmission line basics
2.2.1 Introduction to transmission lines
In this discussion we shall start off using some basic terms which are easily understood
with sound waves. We will then use these terms to show that these properties are also
applicable to electrical transmission systems. Much of the explanation given in these
sections will be based on examples using sinusoids because they are easier to understand.
But this information applies equally well to digital waveforms because digital signals are
composed of sinusoid components combined in a precise amplitude and phase manner.
Therefore, it is vitally important that you do not form the mistaken idea that transmission
line theory only applies to analogue waveforms.
2.2.2 General properties of transmission systems
Transmission systems are used to transfer energy from one point to another. The energy
transferred may be sound power or electrical power, or digital/ analogue/optical signals or
any combination1 of the above.
One easy way of refreshing your memory about signal transmission is to imagine that
you are looking into a deep long straight tunnel with walls on either side of you. When you
speak, you propagate sound energy along a transmission path down the length of the
tunnel. Your voice is restricted to propagation along the length of the tunnel because walls
on either side act as waveguides.
Waves emerging directly from the sender are known as incident waves. As your vocal
cords try to propagate incident waves along the tunnel, they encounter an opposition or
impedance caused by the air mass in the tunnel. The impedance is determined by the
physical characteristics of the tunnel such as its width and height and the manner in which
1 For example, the coaxial cable connecting the domestic satellite receiver to the low noise amplifier on the
satellite dish often carries d.c. power up to the low noise amplifier, radio frequency signals down to the receiver,
and in some cases even digital control signals for positioning the aerial.
46 Transmission lines
it impedes air mass movement within the tunnel. This impedance is therefore called the
characteristic impedance (Z 0) of the tunnel.
Bends or rock protrusions along the tunnel walls cause a change in the effective dimen-
sions of the tunnel. These discontinuities in effective dimensions can cause minor reflec-
tions in the signal propagation path. They also affect the characteristic impedance of the
transmission channel.
You should note that the walls of the tunnel do not take part in the main propagation of
sound waves. However, they do absorb some energy and therefore weaken or attenuate
the propagated sound energy. Amplitude attenuation per unit length is usually represented
by the symbol α.
Moss, lichen and shrubs growing on the walls will tend to absorb high frequency sound
better than low frequency sound, therefore your voice will also suffer frequency attenu-
ation. Frequency attenuation is known as dispersion.
There is also a speed or propagation velocity with which your voice will travel down
the tunnel. This velocity is dependent on the material (air mixture of gases), its density and
temperature within the tunnel. With sound waves, this velocity is about 331 metres per
If the tunnel is infinitely long, your voice will propagate along the tunnel until it is totally
attenuated or absorbed. If the tunnel is not infinitely long, your voice will be reflected when
it reaches the end wall of the tunnel and it will return to you as an echo or reflected wave.
The ratio reflected wave/incident wave is called the reflection coefficient. You can prevent
this reflection if it were physically possible to put some good sound absorption material at the
end of the tunnel which absorbs all the incident sound. In other words, you would be creating
a matching termination or matched load impedance (ZL) which matches the propagation
characteristics of an infinitely long tunnel in a tunnel of finite length. The ratio of the
received sound relative to the incident sound is known as the transmission coefficient.
A signal travelling from a point A to another point B takes time to reach point B. This
time delay is known as propagation time delay for the signal to travel from point A to
point B. In fact, any signal travelling over any distance undergoes a propagation time
Time propagation delay can be specified in three main ways: (i) seconds, (ii) periodic
time (T) and (iii) phase delay. The first way is obvious, one merely has to note the time
in seconds which it has taken for a signal to travel a given distance. Periodic time (T) is an
interval of time; it is equal to [1/(frequency in Hz)] seconds. For example, if a 1000 Hz
sinusoid requires four periodic times (4T) to travel a certain distance, then the time delay
is 4 × (1/1000) seconds or four milliseconds. Phase delay can be used to measure time
because there are 2π radians in a period time (T). For the example of a 1000 Hz signal, a
phase delay of (4 × 2π) radians is equivalent to four periodic times (T) or four millisec-
onds. Phase delay per unit length is usually represented by the symbol (β). It is measured
in radians per metre.
Hence if we were to sum up propagation properties, there would be at least three prop-
erties which are obvious:
• attenuation of the signal as it travels along the line;
• the time or phase delay as the signal travels along the line;
• dispersion which is the different attenuation experienced by different frequencies as it
travels along a line.
Types of electrical transmission lines 47
Finally, if you walked along a tunnel which produces echoes, while a friend whistled at
a constant amplitude and pitch, you would notice the reflected sound interfering with the
incident sound. In some places, the whistle will sound louder (addition of incident and
reflected signal); in other places the whistle will sound weaker (subtraction of incident and
reflected signal). Provided your friend maintains the whistle at a constant amplitude and
pitch, you will find that louder and weaker sounds always occur at the same locations in
the tunnel. In other words, the pattern of louder and weaker sounds remains stationary and
appears to be standing still. This change in sound intensity levels therefore produces a
standing wave pattern along the length of the tunnel. The ratio of the maximum to mini-
mum sound is known as the standing wave ratio (SWR). It will be shown later that the
measurement of standing wave patterns is a very useful technique for describing the prop-
erties of transmission line systems.
In the above discussions, you have used knowledge gained from the university of life
to understand the definitions of many transmission line terms. These definitions are not
trivial because you will soon see that many of the above principles and terms also relate
to electrical transmission lines. In fact, if you can spare the time, re-read the above
paragraphs again just to ensure that you are fully cognisant of the terms shown in bold
2.3 Types of electrical transmission lines
Many of the terms introduced in the last section also apply to electrical transmission lines.
However, you should be aware of the great difference in the velocity of sound waves (331
ms –1) and the velocity of electrical waves (3 × 108 ms –1 in air). There is also a great differ-
ence in frequency because audible sound waves are usually less than 20 kHz whereas radio
frequencies are often in tens of GHz. For example, satellite broadcasting uses frequencies
of about 10–12 GHz. Since wavelength = velocity/frequency, it follows that there will be
a difference in wavelength and that in turn will affect the physical size of transmission
lines. For example, the dimensions of a typical waveguide (Figure 2.3(a)) for use at
frequencies between 10 GHz and 15 GHz are A = 19 mm, B = 9.5 mm, C = 21.6 mm, D =
12.1 mm.
There are many types of transmission lines.2 These range from the two wire lines which
you find in your home for table lamps, and three wire lines used for your electric kettle.
Although these cables work efficiently at power frequencies (50~60 Hz), they become
very inefficient at high frequencies because their inherent construction blocks high
frequency signals and encourages radiation of energy.
2.3.1 Waveguides and coplanar waveguides
Other methods must be used and one method that comes readily to mind is the tunnel or
waveguide described in Section 2.2.2. This waveguide is shown in Figure 2.3(a). It works
efficiently as a high frequency transmission line because of its low attenuation and radia-
tion losses but it is expensive because of its metallic construction (usually copper). It is
also relatively heavy and lacks flexibility in use because special arrangements must be
2 The term ‘transmission line’ is often abbreviated to ‘tx lines’.
48 Transmission lines
Fig. 2.3 (a) Metallic waveguide; (b) coplanar waveguide
used to bend a transmission path. One variant of the waveguide is known as the coplanar
waveguide (Figure 2.3(b)).
2.3.2 Coaxial and strip lines
Another way of carrying high frequency signals is to use a coaxial transmission line simi-
lar to the one that connects your TV set to its antenna. The coaxial line is shown in Figure
2.4(a). This is merely a two wire line but the outer conductor forms a circular shield around
the inner conductor to prevent radiation.
One variation of the coaxial line appears as the strip line (Figure 2.4(b)). The strip line
is similar to a ‘flattened’ coaxial line. It has the advantage that it can be easily constructed
with integrated circuits.
Fig. 2.4 (a) Coaxial cable; (b) strip line
2.3.3 Microstrip and slot lines
The microstrip line (Figure 2.5(a)) is a variant of the stripline with part of the ‘shield’
removed. The slot line (Figure 2.5(b)) is also a useful line for h.f. transmission.
Types of electrical transmission lines 49
Fig. 2.5 (a) Microstrip line; (b) slot line
2.3.4 Twin lines
In Figure 2.6, we show a sketch of a twin line carefully spaced by a polyethylene dielec-
tric. This is used at relatively low frequencies. This twin cable is designed to have a
characteristic impedance (Z 0) of approximately 300 W and it is frequently used as a VHF
cable or as a dipole antenna for FM radio receivers in the FM band. The parallel wire line
arrangement of Figure 2.6 without a dielectric support can also be seen mounted on poles
as overhead telephone lines, overhead power lines, and sometimes as lines connecting high
power, low and medium frequency radio transmitters to their antennas.
Fig. 2.6 Twin parallel wire VHF cable
All seven transmission lines shown in Figures 2.3–2.6 have advantages and disadvan-
tages. For minimum loss, you would use the waveguide, the coaxial line and the strip line
in integrated circuits. However, the latter two lines present difficulties in connecting exter-
nal components to the inner conductor. The coplanar waveguide is better in this respect and
finds favour in monolithic microwave integrated circuits (MMIC) because it allows easy
series and parallel connections to external electrical components. The microstrip line is
also useful for making series connections but not parallel connections because the only
way through to the ground plane is either through or around the edge of the substrate. This
is particularly true when a short circuit is required between the upper conductor and the
ground plane; holes have to be drilled through the substrate. Microstrip also suffers from
radiation losses. Nevertheless, microstrip can be made easily and conveniently and it is
therefore used extensively.
50 Transmission lines
2.3.5 Coupled lines
Coupled lines are lines which are laid alongside each other in order to permit coupling
between the two lines. One example of microstrip coupled lines is shown in the F1 layout
box of Figure 2.2 where three sets of coupled lines are used to couple energy from input
port 1 to output port 4.
2.4 Line characteristic impedances and physical parameters
The characteristic impedance of transmission lines is calculated in two main ways:
• from physical parameters and configuration;
• from distributed electrical parameters of the line.
Some relevant expressions for calculating the impedance of these lines from physical
parameters are given in the following sections.
2.4.1 Coaxial line characteristic impedance (Z0)
The expression for calculating the characteristic impedance of the coaxial transmission
line shown in Figure 2.4(a) is:
138 D
Z 0 = —— log10 — (2.1)
e d
d = outer diameter of the inner conductor
D = inner diameter of the outer conductor
e = dielectric constant of the space between inner and outer conductor (e = 1 for air)
Example 2.1
You will often find two types of flexible coaxial cables: one with a characteristic imped-
ance Z 0 of 50 W which is used mainly for r.f. instrumentation and the other has a charac-
teristic impedance of 75 W used mainly for antennas. The inner diameter of the outer
conductor is the same in both cables. How would you distinguish the impedance of the two
cables using only your eye?
Solution. In general, to save money, both cables are normally made with the same outer
diameter. This is even more evident when the cables are terminated in a type of r.f. connec-
tor known as BNC.3 Since these connectors have the same outer diameter, by using Equa-
tion 2.1 you can deduce that for Z 0 = 75 W, the inner conductor will be smaller than that of
the 50 W cable. In practice, you will be able to recognise this distinction quite easily.
3 BNC is an abbreviation for ‘baby N connector’. It is derived from an earlier, larger threaded connector, the
Type N connector, named after Paul Neill, a Bell Laboratories engineer. BNC uses a bayonet type fixing. There
is also a BNC type connector which uses a thread type fixing; it is called a TNC type connector.
Line characteristic impedances and physical parameters 51
2.4.2 Twin parallel wire characteristic impedance (Z0)
The expression for calculating the characteristic impedance of the type of parallel trans-
mission line shown in Figure 2.6 is:
276 2D
Z 0 ≈ —— log10 —— (2.2)
e d
d = outer diameter of one of the identical conductors
D = distance between the centres of the two conductors
e = relative dielectric constant = 1 for air
Example 2.2
The twin parallel transmission line shown in Figure 2.6 is separated by a distance (D) of
300 mm between the centre lines of the conductors. The diameter (d) of the identical
conductors is 4 mm. What is the characteristic impedance (Z 0) of the line? Assume that the
transmission line is suspended in free space, i.e. e = 1.
Given: D = 300 mm, d = 4 mm, e = 1.
Required: Z 0.
Solution. Using Equation 2.2
276 2D 276 2 × 300
Z 0 ≈ —— log10 —— = —— log10 ———— ≈ 600 W
1 d 1 4
2.4.3 Microstrip line characteristic impedance (Z0)
Before we start, it is best to identify some properties that are used in the calculations on
microstrip. These are shown in Figure 2.7 where w = width of the microstrip, h = thick-
ness of the substrate, t = thickness of the metallisation normally assumed to approach zero
in these calculations, er = dielectric constant of the substrate. Note that there are two
dielectric constants involved in the calculations, the relative bulk dielectric constant er and
the effective dielectric constant ee. The effective dielectric is inevitable because some of
the electric field passes directly from the bottom of the strip width to the ground plane
whereas some of the electric field travels via air and the substrate to the ground plate.
Fig. 2.7 (a) Microstrip line; (b) end view of microstrip line
52 Transmission lines
There are many expressions for calculating microstrip properties4 but we will use two
main methods. These are:
• an analysis method when we know the width/height (w/h) ratio and the bulk dielectric
constant (er) and want to find Z 0;
• a synthesis method when we know the characteristic impedance Z 0 and the bulk dielec-
tric constant (er) and want to find the w/h ratio and the effective dielectric constant (ee).
Analysis formulae
In the analysis case we know w/h and er and want to find Z 0. The expressions which follow
are mainly due to H. Wheeler’s work.5
For narrow strips, i.e. w/h < 3.3
⎧ ⎡ ⎤ 1 ⎛ ε − 1⎞ ⎛ π 1 ⎫
⎪ ⎢ 4h h 2 4⎞⎪
+ 16⎛ ⎞ + 2 ⎥ − ⎜ r
Z0 = ⎨ ln ⎟⎜ ln + ln ⎟ ⎬ (2.3)
2(ε r + 1) ⎝ w⎠ ⎥ 2 ⎝ ε r + 1⎠ ⎝ 2 ε r π ⎠ ⎪
⎪ ⎢w
⎩ ⎣ ⎦ ⎭
For wide strips, i.e. w/h > 3.3
119.9 ⎧ w ln 4 ln (επ 2 16) ⎛ ε r − 1⎞
Z0 = ⎨ + + ⎜ 2 ⎟
2 ε r ⎪ 2h
⎩ π 2π ⎝ εr ⎠
ε + 1 ⎡ πε r
+ r ⎛w ⎞ ⎤⎫
⎢ln 2 + ln⎝ 2 h + 0.94⎠ ⎥ ⎬
2πε r ⎣ ⎦⎭
Synthesis formulae
In the synthesis case we know Z 0 and er and want to find w/h and ee.
For narrow strips, i.e. Z 0 > (44 – 2er) W
w ⎛ exp H 1 ⎞
=⎜ − ⎟ (2.5)
h ⎝ 8 4 exp H ⎠
Z0 2(ε r + 1) 1 ε r − 1 ⎛ π 1 4⎞
H= + ⎜ ln + ln ⎟ (2.6)
119.9 2 εr + 1 ⎝ 2 εr π ⎠
ε +1 ⎡ 1 εr − 1 ⎛ π 1 4⎞⎤
εe = r + ⎢1 − ⎜ ln + ln ⎟ ⎥ (2.7)
2 ⎢
⎣ 2H εr + 1 ⎝ 2 εr π ⎠ ⎦⎥
Note: Equation 2.7 was derived under a slightly different changeover value of
Z 0 > (63 – er) Ω.
4 In practice these are almost always calculated using CAD/CAE programmes.
5 Wheeler, H.A. Transmission lines properties of parallel wide strips separated by a dielectric sheet, IEEE
Trans, MTT-13 No. 3, 1965.
Line characteristic impedances and physical parameters 53
For wide strips, i.e. Z 0 < (44 – 2er)Ω
w 2 ε −1 ⎡ 0.517 ⎤
= [( de − 1) − ln(2 de − 1)] + r ⎢ln ( de − 1) + 0.293 − ⎥ (2.8)
h π πε r ⎣ εr ⎦
de = ———— (2.9)
Z 0 er
and under a slightly different value of Z 0 > (63 – 2er )Ω
εr + 1 εr − 1 ⎛ h −0.555
εe = + 1 + 10 ⎞ (2.10)
2 2 ⎝ w⎠
Equations 2.3 to 2.10 are accurate up to about 2 GHz. For higher frequencies, the effect of
frequency dependence of ee has to be taken into account. An expression often used to eval-
uate ee(ƒ) as frequency (ƒ) varies is
εr − εe
εe ( f ) = εr − 1.33 (2.11)
1 + ( h Z0 ) (0.43 f 2 − 0.009 f 3 )
where h is in millimetres, ƒ is in gigahertz, and ee is the value calculated by either Equa-
tion 2.7 or 2.10.
Example 2.3
Two microstrip lines are printed on the same dielectric substrate. One line has a wider
centre strip than the other. Which line has the lower characteristic impedance? Assume that
there is no coupling between the two lines.
Solution. If you refer to Equation 2.3 and examine the h/w ratio, you will see that Z 0
varies as a function of h/w. Therefore, the line with the lower characteristic impedance will
have a wider centre conductor.
As you can see for yourself, Equations 2.3 to 2.11 are rather complicated and should be
avoided when possible. To avoid these types of calculations, we have included with this book
a computer software program called PUFF. With this program, it is only necessary to decide
on the characteristic impedance of the microstrip or stripline which we require and PUFF will
do the rest. We will return to PUFF when we have explained the basic terms for using it.
Expressions also exist for calculating the characteristic impedance of other lines such
as the strip line, coplanar waveguide, slot line, etc. These are equally complicated but
details of how to calculate them have been compiled by Gupta, Garg and Chadha.6 There
is also a software program called AppCAD7 which calculates these impedances.
6 Gupta, K.C., Garg, R. and Chadha, R., Computer-Aided Design of Microwave Circuits, Artech House Inc,
Norwood MA 02062 USA, ISBN: 0–89006–105–X.
7 AppCAD is a proprietary software program from the Hewlett Packard Co, Page Mill Road, Palo Alto CA,
54 Transmission lines
Note: In the previous sections, I have produced equations which are peculiar to types of
different transmission lines. From now on, and unless stated otherwise, all the equations in
the sections that follow apply to all types of transmission lines.
2.5 Characteristic impedance (Z0) from primary electrical
A typical twin conductor type transmission line is shown in Figure 2.8. Each wire conduc-
tor has resistance and inductance associated with it. The resistance is associated with the
material of the metal conductors, effective conductor cross-sectional area and length. The
inductance is mainly dependent on length and type of material. In addition to these, there
is capacitance between the two conductors. The capacitance is mainly dependent on the
dielectric type, its effective permittivity, the effective cross-sectional area between conduc-
tors, the distance between the conductors and the length of the transmission line. When a
voltage is applied, there is also a leakage current between the two conductors caused by
the non-infinite resistance of the insulation between the two conductors. This non-infinite
resistance is usually expressed in terms of a shunt resistance or parallel conductance.
Therefore, transmission lines possess inherent resistance, inductance, capacitance and
conductance. It is very important to realise that these properties are distributed along the
length of the line and that they are not physically lumped together. The lumped approach
is only applicable when extremely short lengths of line are considered and as a practical
line is made up of many short lengths of these lines, the lumped circuit equivalent of a
transmission line would look more like that shown in Figure 2.8. This is an approximation
but nevertheless it is an extremely useful one because it allows engineers to construct and
simulate the properties of transmission lines.
2.5.1 Representation of primary line constants
In Figure 2.8, let:
R represent the resistance per metre (ohms/metre)
L represent the inductance per metre (henry/metre)
G represent the conductance per metre (siemen/metre)
C represent the capacitance per metre (farad/metre)
It follows that for a short length dl, we would obtain Rdl, Ldl, Gdl, and Cdl respectively.
R jωL ⎞
δl = ⎛ +
δl (2.12)
4 ⎝4 4 ⎠
Z1δl = ( R + jωL)δl (2.13)
Ydl = (G + jwC)dl (2.14)
Characteristic impedance (Z0) from primary electrical parameters 55
Fig. 2.8 Expanded view of a short section of transmission line
Zδl = = (2.15)
Yδ l (G + jωC )δ l
2.5.2 Derivation of line impedance
The input impedance Z in of the short section dl when terminated by a matched line is given
Zδ l ⎛ 1 δ l + Z0 + 1 δ l⎞
Z Z
Z1 ⎝ 4 4 ⎠ Z1
Zin = δl + + δl
4 Z1 Z
Zδ l + δ l + Z0 + 1 δ l 4
Since the line is terminated by another line, Z in = Z 0
Zδl ⎛ 1 δl + Z0 ⎞
Z1 ⎝ 2 ⎠
Zin = Z0 = δl +
2 Z1
Zδl + δl + Z0
8 The Z term in the centre fraction on the right-hand side of the equation is present because the short section
of line (dl) is terminated by an additional line which presents an input impedance of Z 0.
56 Transmission lines
Cross-multiplying, we get
Z0 ⎛ Zδ l + 1 δ l + Z0 ⎞ = 1 δ l ⎛ Zδ l + 1 δ l + Z0 ⎞ + Zδ l ⎛ 1 δ l + Z0 ⎞
Z Z Z Z
⎝ 2 ⎠ 2 ⎝ 2 ⎠ ⎝ 2 ⎠
Z0 Z1 ZZ Z2 Z Z ZZ
Z0 Zδ l + 2
δ l + Z0 = 1 δ l 2 + 1 δ l 2 + 0 1 δ l + 1 δ l 2 + ZZ0δ l
Z1 2
Z0 = ZZ1δ l 2 + δl (2.16)
Substituting for Z and Z1, we get
2 ( R + jωL)δ l ⎛ R jωl ⎞ 2 2
Z0 = + + δl
(G + jωC )δ l ⎝ 4 4 ⎠
In the limit when dl → 0, and taking the positive square root term
R + jωL
Z0 = (2.17)
G + jωC
If you examine the expression for Z 0 a bit more closely, you will see that there are two
regions where Z 0 tends to be resistive and constant. The first region occurs at very low
frequencies when R jwL and G jwC. This results in
Z0 ≈ (2.18)
The second region occurs at very high frequencies when jwL R and jwC G. This
results in
Z0 ≈ (2.19)
The second region is also known as the frequency region where a transmission line is
said to be ‘lossless’ because there are ‘no’ dissipative elements in the line.
Equation 2.19 is also useful because it explains why inductive loading, putting small
lumped element inductors in series with lines, is used to produce a more constant imped-
ance for the line. The frequency regions of operation described by Equations (2.18) and
(2.19) are important because under these conditions, line impedance tends to remain
frequency independent and a state known as ‘distortionless transmission’ exists. The
distortionless condition is very useful for pulse waveform/digital transmissions because in
these regions, frequency dispersion and waveform distortion tend to be minimal.
These statements can also be verified by the following practical example.
Characteristic impedance (Z0) from primary electrical parameters 57
Example 2.4
A transmission line has the following primary constants: R = 23 W km–1, G = 4 mS km–1,
L = 125 µH km–1 and C = 48 nF km–1. Calculate the characteristic impedance, Z0, of the
line at a frequency of (a) 100 Hz, (b) 500 Hz, (c) 15 kHz, (d) 5 MHz and (e) 10 MHz.
Given: R = 23 W km–1, G = 4 mS km–1, L = 125 µH km–1 and C = 48 nF km–1.
Required: Z 0 at (a) 100 Hz, (b) 500 Hz, (c) 15 kHz, (d) 5 MHz and (e) 10 MHz.
Solution. Use Equation 2.17 in the calculations that follow.
(a) At 100 Hz
R + jwL = (23 + j0.08) W km–1, G + jwC = (4 + j0.030) mS km–1
23 + j0.08
Z0 = = 75.83Ω / − 2.06 × 10 −3 rad
( 4 + j0.030) × 10 −3
(b) At 500 Hz
R + jwL = (23 + j0.39) W km–1, G + jwC = (4 + j0.15) mS km–1
23 + j0.39
Z0 = = 75.81Ω / − 10.30 × 10 −3 rad
( 4 + j0.15) × 10 −3
(c) At 15 kHz
R + jwL = (23 + j11.78) W km–1, G + jwC = 4 + j4.52 mS km–1
23 + j11.78
Z0 = = 65.42Ω / − 0.19 rad
( 4 + j4.52) × 10 −3
(d) At 5 MHz
R + jwL = (23 + j3926.99) W km–1, G + jwC = (4 + j1508) mS km–1
23 + j3926.99
Z0 = = 50.03Ω / − 0.00 rad
( 4 + j1508) × 10 −3
(e) At 10 MHz
R + jwL = (23 + j7853.98) W km–1, G + jwC = (4 + j3016) mS km–1
23 + j7853.98
Z0 = = 50.03Ω / − 0.00 rad
( 4 + j3016) × 10 −3
58 Transmission lines
Conclusions from Example 2.4. At low frequencies, i.e. 100–500 Hz, the line impedance
Z 0 tends to remain at about 75 W with very little phase shift over a wide frequency range.
For most purposes, it is resistive and constant in this region. See cases (a) and (b). At high
frequencies, i.e. 5–10 MHz, the line impedance Z 0 tends to remain constant at about 50 W
with little phase shift over a wide frequency range. For most purposes, it is resistive and
constant in this region. See cases (d) and (e). In between the above regions, the line imped-
ance Z0 varies with frequency and tends to be reactive. See case (c). For radio work, we
tend to use transmission lines in the ‘lossless’ condition (Equation 2.19) and this helps
considerably in the matching of line impedances.
2.6 Characteristic impedance (Z0) by measurement
Occasions often arise when the primary constants of a line are unknown yet it is necessary
to find the characteristic impedance (Z0). In this case, Z 0 can be obtained by measuring the
short- and open-circuit impedance of the line. In Figures 2.9 and 2.10 as in Figure 2.8 let:
R represent the resistance per metre (ohms/metre)
L represent the inductance per metre (henry/metre)
G represent the conductance per metre (siemen/metre)
C represent the capacitance per metre (farad/metre)
It follows that for a short length dl, we would obtain Rdl, Ldl, Gdl and Cdl respectively.
Fig. 2.9 Open-circuit equivalent of a short length of transmission line
2.6.1 Open-circuit measurement (Zoc)
Hence defining Zdl as l/Ydl we have
Z1 Z
Zoc = δ l + Zδ l + 1 δ l
4 4 (2.20)
= δ l + Zδ l
Characteristic impedance (Z0) by measurement 59
Fig. 2.10 Short-circuit equivalent of a short length of transmission line
2.6.2 Short-circuit measurement (Zsc )
The short-circuit impedance is
Zδ l ⎛ 1 δ l + 1 δ l⎞
Z Z
Z ⎝ 4 4 ⎠ Z
Zsc = 1 δl + + 1 δl
Zδ l + ⎛ Z1 δ l + Z1 δ l⎞ 4
⎝ 4 4 ⎠
ZZ1 2
Z1 2
= δl +
2 Z
Zδ l + 1 δ l
Z1 ⎛
δ l Zδ l + 1 δ l⎞ + 1 δ l 2
Z ZZ
2 ⎝ 2 ⎠ 2
Zδ l + δ l
Z1 2
ZZ1δ l 2 + δl
= 4
Zδ l + 1 δ l
Using Equation 2.16 to substitute for the numerator and Equation 2.20 to substitute for
the denominator, we have
60 Transmission lines
Z 02
Z sc = ——
Z oc
Z 02 = Z sc Z oc
Z0 = Zsc Zoc (2.21)
Example 2.5
The following measurements have been made on a line at 1.6 MHz where Z oc = 900 W
/–30° and Z sc = 400 W /–10°. What is the characteristic impedance (Z 0) of the line at 1.6
Given: f = 1.6 MHz, Z oc = 900 W /–30° , Z sc = 400 W /–10°.
Required: Z 0 at 1.6 MHz.
Solution. Using Equation 2.21
Z0 = Zsc Zoc
= 900 Ω / − 30° × 400 Ω / − 10°
= 600 Ω / − 20°
2.7 Typical commercial cable impedances
Manufacturers tend to make cables with the following characteristic impedances (Z 0).
These are:
• 50 Ω – This type of cable finds favour in measurement systems and most radio instru-
ments are matched for this impedance. It is also used extensively in amateur radio
Most cable manufacturers make more than one type of 50 Ω line. For example, you
can buy 50 Ω rigid lines (solid outer connector), 50 Ω low loss lines (helical and air
dielectrics), 50 Ω high frequency lines for use up to 50 GHz with minimal loss. The
reason for this is that different uses require different types of lines. Remember that in
Equation 2.1 repeated here for convenience Z0 = 138 log10 ( D d ) ε and the dimen-
sion of the variables can be changed to produce the desired impedance.
• 75 Ω – This type of cable is favoured by the television industry because it provides a
close match to the impedance (73.13 Ω) of a dipole aerial. Most TV aerials are designed
for this impedance and it is almost certain that the cable that joins your TV set to the
external aerial will have this impedance. The comments relating to the different types of
50 Ω lines also apply to 75 Ω lines.
Signal propagation on transmission lines 61
• 140 Ω – This type of cable is used extensively by the telephone industry. The comments
relating to the different types of 50 Ω lines also apply to 140 Ω lines.
• 300 Ω – This type of cable is favoured by both the radio and television industry because
it provides a close match for the impedance (292.5 Ω) of a very popular antenna (folded
dipole antenna) which is used extensively for VHF-FM reception. The comments relat-
ing to the different types of 50 Ω lines also apply to 300 Ω lines.
• 600 Ω – This type of cable is used extensively by the telephone industry and many of
their instruments are matched to this impedance. The comments relating to the different
types of 50 Ω lines also apply to 600 Ω lines.
2.8 Signal propagation on transmission lines
2.8.1 Pulse propagation on an infinitely long or matched transmission
We are now going to use some of the ideas introduced in the previous sections, particularly
Section 2.2.2, to describe qualitatively the propagation of signals along an infinitely long
transmission line. In this description we will only make two assumptions:
• the transmission line is perfectly uniform, that is its electrical properties are identical all
along its length;
• the line extends infinitely in one direction or is perfectly terminated.
To keep the explanation simple, we will initially only consider the propagation of a
single electrical pulse along the line9 shown in Figure 2.11. At the beginning of the line
(top left hand corner) a voltage source (Vs) produces the single pulse shown in Figure 2.11.
The waveforms shown at various planes (plane 1, plane 2, plane 3) on the line illustrate
three of the main properties of signal propagation along a transmission line:
• propagation delay – the pulse appears at each successive point on the line later than at
the preceding point;
• attenuation – the peak value of the pulse is attenuated progressively;
• waveform distortion and frequency dispersion – its shape differs from its original
shape at successive points.
2.8.2 Propagation delay
The pulse appears later and later at successive points on the line because it takes time to
travel over any distance, i.e. there is a propagation delay. As the line is uniform throughout
9 The behaviour of a pulse travelling along an infinitely long transmission line is very similar to the example
you were given in Section 2.2.2 concerning sound travelling down an infinitely long tunnel except that this time
instead of voice sounds, consider the sound to originate from a single drum beat or pulse. You will no doubt
remember from earlier work that a pulse is a waveform which is made up from a fundamental sinusoid and its
harmonics combined together in a precise amplitude, phase and time relationship.
62 Transmission lines
Fig. 2.11 Pulse propagation in a transmission line
its length, the amount of delay at any point is proportional to distance between that point
and the source of the pulse. These time delays are shown as t1, t 2, and t 3 in Figure 2.11.
Another way of describing this is to say that the pulse propagates along the line with a
uniform velocity.
2.8.3 Attenuation
The amplitude of the pulse is attenuated as it propagates down the line because of resis-
tive losses in the wires. The amount of attenuation per unit length is uniform throughout
the line because the line cross-section is uniform throughout the line length. Uniform
attenuation means that the fractional reduction in pulse amplitude is the same on any line
section of a given length. This is more easily understood by referring to Figure 2.11,
where the pulse amplitude at plane 1 has been reduced by a factor of 0.8. At plane 2,
which is twice as far from the source as plane 1, the pulse height has been reduced by a
further factor of 0.8, i.e. a total of 0.82 or 0.64 of its original amplitude. At plane 3, which
is three times as far from the source as plane 1, the reduction is 0.83 or 0.512 of the orig-
inal amplitude.
More generally, at a distance equal to l times the distance from the source to plane 1,
the height is reduced by (0.8)l. Because l is the exponent in this expression this type of
amplitude variation is called exponential. It can also be expressed in the form (eα)l or eαl,
where ea represents the loss per unit length and is 0.8 in this particular example. In fact a
is the natural logarithm of the amplitude reduction per unit length. Its unit is called the
neper and loss (dB) = 8.686 nepers.10
Example 2.6
A transmission has a loss of two nepers per kilometre. What is the loss in dB for a length
of 10 kilometres?
Given: Attenuation constant (α) = 2 nepers per km.
Required: Loss in dB for a length of 10 km.
10 This is because dB = 20 Log (Ratio) = 20 Log (ea) = 20 × a × Log (e) = 20 × a × 0.4343 = 8.686a.
Waveform distortion and frequency dispersion 63
Solution. If 1 km represents a loss of 2 nepers, then 10 km = 10 × 2 = 20 nepers. There-
loss = 8.686 × 20
= 173.72 dB
2.9 Waveform distortion and frequency dispersion
2.9.1 Amplitude distortion
The waveform of the pulse in Figure 2.11 alters as it travels along the line. This shape
alteration is caused by the line constants (inductance, capacitance, resistance and conduc-
tance of the line) affecting each sinusoidal component of the waveform in a different
manner. The high frequency components, which predominate on the edges of the pulse
waveform, suffer greater attenuation because of increased reactive effects; the lower
frequency components, which predominate on the flat portion of the waveform, suffer less
attenuation. The variation of attenuation with frequency is described by the frequency
response of the line.
2.9.2 Frequency distortion
In addition to attenuation, there are also time constants associated with the line compo-
nents (inductance, capacitance, resistance and conductance). These cause high frequency
components to travel at a different velocity from low frequency components. The variation
of velocity with frequency is called the frequency dispersion of the line.
2.9.3 Phase and group velocities
As a pulse consists of sinusoidal components of different frequencies, each component
will therefore be altered differently. Distinction must be made between the velocities
of the sinusoidal components which are called phase velocities, up. The phase velocity
(b) is defined as the change in radians over a wavelength and since there is a phase
change of 2π radians in every wavelength, it follows that b = 2π radians/wavelength (l)
b = —— (2.22)
The velocity of the complete waveform is called the group velocity, ug. The apparent
velocity of the pulse in Figure 2.11 is called its group velocity.
It is important to realise that if the line velocity and line attenuation of all the compo-
nent sinusoids which make up a pulse waveform are not identical then deterioration in
pulse waveform shape will occur. Pulse distortion is particularly critical in high speed data
transmission where a series of distorted pulses can easily merge into one another and cause
pulse detection errors.
If distortion occurs and if it is desired to know how and why a particular waveform
64 Transmission lines
has changed its shape, it will be necessary to examine the propagation of the constituent
sinusoids of the waveform itself and to instigate methods, such as frequency and phase
equalisation, to ensure minimal waveform change during signal propagation through the
2.10 Transmission lines of finite length
2.10.1 Introduction
In Section 2.8.1, we discussed waveforms travelling down infinitely long lines. In practice,
infinitely long lines do not exist but finite lines can be made to behave like infinitely long
lines if they are terminated with the characteristic impedance of the line.11
2.10.2 Matched and unmatched lines
A transmission line which is terminated by its own characteristic impedance, Z 0, is said to
be matched or properly terminated. A line which is terminated in any impedance other
than Z 0 is said to be unmatched or improperly terminated. To prevent reflections it is
usual for a transmission line to be properly terminated and so it is a common condition for
a transmission line to behave electrically as though it was of infinite length.
If a transmission line is to be used for signals with a wide range of frequency compo-
nents, it may be difficult to terminate it properly. In general, the characteristic impedance
of a transmission line will vary with frequency and if the matching load fails to match the
line at all frequencies, then the line will not be properly terminated and reflections will
In practice, it is usual to properly terminate both ends of a transmission line, i.e. both
at the sending end and the receiving end; otherwise any signal reflected from the receiv-
ing end and travelling back towards the sending end will be re-reflected again down the
line to cause further reflections. The sending end can be properly terminated either by
using a source generator with an impedance equal to the characteristic impedance of the
line or by using a matching network to make a source generator present a matched imped-
ance to the transmission line.
2.11 Reflection transmission coefficients and VSWR
2.11.1 Introduction
Reflection coefficients are based on concepts introduced in your childhood. Consider the
case when you throw a ball at a vertical stone wall. The ball with its incident power will
11 This argument is similar to the case mentioned in Section 2.2.2 where it was shown that if our finite tunnel
was terminated with material with the same properties as an infinitely long tunnel which absorbed all the inci-
dent energy then it would also behave like an infinitely long tunnel.
12 You see this reflection effect as multiple images on your television screen when the TV input signal is not
properly terminated by the TV system. TV engineers call this effect ‘ghosting’.
Reflection and transmission coefficients 65
travel towards the wall, hit the wall which will absorb some of its incident power and then
the remaining power (reflected power) will cause the ball to bounce back.
The ratio (reflected power)/(incident power) is called the reflection coefficient. The
reflection coefficient is frequently represented by the Greek letter gamma (G). In mathe-
matical terms, we have
reflected power
incident power
This simple equation is very useful for the following reasons.
• Its value is independent of incident power because if you double incident power,
reflected power will also double.13 If you like, you can say that Γ is normalised to its
incident power.
• It gives you a measure of the hardness (impedance) of the wall to incident power. For
example if the wall is made of stone, it is likely that very little incident power will be
absorbed and most of the incident power will be returned to you as reflected power. You
will get a high reflection coefficient (Γ → 1). If the wall is made of wood, it is likely that
the wood would bend a bit (less resistance), absorb more incident energy and return a
weaker bounce. You will get a lower reflection coefficient (Γ < 1). Similarly if the wall
was made of straw, it is more than likely that most of the incident energy would be
absorbed and there would be little rebounce or reflected energy (Γ → 0). Finally if the
wall was made of air, the ball will simply go through the air wall and carry all its inci-
dent power with it (G = 0). There will be no reflected energy because the incident energy
would simply be expended in carrying the ball further. Note in this case that the trans-
mission medium resistance is air, and it is the same as the air wall resistance which is
the load and we simply say that the load is matched to the transmission medium.
• By measuring the angle of the re-bounce relative to the incident direction, it is possible
to tell whether the wall is vertical and facing the thrower or whether it is at an angle
(phase) to the face-on position. Hence we can determine the direction of the wall.
• The path through which the ball travels is called the transmission path.
• Last but not least, you need not even physically touch the wall to find out some of its
characteristics. In other words, measurement is indirect. This is useful in the measure-
ment of transistors where the elements cannot be directly touched. It is also very useful
when you want to measure the impedance of an aerial on top of a high transmitting tower
when your measuring equipment is at ground level. The justification for this statement
will be proved in Section 2.13.
2.11.2 Voltage reflection coefficient14 (Γv) in transmission lines
The same principles described above can also be applied to electrical energy. This is best
explained by Figure 2.12 where we have a signal generator with a source impedance, Z s,
sending electrical waves through a transmission line whose impedance is Z 0, into a load
impedance, Z L.
13 This of course assumes that the hardness of your wall is independent of the incident power impinging on
Some authors use different symbols for voltage reflection coefficient. Some use Gv , while others use rv. In
this book, where possible, we will use Gv for components and rv for systems.
66 Transmission lines
Fig. 2.12 Incident and reflected waves on a transmission line
If the load impedance (Z L) is exactly equal to Z 0, the incident wave is totally absorbed
in the load and there is no reflected wave. If Z L differs from Z 0, some of the incident wave
is not absorbed in the load and is reflected back towards the source. If the source imped-
ance (Z s) is equal to Z 0, the reflected wave from the load will be absorbed in the source
and no further reflections will occur. If Z s is not equal to Z 0, a portion of the reflected wave
from the load is re-reflected from the source back toward the load and the entire process
repeats itself until all the energy is dissipated. The degree of mis-match between Z 0 and
Z L or Z s determines the amount of the incident wave that is reflected.
By definition
voltage reflection coefficient = ———— = Γv ∠ q (2.23)
i reflected
current reflection coefficient = ———— = Γi ∠ θ (2.24)
i incident
From inspection of the circuit of Figure 2.12
Z0 = (2.25)
Z0 = (2.26)
The minus sign in Equation 2.27 occurs because we use the mathematical convention
that current flows to the right are positive, therefore current flows to the left are negative.
ZL =
vi + v r vi + v r v (1 + Γv )
= = = Z0 i (2.27)
ii − ir vi Z 0 − v r Z 0 vi (1 − Γv )
Reflection and transmission coefficients 67
Sorting out terms in respect of Γv
(1 + Gv)
Z L = Z 0 —— ——
(1 – Gv)
(Z L – Z 0 )
Gv = ——— —— (2.28)
(Z L + Z 0 )
Returning to Equation 2.24 and recalling Equation 2.23
ir – vr /Z 0
Γi = — = ———— = – Γv (2.29)
ii vi /Z 0
As the match between the characteristic impedance of the transmission line Z 0 and the
terminating impedance Z L improves, the reflected wave becomes smaller. Therefore, using
Equation 2.28, the reflection coefficient decreases. When a perfect match exists, there is
no reflected wave and the reflection coefficient is zero. If the load Z L on the other hand is
an open or short circuit, none of the incident power can be absorbed in the load and all of
it will be reflected back toward the source. In this case, the reflection coefficient is equal
to 1, or a perfect mismatch. Thus the normal range of values for the magnitude of the
reflection coefficient is between zero and unity.
Example 2.7
Calculate the voltage reflection coefficient for the case where Z L = (80 – j10) Ω and Z 0 =
50 Ω.
Given: ZL = (80 – j10), Z0 = 50Ω
Required: Γv
Solution. Using Equation 2.28
ZL – Z0 80 – j10 – 50 30 – j10
Γv = ———— = —————— = ————
ZL + Z0 80 – j10 + 50 130 – j10
31.62 ∠ –18.43°
= ——————— = 0.24 ∠ –14.03°
130.38 ∠ –4.40°
Example 2.8
Calculate the voltage reflection coefficients at the terminating end of a transmission
line with a characteristic impedance of 50 Ω when it is terminated by (a) a 50 Ω termi-
nation, (b) an open-circuit termination, (c) a short-circuit termination and (d) a 75 Ω
68 Transmission lines
Given: Z 0 = 50 Ω, Z L = (a) 50 Ω, (b) open-circuit = ∞, (c) short-circuit = 0 Ω, (d) = 75
Required: Γv for (a), (b), (c), (d).
Solution. Use Equation 2.28.
(a) with Z L = 50 Ω
ZL − Z0 50 − 50
Γv = = = 0 / 0°
ZL + Z0 50 + 50
(b) with Z L = open circuit = ∞ Ω
ZL − Z0 ∞ − 50
Γv = = = 1/ 0°
ZL + Z0 ∞ + 50
(c) with Z L = short circuit = 0 Ω
ZL − Z0 0 − 50
Γv = = = −1/ 0° or 1/180°
ZL + Z0 0 + 50
(d) with Z L = 75 Ω
ZL − Z0 75 − 50
Γv = = = 0.2 / 0°
ZL + Z0 75 + 50
Example 2.8 is instructive because it shows the following.
• If you want to transfer an incident voltage wave with no reflections then the terminating
load (Z L) must match the characteristic impedance (Z 0) exactly. See case (a). This is the
desired condition for efficient transfer of power through a transmission line.
• Maximum in-phase voltage reflection occurs with an open circuit and maximum anti-
phase voltage reflection occurs with a short circuit. See cases (b) and (c). This is because
there is no voltage across a short circuit and therefore the reflected wave must cancel the
incident wave.
• Intermediate values of terminating impedances produce intermediate values of reflection
coefficients. See case (d).
2.11.3 Return loss
Incident power (Pinc) and reflected power (Pref) can be related by using the magnitude of
the voltage reflection coefficient (G). Since G = vref /vinc, it follows that
Pref vref Rload
= 2 = Γ2 (2.30)
Pinc Vinc Rload
The return loss gives the amount of power reflected from a load and is calculated from:
return loss (dB) = –10 log Γ 2 = –20 log Γ (2.31)
Reflection and transmission coefficients 69
2.11.4 Mismatched loss
The amount of power transmitted to the load (P L) is determined from
PL = Pinc – Pref = Pinc(1 – Γ 2) (2.32)
The fraction of the incident power not reaching the load because of mismatches and
reflections is
Pload P (2.33)
= L = 1 − Γ2
Pincident Pinc
Hence the mismatch loss (or reflection loss) is calculated from
ML(dB) = –10 log (1 – Γ 2) (2.34)
2.11.5 Transmission coefficient
The transmission coefficient (τv ) is defined as the ratio of the load voltage (vL) to the inci-
dent voltage (vinc) but vL = vinc + vref . Hence
vL v + vref
τv = = inc = 1 + Γv (2.35)
vinc vinc
If we now use Equation 2.28 to substitute for Γv , we obtain
Z L − Z0 2 ZL
τ v = 1 + Γv = 1 + = (2.36)
Z L + Z0 Z L + Z0
Sometimes Equation 2.36 is normalised to Z 0 and when Z L/Z 0 is defined as z, we obtain
τv = (2.36a)
z +1
Equation 2.36a is the form you frequently find in some articles.
2.11.6 Voltage standing wave ratio (VSWR)
Cases often arise when the terminating impedance for a transmission line is not strictly
within the control of the designer. Consider a typical case where a transmitter designed for
operating into a 50 W transmission line is made to feed an antenna with a nominal imped-
ance of 50 W. In the ideal world, apart from a little loss in the transmission line, all the
energy produced by the transmitter will be passed on to the antenna. In the practical world,
an exact antenna match to the transmission line is seldom achieved and most antenna
manufacturers are honest enough to admit the discrepancy and they use a term called the
voltage standing wave ratio15 to indicate the degree of mismatch.
15 This term is based on the Standing Wave Pattern principle which was introduced in Section 2.2.2 where you
walked along a tunnel which produced echoes while your friend whistled at a constant amplitude and pitch. In the
tunnel case, the loudest (maximum intensity) sound occurred where the incident and reflected wave added, while
the weakest sound (minimum intensity) occurred where the incident and reflected sound opposed each other.
70 Transmission lines
VSWR is useful because
• it is relatively easy to measure – it is based on modulus values rather than phasor quan-
tities which enables simple diode detectors to be used for measurement purposes;
• it indicates the degree of mismatch in a termination;
• it is related to the modulus of the reflection coefficient (shown later).
Voltage standing wave ratio is defined as
Vmax Vinc + Vref
VSWR = = (2.37)
Vmin Vinc − Vref
A VSWR of |1| represents the best possible match.16 Any VSWR greater than |1| indi-
cates a mismatch and a large VSWR indicates a greater mismatch than a smaller VSWR.
Typical Figures of VSWRs for good practical terminations range from 1.02 to 1.1.
Example 2.9
In Figure 2.12, the incident voltage measured along the transmission line is 100 V and the
reflected voltage measured on the same line is 10 V. What is its VSWR?
Soloution. Using Equation 2.37
Vinc + Vref 100 + 10
VSWR = = = 1.22
Vinc − Vref 100 − 10
2.11.7 VSWR and reflection coefficient (Γv )
VSWR is related to the voltage reflection coefficient by:
Vinc + Vref Vinc 1 + Γv
VSWR = = = (2.38)
Vinc − Vref V 1 − Γv
1 − ref
VSWR − 1
Γv = (2.38a)
VSWR + 1
Example 2.10
What is the VSWR of a transmission system if its reflection coefficient |Γv| is 0.1?
16 In a properly terminated line, there are no reflections. V ref = 0 and substituting this value into Equation 2.37
|V inc| + |0|
VSWR = ————— = |1|
|V inc| – |0|
Reflection and transmission coefficients 71
Given: |Γv| = 0.1
Required: VSWR
Solution. Using Equation 2.38
1 + Γv 1 + 0.1
VSWR = = = 1.22
1 − Γv 1 − 0.1
Perfect match occurs when VSWR = |1|. This is the optimum condition and examination
of Equation 2.38 shows that this occurs when |Γv | = 0. With this condition, there is no
reflection, optimum power is transferred to the load and there are no standing wave
patterns on the line to cause excessive insulation breakdown or electrical discharges to
surrounding conductors and there are no ‘hot spots’ or excessive currents on sections of the
Example 2.11
A manufacturer quotes a maximum VSWR of 1.07 for a resistive load when it is used to
terminate a 50 Ω transmission line. Calculate the reflected power as a percentage of the
incident power.
Given: VSWR = 1.07, Z0 = 50 W
Required: Reflected power as a percentage of incident power
Solution. Using Equation 2.38a
VSWR – 1
|Γv | = ————— = 0.034
VSWR + 1
Since power is proportional to V2
Pref = (0.034)2 × Pinc
= 0.001 × Pinc
= 0.1% of Pinc
From the answer to Example 2.11, you should now realise that:
• a load with an SWR of 1.07 is a good terminating load;
• there are hardly any reflections when a transmission line is terminated with such a load;
• the transmission line is likely to behave like an infinitely long line.
2.11.8 Summary of Section 2.11
If a transmission line is not properly terminated, reflections will occur in a line. These
reflections can aid or oppose the incident wave. In high voltage lines, it is possible for the
aiding voltages to cause line insulation breakdown. In high current lines, it is possible at
high current points for aiding currents to overheat or even destroy the metallic conductors.
The voltage reflection coefficient (Γv ) can be calculated by
reflected voltage wave ZL − Z0
Γv = =
incident voltage wave ZL + Z0
72 Transmission lines
Manufacturers tend to use VSWRs when quoting the impedances associated with their
equipment. A VSWR of |1| is the optimum condition and indicates that a perfect match is
possible and that there will be no reflections when the equipment is matched to a perfect
transmission line. VSWR can be calculated from the reflection coefficients by
Vmax V + Vref 1 + Γv
VSWR = = inc =
Vmin Vinc − Vref 1 − Γv
The return loss is a way of specifying the power reflected from a load and is equal to –10
log Γ 2. The mismatch loss or reflection loss specifies the fraction of incident power not
reaching the load and is equal to –10 log (1 – Γ 2).
2.12 Propagation constant (γ) of transmission lines
2.12.1 Introduction
In Section 2.8, we saw that signals on transmission lines suffer attenuation, phase or time
delay, and often frequency distortion. In this section, we will show the relationships
between these properties and the primary constants (R, G, L and C) of a transmission line.
2.12.2 The propagation constant (γ) in terms of the primary constants
To find the propagation constant (g) we start with the same equivalent circuit (Figure 2.8)
used for the derivation of Z 0. It is re-drawn in Figure 2.13 with the voltage and current
phasors indicated.
The propagation constant, as defined, relates V2 and V1 by
— = e–γδ l
— (2.39)
Fig. 2.13 Equivalent circuit of a very short length of line
Propagation constant (γ) of transmission lines 73
where δ l is still the short length of line referred to in Figure 2.8. It is easier to find γ using
the current phasors rather than the voltage phasors; so, using I1 = V1/Z 0 and I2 = V2 /Z 0
— = e–γδ l (2.40)
or alternatively
— = eγδl (2.40a)
The current I1 splits into two parts: I2 and a part going through Z 2. By the current divider
rule, the split is
I2 = —————— I1 —
Z 2 + Z 1/2 + Z 0
I1 Z Z
= 1+ 1 + 0
I2 2 Z2 Z2
Substituting the definitions for Z 1 and Z 2 and the formula for Z 0 derived above gives
I1 1
= 1 + ( R + jjωL)(G + jjωL)(δll )2+ (( R++jωLL)(G +jωL))δ l
( R + ωL)(G + ωL )(δ )2 + R jω )(G + jεL δl
I2 2
= 1 + ( R + jωL)(G + jωL)δ l + ( R + jωL)(G + jωL)(δ l )2
Also I1/I2 = eγδ l. To use these two expressions for I1/I2 to find γ, we must first expand eγδ l
into a Taylor series. Since
ex = 1 + x + — + . . .
we can write eγδ l as
γ 2 (δ l )2
eγδ l = 1 + γδ l + +…
Equating the two expressions for I1/I2 gives
1 + γδ l + γ 2 (δ l )2 2 = 1 + ( R + jωL)(G + jωL)δ l
+ ( R + jωL)(G + jωL)(δ l )2
Subtracting 1 from each side and dividing by δ l gives
γδ l + γ 2 (δ l )2 2 = ( R + jωL)(G + jωL)δ l + ( R + jωL)(G + jωL)(δ l )2
74 Transmission lines
and as δ l approaches zero
γ = ( R + jωL)(G + jωL) (2.41)
Since γ is complex consisting of a real term α for amplitude and β for phase, we can also
γ = α + jβ = ( R + jωL)(G + jωL) (2.42)
If the expression for γ (Equation 2.42) is examined more closely, it can be seen that there
are two regions where γ tends to be resistive and constant. The first region occurs at very
low frequencies when R j ωL and G j ωC. This results in
γ ≈ ( R)(G) (2.43)
In this region γ is a real number which does not depend on ω. Since the real part of g is a,
the attenuation index, there is no amplitude distortion in the very low frequency range. The
second region occurs at very high frequencies when j ωL R and j ωC G. This results in
γ ≈ jω ( L)(C ) (2.44)
In this region γ is purely imaginary and is proportional to ω. Since the imaginary part of
γ is β, the phase index, it means that there is no dispersion (because β is proportional to
ω) in the high frequency range. The region is very useful for pulse waveform/digital
transmissions because in it frequency dispersion and waveform distortion tend to be
Equation 2.44 is also useful because it explains why inductive loading, putting small
lumped element inductors in series with lines, is sometimes used to reduce dispersion in
Example 2.12
A transmission line has the following primary constants: R = 23 Ω km–1, G = 4 mS km–1,
L = 125 µH km–1 and C = 48 nF km–1. Calculate the propagation constant γ of the line,
and the characteristic impedance Z 0 of the line at a frequency of (a) 100 Hz, (b) 500 Hz,
(c) 15 kHz, (d) 5 MHz and (e) 10 MHz.
Solution. The characteristic impedance Z 0 will not be calculated here because it has
already been carried out in Example 2.4. However, the results will be copied to allow easy
comparison with the propagation results calculated here for the discussion that follows
after this answer. Equation 2.41 will be used to calculate the propagation constant γ, and
Equation 2.42 will be used to derive the attenuation constant α and the phase constant β
in all the calculations that follow.
(a) At 100 Hz
R + jω L = (23 + j(2π × 100 × 125 µH)) = (23 + j0.08) Ω km–1
G + jwC = (4 mS + j(2π × 100 × 48 nF)) = (4 + j0.030) mS km–1
Propagation constant (γ) of transmission lines 75
γ = (23 + j0.08)( 4 + j0.030) × 10 −3 ) = 0.30 / 0.01
= (0.30 nepers + 1.66 × 10 −3 rad) km −1
23 + j0.08
Z0 = = 75.83Ω / − 2.06 × 10 −3 rad
( 4 + j0.030) × 10 −3
(b) At 500 Hz
R + jwL = (23 + j(2π × 500 × 125 µH)) = (23 + j0.39) Ω km–1
G + j ωC = (4 mS + j(2π × 500 × 48 nF)) = (4 + j0.15) mS km–1
γ = (23 + j0.39)( 4 + j0.15) × 10 −3 = 0.30 / 0.03 rad
= (0.30 nepers + 8.31 × 10 −3 rad) km −1
23 + j0.39
Z0 = = 75.82Ω / − 10.30 × 10 −3 rad
( 4 + j0.15) × 10 −3
(c) At 15 kHz
R + j ω L = (23 + j(2π × 15 × 103 × 125 µH)) = (23 + j11.78) Ω km–1
G + j ω C = (4 mS + j(2π × 15 × 103 × 48 nF)) = (4 + j4.52 × 10–3) mS km–1
γ = (23 + j11.78)( 4 + j4.52) × 10 −3 = 0.40 / 0.66 rad
= (0.31 nepers + 242 × 10 −3 rad) km −1
23 + j11.78
Z0 = = 65.42Ω / − 0.19 rad
( 4 + j4.52) × 10 −3
(d) At 5 MHz
R + j ω L = (23 + j(2π × 5 × 106 × 125 µH)) = (23 + j3926.99) Ω km–1
G + j ω C = (4 mS + j(2π × 5 × 106 × 48 nF)) = (4 + j1508) mS km–1
76 Transmission lines
γ = (23 + 3926.99)( 4 + j1508) × 10 −3 = 76.95/ 1.567 rad
= (0.33 nepers + 76.95 rad) km −1
23 + j3926.99
Z0 = = 50.03Ω / − 0.00 rad
( 4 + j1508) × 10 −3
(e) At 10 MHz
R + j ωL = (23 + j(2π × 10 × 106 × 125 µH) = (23 + j7853.98) Ω km–1
G + j ωC = (4 mS + j(2π × 10 × 106 × 48 nF)) = (4 + j3016) mS km–1
γ = (23 + j7853.98)( 4 + j3016) × 10 −3 = 153.91/ 1.569 rad
= (0.33 nepers + j153.9 rad) km −1
23 + j7853.98
Z0 = = 50.03Ω / − 0.00 rad
( 4 + j3016) × 10 −3
Conclusions from Example 2.12. In the frequency range 100–500 Hz, the attenuation
constant α tends to remain at about 0.30 nepers per km and the phase propagation
β increases linearly with frequency. See cases (a) and (b). If you now compare this
set of results with the same cases from Example 2.4, you will see that in this fre-
quency range, Z 0 and α tend to remain constant and β tends to vary linearly with
What this means is that if you transmit a rectangular pulse or digital signals in this
frequency range, you will find that it will pass through the transmission line attenuated but
with its shape virtually unchanged. The reason for the waveform shape not changing is
because the Fourier amplitude and phase relationships have not been changed.
In the frequency range 5–10 MHz, the attenuation constant α tends to remain at 0.33
nepers per km and the phase propagation β also increases linearly with frequency. See
cases (d) and (e). If you now compare the above set of cases from Example 2.12 with an
identical set from Example 2.4, you will see that within these frequency ranges, Z 0 and α
tend to remain constant and β tends to vary linearly with frequency. Therefore the same
argument in the foregoing paragraphs applies to this frequency range. This is also known
as the ‘distortionless’ range of the transmission line.
In the intermediate frequency range of operation (see case (c) of Examples 2.4 and
2.12), both the propagation constant α and β, and the characteristic impedance of the line
Z 0 vary. Fourier amplitude and phase relations are not maintained as waveforms are trans-
mitted along the line and waveform distortion is the result.
Transmission lines as electrical components 77
2.12.3 Summary of propagation properties of transmission lines
There are two frequency regions where signals can be passed through transmission lines
with minimum distortion; a low frequency region and a high frequency region. The low
frequency region occurs when R ω L, and G ω C. The high frequency region occurs
when ω L R and ω C G. The high frequency region is also sometimes called the ‘loss-
less’ region of transmission.
At both these high and low frequency regions of operation, the simplified expressions for
Z 0 (Equations 2.18 and 2.19) and γ (Equations 2.43 and 2.44) show that there is little distor-
tion and that the transmission line can be more easily terminated by a matched resistor.
Good cables which operate up to 50 GHz are available. They are relatively costly
because of the necessary physical tolerances required in their manufacture.
2.13 Transmission lines as electrical components
Transmission lines can be made to behave like electrical components, such as resistors,
inductors, capacitors, tuned circuits and transformers. These components are usually made
by careful choice of transmission line characteristic impedance (Z 0), line length (l) and
termination (Z L). The properties of these components can be calculated by using well
known expressions for calculating the input impedance of a transmission line.
2.13.1 Impedance relations in transmission lines
We shall now recall some transmission line properties which you learnt in Sections 2.11.2
and 2.12.2 to show you how the input impedance varies along the line and how transmis-
sion lines can be manipulated to produce capacitors, inductors, resistors and tuned circuits.
These are Equations 2.28 and 2.29 which are repeated below for convenience:
Z L − Z0
Γv = (2.28)
Z L + Z0
Γi = –Γv (2.29)
In previous derivations, voltages and currents references have been taken from the input
end of the line. Sometimes, it is more convenient to take voltage and current references
from the terminating or load end of the line. This is shown in Figure 2.14.
From the definition of line attenuation and for a distance l from the load, we have
v i = ViL e+γ l (2.45)
vr = VrL e–γ l (2.46)
and using the definition for voltage reflection coefficient Γv
Vrl VrL e −γ l VrL
Γvl = = = = ΓL e −2γ l (2.47)
Vil VrL e +γ l ViL
78 Transmission lines
Fig. 2.14 Line voltages reference to the load end
l = line length equal to distance l
Γv = voltage reflection coefficient at load
Γvl = voltage reflection coefficient at distance l from load
γ = propagation constant = (α + j β) nepers/m
At any point on a transmission line of distance l from the load
vl = v i + v r = v i + vi Γv e–2γ l (2.48)
i l = i i + i r = i i + i i Γi e–2γ l (2.49)
Dividing Equation 2.48 by Equation 2.49 and using Equations 2.28 and 2.29
vl vi + vi Γv e −2γ l
il ii − ii Γv e −2γ l
vi (1 + Γv e −2γ l )
= (2.50)
ii (1 − Γv e −2γ l )
Defining Z l as impedance at point l, and Z 0 as the line characteristic impedance, Equa-
tion 2.50 becomes
1 + Gv e–2gl
Zl = Z0
[ —————
1 – Gv e–2gl ] (2.51)
Substituting equation 2.28 in equation 2.51 results in
[ ]
ZL – Z0
1 + ———— e–2gl
ZL + Z0
Zl = Z0 —————————
ZL – Z 0
1 – ————— e–2gl
ZL + Z0
Transmission lines as electrical components 79
Multiplying out and simplifying
Z L + Z 0 + (Z L – Z 0) e–2gl
Zl = Z 0
[ ———————————
Z L + Z 0 – (Z L – Z 0) e–2gl ]
Sorting out Z 0 and Z L gives
Z L(1 + e–2gl ) + Z 0 (1 – e–2gl )
Zl = Z 0
[ ————————————
Z L(1 – e–2gl ) + Z 0 (1 + e–2gl ) ]
Multiplying all bracketed terms by (egl/2) results in
egl – e–gl egl + e–gl
Zl = Z0
[ Z 0 ———— + Z L ————
gl + e–gl
gl – e–gl
Z 0 ———— + Z L ————
Bear in mind that by definition
egl – e–gl egl + e–gl
sinh gl = ———— and cosh gl = ————
Substituting for sinh gl and cosh gl in the above equation results in
Z 0 sinh gl + Z L cosh gl
Z l = Z 0 ——————————
Z 0 cosh gl + Z L sinh gl ] (2.52)
Equation 2.52 is a very important equation because it enables us to investigate the proper-
ties of a transmission line. If the total length of the line is l then Z l becomes the input
impedance of the line. Hence, Equation 2.52 becomes
Z 0 sinh gl + Z L cosh gl
Z in = Z 0
[ ——————————
Z 0 cosh gl + Z L sinh gl ] (2.53)
2.13.2 Input impedance of low loss transmission lines
From Equation 2.42, we know that g = a + jb. When a << b, g = jb.
From Equation 2.22, we know that b = 2π/l. From mathematical tables17 we know that
sin (jb) = j sin b and cos(jb) = cos b. If we now substitute the above facts into Equation
2.53, we will get Equation 2.54. The input impedance of a low loss transmission line is
given by the expression
17 You can also check this for yourself if you take the series for sin x and cos x and substitute jb in place of
80 Transmission lines
[ ]
2πl 2πl
jZ 0 sin —— + Z L cos ——
l l
Z in = Z 0 ————————————— (2.54)
2πl 2πl
jZ L sin —— + Z 0 cos ——
l l
Z in = input impedance (ohms)
Z0 = characteristic impedance of line (ohms)
ZL = termination load on line (ohms)
l = electrical wavelength at the operating frequency
l = transmission line length
2.13.3 Reactances using transmission lines
A transmission line can be made to behave like a reactance by making the terminating load
a short circuit (Z L= 0). In this case, Equation 2.54 becomes
[ ][ ]
2πl 2πl
jZ 0 sin —— + 0 j sin ——
l l 2πl
Z in = Z 0 ———————— = Z 0 ————— = jZ 0 tan —— (2.55)
2πl 2πl l
0 + Z 0 cos —— cos ——
l l
When l < l/4
Z in = jZ 0 tan — — (2.55a)
and is inductive.
When l /4 < l > l/2
Z in = –jZ 0 tan — — (2.55b)
and is capacitive.
Equation 2.55 follows a tangent curve and like any tangent curve it will yield positive
and negative values. Therefore Equation 2.55 can be used to calculate inductive and
capacitive reactances. Adjustment of Z 0 and line length, l, will no doubt set the required
Example 2.13
A 377 W transmission line is terminated by a short circuit at one end. Its electrical length
is l/7. Calculate its input impedance at the other end.
Transmission lines as electrical components 81
Solution. Using Equation 2.55a
2π l
[ ]
Z in = jZ 0 tan —— = j377 tan — —
— = j377 × 1.254 = j472.8 W
l l 7
Similar reactive effects can also be produced by using an open-circuited load18 and apply-
ing it to Equation 2.54 to produce inductive and capacitive reactances:
Z in = –jZ 0 cot —— (2.56)
Equation 2.56 follows a cotangent curve and will therefore also produce positive and nega-
tive impedances. Adjustment of Z0 and line length will set the required reactance.
Example 2.14
A 75 W line is left unterminated with an open circuit at one end. Its electrical length is l/5.
Calculate its input impedance at the other end.
Solution. Using Equation 2.56
2πl 2π l
Z in = –jZ 0 cot — = –j75 cot — — = –j75 × 0.325 = –j24.4 W
— —
l l 5
2.13.4 Transmission lines as transformers
An interesting case arises when l = l/2. In this case Equation 2.54 becomes
jZ 0 sin π + Z L cos π
Z in = Z 0
[ ————————— = Z L
jZ L sin π + Z 0 cos π ]
What this means is that the transmission line acts as a 1:1 transformer which is very
useful for transferring the electrical loading effect of a termination which cannot be
placed in a particular physical position. For example, a resistor dissipating a lot of heat
adjacent to a transistor can cause the latter to malfunction. With a 1:1 transformer, the
resistor can be physically moved away from the transistor without upsetting electrical
operating conditions.
Another interesting case arises when l = l/4. In this case, Equation 2.54 becomes
[ ]
π π
jZ 0 sin — + Z L cos —
2 2 Z 02
Z in = Z 0 —————————— = ——
π π ZL
Z L jsin — + Z 0 cos —
18 Purists might argue that an open circuit does not exist at radio frequencies because any unterminated TX
line has stray capacitance associated with an open circuit. We will ignore this stray capacitance temporarily
because, for our frequencies of operation, its reactance is extremely high.
82 Transmission lines
Therefore the transmission line behaves like a transformer where
Z 02
Z in = —— (2.57)
At first glance Equation 2.57 may not seem to be very useful but if you refer back to Figure
2.7, you will see that the characteristic impedance (Z 0) of microstrip transmission lines can
be easily changed by changing its width (w); therefore impedance matching is a very prac-
tical proposition.
Example 2.15
A transmission line has a characteristic impedance (Z 0) of 90 W. Its electrical length is l/4
and it is terminated by a load impedance (Z L) of 20 W. Calculate the input impedance (Z in)
presented by the line.
Given: Z0 = 90 W, ZL = 20 W, l = l/4
Required: Zin
Solution. Using Equation 2.57
Z in = (90)2/20 = 405 W
2.14 Transmission line couplers
Transmission lines can be arranged in special configurations to divide an input signal at the
input port into two separate signals at the output ports. Such an arrangement is often called
a signal splitter. Since the splitter is bi-directional, the same arrangement can also be used
to combine two separate signals into one. These splitter/combiners are often called couplers.
The advantages of couplers are that they are very efficient (low loss), provide good
matching on all ports, and offer reasonable isolation between the output ports so that one
port does not interfere with the other. The greatest disadvantage of these couplers is their
large physical size when used in their distributed forms.
2.14.1 The branch-line coupler
The basic configuration of the branch-line coupler is shown in Figure 2.15. It consists of
transmission lines, each having a length of l/4. Two opposite facing-lines have equal
Fig. 2.15 A branch-line coupler
Transmission line couplers 83
impedances, Z 0, and the remaining opposite facing-lines have an impedance of Z 0/ 2. In
the case of a 50 W coupler, Z 0 = 50 W and Z 0 / 2 = 35.355 W.
Principle of operation
All ports are terminated in Z 0. Signal is applied to port 1 and it is divided equally between
ports 2 and 3. There is no output signal from port 4 because the path from port 1 to 4 is
l/4 while the path from port 1 to 4 via ports 2 and 3 is 3l/4. The path difference is l/2;
hence the signals cancel each other at port 4. The net result is that the signal at port 4 is
zero and it can be considered as a virtual earth point. With this virtual earth point Figure
2.15 becomes Figure 2.16(a). We know from transmission line theory (Equation 2.57) that
for a l/4 length, Z in = Z 02/Zl. In other words, a short circuit at port 4 appears as open
circuits at port 1 and 3. This result is shown in Figure 2.16(b). If we now transform the
impedance at port 3 to port 2, we get
Z(transformed) = Z 02/Z 0 = Z 0
Hence we obtained the transformed Z 0 in parallel with the Z 0 termination of port 2. This
situation is shown in Figure 2.16(c). We now need to transform the Z 0 /2 termination at port
2 to port 1. At port 1
(Z 0 / 2) 2 Z 02 2
Zl = ———— = —— × — = Z 0
(Z 0 /2) 2 Z0
This condition is shown in Figure 2.16(d).
Fig. 2.16(a) Virtual short circuit at port 4
Fig. 2.16(b) Effect of virtual short circuit at port 4
84 Transmission lines
Fig. 2.16(c) Effect of port 3 transferred to port 2
Fig. 2.16(d) Effect of port 2 transferred to port 1
We conclude that the hybrid 3 dB coupler provides a good match to its source imped-
ance Z 0. Assuming lossless lines, it divides its signal equally at ports 2 and 3. Since signal
travels over l/4 to port 2, there is a phase delay of 90° at port 2 and another 90° phase delay
from port 2 to port 3. Thus the signal arriving at port 3 suffers a delay of 180° from the
signal at port 1.
The response of such a coupler designed for 5 GHz is shown in Figure 2.17. As you can
see for yourself, the signal path from port 1 to port 2 (S21) is 3 dB down at 5 GHz. This is
also true of the signal response to port 3 (S31). The signal attenuation to port 4 is theoret-
ically infinite but this value is outside the range of the graph.
A similar analysis will show that power entering at port 2 will be distributed between
ports 1 and 4 and not at port 3. A similar analysis for port 3 will show power dividing
between ports 1 and 4 but not at port 2. The net result of this analysis shows that signals
Fig. 2.17 Unadjusted response of the quadrate 3 dB coupler: S21 = signal attenuation path from port 1 to port 2; S31
= signal attenuation path from port 1 to port 3; S41 = signal attenuation path from port 1 to port 4
Transmission line couplers 85
into port 2 and 3 are isolated from each other. This is a very useful feature in mixer and
amplifier designs.
The advantage of the quadrature coupler is easy construction but the disadvantage of
this coupler is its narrow operational bandwidth because perfect match is only obtained at
the design frequency where each line is exactly l/4 long. At other frequencies, each line
length is no longer l/4 and signal attenuation increases while signal isolation decreases
between the relevant ports.
Finally, you may well ask ‘if port 4 is at virtual earth, why is it necessary to have a
matched resistor at port 4?’ The reason is that signal balance is not perfect and the resistor
helps to absorb unbalanced signals and minimises reflections.
2.14.2 The ring coupler
Ring forms of couplers have been known for many years in waveguide, coaxial and
stripline configurations. The basic design requirements are similar to that of the quadrature
coupler except that curved lines are used instead of straight lines. One such coupler is
shown in Figure 2.18. The principle of operation of ring couplers is similar to that of
branch-line or quadrature couplers.
Fig. 2.18 The 3 dB ring-form branch-line directional coupler
2.14.3 The ‘rat-race’ coupler
A sketch of a ‘rat-race’ coupler is shown in Figure 2.19. The mean circumference of the
ring is 1.5l. This coupler is easy to construct and provides good performance for narrow
band frequencies. The characteristic impedance of the coupler ring is Z 0 2 W which in
the case of Z 0 = 50 W is a circular 70.7 W transmission line which is 1.5l in circumference.
The four Z 0 ports are connected to the ring in such a manner that ports 2 to 3, ports 3 to 1
and ports 1 to 4 are each separated by l/4. Port 4 to port 2 is separated by 0.75l. The oper-
ation of this device is illustrated in Figure 2.20.
If a signal is injected at port 1, the voltage appearing at port 2 is zero, since the path
lengths differ by 0.5l; thus port 2 can be treated as a virtual ground. Hence the transmis-
sion-line portions of the ring between ports 2 and 3, and ports 2 and 4, act as short-
circuited stubs connected across the loads presented at ports 3 and 4. For centre frequency
86 Transmission lines
Fig. 2.19 The ‘rat-race’ or ‘hybrid ring’ coupler
Fig. 2.20 (a) Equivalent circuit of ring hybrid with port 1 as input and ports 2 and 4 as outputs (transmission-line
model with port 3 as virtual ground); (b) equivalent circuit at centre frequency
operation, these stubs appear as open circuits. Similarly, the transmission line lengths
between ports 3 and 1, and ports 4 and 1, transform the 50 W load impedances at ports 3
and 4 to 100 W (2 Z 0) at port 1. When combined at port 1, these transformed impedances
produce the 50 W impedance at port 1. See Figure 2.20(a) and (b).
A similar analysis can be applied at each port, showing that the hybrid exhibits a matched
Summary 87
impedance of 50 W or Z 0 at all nodes. It should be noted that when port 1 is driven, the
outputs at ports 3 and 4 are equal and in phase, while ideally there is no signal at port 2.
However, when port 2 is driven, the output signals appearing at ports 3 and 4 are equal but
exactly out of phase. Also there is no signal at port 1. Hence ports 1 and 2 are isolated. This
is very useful especially in signal mixing circuits because it enables two slightly different
frequencies, for example ƒ1 at port 1 and ƒ2 at port 2, to be applied to a balanced mixer whose
diodes may be connected to ports 3 and 4 without coupling between the sources at port 1 and
port 3. It also helps to combine the inputs or outputs of two amplifiers without mutual inter-
ference. The unfortunate thing about the ring is that it is a relatively narrow-band device.
2.15 Summary
Chapter 2 has provided you with a thorough basic knowledge of transmission lines and
their properties which you will find very useful in circuit and system design of radio and
microwave systems. You have been introduced to many properties of transmission lines in
this chapter.
In Section 2.3, you were introduced to some of the more frequently used types of trans-
mission lines. These included waveguides, coplanar waveguides, coaxial lines, microstrip
and strip lines, slot lines, twin lines and finally coupled microstrip lines.
In Section 2.4, you were shown how the characteristic impedance of the coaxial, twin
line and microstrip line can be calculated from its physical parameters. The information
demonstrated what properties you should look for if the characteristic impedance of a line
does not behave as expected.
Sections 2.5 and 2.6 demonstrated how the characteristic impedance can be calculated
and measured from primary constants of the line. Section 2.7 mentions some of the more
common impedances associated with commercial transmission lines but it also brought to
your attention that there are many types of lines with the same impedance.
Section 2.8 explained how propagation delay, attenuation and frequency dispersion
affect waveforms as they travel along transmission lines. This was followed by more
discussions on the effects of these properties in Section 2.9.
In Section 2.10, we introduced the concepts of matched and unmatched lines. This was
followed by a thorough discussion of reflection coefficients and voltage standing wave
ratios in Section 2.11.
Section 2.12 dealt with the propagation properties of lines and showed how these can
be derived from the primary constants of the line. The section also showed how optimum
transmission can be achieved.
Section 2.13 showed how transmission lines can be used as transformers, impedance
matching devices, inductive and capacitive reactances which in turn can be used to produce
filters. Microstrip lines are particularly useful for making filters at the higher frequencies
because of their versatility in allowing characteristic impedance changes to be made easily.
There is also a greater tendency to use transmission lines as the tuning elements.
Section 2.14 showed you how transmission lines can be connected to act as signal
Finally, you will come across these examples again when we introduce the software
program PUFF in Chapter 4 and carry out some examples to further clarify your theory
and the use of the program.
Smith charts and scattering
3.1 Introduction
3.1.1 Aims
The aims of this part are to familiarise you with two fundamental necessities in radio engi-
neering; Smith charts and scattering parameters. We start with Smith charts because
they are very useful for amplifier design, gain circles, noise circles, matching network
design, impedance and admittance determination and finding reflection coefficients and
voltage standing wave ratios.
Scattering parameters are important because many basic items such as filters, hybrid
transformers, matching networks, transistors, amplifiers, gain blocks and monolithic
microwave integrated circuits (MMIC) are used and described by manufacturers in terms
of two port networks. You also need s-parameters for circuit and system design.
In this part, Sections 3.2 to 3.4 have been devoted to a description of the Smith chart.
Section 3.5 covers its theory and Sections 3.6 to 3.9 provide examples on Smith chart
applications. Section 3.10 deals with the fundamentals of scattering parameters and
Section 3.11 gives examples of its applications.
3.1.2 Objectives
After reading the section on Smith charts, you should be able to:
• evaluate impedance and admittance networks;
• design impedance and admittance networks;
• design matching networks.
After reading the section on scattering parameters, you should be able to:
• understand scattering parameters;
• use two port scattering parameters efficiently;
• design two port scattering networks;
• evaluate two port scattering networks;
• calculate the frequency response of networks;
• calculate the gain of two port networks or an amplifier;
• calculate the input impedance of two port networks or an amplifier;
• manipulate two port data into other types of parameter data.
Smith charts 89
3.2 Smith charts
3.2.1 Introduction
The Smith chart is intended to provide a graphical method for displaying information, for
impedance matching using lumped and distributed elements and is particularly useful in
solving transmission line problems. It avoids the tedious calculations involved in applying
the expressions obtained and proved in Part 2. It is an alternative check to your calcula-
tions and provides a picture of circuit behaviour to help you visualise what happens in a
Fig. 3.1 Smith chart
90 Smith charts and scattering parameters
circuit. The Smith chart1 was devised by P.H. Smith in the late 1930s and an improved
version was published in 1944. It is shown in Figure 3.1.
The Smith chart is based on two sets of circles which cut each other at right-angles. One
set (Figure 3.2) represents the ratio R/Z 0, where R is the resistive component of the line
impedance Zx = R + jX. Z 0 is usually taken as the characteristic impedance of a transmis-
sion line. Sometimes Z 0 is just chosen to be a number that will provide a convenient
display on the Smith chart. The other set of circles (Figure 3.3) represents the ratio jX/Z 0,
Fig. 3.2 Resistive circles of 0, 0.2, 0.4, 1, 2, 4 and 20 are shown in bold
1 The Smith chart is a copyright of Analog Instruments Co., P.O. Box 808, New Providence, NJ07974, USA.
Smith charts 91
where X is the reactive component of the line impedance Zx = R + jX. Z 0 is usually taken
as the characteristic impedance of a transmission line. Sometimes Z 0 is just chosen to be
a number that will provide a convenient display on the Smith chart. In both cases of
normalisation, the same number must be used for both resistance and reactance normali-
sation. The circles have been designed so that conditions on a lossless line with a given
VSWR can be found by drawing a circle with its centre at the centre of the chart.
Figures 3.1, 3.2 and 3.3 are full detailed versions of the Smith chart. When electronic
versions of the chart are used, such detail tends to clutter a small computer screen. In such
cases, it is best to show an outline Smith chart and to provide the actual coordinates in
complex numbers. This is the system which has been used by the software program PUFF,
Fig. 3.3 Reactance arcs of circles for j = ±0, 0.2, 0.4, 1, 2, 4
92 Smith charts and scattering parameters
provided with this book. There is also another Smith chart program called MIMP2
(Motorola Impedance Matching Program).
3.2.2 Plotting impedance values
Any point on the Smith chart represents a series combination of resistance and reactance of
the form Z = R + jX. Thus, to locate the impedance Z = 1 + jl, you would find the R = 1
constant resistance circle and follow it until it crosses the X = 1 constant reactance circle.
The junction of these two circles would then represent the needed impedance value. This
particular point, A shown in Figure 3.4, is located in the upper half of the chart because X is
a positive reactance or inductive reactance. On the other hand, the point B which is 1 – jl is
Fig. 3.4 Some values on a simplified Smith chart
2 At the time of writing, this program can be obtained free from some authorised Motorola agents. This
program provides active variable component matching facilities.
Smith charts 93
located in the lower half of the chart because, in this instance, X is a negative quantity and
represents capacitive reactance. Thus, the junction of the R = 1 constant resistance circle
and the X = –1 constant reactance circle defines that point. In general, then, to find any
series impedance of the form R + jX on a Smith chart, you simply find the junction of the
R = constant and X = constant circles.
In order to give you a clearer picture of impedance values on the Smith chart, we plot
additional impedance values in Figure 3.4. These values are shown in Table 3.1.
Table 3.1 Impedance values for points plotted in Figure 3.4
A = 1 + j1 B = 1 – j1 C = 0 + j0
D = 0.2 + j0.2 E = 0.2 + j0.7 F = 0.2 + j1.2
G = 0.5 + j0.5 H = 1 + j0 I = ∞ + j0
J = 6 + j2 K = 0.2 – j0.6 L = 0.5 – j0.2
M = 0.6 – j2 N = 5 – j5
Try to plot these values on Figure 3.4 and check if you get the correct values.
In some cases, you will not find the circles that you want; when this happens, you will
have to interpolate between the two nearest values that are shown. Hence, plotting imped-
ances on the Smith chart produces a plotting error. However, the error introduced is rela-
tively small and is negligible for practical work.
If you try to plot an impedance of Z = 20 + j20 W, you will not be able to do it accu-
rately because the R = 20 and X = 20 W circles would be (if they were drawn) on the
extreme right edge of the chart – very close to infinity. In order to facilitate the plotting of
larger impedances, normalisation is used. That is, each impedance to be plotted is divided
by a convenient number that will place the new normalised impedance near the centre of
the chart where increased accuracy in plotting is obtained. For the preceding example,
where Z = 20 + j20 W, it would be convenient to divide Z by 100, which yields the value
Z = 0.2 + j0.2. This is very easily found on the chart.
The important thing to realise is that if normalisation is carried out for one impedance
then all impedances plotted on that chart must be divided by the same number in the
normalisation process. Otherwise, you will not be able to use the chart. Last but not least,
when you have finished with your chart manipulations, you must then re-normalise
(multiply by the same number used previously) to get your true values.
Example 3.1
Plot the points (0.7 – j0.2), (0.7 + j0.3), (0.3 – j0.5) and (0.3 + j0.3) on the Smith chart in
Figure 3.4.
Solution. The above points are all shown on the Smith chart in Figure 3.5. Check your
plotting points in Figure 3.4 against those in Figure 3.5.
3.2.3 Q of points on a Smith chart
Since the quality factor Q is defined as reactance/resistance, it follows that every point on
the Smith chart has a value of Q associated with it. For example, the plotted points of
Example 3.1 are shown in Table 3.2.
94 Smith charts and scattering parameters
Table 3.2 Q values of the points in Example 3.1
Resistance Reactance Q = |X|/R
(R) (X)
0.7 –0.2 0.286
0.7 +0.3 0.429
0.3 –0.5 1.667
0.3 +0.3 1.000
At the moment, it is not clear as to what can be done with these Q values but you will
understand their validity later when we investigate broadband matching techniques in
Section 3.8.2.
3.2.4 Impedance manipulation on the chart
Fig. 3.5 A = 0.7 + j0.3, B = 0.7 – j0.2, C = 0.3 – j0.5, D = 0.3 + j0.3
Figure 3.5 indicates graphically what happens when a series capacitive reactance of –j0.5
W is added to an impedance of Z = (0.7 + j0.3) W. Mathematically, the result is
Smith charts 95
Z = (0.7 + j0.3 – j0.5) W = (0.7 – j0.2) W
which represents a series RC quantity. Graphically, we have plotted (0.7 + j0.3) as point A
in Figure 3.5. You then read the reactance scale on the periphery of the chart and move
anti-clockwise along the R = 0.7 W constant resistance circle for a distance of X = –j0.5 W.
This is the plotted impedance point of Z = (0.7 – j0.2) W, shown as point B in Figure 3.5.
Adding a series inductance to a plotted impedance value simply causes a move clock-
wise along a constant resistance circle to the new impedance value. Consider the case in
Figure 3.5 where a series inductance j0.8 W is added to an impedance of (0.3 – j0.5) W.
Mathematically the result is
Z = (0.3 – j0.5 + j0.8) W = (0.3 + j0.3) W
Graphically, we have plotted point (0.3 – j0.5) in Figure 3.5 as point C then moved along
the 0.3 resistance circle and added j0.8 to that point to arrive at point D.
In general the addition of a series capacitor to a plotted impedance moves that imped-
ance counter-clockwise along a constant resistance circle for a distance that is equal to the
reactance of the capacitor. The addition of a series inductor to a plotted impedance moves
that impedance clockwise along a constant resistance circle for a distance that is equal to
the reactance of the inductor.
3.2.5 Conversion of impedance to admittance
The Smith chart, described so far as a family of impedance coordinates, can easily be used
to convert any impedance (Z) to an admittance (Y), and vice-versa. In mathematical terms,
an admittance is simply the inverse of an impedance, or
Y=— (3.1)
where, the admittance (Y) contains both a real and an imaginary part, similar to the imped-
ance (Z). Thus
Y = G ± jB (3.2)
G = conductance in Siemens (S)
B = susceptance in Siemens (S)
To find the inverse of a series impedance of the form Z = R + jX mathematically, you would
simply use Equation 3.1 and perform the necessary calculation. But, how can you use the
Smith chart to perform the calculation for you without the need for a calculator? The easi-
est way of describing the use of the chart in performing this function is to first work a prob-
lem out mathematically, and then plot the results on the chart to see how the two functions
are related. Take, for example, the series impedance Z = (1 + jl) W. The inverse of Z is
Y = ———— = —————— = 0.7071 S ∠ –45° = (0.5 –j0.5) S
1 + j1 W 1.414 W ∠45°
If we plot the points (1 + jl) and (0.5 – j0.5) on the Smith chart, we can easily see the
graphical relationship between the two. This construction is shown in Figure 3.6. Note that
96 Smith charts and scattering parameters
Fig. 3.6 Changing an impedance to admittance: A = (1 + j1) Ω, B = (0.5 – j0.5) S
the two points are located at exactly the same distance (d) from the centre of the chart but
in opposite directions (180°) from each other. Indeed, the same relationship holds true for
any impedance and its inverse. Therefore, without the aid of a calculator, you can find the
reciprocal of an impedance or an admittance by simply plotting the point on the chart,
measuring the distance (d) from the centre of the chart to that point, and then plotting the
measured result the same distance from the centre but in the opposite direction (180°) from
the original point. This is a very simple construction technique that can be done in seconds.
Example 3.2
Use the Smith chart in Figure 3.6 to find the admittance of the impedance (0.8 – j1.6).
Given: Z = (0.8 – j1.6)
Required: Admittance value Y
Solution. The admittance value is located at the point (0.25 + j0.5). You can verify this
yourself by entering the point (0.8 – j1.6) in Figure 3.6. Measure the distance from your
The immittance Smith chart 97
point to the chart centre. Call this distance d. Draw a line of length 2d, from your point
through the centre of the chart. Read off the coordinates at the end of this line. You should
now get (0.25 + j0.5).
3.3 The immittance Smith chart
Fig. 3.7 Immittance Smith chart – reading a point as an impedance or admittance
Alternatively, we can use another Smith chart, rotate it by 180°, and overlay it on top of
a conventional Smith chart. Such an arrangement is shown in Figure 3.7. The chart that
you see in Figure 3.7 is one which I have prepared for you. Detailed charts are also
obtainable commercially as Smith chart3 Form ZY-01-N. With these charts, we can plot
3 Smith chart Form ZY-01-N is a copyright of Analog Instruments Co, P.O. Box 808, New Providence, NJ
07974, USA.
98 Smith charts and scattering parameters
the coordinates (1 + j1) directly on the impedance chart and read its admittance equivalent
(0.5 – j0.5) on the rotated admittance chart directly. Another approach that we could take
(if we are working solely with admittances) is to just rotate the chart itself 180° and manip-
ulate values on the chart as admittances. This will be shown more clearly in the next section.
3.4 Admittance manipulation on the chart
Fig. 3.8 Impedance chart used as an admittance chart: A = (0.2 – j0.6) S, B = (0.2 + j0.3) S, C = (1.2 + j0.4) S, D =
(1.2 – j0.6) S
In this section, we want to present a visual indication of what happens when a shunt
element is added to an admittance. The addition of a shunt capacitor is shown in Figure
3.8. For an example we will choose an admittance of Y = (0.2 – j0.6) S and add a shunt
capacitor with a susceptance (reciprocal of reactance) of +j0.9 S. Mathematically, we
Smith chart theory and applications 99
know that parallel susceptances are simply added together to find the equivalent suscep-
tance. When this is done, the result becomes:
Y = (0.2 – j0.6) S + j0.9 S = (0.2 + j0.3) S
If this point is plotted on the admittance chart, we quickly recognise that all we have done
is to move along a constant conductance circle (G) clockwise a distance of jB = 0.9 S. In
other words, the real part of the admittance has not changed, only the imaginary part.
Similarly, if we had a point (1.2 + j0.4) S and added an inductive susceptance of
(–j10) S to it, we would get (1.2 + j0.4) S – j10 S = (1.2 – j0.6) S. This is also shown in
Figure 3.8. Hence, adding a shunt inductance to an admittance moves the point along a
constant conductance circle counter-clockwise a distance (–jB) equal to the value of its
susceptance, –j10 S, as shown in Figure 3.8.
3.5 Smith chart theory and applications
This section deals with the derivation of the resistance (R) and reactance (X) circles of the
Smith chart (see Figure 3.9). If you are prepared to accept the Smith chart and are not inter-
ested in the theory of the chart, then you may ignore all of Section 3.5.1 because it will not
stop you from using the chart.
3.5.1 Derivations of circles
Our definitions are:
(i) the normalised load impedance is
Z R + jX
z= L = = r + jx
Z0 Z0
(ii) the reflection coefficient is
G = p + jq (3.4)
Here p (for in-phase) is the real part of G and q (for quadrature) is the imaginary part. From
Part 2, we use the definition for G = (Z L– Z 0)/(Z L+ Z 0) and changing into normalised
values by using Equation 3.3, we obtain
z −1
z +1
and by transposing
z = ——
Substituting z from Equation 3.3 and Γ from Equation 3.4 gives
1 + p + jq
r + jx = ————
1 – p – jq
100 Smith charts and scattering parameters
Fig. 3.9 Resistive and reactive circles in the Smith chart
To rationalise the denominator, multiply through by [(1 – p) + jq]. This gives
[(1 + p) + jq][(1 − p) + jq]
r + jx =
[(1 − p) − jq][(1 − p) + jq]
(1 − p 2 − q 2 ) + j2 q
(1 − p)2 + q 2
Equating real parts
(1 − p 2 − q 2 )
r= (3.5)
(1 − p)2 + q 2
Equating imaginary parts
x= (3.6)
(1 − p)2 + q 2
Smith chart theory and applications 101
First, we derive the equation of an r circle. From Equation 3.5
r(1 – 2p + p2 + q 2) = (1 – p2 – q 2)
(r + 1)p 2 – 2pr + (r + 1)q 2 = 1 – r
Dividing throughout by (r + 1)
⎛ p 2 − 2 pr ⎞ + q 2 = 1 − r × (1 + r ) = 1 − r
⎝ r + 1⎠ r + 1 (1 + r ) (r + 1)2
To complete the square in p, add r2/(r + 1)2 to both sides. This gives
⎛ 2 2 pr r2 ⎞ 1 − r2 r2
⎜ p − r +1 + 2⎟
+ q2 = 2
⎝ (r + 1) ⎠ (r + 1) (r + 1)2
⎛ p − r ⎞ + q2 = ⎛ 1 ⎞ (3.7)
⎝ r + 1⎠ ⎝ r + 1⎠
This is the equation of a circle, with centre at [p = r/(r +1), q = 0] and radius 1/(r + 1). For
any given value of r, the resultant is called a constant-r circle, or just an r circle.
Next, we derive an equation of an x circle. From Equation 3.6
x(1 – 2p + p 2 + q 2) = 2q
Dividing throughout by x
( p 2 − 2 p + 1) + ⎛ q 2 − ⎞ = 0
⎝ x⎠
To complete the square, we add 1/x 2 to both sides:
⎛ 2q 1 ⎞ 1
( p 2 − 2 p + 1) + ⎜ q 2 − + 2⎟ = 2
⎝ x x ⎠ x
( p − 1)2 + ⎛ q − ⎞ = ⎛ ⎞
⎝ x⎠ ⎝ x⎠
This is the equation of a circle, with centre at [ p = 1, q = 1/x] and radius, 1/x. For any
given value of x, the resultant circle is called a constant-x circle, or just an x circle.
3.5.2 Smith chart applications
As the proof and theory of the Smith chart has already been explained in Section 3.5.1, we
will now concentrate on using the chart. We will commence by using the chart to find
reflection coefficients and impedances of networks, then progress on to using the chart for
matching with λ/4 transformers and tuning stubs.
102 Smith charts and scattering parameters
3.6 Reflection coefficients and impedance networks
3.6.1 Reflection coefficients
The Smith chart can be used to find the reflection coefficient at any point, in modulus-and-
angle form. If we plot the point (0.8 – j1.6) denoted by point A on the Smith chart of Figure
3.10 and extend the line OA to B, we will find by measurement that the angle BOC is about
–55.5°. You will not see this angle scale in Figure 3.10 because our condensed Smith chart
does not have an angle scale.4
The modulus of the reflection coefficient can be found from Equation 2.38 which states
1 + Γv
VSWR =
1 − Γv
Fig. 3.10 Reflection coefficient: A = (0.8 – j1.6), angle BOC = –55.5°
4 The full Smith chart also carries an angle scale. See Figure 3.1.
Reflection coefficients and impedance networks 103
So, rearranging
VSWR − 1
Γv =
VSWR + 1
In our example, VSWR5 ≈ 5, so |Γv | ≈ (5 – 1)/(5 + 1) = 0.667 and hence the reflection coef-
ficient = 0.667/–55.5°.
Some Smith charts such as Figure 3.1 do have ‘radially scaled parameters’ scales, and
it is possible to obtain the reflection coefficient magnitude directly by simply measuring
the radius of the VSWR circle and reading off the same distance on the voltage reflection
coefficient scale. Alternatively, use Equation 2.38a to calculate it.
Example 3.3
In Figure 3.11, the VSWR circle has a radius of 0.667. An impedance is shown on the
Smith chart as point B which is (0.25 – j0.5). What is its reflection coefficient?
Given: VSWR circle radius = 0.667.
Impedance at point B = (0.25 – j0.5).
Required: Voltage reflection coefficient Γ.
Solution. The answer is 0.667 /–124°. This is shown as point C in Figure 3.11.
Fig. 3.11 Smith chart for Examples 3.3 and 3.4
5 VSWR is obtained by completing the circle enclosing the point A (Figure 3.10). It is then read off the inter-
section between the circle and the real axis and in this case = 5. Proof of this will be given when we derive
Equation 3.14.
104 Smith charts and scattering parameters
Example 3.4
In Figure 3.11, the VSWR circle has a radius of 0.667. If the angle AOD is +156°, what
impedance does the point E represent on the Smith chart?
Given: VSWR circle radius = 0.667
AOD = +156°.
Required: Impedance at point E.
Solution. The answer is approximately 0.21 + j0.21. This is shown as point E in Figure
3.6.2 Impedance of multi-element circuits
The impedance and/or admittance of multi-element networks can be found on the Smith
chart without any calculations.
Example 3.5
What is (a) the impedance and (b) the reflection coefficient looking into the network
shown in Figure 3.12?
Given: Network shown in Figure 3.12.
Required: (a) Input impedance, (b) reflection coefficient.
Fig. 3.12 Network to be analysed
Fig. 3.13 Circuit of Figure 3.12 dis-assembled for analysis
Reflection coefficients and impedance networks 105
Solution. The problem is easily handled on a Smith chart and not a single calculation
needs to be performed. The solution is shown by using Figure 3.13.
(a) To find the impedance, proceed as follows.
(1) Separate the circuit down into individual branches as shown in Figure 3.13. Plot
the series branch where Z = (1 + j1.2) W. This is point A in Figures 3.13 and
(2) Add each component back into the circuit – one at a time. The following rule is
particularly important. Every time you add an impedance, use the impedance part
of the chart. Every time you add an admittance, use the admittance part of the
chart. If you observe the above rule, you will have no difficulty following the
construction order below:
Fig. 3.14 Plot of Example 3.5: A = (1 + j1.2 ) Ω or (0.41 – j0.492) S, B = (0.3 + j0.8) Ω or (0.41 – j1.092) S, C = (0.3
– j0.2) Ω or (2.3326 + j1.522) S, D = (0.206 – j0.215) Ω or (2.326 + j2.422) S, E = (0.206 + j0.635) Ω or (0.462 –
j1.425) S, F = 0.746 ∠ 113.58°
106 Smith charts and scattering parameters
Arc AB = shunt L = – jB = –0.6 S
Arc BC = series C = –jX = –1.0 Ω
Arc CD = shunt L = +jB = +0.9 S
Arc DE = series C = +jX = +0.85 Ω
(3) The impedance value (point E) can then be read directly from Figure 3.14. It is
Z = (0.2 + j0.63).
(b) To find the reflection coefficient, proceed as follows.
(1) Draw a line from the centre of the chart (Figure 3.14) through the point E to cut
the chart periphery at point F. Measure the distance from the chart centre O to
point E and transfer this distance to the reflection coefficient scale (if you have
one) to obtain its value.6 This value is 0.74.
(2) Read the angle of intersection of the line OE and the periphery. This angle is 114°.
(3) Hence the value of the reflection coefficient is 0.74 ∠114°.
3.7 Impedance of distributed circuits
In the previous example, we showed you how the Smith chart can be used for lumped
circuit elements. In Sections 3.7 and 3.8, we will show you how the Smith chart can also
be used with distributed circuit elements like transmission lines.
For ease of verifying Smith chart results, some of the transmission line expressions
derived earlier will now be repeated without proof. These are:
⎡1 + Γv e −2γl ⎤
Zl = Z 0 ⎢ −2γl ⎥ (3.9)
⎣1 − Γv e
⎢ ⎥
For a lossless line, with g = jb, this becomes
⎡1 + Γv e − j2 βl ⎤
Zl = Z 0 ⎢ − j2 βl ⎥
⎣1 − Γv e
⎢ ⎥
Equation 3.9 has been shown to be
⎡ Z sinh γl + ZL cosh γl ⎤ (3.11)
Zin = Z0 ⎢ 0 ⎥
⎣ Z0 cosh γl + ZL sinh γl ⎦
Equation 3.11 can also be written in the form
Zin ⎡ Z0 sinh γl + ZL cosh γl ⎤
=⎢ ⎥ (3.12)
Z0 ⎣ Z0 cosh γl + ZL sinh γl ⎦
It is also possible to divide each term in Equation 3.12 by (Z 0 cosh gl) and, remembering
that for a lossless line that tanh gl = j tan bl, we get
6 The unfortunate part of this condensed Smith chart is that we do not have a voltage reflection coefficient
scale. So please accept my answer for now until we use PUFF. Alternatively, use Equation 2.38a to calculate it.
Impedance of distributed circuits 107
Zin ⎡ j tan βl + ZL Z0 ⎤
=⎢ ⎥ (3.13)
Z0 ⎣ j ( ZL Z0 ) tan βl + 1 ⎦
Note Equations 3.12 and 3.13 have been normalised with respect to Z 0. Although stated
earlier, we will re-iterate that normalisation is used on Smith charts because it enables a
larger range of values to be covered with greater accuracy on the same chart. However, if
we divide a value by Z 0 (normalised) before entering it on the Smith chart, it stands to
reason that we must then multiply the Smith chart result by Z 0 (re-normalised) to gets its
true value.
Nowadays, the above mathematical calculations can be easily programmed into a
computer, and you may well question the necessity of the Smith chart for line calculations.
However, as you will see shortly the Smith chart is more than just a tool for line calcula-
tions; it is invaluable for gaining an insight into line conditions for matching purposes,
parameter presentation, constant gain circles, stability and instability circles, Qs of
elements, standing-wave ratios, and voltage maxima and minima positions.
The input and output impedances of transistors are usually complex, with a significant
reactive component. The Smith chart is particularly useful for matching purposes when a
transmission line is terminated by the input of a transistor amplifier, or when the line is
driven from an amplifier. Matching is necessary because it avoids reflections on the lines
and ensures maximum power transfer from source to load.
3.7.1 Finding line impedances
Smith charts can be used to find line impedances as demonstrated by the following ex-
Example 3.6
A transmission line with a characteristic impedance Z 0 = 50 Ω is terminated with a load
impedance of Z L = (40 – j80) Ω. What is its input impedance when the line is (a) 0.096λ,
(b) 0.173λ and (c) 0.206λ?
Given: Z 0 = 50 Ω, Z L = (40 – j80) Ω.
Required: The input impedance Z in of the terminated line when (a) the line is 0.096λ, (b)
0.173λ and (c) 0.206λ .
Solution. First, the load impedance is expressed in normalised form as
ZL ( 40 − j80)Ω
= = 0.8 − j1.6
Z0 50Ω
This value is plotted on the chart as point A in Figure 3.15. Note that this point is the inter-
section of the arcs of two circles. The first is that which cuts the horizontal axis labelled
‘resistance component (R/Z 0)’ at the value 0.8. Because the reactive component is nega-
tive, the second circle is that which cuts the circular axis on the periphery labelled ‘capac-
itance reactance component (–jX/Z 0)’ at the value 1.6. A line is now drawn from the centre
of the Smith chart (point 1 + j0) through the point 0.8 – j1.6 and projected to the periph-
ery of the circle to cut the ‘wavelengths towards generator’ circle at 0.327l. This is shown
as point B.
108 Smith charts and scattering parameters
Fig. 3.15 Finding input impedance of a line: A = 0.8 – j1.6, B = 0.327l, C = 0.423l, D = 0.25 – j0.5 E = 0.5l, or
0l, F = 0.2 + j0, G = 0.033l, H = 0.2 + j02, I = 0.077l, J = 0.25 – j0.5, K = 5 + j0
Next, a circle is drawn with its centre at the centre of the chart (point 1 + j0), passing
through the load impedance value (point A) in Figure 3.15. This circle represents all possi-
ble line impedances along the transmission line and we will call it our impedance circle.7
(a) We wish to know Z in when the TX line is 0.096l long. For this, we start at the point
B, 0.327l on the ‘wavelengths towards generator’ circle and move 0.096l in the direc-
tion of the generator (i.e. away from the load). Since 0.327l + 0.096l = 0.423l, the
point we want is 0.423l on the ‘wavelengths towards generator’ circle. I have denoted
this point as point C. From C, draw a straight line to the Smith chart centre. Where this
line cuts our impedance circle, read out the value at the intersection. I read it as (0.25
– j0.5) and have marked it as point D. To get the actual value, I must re-normalise the
chart value by 50 W and get (0.25 – j0.5) × 50 W = (12.5 – j25) W.
7 This circle is also called the voltage standing wave ratio (VSWR) circle.
Impedance of distributed circuits 109
(b) For the case where the line length is 0.173l, I must move 0.173l on the ‘towards the
generator’ scale from the same starting point of 0.327l, i.e. point B. So I must get to
the point (0.327 + 0.173)l or 0.5l on the ‘wavelengths towards generator’ scale. This
is shown as point E in Figure 3.15. Joining point E with the Smith chart centre shows
that I cut our original impedance circle at the point F, where the value is (0.2 + j0). To
get the true value, I must re-normalise, (0.2 + j0) × 50 W = (10 + j0) W.
Note: It is possible to get a pure resistance from a complex load by choosing the right
length of transmission line. This is very important as you will see later when we do designs
on matching complex transistor loads to pure resistances.
(c) For the case where the line length is 0.206l, I must move 0.206l on the ‘towards the
generator’ scale from the same starting point of 0.327λ, i.e. point B. This scale only
goes up to 0.5l and then restarts again. So I must get to 0.5l which is a movement of
(0.5 – 0.327)l or 0.173l then add another (0.206 – 0.173)λ or 0.033l to complete the
full movement of 0.206l to arrive at point G. Joining point G with the Smith
chart centre shows that I cut our original impedance circle at the point H, where the
value is (0.2 + j0.2). To get the true value, I must re-normalise, (0.2 + j0.2) × 50 W
= (10 + j10) W.
If you wish, you may confirm these three values (12.5 – j25) W, 10 W and (10 + j10) W by
calculations using Equation 3.11 but bear in mind that you will have to change the wave-
length8 into radians before applying the equation.
Example 3.7
Show that, when Z L = R L + j0 on a lossless line, the VSWR equals R L/Z 0. (Hint: First find
the reflection coefficient of the load.)
Solution. Using Equation 2.28
ZL – Z0
Γv = ————
ZL + Z0
When Z L = R L, this becomes
RL – Z0
Γv = ————
RL + Z0
Here, both R L and Z 0 are purely resistive, so Γv is a real number. Using Equation
1 + Γv 1 + Γv
VSWR = =
1 − Γv 1 − Γv
because Γv is real. So
8 One wavelength = 2π radians.
110 Smith charts and scattering parameters
1 + (R L − Z0 ) ( RL + Z0 )
VSWR =
1 − ( RL − Z0 ) ( RL + Z0 )
( RL + Z0 ) + ( RL − Z0 )
( RL + Z0 ) − ( RL − Z0 )
2 RL
2 Z0
VSWR = (3.14)
It has been shown in Example 3.7 that when Z L = R L, and Z 0 is resistive, the VSWR is
given simply by VSWR = R L/Z 0. In this case R L/Z 0 = 5.0 (point K in Figure 3.15), so the
VSWR is 5.0. Thus the point where the circle cuts the right-hand horizontal axis gives the
VSWR on the line.
Looking at successive points around the standing-wave circle drawn through the load
impedance is equivalent to looking at successive points along a lossless line on which the
VSWR equals that of the circle. The successive values of input line impedances at points D,
F, H, K around the circle correspond to line impedances at successive points along the line.
The distance along the line is directly proportional to the angle around the standing-wave
circle. One complete revolution takes us from a voltage minimum at the point F in Figure 3.15,
where Z in = 0.2 Z 0, to the point opposite this on the circle with a voltage maximum where Z in
= 5 Z 0 at point K, and back to the first minimum. Since standing-wave minima repeat every
half wavelength, one complete revolution corresponds to λ /2. The peripheral scales marked
‘wavelengths towards generator’ and ‘wavelengths towards load’ are calibrated accordingly.
Figure 3.15 shows that, for our example, the position (point F) of the first voltage mini-
mum is at 0.173λ back from the load, clockwise around the scale ‘wavelengths toward
generator’. The first maximum (point K) is shown at [(0.5 – 0.327) + 0.25]λ or 0.423λ from
the load, clockwise around the ‘wavelengths towards generator’ scale. The distance between
voltage maxima and minima is 0.25λ as you should expect from transmission line theory.
Example 3.8
Use the Smith chart in Figure 3.15 to find the line impedance at a point one quarter wave-
length from a load of (40 – j80) Ω.
Given: ZL = (40 – j80)Ω, l = λ/4.
Required: Z @ λ/4 from ZL.
Solution. Moving around the chart away from the load and towards the generator (that is
clockwise) through 0.25l (that is 180°) brings us to the normalised impedance (0.25 +
j0.5), point J in Figure 3.15. Since the chart was normalised to 50 Ω, to get the true value
we must multiply (0.25 + j0.5) × 50 = (12.5 + j25) Ω.
3.8 Impedance matching
When a load such as a transistor input, with a substantial reactive component, is driven from
a transmission line, it is necessary to match the load to the line to avoid reflections and to
Impedance matching 111
transfer the most power from source to load. The reactive component of the load can be
tuned out, and the resistive component matched to the line, using a matching network of
inductors and capacitors. However, at UHF and microwave frequencies, lumped inductors
and capacitors can be very lossy, and much higher Q values can be obtained by using addi-
tional sections of transmission line instead. There are two common techniques used, the
quarter-wavelength transformer and the stub tuner. These are described below.
3.8.1 Impedance matching using a λ/4 transformer
This is shown by Example 3.9 which first uses a line length to convert a complex load to
a resistive load and then uses a 1/4 line transformer to transform the resistive load to match
the desired source impedance.
Example 3.9
Figure 3.15 shows how a quarter-wavelength section of line can be used to match a load, such
as the input of a transistor, to a line. Suppose the load has the normalised value of the previ-
ous example, that is (0.8 – j1.6), point A in Figure 3.15. First, the Smith chart is used to choose
a length l of line which, when connected to the load, will have a purely resistive input imped-
ance. In this example, the length could be either 0.173λ, with a normalised input impedance
of 0.2 (point F in Figure 3.15) or 0.432λ with a normalised input impedance of 5.0 (point K
in Figure 3.15). Suppose we choose the higher value, with l = 0.432λ and Z /Z 0 = 5.0.
Next, we calculate the characteristic impedance of the quarter-wave section. It is required
to match Z in to the input line which is Z 0, so its input impedance Z in must equal Z 0.
The λ/4 transformer. The equation for a quarter-wave transformer has been derived in
Chapter 2 as Equation 2.57. To distinguish the main transmission line impedance (Z 0)
from the λ/4 transformer line impedance, we shall denote the latter as Z 0t. Therefore
Z0 t
Zin = or Z0 t = Zin ZL
Since we wish to make Z in of the l/4 transformer match Z 0, we therefore get
Z0 t = Zin ZL = Z0 ZL
Normalising, this becomes
Z0 t Z0 Z L ZL
= =
Z0 Z0 Z0
In the example, Z L/Z 0 = 5.0. Therefore
Z0 t
= 5.0 ≈ 2.24
Therefore, the λ/4 transformer characteristic impedance, Z 0t = 2.24 Z 0. With Z 0 = 50 Ω
Z 0t = 2.24 × 50 Ω ≈ 112 Ω
In microstrip, an impedance of 112 Ω is possible but the microstrip itself is beginning
to get narrow and fabrication accuracy of the 112 Ω line might be more difficult.
112 Smith charts and scattering parameters
Example 3.10
Find the characteristic impedance Z 0t of a λ/4 transformer required for the case when
Z L/Z 0 = 0.2, and Z 0 = 50 Ω.
Solution. Again using Equation 2.57
Z0 t Z0 = ZL Z0 = 0.2 = 0.447
So, with Z 0 = 50 Ω
Z 0t = 0.447 × 50 ≈ 22 Ω
The answer to Example 3.7 shows that the required λ/4 line’s characteristic impedance
turns out to be low. In microstrip, a low line impedance means that the microstrip itself is
wider and in some cases this might be an easier fabrication process.
In general, a higher value of Z 0t results if a λ/4 transformer is coupled to a standing-
wave voltage maxima on the matching line, and a lower value of Z 0t results if a λ/4 trans-
former is coupled to a standing-wave voltage minima on the matching line. The choice as
to where the λ/4 transformer is coupled will depend on circumstances such as physical
size, dielectric constants of microstrip line, etc.
3.8.2 Broadband matching using λ/4 transformers
Broadband matching is used when we want to achieve the best compromise match between
source and load across a given bandwidth. Compromise is necessary because perfect
matching at all individual frequencies is not feasible. The term ‘broadband’ implies that
impedance compensation should be achieved over frequency ranges larger than 50% of the
central frequency.
Distributed elements, due to their fixed geometric characteristics, are usually poor
performers when broadband performance is required. There are, nevertheless, distributed
networks that exhibit better broadband performance than others. For example, the λ/4 line
transformer of Figure 3.16(a) allows matching over a small frequency band, while two λ/4
lines in cascade (Figure 3.16(b)) provide a greater matching bandwidth and, of course, the
Fig. 3.16 Layout of line transformers: (a) one l/4; (b) two l/4 lines in cascade; (c) five l/4 lines cascaded
Impedance matching 113
Fig. 3.17 Constant-Q arcs on the Smith chart
five λ/4 cascaded transformers of Figure 3.16(c) would provide an even greater bandwidth.
In general cascading more quarter-wave transformers provides greater bandwidth.
However, you should be aware that λ/4 line transformers can only be used in the GHz
range because below these frequencies λ/4 lengths can be very long. For example at 30
MHz in air, λ/4 = 2.5 m. When quarter-wave length lines are required at low frequencies,
the lumped circuit equivalent of a quarter-wave line length (often a π circuit) is used.
In the next example, we will investigate the difference in bandwidth obtained between
using a single λ/4 transformer and a transformer produced by two λ/4 lines in cascade.
However, before commencing, it is worth investigating how the Smith chart can be used to
help design. In Section 3.2.3, we showed that any point on the Smith chart has a Q value asso-
ciated with it. The locus of impedances on the chart with the same Q is an arc that crosses the
open-circuit and short-circuit loads. Several Q arcs are shown in Figure 3.17. The Q arcs in
the Smith chart can be used to provide the limits within which the matching network should
remain in order to provide a larger operational bandwidth. Remembering that Q = fcentre/fband-
width, it follows that for a given centre frequency a wider bandwidth requires a lower Q.
Example 3.11
A source impedance Z S of (50 + j0) is to be matched to a load of (100 + j0) over
a frequency range of 600–1400 MHz. Match the source and load by using (a) one
114 Smith charts and scattering parameters
quarter-wave transformer and (b) two quarter-wave transformers. (c) Sketch a graph of
the reflection coefficient against frequency.
Given: Z S = (50 + j0) Ω, Z L = (100 + j0) Ω, bandwidth = 600–1400 MHz.
Required: (a) Matching network using one λ/4 transformer, (b) matching network using
two λ/4 transformers, (c) a sketch of their network reflection coefficient against bandwidth.
(a) Use one quarter-wave transformer as in the circuit of Figure 3.16(a). We start by using
Equation 2.57 which is Z in = Z 02/Z L which yields
Z0 = Zin ZL = (50 + j0)(100 + j0) = 70.71Ω
(b) Use two quarter-wave transformers as in Figure 3.18. Note in this case, I have called
the characteristic impedance of the first λ /4 line from the load, Z 0t1 and the character-
istic impedance of the second λ /4 line from the load, Z 0t2.
Fig. 3.18 Matching with two quarter-wave transformers
From Equation 2.57
Z A = (Z 0t1) 2/Z L
Again using Equation 2.57, and substituting for ZA
Z in = (Z 0t2) 2/ZA = (Z 0t2) 2/(Z 0t1)2/Z L
Sorting out, we get
Zin ( Z0 t2 )2 Z0 t1 ZL
= or =
ZL ( Z0 t1 )2 Z0 t2 Zin
Bearing in mind that Z in must match the Z s and substituting in values
Z0 t1 100 + j0
= = 1.414
Z0 t2 50 + j0
If I choose a value of 60 Ω for Z 0t2 then Z 0t1 = 1.414 × 60 = 84.85 Ω
(c) If the reflection coefficients of the network 1 and network 2 are plotted against
frequency, you will get Figure 3.19. I have also included Table 3.3 to give you some
idea of the difference between the networks. From both the table and the graph, you
should note that the two λ/4 network has lowered the reflection coefficient by approx-
imately 6 dB at 600 MHz and 1400 MHz. There has also been a reduction of about
12.8 dB at 800 MHz and 1200 MHz.
Impedance matching 115
Fig. 3.19 Matching with λ/4 line transformers: using one l/4 transformer; using two l/4 transformers
Table 3.3 Reflection coefficient (dB) against frequency (GHz)
(GHz) 0.6 0.8 1.0 1.2 1.4
One l/4 TX –13.83 –19.28 > –60 –19.28 –13.83
Two l/4 TXs –18.81 –32.09 –38.69 –32.09 –18.81
Note: The Smith chart graphics and calculations to obtain this graph are quite long; to
save work, I have used the PUFF software supplied with this book.
3.8.3 Impedance matching using a stub tuner
An alternative method to impedance matching by λ/4 transformers is shown in Figure 3.20
where a transmission line stub (stub matching) is tapped into the main transmission line
R 0 at a distance l 1 from the load Z L to provide a good match between Z L and a source
Fig. 3.20 Principle of stub matching
116 Smith charts and scattering parameters
generator, R 0. The process appears simple enough but the line length l 1 and the stub length
are critical and must be carefully controlled for a good match.
There are several methods of calculating these line lengths and we will give you two
methods at this stage. These are explained and illustrated in Tables 3.4 and 3.5 . Examples
of how these methods are used in calculating line lengths are given in Example 3.11 which
follows after the explanation.
Example 3.12 is an example of how stub matching is carried out on a Smith
Table 3.4 (Matching method 1)
Step 1
Z L is the load which is to be matched to the
transmission line, R 0 , and its generator, R 0 for
maximum power transfer and a good match. Line
l1 is the line length which will be used to trans-
form the load to the plane AA′.
Step 2
Convert the load Z L into its admittance form, i.e.
conductance G L and susceptance B L.
Step 3
Transform via line length l1, conductance G L and
susceptance B L to G L′ and B L′. Choose line
length l1 so that G L′ = 1/R 0, i.e. 1 on the conduc-
tance circle. Ignore the reactive component B L′
for the time being.
Step 4
Introduce a reactive conjugate component (B L′)*
to tune out B L′. The net result is that they cancel
out the effect of each other.
Step 5
Since the effects of the reactive elements have
been cancelled, the net result is a conductance
G L′.
Step 6
This figure results when G L′ is mathematically
transformed back to R 0 . We now have a good
matched system for maximum power transfer.
Impedance matching 117
Table 3.5 (Matching method 2)
Step 1
Z L is the load which is to be matched to the
transmission line, R 0 , and its generator, R 0 for
maximum power transfer and a good match. Line
l1 is the line length which will be used to trans-
form Z L to the plane AA′.
Step 2
At plane AA′, the load Z L has been transformed
via line length l1 to Z L′.
Step 3
Z L′ has been converted into G L′ and B L′. Choose
line length l1 so that G L′ = 1/R 0 . Ignore the reac-
tive component B L′ for the time being.
Step 4
A reactive conjugate component (B L′)* is now
introduced to tune out B L′. The total effect of
these reactive elements is that they ‘cancel out’
the effect of each other.
Step 5
Since the effects of the reactive elements have
cancelled each other out, the circuit is terminated
in a conductance G L′.
Step 6
This figure results when G L′ is mathematically
transformed back to R 0 . We now have a good
matched system for maximum power transfer.
Example 3.12
A series impedance load (40 – j80) W is to be matched to a generator source of 50 W via
a 50 W transmission line. Design a single stub matching system to provide this result.
Given: Z L = (40 – j80) Ω, Z g = 50 Ω, Z 0 = 50 Ω.
Required: A single stub matching system.
Solution. Two methods will be given. Method 1 is depicted in Figure 3.21 and Method 2
is shown in Figure 3.22.
118 Smith charts and scattering parameters
Fig. 3.21 Using matching method 1: A = (0.25 + j0.50) Ω , B = 0.077072λ, C = (1 + j1.8027) S, D =
0.183377λ, E = (∞ + j ∞) S, G = (0 – j1.8027) S, arc BD = 0.106 305l, arc FH = 0.080 606l
Method 1 based on Table 3.4. To simplify this example, I will use an ordinary Smith
chart in its admittance form.
1 Normalise the load impedance with respect to 50 Ω. This gives (40 – j80)/50 = 0.8 –
j1.6. (See step 1 of Table 3.4.)
2 Convert the normalised impedance into its admittance form by calculation. This gives
1/(0.8 – j1.6) = 0.25 + j0.5. (See step 2 of Table 3.4). The value 0.25 + j0.5 is shown as
point A in Figure 3.21. Draw a constant SWR circle, using the centre of the Smith chart
(point O) and a radius equal to OA.
3 Project the line from the centre of the Smith chart (point O), through point A to point B.
Read point B on the ‘wavelengths towards generator’ scale. This is 0.077λ.
4 Transform point A along the SWR circle in the ‘towards generator’ direction until
you obtain a conductance of 1 or unity. (Step 3 of Table 3.4.) This is shown as point
C in Figure 3.21. At point C, the admittance is (1 + j1.8) S. This tells you that the
Impedance matching 119
conductance of the circle is matched but that you must get rid of a susceptance of
5 Project the line OC to point D. Read this value on the wavelength towards generator
scale. It is 0.183λ. Subtract this value from the value at point B, i.e. (0.183 – 0.077) λ =
0.106λ. This is the length of the line l1.
6 The unwanted susceptance of +j1.8 obtained in step 4 above must be cancelled out. For
this, we must introduce a susceptance of –j1.8 to tune out the unwanted susceptance of
+j1.8. (See step 4 of Table 3.4.) Bear in mind that a short circuit (0 Ω) has a conduc-
tance of ∞ Ω. We start at point E and increase the stub length until we obtain a suscep-
tance of –j1.8. This is point G in Figure 3.21.
7 Project line OG to H and line OE to F. Measure the wavelength distance on the ‘towards
generator scale’ between the points F and H to obtain the length of the stub. In this case,
it is 0.331 λ – 0.250λ = 0.081λ. Hence the length of the short-circuited tuning stub to
produce a susceptance of – j1.8 to cancel out the unwanted +j1.8 is 0.081λ. (See steps
5 and 6 of Table 3.4.) The generator, line and load are now all matched to each other.
Results using method 1
• The position of the stub l 1 from the load is 0.106λ.
• The length of the short-circuited stub line is 0.081λ.
These results are obtained from the chart.
Just to convince you that the graphical method is correct, I have also calculated out all
the values to several decimal points by using a spreadsheet. These numbers are given in
the caption in Figure 3.21; however, you should be aware that in practice, accuracy beyond
two decimal points is seldom necessary. In other words, the accuracy of the Smith chart is
sufficient for practical design.
Method 2 based on Table 3.5. To simplify this example, we will use Smith chart type ZY-
01-N which was first introduced to you in Figure 3.7. This chart is used because it affords
easy conversion from impedance to admittance plots and vice-versa. If you have forgotten
how to use this chart refer back to Figure 3.7.
1 Normalising Z L with respect to 50 Ω gives (40 – j80)/50 = 0.8 – j1.6. Since this value
is impedance, the solid coordinate (impedance) lines are used to locate the point in
Figure 3.22. This is plotted as point A in the figure. (See step 1 of Table 3.5.) Extend the
line OA to the outer periphery (point B). Note the reading on the ‘wavelengths towards
generator scale’. This is 0.327λ and is denoted by point B.
2 Draw a constant SWR circle with its centre at the Smith chart centre (point O) and a
radius equal to the distance from the chart centre to point A.
3 Move clockwise (towards the generator) along the constant SWR circle until you come
to the unity conductance circle (admittance coordinates) at point C. The reason why
point C is chosen is because at the unity conductance circle, the conductance element is
matched to the system. (See step 2 of Table 3.5.) Draw a line from the Smith chart centre
(point O) through point C to the same ‘wavelengths scale’. This line cuts the circle
(point D) at 0.433λ. Subtracting the wavelength value of D from C (0.433–0.327)λ =
0.106λ. This distance is designated as ‘length l 1’ in Figure 3.20. It tells you that the stub
must be placed at a point 0.106λ from the load.
120 Smith charts and scattering parameters
Fig. 3.22 Using matching method 2
4 Return to point C and read its admittance value which is (1 + j1.8). The conductance
value is 1 and it is telling you that at this point the transformed conductive element is
already matched to Yo. (See step 3 of Table 3.5.) However, at point C, we also have a
susceptance of +j1.8. We want only a conductance element and do not want any suscep-
tance and will nullify the unwanted susceptance effect by tuning it out with –j1.8 from
the stub line. (See steps 4, 5 and 6 of Table 3.5.)
5 The stub line used in Figure 3.20 is a short-circuit line which means that Z L = 0 + j0
and that its admittance load is (∞ – j∞ ). This is shown as point E on the admittance scale
in Figure 3.22. The extended line OE also cuts the ‘wavelengths towards generator’
circle at 0.00λ (point F). We now have to move clockwise (towards the generator away
from the short-circuit load) until we generate a susceptance of –j1.8. This is shown as
point G in Figure 3.22. The extended line OH from the Smith chart centre through the
–j1.8 point is 0.081λ. Therefore the stub length is (0.081 – 0.00)λ = 0.081λ. The gener-
ator, line and load are now all matched to each other.
Results using method 2
• The position of the stub from the load is 0.106λ.
• The length of the short-circuited stub line is 0.081λ.
Impedance matching 121
Summing up. Within the limits of graphical accuracy, both methods produce the same
• The position of the stub from the load is 0.106λ.
• The length of the short-circuited stub line is 0.081λ.
Example 3.13
A transistor amplifier has an input resistance of 100 Ω shunted by a capacitance of 5 pF.
Find (a) the position of short-circuit stub on the line required to match the amplifier input
to a 50 Ω line at 1 GHz and (b) its length.
Given: Transistor input impedance = 100 Ω shunted by 5 pF, frequency = 1 GHz.
Required: Matching circuit to a 50 Ω line (a) determine length of short-circuit stub, (b)
determine its position from the load.
Fig. 3.23 Matching of transistor input impedance: A (0.5 + j1.57), B (0.165l), C (1 + j2.3), D (0.193l), E (–j2.3),
F (0.315l), G (0.25l)
122 Smith charts and scattering parameters
(a) Yo = 1/50 W = 20 mS
YL = GL + jwCL = 1/100 W + j2p × 1 GHz × 5 pF
= 10 mS + j31.4 mS
YL/Yo = 0.5 + j1.57
This is plotted in Figure 3.23 as point A. The radius through this cuts the ‘wavelengths
towards generator’ scale at 0.165λ (point B). The VSWR circle cuts the G/Yo = 1 circle
at 1 + j2.3 (point C), which corresponds to 0.193λ (point D) toward the generator. So the
stub connection point should be at (0.193λ – 0.165)λ = 0.028λ from the transistor input.
(b) The required normalised stub susceptance is –j2.3. This is plotted as point E. The
radius through this cuts the ‘wavelengths towards generator’ scale at 0.315λ (point F).
The short-circuit stub length should be 0.315λ – 0.25λ = 0.065λ.
Summing up
• The stub connection point is 0.028λ from the transistor input.
• The short-circuit stub length is 0.065λ at the connection point.
The program PUFF issued with this book has facilities for single stub matching.
3.8.4 Impedance matching using multiple stubs
In single stub matching (Figure 3.20) the distance from the load to the stub and the length
of the stub must be accurately controlled. In some situations, for example an antenna
mounted on a tower, you cannot easily control the distance from antenna to stub, therefore
we add one or more stubs to provide matching. One arrangement of double stub matching
is shown in Figure 3.24.
Fig. 3.24 Double stub matching network
Impedance matching 123
In double stub impedance matching, two stubs are shunted at fixed positions across the
main transmission line. Each stub may be either short-circuited or open-circuited. Its lengths
are given by l1 and l2 respectively. The distance, d2, between the stubs is usually fixed at 1/8,
3/8, or 5/8 of a wavelength, whereas the position of the nearest stub from the load, d1, is
determined by the distance from the load. Explanation is best given by an example.
Example 3.14
A system similar to that shown in Figure 3.24 has a load Z L = (50 + j50) Ω which is to be
matched to a transmission line and source system with a characteristic impedance of 50 Ω.
The distance, d1, between the load and the first stub is 0.2λ at the operating frequency. The
distance, d2, between the two stubs is 0.125λ at the operating frequency. Use a Smith chart
to estimate the lengths of l1 and l 2 of the stubs.
Solution. This problem will be solved by normalising all values in the question by 50 W.
Hence (50 + j50) Ω and 50 Ω will become (1 + j1) n and 1 n respectively; the ‘n’ is used
to denote normalised values. Values will then be plotted on the immittance Smith chart of
Figure 3.25 and the chart result will be re-normalised to produce the correct answer.
Fig. 3.25 Double stub matching: A = (1 + j1) n, Arc BC = 0.2l, D = (0.759 – j0.838) n, E = (1.653 – j0.224) n, Arc
FG = 0.125l, H = (0.781 – j0.421) n, O = (1.008 – j0.001) n
124 Smith charts and scattering parameters
Before starting, you should realise that distance d1 and d 2 are out of your control; d1
is fixed by the system structure, d2 is fixed after you have selected your double tuning
stub device which is 0.125λ in this case. You can only vary the susceptance of the stubs.
Therefore, you vary stub 1 to a convenient point E in Figure 3.25 so that when that value
is moved distance d 2 (0.125λ), the new point will be on the unit conductance circle
where you can use stub 2 to vary the susceptance value until it reaches the impedance
(1 + j0) n.
1 The load is plotted at (1 + j1) using the impedance coordinates. This is shown as A in
Figure 3.25. Project OA until it cuts the wavelengths to generator scale at point B which
reads 0.161λ. Move along this scale for 0.2λ. This is denoted by the arc BC. Point C is
0.361λ. Draw a line from C to O.
2 An arc of a constant |Γ| circle with radius OA is drawn clockwise from the load point A
to D which cuts the line OC. Point D is the value of the transferred load through 0.2λ.
From point D, the first susceptance stub moves the transferred load to E. Point E was
found experimentally by altering the length of stub 1. Extend line OE to F on the periph-
ery. Move the arc along a periphery distance of 0.125λ to G. Arc FG represents the
0.125λ distance between the two stubs. Join O to G.
3 An arc of a constant |Γ| circle with radius OE is drawn clockwise until it cuts the line
OG at H. Point H is the transferred load from E after moving through 0.125λ. Ideally,
point H should be on the unit conductance circle which means that the resistive element
is matched and that stub 2 can now be used to move point H to point O which is the
desired point (1 + j0).
4 The normalised values of all the points are given in the annotation for Figure 3.25. From
these values, we can now calculate the susceptance which each stub must provide.
As before, for stub 1, we need (–j0.1 at E) – (–j0.65 at D) = +j0.55. The length l1 of
the stub is found by plotting its load impedance at the point S and following round the
‘wavelengths towards generator’ scale to the point where the 0.55 susceptance circle
cuts the perimeter, at about 0.17λ. (The calculated value is 0.167λ.)
For stub 2, we need (j0 at O) – (–j0.55 at H) = +j0.55. The length l2 of this stub is found
as for stub 1 yielding, again, a value of about 0.17λ. (The calculated value is 0.172λ.)
Finally when you are faced with trial and error methods such as selecting point E in the
above example, it is much easier if you have a dynamic impedance matching computer
program. One such program is the Motorola Impedance Matching Program often called
MIMP. This program allows you to alter values and see results instantaneously. MIMP has
been written by Dan Moline and at the time of writing this book, Motorola generally issues
a copy of it free of charge to bona fide engineers.
The program PUFF issued with this book has facilities for checking multiple stub
matching. In fact, Example 3.14 is repeated electronically in Secton 4.13.4.
Scattering parameters (s-parameters) 125
3.9 Summary of Smith charts
The Smith chart is a phasor diagram of the reflection coefficient, Γ, on which constant-r
and constant-x circles are drawn, where r and x are the normalised values of the series
resistive and reactive parts of the load impedance. The horizontal and vertical axes of the
chart are the real and imaginary axes of the reflection coefficient, but they are not labelled
as such.
Any circle centred on the Smith chart centre is a constant-|Γ| circle and a constant
VSWR circle, too.
A load impedance, or the impedance looking into a line towards the load, is represented
by the intersection of an r circle and an x circle.
If a series lumped reactance is added to the load, the r circle through the load imped-
ance point is followed and the added normalised reactance is represented by the increase
or decrease in the corresponding value of the x circle crossed.
If a series line is added at any point then a constant-|Γ| circle is followed, clockwise
‘towards the generator’ through an angle on the chart corresponding to its length in wave-
The admittance chart is a version of the impedance Smith chart rotated through 180°.
The r and x circles become g and b circles and their intersections represent admittances.
The immittance chart is a combination of both the impedance chart and the admittance
If a lumped susceptance is shunted across the load, the g circle through the load admit-
tance point is followed and the added normalised susceptance is represented by the
increase or decrease in the corresponding value of the b circle crossed.
If a short-circuited shunt line (stub) is shunted across the line at any point, then the
g circle through the point is followed, through a susceptance change corresponding to
the stub length in wavelengths. For lengths less than a quarter wavelength, the short-
circuit stub appears capacitive and rotation is clockwise round the g circle. For lengths
up to three-quarters of a wavelength, the stub appears inductive and rotation is
Open-circuit stubs have the opposite susceptance, with rotation in the opposite direc-
tion around the g circle.
Double stubs are useful when loads are variable. Usually the stub spacing is kept fixed,
but the stub lengths are adjustable to achieve matching.
3.10 Scattering parameters (s-parameters)
3.10.1 Introduction
Voltages and currents are difficult to measure in microwave structures because they are
distributed values and vary with their position in microwave structures. In fact, the widely
spread current in a waveguide is virtually impossible to measure directly.
Waves are more easily measured in microwave networks. One method of describing the
behaviour of a two port network is in terms of incident and reflected waves. This is shown
126 Smith charts and scattering parameters
in Figure 3.26. This method is known as scattering parameters or usually denoted as s-
parameters. The s-parameter approach avoids many voltage and current problems particu-
larly in the measurement of transistors where short- and open-circuit terminations can
cause transistor instability and in some cases failure. In many cases, measurement is
carried out in s-parameters using an automated computer corrected network analyser. This
method is fast and accurate and the results obtained are then mathematically converted into
the requisite z, h, y and ABCD parameters. The converted information can be trusted
because the accuracy of the original measured data is high.
v1 v2
Fig. 3.26 Two port scattering network with source and load
3.10.2 Overall view of scattering parameters
Figure 3.26 represents a scattering parameter two-port network driven from a source with
impedance Z 0, and driving a load of impedance Z L. In the figure, a1 and a2 represent inci-
dent voltage waves; and b1 and b2 represent reflected voltage waves. These four waves are
related by the following equations where s11, s12, s21 and s22 are the ‘scattering’, or s-para-
b1 = s11a1 + s12a2 (3.15)
b2 = s21a1 + s12a2 (3.16)
Equations 3.15 and 3.16 are also written in matrix form as
⎡ b1 ⎤ ⎡ s11 s12 ⎤ ⎡ a1 ⎤
⎢b ⎥ = ⎢ s ⎥⎢ ⎥ (3.17)
⎣ 2 ⎦ ⎣ 21 s22 ⎦ ⎣a2 ⎦
When scattering parameters are to be measured, the applied source is a generator
which has the source impedance Z 0 equal to the system characteristic impedance and
this generator is connected to the system by a line of characteristic impedance Z 0, as
in Figure 3.27. The load is purely resistive, with impedance Z 0, and connected by a
line of impedance Z 0. So the source seen by the two port’s input is Z 0, and the load seen
by its output is also Z 0. In this case, there is no power reflected from the load, so a 2 =
From Equation 3.15, if a 2 = 0, then b1 = s11a1. So s11 can be defined as
Scattering parameters (s-parameters) 127
Fig. 3.27 Measurement of s-parameters
s11 = (3.18)
a1 a2 = 0
and s11 is the reflection coefficient at the input port (port 1) of the network.
From Equation 3.16, with a2 = 0, b2 = s21a1. So s21 can be defined as
s21 = (3.19)
a1 a2 = 0
Since this is the ratio of the output wave voltage to the incident wave voltage, |s21|2 is
the insertion power gain of the network.
The other two s-parameters, s12 and s22, are found by inter-changing the electrical
connections to the two ports, so that port 2 is driven from the source, and port 1 is loaded
by Z 0. Now a1 = 0, and
s12 = (3.20)
a2 a1 = 0
s22 = (3.21)
a2 a1 = 0
|s12| 2 is the reverse insertion power gain, and s22 is the output port reflection coefficient.
It should be clear, but two points are worth stressing.
• The scattering parameters are defined, and measured, relative to a fixed system imped-
ance Z 0. In practice, the chosen value is nearly always 50 Ω resistive.
• The scattering parameters are complex quantities, representing ratios of phasors at a
defined plane at each port.
It is now necessary to define symbols a1, a 2, b1 and b2 in terms of voltages and currents.
3.10.3 Incident and reflected waves in scattering parameters
It will ease understanding if the explanation of incident and reflected waves is taken in two
parts. We will begin with the ‘ideal’ situation where there is complete match within the
128 Smith charts and scattering parameters
Fig. 3.28 ‘Ideal’ two-port network
system, i.e. where the source generator, connecting lines, two-port network and load
impedance all have characteristic impedances of Z 0. Then, we will proceed with the real
life practical situation where the two-port network does not match the measuring system.
In Figure 3.28, we consider the ideal situation where a generator (vs ) with an internal
impedance Z 0 feeds a transmission line whose impedance is Z 0 which in turn feeds a two-
port network whose input and output impedances are Z 0. The output from the two-port
network is then fed through another transmission line of Z 0 to a termination load where Z L
= Z 0. In other words because everything in the system matches, we have no reflected
power, therefore the incident voltage (vi ) represents the input voltage to the network and
the incident current (i i ) represents the current flowing into the network.
Now consider the practical case (Figure 3.29) where the-two port network is not
matched to the same system. Due to the mismatch, we will now have reflected power. This
reflected power will produce a reflected voltage vr and a reflected current i r. If we now
defined v1 as being the sum of the incident and reflected voltages and i1 as the difference
of the incident and reflected currents, we have
v 1 = v i + vr (3.22)
i1 = ii – ir (3.23)
By the definition of impedances, we have
vi vr
Z0 = — = — (3.24)
ii ir
Fig. 3.29 Practical two-port network
Scattering parameters (s-parameters) 129
Equation 3.24 can be re-written to yield
ii = (3.25)
ir = (3.25a)
Substituting Equations 3.25 and 3.25a into Equation 3.23 gives
vi v
il = − r
Z0 Z0
Z 0 i1 = v i – v r (3.26)
Adding Equations 3.22 and 3.26 yields
2v i = v1 + Z 0 i1
v i = — [v1 + Z 0 i1] (3.27)
Subtracting Equation 3.26 from Equation 3.22 yields
2v r = v1 – Z 0i1
v r = — [v1 – Z 0 i 1] (3.28)
The incident wave v i is defined as the square root of the incident power. Therefore,
vi2 v
a1 = = i (3.29)
Z0 Z0
Using Equation 3.27 to substitute for v i in Equation 3.29 and dividing by Z0 , we get
vi 1⎡ v ⎤
a1 = = ⎢ l + Z0 i1 ⎥ (3.30)
Z0 2 ⎢ Z0
⎣ ⎥
Similarly, the reflected wave v r is defined as the square root of the reflected power.
vr v
b1 = = r (3.31)
Z0 Z0
130 Smith charts and scattering parameters
Using Equation 3.28 to substitute for v r in Equation 3.31 and dividing by Z0 , we
vr 1⎡ v ⎤
b1 = = ⎢ 1 − Z0 i1 ⎥ (3.32)
Z0 2 ⎢ Z0
⎣ ⎥
Again using similar arguments, it can be shown that
1 ⎡ v2 ⎤
a2 = ⎢ + Z0 i2 ⎥ (3.33)
2 ⎢ Z0
⎣ ⎥
1 ⎡ v2 ⎤
b2 = ⎢ − Z0 i2 ⎥ (3.34)
2 ⎣ Z0 ⎥
Thus, we have now evaluated a1, a 2, b1 and b 2 in terms of incident voltages and currents
and the characteristic impedance of the measuring system.
3.10.4 S-parameters in terms of impedances
From Equations 3.18, 3.27, 3.28, 3.30 and 3.32 we write
1 i1 ⎡ v1 ⎤
[v1 − Z0 i1 ] 2 ⎢ i − Z0 ⎥
= ⎣1 ⎦
s11 = 1 = 2
a1 a = 0 1 [v + Z i ] i1 ⎡ v1 ⎤
1 01 ⎢ + Z0 ⎥
2 ⎣ i1 ⎦
and since v1/i1 = input impedance at port 1 of the two-port network which we will call Z1,
we have
b1 Z1 − Z0
s11 = = (3.35)
a1 a2 = 0
Z1 + Z0
Note that Z 1 is really the load for the signal generator in this case; in some cases, it is
common to write Z L instead of Z 1 which makes Equation 3.35 identical to the transmis-
sion line reflection coefficient (Γ1) so that we have
b1 Z L − Z0
s11 = = = Γ1 (3.36)
a1 a2 = 0
Z L + Z0
Equation 3.36 also confirms what we have shown in Figure 3.3 that when the input
impedance of the two-port network = Z 0, the reflection coefficient is zero and that there is
no reflected wave.
Using the same process as above, it is possible to show that
Applied examples of s-parameters in two port networks 131
b2 Z2 − Z0
s22 = = = Γ2 (3.37)
a2 a1 = 0
Z2 + Z0
where Z 2 is the driving impedance of output port (port 2) of the two-port network.
3.10.5 Conversion between s-parameters and y-parameters
Most radio frequency measurements are now carried out using automated computer
controlled network analysers with error correction. The measurements are then converted
from s-parameters to other types of parameters such as transmission parameters (ABCD),
hybrid h-parameters, and admittance y-parameters. We provide you with Table 3.6 to
enable conversion between s- and y-parameters.
Table 3.6 Conversion between scattering s-parameters and y-parameters
(1 − y11 )(1 − y22 ) + y12 y21 ⎛ (1 + s22 )(1 − s11 ) + s12 s21 ⎞ 1
s11 = † y11 = ⎜ ⎟ *
(1 + y11 )(1 + y22 ) − y12 y21 ⎝ (1 + s11 )(1 + s22 ) − s12 s21 ⎠ Z0
−2 y12 ⎛ −2 s12 ⎞ 1
s12 = † y12 = ⎜ *
(1 + y11 )(1 + y22 ) − y12 y21 ⎟
⎝ (1 + s11 )(1 + s22 ) − s12 s21 ⎠ Z0
−2 y21 ⎛ −2 s21 ⎞ 1
s21 = † y21 = ⎜ ⎟ *
(1 + y11 )(1 + y22 ) − y12 y21 ⎝ (1 + s11 )(1 + s22 ) − s12 s21 ⎠ Z0
(1 + y11 )(1 − y22 ) + y12 y21 ⎛ (1 + s11 )(1 − s22 ) + s12 s21 ⎞ 1
s22 = † y22 = ⎜ ⎟ *
(1 + y11 )(1 + y22 ) − y12 y21 ⎝ (1 + s11 )(1 + s22 ) − s12 s21 ⎠ Z0
* where Z 0 = the characteristic impedance of the transmission lines used in the scattering parameter system, usually 50 W.
† Notice that when you are converting from admittance (Y) to s-parameters, (left hand column of Table 3.6), each
individual Y parameter must first be multiplied by Z 0 before being substituted into the equations.
3.11 Applied examples of s-parameters in two-port networks
Most people who have not encountered s-parameters earlier tend to find s-parameter topics
a bit abstract because they have only been used to tangible voltages, currents and lumped
circuitry. In order to encourage familiarity with this topic, we offer some examples.
3.11.1 Use of s-parameters for series elements
Example 3.15
Calculate the s-parameters for the two-port network shown in Figure 3.30 for the case
where Z 0 = 50 Ω.
Given: Network of Figure 3.30 with Z 0 = Z L = 50 Ω.
Required: s-parameters.
132 Smith charts and scattering parameters
Fig. 3.30 Resistive network
s11: Terminate the output in Z 0 and determine Γ1 at the input. See Figure 3.31(a).
By inspection:
Z 1 = 50 W + 50 Ω = 100 Ω
From Equation 3.36
Z1 − Z0 100 − 50 1
s11 = Γ1 = = =
Z1 + Z0 100 + 50 3
s11 = 0.333 ∠ 0° or –9.551 dB ∠ 0°
s22: Terminate the input with Z 0 and determine Γ2 at the output. See Figure 3.31(b).
By inspection:
Z 2 = 50 Ω + 50 Ω = 100 Ω
From Equation 3.37
Z2 − Z0 100 − 50 1
s22 = Γ2 = = =
Z2 + Z0 100 + 50 3
s22 = 0.333 ∠ 0° or –9.551 dB ∠ 0°
Note: s11 and s 22 are identical. This is what you would expect because the network is
Fig. 3.31(a) Calculating s 11 Fig. 3.31(b) Calculating s 22
Applied examples of s-parameters in two port networks 133
2v1+ v2 v1 2v2+
Fig. 3.31(c) Calculating s 21 Fig. 3.31(d) Calculating s 12
s21: Drive port 1 with the 50 Ω generator and open-circuit voltage of 2V1+. The multi-
plication factor of 2 is used for mathematical convenience as it ensures that the voltage
incident on the matched load will be V1+. The superscript + sign following the voltage is
meant to indicate that the voltage is incident on a particular port. Calculate voltage V2. See
Figure 3.31(c).
V0 V2
s21 = =
V1+ V1+
By inspection:
V2 + = (2V1+ ) = (2V1+ ) = V1+
50 + (50 + 50) 150 3
V0 V2 2V + 2
s21 = = = 1+ =
V1+ V1+ 3V1 3
s 21 = 0.667 ∠ 0° or –3.517 dB ∠ 0°
s12: See Figure 3.31(d). By inspection:
V1 = (2V2+ ) = V2+
50 + (50 + 50) 3
V1 2V + 2
s12 = = 2 =
V2+ 3V2+ 3
s12 = 0.667∠0° or − 3.517 dB ∠0°
Note: s12 and s 21 are the same because the network is symmetrical.
Summing up
For s-parameters:
s11 = 0.333 ∠ 0° s 12 = 0.667 ∠ 0°
s 21 = 0.667 ∠ 0° s 22 = 0.333 ∠ 0°
134 Smith charts and scattering parameters
or in matrix notation
[s] = [ 0.333 ∠ 0°
0.667 ∠ 0°
0.667 ∠ 0°
0.333 ∠ 0° ]
Later on, you will see that this symmetry in a network often leads to considerable simpli-
fication in manipulating networks.
3.11.2 Use of s-parameters for shunt elements
Example 3.16
Calculate the s-parameters for the two port network shown in Figure 3.32 for the case
where Z 0 = 50 Ω.
Given: Network of Figure 3.32 with Z 0 = Z L = 50 Ω.
Required: s-parameters.
Fig. 3.32 Resistive network
s11: Terminate the output in Z 0 and determine Γ1 at the input. See Figure 3.33(a). By
(50 × 50)Ω
Z1 = = 25 Ω
(50 + 50)Ω
From Equation 3.36
Z1 − Z0 25 − 50 −1
S11 = Γ1 = = =
Z1 + Z0 25 + 50 3
s11 = 0.333 ∠ 180° or –9.551 dB ∠ 180°
s22: Terminate the input with Z 0 and determine Γ2 at the output. See Figure 3.33(b). By
(50 × 50) Ω
Z 2 = ————— = 25 Ω
(50 + 50) Ω
Applied examples of s-parameters in two port networks 135
Fig. 3.33(a) Calculating s11 Fig. 3.33 (b) Calculating s 22
From Equation 3.37
Z1 – Z 0 25 – 50 –1
s22 = r 2 = ———— = ———— = ——
Z1 + Z 0 25 + 50 3
s22 = 0.333 ∠ 180° or –9.551 dB ∠ 180°
Note: s11 and s 22 are identical. This is what you would expect because the network is
s21: Drive port 1 with the 50 Ω generator and open-circuit voltage of 2V1+. The multi-
plication factor of 2 is used for mathematical convenience as it ensures that the voltage
incident on the matched load will be V1+. The superscript + sign following the voltage is
meant to indicate that the voltage is incident on a particular port. Calculate voltage V2. See
Figure 3.33(c).
V0 V2
s21 = —— = ——
V1+ V1+
By inspection and bearing in mind that two 50 W resistors in parallel = 25 Ω
V2 = (2V1+ ) = (2V1+ ) = V1+
50 + 25 75 3
2V1+ 2
s21 = =
3V1+ 3
s 21 = 0.667 ∠ 0° or –3.517 dB ∠ 0°
s12: See Figure 3.33(d). By inspection, and bearing in mind that two 50 Ω resistors in
parallel = 25 Ω
V1 = (2V2+ ) = V2+
50 + 25 3
136 Smith charts and scattering parameters
2v1+ v2 v1 2v2+
Fig. 3.33(c) Calculating s 21 Fig. 3.33(d) Calculating s12
V1 2V + 2
s12 = = 2 =
V2+ 3V2+ 3
s 12 = 0.667∠ 0° or –3.517 dB ∠ 0°
Note: s12 and s22 are the same because the network is symmetrical.
Summing up
For s-parameters:
s11 = 0.333 ∠ 180° s12 = 0.667 ∠ 0°
s 21 = 0.667 ∠ 0° s 22 = 0.333 ∠ 180°
3.11.3 Use of s-parameters for series and shunt elements
Example 3.17
(a) Calculate the s-parameters for the two-port network shown in Figure 3.34 for the case
where Z 0 = 50 Ω.
(b) Find the return loss at the input with Z L = Z 0.
(c) Determine the insertion loss for the network when the generator and the termination
are both 50 Ω.
Given: Network of Figure 3.34 with Z 0 = Z L.
Required: (a) s-parameters, (b) return loss, (c) insertion loss.
Fig. 3.34 Complex network
Applied examples of s-parameters in two port networks 137
(a) s11: Terminate the output in Z 0 and determine ρ at the input. See Figure 3.35(a).
By inspection the combined value of the 30 Ω and 50 Ω resistors is:
(30 × 50)/(30 + 50) = 18.75 Ω
Z 1 = 18.75 Ω + j20 Ω
From Equation 3.36
Z1 − Z0 (18.75 + j20) − 50
s11 = ρ1 = =
Z1 + Z0 (18.75 + j20) + 50
−31.25 + j20 37.10∠147.38°
= =
68.75 + j20 71.60∠16.22°
s11 = 0.518 ∠ 131.16° or –5.713 dB ∠ 131.16°
s22: Terminate the input with Z 0 and determine ρ at the output. See Figure 3.35(b). By
(50 + j20)(30) 1500 + j600
Z2 = =
(50 + j20 + 30) 80 + j20
= = 19.591 ∠ 7.765° or 19.412 + j2.647
From Equation 3.37
Z2 − Z0 (19.412 + j2.647) − 50 −30.588 + j2.647
s22 = ρ2 = = =
Z2 + Z0 (19.412 + j2.647) + 50 69.412 + j2.647
30.702 ∠ 175.054°
= = 0.442 ∠ 172.870°
69.462 ∠ 2.184°
s22 = 0.442 ∠ 172.87° or –7.092 dB ∠ 172.87°
Fig. 3.35(a) Calculating s11 Fig. 3.35(b) Calculating s22
138 Smith charts and scattering parameters
2v1+ v2
Fig. 3.35(c) Calculating s21
s21: Drive port 1 with the 50 Ω generator and open-circuit voltage of 2V1+. The multi-
plication factor of 2 is used for mathematical convenience as it ensures that the voltage
incident on the matched load will be V1+. The superscript + sign following the voltage is
meant to indicate that the voltage is incident on a particular port. Measure voltage across
V2. See Figure 3.35(c).
V0 V2
s21 = =
V1+ V1+
By inspection:
30 // 50 18.75
V2 = (2V1+ ) = (2V1+ )
30 // 50 + (50 + j20) 68.75 + j20
(18.75)(2V1+ )
= = (0.524 ∠ − 16.22°)V1+
71.60 ∠ 16.22°
(0.524 ∠ –16.22°)V1+
s21 = —————————
s21 = 0.524 ∠ –16.22° or – 5.613 dB ∠ –16.22°
s12: The Thévenin equivalent of the generator to the right of the plane ‘c c’ is obtained
first. See Figures 3.35(d) and 3.35(e).
v2 2v2+
Fig. 3.35(d) Circuit before applying Thévenin’s theorem
Applied examples of s-parameters in two port networks 139
v2 0.75v2+
Fig. 3.35(e) Circuit after applying Thévenin’s theorem
The Thévenin equivalent generator voltage is
(2V2+ ) = 0.75V2+
30 + 50
The Thévenin equivalent internal resistance is
30 × 50
= 18.75 Ω
30 + 50
By inspection:
50 37.5
V1 = (0.75V2+ ) = V2+
(18.75 + 50 + j20) 71.60 ∠ 16.22°
= 0.524 ∠ − 16.22°
V1 (0.524 ∠ − 16.22°)V2+
s12 = = = 0.524 ∠ − 16.22°
V2+ V2+
s12 = 0.524 ∠ –16.22° or –5.613 dB ∠ –16.22°
To sum up for s-parameters
(a) s11 = 0.518 ∠ 131.16° s12 = 0.524 ∠ –16.22°
s21 = 0.524 ∠ –16.22° s22 = 0.442 ∠ 172.87°
(b) From part (a), Γ ∠ θ = 0.518 ∠ 131.16°. Therefore
return loss (dB) = –20 log10 |0.518| = –20 × (–0.286) = 5.713 dB
(c) The forward power gain of the network will be |s21| 2.
|s21| 2 = (0.524)2 = 0.275
This represents a loss of –10 log10 0.275 = 5.61 dB.
140 Smith charts and scattering parameters
3.11.4 Use of s-parameters for active elements
Example 3.18
A 50 Ω microwave integrated circuit (MIC) amplifier has the following s-parameters:
s11 = 0.12 ∠ –10° s12 = 0.002 ∠ –78°
s21 = 9.8 ∠ 160° s22 = 0.01 ∠ –15°
Calculate: (a) input VSWR, (b) return loss, (c) forward insertion power gain and (d)
reverse insertion power loss.
Given: s11 = 0.12 ∠ –10° s12 = 0.002 ∠ –78°
s21 = 9.8 ∠ 160° s22 = 0.01 ∠ –15°
Required: (a) Input VSWR, (b) return loss, (c) forward insertion power gain, (d) reverse
insertion power loss.
(a) From Equation 2.38
1 + Γ 1 + s11 1 + 0.12
VSWR = = =
1 − Γ 1 − s11 1 − 0.12
= 1.27
(b) Return loss (dB) = –20 log10 0.12 = 18.42 dB
(c) Forward insertion gain = |s21| 2 = (9.8)2 = 96.04 or
Forward insertion gain = 10 log10 (9.8)2 dB = 19.83 dB
(d) Reverse insertion gain = |s12| 2 = (0.002)2 = 4 × 10–6 or
Reverse insertion gain = 10 log10 (0.002)2 dB = –53.98 dB
The amplifier is virtually unilateral with (53.98 – 19.83) or 34.15 dB of output to input
3.12 Summary of scattering parameters
Section 3.10 has been devoted to the understanding of two-port scattering networks.
Section 3.11 has been devoted to the use of two-port scattering networks. You should now
be able to manipulate two-port networks skilfully and have the ability to change two-port
parameters given in one parameter set to another parameter set.
An excellent understanding of scattering parameters is vitally important in microwave
engineering because most data given by manufacturers are in terms of these parameters. In
fact, you will find it difficult to proceed without a knowledge of s-parameters. This is the
reason why we have provided you with several examples of s-parameter applications. The
examples will be repeated using a software program called PUFF which has been supplied
to you with this book. The purpose of these software exercises is to reinforce the concepts
you have learnt and also to convince you that what we have been doing is correct.
Do not be unduly perturbed if you initially found s-parameters difficult to understand.
Summary of scattering parameters 141
Understanding of s-parameters is slightly more difficult because they deal with waves,
which is a very different concept from the steady-state voltages and currents which we
have used in the past.
Finally, the information you have acquired is very important because most information
in radio and microwave engineering is given in terms of scattering and admittance para-
meters. We have devoted particular attention to s-parameters because, later on, when you
start analysing microwave components, filters, amplifiers, oscillators and measurements,
you will be confronted with scattering parameters again and again. This is the reason why
we have provided you with many examples on the use of scattering parameters.
PUFF software
4.1 Introduction
4.1.1 Aims
The aims of this chapter are threefold: (i) to help you install software program PUFF 2.1
on your computer; (ii) to use PUFF (software program supplied with this book) to verify
the examples which you had worked with in the earlier chapters; and (iii) to give you confi-
dence in using the software.
The software program that we introduce here is known as PUFF Version 2.1 It is a
radio and microwave design and layout computer program developed by the California
Institute of Technology (CalTech) Pasadena, USA. The program has been licensed to the
publishers for use with this book. The conditions of the licence are that you use it for
private study and experimental designs, and that copies of the program must not be made
available to the general public and on the Internet, World Wide Web, etc. For ease of
understanding, in all the discussions that follow, we will refer to PUFF Version 2.1 as
Note: When installed on a computer with a colour monitor, PUFF displays a default
colour screen. In the descriptions that follow, it is not feasible to display colour pictures,
and every effort has been made to annotate the graphs. However, if you still have difficulty,
run PUFF on your computer using the PUFF examples supplied on the PUFF disk. In fact,
set up each example and check the results for yourself. It will give you confidence in using
To install your software and to be able to manipulate the program, read Sections 4.2 to 4.10.
4.1.2 Objectives
The objectives of Part 4 are to teach you how to use PUFF for the following topics:
• amplifier designs
• calculating s-parameters for circuits
• calculating s-parameters for distributed components
• calculating s-parameters for lumped components
• circuit layout
Installation of PUFF 143
• coupled circuits
• filter frequency response and matching
• line transformer matching and frequency response
• stub line matching and frequency response
• transistor matching and frequency response
4.2 CalTech’s PUFF Version 2.1
CalTech’s PUFF is a computer program that allows you to design a circuit using pre-
selected lumped components (R, L, C and transistors) and/or distributed components
(microstrip lines and striplines variants). The components may be arranged to form
circuits. The layout of the circuit can be computer magnified, printed to provide a template
for printed circuit layout and construction. The program provides facilities for calculating
the scattering parameters (s-parameters) of the designed circuit layout with respect to
frequency. The results can be read directly, plotted in Cartesian coordinates (X-Y plots) and
in Smith chart impedance or admittance form. The Smith chart can also be used for match-
ing purposes and for oscillator design.
4.3 Installation of PUFF
4.3.1 Introduction
The following notes are written with the express objective of explaining how to install
PUFF on a personal computer. CalTech’s PUFF is supplied on a compact disk. The
program is designed to work on all PC, IBM PC/XT/AT and compatible computers. The
minimum hardware requirements for the computer are a CD drive, a hard disk, 640K RAM
and a 80286 or higher processor. If you do not have a CD drive, then ask a friend to copy
all the PUFF folder files to a high density 3.5″ disk for transfer to your hard disk. The
processor should have its matching coprocessor; if no coprocessor is present PUFF will
run in its floating point mode and operate less rapidly. MS-DOS(R) versions 3.0 or higher
should be used.
All printed outputs are directed to the parallel port LPT1. Printing graphic screens
requires an Epson or a Laser compatible printer. PUFF provides six printer drivers for
‘screen dumps’.
4.3.2 Installation
In the instructions which follow, I will use these conventions.
• Bold letters for what you have to type.
• Plus signs to indicate that two or more keys must be pressed together. For example, ctrl
+ f means that you keep the ctrl key pressed down while you type f. Similarly alt + shift
+ g means that you keep the alt and shift keys pressed down while you type g.
• The output from the screen is shown in italics.
• To simplify explanation I shall assume that your hard disk is c: and that you are installing
from drive a:. If your hard disk drive is not c: then substitute the drive letter of your hard
144 PUFF software
disk drive whenever you see c:. If you are not installing from drive a:, then substitute the
drive letter of your drive whenever you see a:.
4.3.3 Installing PUFF
This program is relatively small. It is not compressed. It can be run directly from the
supplied disk initially. We do not recommend that you do this because if you damage your
original disk, then you will not have a ‘back-up’ program. We recommend that you run
PUFF from a directory on your hard disk. You install it simply by making a directory
called PUFF on your hard disk and copy all programs including the sub-directory
VGA_eps from your diskette to your PUFF directory. Details are given in the following
Windows installation
1 Insert your PUFF disk in drive a:.
2 Start Windows in the usual manner and ensure that the Program Manager is
3 From Program Manager, click on the main menu and select File Manager.
4 From the File menu, select create a directory. In the Create Directory box type c:puff.
Press the RETURN key.
5 Keep in the File Manager window. Click on disk a on the menu bars to display all the
files on disk a:.
6 Select all items on disk a: and copy them all to the new directory PUFF which you have
just created. You select copy from the File menu in File Manager. Press the RETURN
key. When presented with the Copy box, type c:puff. Press the RETURN key.
7 All PUFF folder files from disk a: should be copied in your directory c:PUFF.
8 Installation is now complete.
DOS installation
1 Insert the PUFF disk in drive a:.
2 Create a directory PUFF using the DOS MKDIR command. At the DOS prompt C:\>,
type MKDIR PUFF and press the RETURN key.
3 To copy all the files from the PUFF disk in drive a: into the directory C :\PUFF. type
COPY A: *.* C:\PUFF\*.* and press the RETURN key.
4 Check that all the files including subdirectory VGA_eps have been copied into your
PUFF directory by typing DIR C:\PUFF\*.* and press the RETURN key.
4.4 Running PUFF
4.4.1 Running under Windows
PUFF will run in a DOS window if accessed via File Manager in Windows 3.1 or My
Computer or Windows Explorer or File Manager in Windows 95, 98 and ME. The ability
of Windows to support PUFF is dependent on computer memory. If you attempt unsuc-
cessfully the techniques below, then close down Windows and follow the DOS instructions.
Running PUFF 145
Running under Windows 3.1
Double click on the c:PUFF directory in File Manager. Double click on PUFF.exe. PUFF
should load into a DOS window.
An alternative to this is to double click on the MSDOS icon and follow the DOS
instructions in Section 4.4.2.
Running under Windows 95/98/2000
My Computer
1 Double click on the My Computer icon on the Windows Desktop.
2 Double click on [c:].
3 Double click on the PUFF folder icon.
4 Double click on the PUFF program icon. PUFF should now run in a DOS window.
Windows Explorer
1 Select Programs from the Start menu.
2 Click on Windows Explorer.
3 Double click on the c:PUFF folder.
4 Double click on the PUFF application. PUFF should load into a DOS window.
An alternative to either of these methods is to select Programs from the Start menu, then
choose the MS-DOS prompt and follow the DOS instructions.
4.4.2 DOS instructions
To run PUFF, you must first call up the directory PUFF and then type PUFF to start the
program. Follow this procedure.
1 Type CD C:\PUFF and press the RETURN key. On the screen you should see
2 Type PUFF and press the RETURN key.
The program will start with an information text screen giving details of the program and
your computer. This is essentially an information sheet. At the last line, it will say Press
ESC to leave the program or any other key to continue. Type any key.
If your computer has a colour monitor then you will see Figure 4.1 in colour. Otherwise
you will see it in monochrome. Throughout this block, I will use monochrome but where
or when necessary, I will add annotation to enable easy identification of program parame-
ters. The Figure 4.1 screen has been set by a template called setup.puf which is called up
automatically when you start the program without defining a particular template.
Templates are used to keep the program size small so that PUFF remains versatile and will
run on personal computers. This template specifies the physical properties (size, thickness,
dielectric constant, terminal connections) and the electrical parameters (dielectric
constant, dielectric loss tangent, metal conductivity, etc.) of the board which will be used
to construct the circuit. Frequency ranges and components are also defined by the
template. Some of these properties can be changed directly within the program, others will
have to be changed by modifying the template. We will show you how to change these
properties later, but for now just accept the default template.
146 PUFF software
Fig. 4.1 Default screen for PUFF (words in italics have been added for easy identification)
We will now examine the elements of Figure 4.1.
1 Box F1: LAYOUT is a blank board on which we design our circuit. We can insert and
join up many types of components on it.
2 Box F2: PLOT specifies the parameters which will be used for plotting the circuit. You
can change these parameters and plot up to four s-parameters simultaneously. Within
limits, you can also change the frequency and the number of plotting points.
3 Message Box between F2 and F3. This box is an information box used for communi-
cations between the user and the program. It is normally blank when the program is
started. It can also be used to yield S11, S22, S33, S44 in terms of impedance or admit-
4 Box F3: PARTS specifies the components which can be used in the design. At the
moment, we have only shown seven components but you can specify up to 18 different
components in the design. These can be resistors, capacitors, inductors, transistors,
transformers, attenuators, lossless transmission lines (tlines), lossy transmission lines
(qlines) and coupled lines (clines). If on your screen, you see Ü instead of Ω, then your
machine has been configured differently to the way expected by the program. This will
not affect your work but just make a note that Ü means Ω.
5 Box F4: BOARD displays some of the layout board properties. It tells you that the
impedance (zd) of the connecting lines, its source impedance and load impedance are
50 W. It specifies the dielectric constant (er) of the board as 10.2 at 5 GHz. The
Examples 147
thickness (h) of the board is 1.27 mm, its size (s) is 25.4 mm square and the distance
between connectors (c) is 19.00 mm. The board is configured for microstrip layout.
6 The Smith chart on the top right side shows you the s-parameters of the designed
circuit. It can also be expanded and changed into admittance form.
7 The Rectangular plot (amplitude vs frequency graph) at the bottom of the screen plots
s-parameters in dB against frequency. In future, I will simply call this plot the rectan-
gular plot.
8 Pressing the F10 key will give you a small help screen.
4.5 Examples
The use of the above properties will now be illustrated.
4.5.1 Example 4.1
Example 4.1 is relatively simple but it does help you gain confidence in using the program.
We will start by constructing a 50 W transmission line on the layout board. This is shown
in Figure 4.2. To carry out the above construction proceed as follows.
1 If necessary switch your computer on, call your PUFF directory, type PUFF, press the
RETURN key and you should get Figure 4.1.
Fig. 4.2 Diagram showing a 50 W transmission line
148 PUFF software
2 Press the F4 key. F4 will now be highlighted and you will be permitted to change
values in the F4 box. An underlined cursor will appear on the zd line. Press the down
arrow key (five times) until you get to the c 19.00 mm line. Throughout this set of
instructions, keep looking at Figure 4.2 for guidance and confirmation of your actions.
3 Press the right arrow key until the cursor is under the 1 on the line, then type 00
(zero). Line c should now read c 00.000 mm. What you have effectively done is reduce
the spacing between connectors 1 and 3, and 2 and 4 to zero. You will not see the effect
as yet.
4 Press the F1 key. F1 will now be highlighted and you will be permitted to lay compo-
nents in the F1 box. An X will appear in the centre of the board. Note that there are
now only two connecting terminals on the centre edges of the board. This is because
of the action carried out in step 3. Notice also that in the F3 box, line a is highlighted.
This means that you have selected a lumped 150 W resistor to be put on the board. We
do not want this; we only want the 50 W transmission line on line b to be selected, so
type b. Line b tline 50 W 90° will be highlighted.
5 Press the left arrow key and you will see the circuit of Figure 4.2 emerging. (If you
make any mistake in carrying out these instructions, erase by retracing your step with
the shift key pressed down. For example, if you want to erase what you have just done,
press shift+right arrow keys. You can also erase the entire circuit by pressing ctrl+e
6 Type 1 and the tline will be joined to terminal 1.
7 Press right arrow key twice. See Figure 4.2.
8 Type 2 to join the right-hand section of line to terminal 2. Our circuit is now complete
and we have put two sections of 50 transmission line (b) between a 50 W generator and
load. We are now in a position to investigate its electrical properties.
9 Press the F2 key. F2 will now be highlighted and you will be permitted to specify your
measurement parameters in the F1 box.
10 Press the down arrow key three times. This will produce a new line XS.
11 Type 21 because we want to measure the parameters S11 and S21. Again refer to
Figure 4.2 for guidance.
12 Type p to plot your parameters. You will now get Figure 4.2.
13 If you do not get this figure then repeat the above steps.
14 To save Figure 4.2, press the F2 key. Type Ctrl+s. In the message box you will see File
to save? Type TX50 and press the RETURN key. Figure 4.2 is now saved under the
file name TX50.
I will now explain to you the meaning of Figure 4.2. From the F2 box, you can see that
we have measured S11 and S21 over 21 frequency points within the frequency range from
0–10 GHz. This frequency range is marked on the rectangular plot (amplitude vs
frequency graph) on the bottom right-hand side of the screen. S21 is indicated on the graph
and on the outer periphery of the Smith chart. (If you have a colour monitor, it is the blue
line.) S11 is indicated on the centre of the Smith chart. S11 cannot be shown on the rectan-
gular plot because its value is minus infinity and outside the range of the plot. PUFF
reports any magnitude as small as –100 dB as zero and any magnitude greater than 100 dB
as zero.
You can also obtain the values of S11 and S21 at discrete frequency points by referring
to the F2 box. At the moment, it is showing that at 5 GHz, S11 = 0 and S21 = 0 dB (ratio
Examples 149
of 1). This is expected because the reflection coefficient (S11) is zero and as tlines are
considered as lossless lines in the program, the gain (S21) is also zero.
You can also check the input impedance of the line by moving the cursor to the S11 line
and typing = . The actual value of the input line is shown on the Message box as Rs = 50
and Xs = 0. This tells you that the input impedance of the line is 50 W.
You should still be in the F2 mode. Press the PageUp key and watch the changes of
frequency, S11, and S21 in the F2 box. In addition the symbols for S11 and X for S21 also
move on the Smith chart and the rectangular plot. Press this again and watch the same
movement. To lower the frequency, press the PageDown key and watch the same parame-
ters noted previously. Note the PageUp and PageDown keys will only function when F2
is highlighted. The reason why you do not see drastic changes in the S-parameters is
because from theory, we know that a lossless transmission with a characteristic impedance
of 50 W inserted within a matched 50 system will only produce phase changes with
4.5.2 Smith chart expansion
While we are in F2 mode, it is also possible to expand the Smith chart to get a clearer view.
Press alt+s. You will now get Figure 4.3. You can also press the TAB key to change the
Smith chart form into an admittance display. See Figure 4.4.
When you have finished, use alt+s to toggle back to the normal display. Remember the
Fig. 4.3 An expanded view of the Smith chart in impedance mode
150 PUFF software
Fig. 4.4 An expanded view of the Smith chart in admittance mode
TAB and alt+s actions act as toggle switches and only perform these functions when the
F2 box is active.
Another point that you should be aware of is that in the F2 mode, if you press the TAB
key to change the Smith chart from impedance to admittance and if you move the cursor
to S11 and type = , you will get the parallel values of the line rather than the series values
mentioned earlier.
4.5.3 Printing and fabrication of artwork
If you have the proper printer connected, you should be able to print out the layout for
photo-etching purposes to make the actual printed circuit board. The print-out shown in
Figure 4.5 is five times the actual physical size of the layout. The magnification of the
layout is chosen to reduced the ‘jagged edges’ (constant in a printer) to an insignificant
width of the line. The print-out is then photographed, and reduced back to the original
layout size. In the photographic reduction process, these jagged edges are also reduced by
five, so that its effect on the true width of the line is less. The net result is that the line
impedance is reproduced more accurately.
Note: Do not attempt to print at this time; it will be covered later.
Examples 151
Fig. 4.5 Print-out of microstrip line
4.5.4 Summary of Example 4.1
From Example 4.1 you have learnt how to use PUFF to:
1 change the position of the terminal connections (c in F4 mode);
2 select parts from the Parts Board (b in F3 mode);
3 layout and connect selected components to the board terminals; erase your layout circuit
or components (ctrl+e, shift+arrow, in F1 mode);
4 plot and read the results of your circuit layout (S11 and S21 in F2 mode);
5 expand the Z- and Y-display of the Smith chart for more accurate readings (alt+s and
TAB in F2 mode);
6 be able to obtain series and parallel values using S11 and the = sign;
152 PUFF software
7 save a file;
8 print out your layout for subsequent circuit construction.
Now try Example 4.1 on your own to see if you have remembered the procedures.
4.6 Bandpass filter
The electrical results of Example 4.1 have not been too interesting because it is commonly
known that if a lossless 50 W line is inserted into a 50 W system, then little change, other
than phase, takes place. However, it was deliberately chosen to produce minimum confu-
sion in learning how to use the program and also to show that the program actually
produces a well known and expected result.
4.6.1 Example 4.2
In Example 4.2, we will become a bit more adventurous and introduce some quarter-wave
line short-circuited stubs across the junctions A, B, and C in the system. This case is shown
in Figure 4.6. The procedure for Example 4.2 is almost identical to that for Example 4.1
except that you will have to remember (i) that to short-circuit a component to the ground
plane, you must type the equal (=) sign at the end you want grounded; and (ii) to lay a
component in the vertical direction, you must type either the up arrow key or the down
arrow key.
Fig. 4.6 Bandpass filter using quarter-wave lengths of line
Bandpass filter 153
In Figure 4.6, we have produced a bandpass filter centred at the centre frequency ƒ0
(5 GHz in this case) where the lines are exactly a quarter-wave long. You will no doubt
remember from previous transmission line theory that the transformation of a l/4 line
is Z in = Z 02 /Z load. At ƒ0, all three short-circuited lines produce an infinite impedance
across junctions A, B, and C. This means that signal transmission is unimpaired at
ƒ0, and you will get an identical result to that of Example 4.1, i.e. the 180° phase shift
and, since tlines in this program are assumed to be lossless, you will also obtain zero
When the frequency is not ƒ0, then the shorted stubs do not present infinite impedances1
at the junctions. At ƒ < ƒ0, the shorted stubs will be inductive and will shunt signal to
ground. At ƒ > ƒ0, the shorted stubs will be capacitive and shunt signal to ground. The
result is shown in Figure 4.6.
I suggest that you try to reproduce Figure 4.6 on your own but do not despair if you do
not succeed because the details are given below. However, here are a few hints which
might prove useful before you begin.
Hint: From F1 use the down arrow keys when you want to lay a component downwards,
an up arrow key when you want to move upwards and use the equal sign (=) key when
you want to ground a component. Now carry out the relevant procedures of Example 4.2.
If you are still unable to get Figure 4.6, then carry out the instructions given below but
throughout this set of instructions, keep looking at Figure 4.6 for guidance and confirma-
tion of your actions.
1 If necessary switch on your computer. Call your PUFF directory, type PUFF, press the
RETURN key and you should get Figure 4.1.
2 Press the F4 key. F4 will now be highlighted and you will be permitted to change
values in the F4 box. An underlined cursor will appear on the zd line. Press the down
arrow key (five times) until you get to the c 19.00 mm line.
3 Press the right arrow key until the cursor is under the 1 on the line, type 00 (zero).
Line c should now read c 00.000 mm as in Figure 4.6. What you have effectively done
is reduce the space between connectors 1 and 3, and 2 and 4 to zero. You will not see
the effect as yet.
4 Press the F1 key. F1 will now be highlighted and you will be permitted to lay
components in the F1 box. An X will appear in the centre of the board. Note that
there are now only two connecting terminals on the centre edges of the board. This
is because of the action carried out in step 3. Notice also that in the F3 box, line a
is highlighted. This means that you have selected a lumped 50 W resistor to be put
on the board. We do not want this; we only want a 50 W transmission line 90° long
to be inserted into the system which means that we want to use part b for our
5 Press the F3 key. Press the down arrow key. This will get us to line b. Confirm that
this line reads b tline 50 W 90°. If this is not the case, then overtype the line to correct
6 Press the F1 key to return to the Layout box. An X will appear in the centre of the board.
If line b of F3 is not already selected, type b to select the 50 W 90° transmission line.
1 Remember the expression for a short-circuited transmission line Z = jZ tan (b l). When b l = 90°, Z =
in 0 in
infinity; when bl < 90°, Z in is inductive; when bl > 90°, Z in is capacitive.
154 PUFF software
Press the left arrow key once, then the down arrow key followed by an = key. Press
the up arrow key. This will get you back to the line junction A (not shown on computer
screen), press the 1 key. Your construction should now look like the left-hand side of
the layout board of Figure 4.6. (If you make any mistake in carrying out these instruc-
tions, erase by retracing your step with the shift key pressed down. For example, if you
want to erase the horizontal transmission line, press shift+right arrow keys. You can
also erase the entire circuit by pressing ctrl+e keys.)
7 Press the right arrow key once; it will now return your cursor to the centre of the
board (see junction B of Figure 4.6). Press the down arrow key once and press the =
key. This should now give you the centre of Figure 4.6.
8 Press the up arrow key once to return to the centre of the board, press the right arrow
key once to get to junction C. See Figure 4.6. Press the down arrow key once and press
the = key. Press the up arrow key once; follow by typing the 2 key. You should now
get the complete construction of Figure 4.6.
9 Our circuit is now complete and we have put two l/4 sections of 50 W transmission
line sandwiched between a 50 W generator and a 50 W load. We also have three short-
circuited l/4 lines between junctions A, B, C and ground. We are now in a position to
investigate the electrical properties of the filter.
10 Press the F2 key. F2 will now be highlighted and you will be permitted to specify your
measurement parameters in the F2 box.
11 Press the down arrow key three times. This will produce a new line X S.
12 Type 21 because we want to measure the parameters S11 and S21. Again refer to
Figure 4.6 for guidance.
13 Type p to plot your parameters. You will now get the entire picture of Figure 4.6.
14 If you do not get this figure then repeat the above steps again.
15 To save Figure 4.6, press the F2 key. Type ctrl+s. In the Message box you will see File
to save? Type distbpf and press the RETURN key. Figure 4.6 is now saved under the
file name distbpf.
4.6.2 Printing and fabrication of artwork
If you have the proper printer connected, you should be able to print out the layout for
photo-etching purposes to make the actual printed circuit board. The print-out shown in
Figure 4.7 is five times the actual physical size of the layout. The magnification of the
layout is chosen to reduced the ‘jagged edges’ (constant in a printer) to an insignificant
width of the line. The print-out is then photographed and reduced back to the original
layout size. In the photographic reduction process, these jagged edges are also reduced by
five, so that its effect on the true width of the line is less. The net result is that the line
impedance is reproduced more accurately. Note that in the print-out, there is no distinction
between the width of the 50 W lines. Ground points are also not shown as these have to be
drilled through the board. Do not attempt to print at this time; it will be covered later in
the guide.
4.6.3 Summary of Example 4.2
In Example 4.2, you have:
Bandpass filter 155
Fig. 4.7 Print-out of a bandpass filter in a 50 Ω system
1 learnt how a bandpass filter can be constructed from l/4 lines;
2 reinforced your ideas of how to use PUFF;
3 read and interpreted the rectangular plot and Smith chart of the bandpass filter intro-
duced in a 50 W system;
4 saved another file;
5 understood the artwork for Example 4.2.
Self test question 4.1
What do the S11 and S21 rectangular plots (amplitude vs frequency graph) tell you?
Answer. The S11 rectangular plot shows that a very good match exists in the passband of
the filter and that poor match occurs outside the filter passband. You can check this by
pressing the F2 key to enter the plot mode, and by pressing the PageUp and PageDown
keys to change the frequency to read S11 at 5 GHz, where the return loss tends toward
infinity. You cannot see this on the rectangular plot because, for practical reasons, PUFF
reports any magnitude as small as –100 dB as zero and any magnitude greater than 100 dB
as zero. At frequencies 3 GHz and 7 GHz, S11 is only about –4 dB. The S21 plot shows
156 PUFF software
the transmission plot loss as varying between 0 dB at 5 GHz to about 12 dB at 2 GHz and
8 GHz.
4.7 PUFF commands
At this stage, it is becoming increasingly difficult to remember all the commands that you
have been shown. To facilitate your work, I have tabulated some commands in Tables 4.1
to 4.4.
Table 4.1
F1 box Function
right arrow to lay a previously selected component to the right
left arrow to lay a previously selected component to the left
up arrow to lay a previously selected component above
down arrow to lay a previously selected component below
1 to connect a point to connector 1
2 to connect a point to connector 2
3 to connect a point to connector 3
4 to connect a point to connector 4
shift+right arrow to erase a component inserted by the left arrow key
shift+left arrow to erase a component inserted by the right arrow key
shift+up arrow to erase a component inserted by the down arrow key
shift+down arrow to erase a component inserted by the up arrow key
shift+e to erase the entire circuit
shift+n to move between nodes
shift + 1 to move selector to port 1
shift + 2 to move selector to port 2
shift + 3 to move selector to port 3
shift + 4 to move selector to port 4
= to earth a point
Table 4.2
F2 box Function
p plot
ctrl+p plot new modified parameters and keep previous plot
page up move measurement up in frequency
page down move measurement down in frequency
arrow to Points retyping changes number of measurement points
arrow to Smith retyping changes radius of Smith chart
arrow to S lines to add additional S-parameter measurements
TAB toggles Smith chart between Y- and Z-parameters
alt+s to toggle an enlarged Smith chart
alt+s then TAB toggles an enlarged Smith chart to Y- or Z-parameters
ctrl+a prints board artwork on appropriate printer
ctrl+s saves file
= cursor on Sxx and Smith chart on impedance yields series resistance and
reactance of the circuit at port xx
= cursor on Sxx and Smith chart on admittance yields parallel resistance and
reactance of the circuit at port xx
Templates 157
Table 4.3
F3 box Function
up arrow move up a line
down arrow move down a line
right arrow move a space to the right
left arrow move a space to the left
insert key allows the insertion of characters
alt+d inserts the symbol for degrees
alt+o inserts the symbol for ohm
ctrl+r reads a file
j inserts symbol for positive reactance
–j inserts symbol for negative reactance
S symbol for susceptance
mm denotes component size in millimetres
M megohms
+ used for series connections of components, e.g. R + jxx – jxx means
resistance + inductance + capacitance in series
alt p (//) used when you want components in parallel, e.g. R//jxx//–jxx means
resistor, inductance and capacitance are in parallel
TAB elongates F3 list to accommodate 18 different components
Table 4.4
F4 box Function
zd XXX allows change of system impedance
fd XXX allow change of central frequency
er XXX allows change of board dielectric constant
h XXX changes thickness of substrate board
s XXX changes size of substrate board
c XXX changes distances between connectors
TAB toggles layout between microstrip, stripline & Manhattan modes.
Microstrip and stripline modes are scaled. Manhattan mode is not scaled
but allows PUFF to be used for evaluation and plotting of circuits
4.8 Templates
4.8.1 Introduction
At the beginning, we told you that the reason why PUFF is a powerful but relatively small
program is because it uses templates to store information. We also asked you to temporar-
ily accept the default templates introduced earlier.
4.8.2 Setup templates
We are now in a position to show you the default template, setup.puf, which is automati-
cally called up when you start up PUFF.2
2 If you want to start up PUFF with a specified template called another, then you must specify it, when you
start PUFF by typing PUFF another. PUFF will then start up using the template called another.
158 PUFF software
Default template: setup.puf
\b{oard} {.puf file for PUFF, version 2.1}
d 0 {display: 0 VGA or PUFF chooses, 1 EGA, 2 CGA, 3 One-color}
o 1 {artwork output format: 0 dot-matrix, 1 LaserJet, 2 HPGL file}
t 0 {type: 0 for microstrip, 1 for stripline, 2 for Manhattan}
zd 50.000 Ohms {normalizing impedance. 0<zd}
fd 5.000 GHz {design frequency. 0<fd}
er 10.200 {dielectric constant. er>0}
h 1.270 mm {dielectric thickness. h>0}
s 25.400 mm {circuit-board side length. s>0}
c 00.000 mm {connector separation. c>=0}
r 0.200 mm {circuit resolution, r>0, use Um for micrometers}
a 0.000 mm {artwork width correction.}
mt 0.010 mm {metal thickness, use Um for micrometers.}
sr 0.000 Um {metal surface roughness, use Um for micrometers.}
lt 0.0E+0000 {dielectric loss tangent.}
cd 5.8E+0007 {conductivity of metal in mhos/meter.}
p 5.000 {photographic reduction ratio. p<=203.2mm/s}
m 0.600 {mitering fraction. 0<=m<1}
\k{ey for plot window}
du 0 {upper dB-axis limit}
dl –20 {lower dB-axis limit}
fl 0 {lower frequency limit in GHZ. fl>=0}
fu 10 {upper frequency limit in GHZ. fu>fl}
pts 91 {number of points, positive integer}
sr 1 {Smith-chart radius. sr>0}
S 11 {subscripts must be 1, 2, 3, or 4}
S . . . 21
\p{arts window} {O = Ohms, D = degrees, U = micro, |=parallel}
lumped 150O
tline 50O 90D
qline 50O 130D
xformer 1.73:1
atten 4dB
device fhx04
clines 60O 40O 90D
{Blank at Part h }
{Blank at Part i }
{Blank at Part j }
{Blank at Part k }
{Blank at Part l }
{Blank at Part m }
{Blank at Part n }
{Blank at Part o }
{Blank at Part p }
{Blank at Part q }
{Blank at Part r }
Templates 159
As you can see for yourself, the template contains a lot of information. First and foremost,
we shall examine the contents in the default template. Later we will create our own
In viewing the template, you will have to constantly refer to the first characters of a line
to follow my discussion.
\b This line is for identification purposes to enable PUFF to know that it is a template.
d The number specifies your display screen. For a VGA screen use 0.
o The number specifies the type of printer which you will use to produce the artwork.3
For a bubblejet or deskjet printer try using 1.
zd Specifies the normalising impedance of your system. Some antenna systems use 75
normalising impedance. If you use PUFF for other impedances then you must change
the value accordingly.
fd Specifies the design frequency of your system. Change it if you want a different
design frequency.
er The bulk dielectric constant of your substrate. Alumina and some Duroids have high
dielectric constants. The dielectric constant determines the physical size of your
distributed components. Lay your components only on the board with the dielectric
constant which you will use in your construction.
h Dielectric thickness also affects the physical size of your distributed components.
Ensure that your dielectric thickness is that which you will use in your construction.
s The square size of your board. The standard default size is 25.4 mm (1 inch square).
This is a common size for experimental work but you can choose a size (within limits)
to suit yourself. Remember though that your artwork will be several times larger and
your printer might not be able to handle it.
c The ability to move connector spacing is important because it enables short leads to
the connector.
r If the distance between components is less than the circuit resolution PUFF will
connect the parts together.
a Artwork width correction. This is used when you want to alter line width from that
calculated from the program.
mt The metal thickness is needed for calculating line losses.
sr Surface roughness is required for calculating line losses.
lt Dielectric loss tangent is also required for calculating line losses.
cd Metallic conductivity is also required for calculating line losses.
p The photographic reduction ratio used in printing the artwork. You can alter it for
greater accuracy but bear in mind your maximum printing size.
m Mitreing function is used because when a tx line changes direction or when an open-
circuited tx line ends, there is associated stray capacitance. To minimise these effects,
corner joints and end joints are frequently ‘thinned’ or mitred.
\k Informs PUFF what limits to display on your Smith chart and rectangular plot.
du Sets your upper limit in dB (y axis).
3 The reason why I told you not to attempt printing the artwork for the earlier examples is because I have no
way of knowing the type (dot-matrix, Laser jet, HPGL file) of your printer. In my case, I have successfully used
the Laserjet driver for my Hewlett-Packard bubble-jet printer. Some bubble-jet printers do not respond in this
manner and you may be left with half a printed artwork, etc. However, as some of those files were saved, you can
print them if you have a printer that will respond to one of the software drivers.
160 PUFF software
dl Sets your lower limit in dB (y axis).
fl Sets the lower limit in GHz (x axis).
fu Sets the upper limit in GHz (x axis).
pts Sets the number of points to which you want the circuit calculated and plotted. A
larger number gives greater accuracy but takes a longer time. To make your display
symmetrical about the centre frequency, you should choose an odd number of
sr Determines the Smith chart radius. In most cases you would want a factor of 1, but in
oscillator design it is usual to use a Smith chart radius greater than 1.
S11 Determines the subscripts you want in your scattering parameter measurements. You
can simultaneously measure up to four scattering parameters.
\p Informs PUFF that what follows relates to the F3 parts window and that O = ohms, D
= degrees, U = micro, | = parallel.
lumped 150O = 150 W lumped resistor.
tline 50O 90D = 50 W lossless tx line 90° length at centre frequency
qline 50O 130D = 50 W lossy tx line 130° length at centre frequency
xformer 1.73:1 = transformer with a transformation ratio of 1.73:1
atten 4dB = attenuator with 4 dB attenuation
device fhx04 = a transistor called fhx04 whose s-parameters are enclosed
clines 60O 40O 90D = coupled lines, one of 60 W, one of 40 W, length = 90°
{Blank at Part h} etc. = no parts specified.
Note: It is very important that you realise that the template is written in ASCII text and
can only be read in ASCII text by PUFF. Do not under any circumstances try to change the
default template directly because if you mess it up, PUFF will not run. Follow this proce-
1 always make a copy of the template before altering it;
2 alter the copied template with an ASCII text editor;
3 save your new template in ASCII TEXT.
Most DOS provide an ASCII text editor. For example, MSDOS supplies edlin.com for text
editing. If you do not have one, then proceed very carefully with a word processor only if
it can handle textfiles. Open the template in ASCII text. Make all your changes in ASCII
text and save the file in ASCII text. Then and only then will PUFF read the file.
When you save a file, PUFF saves the file on a template and includes calculated
answers and drawing instructions for the file. You can obtain a full version of it from
your distbpf .puf file. I offer an abridged version here for those of you who cannot read
the file.
Example 4.2 saved template (abridged)
\b{oard} {.puf file for PUFF, version 2.1}
d 0 {display: 0 VGA or PUFF chooses, 1 EGA, 2 CGA, 3 One-color}
o 1 {artwork output format: 0 dot-matrix, 1 LaserJet, 2 HPGL file}
t 0 {type: 0 for microstrip, 1 for stripline, 2 for Manhattan}
zd 50.000 Ohms {normalizing impedance. 0<zd}
fd 5.000 GHz {design frequency. 0<fd}
er 10.200 {dielectric constant. er>0}
Templates 161
h 1.270 mm {dielectric thickness. h>0}
s 25.400 mm {circuit-board side length. s>0}
c 19.000 mm {connector separation. c>=0}
r 0.200 mm {circuit resolution, r>0, use Um for micrometers}
a 0.000 mm {artwork width correction.}
mt 0.000 mm {metal thickness, use Um for micrometers.}
sr 0.000 Um {metal surface roughness, use Um for micrometers.}
lt 0.0E+0000 {dielectric loss tangent.}
cd 5.8E+0007 {conductivity of metal in mhos/meter.}
p 5.000 {photographic reduction ratio. p<=203.2mm/s}
m 0.600 {mitering fraction. 0<=m<1}
\k{ey for plot window}
du 0 {upper dB-axis limit}
dl –20 {lower dB-axis limit}
fl 0 {lower frequency limit. fl>=0}
fu 10 {upper frequency limit. fu>fl}
pts 91 {number of points, positive integer}
sr 1 {Smith-chart radius. sr>0}
S 11 {subscripts must be 1, 2, 3, or 4}
S 21
\p{arts window} {O = Ohms, D = degrees, U = micro, |=parallel}
lumped 150O
tline 50O 90D
qline 50O 130D
xformer 1.73:1
atten 4dB
device fhx04
clines 60O 40O 90D
{Blank at Part h }
. . . . . .(abridged)
{Blank at Part r }
f S11 S21
0.00000 1.00000 –180.0 2.4E–0013 0.0
0.11111 0.99996 177.5 8.7E–0003 87.5
0.22222 0.99985 175.0 1.8E–0002 85.0
0.33333 0.99965 172.5 2.7E–0002 82.5
0.44444 0.99936 169.9 3.6E–0002 79.9
0.55556 0.99896 167.3 4.6E–0002 77.3
. . . abridged
4.00000 2.3E–0002 154.7 0.99974 –115.3
4.11111 4.5E–0002 147.1 0.99897 –122.9
4.22222 5.9E–0002 139.6 0.99826 –130.4
4.33333 6.5E–0002 132.3 0.99790 –137.7
4.44444 6.4E–0002 125.2 0.99794 –144.8
4.55556 5.8E–0002 118.1 0.99832 –151.9
4.66667 4.7E–0002 111.0 0.99888 –159.0
162 PUFF software
4.77778 3.3E–0002 104.0 0.99944 –166.0
4.88889 1.7E–0002 97.0 0.99985 –173.0
5.00000 1.5E–0010 90.2 1.00000 –180.0
5.11111 1.7E–0002 –97.0 0.99985 173.0
5.22222 3.3E–0002 –104.0 0.99944 166.0
5.33333 4.7E–0002 –111.0 0.99888 159.0
5.44444 5.8E–0002 –118.1 0.99832 151.9
5.55556 6.4E–0002 –125.2 0.99794 144.8
5.66667 6.5E–0002 –132.3 0.99790 137.7
5.77778 5.9E–0002 –139.6 0.99826 130.4
5.88889 4.5E–0002 –147.1 0.99897 122.9
6.00000 2.3E–0002 –154.7 0.99974 115.3
. . . abridged
9.00000 0.99588 –156.3 9.1E–0002 –66.3
9.11111 0.99694 –159.2 7.8E–0002 –69.2
9.22222 0.99778 –162.0 6.7E–0002 –72.0
9.33333 0.99844 –164.7 5.6E–0002 –74.7
9.44444 0.99896 –167.3 4.6E–0002 –77.3
9.55556 0.99936 –169.9 3.6E–0002 –79.9
9.66667 0.99965 –172.5 2.7E–0002 –82.5
9.77778 0.99985 –175.0 1.8E–0002 –85.0
9.88889 0.99996 –177.5 8.7E–0003 –87.5
10.00000 1.00000 –180.0 1.5E–0010 –89.9
\c{ircuit} (instructions for layout)
98 0 b
203 2 left
208 3 down
61 3 =
200 2 up
205 1 right
208 4 down
61 4 =
200 1 up
205 5 right
208 6 down
61 6 =
200 5 up
From the above two examples, you should now be able to specify a template for different
Self test question 4.2
The measurement frequency range for Figure 4.1 is 0–10 GHz. If the measurement
frequency range for Figure 4.1 is to be changed from 4.5 to 5.5 GHz, how would you
modify the template to measure the frequency range 4.5–5.5 GHz?
Modification of transistor templates 163
1 Copy the setup template into an ASCII text editor.
2 Find the line beginning with fl 0. Change the value 0 to 4.5.
3 Find the line beginning with fu 10. Change the value 10 to 5.5. This will make the
rectangular plot display its frequency axis as 4.5 to 5.5 GHZ.
4 Save the file in ASCII text format. For ease of explanation, we will call the saved file
5 When you restart PUFF, type PUFF setup1. PUFF will now start using setup1 as the
default template.
6 Alternatively, you can start PUFF in the conventional manner. If necessary, press the
F3 key to enter the F3 box. Type ctrl+r. You will obtain the reply File to read. Type
Alternative method Press F2 key. Press ‘down arrow’ key until cursor is under 0.
Overtype 4.5. Next, press ‘down arrow’ key until cursor is under 10. Overtype 5.5.
Self test question 4.3
How would you change the amplitude axis of PUFF’s rectangular plot to display an ampli-
tude of +20–40 dB?
1 Copy the setup template into an ASCII text editor.
2 Find the line beginning with du 0. Change the value 0 to 20.
3 Find the line beginning with dl –20. Change the value –20 to –40. This will make the
rectangular plot display its amplitude axis as 20 to –40.
4 Save the file in ASCII text format. For ease of explanation, we will call the saved file
5 When you restart PUFF, type PUFF setup2. PUFF will now start using setup2 as the
default template.
Example 4.3
A template is required to make PUFF operate in the range 0–300 MHz and over a dynamic
range of 0–40 dB. Make such a template and name it setup300.
1 Copy the setup template into an ASCII text editor.
2 Find the line beginning with dl –20. Move the cursor under the hyphen in 20, type –40.
This will make the rectangular plot display its amplitude axis as 0 to –40 dB.
3 Find the line beginning with fu 10. Move the cursor under the 1 in 10, type .3 . This
will make the rectangular plot display its frequency axis as 0 to 0.3 GHz.
4 Save the file as setup300 in ASCII text format.
5 Restart PUFF. Type PUFF setup300. PUFF will now start using setup300 as the default
Alternative method Press F2 key. Press ‘down arrow’ key until cursor is under –20 posi-
tion. Overtype –40. Next, press ‘down arrow’ key until cursor is under 10. Overtype 0.3.
164 PUFF software
4.9 Modification of transistor templates
PUFF comes with the template for a HEMT (high electron mobility transistor) called
FHX04.dev. It is possible to use other transistor templates provided they are in ASCII text
and provided they follow the exact layout for the FHX04.dev detailed below. Again, if you
want to make your own template we suggest that you copy the transistor template below
and modify it.
Template for FHX04.dev
{FHX04FA/LG Fujitsu HEMT (89/90), f=0 extrapolated; Vds=2V, Ids=10mA}
f s11 s21 s12 s22
0.0 1.000 0.0 4.375 180.0 0.000 0.0 0.625 0.0
1.0 0.982 –20.0 4.257 160.4 0.018 74.8 0.620 –15.2
2.0 0.952 –39.0 4.113 142.0 0.033 62.9 0.604 –28.9
3.0 0.910 –57.3 3.934 124.3 0.046 51.5 0.585 –42.4
4.0 0.863 –75.2 3.735 107.0 0.057 40.3 0.564 –55.8
5.0 0.809 –92.3 3.487 90.4 0.065 30.3 0.541 –69.2
6.0 0.760 –108.1 3.231 75.0 0.069 21.0 0.524 –82.0
7.0 0.727 –122.4 3.018 60.9 0.072 14.1 0.521 –93.6
8.0 0.701 –135.5 2.817 47.3 0.073 7.9 0.524 –104.7
9.0 0.678 –147.9 2.656 33.8 0.074 1.6 0.538 –115.4
10.0 0.653 –159.8 2.512 20.2 0.076 –4.0 0.552 –125.7
11.0 0.623 –171.1 2.367 7.1 0.076 –10.1 0.568 –136.4
12.0 0.601 178.5 2.245 –5.7 0.076 –15.9 0.587 –146.4
13.0 0.582 168.8 2.153 –18.4 0.076 –21.9 0.611 –156.2
14.0 0.564 160.2 2.065 –31.2 0.077 –28.6 0.644 –165.4
15.0 0.533 151.6 2.001 –44.5 0.079 –36.8 0.676 –174.8
16.0 0.500 142.8 1.938 –58.8 0.082 –48.5 0.707 174.2
17.0 0.461 134.3 1.884 –73.7 0.083 –61.7 0.733 163.6
18.0 0.424 126.6 1.817 –89.7 0.085 –77.9 0.758 150.9
19.0 0.385 121.7 1.708 –106.5 0.087 –97.2 0.783 139.1
20.0 0.347 119.9 1.613 –123.7 0.098 –119.9 0.793 126.6
Note that each S-parameter is denoted by an amplitude ratio and angle in degrees. For
example at 1 GHz
S11 = 0.982 ∠ –20.0° S21 = 4.257 ∠ 160.4°
S12 = 0.018 ∠74.8° S22 = 0.620 ∠ –15.2°
Self test question 4.4
If you wanted to change the S12 parameter for HEMT FHX04 at 5 GHz to 0.063 31.4,
how would you do it?
1 Copy the transistor template (FHX04.dev) into an ASCII text editor.
2 Find the line beginning with 5.0. It should read:
5.0 0.809 –92.3 3.487 90.4 0.065 30.3 0.541 –69.2
3 Change it to read:
5.0 0.809 –92.3 3.487 90.4 0.063 31.4 0.541 –69.2
Verification of some examples given in Chapters 2 and 3 165
4 Save the file in ASCII text format. For ease of explanation, we will call the saved file
You can then insert or change this part name in your chosen set-up template so that when-
ever you start PUFF, the device will be shown in the F3 box. Alternatively you can insert
the part directly into the F3 box whenever you want to use the transistor.
4.10 Verification of some examples given in Chapters 2 and 3
4.10.1 Verification of microstrip line
We can now use PUFF to verify some of the conclusions reached in Chapters 2 and 3. In
doing so, I will only quote the example number, and where appropriate its question and
answer. I will then use PUFF to show that the conclusion is correct.
Example 2.3
Two microstrip lines are printed on the same dielectric substrate. One line has a wider
centre strip than the other. Which line has the lower characteristic impedance? Assume that
there is no coupling between the two lines.
Solution. The broader microstrip has the lower characteristic impedance. Using PUFF
this is confirmed in Figure 4.8 where the broad microstrip has a characteristic impedance
of 20 W and the narrow microstrip has a Z 0 of 90 W.
Fig. 4.8 PUFF results showing a 20 Ω microstrip and a 90 Ω microstrip
166 PUFF software
4.10.2 Verification of reflection coefficient
Example 2.7
Calculate the reflection coefficient for the case at 5 GHz where Z L = (80 – j10) W and Z 0
= 50 W.
ZL − Z0 80 − j10 − 50 30 − j10
Γ= = =
ZL + Z0 80 − j10 + 50 130 − j10
31.62 ∠ − 18.43°
= = 0.24 ∠ − 14.03° or − 12.305 dB ∠ − 14.03°
130.38 ∠ − 4.40°
Solution. Using Equation 2.24
When the above answer is written in dB, we get –12.3 dB ∠ –14.03°. Compare this answer
with S11 in the F2 box of Figure 4.9.
Fig. 4.9 Verification of S11 for Example 2.7
Example 2.8
Calculate the voltage reflection coefficients at the terminating end of a transmission line
with a characteristic impedance of 50 W when it is terminated by (a) a 50 W termination,
(b) an open-circuit termination, (c) a short-circuit termination, and (d) a 75 W termination.
Given: Z 0 = 50 W, Z L = (a) 50 W, (b) open-circuit = ∞, (c) short-circuit = 0 W, (d) = 75 W.
Required: Gv for (a), (b), (c), (d).
Verification of some examples given in Chapters 2 and 3 167
Fig. 4.10 Verification of Example 2.8
Solution. Use Equation 2.24.
(a) Z L = 50/0°
ZL – Z0 50/0° – 50/0°
Gv = ——— = —————— = 0/0° 0 dB
ZL + Z0 50/0° + 50/0°
(b) Z L = open-circuit = ∞ /0°
ZL – Z0 ∞ /0° – 50/0°
Gv = ——— = —————— = 1/0° = 0 dB ∠ 0°
ZL + Z0 ∞ /0° + 50/0°
(c) Z L = short-circuit = 0/0°
Z L – Z 0 0/0° – 50/0°
Gv = ——— = ————— = –1/0°
— — or 1/180° = 0 dB ∠ 180°
Z L + Z 0 0/0° + 50/0°
(d) Z L = 75/0°
Z L – Z 0 75/0° – 50/0°
Gv = ——— = ————— = 0.2/0° = –13.98 dB ∠ 0°
— —
Z L + Z 0 75/0° + 50/0°
In Figure 4.10, I have plotted case (a) as S11, case (b) as S33,case (c) as S22, case (d)
as S44. For S33, I have used a 1000 MW resistor to represent a resistor of infinite ohms.
168 PUFF software
Alternatively, you could have used nothing to represent an open circuit. Most r.f.
designers do not like open circuits because open circuits can pickup static charges
which can destroy a circuit. It is far better to have some very high resistance so that
static charges can be discharged to earth. Note how all the answers agree with the
calculated ones.
4.10.3 Verification of input impedance
Example 2.13
A 377 W transmission line is terminated by a short circuit at one end. Its electrical length
is l/7. Calculate its input impedance at the other end.
Solution. Using Equation 2.55
2π l ⎡ 2π λ ⎤
Zin = jZ0 tan = j377 tan ⎢ = j377 × 1.254 = j472.8 Ω
λ ⎣ λ 7⎥ ⎦
Remembering that l/7 = 51.43° and using PUFF, we see from the Message box that Rs =
0 and Xs = 472.767 W (see Figure 4.11). This confirms the calculated value.
Fig. 4.11 Verification of Example 2.13
Verification of some examples given in Chapters 2 and 3 169
Example 2.14
A 75 W line is left unterminated with an open circuit at one end. Its electrical length is l/5.
Calculate its input impedance at the other end.
Solution. Using Equation 2.56
2π l ⎡ 2π λ ⎤
Zin = jZ0 cot = − j75 cot ⎢ = − j75 × 0.325 = − j24.4 Ω
λ ⎣ λ 5⎥ ⎦
Bearing in mind that l/5 = 72° and using 1000 MW to simulate an open circuit, PUFF
gives the answer in the Message box as –j24.369 W (see Figure 4.12).
Fig. 4.12 Verification of Example 2.14
Example 2.15
A transmission line has a characteristic impedance (Z 0) of 90 W. Its electrical length is l/4
and it is terminated by a load impedance (Z L ) of 20 W. Calculate the input impedance (Z in)
presented by the line.
Solution. Using Equation 2.57
Z in = (90) 2/20 = 405 W
Using PUFF and reading the Message box, we get Rs = 405 W, Xs = 0 W (see Figure 4.13).
The above examples should now convince you that much of the transmission line theory
covered in Chapter 2 has been proven.
170 PUFF software
Fig. 4.13 Verification of Example 2.15
4.11 Using PUFF to evaluate couplers
In Section 2.14, we investigated the theory of two popular couplers. These are (a) the
branch-line coupler and (b) the rat-race coupler.
4.11.1 Branch-line coupler
The theory for the branch-line coupler was covered in Section 2.14.1. For the branch-line
coupler, I have used vertical lines with Z 0 of 50 W and horizontal lines with Z 0 of 35.55
W. In the F2 box (Figure 4.14), you can see that the match (S11 to a 50 W system) is excel-
lent. S21 and S41 show that the input power from S11 is divided equally between the two
ports but that there is a phase change as explained in the text. S31 shows you that there is
very little or no transmission to port 3. All the above statements confirm the theory
presented in Section 2.14.1.
4.11.2 Rat-race coupler
The theory for the rat-race coupler was covered in Section 2.14.3. Here PUFF confirms
what we have discussed. In using PUFF, I have chosen Z 0 for the ring as 70.711 W and
used a rectangle to represent the ring but all distances between the ports have been kept as
before. This is shown in Figure 4.15.
Using PUFF to evaluate couplers 171
Fig. 4.14 Verification of the branch-line coupler theory
Fig. 4.15 Verification of the rat-race coupler theory
172 PUFF software
4.12 Verification of Smith chart applications
In Chapter 3, we used the Smith chart to derive admittances from impedances, calculate
line input impedance, and solve matching networks. In this section, we show you how it
can also be done with PUFF but note that the intermediate steps in the solutions are not
given and sometimes it can be difficult to visualise what is actually happening in a circuit.
We shall begin with Example 3.2. As usual we will simply use a numbered example, intro-
duce its context and solution and then show how it can be solved with PUFF.
4.12.1 Admittance
Example 3.2
Use the Smith chart in Figure 3.6 to find the admittance of the impedance (0.8 – j1.6).
Solution. The admittance value is located at the point (0.25 + j0.5). You can verify this
yourself by entering the point (0.8 – j1.6) in Figure 3.6. Measure the distance from your
point to the chart centre and call this distance d. Draw a line of length 2d from your point
through the centre of the chart. Read off the coordinates at the end of this line. You should
now get (0.25 + j0.5) S.
With PUFF, we simply insert its value in the F3 box, and draw it in the F1 box. In
the F2 box, press the TAB key to change the Smith chart into its admittance form. Move
the cursor to S11, press p for plot and the equals sign (=) to read its value in the
Fig. 4.16 Verification of Example 3.2
Verification of Smith chart applications 173
Message box. Note that Rp and Xp are given as parallel elements and its units are in
ohms. However, remembering that (0.25 + j0.5) S is a combination of a conductance of
0.25 S to represent a resistor and j0.5 S to represent a capacitor, we simply take the
reciprocal of each element to get the desired answer for each element. This is shown in
Figure 4.16.
4.12.2 Verification of network impedances
For Example 3.5, we simply draw the network in the F1 box and seek its results in the F2
box and Message boxes.
Example 3.5
What is (a) the impedance and (b) the reflection coefficient looking into the network
shown in Figure 3.12?
Solution. The solution to this problem was given in the annotation to Figure 3.14 as:
(a) impedance Z = (0.206 + j0.635) W
(b) reflection coefficient G = 0.746 ∠ 113.58°
The PUFF solutions (Figure 4.17) give:
(a) impedance Z = (0.206 + j0.635) W
(b) reflection coefficient G = –2.55 dB ∠ 113.6° which is 0.746 ∠ 113.6°
Fig. 4.17 Verification of Example 3.5
174 PUFF software
• In Figure 4.17 the drive impedance (zd) in PUFF has to be reduced to 1 W instead of the
usual 50 W because the values in the circuit have been normalised.
• PUFF only gives the overall input impedance. If you had wanted intermediate values,
then you would have to add one immittance at a time and read out its value.
4.12.3 Verification of input impedance of line
For this we will use Example 3.6.
Example 3.6
A transmission line with a characteristic impedance Z 0 = 50 W is terminated with a load
impedance of Z L = (40 – j80) W. What is its input impedance when the line is (a) 0.096l,
(b) 0.173l, and (c) 0.206l?
Solution. The answers calculated previously are:
(a) (0.25 – j0.5) W which after re-normalisation yields (12.5 – j25) W
(b) (0.20 – j0.0) W which after re-normalisation yields (10.0 – j0) W
(c) (0.21 + j0.2) W which after re-normalisation yields (10.5 + j10) W
With PUFF, we obtain:
(a) (12.489 – j24.947) W – shown in Figure 4.18.
(b) (9.897 + j0.022) W – not shown in Figure 4.18.
(c) (10.319 + j10.111) W – not shown in Figure 4.18.
Fig. 4.18 Verification of Example 3.6 results
Verification of Smith chart applications 175
Items (b) and (c) were obtained from PUFF by moving the cursor in the F2 box to S22 and
pressing the = key, and then to S33 and pressing the = key.
4.12.4 Verification of quarter-wave transformers
PUFF can be used to investigate the effect of quarter-wave line transformer matching. For
this we will use Example 3.11.
Example 3.11
A source impedance of (50 + j0) is to be matched to a load of (100 + j0) over a frequency
range of 600–1400 MHz. Match the source and load by using (a) one quarter-wave trans-
former, and (b) two quarter-wave transformers. Sketch a graph of the reflection coefficient
against frequency.
Solution. Use Equation 2.57. For the one l/4 transformer, we had previously calculated
Z 0t1 as 70.711 W. For the two l/4 transformers, we had previously calculated Z 0t1 as 60 W
and Z 0t2 as 84.85 W. We had also obtained the following table.
(GHz) 0.6 0.8 1.0 1.2 1.4
One l/4 TX – 13.83 – 19.28 > – 60 – 19.28 – 13.83
Two l/4 TXs 18.81 – 32.09 38.69 32.09 – 18.81
Using PUFF, we show the two results in Figure 4.19.
Fig. 4.19 Verification of reflection coefficient using λ/4 line transformers
176 PUFF software
4.13 Verification of stub matching
Stub matching is very important. PUFF can be used to match both passive and active
networks. To give you an idea of how this is achieved, I will detail the matching of
Example 3.12 which was carried out manually in Section 3.8.3.
Example 3.12
Use microstrip lines to match a series impedance of (40 – j80) W to 50 W at 1 GHz.
1 Switch on PUFF. Press the F3 key. Type ctrl+r. The program will reply File to read:?
2 Type match1. Press the RETURN key. You will obtain Figure 4.20(a). Note that in the
F3 box, we have:
• the series impedance we wish to match, (40 – j80);
• a transmission line (b) whose length is designated as ?50°;
• a transmission line (c) whose length is designated as 50°;
• note also that the rectangular plot x axis is marked in degrees and not frequency.
The next step is to construct the circuit shown in Figure 4.20(b).
3 Press the F1 key. Type a. Press shift+right arrow key seven times until the cursor is
positioned to the right of the layout board.
Fig. 4.20(a) Blank matching screen
Verification of stub matching 177
Fig. 4.20(b) Second stage of matching
4 Press the down arrow key. Type =. This grounds part a.
5 Press the up arrow key. Type b. Press left arrow key. Type 1. The layout board is now
6 Press the F2 key. Type p. Press the TAB key to change the Smith chart coordinates to
an admittance display. You will now obtain Figure 4.20(b). Note that the little square
marker is on the lower right of the Smith chart.
7 Type alt+s. You will get an expanded view of the Smith chart similar to that of Figure
8 We now want to move the marker until the square marker intersects the unity circle.
Press the page up key several times and the marker will begin to move towards the
intersection point with the unity circle. At the same time, look in the F2 box and you
will see part b length increasing in degrees. The square marker should reach the unity
circle when line b is approximately 38°. At the intersection point, the conductive part
of the line is matched to the input, but there is also a reactive part which must be ‘tuned
out’. At this stage, we do not know its value but from its chart position, we know that
the reactive part is capacitive.
9 Type alt+s to revert back to the display of Figure 4.20(b).
10 Press the F3 key. Use arrow keys to move the cursor to line b and overtype ?50 with
38. You have now fixed line length b at 38°. Use arrow keys to move the cursor to line
c and overtype 50 with ?50. See Figure 4.20(d) for both actions.
11 Press the F1 key. Note how line length b has changed. If not already there, use arrow
keys to move the cursor until it is at the junction of line 1 and line b.
178 PUFF software
Fig. 4.20(c) Intersection point in Smith chart
Fig. 4.20(d) Connecting the tuning stub line
Verification of stub matching 179
12 Type c to select the other line. To layout line c in the position shown in Figure 4.20(d),
press the down arrow key once. Type =.
13 Press the F2 key. Type p. Press the TAB key. You will now get Figure 4.20(d).
14 Type alt+s. You will now obtain the expanded Smith chart shown in Figure 4.20(e) but
with the exception that the square marker is on the left edge of the Smith chart.
15 We now want to move the square marker on the unity circle until it reaches the match
point in the centre of the Smith chart. Press the page up key several times and the
marker will begin to move towards the match point at the centre of the Smith chart. At
the same time, look in the F2 box and you will see part c length decreasing in degrees.
16 The square marker should reach the unity circle match point when line c is approxi-
mately 29°. Note this length. At the match point, the reactive part will have been ‘tuned
out’ by the stub. We can now leave the Smith chart.
17 Summing up: we now know the two line lengths required; line b is 38° and line c is a
short-circuited stub of 29°.
18 Press the F3 key. Type ctrl+r. When the program replies File to read? type match6.
Press the RETURN key.
19 You will get Figure 4.20(f) where the previously calculated line lengths have been used.
Note that the match is best (return loss ≈ 38 dB) at our design frequency of 1 GHz.
It is possible to get Figure 4.20(f) directly from Figure 4.20(d). After obtaining and
entering line lengths b as 38° and c as 29°, press the F2 key. Press the down-arrow key nine
times. Retype over 120 to show 2. Press p and you will obtain the file MATCH6.
Fig. 4.20(e) Expanded Smith chart for tuning stub match
180 PUFF software
Fig. 4.20(f) Showing the effect of the matching stubs at 1 GHz
4.13.1 Summary of matching methods
I would now like to summarise the methods we have used for matching. For easy compar-
ison remember that 1 wavelength = 360°, so 29° = 0.081l and 38° = 0.106l. The results
for three methods are given in Table 4.5.
From the above, you will see that there is little difference whichever method you use.
The direct calculations have been found through a ‘goal seek’ program in an Excel spread-
sheet. The answers are definitely more accurate, but in practical situations we do not
require such accuracy. Also do not worry unduly if you find that when you repeat the same
calculations on PUFF your answers may differ slightly (≈0.1 dB). This is due to truncation
errors in the program.
The graphical methods give an insight into what can be achieved more easily. For exam-
ple, in microstrip, a short-circuited stub is not easy to manufacture. In this particular case,
we could have increased the length of line b to move the matching point into the inductive
part of the Smith chart and then used a capacitive stub for tuning out the inductive part of
the circuit. I will not show you how this is done in this example but it is done in Example
4.4 which follows immediately.
Table 4.5
PUFF matching Smith chart Direct calculations
Example 3.12 (PUFF) Example 3.12 of Chapter 3 Eqn 2.54
l 1 = 38° = 0.106l l 1 = 0.106l l 1 = 0.106 305l
Stub = 29° = 0.081l Stub = 0.081l Stub = 0.0 806 036l
Verification of stub matching 181
4.13.2 Matching transistor impedances
Example 4.4 shows how the input impedance of a transistor (type fhx04) can be matched
to 50 W at 5 GHz. In this example, I will only provide you with intermediate diagrams
because the methodology is identical to that of Example 3.12.
Two methods are shown:
• a capacitive stub matching system shown in Figures 4.21(a) to (d);
• an inductive stub matching system shown in Figures 4.22(a) to (c).
Either method is suitable, but in microstrip circuits it is much easier to make an open-
circuited stub than a short-circuited stub; therefore the capacitive stub matching method is
preferred. However, you should compare Figure 4.21(d) and Figure 4.22(c) and note that
although matching is achieved at our desired frequency of 5 GHz, there are matching
differences on ‘off-frequency’ matching.
If you wish to try these matching networks out for yourself, Figures 4.21(a) to (d) are given
as files sweep1 to sweep4 respectively on your disk. Figures 4.22(a) to (c) are given on your
disk as files sweep22, sweep33 and sweep44 respectively. As before, I suggest that you start
off with Figure 4.21(a) and build up the capacitive stub matching system. This ensures that if
you run into trouble, you will have the other templates available. Similarly, start off with
Figure 4.22(a) for the inductive stub matching system and build up the circuit accordingly.
Transistor matching using a capacitive stub
Fig. 4.21(a) First matching line
182 PUFF software
Fig. 4.21(b) Determining first matching line length for a capacitive stub
Fig. 4.21(c) Determining length of a capacitive tuning stub
Verification of stub matching 183
Fig. 4.21(d) Matching using a capacitive stub
Transistor matching using an inductive stub
Fig. 4.22(a) Determining first matching line length for an inductive matching stub
184 PUFF software
Fig. 4.22(b) Determining length of an inductive tuning stub
Fig. 4.22(c) Matching at 5 GHz using an inductive stub
Verification of stub matching 185
Verification of output impedance matching
If you want to match the output impedance of any active or passive device use similar
procedures to that of Example 4.4. The detailed matching procedure is again identical to
Example 3.12.
4.13.3 Stub matching second example
Here is another example of stub matching. Try to see if you can carry out the stub match-
ing for Example 3.13 which was done manually in Chapter 3.
Example 3.13
A transistor amplifier has an h.f. input resistance of 100 W shunted by a capacitance of
5 pF. Find the length of short-circuit stub, and its position on the line, required to match
the amplifier input to a 50 W line at 1 GHz.
• The stub connection point is 0.028l from the transistor input.
• The short-circuit stub length is 0.065l at the connection point.
Example 3.13 is another example of stub matching. You already know the answer. See if
you can use PUFF to match the circuit on your own. Hint: if you cannot do it, look at
Figure 4.23.
Fig. 4.23 Verification of Example 3.13
186 PUFF software
4.13.4 Double stub tuning verification
PUFF can also be used for verifying double stub matching. We will use Example 3.14
which was carried out manually in Chapter 3.
Example 3.14
A system similar to the double stub matching system shown in Figure 3.16 has a load Z L =
(50 + j50) W which is to be matched to a transmission line and source system with a char-
acteristic impedance of 50 W. The distance, d 1, between the load and the first stub is 0.2l
(72°) at the operating frequency. The distance, d 2, between the two stubs is 0.125l (45°) at
the operating frequency. Use a Smith chart to estimate the lengths l1 and l 2 of the stubs.
Solution. In Example 3.14, we found that:
• stub 1 required an electrical length of 0.167l or 60.12°;
• stub 2 required an electrical length of 0.172l or 61.92°;
The construction using PUFF is shown in Figure 4.24.
Fig. 4.24 Verification of Example 3.14
4.14 Scattering parameters
PUFF can also be used for calculating S-parameters. We demonstrate this with Examples
3.15, 3.16 and 3.17.
Scattering parameters 187
4.14.1 Series elements
Example 3.15
Calculate the S-parameters for the two-port network shown in Figure 3.30 for the case
where Z 0 = 50 W.
Solution. Summing up for S-parameters:
S11 = 0.333 ∠ 0° or –9.551 dB ∠ 0° S12 = 0.667 ∠ 0° or –3.517 dB ∠ 0°
S21 = 0.667 ∠ 0° or –3.517 dB ∠ 0° S22 = 0.333 ∠ 0° or –9.551 dB ∠ 0°
This circuit is shown in Figure 4.25.
Fig. 4.25 Verification of Example 3.15
4.14.2 Shunt elements
PUFF can be used to calculate shunt elements as shown in Example 3.16.
Example 3.16
Calculate the S-parameters for the two port network shown in Figure 3.32 for the case
where Z 0 = 50 W.
Solution. Summing up for S-parameters:
S11 = 0.333 ∠ 180° or –9.551 dB ∠ 180° S12 = 0.667 ∠ 0° or –3.517 dB ∠ 180°
S21 = 0.667 ∠ 0° or –3.517 dB ∠ 180° S22 = 0.333 ∠ 180° or –9.551 dB ∠ 180°
188 PUFF software
Fig. 4.26 Verification of Example 3.16
The values calculated by PUFF are shown in Figure 4.26. You can see that they are
4.14.3 Ladder network
PUFF can be used for calculating ladder networks. This is shown by Example 3.17 which
was carried out manually in Chapter 3.
Example 3.17
(a) Calculate the S-parameters for the two-port network shown in Figure 3.32 for the case
where Z 0= 50 W.
(b) Find the return loss at the input with Z L = Z 0.
(c) Determine the insertion loss for the network when the generator and the termination
are both 50 W.
Solution. To sum up for S-parameters
(a) S11 = 0.518 ∠ 131.16° or –5.713 dB ∠ 131.16°
S12 = 0.524 ∠ –16.22° or –5.613 dB ∠ –16.22°
S21 = 0.524 ∠ –16.22° or –5.613 dB ∠ –16.22°
S22 = 0.442 ∠ 172.87° or –7.092 dB ∠ 131.16°
Discontinuities: physical and electrical line lengths 189
Fig. 4.27 Verification of Example 3.17
(b) From part (a), G ∠ q = 0.518 ∠ 131.16°. Hence
return loss (dB) = –20 log10 |0.518| = –20 × (–0.286) = 5.71 dB
(c) The forward power gain of the network will be |S21| 2.
|S21| 2 = (0.524)2 = 0.275
This represents a loss of –10 log10 0.275 = 5.61 dB.
Figure 4.27 gives the answers for the parameters. To derive the other two answers,
namely (b) return loss and (c) forward power, you merely read off S11 and S21 and
calculate to get the values.
4.15 Discontinuities: physical and electrical line lengths
This section is vitally important in the construction of your circuits. In all the problem
solving given earlier, the theoretical (electrical) line length has been assumed. In other
words, we have taken a physical line length of one wavelength and assumed it to be 360
electrical degrees. In the practical case, if you were to do this with a transmission line, you
would find that a physical length of one wavelength is not likely to be 360 electrical
190 PUFF software
Fig. 4.28 Effect of fringing fields in transmission lines
degrees. This is due to end effects and line fringing effects, shown in Figure 4.28. The
general name for these effects is discontinuities.
PUFF does not take these effects into account when drawing the artwork; therefore you
must compensate for them when you use PUFF to draw the artwork. UCLA (PUFF
program writers)1 suggest that we consider four dominant discontinuities in microstrip.
These are:
• excess capacitance of a corner;
• capacitive end effects for an open circuit;
• step change in width;
• length correction for the shunt arm of a tee junction.
4.15.1 Excess capacitance of a corner
When a sharp right-angle bend occurs in a circuit (Figure 4.29(a)) there will be a large
reflection from the corner capacitance. PUFF mitres corners to reduce the capacitance and
minimise this reflection as shown in Figure 4.29(b). You can change the value of the mitre
fraction (m) set in the setup.puf template as 0.6. The mitre fraction (m) is defined as:
m = 1− b w1 + w2 (4.1)
Fig. 4.29 Chamfering (mitreing) of corners
1 This is also on the CD-ROM PUFF manual.
Discontinuities: physical and electrical line lengths 191
Fig. 4.30 Line length compensation for end effects
4.15.2 Capacitance end effects for an open circuit
In an open-circuit line, the electric fields extend beyond the end of the line. This excess
capacitance makes the electrical length longer than the nominal length of the line, typically
by a third to a half of the substrate thickness. To compensate for this effect in the artwork,
a negative length correction can be added to the parts list. Hammerstad and Bekkadal4 give
an empirical formula for the length extension l in microstrip:
l ⎛ ε + 0.3 ⎞ ⎛ w h + 0.262 ⎞
= 0.412⎜ eff ⎟⎜ ⎟ (4.2)
h ⎝ ε eff − 0.258 ⎠ ⎝ w h + 0.813 ⎠
where eeff is the effective dielectric constant of the through arm.
Note: In the above correction, the length l must be negative, i.e. the length l must be
subtracted from the desired length in the parts list.
4.15.3 Step change in width of microstrip
A similar method may be used to compensate for a step change in width between high
and low impedance lines. This is shown in Figure 4.31. The discontinuity capacitance
at the end of the low impedance line will have the effect of increasing its electrical
4 E.O. Hammerstad and F. Bekkadal, A Microstrip Handbook, ELAB Report, STF 44 A74169, N7034,
University of Trondheim, Norway, 1975.
192 PUFF software
Fig. 4.31 Step change in width of microstrip
Assuming the wider low impedance line has width w2, and the narrow high impedance
line has width w1, compensate using the expression suggested by Edwards5
ls ⎛ ε + 0.3 ⎞ ⎛ w h + 0.262 ⎞ ⎡ w1 ⎤
= 0.412⎜ eff ⎟⎜ ⎟ ⎢1 − ⎥ (4.3)
h ⎝ ε eff − 0.258 ⎠ ⎝ w h + 0.813 ⎠ ⎣ w2 ⎦
ls l ⎡ w1 ⎤
= ⎢1 − ⎥ (4.3a)
h h ⎣ w2 ⎦
where ls is the step length correction for line w2 and l/h is the value obtained from Equation
4.2 and Figure 4.30.
4.15.4 Length correction for the shunt arm of a tee-junction
In the tee-junction shown in Figure 4.32, the electrical length of the shunt arm is short-
ened by distance d 2. The currents effectively take a short cut, passing close to the
corner. It is particularly noticeable in the branch-line coupler because there are four tee-
Fig. 4.32 Length correction for the shunt arm of a tee-junction
5 T.C. Edwards, Foundations for microstrip circuit design, second edition, John Wiley & Sons, ISBN 0 471
93062 8, 1992.
Summary 193
junctions. Hammerstad and Bekkadal (see footnote 4) give an empirical formula for d 2
in microstrip:
d2 120π ⎧ Z1 ⎫
= ⎨0.5 − 0.16 [1 − 2 ln( Z1 Z2 )]⎬ (4.4)
h Ζ 1 ε eff ⎩ Z2 ⎭
where εeff is the effective dielectric constant of the through arm. Equation 4.4 is plotted in
Figure 4.32 for a 50 W through line. Additional help on discontinuity modelling for both
microstrip and stripline can be found in a book by Gupta.6
4.16 Summary
By now, I am sure you will agree that your studies in Parts 2 and 3 on transmission lines,
Smith charts and S-parameters are beginning to bring rewards and help you toward the goal
of being a good h.f. and microwave engineer.
Sections 4.1 to 4.3 of this part have been devoted to the installation of PUFF. In Section
4.4, we covered the principles of using PUFF. Section 4.5 provided some simple examples
of how to use the facilities provided by PUFF for printing and production of artwork. In
Section 4.6, we designed and produced the artwork, and measured the frequency response
of a bandpass filter using transmission lines.
Section 4.7 was used to collect and collate all the PUFF commands that you had learnt
previously. The use, design and modification of templates for the PUFF system were
discussed in Section 4.8. In Section 4.9, we learnt how to manipulate and alter transistor
templates for PUFF.
In Section 4.10, we achieved our goal of verifying and checking that the work carried
out in Parts 2 and 3 was valid. Fifteen examples (2.3, 2.7, 2.8, 2.13, 2.14, 2.15, 3.2, 3.5,
3.6, 3.11, 3.12, 3.13, 3.14, 3.15, 3.16 and 3.17) were entered into PUFF. Their results were
compared with the examples produced manually in Parts 2 and 3. The fact that both sets
of results agree should give you confidence in the use of either method.
In Section 4.11, we used PUFF to investigate the properties of the branch-line coupler
and the rat-race coupler. We evaluated their transmission and matching properties.
Section 4.12 was used to show how PUFF can be used to find admittances (Example
3.2), network impedance and reflection coefficient (Example 3.5), input impedance of
transmission lines (Example 3.6), quarter-wave transformers (Example 3.11), and cascad-
ing of quarter-wave transformers. The examples created manually in Part 3 all agree with
the solutions provided by PUFF.
The very important technique of stub matching was detailed and demonstrated in
Section 4.13. It provided details on how single stub matching can be carried out with
PUFF. The PUFF answer agreed well with Example 3.12 which was previously carried out
manually and also with direct calculations. See Table 4.5 in Section 4.13.1 for details.
Section 4.13.2 provided details on how matching can be carried out using inductive or
capacitive tuning stubs. Section 4.13.3 provided an electronic matching of Example 3.13.
Double stub electronic matching of Example 3.14 was verified in Section 4.13.4.
6 K.C. Gupta, R. Garg and R. Chadha, Computer-aided design of microwave circuits, Artech House, Deham,
Mass. USA, ISBN 0–89006–105–X, 1981.
194 PUFF software
Section 4.14 demonstrated how PUFF can be used to calculate the S-parameters of
series elements (Example 3.15), shunt elements (Example 3.16) and networks (Example
The important subject of discontinuities in microstrip and how they may be compen-
sated for in the PUFF artwork was covered in Section 4.15. Four types were discussed, and
compensation methods for minimising these effects were shown.
Now that you are familiar with many passive networks and their solutions, we will be
moving on to active circuits, mainly the design of amplifiers, in the following parts.
However, this is not the last of PUFF because we will be using it in the design of filters
and amplifiers.
Last but not least, the use of PUFF in the design and layout of circuits detailed in an
article called ‘Practical Circuit Design’ is reproduced on the disk accompanying this book.
However, I advise you to defer reading it until you have reached the end of Part 7 because
many of the principles and techniques used in the article have yet to be explained.
Amplifier basics
5.1 Introduction
The information gained in the previous parts has now allowed us to move into the realms
of amplifier design. Small signal r.f. amplifiers assume many configurations. We show two
common configurations. In Figure 5.1, we show the circuit of a single stage amplifier. It
consists of five main sections:
• input source with a source impedance Z s;
• an input tuned/matching circuit comprising C1, L1 and C 2;
• a transistor amplifier (transistor biasing is not shown);
• an output tuned/matching circuit comprising L 2 and C 3;
• load (Z L).
In Figure 5.2, we show the circuit of a multi-stage integrated amplifier circuit. It
consists of five main sections:
• input source with a source impedance Z s;
• a multi-stage amplifier gain block (sometimes called ‘gain blok’);
Fig. 5.1 Single stage amplifier
196 Amplifier basics
Fig. 5.2 Multi-stage amplifier
• a multi-tuned/matching filter circuit comprising C1, L 1, C 2, L 2 and C3;
• a multi-stage amplifier gain block (sometimes called ‘gain blok’);
• load (Z L).
From the above figures, it is clear that to design a circuit, we must understand:
• tuned circuits
• filters
• matching techniques
• amplifier parameters
• gain block parameters
Much material will be devoted to matching circuits in this chapter because after selection
of a transistor or gain block for a particular design, there is not much you can do within
the active device other than present efficient ways in which energy can be coupled in and
out of the device. This in turn calls for efficient matching circuits for the intended
In this chapter, we will investigate tuned circuits, filters and impedance matching tech-
niques. This will enable us to deal with transistors, and semiconductor devices in the next
5.1.1 Aims
The aims of this chapter are to introduce you to the passive elements and/or devices which
are used in conjunction with active devices (transistors, etc.) to design complete circuits.
5.1.2 Objectives
The objectives of this chapter are to show how passive, discrete and distributed elements
can be used in the design of tuned circuits, filters and impedance matching networks.
5.2 Tuned circuits
As these equations are readily available in any elementary circuit theory book, we shall
simply state the equations associated on single series and parallel tuned circuits.
Tuned circuits 197
5.2.1 Series circuits
The series C, L and R circuit is shown in Figure 5.3. The fundamental equations relating
to the series circuit are:
Z = R + jωL + 1 ( jωC )
− (5.1)
ωo = 1 LC (5.2)
vr R
= (5.3)
vs R + j (ωL − 1 ωC )
Q = ωoL R or 1 (ω o CR) (5.4)
Q = ω o (ω 2 − ω1 ) (5.5)
Z = input impedance with R (ohms), L (Henries), C (Farads)
wo = resonant frequency in radians per second
vr = voltage across resistor R
vs = open-circuit source voltage
Q = quality factor
w2 = upper frequency (rads/sec) where the response has fallen by 3 dB
w1 = lower frequency (rads/sec) where the response has fallen by 3 dB
w2 – w1 = 3 dB bandwidth of the circuit
Fig. 5.3 Series circuit
Example 5.1
A series CLR circuit has R = 3 W, L = 20 nH and a resonant frequency ( f0) of 500 MHz.
Estimate (a) its impedance at resonance, (b) the value of the capacitance needed for reso-
nance at 500 MHz, (c) Q of the circuit at resonance, and (d) the 3 dB bandwidth.
Given: R = 3 W, L = 20 nH and f 0 = 500 MHz.
Required: (a) Impedance at resonance, (b) value of series capacitance for resonance at
500 MHz, (c) Q of the circuit at resonance and (d) the 3 dB bandwidth of the circuit.
(a) Using Equation 5.1
Z = R + jwL + (1/( jwC) = 3 + j(X L) – j(XC)
198 Amplifier basics
Since XL = XC at resonance
(b) Using Equation 5.2
ωo = 1 LC = 2π × 500 MHz = (20 nH × C pF) −0.5
C = 5.066 pF
(c) Using Equation 5.4
Q = woL/R or 1/(woCR) = (2p × 500 MHz × 20 nH)/3 = 62.832/3 = 20.944
(d) Using Equation 5.5
Q = wo/(w2 – w1)
Therefore 20.944 = 500 MHz/bandwidth MHz. Hence
3 dB bandwidth ≈ 23.873 MHz
Using PUFF
To find C, we invoke the simple optimiser in PUFF which is called the component sweep.
Instead of sweeping with frequency, a circuit’s scattering parameters may be swept with
respect to a changing component parameter. This feature is invoked by placing a question
mark (?) in front of the parameter to be swept in the appropriate position of a part description
in the F3 box. This is shown in Figure 5.4 where a question mark (?) has been placed in front
Fig. 5.4 Using PUFF to sweep-change the value of C
Tuned circuits 199
Fig. 5.5 Equivalent circuit used by PUFF
of the 10 pF in the F3 box.1 PUFF now knows that we wish to sweep-change the value of C
until we get resonance at the fd frequency which has been set to 0.5 GHz or 500 MHz in the
F4 box.
If you now press the F2 key and press the p key, you will get the plot shown in Figure
5.4. Note that in the third sentence in the F2 box, we have the Part b 5.0625 pF. You will
not get this value initially, because the first displayed value is not in resonance; however,
if while in the p plot mode, you press the page up and/or page down keys, you will find
the value of Part b changing. You will also see the X mark on the rectangular plot move
simultaneously. Keep pressing the page up or page down keys, until S11 shown in the F2
box is reading the largest negative number, in this case, S11 ≈ –67 dB. Now Part b will
show 5.06 pF approximately. This is the value of C which will resonate with the 20 nH to
produce resonance at 500 MHz.
The reason why I have asked you to use the S11 indicator rather than the S21 indicator
is because it is easier to locate the minimum point. This is best explained by Figure 5.5
where I have shown the equivalent circuit used by PUFF. You know that you only get
perfect match (S11 = 0) when Z d (source) is terminated by Zd (load). This will only occur
when j(wL – 1/wC) = 0. Having found the value of C as 5.06 pF, we can now plot the circuit
conventionally and obtain the response of Figure 5.6. To obtain better scaling in Figure 5.6,
I have copied and modified the PUFF set-up template and changed the frequency range
from 400 MHz to 600 MHz, fd to 500 MHz and Zd to 1.5 W so that the total resistance in
the circuit is 1.5 W + 1.5 W (see Figure 5.5), i.e. 3 W.
If we now press the F2 key and p, we will get the response of the Q curve and by using
the page up and page down keys, we can observe that the upper –3 dB point occurs at
approximately 512.5 MHz. The low –3 dB point occurs at 488.5 MHz. Hence the 3 dB
bandwidth is (512.5 – 488.5) MHz = 24 MHz. To sum up:
Item Calculation PUFF
Value of capacitance 5.066 pF 5.0625 pF
3 dB bandwidth 23.873 MHz 24 MHz
You may well ask whether doing it by PUFF is worth the effort when you can obtain
good results by calculation. It is worth it, because PUFF gives you a picture of the circuit
1 Swept lumped components are restricted to single resistors, capacitors or inductors. A description such as
lumped ?1+5j–5j W is not allowed since it is a series CLR circuit. In addition the parallel sign | | cannot be used
in the lumped specification. The unit and prefix given in the part description (following the ?) is inherited by the
component sweep.
200 Amplifier basics
Fig. 5.6 Response of tuned circuit
response. It tells you the amount of rejection (attenuation) for frequencies outside the reso-
nance frequency. It also shows you how the Q must be modified to get the results you want.
Last, but not least, using PUFF gives you confidence for more complicated circuits later in
the book.
5.2.2 Parallel circuits
Similar results can also be obtained either by calculation or by using PUFF for parallel
circuits which are normally used for the load impedance of amplifiers. The fundamental
equations relating to the parallel circuit are:
Z=1 Y (5.6)
Y = G + jwC – j(1 wL) (5.7)
ωo = 1 LC (5.8)
ir G
= (5.9)
is G + jωC − j1 ωL
Q = R wo L or w oCR (5.10)
Q = wo (w 2 – w1) (5.11)
Tuned circuits 201
Z = input impedance with R (ohms), L (henries), C (farads)
Y = input admittance with G (Siemens), L (henries), C (farads)
wo = resonant frequency in radians per second
ir = current across conductance G
is = total current through admittance
Q = quality factor
w2 = upper frequency (rads/sec) where the response has fallen by 3 dB
w1 = lower frequency (rads/sec) where the response has fallen by 3 dB
w 2 – w1 = 3 dB bandwidth of the circuit
Example 5.2
A parallel circuit (Figure 5.7) consists of an inductor of 20 nH, a capacitance of 5.06 pF
and a resistance across the tuned circuit of 2.5 kW. It is driven from a current source. Plot
its frequency response from 400 MHz to 600 MHz.
Fig. 5.7 Parallel tuned circuit
Solution. Using PUFF, we obtain Figure 5.8. In the F3 box, we have used a symbol | |
which signifies that two elements are in parallel and changed the scaled microstrip lines in
the F4 box to the Manhattan mode. The Manhattan mode allows PUFF to carry out calcu-
lations without bothering about the physical size of components. In the F4 box, I have also
set Zd to 5000 W and, bearing in mind Figure 5.7, it is readily seen that this is equivalent
to having a combined resistance of 2.5 kW across the tuned circuit. From Figure 5.8, the
–3 dB bandwidth of the circuit is measured to be (506.5 – 494.0) MHz or 12.5 MHz. The
Q of the circuit is 500 MHz/12.5 MHz = 40.
5.2.3 Cascading of tuned circuits
Most radio frequency systems use a number of tuned circuits in cascade to achieve the
required selectivity (tuning response). One such arrangement commonly used in broadcast
receivers is shown in Figure 5.9. You should note that amplifiers are placed in between the
tuned circuits so that they do not interact directly with each other.
One way would be to derive the response of each individual tuned circuit, then multi-
ply their individual responses to obtain the overall response. Another way would be to
take the individual responses in dB and add them together. I have done this for you. The
202 Amplifier basics
Fig. 5.8 Frequency response of a parallel tuned circuit
calculations are shown in Table 5.1. The results can then be plotted as shown in Figure
5.10. Although there are two tuned circuits each with a Q of 35, I have only plotted one
for the sake of clarity. You should also note that the frequency scale (w/wo) has been plot-
ted linearly this time instead of logarithmic. This is to allow a better view of the response
near the resonant frequency.
Q = 15
Fig. 5.9 Tuned circuit arrangement of a broadcast radio receiver
Filter design 203
Table 5.1 Q factor responses
Fractional frequency (rad/s) Attenuation is given in dB
w/wo Q = 15 Q = 35 Q = 35 Total Q
0.85 –13.98 –21.19 –21.19 –56.4
0.9 –10.42 –17.45 –17.45 –45.3
0.93 –7.595 –14.29 –14.29 –36.2
0.95 –5.276 –11.43 –11.43 –28.1
0.97 –2.637 –7.441 –7.441 –17.5
0.98 –1.359 –4.772 –4.772 –10.9
0.99 –0.378 –1.746 –1.746 –3.9
0.995 –0.097 –0.504 –0.504 –1.1
1.0 0.0 0.0 0.0 0.0
1.005 –0.096 –0.5 –0.5 –1.1
1.01 –0.371 –1.718 –1.718 –3.8
1.015 –0.79 –3.194 –3.194 –7.2
1.02 –1.313 –4.656 –4.656 –10.6
1.05 –4.975 –11.03 –11.03 –27.0
1.08 –8.022 –14.78 –14.78 –37.6
1.11 –10.35 –17.37 –17.37 –45.1
1.13 –11.62 –18.72 –18.72 –49.1
1.15 –12.72 –19.88 –19.88 –52.5
1.18 –14.13 –21.35 –21.35 –56.8
Fig. 5.10 Graph of the r.f. response curve of a broadcast radio receiver
5.3 Filter design
I will now refer you to Figure 5.2 where filters are interspersed between amplifiers to
determine the frequency response of an amplifier block. We will commence by discussing
the main types of filters and later provide details of how these can be designed.
204 Amplifier basics
5.3.1 Introduction
Figure 5.11 shows a multi-element low pass filter which is used at low frequencies. As
frequency increases, the circuit elements C1, L 2, C3, L4 and C5 decrease and at microwave
frequencies these element values become very small. In fact, in many cases, these calcu-
lated values are simply too small in value to implement as lumped elements and transmis-
sion line elements are used to provide the equivalent. Such a microstrip filter is shown in
Figure 5.12.
Comparing the two figures, it can be seen that capacitors are represented by low imped-
ance lines while inductors are represented by high impedance lines. You should not be
surprised by this innovation because in Section 2.13.3 we have already shown you how
transmission lines can be used to construct inductors and capacitors.
Fig. 5.11 Low pass filter
Fig. 5.12 Microstrip low pass filter
An example of how a high pass filter (Figure 5.13) can be implemented in microstrip
line is shown in Figure 5.14. In this case similar microstrip type configurations are used to
represent inductors and capacitors.
Fig. 5.13 High pass filter
Filter design 205
Fig. 5.14 Microstrip high pass filter
Finally in Figures 5.15 and 5.16, we show how coupled filter circuits can be constructed
in microstrip configuration.
Fig. 5.15 Coupled tuned circuits
Fig. 5.16 Microstrip coupled circuits
In general, most microwave filters are first designed as conventional filters and then the
calculated values are translated into microwave elements. In the above figures, we have
only shown microstrip filters but there is no reason why microwave filters cannot be made
in other configurations such as transmission lines and waveguides. Microstrip lines are
more popular because they can be made easily and are relatively cheap.
5.3.2 Overview of filters
In this section we will show you how to select and design various types of multi-element
filters for dedicated purposes. Once these methods have been learnt then it becomes
comparatively easy to design microwave filters. Hence the following sections will concen-
trate on:
• formulating your filter performance requirements;
• deciding which type of filter network you need to meet these requirements;
• calculating or finding out where the normalised element values are published;
• performing simple multiplication and/or division to obtain the component values.
206 Amplifier basics
5.3.3 Specifying filters
The important thing to bear in mind is that although the discussion on filters starts off by
describing low pass filters, we will show you later by examples how easy it is to change a
low pass filter into a high pass, a bandpass or a bandstop filter.
Figure 5.17(a) shows the transmission characteristics of an ideal low pass filter on a
normalised frequency scale, i.e. the frequency variable (f) has been divided by the pass-
band line frequency ( fp). Such an ideal filter cannot, of course, be realised in practice. For
a practical filter, tolerance limits have to be imposed and it may be represented pictorially
as in Figure 5.17(b).
Fig. 5.17 (a) Ideal filter
Fig. 5.17 (b) Practical filter
The frequency spectrum is divided into three parts, first the passband in which the inser-
tion loss (A p) is to be less than a prescribed maximum loss up to a prescribed minimum
frequency ( fp). The second part is the transition limit of the passband frequency limit fp
and a frequency Ws in which the transition band attenuation must be greater than its design
attenuation. The third part is the stopband limit in which the insertion loss or attenuation
is to be greater than a prescribed minimum number of decibels.
Hence, the performance requirement can be specified by five parameters:
• the filter impedance Z 0
• the passband maximum insertion loss (A p)
• the passband frequency limit ( fp)
• the stopband minimum attenuation (A s)
• the lower stopband frequency limit (Ws)
Filter design 207
Table 5.2 Equivalence between reflection coefficient, RLR, Ap and VSWR
Maximum reflection Minimum return loss Maximum passband Zout
coefficient r% ratio RLR(dB) insertion loss VSWR = ——
Ap (dB) Zin
1 40 0.00043 1.020
1.7 35 0.001 1.033
2 34 0.0017 1.041
3 30 0.0043 1.062
4 28 0.007 1.083
5 26 0.01 1.105
8 22 0.028 1.174
10 20 0.043 1.222
15 16 0.1 1.353
20 14 0.18 1.50
25 12 0.28 1.667
33 10 0.5 1.984
45 7 1 2.661
50 6 1.25 3
61 4.3 2 4.12
71 3 3 5.8
Sometimes, manufacturers prefer to specify passband loss in terms of return loss ratio
(RLR) or reflection coefficient (r). We provide Table 5.2 to show you the relationship
between these parameters. If the values that you require are not in the table, then use the
set of formulae we have provided to calculate your own values.
These parameters are inter-related by the following equations, assuming loss-less reac-
RLR = –20 log |r| (5.12)
Ap = 10 log (1 – | r|2) (5.13)
Zout 1 + |r|
VSWR = —— = ———— (5.14)
Z in 1 – | ρ|
5.3.4 Types of filters
There are many types of filter. The more popular ones are:
• Butterworth or maximally flat filter;
• Tchebyscheff (also known as Chebishev) filter;
• Cauer (or elliptical) filter for steeper attenuation slopes;
• Bessel or maximally flat group delay filter.
All of these filters have advantages and disadvantages and the one usually chosen is the
filter type that suits the designer’s needs best. You should bear in mind that each of these
filter types is also available in low pass, high pass, bandpass and stopband configurations.
We will discuss in detail the Butterworth and the Tchebyscheff filters.
208 Amplifier basics
Fig. 5.18 Butterworth filter
5.4 Butterworth filter
Figure 5.18 shows the response of the maximally flat, power law or Butterworth type
which is used when a fairly flat attenuation in the passband is required. The Butterworth
filter achieves the ideal situation only at the ends of the frequency spectrum. At zero
frequency the insertion loss is zero, at low frequencies the attenuation increases very grad-
ually, the curve being virtually flat. With increasing frequency the attenuation rises until it
reaches the prescribed limit.
At the 3 dB frequency there is a point of inflexion, and thereafter the cut-off rate
increases to an asymptotic value of 6n dB/octave, where n is the number of arms. As the
number of arms is increased the approximation to the ideal improves, the passband
response becomes flatter and the transition sharper. For instance, for 3, 5 and 7 arms, the
1 dB loss passband frequencies are 0.8, 0.875 and 0.91 respectively of the 3 dB
frequency, and the corresponding 40 dB frequencies are 4.6, 2.6 and 1.9 times the 3 dB
For values of As greater than 20 dB and RLR greater than 10 dB, you can use Equation
5.15 to calculate the number of arms (n) required in the filter for a given attenuation at a
given frequency:
A s + RLR
n = ———— — (5.15)
[20 log Ws]
Alternatively if you prefer, you can use the ABAC of Table 5.3 to get the same result. To
use the ABAC, simply lay a ruler across any two parameters and read the third parameter.
This is best demonstrated by an example.
Example 5.3
A low pass Butterworth filter is to have a cut-off frequency of 100 MHz. At 260 MHz, the
minimum attenuation in the stopband must be greater than 40 dB. Estimate the number of
arms required for the filter.
Solution. In terms of normalised units (Ws), 260 MHz/100 MHz = 2.6. Using Equation
Butterworth filter 209
As + RLR 40
n= = = 4.82 ≈ 5 arms
20 log Ωs 20 log 2.6
Alternatively, using the ABAC of Table 5.3 and drawing a straight line between 40 on the
left and 2.6 on the right will also give an answer of n < 5 arms.
Table 5.3 ABAC for estimating the number of arms required for a given return loss and attenuation
210 Amplifier basics
5.4.1 Normalised parameters
Each type of filter also has one or more sets of normalised parameters which are used for
calculating its component values. Normalised parameters are values which a low pass filter
would assume for its components if it was designed for (i) 1 W termination and (ii) an oper-
ating angular frequency of 1 rad/s. The reason for choosing 1 W and 1 rad/s is that it
enables easy scaling for different filter impedances and operating frequencies. Another
point you should note about Figure 5.19 is that the same configuration can be used for a
high pass filter by simply interchanging L with C, etc. Bandpass and bandstop filters can
also be produced in this manner. We shall carry out all these manipulations later in the
design examples.
Fig. 5.19 Schematic of a Butterworth normalised low pass filter
The set of normalised parameters for a Butterworth filter can be calculated from
Equation 5.16:
gk = 2 sin
[ (2k – 1)p
2n ] (5.16)
k = position of element in array [k = 1, 2, 3, . . ., n – 1, n]
n = number of elements for the filter
Note: Before you take the sine value, check that your calculator is set to read radians.
To save you the problem of calculating values, we attach a set of values calculated on
a spreadsheet. These are shown in Table 5.4. In this table, n signifies the number of
components that you are going to use in your filter; k signifies the position of the
element. You can get a diagrammatic view of the system by referring to Figure 5.19.
The values in the tables are constants but they represent Henries or Farads according to
Table 5.4 Butterworth normalised values
k/n 2 3 4 5 6 7 8
1 1.4142 1.0000 0.7654 0.6180 0.5176 0.4550 0.3902
2 1.4142 2.0000 1.8478 1.6180 1.4142 1.2470 1.1111
3 1.0000 1.8478 2.0000 1.9319 1.8019 1.6629
4 0.7654 1.6180 1.9319 2.0000 1.9616
5 0.6180 1.4142 1.8019 1.9616
6 0.5176 1.2470 1.6629
7 0.4450 1.1111
8 0.3902
Butterworth filter 211
the configuration in which they are used. This is best demonstrated by using simple
5.4.2 Low pass filter design
A typical procedure for low pass filter design is given below, followed by a design example.
Procedure to design a low pass filter
1 Decide the passband and stopband frequencies.
2 Decide on the stopband attenuation.
3 Decide on the type of filter (Butterworth, etc.) you want to use, bearing in mind that
some types such as the Butterworth filter give better amplitude characteristics while the
Bessel filter gives better group delay.
4 Calculate the number of arms you need to achieve your requirements.
5 Calculate or use normalised tables to find the values of the filter elements. These
normalised values (in farads and henries) are the component values required to make a
low pass filter with an impedance of 1 W and a passband limit frequency ( fp) of one
6 To make a filter having a different impedance, e.g. 50 W, the impedance of each compo-
nent must be increased by the impedance ratio, i.e. all inductances must be multiplied
and all capacitances must be divided by the impedance ratio.
7 To make a filter having a higher band limit than the normalised 1 rad/s, divide the value
of each component by (2p times the frequency).
Example 5.4
A five element maximally flat (Butterworth) low pass filter is to be designed for use in a
50 W circuit. Its 3 dB point is 500 MHz. Calculate its component values.
Given: Five element low pass Butterworth filter, fp = 500 MHz, Z0 = 50 Ω.
Required: Calculation of five elements for a low pass filter.
Solution. The required low pass filter circuit is shown in Figure 5.20. The normalised
values for this filter will be taken from column 5 of Table 5.4 because we want a five
element Butterworth filter.
Fig. 5.20 Low pass configuration
212 Amplifier basics
Table 5.5 shows how the filter design is carried out.
Table 5.5 Calculated values for a low pass Butterworth filter
Circuit reference Normalised Z0 = 50 W Z0 = 50 W
Z0 = 1 W f = 1/(2p) Hz fp = 500 MHz
w = 1 rad/s
0.6180 0.6180
g1 or C1 0.6180 F ——— F ———————— F or 3.93 pF
50 50 × 2p × 500 MHz
1.6180 × 50
g2 or L2 1.6180 H 1.6180 × 50 H —————— H or 25.75 nH
2p × 500 MHz
2.0000 2.0000
g3 or C3 2.0000 F ——— F ———————— F or 12.73 pF
50 50 × 2p × 500 MHz
1.6180 × 50
g4 or L4 1.6180 H 1.6180 × 50 H —————— H or 25.75 nH
2p × 500 MHz
0.6180 0.6180
g5 or C5 0.6180 F ——— F ———————— F or 3.93 pF
50 50 × 2p × 500 MHz
Hence, the calculated values for a five element low pass Butterworth filter with a nomi-
nal impedance of 50 W and a 3 dB cut-off frequency at 500 MHz are:
C1 = 3.93 pF, L2 = 25.75 nH, C3 = 12.73 pF, L4 = 25.75 nH and C5 = 3.93 pF
The response of the filter designed in the above example is shown in Figure 5.21.
Fig. 5.21 Results of Example 5.4
Alternatively to save yourself work, you can use PUFF for the result. This is shown in
Figure 5.22.
Butterworth filter 213
Fig. 5.22 Results of low pass filter using PUFF
5.4.3 Low pass filter example
Low pass filters can also be designed using transmission lines. We will not do it here because
(i) the circuit elements must first be translated into electrical line lengths, and their end
capacitances and discontinuities must be calculated, (ii) the effect on the other components
must be compensated which means altering line lengths again, and (iii) changing line lengths
of each element in turn, then again compensating for the effect of each line length on the
other elements. The whole process is rather laborious and is best done by computer design.
You can find more detail in filter design from two well known books, the first by Matthei,
Young and Jones and the other by T. C. Edwards.2
5.4.4 Low pass filter using microstrip lines
We use PUFF to show an example of a distributed low pass filter.
Example 5.5
Figure 5.23 shows how a low pass filter can be produced using transmission lines of varying
impedances and lengths. You should note that distributed line filters of this type also conduct
2 G. L. Matthei, L. Young and E. M. T. Jones, Microwave filters, impedance matching networks and coupling
structures, McGraw-Hill, New York NY, 1964 and T. C. Edwards, Foundations for microstrip design, second
edition, John Wiley and Sons, 1992.
214 Amplifier basics
Fig. 5.23 Low pass filter construction using microstrip
d.c. and that if you want to block d.c. then you should use d.c. blocking components such as
series capacitors. For details of the filter elements, refer to the F3 box of Figure 5.23.
5.4.5 High pass filter
A typical procedure for high pass filter design is given below. It is immediately followed
by a design example.
Procedure to design a high pass filter
1 Decide the passband and stopband frequencies.
2 Decide on the stopband attenuation.
3 Decide on the type of filter (Butterworth, etc.) you want to use, bearing in mind that
some types such as the Butterworth filter give better amplitude characteristics while the
Bessel filter gives better group delay.
4 Calculate the number of arms you need to achieve your requirements.
5 Calculate or use normalised tables to find the values of the filter elements. These
normalised values (in farads and henries) are the component values required to make a
low pass filter with an impedance of 1 W and a passband limit frequency ( fp) of 1 rad/s.
6 To calculate the corresponding high pass filter, we must (a) replace each capacitor by an
inductor and each inductor by a capacitor and (b) give each component a normalised
value equal to the reciprocal of the normalised component it replaces.
Butterworth filter 215
7 To make a filter having a different impedance, e.g. 50 W, the impedance of each compo-
nent must be increased by the impedance ratio, i.e. all inductances must be multiplied,
all capacitances must be divided by the impedance ratio.
8 To make a filter having a higher band limit than the normalised 1 rad/s, divide the value
of each component by (2p times the frequency).
Example 5.6 High pass filter design
A five element maximally flat (Butterworth) high pass filter is to be designed for use in a
50 W circuit. Its 3 dB point is 500 MHz. Calculate its component values. Hint: note that
this is the high pass equivalent of the low pass filter designed previously.
Given: Five element high pass Butterworth filter, fp = 500 MHz, Z 0 = 50 W.
Required: Calculation of five elements for a high pass filter.
Solution. The circuit is shown in Figure 5.24.
Fig. 5.24 High pass configuration
The normalised values for this filter will be taken from column 5 of Table 5.4 because
we want a five element Butterworth filter. Table 5.6a shows how this is carried out.
Table 5.6a Calculated values for a high pass Butterworth filter
Circuit reference Normalised Z0 = 50 W Z0 = 50 W
Z0 = 1 W f = 1/(2p) Hz fp = 500 MHz
w = 1 rad/s
1 50 50 × 109
g1 or L1 ——— H ——— H —————————— = 25.75 nH
0.6180 0.6180 0.6180 × 2p × 500 × 106
1 1 1 × 1012
g2 or C2 ——— F ————— F ———————————— = 3.93 pF
1.6180 1.6180 × 50 1.618 × 50 × 2p × 500 × 106
1 50 50 × 109
g3 or L3 ——— H ——— H —————————— 7.95 nH
2.0000 2.0000 2.000 × 2p × 500 × 106
1 1 1 × 1012
g4 or C4 ——— F ————— F ———————————— = 3.93 pF
1.6180 1.6180 × 50 1.618 × 50 × 2p × 500 × 106
1 50 50 × 109
g5 or L5 ——— H ——— H —————————— = 25.75 nH
0.6180 0.6180 0.6180 × 2p × 500 × 106
216 Amplifier basics
Hence, the calculated values for a five element high pass Butterworth filter with a nomi-
nal impedance of 50 W and a 3 dB cut-off frequency at 500 MHz are:
L 1 = 25.75 nH, C 2 = 3.93 pF, L 3 = 7.95 nH, C4 = 3.93 pF and L 5 = 25.75 nH
The response of the filter designed in the above example is shown in Figure 5.25.
Fig. 5.25 High pass filter
You can verify this design for yourself by using PUFF. Details of this are given in
Figure 5.26.
Fig. 5.26 High pass filter using PUFF
Butterworth filter 217
5.4.6 High pass filter using microstrip lines
An example of high pass filter design using microstrip lines is given below.
Example 5.7
The construction of a high pass filter using microstrip lines is shown in Figure 5.27.
Details of this filter can be found in the same figure.
Fig. 5.27 Construction of a high pass filter using transmission lines
5.4.7 Bandpass filter
For a bandpass filter, the performance must be specified in terms of bandwidth (see Figure
5.28). The passband limit ( fp ) becomes the difference between the upper frequency limit
( fb ) and the lower frequency limit ( fa ) of the passband, i.e. fp = fb – fa. Similarly the
frequency variable ( f ) becomes the frequency difference between any two points on the
response curve at the same level: and the stopband limit ( fs ) becomes the frequency differ-
ence between the two frequencies ( fx and fy ) outside of which the required minimum stop-
band attenuation (A s ) is to be achieved.
The response curve will have geometric symmetry about the centre frequency ( f0), i.e.
f0 = fa . fb = fx . fy . This means that the cut-off rate in dB/Hz will be greater on the
218 Amplifier basics
Fig. 5.28 Bandpass characteristics
low frequency side and usually the number of arms required in the filter will be dependent
on the cut-off rate of the high frequency side. The normalised stopband limit is given by
Ws = (fs/fp ) = (fy – fx )/( fb – fa )
Procedure to design a bandpass filter
To evaluate a bandpass filter having a passband from fa to fb:
1 define the pass bandwidth fp = fb – fa;
2 calculate the geometric centre frequency f0 = fa fb ;
3 evaluate as previously the lowpass filter having its passband limit frequency equal to fp;
4 add in series with each inductance (L) a capacitance of value (1/(wo2L) and in parallel
with each capacitance (C) an inductance of value (1/(wo2C), i.e. the added component
resonates the original component at the band centre frequency f0.
Example 5.8
A five element maximally flat (Butterworth) bandpass filter is to be designed for use in a
50 W circuit. Its upper passband frequency limit ( fb ) is 525 MHz and its lower passband
frequency limit is 475 MHz. Calculate its component values. Hint: calculate the low pass
filter for the passband design bandwidth, then ‘translate’ the circuit for operation at
f0 = fa × fb .
Given: Five element bandpass Butterworth filter, fp = 50 MHz, Z 0 = 50 W.
Required: Calculation of five elements for a high pass filter.
Solution. The passband filter components are shown in Figure 5.29.
1 Define the passband frequency;
fp = fb – fa = (525 – 475) MHz = 50 MHz
2 Calculate the geometric mean frequency;
f0 = fb × fa = 525 × 475 MHz ≈ 499.4 MHz
Butterworth filter 219
Fig. 5.29 Bandpass configuration
3 Evaluate the low pass filter having its passband limit frequency equal to fp. Use column
5 of Table 5.4.
Table 5.7 Calculated primary values for a bandpass Butterworth filter
Circuit reference Normalised Z0 = 50 W Z0 = 50 W
Z0 = 1 W f = 1/(2p) Hz fp = 50 MHz
w = 1 rad/s
0.6180 0.6180
g1 or C1 0.6180 F ——— F ———————— F or 39.343 pF
50 50 × 2p × 50 MHz
1.6180 × 50
g2 or L2 1.6180 H 1.6180 × 50 H —————— H or 257.513 nH
2p × 50 MHz
2.0000 2.0000
g3 or C3 2.0000 F ——— F ———————— F or 127.324 pF
50 50 × 2p × 50 MHz
1.6180 × 50
g4 or L4 1.6180 H 1.6180 × 50 H —————— H or 257.513 nH
2p × 50 MHz
0.6180 0.6180
g5 or C5 0.6180 F ——— F ———————— F or 39.343 pF
50 50 × 2p × 50 MHz
4 Add in series with each inductance (L) a capacitance of value (1/wo2L) and in parallel
with each capacitance (C) an inductance of value (1/wo2C), i.e. the added component
resonates with the original component at the band centre frequency ( f0). In this case,
f0 = 525 × 475 MHz ≈ 499.4 MHz.
Table 5.7(a) Resonating values of a bandpass Butterworth filter
Low pass values Resonating values for fo.
C1 39.343 pF L1 2.582 nH
L2 257.513 nH C2 0.394 pF
C3 127.324 pF L3 0.797 nH
L4 257.513 nH C4 0.394 pF
C5 39.343 pF L5 2.582 nH
The response of the filter designed in the above example is shown in Figure 5.30.
220 Amplifier basics
Fig. 5.30 Bandpass filter
Alternatively, you can use PUFF to give you the results shown in Figure 5.31.
Fig. 5.31 Using PUFF to plot results
Butterworth filter 221
5.4.8 Bandpass filter design using microstrip lines
Example 5.9
An example of a coupled line passband filter using microstrip is shown in Figure 5.32. A
coupled filter provides d.c. isolation between the input and output ports.
Fig. 5.32 Bandpass filter using coupled microstrip lines
Design modifications
PUFF permits modification and on-screen comparison of different designs. For this
demonstration, we will use Figure 5.32 as a template and produce Figure 5.33. This is
carried out by using Figure 5.32, going into the F3 box to change values and then re-plot-
ting by pressing the F2 key followed by pressing ctrl + p.
Bandpass filter modification
Figure 5.33 shows clearly how comparisons can be made to a design prior to final choice
and fabrication.
222 Amplifier basics
Fig. 5.33 Showing how modifications to a design can be compared to an original design to see if a change is desir-
5.4.9 Bandstop filter
For a bandstop filter, the performance must be specified in terms of bandwidth (see Figure
5.34). The stopband limit ( fp ) becomes the difference between the upper frequency limit
( fb ) and the lower frequency limit (fa ) of the passband, i.e. fp = fb – fa. Similarly the
frequency variable (f ) becomes the frequency difference between any two points on the
response curve at the same level: the stopband limit (fs ) becomes the frequency difference
Fig. 5.34 Stoppass characteristics
Butterworth filter 223
between the two frequencies (fx and fy ) inside of which the required minimum stopband
attenuation (As ) is to be achieved.
The response curve will have geometric symmetry about the centre frequency ( f0 ), i.e.
f0 = fa fb = fx fy . This means that the cut-off rate in dB/Hz will be greater on the low
frequency side and usually the number of arms required in the filter will be dependent on
the cut-off rate of the high frequency side. The normalised stopband limit is given by
Ws = (fs/fp ) = ( fy – fx)/( fb – fa )
Procedure to design a bandstop filter
To evaluate a bandstop filter having a stopband from fa to fb:
1 define the stop bandwidth fp = fb – fa;
2 calculate the geometric centre frequency f0 = fa fb .
3 evaluate as previously the high pass filter having its stopband limit frequency equal to
4 add in series with each inductance (L) a capacitance of value (1/wo2L) and in parallel
with each capacitance (C) an inductance of value (1/wo2C), i.e. the added component
resonates the original at the band centre f0.
Example 5.10
A five element maximally flat (Butterworth) bandstop filter is to be designed for use in a
50 W circuit. Its upper stopband frequency limit ( fb) is 525 MHz and its lower stopband
frequency limit is 475 MHz. Calculate its component values. Hint: calculate the high pass
filter for the stopband design bandwidth, then ‘translate’ the circuit for operation at
f0 = f b f a .
Given: Five element band-stop Butterworth filter, fp = 50 MHz, Z 0 = 50 W.
Required: Calculation of ten elements for a bandstop filter.
Solution. The bandstop filter components are shown in Figure 5.35.
Fig. 5.35 Bandstop configuration
1 Define the stopband frequency:
fp = fb – fa = (525 – 475) MHz = 50 MHz
2 Calculate the geometric mean frequency
f0 = fb × fa = 525 × 475 MHz ≈ 499.4 Mhz
224 Amplifier basics
3 Evaluate the high pass filter having its passband limit frequency equal to fp. The
normalised values have originally been taken from column 5 of Table 5.4 but because
we are evaluating a high pass filter, the reciprocal values have been used.
Table 5.8 Calculated primary values for a bandstop Butterworth filter
Circuit reference Normalised Z0 = 50 W Z0 = 50 W
Z0 = 1 W f = 1/(2p) Hz fp = 50 MHz
w = 1 rad/s
1 50 50 × 109
g1 or L1 ——— H ——— H ————————— = 257.5 nH—
0.6180 0.6180 0.6180 × 2p × 50 × 106
1 1 1 × 1012
g2 or C2 ——— F ————— F ——————————— = 39.3 pF
1.6180 1.6180 × 50 1.618 × 50 × 2p × 50 × 106
1 50 50 × 109
g3 or L3 ——— H ——— H ————————— 79.5 nH —
2.0000 2.0000 2.000 × 2p × 50 × 106
1 1 1 × 1012
g4 or C4 ——— F ————— F ——————————— = 39.3 pF
1.6180 1.6180 × 50 1.618 × 50 × 2p × 50 × 106
1 50 50 × 109
g5 or L5 ——— H ——— H ————————— = 257.5 nH—
0.6180 0.6180 0.6180 × 2p × 50 × 106
4 Add in series with each inductance (L) a capacitance of value 1/(wo2L) and in parallel
with each capacitance (C) an inductance of value 1/(wo2C), i.e. the added component
resonates with the original components at the band centre f0. In this case,
f0 = 525 × 475 MHz ≈ 499.4 MHz.
Table 5.9 Resonating values for a bandstop Butterworth filter
High pass values Resonating values for f0
L1 257.53 nH C1 0.39 pF
C2 39.35 pF L2 2.58 nH
L3 79.58 nH C3 1.27 pF
C4 39.35 pF L4 2.58 nH
L5 257.53 nH C5 0.39 pF
The response of the filter designed in the above example is shown in Figure 5.36.
Alternatively, you can use PUFF to plot the result as shown in Figure 5.37.
Fig. 5.36 Bandstop filter
Tchebyscheff filter 225
Fig. 5.37 Bandstop filter using PUFF
5.5 Tchebyscheff filter
Figure 5.38 shows the response of a filter with a Tchebyscheff or equal ripple type of char-
acteristic. Its response differs from the Butterworth filter in that (i) there is a ripple in the
passband response and (ii) the transition region from passband to stopband is more
pronounced. In short, the filter ‘trades off’ passband ripple (Am) to achieve a greater skirt
loss for the same number of filter components. The penalty for using a Tchebyscheff filter
is that there is also greater group-delay distortion.
The passband approaches the ideal filter at a number of frequencies, between which
the insertion loss is allowed to reach the design limit (Ap). This results in a better
approximation and the transition becomes steeper than that of a Butterworth filter with
the same number of arms, e.g. for a filter with 0.1 dB passband ripple and having 3,
5 or 7 arms, the 1 dB frequency is 0.86, 0.945 or 0.97 respectively times the 3 dB
frequency and the corresponding 40 dB frequency is 3.8, 2.0 or 1.5 times the
3 dB frequency. For As > 20 dB and RLR > 10 dB the number of arms required may be
estimated from Figure 5.39.
Note that Tchebyscheff type filters, having an even number of arms, may have differ-
ent terminal impedances, the ratio of which is a function of Ap, as shown in Figure
5.39. These filters are sometimes designated by ‘b’. Tchebyscheff filters, having an
even number of arms and equal terminating impedances, may be designated by ‘c’.
226 Amplifier basics
Fig. 5.38 Tchebyscheff filter response
Fig. 5.39 Estimate of number of arms for Tchebyscheff filter
Tchebyscheff filter 227
The ‘b’ sub-type filters have a slightly greater cut-off rate than the corresponding ‘c’ sub-
5.5.1 Normalised Tchebyscheff tables
Normalised tables for the Tchebyscheff filter may be calculated from Equations 5.17 to
5.23. These are:
[ ]
b = ln coth ——— (5.17)
where Am = maximum amplitude of passband ripple in dB
g = sinh
[ ]——
where n = total number of arms in the filter
[ ]
(2k – 1)p
ak = sin — —
— — — , k = 1, 2, . . ., n (5.19)
bk = g 2 + sin2
[ ]
—— , k = 1, 2, . . ., n
g = 1 for n odd, g = tanh2 (b/4) for n even (5.21)
g1 = 2a1/g (5.22)
4(ak–1)(ak )
gk = ————— , k = 2, 3, . . ., n (5.23)
As you can see the calculations for normalised Tchebyscheff filters are quite formidable.
To save you time, we provide Tables 5.10–5.12.
Table 5.10 Tchebyscheff normalised values (Am = 0.01 dB)
k\n 2 3 4 5 6 7 8
1 0.4488 0.6291 0.7128 0.7563 0.7813 0.7969 0.8072
2 0.4077 0.9702 1.2003 1.3049 1.3600 1.3924 1.4130
3 0.6291 1.3212 1.5773 1.6896 1.7481 1.7824
4 0.6476 1.3049 1.5350 1.6331 1.6833
5 0.7563 1.4970 1.7481 1.8529
6 0.7098 1.3924 1.6193
7 0.7969 1.5554
8 0.7333
228 Amplifier basics
Table 5.11 Tchebyscheff normalised values (Am = 0.1 dB)
k\n 2 3 4 5 6 7 8
1 0.8430 1.0315 1.1088 1.1468 1.1681 1.1811 1.1897
2 0.6220 1.1474 1.3061 1.3712 1.4039 1.4228 1.4346
3 1.0315 1.7703 1.9750 2.0562 2.0966 2.1199
4 0.8180 1.3712 1.5170 1.5733 1.6010
5 1.1468 1.9029 2.0966 2.1699
6 0.8613 1.4228 1.5640
7 1.1811 1.9444
8 0.8778
Table 5.12 Tchebyscheff normalised values (Am = 0.25 dB)3
k\n 2 3 4 5 6 7 8
1 1.113 1.303 1.378 1.382 1.437 1.447 1.454
2 0.688 1.146 1.269 1.326 1.341 1.356 1.365
3 1.303 2.056 2.209 2.316 2.348 2.367
4 0.851 1.326 1.462 1.469 1.489
5 1.382 2.178 2.348 2.411
6 0.885 1.356 1.462
7 1.447 2.210
8 0.898
5.5.2 Design procedure for Tchebyscheff filters
1 Decide on the number of arms you require to achieve your passband ripple and desired
2 Obtain the normalised values from the Tchebyscheff tables.
3 Follow the same procedures as for the Butterworth design examples for low pass, high-
pass, bandpass and stopband filters.
Example 5.11
A Tchebyscheff 50 W low pass filter is to be designed with its 3 dB cut-off frequency at 50
MHz. The passband ripple is not to exceed 0.1 dB. The filter must offer a minimum of 30
dB attenuation at 100 MHz. Find (a) the number of arms.
Given: 50 W low pass Tchebyscheff filter, passband ripple ≤ 0.1 dB, minimum attenua-
tion at 100 MHz ≥ 30 dB.
Required: (a) Number of arms of low pass filter, (b) component values for filter.
(a) Since 100 MHz/50 MHz = 2 and its reciprocal is 0.5 and assuming a return loss of 20
dB and a passband attenuation of 30 dB from Figure 5.39, it is seen that about five
arms will be required.
(b) Therefore the filter follows the configuration shown below in Figure 5.40a. The perti-
nent values are taken from Table 5.11 and the method of calculation is similar to that
of Example 5.4.
3 Further tables may be obtained from Reference Data for Radio Engineers, International Telephone and
Telegraph Corporation, 320 Park Avenue, New York 22.
Tchebyscheff filter 229
Fig. 5.40a Low pass configuration
Table 5.13 Calculating values for a low pass Tchebyscheff filter
Circuit reference Normalised Z0 = 50 W Z0 = 50 W
Z0 = 1 W f = 1/(2p) Hz fp = 50 MHz
w = 1 rad/s
1.1468 1.1468
g1 or C1 0.1468 F ——— F ———————— F or 73.01 pF
50 50 × 2p × 50 MHz
1.3712 × 50
g2 or L2 1.3712 H 1.3712 × 50 H —————— H or 218.23 nH
2p × 50 MHz
1.9760 1.9760
g3 or C3 1.9760 F ——— F ———————— F or 125.73 pF
50 50 × 2p × 50 MHz
1.3712 × 50
g4 or L4 1.3712 H 1.3712 × 50 H —————— H or 218.23 nH
2p × 50 MHz
1.1468 1.1468
g5 or C5 1.1468 F ——— F ———————— F or 73.01 pF
50 50 × 2p × 50 MHz
Hence the calculated values for a five element low pass Tchebyscheff filter (Am ≤ 0.1
dB) with a nominal impedance of 50 W and a 3 dB cut-off frequency at 50 MHz are:
C1 = 73.01 pF, L2 = 218.23 nH, C3 = 125.73 pF, L4 = 218.23 nH, C5 = 73.01 pF
The response of this filter is shown in Figure 5.40b. It has been obtained by using
PUFF. Notice that there is a ripple in the passband but, because of the scale we have used,
it is not clear. However, you can try this on PUFF and set a small attenuation scale and
see the ripple. Alternatively, you can sweep the frequency and notice the variation in S21
in PUFF.
230 Amplifier basics
Fig. 5.40b Tchebyscheff low pass filter
Example 5.12
A Tchebyscheff 75 W high pass filter is to be designed with its 3 dB cut-off frequency at
500 MHz. The passband ripple is not to exceed 0.25 dB. The filter must offer a minimum
of 30 dB passband attenuation at 250 MHz. Find (a) the number of arms required, and (b)
the component values.
Given: 75 W high pass Tchebyscheff filter, passband ripple ≤ 0.25 dB, minimum attenu-
ation at 250 MHz ≥ 30 dB.
Required: (a) Number of arms of high pass filter, (b) component values for filter.
(a) Since 500 MHz/250 MHz = 2 and its reciprocal is 0.5 and assuming a return loss of 20
dB and a passband attenuation of 30 dB from Figure 5.39, it is seen that about five
arms will be required.
(b) Therefore the filter follows the configuration shown in Figure 5.41.
Fig. 5.41 High pass configuration
Summary on filters 231
The pertinent values are taken from Table 5.12 and the method of calculation is similar
to that of Example 5.4. Table 5.14 shows how this is carried out.
Table 5.14 Calculated values for a low pass Tchebyscheff filter
Circuit reference Normalised Z0 = 75 W Z0 = 75 W
Z0 = 1 W f = 1/(2p) Hz fp = 500 MHz
w = 1 rad/s
1 75 75 × 109
g1 or L1 ——— H ——— H —————————— = 17.27 nH
1.382 1.382 1.382 × 2p × 500 × 106
1 1 1 × 1012
g2 or C2 ——— F ————— F ———————————— = 3.20 pF
1.326 1.326 × 75 1.326 × 75 × 2p × 500 × 106
1 75 75 × 109
g3 or L3 —— —H ——— H —————————— = 10.80 nH
2.209 2.209 2.209 × 2p × 500 × 106
1 1 1 × 1012
g4 or C4 —
— —F ————— F ———————————— = 3.20 pF
1.326 1.326 × 75 1.326 × 75 × 2p × 500 × 106
1 75 75 × 109
g5 or L5 —
— —H ———H —————————— = 17.27 nH
1.382 1.382 1.382 × 2p × 500 × 106
Hence the calculated values for a five element high pass Tchebyscheff filter (Am ≤
0.25 dB) with a nominal impedance of 75 W and a 3 dB cut-off frequency at 500 MHz
L1 = 17.27 nH, C2 = 3.20 pF, L 3 = 10.80 nH, C4 = 3.20 pF, and L5 = 17.27 nH
The response of this filter is shown in Figure 5.42. It has been plotted by using PUFF.
Notice the ripple in the passband.
232 Amplifier basics
Fig. 5.42 Response of high pass filter using PUFF
5.6 Summary on filters
In the previous sections, we have shown you how to synthesise or design low pass, high
pass, bandpass and stopband filters using normalised tables for the Butterworth and
Tchebyscheff type filters. We have assumed that the unloaded Q of the elements are rela-
tively high when compared with the loaded Q of the filter.
The calculations for these design examples have been carried out using a spreadsheet.
There are computer programs available which will compute filter components and in some
cases even produce the microwave circuit layout.
Space limitations prevent us from showing you the synthesis of many other filter types.
However, the design procedures are similar. Many filter designers have produced
normalised tables for various types of filters. If you wish to pursue this topic further, the
classical microwave filter design book is by Matthei, Young and Jones.4 Last but not least,
you should realise that many programs (e.g. SPICE, AppCAD, PUFF) exist to help you
with the calculation and response of these filters.
4 G. Matthei, L. Young and E. Jones, Design of microwave filters, impedance matching networks and
coupling structures, McGraw-Hill, New York NY, 1964.
Impedance matching 233
5.7 Impedance matching
Impedance matching is a vitally important part of amplifier design. There are four main
reasons for impedance matching:
• to match an impedance to the conjugate impedance of a source or load for maximum
power transfer;
• to match an amplifier to a certain load value to provide a required transistor gain;
• to match an amplifier to a load that does not cause transistor instability;
• to provide a load for an oscillator that will cause instability and hence oscillations.
5.7.1 Matching methods
There are many means of impedance matching in h.f. and r.f. design. These include:
• quarter-wave transmission line matching • capacitive matching
• single stub matching • L network matching
• double stub matching • pi network matching
• transformer matching • T network matching
• auto-transformer matching
You have already been shown the first three methods. We will now show you the rest.
5.7.2 Transformer matching
The schematic of a typical i.f. (intermediate frequency) transformer is shown in Figure
5.43. The primary coil (terminals 1 and 2) consists of a number (N1) turns of wire wound
in close magnetic proximity to a secondary winding (terminals 3 and 4) with a number (N2)
turns of wire. Both primary and secondary windings are normally wound on to a coil
former consisting of magnetic material (iron, ferrite or special magnetic compounds).
Voltage V1 is the voltage applied to the primary and V2 is the voltage induced in the
secondary by magnetic action. The currents i1 and i2 represent the currents flowing in the
primary and secondary windings respectively.
Fig. 5.43 Schematic of a typical i.f. transformer
Magnetic flux linkage between primary and secondary is nearly perfect and for practi-
cal purposes the coupling coefficient between primary and secondary coils is unity.5
5 A transformer is said to have a coupling coefficient of 1 if all the flux produced by one winding links with
the other windings.
234 Amplifier basics
Manufacturers tend to quote coupling coefficients greater than 0.95 for transformers used
in 465 kHz i.f. amplifiers. Unless stated otherwise, from now on it will be assumed that the
coupling coefficient is unity.
Operating principles
The operating principles of these transformers can be easily understood by using Michael
Faraday’s law, which states that the voltage (V) induced in a conductor is directly propor-
tional to the rate of change of the effective magnetic flux (∂ø/∂t) across it.
If we define N as the number of turns of the conductor, and ø as the magnetic field, the
induced voltage (V) can be calculated by the following expressions:
V1 = N1 —— (5.24)
V2 = N2 — (5.25)
To calculate the voltage ratio of the transformer we simply divide Equation 5.25 by
Equation 5.24 giving
V2 N2
— =— — (5.26)
V1 N1
V1 N1
— =— — (5.27)
V2 N2
Current ratio
If we define N as the number of turns, i as the current flowing in a circuit and k as a
constant of that circuit, then according to Biot Savart’s law the magnetic field (ø) produced
can be written as:
ø1 = kN1i1 (5.28)
ø2 = kN2i 2 (5.29)
Since the magnetic field is the same for both windings in our transformer, we can
combine Equations 5.28 and 5.29 to give
N1i1 = N2i 2
and by transposing we get
i2 N1
—= —— (5.30)
i1 N2
If we define Z1 = V1/i1 and Z 2 = V2/i 2, we can obtain the input impedance of a trans-
former by dividing Equation 5.26 by Equation 5.30 to give
Impedance matching 235
V2/V1 N2/N1
——— = ———
i2/i1 N1/N2
Transposing the above and substituting for Z1 = V1/i1 and Z 2 = V2 /i2 yields
[ ]
Z 2 = Z1 —— (5.31)
Similarly, by transposing
Z1 = Z 2 — (5.32)
Equations 5.26, 5.30, 5.31 and 5.32 are suitable for use with ‘ideal’ transformers.6
Example 5.12
The primary winding of a two winding transformer is wound with 16 turns while its
secondary has 8 turns. The terminating resistance on the secondary is 16 W. What is its
effective resistance at the primary? Assume that the transformer is ‘ideal’ and that the coef-
ficient of coupling between the primary and the secondary is 1.
Solution. Since the transformer is ‘ideal’ with a coupling coefficient of unity, the answer can
be found by applying Equation 5.32 and remembering that Z p = Z 1 and Z s = Z 2. This gives
[ ] [ ]
N1 16
Zp = Zs —— = 16 × —— = 64 W
N2 8
Finally before leaving the subject of transformers for now, I would like to mention the
5.7.3 Auto-transformers
The auto-transformer is a transformer in which the secondary winding is tapped off the
primary winding. It has the advantage that less copper wire is required for the windings
but suffers from the fact that the primary and secondary windings are not directly isolated.
Apart from constructional details, Equations 5.24–5.32 apply when the auto-transformer
is ideal. The application of transformer action in r.f. design is given in Section 5.7.4.
Fig. 5.44 Auto-transformer
6 An ideal transformer is a transformer which has negligible losses and is one in which all powers coupled
between windings is purely due to the magnetic field.
236 Amplifier basics
5.7.4 Intermediate frequency (i.f.) amplifier with transformers
A schematic of a typical intermediate frequency (i.f.) amplifier is shown in Figure 5.45.
This amplifier is designed to operate efficiently at one frequency. The operational
frequency of the amplifier is determined by the tuned circuit components, C T and L T. CT
represents the total capacitance of the tuned circuit. It includes circuit tuning capacitor,
effective output capacitance of the transistor and all stray capacitances. L T represents the
effective inductance of the circuit. It is mainly due to the primary winding inductance of
T 2. The tapping point on the primary winding is at r.f. earth potential because this point is
effectively short-circuited to earth through the power supply decoupling capacitors. The
position of the tapping point is very important because it determines the working Q and
bandwidth of the amplifier circuit. This will be explained shortly.
Intermediate frequency amplifiers are used very extensively in superhet receivers and,
to keep costs minimal, standardised construction methods are used in the design of their
tuned circuits. Figure 5.46 shows a sectional view of a typical 465 kHz i.f. tuned trans-
former. Tuned transformers are available in two main base sizes, 7 × 7 mm and 10 × 10
mm. The tuning capacitor (180 pF for the 7 mm size and 150 pF for the 10 mm size) is
mounted within the plastic base of the transformer. The primary tuning coil consists of
approximately 200 turns of 0.065 mm diameter wire wound on a ferrite bobbin mounted
Fig. 5.45 Schematic diagram of a typical 465 kHz i.f. amplifier
Fig. 5.46 Sectioned view of a typical 465 kHz i.f. transformer
Impedance matching 237
on a plastic base. This bobbin is shaped like a dumbbell. Unloaded primary coil Qs tend
to be standardised at 70, 100 or 130. The secondary winding consists of about five to
eight turns of wire. Magnetic flux linkage between primary and secondary is nearly
perfect and for practical purposes the coupling coefficient between primary and
secondary coils is unity.7 Manufacturers tend to quote coupling coefficients greater than
0.95. Unless stated otherwise, from now on it will be assumed that the coupling coeffi-
cient is unity.
All coil leads are welded to pins on the base. Welding is used to ensure that coil connec-
tions do not become detached during external soldering operations. Circuit resonance is
adjusted by varying tuning inductance. This is done by altering the position of the ferrite
cap relative to the winding bobbin.
The tuned circuit load impedance presented by transformer T2 to its driving transistor
is set by using the primary winding of T2 as an auto-transformer and by careful choice of
winding ratios between n1, n2 and n3. See Figures 5.45 and 5.47. Three resistances are
reflected into the primary of T2. These are shown in Figure 5.47.
Fig. 5.47 Equivalent tuned circuit of Figure 5.45
The impedance (R ′L) reflected into the primary circuit of T1 by R L in the secondary is
given by
[ ]
n1 + n2
R′L = ———— × RL (5.33)
Rcircuit represents the resistive losses associated with the use of non-perfect capacitors,
inductors and transformers. It is
Rcircuit = Qunloaded woLT or Qunloaded/woCT (5.34)
R tr represents the output resistance of the transistor transformed across the tuned circuit,
Rtr = [(n1 + n2)/n 2]2 × Rtransistor (5.35)
These three resistances in parallel form Reqv . Therefore
Reqv = R′ L //Rcircuit //R′tr (5.36)
The collector load of the transistor is the ratio {n2/(n1 + n 2 )}2 × R′ L //Rcircuit across the
total primary winding. Therefore
7 A transformer is said to have a coupling coefficient of 1 if all the flux produced by one winding links with
the other windings.
238 Amplifier basics
[ ]
Transistor load = ——— × RL //Rcircuit (5.37)
n1 + n2
Example 5.14
In Figure 5.47, n 1 = 160, n 2 = 40, n3 = 8 and RL = 2 kW. The resistive losses associated with
the tuning capacitor and inductors can be assumed to be negligible and the transistor output
resistance reflected across the primary of the tuned circuit is so large that it can be neglected.
The magnetic coupling coefficients between coils may be assumed to be unity. (a) What is
the transistor load impedance at resonance? (b) If the tapping point on L T is changed so that
n 1 = 150 and n 2 = 50, what is the new transistor load impedance at resonance?
(a) At resonance, the tuned circuit impedance is very high when compared to the reflected
load R′L from the secondary. Using Equation 5.33
n 1+ n 2 2 160 + 40 2
R′L = ———
n3 ] [
— × R L = ———— × 2000 = 1 250 000 W
8 ]
Using Equation 5.37 and noting in this case that Reqv = R′L, the transistor load impedance
[ ] [ ]
n2 40
——— × R′L = ———— × 1 250 000 = 50 kW
n1 + n 2 160 + 40
(b) When n1 = 150 and n2 = 50, using Equation 5.33
[ ] [ ]
n1 + n2 150 + 50
R′L = ——— — × R L = ———— × 2000 = 1 250 000 W
n3 8
Using Equation 5.33 and noting in this case that R eqv = R′L, the transistor load impedance
[ ] [ ]
n2 50
——— × R′L = ———— × 1 250 000 = 78.5 kW
n1 + n2 150 + 50
Example 5.14 shows clearly that different collector load impedances can be obtained from
a standard tuned i.f. transformer simply by altering the tapping point on the primary coil.
For clarity in understanding the previous example, it was assumed that tuned circuits
losses (Rcircuit) were negligible and that the reflected output resistance of the transistor (R′tr)
was so high that it did not affect the value of R eqv. In practice, the additional resistive losses
across the tuned circuit are not negligible and must be taken into account in designing the
amplifier. The effect of these additional losses is demonstrated in Example 5.14.
Example 5.15
In Figure 5.45, n1 = 160, n2 = 40, n3 = 8 and R L = 2 kW. The unloaded Q of the tuned
circuit is 100. The value of the tuning capacitor is 180 pF and the circuit is resonant at 465
Impedance matching 239
kHz. If the coupling coefficient between coils is unity, what is the transistor load imped-
ance at resonance? Assume the output impedance of the transistor to be 100 kW.
Solution. At resonance, the effective tuned circuit impedance is the parallel value of the
reflected load (R′L) from the secondary, the equivalent loss resistance of the tuned circuit
(Rcircuit) and the reflected output resistance of the transistor (Rtr). Using Equation 5.33, the
reflected load
[ ] [ ]
n1 + n2 160 + 40
R′L = ——— × R L = ———— × 2000 = 1.25 MW
n3 8
The effective loss resistance of the tuned circuit is obtained by using Equation 5.34:
R circuit = Q/(woCT) = 100 /(2p × 465 × 103 × 180 × 10–12) = 190.143 kW
Using Equation 5.35
[ ]
[ ]
n1 + n2 160 + 40 2
R tr = ——— × R transistor = ———— × 100 000 = 2.5 MW
n2 40
Using Equation 5.37, the transistor load impedance is
[ ] [ ]
n2 40
——— × R′L//R circuit = ———— × 154.8 kW ≈ 6.19 kW
n1 + n2 160 + 40
The answer to Example 5.15 clearly indicates that the unloaded Q of the tuned circuit
affects the transistor collector load and that the transformation ratio must be changed if the
original load impedance of 50 kW is desired. Another interesting point is that the loaded Q
of the circuit has also fallen drastically because the effective resistance (R eqv) across the
tuned circuit has been reduced.
Example 5.16
The unloaded Q of a tuned circuit is 100. The value of its tuning inductance is 650 mH and
the circuit is resonant at 465 kHz. When the circuit is loaded by the input resistance of a
transistor, the effective resistance across the ends of the tuned circuit is 125 kW. What is
its loaded Q and 3 dB bandwidth?
Solution. Using Equation 5.10
R eqv = Qloaded woL
R eqv 125 000
Q loaded = ——— = ——————————— = 65.8
ωoL 2π × 465 000 × 650 × 10–6
From Equation 5.11
Qloaded = —————
240 Amplifier basics
f0 465 000
bandwidth = ——— = ———— = 7066 Hz
Qloaded 65.8
Examples 5.15 and 5.16 have brought out a very important point. It is that transistor
collector load impedance and circuit bandwidth can be altered independently (within
limits) by careful choice of the ratios between n1, n 2, and n 3. This independence of the two
is possible even when standard i.f. transformers are used.
5.7.5 Capacitive matching
For tuned circuits at the higher frequencies, smaller inductance values are required.
Smaller inductance values mean fewer turns and trying to make impedance transformers
with the correct transformation ratio is difficult. In cases like that, it is often more conve-
nient to use capacitors as the matching element. One such arrangement is shown in Figure
Fig. 5.48 Capacitive divider method
Capacitive divider matching
Capacitive divider matching (Figure 5.48) is particularly useful at very high frequencies
(VHF) where the number of turns on an inductor is small and where location of a connec-
tion to produce an auto-transformer from the inductor is not practical. The capacitive
matching network of Figure 5.48 can be easily obtained by defining
Z1 = V1/I1 and Z2 = V2/I2
For ease of analysis, we shall assume that XL (Xc1 + Xc2), and that Z 2 is not loading the
circuit. Then by inspection
V1 V2
I1 = ———— and I2 = ——
Xc1 + Xc2 Xc2
I2 V2 Xc1 + Xc2
—— = —— × ————
I1 Xc2 V1
Impedance matching 241
and transposing
V1 V2 Xc1 + Xc2
—— = —— × ————
I1 I2 Xc2
and substituting for Z 1, Z 2 and reactances
[ ]
1/jwC1 + 1/jwC 2
Z1 = Z 2 ———————
1/jwC 2
and multiplying all terms by jwC1C 2
[ ]
C1 + C 2
Z1 = Z 2 ———— (5.38)
Example 5.17
In the circuit shown in Figure 5.48, C1 = 10 pF, C 2 = 100 pF and Z 1 = 22 kW. If the reac-
tance of the inductor is very much greater than the combined series reactance of C1 and
C 2, calculate the transformed value shown as Z 2.
Solution. Using equation 5.38:
[ ] [ ]
C1 10
Z 2 = Z 1 ———— = 22 000 ———— = 2 kW
C1 + C 2 10 + 100
5.7.6 Impedance matching using circuit configurations
Many circuit configurations can be used as matching networks. In Figures 5.49, 5.50 and
5.51 we show three circuits which are often used for impedance matching. In Figure 5.49,
Z1 and Z 2 are used to match the load Z L to the source Z s. In Figure 5.50, Z a and part of Z b
are used for matching to Z1, while the remaining half of Z b and Z c are used for matching
to Z 2. In Figure 5.51, Z a and part of Z b are used for matching to Z1 while the remaining
part of Z b and Z c are used to match to Z 2.
Fig. 5.49 Matching L network
242 Amplifier basics
Fig. 5.50 Matching π network
Fig. 5.51 Matching T network
The details of how these circuits can be used to provide matching will be explained
shortly but for ease of understanding, we shall first review some fundamental concepts on
the series, parallel and Q equivalents of components.
5.7.7 Series and parallel equivalents and quality factor of components
Before we can commence network matching methods, it is best to revise some fundamen-
tal concepts on the representation of capacitors and inductors.
Series and parallel forms and Qs of capacitors
Capacitors are not perfect because their conducting plates contain resistance and their
dielectric materials are not perfect insulators. The combined losses can be taken into
account by an equivalent series resistance (R s ) in the series case or by an equivalent paral-
lel resistance (R p) in the parallel case as shown in Figure 5.52. In this diagram, R s and Cs
represent the series equivalent circuit of the capacitor while R p and Cp represent the paral-
lel equivalent circuit. The parallel representation is preferred when dealing with circuits
where elements are connected in parallel.
Quality factor (Q) of a capacitor. We will follow normal convention and define the series
quality factor (Qs ) as
reactance 1/wCs 1
Qs = ————— = ——— = ——— (5.39)
resistance Rs wCsR s
w = angular frequency in radians
Cs = capacitance in Farads
R s = equivalent series resistance (ESR) of a capacitor in ohms
Impedance matching 243
Fig. 5.52 Series and parallel form of a ‘practical’ capacitor
Similarly, we will define the parallel quality factor (Qp) as
susceptance ωCp
Qp = —————— = —— = ωCpR p
— (5.40)
conductance 1/R p
ω = angular frequency in radians
Cp = capacitance in Farads
R p = equivalent parallel resistance (EPR) of a capacitor in ohms
Example 5.18
A capacitor has an equivalent parallel resistance of 15 000 W and a capacity of 100 pF.
Calculate its quality factor (Qp) at 100 MHz.
Solution. Using Equation 5.40:
Qp = wCpR p = 2p × 100 MHz × 100 pF × 15 000 = 942.48 ≈ 943
Equivalence of the series and parallel representations. The purpose of this section is to
show the relationships between series and parallel circuit representations. Using the same
symbols as before and referring to Figure 5.52
{ }
1 1 R s – 1/ jwCs
Y = ——————— = ————— × —————
impedance (Z) R s + 1/jwCs R s – 1/jwCs
Rs 1/jwCs
G + jB = ————— – ——————
R 2 + ———
s R 2 + ———
w 2C 2
s w 2C 2
Using Equation 5.39:
R s/R 2
s Qs /R s
G + jB = ———— + j ————
1 + Q2 s 1 + Q2 s
244 Amplifier basics
1 Qs /R s
G + jB = ————— + j ———— (5.41)
R s(1 + Q 2)
s 1 + Q2 s
Equating ‘real’ parts of Equation 5.41:
G = —— = —————
Rp R s(1 + Q 2)
R p = R s(1 + Q 2 )
s (5.42)
Qs = —— – 1 (5.43)
Equation 5.43 is used extensively in the matching of L networks which will be discussed
If Q > 10, then from Equation 5.42
R p ≈ R sQ 2
s (5.44)
Equating imaginary parts of Equation 5.41
Qs /Rs
B = ωCp = ———
1 + Q2 s
and using Equation 5.42 gives
wCp = ————————
———— (1 + Q 2)
(1 + Q 2)
Transposing R p and using Equation 5.40 for Qp
wCpR p = Qp = Qs (5.45)
Substituting Equation 5.39 for Qs in Equation 5.45 and transposing w and R p
1 Cs
Cp = ————— × ——
w 2Cs Rs Rp Cs
and substituting Equation 5.42 for R p
Q 2Cs
Cp = ———— (5.46)
(1 + Q 2)
Impedance matching 245
If Qs > 10, then from Equation 5.46
Cp ≈ —— Cs ≈ Cs (5.47)
Series and parallel forms and Qs of inductors
Quality factor (Qs) of an inductor. The ‘goodness’ or quality factor (Qs ) of an inductor
is defined as:
Qs = —— (5.48)
w = angular frequency in radians
Ls = inductance in Henries
R s = series resistance of inductor in ohms
Example 5.19
An inductor has a series resistance of 8 W and an inductance of 365 mH. Calculate its qual-
ity factor (Q) at 800 kHz.
Solution. Using Equation 5.48
wLs 2p × 800 × 103 × 365 × 10–6
Qs = —— = ———————————— ≈ 229
Rs 8
Equivalence of the series and parallel representations of inductors. Sometimes, it is
more convenient to represent the quality factor (Qp) of an inductor in its parallel form as
shown in Figure 5.53. In this diagram, Rs and L s represent the series equivalent circuit,
while Rp and L p represent the parallel equivalent circuit.
The equivalent values can be calculated by taking the admittance form of the series circuit:
{ }
1 1 Rs – jwLs
Y = —————— = ———— × —————
impedance (Z) Rs + jwLs Rs – jwL s
Fig. 5.53 Series and parallel form of an inductor
246 Amplifier basics
Rs jwLs
G – jBL = ———— – ———— —
R 2 + w 2L 2 R 2 + w 2L 2
s s s s
Using Equation 5.48
Rs/R 2
s Qs/Rs
G – jBL = ——— – j ———
1 + Q2 s 1 + Q2s
1 1 + Q2 s
R p = — = ——— = Rs(1 + Q 2)
s (5.49)
G 1/Rs
Qs = —— – 1 (5.50)
Equation 5.50 is used extensively in the matching of L networks which will be discussed
If Q > 10, then from Equation 5.49
R p ≈ Rs Q 2
s (5.51)
–j –jwLs
–jBL = —— = ———— —
wL p R 2 + w 2L2
s s
Transposing and dividing by w
R 2 + w 2L2
s s
Lp = ———— —
w s
1 + Q2 1 + Q2
= ——— =
Qs s
{ }
Ls (5.52)
If Qs > 10, then from Equation 5.52
L p ≈ —— Ls ≈ Ls (5.53)
Finally, defining the parallel equivalent quality factor (Qp) = susceptance/conductance and
using Equations 5.49 and 5.52, it can be shown that Qp = Qs:
Rp Rs (1 + Qs ) Q2
Qp = = = s = Qs
ωLp ⎧1 + Q 2 ⎫
⎪ ⎪ Qs (5.54)
ω ⎨ 2 s ⎬ Ls
⎪ Qs ⎪
⎩ ⎭
Impedance matching 247
Example 5.20
The inductor shown in Figure 5.53 has a series resistance (R s) of 2 W and inductance (Ls)
of 15 mH. Calculate its equivalent parallel resistance (R p), parallel inductance (L p) and its
equivalent quality factor (Qp), at 10 MHz.
Solution. From Equation 5.48
wLs 2p × 10 × 106 × 15 × 10–6 942.5
Qs = —— = ——————————— = ——— ≈ 471
Rs 2 2
From Equation 5.49
R p = R s(1 + Q 2) = 2 × (1 + 4712) ≈ 444 kW
From Equation 5.52
1 + Q2 1 + 4712
Lp =
{ }———
Ls = 15 × 10–6
{ }
———— ≈ 15 µH
From Equation 5.54
Qp = Qs = 471
Note: Since Qs is very high in this case, the approximate formulae in Equations 5.51 and
5.53 could have been used.
5.7.8 L matching network
L matching networks are frequently used to match one impedance to another. They are
called L matching networks because the two reactances used (Xa and Xb) are arranged in
the form of a letter L.
Figure 5.54 shows a common L type network used when it is desired to match the input
impedance of a transistor to a network. In this circuit, Cin and R in represent the input
impedance of the transistor. C2 is put in parallel with Cin to form a total capacitance Ct;
therefore Xb = 1/jwCt. Xa = 1/jwC1. Z s is the input impedance looking into the circuit. It
follows that
Fig. 5.54 Matching L network
248 Amplifier basics
1 Rin(1/jwCt )
Z s = ——— + ——————
jwC1 R in + (1/jwCt )
Multiplying the numerator and denominator of the second term by jwCt and normalising
[ ]
1 R in 1 – jwCt R in
Z s = —— + ———— × —————
jwC1 1 + jwCt R in 1 – jwCt R in
which when multiplied out results in
wCt R in2
[ ]
R in 1
Z s = —————— – j —— + ————— — (5.55)
1 + (wCt R in)2 wC1 1 + (wCt R in)2
The first term is real and shows that the resistance R in has been transformed. The reactive
term is incorporated into the tuning capacitance of the input tuned circuit. In this particu-
lar case, Xa and Xb have turned out to be capacitive, but this is not always the case. In some
cases you may find that Xa and/or Xb may be inductive.
In Figure 5.55, we wish to use an L network to match a load resistance (R L) of 500 W to a
generator resistance (R g) of 50 W.
1 We begin by deciding on the type of reactance (inductive or capacitive) we would like
to use for X p.
2 We then calculate the equivalent series combination of series resistance (Rs) and series
reactance (Xs) from the parallel combination of X p and (R p = R L) . You already know
how to do this. See Figure 5.54 and Equations 5.49 to 5.54. However, here we must
make the transformed R s = R generator = R g to ensure matching conditions for maximum
power transfer from the generator to the transformed load.
3 We evaluate Xa so that it cancels out the transformed reactance (Xs) in the circuit.
4 Then we calculate the value Xb.
An example will help to clarify the method.
Fig. 5.55 Using a L network for matching
Impedance matching 249
Example 5.21
Calculate a matching network for the circuit shown in Figure 5.55.
Solution. We begin by choosing X p to be inductive. Next, we use Equation 5.50 which is
repeated below for convenience:
Qs = Rp Rs − 1
Note in this case that R p is the load resistance (R L) of 500 W and that we want R s = 50 W
to match the generator resistance (R g) of 50 W. Substituting these values, we obtain,
Qs = 500 50 − 1 = 3
Using Equation 5.48, and again remembering that R g = R s, we have
Qs = wLs/Rs
and transposing yields
R sQ s 50 × 3
L s = —— = —————— = 239 nH
w 2p × 100 MHz
Xs = wL s = 2p × 100 MHz × 239 nH = 150 W
Xa must be chosen to cancel out this inductive reactance of 150 W so X a must be a capac-
itor whose reactance X a = 150 W = 1/wCa. Therefore at 100 MHz
Ca = 1/(2p × 100 MHz × 150) = 10.6 pF
All that remains now is to calculate the parallel value of L p. For this we make use of
Equation 5.52, where
1 + Q2 1 + 32
Lp =
{ ———
} {
Ls = ———
} 239 nH = 265 nH
The L network is now complete and its values are shown in Figure 5.56.
Using PUFF software to check Example 5.21
The results are shown in Figure 5.57. Note that since the matching circuit has been
designed to match a 50 W generator, it follows that S11 will be zero when the transformed
network matches the 50 W generator at the match frequency.
Fig. 5.56 Completed L matching network
250 Amplifier basics
Fig. 5.57 PUFF plot of Example 5.21 matching network
In the real world, R s and/or R L has associated reactances. Suppose R L is 500 W with a
shunt capacitance of 42 pF. How do we solve this problem? This is best explained by using
another example.
Example 5.22
Calculate a matching network for the circuit shown in Figure 5.58.
Solution. This is similar to Example 5.20, except that this time the load resistance (R L) is
shunted by a capacitance of 42 pF (see Figure 5.58). The solution proceeds as follows:
1 Introduce an inductance (L) to negate the effect of Cshunt at the frequency of operation,
that is, choose L so that XL = XC (see Figure 5.59).
Fig. 5.58 L matching network
Impedance matching 251
Fig. 5.59 Using an inductor of 60 nH to resonate with the 42 pF
2 Solve for Xa and Xp as in Example 5.21.
3 Solve for the combined value of the two shunt inductors shown in Figure 5.60.
Fig. 5.60 The intermediate L network
The required value of L is
L = ——— —
= ————————— = 60.3 nH ≈ 60 nH
[2p(100 MHz)]2(42 pF)
See Figure 5.59.
4 Calculate the network as in example 5.21. I shall take the values directly from it. See
Figure 5.60.
5 Solve for the combined value of the equivalent inductor:
Lcombined = ———— ≈ 49 nH
265 + 60
See Figure 5.61 for the final network.
Fig. 5.61 Completed L network
252 Amplifier basics
In Example 5.22, Xa has been chosen to be capacitive while Xp has been chosen to be
inductive. The network could equally have been designed with Xa inductive and Xp capac-
itive. In this case, the shunt capacitance of the load should be subtracted from the calcu-
lated value of the capacitance forming Xp. For example, if the calculated value of the
matching shunt capacitance is 500 pF, then all you need do is subtract the load shunt
capacitance (42 pF) from the calculated shunt value (500 pF) and use a value of (500 – 42)
pF or 458 pF for the total shunt capacitance and use this value to calculate Xa.
Using PUFF software to check Example 5.22
The results are shown in Figure 5.62. Note that since the matching circuit has been
designed to match to a 50 W source, it follows that S11 will be zero at the match
Fig. 5.62 PUFF plot of Example 5.21 matching network
5.8 Three element matching networks
Three or more element matching networks are used when we wish to match and also
control the Q of a circuit. If you examine Equation 5.50 which is repeated for convenience,
you will see that when R p and Rs are fixed you are forced to accept the value of Q calcu-
lated. However, if you can vary either Rp or Rs then you are in a position to set Qs. Using
Equation 5.50 again
Three element matching networks 253
Qs = —— – 1 (5.50)
Varying either R s or R p will no doubt cause a mismatch with your original matching aims.
However, you can overcome this by using a second matching L network to match your
design back to the source or load.
If you examine the p or T network (Figures 5.63 and 5.71) you will see that these
networks are made up from two L type networks. Therefore, it is possible to choose
(within limits) the Q of the first L network and match it to a virtual value R v then use the
second L network to match R v to the load R L.
5.8.1 The p network
The p network (Figure 5.63) can be described as two ‘back to back’ L networks (Figure
5.64) that are both configured to match the source and the load to a virtual resistance R v
located at the junction between the two networks.
More details are provided in Figure 5.65. The significance of the negative signs for –Xs1
and –Xs2 is symbolic. They are used merely to indicate that the Xs values are the opposite
type of reactance from Xp1 and Xp2 respectively. Thus, if Xp1 is a capacitor, Xs1 must be an
inductor and vice-versa. Similarly if Xp2 is an inductor, Xs2 must be a capacitor, and vice-
versa. They do not indicate negative reactances (capacitors).
The design of each section of the p network proceeds exactly as was done for the L
networks in the previous section. The virtual impedance or resistance Rv in Figure 5.65
must be smaller than either Z1 or Z 2 because it is connected to the series arm of each L
section, but otherwise it can be any value of your choice. Most of the time, Rv is deter-
mined by the desired loaded Q of the circuit that you specify at the beginning of the design
Fig. 5.63 π network
Fig. 5.64 π network made up from two L networks
254 Amplifier basics
For our purposes, the loaded Q of the network will be defined as:
Q = ( Rh Rv ) − 1 (5.56)
Rh = the largest terminating value of Z1 or Z 2
Rv = virtual impedance or resistance
Although not entirely accurate, it is a widely accepted Q-determining formula for this
circuit, and it is certainly close enough for most practical work. Example 5.23 illustrates
the procedure.
Example 5.23
A source impedance of (100 + j0) W is to be matched to a load impedance of
(1000 + j0) Ω. Design four p networks with a minimum Q of 15 to match the source and
load impedances.
Given: R s = 100 W, R L = 1000 W, Q = 15.
Required: To design four types of p matching networks.
Solution. Take the output L network on the load side of the p network. From Equation
5.56, we can find the virtual resistance (R v) that we will be matching:
Rh 1000
R v = ——— = ——— = 4.425 W
Q 2+ 1 152 + 1
To find X p2 we use Equation 5.10:
Rp RL 1000
Xp2 = —— = —— = ——— = 66.667 W
Qp Qp 15
Similarly to find Xs2, we use Equation 5.4:
Xs2 = Q × R series = 15(R v) = (15) (4.425) = 66.375 W
This completes the design of the L section on the load side of the network. Note that Rseries
in the above equation was substituted for the virtual resistor (R v) which by definition is in
the series arms of the L section.
The Q for the input (source) L section network is defined by the ratio of Rs to R v, as per
Equation 5.56, where:
Rs 100
Q1 = −1 = − 1 = 4.647
Rv 4.425
Notice here that the source resistor is now considered to be in the shunt leg of the L
network. Therefore R s is defined as R p, and using Equation 5.10
Rp 100
Xp1 = —— = ——— = 21.612 W
Q1 4.627
Three element matching networks 255
Fig. 5.65 Practical details of a π network
Similarly, using Equation 5.4:
Xs1 = Q1R series = (4.647)(4.425) = 20.563 W
The actual network is now complete and is shown in Figure 5.65. Remember that the
virtual resistor (R) is not really in the circuit and therefore is not shown. Reactances –Xs1
and –Xs2 are now in series and can simply be added together to form a single component.
So far in this design, we have dealt only with reactances and have not yet computed
actual component values. This is because of the need to maintain a general design
approach so that the four final networks requested in the example can be generated
Note that Xp1, Xs1, Xp2 and Xs2 can all be either capacitive or inductive reactances. The
only constraint is that Xp1 and Xs1 are of opposite types, and Xp2 and Xs2 are of opposite types.
This yields the four networks shown in Figures 5.66 to 5.69. Note that both the source
and load have been omitted in these figures. Each component in Figures 5.66 to 5.69 is
shown as a reactance in ohms. Therefore to perform the transformation from dual L to p
network, the two series components are merely added if they are the same type, and
subtracted if the reactances are of opposite type. The final step is to change each reac-
tance into a component value of capacitance and inductance at the frequency of opera-
Fig. 5.66 Matching π network used as a low pass filter
Fig. 5.67 Matching π network used as a high pass filter
256 Amplifier basics
Fig. 5.68 Matching π network using inductors
Fig. 5.69 Matching π network using capacitors
Using PUFF software to check Example 5.23
The results are shown in Figure 5.70. For convenience, all four networks have been plot-
ted. S11, S22, S33 and S44 are the results of Figures 5.66, 5.67, 5.68 and 5.69 respectively.
Note that each of the networks produce the desired input impedance of approximately 100
W.1 You can check this in the Message box of Figure 5.70.
Fig. 5.70 PUFF verification of Example 5.23
1 In this particular case, I have carried out the solution at 500 MHz but the components can be selected for oper-
ation at other frequencies.
Three element matching networks 257
5.8.2 The T network
The T network is often used to match two low impedance values when a high Q arrange-
ment is needed. The design of the T network is similar to the design for the p network
except that with the T network, you match the source and the load with two L type
networks to a virtual resistance (R v) which is larger than either of the load or source resis-
tance. This means that the two L type networks will then have their shunt arms connected
together as in Figure 5.71.
Fig. 5.71 T network shown as two back-to-back L networks
As mentioned earlier, the T network is often used to match two low-valued impedances
when a high Q arrangement is desired. The loaded Q of the T network is determined
mainly by the L section that has the highest Q. By definition, the L section with the high-
est Q will occur at the end which has the smallest terminating resistor. Each terminating
resistor is in the series leg of each network. Therefore the formula for determining the
loaded Q of the T network is
Q= −1 (5.57)
Rv = virtual resistance
R small = the smallest terminating resistance
The above expression is similar to the Q formula that was previously given for the p
network. However, since we have reversed the L sections to produce the T network, we
must ensure that we redefine the Q expression to account for the new resistor placement
in relation to those L networks. In other words, Equations 5.56 and 5.57 are only special
applications for the general formula that is given in Equation 5.50 and repeated here for
Qs = −1 (5.58)
R p = resistance in the shunt arm of the L network
R s = resistance in series arm of the L network
Do not be confused with the different definitions of Q because they are all the same in this
case. Each L network is calculated in exactly the same manner as was given for the p
network previously. We will now show this with Example 5.24.
258 Amplifier basics
Example 5.24
Using the configuration shown in Figure 5.71 as a reference, design four different
networks to match a 10 W source to a 50 W load. Design each network for a loaded Q of
Solution. Using Equation 5.57, we can find the required R v for the match for the required
R v = R small(Q 2 + 1) = 10(102 + 1) = 1010 W
Using Equation 5.4
Xs1 = QR s = 10(10) = 100 W
Using Equation 5.10,
Xp2 = R/Q = 1010/10 = 101 W
Now for the L network on the load end, the Q is defined by the virtual resistor and the load
resistor. Thus
Q2 = R RL − 1 = 1010 50 – 1 = 4.382 Ω
Xp2 = R/Q 2 = 1010/4.382 = 230.488 W
Xs2 = Q sR L = (4.382)(50) = 219.100 W
The network is now complete and is shown in Figure 5.72 without the virtual resistor. The
two shunt reactances of Figure 5.72 can again be combined to form a single element by
simply substituting a value that is equal to the combined parallel reactance of the two.
Fig. 5.72 Calculated values for the general T network
The four possible T type networks that can be used are shown in Figures 5.73 to 5.76.
Fig. 5.73 Low pass T configuration
Broadband matching networks 259
Fig. 5.74 High pass T configuration
Fig. 5.75 Inductive matched T section
Fig. 5.76 Capacitive matched T section
Using PUFF software to check Example 5.24
Fig. 5.77 PUFF verification of Example 5.23
260 Amplifier basics
Using PUFF software to check Example 5.24
The results are shown in Figure 5.77. For convenience, all four networks have been plot-
ted. S11, S22, S33 and S44 are the results of Figures 5.73, 5.74, 5.75 and 5.76 respectively.
Note that each of the networks produces the desired input impedance of approximately 10
W. You can read this in the Message box of Figure 5.77.
5.9 Broadband matching networks
With regard to the L network, we have noted that the circuit Q is automatically defined
when the source and load are selected. With the p and T networks, we can choose a circuit
Q provided that the Q chosen is larger than that which is available with the L network. This
indicates that the p and T networks are useful for narrow band matching. However, to
provide a broadband match, we use two L sections in still another configuration. This is
shown in Figures 5.78 and 5.79 where R v is in the shunt arm of one L section and in the
series arm of the other L section. We therefore have two series-connected L sections rather
than the back-to-back connection of the p and T networks. In this new configuration, the
value of R v must be larger than the smallest termination impedance but also smaller than
the largest termination impedance. The net result is a range of loaded Q values that is less
than the range of Q values obtainable from either a single L section, or the p or T networks
previously described.
The maximum bandwidth (minimum Q) available from the networks of Figures 5.78
and 5.79 occurs when R v is made equal to the geometric mean of the two impedances
being matched:
Rv = Rs RL (5.58)
The loaded Q of the above networks is defined as:
Rv Rlarger
Q= −1 = −1 (5.59)
Rsmaller Rv
Fig. 5.78 Series connected L networks for lower Q applications. Rv is shunt leg
Fig. 5.79 Series connected L networks for lower Q applications. R is series leg
Summary of matching networks 261
Fig. 5.80 Expanded version of Figure 5.80 for even wider bandwidth
Rv = the virtual resistance
R smaller = smallest terminating resistance
R larger = largest terminating resistance
For wider bandwidths, more L networks may be cascaded with virtual resistances
between each network as shown in Figure 5.80.
Optimum bandwidths in these cases are obtained if the ratios of each of the two
succeeding resistances are equal:
R v1 R v2 R v3 R larger
——— = —— = —— = ——— (5.60)
R smaller R v1 R v2 . . . Rn
R smaller = smallest terminating resistance
R larger = largest terminating resistance
R v1, R v2, . . ., R vn = the virtual resistances
The design procedure for these wideband matching networks is much the same as was
given for the previous examples. For the configurations of Figures 5.78 and 5.79, use
Equation 5.58 to solve for R v to design for an optimally wideband. For the configurations
of Figures 5.78 and 5.79, use Equation 5.58 to solve for R v to design for a specific low Q.
For the configuration of Figure 5.80, use Equation 5.60 to solve for the different values of
R v. In all three cases after you have determined R v, you can proceed as before.
5.10 Summary of matching networks
You should now be able to use several types of matching networks. These include trans-
former, capacitor-divider, L, π and T networks. Matching is vitally important in amplifier
design because without this ability, it is almost impossible to design good amplifiers and
High frequency transistor
6.1 Introduction
In this part, it is assumed that you are already familiar with transistors and their opera-
tion in low frequency circuits. We will introduce basic principles, biasing, and its effects
on the a.c. equivalent circuit of transistors, and the understanding of manufacturers’ tran-
sistor data in the early parts of the chapter. The latter half of the chapter is devoted to the
design of amplifier circuits.
6.1.1 Aims
The aims of this part are to review:
• basic principles of transistors
• biasing of transistors
• a.c. equivalent circuit of transistors
• manufacturers’ admittance parameters transistor data
• manufacturers’ scattering parameters transistor data
• manufacturers’ transistor data in graphical form
• manufacturers’ transistor data in electronic form
6.1.2 Objectives
After reading this part, you should be able to:
• understand basic operating principles of transistors
• understand manufacturers’ transistor data
• apply manufacturers’ transistor data in amplifier design
• bias a transistor for proper operation
• check for transistor stability
• design amplifiers using admittance parameters
6.2 Bi-polar transistors
The word transistor is an abbreviation of two words transferring resistor.
Bipolar transistors 263
Fig. 6.1(a) Basic construction of an NPN transistor and Fig. 6.1(b) Basic construction of a PNP transistor and its
its symbolic representation symbolic representation
Fig. 6.2 The Ebers Moll model of a transistor
6.2.1 Basic construction
The basic construction of bi-polar transistors and their electrical symbolic representations
are shown in Figure 6.1. The arrow indicates the direction of current flow in the transistor.
Many transistors are made in complementary pairs. Typical examples are the well known
NPN and PNP industrial and military types, 2N2222 and 2N2907, which have been used
for over four decades and are still being used in many designs.
6.2.2 Transistor action1
For explanation purposes, a transistor may be considered as two diodes connected back to
back. This is the well known Ebers Moll model and it is shown in Figure 6.2 for the NPN
transistor. In the Ebers Moll model, a current generator is included to show the relation-
ship (Ic = aIe) between the emitter current (Ie) and the collector current (Ic). In a good tran-
sistor, a ranges from 0.99 to about 0.999.
1 Although the description is mainly for NPN transistors, the same principles apply for PNP transistors except
that, in the latter case, positive charges (holes) are used instead of electrons.
264 High frequency transistor amplifiers
Fig. 6.3 Bi-polar transistor action
The action that takes place for an NPN transistor can be explained by the diagram in
Figure 6.3. In this diagram, emitter, base and collector are diffused together and a
base–emitter depletion layer is set up between base and emitter, and a collector–base
depletion layer is set up between the collector and base. These depletion layers are set up
in the same way as p–n junctions.
In normal transistor operation, the base–emitter junction is forward biased and the
base–collector junction is reversed biased. This results in a narrow depletion (low resis-
tance) at the base–emitter junction and a wide depletion layer (high resistance) at the
collector–base junction.
Electrons from the emitter (Ie) are attracted to the base by the positive potential Vbe. By
the time these electrons arrive in the base region, they will have acquired relatively high
mobility and momentum. Some of these electrons will be attracted towards the positive
potential of Vbe but most of them (>99%) will keep moving across the base region which
is extremely thin (≈0.2 – 15 microns2) and will penetrate the collector–base junction. The
electrons (Ic) will be swept into the collector region where they will be attracted by the
positive potential of Vcb.
Relatively little d.c. energy is required to attract electrons into the base region because
it is forward biased (low resistance – Rbe). Relatively larger amounts of d.c. energy will be
required in the collector–base region because the junction is reversed biased (high resis-
tance – Rcb).
2 1 micron is one millionth of a metre.
Bi-polar transistors 265
Nevertheless we have transferred current flowing in a low resistance region into current
flowing in a high resistance region. The ratio of the powers dissipated in these two regions
power in the collector–base region 2
I c Rcb
——————————————— = ———
power in the base–emitter region 2
I e Rbe
Ic = collector current
Ie = emitter current
Rcb = resistance between collector and base
Rbe = resistance between base and emitter
Since by design, Ic ≈ Ie and since Rcb >> Rbe
I 2Rcb
c Rcb
——— ≈ ——— (6.1)
I c be Rbe
We have a power gain because Rcb >> Rbe. Hence if signal energy is placed between the
base–emitter junction, it will appear at a much higher energy level in the collector region
and amplification has been achieved.
6.2.3 Collector current characteristics
If you were to plot collector current (IC) of an NPN transistor against collector–emitter
(VCE) for various values of the base current (IB), you will get the graph shown in
Figure 6.4. In practice, the graph is either released by transistor manufacturers or you can
obtain it by using an automatic transistor curve plotter.
The points to note about these characteristics are:
• the ‘knee’ of these curves occurs when VCE is at about 0.3–0.7 V;
• collector current (IC) increases with base current (IB) above the ‘knee’ voltage.
Fig. 6.4 Collector current characteristics
266 High frequency transistor amplifiers
6.2.4 Current gain
Because of the slightly non-linear relation between collector current and base current, there
are two ways of specifying the current gain of a transistor in the common emitter circuit.
The d.c. current gain (hFE) is simply obtained by dividing the collector current by the
base current. This value is important in switching circuits. At point P1 of Figure 6.4, when
VCE = 10 V and IB = 40 µA, Ic = 25 mA. Therefore
collector current
hFE = ——————— = 25 mA/40 mA = 625 (6.2)
base current
For most amplification purposes, we are only concerned with small variations in collector
current, and a more appropriate way of specifying current gain is to divide the change in
collector current by the change in base current and obtain the small signal current gain hFE
or b. At point P2 of Figure 6.4, when the operating point is chosen to be at around VCE =
5 V, and I = 20 mA
DIc (14 – 12) mA
hFE = —— = ——————— = 1000 (6.3)
DIb (21 – 19) mA
6.2.5 Operating point
The point at which a transistor operates is very important. For example at point P2 of
Figure 6.4 (see inset), if we choose the operating point to be at VCE = 5 V and IC = 13 mA,
it is immediately clear that you will not get VCE excursions è 5 V because the transistor
will not function when VCE = 0. The same argument is true with current because you will
not get current excursions less than zero. Therefore the operating point must be carefully
chosen for your intended purpose. This act of choosing the operating point is called bias-
ing. The importance of biasing cannot be over-emphasised because, as you will see later,
d.c. biasing also alters the a.c. parameters of a transistor. If the a.c. parameters of your
transistor cannot be held constant (within limits) then your r.f. design will not be stable.
6.2.6 Transistor biasing
The objectives of transistor biasing are:
• to select a suitable operating point for the transistor;
• to maintain the chosen operating point with changes in temperature;
• to maintain the chosen operating point with changes in transistor current gain with
• to maintain the chosen operating point to minimise changes in the a.c. parameters of the
operating transistor;
• to prevent thermal runaway, where an increase in collector current with temperature
causes overheating, burning and self-destruction;
• to try to maintain the chosen operating point with changes in supply voltages – this is
particularly true of battery operated equipment where the supply voltage changes
considerably as the battery discharges;
Bi-polar transistors 267
• to maintain the chosen operating point with changes in b when a transistor of one type
is replaced by another of the same type – it is common to find that b varies from 50% to
300% of its nominal value for the same type of transistor.
There are two basic internal characteristics that have a serious effect upon a transistor’s
d.c. operating point over temperature. They are changes in the base–emitter voltage (DVBE)
and changes in current gain (Db). As temperature increases, the required base-emitter volt-
age (VBE) of a silicon transistor for the same collector current decreases at the rate of about
2.3 mV/°C. This means that if VBE was 0.7 V for a given collector current before a temper-
ature rise, then the same VBE of 0.7 V after a temperature rise will now produce an increase
in base current and more collector current; that in turn causes a further increment in tran-
sistor temperature, more base and collector current, and so on, until the transistor eventu-
ally overheats and burns itself out in a process known as thermal runaway. To prevent
this cyclic action we must reduce the effective VBE with temperature.
Sections 6.2.7 to 6.2.9 indicate several ways of biasing bi-polar transistors in order to
increase bias stability. Complete step-by-step design instructions are included with each
circuit configuration. For ease of understanding, a.c. components such as tuned circuits,
inductors and capacitors have deliberately been left out of the circuits because they play
little part in setting the operating bias point. However, a.c. components will be considered
at a later stage when we come to design r.f. amplifiers.
6.2.7 Voltage feedback bias circuit
One circuit that will compensate for VBE is shown in Figure 6.5. Any increase in the quies-
cent3 collector current, (IC) causes a larger voltage drop across RC which reduces VC which
in turn reduces IB and IC. This can be shown by:
Fig. 6.5 Voltage feedback biasing
3 Quiescent collector current is defined as the collector current which is desired at a given temperature and
with no signal input to the transistor.
268 High frequency transistor amplifiers
IB = (VC – VBE)/RF (6.4)
VC = VCC – (IC + IB)RC (6.5)
Substituting Equation 6.4 in Equation 6.5 yields
VC = VCC – ICRC – RC(VC – VBE)/RF
ICRC = VCC – VC – RCVC/RF + RCVBE /RF
and differentiating IC with respect to VBE and cancelling RC from both sides gives
∂IC = ∂VBE/RF (6.6)
Equation 6.6 shows clearly that the effect of variations in VBE on IC is reduced by a
factor of 1/RF in Figure 6.5.
Example 6.1
Given the transistor circuit of Figure 6.5 with hFE = 200 and VCC = 10 V find an operating
point of IC = 1 mA and VC = 5 V.
Given: Transistor circuit of Figure 6.5 with hFE = 200, VCC = 10 V.
Required: An operating point of IC = 1 mA and VC = 5 V.
Solution Using Equation 6.2
IB = IC/hFE = 1000/200 = 5 µA
Assuming VBE and transposing Equation 6.4
RF = (VC – VBE)/IB = (5 – 0.7) V/ 5 mA = 860 kW
Transposing Equation 6.5
RC = (VCC – VC)/(IC + IB)
= (10 – 5) V/(1000 + 5) µA = 4.97 kΩ
Three points should be noted about the solution in Example 6.1.
• The values calculated are theoretical resistor values, so you must use the nearest avail-
able commercial values.
• Manufacturers often only quote the minimum and maximum values of hFE. In this case,
simply take the geometric mean. For example, if hFE(min) = 100 and hFE(max) = 400, the
geometric mean = 100 × 400 = 200.
• Thermal runaway is prevented when the half-power supply principle is used; that is
when Vc = 0.5VCC. This can be shown as follows.
At temperature T0 collector power (P) is given by
VC IC = [VCC – ICRC]IC
= VCCIC – I CRC (6.7)
Bi-polar transistors 269
At a higher temperature (T1) new collector power (P + ∆P) is given by
(VC – ∆VC)(IC + ∆IC) ≈ VC(IC + ∆IC)
because ∆VC << VC
P + DP = [VCC – (IC + ∆IC)RC](IC + ∆IC)
= VCC(IC + ∆IC) – (IC + ∆IC)2RC (6.7a)
Subtracting Equation 6.7 from Equation 6.7a yields
∆P = VCCIC + VCC ∆IC – (IC2 + 2IC∆IC + ∆IC2)RC – VCCIC + IC2RC
Simplifying and discarding ∆IC2RC because it is very small, gives
∆P = ∆IC(VCC – 2 ICRC)
Since IC = (VCC – VC)/RC
∆P = ∆IC(– VCC + 2VC)
For ∆P to equal zero, we get
VCC = 2VC or VC = 0.5VCC (6.8)
Equation 6.8 is the basis for the half-power supply principle and it should be used when-
ever possible to prevent thermal runaway.
For reasons which will become clearer when we discuss a.c. feedback, you will some-
times find that RF is split into two resistors, RF1 and RF2, with a capacitor CF1 connected
between its junction and chassis ground as shown in Figure 6.6. The purpose of CF1 is to
prevent any output a.c. or r.f. signal from travelling back to the input circuit. RF1 is used to
prevent short-circuiting the collector output signal via CF1 and RF2 is to prevent short-
circuiting the base signal through CF1.
Finally before leaving the bias circuits of Figures 6.5 and 6.6, the great advantage of the
Fig. 6.6 Split feedback bias circuit
270 High frequency transistor amplifiers
voltage feedback circuit is that it enables the emitter to be earthed directly. This is very
important at microwave frequencies because it helps to prevent unwanted feedback which
may affect amplifier stability.
Example 6.2
A transistor has hFE = 250 when VC = 12 V and IC = 2 mA. If your power supply (VCC) is
24 V, design a bias circuit like that shown in Figure 6.5 for operating the transistor with VC
= 12 V and IC = 2 mA.
Solution For this solution use the outlines of Example 6.1. Using Equation 6.2, hFE =
collector current/base current and
IB = IC/hFE = 2000 µA/250 = 8 µA
Transposing Equations 6.4 and 6.5
RF = (VC – VBE)/IB = (12 – 0.7) V/8 µA = 1.41 MW
RC = (VCC – VC)/(IC + IB)
= (24 – 12) V/(2000 + 8) µA = 5.97 kW
6.2.8 Voltage feedback and constant current bias circuit
Another bias circuit frequently used for r.f. amplifiers is shown in Figure 6.7. This circuit
is similar to that shown in Figure 6.5 except that the base current is fed from a more stable
source. Any increase in collector current (∆IC) results in a decrease in VC, VBB and IB
which in turn counteracts any further increase in IC. The design of this circuit is shown in
Example 6.3.
Fig. 6.7 Voltage feedback and constant current bias circuit
Bi-polar transistors 271
Example 6.3
Using the biasing arrangement shown in Figure 6.7, calculate the biasing resistors for a
transistor operating with VC = 10 V, IC = 5 mA and a supply voltage VCC = 20 V. The tran-
sistor has a d.c. gain of hFE = 150.
Given: VCC = 20 V, VC = 10 V, IC = 5 mA and hFE = 150.
Required: R1, RB, RC, RF.
1 Assume values for VBB and IBB to supply a constant current IB:
VBB = 2 V IBB = 1 mA
2 Knowing IC and hFE, calculate IB:
IC 5 mA
IB = —— = ——— = 0.0333 mA
hFE 150
3 Knowing VBB and IB, and assuming that VBE = 0.7 V, calculate RB:
VBB – VBE (2 – 0.7) V
RB = ————— = ————— = 39.39 kW
IB 0.0333 mA
4 Knowing VBB and IBB, calculate R1:
VBB 2V
— —
R1 = —— = —— = 2 kW
IBB 1 mA
5 Knowing VBB, IBB, IB and VC, calculate RF:
VC – VBB (10 – 2) V
RF = ————— = ————— = 7.74 kW
IBB + IB 1.033 mA
6 Knowing VCC, VC, IC, IB and IBB, calculate RC:
VCC – VC (20 – 10) V
RC = —————— = ————— = 1.66 kW
IC + IB + IBB 6.033 mA
Example 6.4
Use the bias circuit shown in Figure 6.7 to set the operating point of a transistor at IC =
1 mA, VC = 6 V. The current gain of the transistor ranges from 100 to 250. The circuit
supply voltage is 12 V.
Note: In my solution I have chosen VBB = 1.5 V and IBB = 0.5 mA; within reason you
may choose other values but if you do then your solutions will obviously differ from
272 High frequency transistor amplifiers
Using Example 6.3 as a basis for the solution, I have calculated the values of resistors but
you must use the closest commercial value in the circuit.
1 The operating point for the transistor is
IC = 1 mA, VC = 6 V, VCC = 12 V and
hFE = 100 × 250 ≈ 158
2 Assume values for VBB and IBB to supply a constant current, IB:
VBB = 1.5 V IBB = 0.5 mA
3 Knowing IC and hFE, calculate IB:
IC 1 mA
IB = —— = ——— ≈ 6.3 mA
hFE 158
4 Knowing VBB and IB, and assuming that VBE = 0.7 V, calculate RB:
VBB – VBE (1.5 – 0.7) V
RB = ————— = —————— ≈ 126.9 kW
IB 6.3 mA
5 Knowing VBB and IBB, calculate R1:
VBB 1.5 V
R1 = —— = ——— = 3 kW
IBB 0.5 mA
6 Knowing VBB, IBB, IB and VC, calculate RF:
VC – VBB (6 – 1.5) V
RF = ———— = ———————— = 8.88 kW
IBB + IB (0.5 mA + 6.3 mA)
7 Knowing VCC, VC, IC, IB and IBB, calculate RC:
VCC – VC (12 – 6) V
RC = —————— = —————————— = 3.98 kW
IC + IB + IBB (1 + 0.0063 + 0.5) mA
6.2.9 Base-voltage potential divider bias circuit
Another bias circuit that is commonly used is the base-voltage potential divider bias circuit
shown in Figure 6.8. In this circuit, VBB is held approximately constant by the voltage
divider network of R1 and R2. VBE is the voltage difference between VBB and VE which is
the product of IE and RE.
Since IE = IC + IB, any collector current rise ∆IC is followed by a ∆IE rise which
increases VE. This increase in VE is a form of negative feedback that tends to reduce bias
on the base–emitter junction and, therefore, decrease the collector current. Example 6.5
shows how to design the bias circuit of Figure 6.8.
Bi-polar transistors 273
Fig. 6.8 Base-voltage potential divider bias circuit
Example 6.5
Using the biasing arrangement shown in Figure 6.8, calculate the biasing resistors for a
transistor operating with VC = 10 V, IC = 10 mA and a supply voltage VCC = 20 V. The tran-
sistor has a d.c. gain of hFE = 50.
Given: VCC = 20 V, VC = 10 V, IC = 10 mA and hFE = 50.
Required: R1, R2, RC, RE.
1 Choose VE to be approximately 10% of VCC. Make
VE = 10% of 20 V = 2 V
2 Assume IE ≈ IC for high gain transistors.
3 Knowing IE and VE, calculate RE:
RE = ——— = 200 W
10 mA
4 Knowing VCC, VC and IC, calculate RC:
VCC – VC (20 – 10) V
RC = ——— — = ————— = 1000 W
— —
IC 10 mA
5 Knowing IC and hFE, calculate IB:
IC 10 mA
IB = —— = ———— = 0.2 mA
hFE 50
274 High frequency transistor amplifiers
6 Knowing VE and VBE, calculate VBB:
VBB = VE + VBE = 2.0 V + 0.7 V = 2.7 V
7 Choose a value for IBB; a normal rule of thumb is that IBB ≈ 10IB. Hence
IBB = 2 mA
8 Knowing IBB and VBB, calculate R2:
VBB 2.7 V
R2 = —— = ——— = 1350 W
IBB 2 mA
9 Knowing VCC, VBB, IBB and IB, calculate R1:
VCC – VBB (20 – 2.7) V
R1 = ————— = —————— = 7864 W
IBB + IB (2 + 0.2)
Example 6.6
Use the bias circuit shown in Figure 6.8 to set the operating point of a transistor at IC = 1
mA, VC = 6 V. The current gain of the transistor ranges from 100 to 250. The circuit supply
voltage is 12 V.
Note: In my solution I have chosen VE = 1.2 V and IBB = 0.5 mA; within reason you
may choose other values but if you do then your solutions will obviously differ from
Using Example 6.5 as a basis for the solution, I have calculated the values of resistors but
you must use the closest commercial value in the circuit.
1 The operating point for the transistor is:
IC = 1 mA, VC = 6 V, VCC = 12 V and
hFE = 100 × 250 ≈ 158
2 Assume a value for VE that considers bias stability: choosing VE to be approximately
10% of VCC
VE = 10% of 12 V = 1.2 V
3 Assume IE ≈ IC for high hFE transistors.
4 Knowing IE and VE, calculate RE:
1.2 V
RE = —— = 1200 W
1 mA
5 Knowing VCC, VC and IC, calculate RC:
VCC – VC (12 – 6) V
RC = ——— = ————— = 6000 W
IC 1 mA
Bi-polar transistors 275
6 Knowing IC and hFE, calculate IB:
IC 1 mA
IB = —— = —— = 6.3 mA
hFE 158
7 Knowing VE and VBE, calculate VBB:
VBB = VE + VBE = 1.2 V + 0.7 V = 1.9 V
8 I have chosen IBB to be:
IBB = 0.5 mA
9 Knowing IBB and VBB, calculate R2:
VBB 1.9 V
R2 = —— = ———— = 3.8 kW
IBB 0.5 mA
10 Knowing VCC, VBB, IBB and IB, calculate R1:
VCC – VBB (12 – 1.9) V
R1 = ————— = ———————— = 19.95 kW
IBB + IB (0.5 + 0.0063) mA
6.2.10 Summary on biasing of bi-polar transistors
In the voltage feedback circuits of Figures 6.5 and 6.6 and the voltage feedback and
constant current circuit of Figure 6.7, increases of collector current with temperature are
kept in check by introducing a form of negative feedback to decrease the effective
base–emitter voltage (VBE). If we were to use upward pointing arrows to indicate increases
and downward pointing arrows to indicate decreases, for the circuit of Figures 6.5 to 6.7,
we would have
IC ↑ ; VBE ↓ ; IB ↓ ; → IC constant
In the base–voltage potential divider circuit of Figure 6.8, the effective base–emitter volt-
age (VBE) is reduced by increasing the emitter voltage (VE) against a quasi fixed potential
(VBB). It is a better bias circuit than the earlier ones because VE can be made to almost track
and compensate for changes in VBE. Using arrows as before, for the circuit of Figure 6.8,
we would have
IC ↑ ; VE≠ ↑; (VBE = VBB – VE) ↓ ; IB ↓ ; → IC constant
The manufacturing tolerance for hFE or b in transistors of the same part number is typi-
cally poor. It is not uncommon for a manufacturer to specify a 10:1 range for b on the data
sheet (such as 30 to 300). This of course makes it extremely difficult to design a bias
network for the device in question when it is used from a production standpoint as well as
a temperature standpoint.
However, the base-voltage potential divider circuit works remarkably well on both
accounts because a high hFE produces a high VE which counteracts the quasi fixed
276 High frequency transistor amplifiers
Fig. 6.9 Active bias circuit for an r.f. amplifier
potential VBB. This bias circuit is therefore widely used in production circuits. One draw-
back of this circuit is that the resistor RE must be bypassed by a capacitor in order not to
affect the a.c. operation of the amplifier. At frequencies less than 100 MHz, this is no prob-
lem but at microwave frequencies effective bypassing is difficult and you will see that the
earlier circuits are often used.
In the discussions on biasing, I have only concentrated on direct biasing where resis-
tors are used to control IB and VBE. However, there are active-bias-circuits where one or
more additional transistors are used to bias the main r.f. amplifier. One well known exam-
ple is shown in Figure 6.9. In this circuit r.f. signal is applied through the input tuned
transformer (T1) to the base of the r.f. amplifier (tr1). The r.f. output signal is taken from
the collector tuned transformer (T2). Capacitors C1 and C2 are bypass capacitors used to
allow r.f. signals to reach the emitter of tr1 easily and to keep r.f. signal out of the bias-
ing circuit.
Biasing is carried out by supplying base current to tr1 via the secondary of T1 from R4
which in turn is fed from the collector current (IC2) of tr2. The base voltage (VB2) of tr2 is
fixed by the potential divider consisting of R1 and R2 which sets the base voltage of tr2
approximately 0.7 V below its emitter voltage (VE2). The emitter voltage (VE2) for tr2 is
provided by the voltage drop across R3 which in turn is caused by the collector current (IC1)
of tr1. If IC1 increases, then VE2 decreases and the effective base–emitter voltage (VB2) of
tr2 decreases, IC2 decreases, which in turn reduces IB1. IC1 then decreases to try and main-
tain its old value. Again, if we were to use upward pointing arrows to indicate increases
and downward pointing arrows to indicate decreases, for the circuit of Figure 6.9, we
would have
IC1≠ ↑ ; VE2 ↓ ; VBE2 ↓ ; IC2 ↓ ; IB1 ↓ ; IC1 ↓ ; → IC1 constant
The biasing of this circuit can be made very stable and I have used this arrangement for
r.f. amplifers operating from –15°C to +75°C. Another advantage of this biasing circuit is
that although tr1 is an expensive r.f. transistor, tr2 can be a cheap low frequency transistor.
The only requirement is that tr1 and tr2 should be made from similar material to enable
easier tracking for VBE and ICO. For example if tr1 is a silicon transistor then tr2 should
also be a silicon transistor.
Review of field effect transistors 277
6.3 Review of field effect transistors
There are two main types of field effect transistors; the junction FET or JFET, and the
metal oxide silicon FET or MOSFET. MOSFETs are sometimes also referred to as ‘insu-
lated gate field effect transistors’, IGFETs. A high electron mobility FET is also known
as a HEMFET. In this discussion, we will look briefly at FET structures, their operating
modes and methods of biasing to enable us to use FETs efficiently at high frequencies.
6.3.1 A brief review of field effect transistor (JFET or FET) construction
An elementary view of field effect transistor construction will aid understanding when we
analyse a.c. equivalent circuits and use them in practical amplifier and oscillator circuits.
FET (n-channel)
The basic construction of an n-channel type field effect transistor is shown in Figure 6.10.
The device has three terminals, a source terminal (S), a gate terminal (G), and a drain
terminal (D). In an n-channel depletion mode FET, n-type material is used as the conduct-
ing channel between source and drain. P type material is placed on either side of the chan-
nel. The effective electrical width of the channel is dependent on the voltage potential
(VGS) between gate and source.
When electrical supplies are connected in the manner shown in Figure 6.11, electrons
flow from the source, past the gate, to the drain. If a negative voltage is applied between
the gate and source, its negative electric field will try to ‘pinch’ the electron flow and
confine it to a smaller cross-section of the n-channel of the FET. This affects the resistance
of the n-channel and restricts the current flowing through it. Hence, by varying the
gate–source voltage (VGS), it is possible to control current flow.
Power gain is obtained because very little energy is required to control the input signal
(VGS) while relatively large amounts of power can be obtained from the variations in
drain–source current (IDS). The symbols for an n-channel depletion mode FET and its
output current characteristics are shown in Figure 6.12. Drain–source current (ID) is maxi-
mum when there is zero voltage on VGS and ID decreases as VGS becomes more negative.
The pinch-off voltage (VP) is the gate–source voltage required to reduce the effective
cross-section of the n-channel to zero. For practical purposes, it is VGS which causes IDS to
Fig. 6.10 An n-channel depletion mode field effect transistor
278 High frequency transistor amplifiers
Fig. 6.11 n-channel FET
Fig. 6.12 Symbols and output current characteristics for an n-channel FET
become zero. IDSS is the drain–source current when the gate and source are shorted
together (VGS = 0) for your particular VDS.
Operating point and biasing of an n-channel FET
Selection of the operating point is similar to that explained for the bi-polar transistor, but
the biasing method is different. An example of how to bias an n-channel FET is given in
Example 6.7.
Example 6.7 Biasing an n-channel FET
Using the biasing arrangement shown in Figure 6.13, calculate the biasing resistors for an
FET operating with VDS = 10 V, IDS = 5 mA and a supply voltage VCC = 24 V. From manu-
facturer’s data, for IDS = 5 mA and VDS = 10 V, VGS = –2.3 V.
Given: VCC = 24 V, VDS = 10 V, IDS = 5 mA and VGS = 2.3 V.
Required: Rs, RD, RG, R.
1 For our particular transistor, the manufacturer’s d.c. curves show that for an operating
current of 5 mA with VDS = 10 V, we require a VGS of –2.3 V which means that the gate
must be 2.3 V negative with respect to the source. We do not have a negative supply but
we can simulate this supply by making the source positive with respect to the gate which
Review of field effect transistors 279
Fig. 6.13 n-channel FET biasing
in turn means that the gate will be negative with respect to the source. This is carried out
by placing a resistor (RS) in series with ID to produce a positive voltage (VS) which is equal
to VGS. We now calculate RS.
2 Since IG = 0, |VGS| = |VS|, and knowing ID, calculating RS gives
|VS| |VGS|
RS = —— = ———
ID ID
2.3 V
= ——— = 460 W
5 mA
Note particularly that RS provides a form of negative feedback to stabilise changes in
FET parameters with temperature. It also stabilises current when FETs of the same type
number are changed, because any increase in IDS immediately produces a correspond-
ing decrease in VGS. The ratio
change in IDS(∆IDS)
change in VGS(∆VGS)
is known as the transconductance (gm) of the transistor.
3 Since IG = 0, RG can be chosen to be any convenient large value of resistor – approxi-
mately 1 MW. This value is useful because it does not appreciably shunt the desirable
high input impedance of the transistor.
4 Knowing VCC, VS, VDS and ID, we can now calculate VD and then RD. Since we require
VDS = 10 V and VS has already been chosen as 2.3 V
VD = VDS + VS = 10 V + 2.3 V = 12.3 V
280 High frequency transistor amplifiers
VCC – VD (24 – 12.3) V
RD = ————— = —————— = 2340 W
ID 5 mA
Our bias circuit is now complete.4
The above biasing circuit is easy to design but complications often arise when the manu-
facturer does not supply the d.c. curves for a particular transistor5 or when you cannot
obtain the characteristic curve from a transistor curve plotter. In this case, refer to the
manufacturer’s FET data sheets for values of VP and IDSS. With these two values known,
you can use the well known FET expression for calculating VGS. It is
[ ]
ID = IDSS 1 – —— (6.9)
For our particular case, the manufacturer states that VP = –8 V and IDSS = 10 mA.
Substituting the values in Equation 6.9 and transposing yields
⎪ ID ⎫⎪
VGS = VP ⎨1 − ⎬
⎩ IDSS ⎪
⎧ 5 mA ⎫
= −8 V ⎨1 − ⎬
⎩ 10 mA ⎭
= −2.34 V
We can then proceed as in steps 3 to 5 above.
Example 6.8
An n-channel JFET has VP = –6 V and IDSS = 8 mA. The desired operating point is ID = 2
mA and VDS = 12 V. The supply voltage (VCC) is 24 V. Design a bias circuit for this oper-
ating point.
Solution: This circuit will be designed following the method given in Example 6.7.
1 The operating point for the transistor is
ID = 2 mA, VD = 12 V, VCC = 24 V
2 Vp and IDSS from the data sheet are
Vp = –6 V
IDSS = 8 mA
4 Older readers might well recognise that this method of biasing is similar to that used for thermionic valves.
The only difference is that in thermionic valves, VS is such a small fraction of the anode–cathode voltage that it
may be neglected.
5 This seems to be the case with many r.f. transistors.
Review of field effect transistors 281
3 Knowing ID, IDSS, and Vp, VGS can be calculated from:
⎪ ID ⎫⎪
VGS = VP ⎨1 − ⎬
⎩ ⎪
IDSS ⎭
⎧ 2 mA ⎫
= −6 ⎨1 − ⎬
⎩ 8 mA ⎭
= −3.0 V
4 Since IG = 0, |VGS| = |VS|, and knowing ID, RS can be calculated:
|VS| |VGS| 3 V
RS = —— = —— = —— = 1500 W
ID ID 2 mA
5 Since IG = 0, RG can be chosen to be any large value of resistor – approximately 1 M W.
6 Knowing VCC, VS, VDS and ID, we can now calculate VD and then RD. Since we require
VDS = 12 V and VS has already been calculated as 3 V
VD = VDS + VS = 12 V + 3 V = 15 V
VCC – VD (24 – 15) V
RD = ————— = —————— = 4500 W
ID 2 mA
The bias circuit is now complete.
FET (p-channel )
It is also possible to make p-channel type depletion mode FETs by substituting p-type
material for n-type and vice-versa in the simplified construction illustrated in Figure 6.10.
However, the voltage supplies must also be reversed to that shown in Figure 6.11, and this
time the current flow is ‘hole’ current instead of electrons. The net result is the same and
gain can be obtained from the FET. Figure 6.14 shows the symbol for a p-channel FET and
Fig. 6.14 Symbols and output current characteristics for a p-channel FET
282 High frequency transistor amplifiers
its output current characteristics. Note that VDS and ID are reversed to that for the n-chan-
nel characteristics given earlier, and that VGS must be positive to decrease ID. Selection of
the operating point is similar to that explained for the bi-polar transistor, but the biasing
method is different. An example of how to bias a p-channel FET is given in Example 6.9.
Example 6.9
Using the biasing arrangement shown in Figure 6.15, calculate the biasing resistors for a
FET operating with VDS = –10 V, IDS = –5 mA and a supply voltage VCC = –24 V. From
manufacturer’s data, for IDS = 5 mA and VDS = –10 V, VGS = +2.3 V.
Given: VCC = –24 V, VDS = –10 V, VGS = +2.3 V, IDS = –5 mA.
Required: RG, RD, RS.
Fig. 6.15 Bias circuit for a p-channel depletion mode FET
1 For our particular transistor, the manufacturer’s d.c. curves show that for an operating
current of –5 mA with VDS = –10 V, we require a VGS of +2.3 V which means that the
gate must be 2.3 V positive with respect to the source. We do not have another power
supply but we can simulate this supply by making the source negative with respect to
the gate which in turn means that the gate will then be positive with respect to the
source. This is carried out by placing a resistor (RS) in series with ID to produce a nega-
tive voltage (–VS), which is equal to VGS. We now calculate RS.
2 Since IG = 0, |VGS| = |VS|, and knowing ID, calculating RS gives
|VS| |VGS| –2.3 V
RS = —— = —— = ——— = 460 W
ID ID –5 mA
3 Since IG = 0, RG can be chosen to be any convenient large value of resistor – approxi-
mately 1 MW. This value is useful because it does not appreciably shunt the desirable
high input impedance of the transistor.
Review of field effect transistors 283
4 Knowing VCC, VS, VDS and ID, we can now calculate VD and then RD. Since we require
VDS = –10 V and VS has already been chosen as –2.3 V
–VD = –VDS – VS = –10 V – 2.3 V = –12.3 V
VCC – VD (–24 V) – (–12.3 V)
RD = ———— = ———————— = 2340 W
ID –5 mA
Our bias circuit is now complete.
The above biasing circuit is easy to design but complications often arise when the manu-
facturer does not supply the d.c. curves for a particular transistor and when you cannot
obtain the characteristic curve from a transistor curve plotter. In this case, refer to the
manufacturer’s FET data sheets for values of VP and IDSS and use Equation 6.9 as before.
With these two values known, you can calculate VGS.
Example 6.10
A p-channel JFET is to be operated with ID = –2 mA, VDS = –8 V. The power supply volt-
age (VCC) is –14 V. The data sheet gives VP = 4 V and IDSS = –6 mA.
Solution: The bias circuit will be designed using Example 6.9 as a guide.
1 The operating point for the transistor is
ID = –2 mA, VDS = –8 V, VCC = –14 V
2 For our particular transistor, VP = 5 V and IDSS = –6 mA.
3 Knowing ID, IDSS and VP, we can calculate VGS:
⎪ ID ⎫⎪
VGS = VP ⎨1 − ⎬
⎩ IDSS ⎪
⎧ −2 mA ⎫
= 5 ⎨1 − ⎬
⎩ −6 mA ⎭
= 2.1 V
4 Since IG = 0, |VGS| = |VS|, and knowing ID, calculating RS gives
|VS| |VGS| –2.1 V
RS = —— = —— = ——— = 1050 W
ID ID –2 mA
5 Since IG = 0, RG can be chosen to be any large value of resistor – approximately
1 M Ω.
6 Knowing VCC, VS, VDS and ID, we can now calculate VD and then RD. Since we require
VDS = –8 V and VS has already been chosen as –2.1 V:
–VD = –VDS – VS = –8 V – 2.1 V = –10.1 V
284 High frequency transistor amplifiers
VCC – VD (–14 + 10.1) V
RD = ————— = ——————— = 1950 W
ID –2 mA
Our bias circuit is now complete.
6.3.2 A brief review of metal oxide silicon field effect transistors
MOSFETs (n-channel enhancement-mode type)
In MOSFETs, the drain and source are p–n junctions formed side by side in the surface of
a silicon substrate as illustrated in Figure 6.16. This time the gate is a conductor, originally
a metal film (hence the name of the device) but nowadays it is usually a layer of well doped
silicon. This gate is separated from the silicon substrate by a film of oxide thus forming an
input capacitance.
The application of a voltage between gate and source induces carriers in the silicon
under the gate – as indicated in Figure 6.17 – the amount of charge induced in the chan-
nel being dependent on the gate voltage. When a drain–source voltage is applied (VDS),
Fig. 6.16 Basic construction of an n-channel MOSFET
Channel of Edge of transition
electrons region
Fig. 6.17 Basic action in an n-channel MOSFET
Review of field effect transistors 285
Fig. 6.18 Symbols and characteristic curves for an n-channel enhancement-mode MOSFET
these induced carriers flow between source and drain – the larger the induced charge, the
greater the drain current ID. In short, VGS controls the current flowing through the channel
for a fixed value of VDS.
Power gain is obtained because very little energy is required to control the input signal
(VGS) while relative large amounts of power can be obtained from the variation in
drain–source current (IDS). The symbols for an n-channel enhancement-mode MOSFET
and its output current characteristics are shown in Figure 6.18. Note that in the first
symbol, the substrate is marked ‘B’ for bulk to distinguish it from the ‘S’ for source.
Sometimes the substrate is joined to the source internally to give a device with three rather
than four terminals. The line connecting source and drain is shown as a dashed line to indi-
cate that in an n-channel enhancement-mode MOSFET, the conduction channel is not
established with zero gate–source voltage.
MOSFETs (p-channel enhancement-mode type)
It is also possible to make p-channel enhancement-mode MOSFETs by substituting p-type
material for n-type and vice-versa to that shown in Figure 6.16. However, the voltage
supplies must also be reversed to that shown in Figure 6.17, and this time the current flow
is hole current instead of electrons. The net result is the same and gain can be obtained
from the MOSFET. Figure 6.19 shows the symbol for a p-channel MOSFET and its output
current characteristics. Note that VDS and ID are reversed to that for the n-channel charac-
teristics given above.
Fig. 6.19 Symbols and characteristic curves for a p-channel enhancement-mode MOSFET
286 High frequency transistor amplifiers
Here again, in the first symbol, the substrate is marked ‘B’ for bulk to distinguish it
from the ‘S’ for source. However, the arrow is pointing away this time to indicate a p-chan-
nel device. Sometimes the substrate is joined to the source internally to give a device with
three rather than four terminals. The line connecting source and drain is shown as a dashed
line to indicate that in a p-channel enhancement-mode MOSFET, the conduction channel
is not established with zero gate–source voltage.
Biasing of MOSFETs
Selection of the operating point is similar to that explained for the bi-polar transistor. The
biasing methods are also similar (see section 6.3.1) but the calculations are simpler
because IG = 0. One method of biasing is shown in Example 6.11.
Example 6.11
The n-channel enhancement-mode MOSFET shown in Figure 6.20 is to be operated with
ID = 5 mA, VDS = 10 V with a supply voltage (VCC) of 18 V. Manufacturer’s data sheets
show that this MOSFET requires a positive bias of 3.2 V for a current of 5 mA when VDS
is 10 V. Calculate the values of resistors required for the circuit in Figure 6.20.
1 The operating point for the transistor is
ID = 5 mA, VDS = 10 V, VCC = 18 V
2 From the manufacturer’s data sheet VGS = +3.2 V for a current of 5 mA.
3 Choose VS to be approximately 10% of VCC:
VS = 10% of 18 V = 1.8 V
4 Knowing VS and ID, calculate RS:
VS 1.8 V
RS = — = —— = 360 W
ID 5 mA
Fig. 6.20 Bias circuit for an n-channel enhancement-mode MOSFET
Review of field effect transistors 287
5 Knowing VS and VGS, calculate VG:
VG = VGS + VS = (3.2 + 1.8) V = 5.0 V
6 Assume a value for R2 based upon d.c. input resistance needs:
R2 = 220 kW
7 Knowing R2, VG and VCC, calculate R1:
R2(VCC – VG)
R1 = ——————
220 kW (18–5) V
= ———————— = 572 kW
8 Knowing VS and VDS, calculate VD and then RD:
VD = VS + VDS = 1.8 V + 10 V = 11.8 V
VCC – VD (18 – 11.8) V
RD = ———— = —————— = 1240 W
ID 5 mA
The bias circuit is now complete.
Example 6.12
An n-channel enhancement-mode MOSFET is to be operated with ID = 2 mA, VDS = 6 V
with a supply voltage (VCC) of 12 V. Manufacturer’s data sheets show that this MOSFET
requires a positive bias of 1.8 V for a current of 2 mA when VDS is 6 V. Calculate the
values of resistors required for the circuit in Figure 6.20.
1 The operating point for the transistor is
ID = 2 mA, VDS = 6 V, VCC = 12 V
2 From the manufacturer’s data sheet VGS = +1.8 V for a current of 2 mA.
3 Choose VS to be approximately 10% of VCC:
VS = 10% of 12 V = 1.2 V
4 Knowing VS and ID, calculate RS:
VS 1.2 V
RS = —— = ——— = 600 W
ID 2 mA
5 Knowing VS and VGS, calculate VG:
VG = VGS + VS = (1.8 + 1.2) V = 3.0 V
288 High frequency transistor amplifiers
6 Assume a value for R2 based upon d.c. input resistance needs:
R2 = 220 kW
7 Knowing R2, VG and VCC, calculate R1:
R2(VCC – VG)
R1 = ———————
220 kW (12 – 3) V
= ———————— = 660 kW
8 Knowing VS and VDS, calculate VD and then RD:
VD = VS + VDS = 1.2 V + 6 V = 7.2 V
VCC – VD (12 – 7.2) V
RD = ———— = ————— = 2400 W
ID 2 mA
The bias circuit is now complete.
6.3.3 Depletion-mode MOSFETs
Depletion-mode MOSFETs are also known in some books as depletion/enhancement-
mode (DE) MOSFETs because, as you will see shortly, they can be biased to operate in
both the depletion and enhancement mode. For our purposes, we will refer to them as
depletion-mode types to avoid unnecessary confusion.
MOSFETs (n-channel depletion-mode type)
Depletion-mode MOSFETs are made in a similar way to that of enhancement MOSFETs
except that a very thin layer of donors is implanted in the surface of the p-type substrate
just under the gate as indicated in Figure 6.21. This is done simply by firing donor atoms
Fig. 6.21 A cross-sectional diagram of an n-channel depletion-mode MOSFET showing the layer of donors implanted
in the surface of the p-type substrate to form a channel of electrons even when VGS = 0
Review of field effect transistors 289
Fig. 6.22 Symbols and output current characteristics for an n-channel depletion-mode MOSFET
in a vacuum at the silicon surface. If the implanted donor density exceeds the density of
holes already there, a channel of electrons will be formed even when VGS = 0 and a drain
current will flow as soon as VDS is applied. The current which flows in a depletion-mode
MOSFET when VGS = 0 is called IDSS and it is quoted in data sheets.
Figure 6.22 shows the symbols and electrical characteristics of an n-channel depletion-
mode MOSFET. Note that VGS can be positive, zero or negative. The VGS required to
reduce ID to zero is called the threshold voltage (VT). In the first symbol of Figure 6.22,
the substrate is marked ‘B’ for bulk to distinguish it from the ‘S’ for source. Sometimes the
substrate is joined to the source internally to give a device with three rather than four termi-
nals. The line connecting source and drain is shown as a full line to indicate that in an
n-channel DE-mode MOSFET, a conduction channel is present even with zero gate–source
MOSFETs (p-channel depletion-mode type)
MOSFETs (p-channel depletion-mode) are made in a similar manner to that shown for the
n-channel depletion-mode MOSFET except that p-type material has been substituted for
n-type material and operating voltages must be reversed. The symbols and current charac-
teristics of a p-channel depletion-mode MOSFET are shown in Figure 6.23.
The explanation of the operation of p-channel MOSFETs is identical to that of the
n-channel MOSFET just described except that all region types, carrier types, voltages
Fig. 6.23 Symbols and output current characteristics for a p-channel depletion-mode MOSFET
290 High frequency transistor amplifiers
and currents are reversed. For example the gate is made increasingly negative to cause
an increase in drain current, and the threshold voltage of an enhancement-mode p-
type channel device is positive. In the first symbol of Figure 6.23, the substrate is
marked ‘B’ for bulk to distinguish it from the ‘S’ for source. Sometimes the substrate
is joined to the source internally to give a device with three rather than four terminals.
The line connecting source and drain is shown as a full line to indicate that in a p-chan-
nel DE-mode MOSFET, a conduction channel is present even with zero gate–source
Biasing of depletion-mode MOSFETs
These MOSFETs can be biased according to the methods given in Examples 6.7, 6.9 and
6.11. The circuit you choose will depend on whether you wish to operate the circuit with
positive or negative VGS.
6.3.4 Summary of the properties of FETs and MOSFETs
Field effect transistors (FETs) can be regarded as three terminal devices whose terminals
are called source, drain and gate. There are two types of field effect transistor; the junction
FET or JFET, and the metal oxide silicon FET or MOSFET. In a JFET current flows
through a channel of silicon whose cross-sectional area is controlled by a p–n junction
whose width is varied by the application of a voltage between gate and source, as illus-
trated in Figure 6.11. In a MOSFET the drain and source are p–n junctions formed side by
side in the surface of a silicon substrate as illustrated in Figure 6.16. The gate is separated
from the silicon substrate by a film of oxide. The application of a voltage between gate and
source induces carriers in the silicon under the gate which then forms the channel between
source and drain. See Figure 6.17.
There are complementary forms of both types of FET: namely n-channel and p-channel
devices. In n-channel devices the drain current is carried by electrons, whilst in p-channel
devices it is carried by holes. There are two variants of MOSFET called enhancement-
mode devices and depletion-mode devices. In enhancement-mode devices ID = 0 when VGS
= 0. In depletion-mode devices (and in JFETs) ID = IDSS when VGS = 0.
The families of output characteristics of all FETs have the same general form and have
been shown in Figures 6.12, 6.14, 6.18, 6.19, 6.22 and 6.23. Note particularly that the
polarities of the voltage and directions of currents of p-channel devices are the reverse of
those of n-channel ones; and that the differences between the three types of n-channel
devices, or between the three p-channel devices, is in their range of values for VGS. The
polarity of the threshold voltages (VT) distinguishes between enhancement-mode and
depletion-mode MOSFETs. (The threshold voltages are the gate voltages at which the
drain current ID just begins to flow.)
There are alternative graphical symbols for MOSFETs in common use, which are also
shown in Figures 6.12, 6.14, 6.18, 6.19, 6.22 and 6.23. Note that if the (longer) central line
represents the piece of silicon, an arrow in a diagram always points at a p-region or away
from an n-region, as with bi-polar transistors. However, the arrow indicates the source in
a MOSFET, but indicates the gate in a JFET. Note the extra line added to the MOSFET
symbols for depletion-mode operation – it is intended to indicate the existence of a chan-
nel when VGS = 0. Other symbols may be found in data sheets and in textbooks, so be sure
to check the meanings of symbols in other publications. In particular a more complicated
A.C. equivalent circuits of transistors 291
standard symbol is used to represent the four terminal nature of MOSFETs, but we shall
not be referring to this in this text.
The carriers in the channel of either type of FET flow from source to drain under the
influence of an electric field, so the currents are drift currents rather than diffusion
The d.c. output characteristics of a JFET can be calculated by Equation 6.7.
The main electrical advantage of MOSFETs over bi-polar transistors is that their d.c.
gate current is virtually zero. This is due to the presence of the oxide layer between the
gate electrode and the substrate. The current gain, if it was ever referred to, would be
approaching infinity. This means that the input power to the device is very small indeed.
The advantage of MOSFETs from a production point of view is that, in general, they are
smaller and cheaper to manufacture than bi-polars. Their main disadvantage is that their
transconductance6 (gm) is normally much less than that of bi-polars at the same operating
current; that is, the control of the output current by the input voltage is normally less effec-
tive. It is possible, however, using special construction methods to produce MOSFETs
which are capable of operating more efficiently than bi-polars at microwave frequencies.
This is particularly true of the high electron mobility field effect transistor known as the
HEMFET which will be discussed later in the design of microwave amplifiers.
6.4 A.C. equivalent circuits of transistors
Some of the material described in Section 6.4 has already been covered in earlier sections
but I need to emphasise some relevant facts to help you understand how a.c. equivalent
circuits are derived. In the analyses of a.c. equivalent circuits, apart from a very brief intro-
duction to FETs, I will concentrate mainly on the bi-polar transistor because many of the
a.c. equivalent circuits (apart from circuit values) apply to JFETs and MOSFETs as well.
6.4.1 A brief review of bi-polar transistor construction
The NPN type bi-polar transistor shown in Figure 6.24 is constructed by using a crystalline
layer of silicon7 into which carefully controlled amounts of impurities such as arsenic,
phosphorus or antimony have been added so that the silicon may be made to provide rela-
tively easy movement for electrons. This layer is known as an n-type material because it
contains ‘free’ negative electrical charges (electrons). In the NPN transistor, this layer is
called the emitter because it has the ability to ‘emit’ electrons under the influence of a
voltage potential. A very thin layer of material about 0.2–10 microns8 thick is then laid
over the emitter. This layer is called the base layer. It is usually made of silicon with care-
fully controlled amounts of impurities such as aluminium, boron, gallium or indium. This
6 Transconductance (gm) is defined as
small change in output current (∆IDS)
small change in input voltage (∆VGS)
7 Germanium can also be used but its electrical characteristics with temperature are less stable than that of
silicon. Therefore, it is becoming obsolete and is only used for special functions or as replacement transistors for
older designs.
8 1 micron = 1 × 10–6 metres.
292 High frequency transistor amplifiers
Fig. 6.24 Basic construction of a bi-polar transistor
layer is known as a p-type layer because there are ‘free’ positive electric charges (holes)
in the material. Finally another layer of n-type material is placed over the base layer. This
layer is known as the collector because it collects all the current.
With suitable operating conditions, and when the transistor is connected to a battery,
electrons from the emitter are made to pass through the base which controls current flow to
the collector. This type of action occurs in an NPN type transistor. The same type of action
described above is possible with a transistor made with PNP type construction, that is with
the emitter and collector constructed of p-type material and the base of n-type material.
6.4.2 A brief review of field effect transistor construction
The basic construction of an n-channel type field effect transistor is shown in Figure 6.25.
In this case, n-type material is used as the conducting channel between source and drain
and p-type material is placed on either side of the channel. The effective electrical width
of the channel is dependent on the voltage potential between gate and source.
When a power supply is connected in the appropriate manner (+ at the drain and – at
the source), electrons flow from the source past the gate to the drain. A negative potential
applied to the gate alters the channel width of the transistor. This in turn affects the resis-
tance of the channel and its current flow. Power gain is obtained because the input power
applied to the gate is very much less than the output power.
Fig. 6.25 An n-channel depletion-mode field effect transistor
6.4.3 Basic a.c. equivalent circuits
As stated earlier and for the sake of clarity, I will concentrate mainly on the bi-polar tran-
sistor. However, you should be aware that much of what is said applies to the FET as well
A.C. equivalent circuits of transistors 293
Fig. 6.26 Approximate equivalent electrical circuit of a transistor
Fig. 6.27 ‘π ’ equivalent circuit of a transistor
because its physical construction also produces inter-electrode resistances and capacitances.
If you examine Figure 6.24 more closely, you will see that due to the proximity of the emit-
ter, base and collector or the source, gate and drain in Figure 6.25, there is bound to be resis-
tance and capacitance between the layers. The approximate9 electrical equivalent circuit for
Figure 6.24 has been drawn for you in Figure 6.26. The abbreviations used are:
Cbe = capacitance between the base and emitter
Rbe = resistance between the base and emitter
Ccb = capacitance between the collector and base
Rcb = resistance between the collector and base
Cce = capacitance between the collector and emitter
Rce = resistance between the collector and the emitter
gmVbe = a current generator which is controlled by the voltage between base and emitter
(Vbe) – this generator is present because there is current gain in a transistor
gm = a constant of the transistor and its operating point – it is defined as change in
collector current (DIc)/change in base–emitter voltage (DVbe)
In Figure 6.27, I have simplified the circuit by removing the bulk diagram of the tran-
sistor and it now becomes more recognisable as the p equivalent circuit of a transistor. The
circuit is called a p equivalent circuit because the components appear in the form of the
Greek letter p. The resistances and capacitances in Figure 6.27 are not fixed. They are
dependent on the d.c. operating conditions of the transistor.10 For example, if the d.c.
current through the transistor increases then Rbe will decrease and vice-versa. Similarly if
the voltage across the transistor increases then Cce will decrease and vice-versa.11 These
variations are inevitable because d.c. operating voltages and currents affect the physical
nature of transistor junctions.
9 The circuit is only approximate because I have not taken into account the resistance and reactances of the
lead and its connections.
10 You will no doubt recall from Section 6.2 that the base–emitter junction of a bi-polar transistor is a forward
biased p–n junction diode and that the resistance of this junction (Rdiode) varies with current.
11 This is due to the change in the width of the depletion layer described in Section 6.2.
294 High frequency transistor amplifiers
Table 6.1 Typical values for a bi-polar transistor
Resistance value Capacitance values Reactances at 100 kHz
Rbe ≈ 1–3 kW Cbe ≈ 10–30 pF Xbe ≈ 159–53 kW
Rbc ≈ 2–5 MW Cbc ≈ 2–5 pF Xbc ≈ 796–318 kW
Rce ≈ 20–50 kW Cce ≈ 2–10 pF Xce ≈ 796–159 kW
gm 40 mA per volt when the collector current is 1 mA
In Table 6.1, I have listed some typical values of components in a p equivalent circuit
when a bi-polar transistor is operated as a small signal amplifier with a collector–emitter
voltage of 6 V and a current of 1 mA. Much of the discussion that follows is dependent on
the relative values of these components to each other.
Low frequency equivalent circuit of a transistor
If you examine the first and the third columns of Table 6.1, you will see that at 100 kHz
• Xbe >> Rbe so that its effect on Rbe is negligible.
• Xce >> Rce so that its effect on Rce is negligible.
• If you refer to Figure 6.27, you will see that the fraction of the output voltage between
the collector and the emitter (Vce) fed back through the feedback path formed by the
parallel combination of Rbc and Ccb the parallel combination of Rbe and Cbe is also very
small because the parallel combination (Rbc // Xcb) >> (Rbe // Xbe). Therefore it can also
be neglected at low frequencies. If you have difficulty understanding this part, refer back
to Figure 6.27.
The effects of the above mean that the circuit of Figure 6.27 can be re-drawn at low
frequencies to be that shown in Figure 6.28. This equivalent circuit is reasonably accurate
for frequencies less than 100 kHz.
Fig. 6.28 Low frequency equivalent circuit of a transistor
High frequency equivalent circuit of a transistor
Returning to the p equivalent circuit of Figure 6.27, you will recall that I mentioned earlier
that the circuit was only approximate because I did not take lead resistances and reactances
into account. When these are added the circuit becomes that shown in Figure 6.29. This
circuit is called the hybrid configuration because it is a hybrid of the p circuit.
Another resistance (Rbb) known as the base spreading resistance emerges. This is the
inevitable resistance that occurs at the junction between the base terminal or contact and
the semiconductor material that composes the base. Its value is usually in tens of ohms.
Smaller transistors tend to exhibit larger values of Rbb because of the greater difficulty of
A.C. equivalent circuits of transistors 295
Fig. 6.29 Hybrid π equivalent circuit
connecting leads to smaller surfaces. Inductors Lb, Lc and Le are the inductances of the
base, collector and emitter leads respectively.
Of the three inductors, Le has the most pronounced effect on circuit performance
because of its feedback effect. This is caused by the input current flowing from B via Lb,
Rbb, Cbe and Rbe in parallel, and out via Le while the output current opposes the input
current since it flows outwards via Lc, the external load, and in again via Le. As frequency
increases, the reactance of Le increases and its effect is to produce a larger (Iout × XLe) volt-
age to oppose input current flow. Manufacturers tend to minimise this effect by providing
two leads for the emitter; one for the input current and the other for the output current. This
is the reason why some r.f. transistors have two emitter leads.
An increase in operating frequency causes reactances Xbe, Xcb and Xce to decrease and
this action will increase their shunting effect on resistances Rbe, Rcb and Rce and eventually
the gain of the transistor will begin to fall. The most serious shunting effect is caused by
Xcb because it affects the negative feedback path from collector to base and a frequency
will be reached when the gain of the transistor is reduced to unity. The unity gain
frequency (fT) is also known as the cut-off frequency of the transistor.
6.4.4 Summary
From the foregoing discussions, you should have realised that:
• transistors come in different shapes and sizes;
• they have d.c. parameters as well as a.c. parameters;
• a.c. parameters vary with d.c. operating conditions;
• a.c. parameters vary with frequency;
• a transistor operating under the same d.c. conditions can be represented by different a.c.
equivalent circuits at different frequencies;
• transistor data given by manufacturers may appear in several ways, namely p, hybrid p,
and other yet to be introduced parameters such as admittance (Y) parameters and scat-
tering (S) parameters.
6.4.5 The transistor as a two-port network
The transistor is obviously a three terminal device consisting of an emitter, base and
collector. In most applications, however, one of the terminals is common to both the input
296 High frequency transistor amplifiers
Fig. 6.30 (a) Common emitter configuration; (b) common base configuration; (c) common collector configuration
and the output network as shown in Figure 6.30. In the common emitter configuration of
Figure 6.30(a), the emitter is grounded and is common to both the input and output
network. So, rather than describe the device as a three terminal network, it is convenient
to describe the transistor as a ‘black box’ by calling it a two port network. One port is
described as the input port while the other is described as the output port. These configu-
rations are shown in Figure 6.30(a, b, and c). Once the two port realisation is made, the
transistor can be completely characterised by observing its behaviour at the two ports.
6.4.6 Two-port networks
Manufacturers are also aware that in many cases, knowing the actual value of components
in a transistor is of little value to the circuit design engineer because there is little that can
be done to alter its internal values after the transistor has been manufactured. Therefore,
manufacturers resort to giving transistor electrical parameters in another manner, namely
the ‘two port’ approach. This is shown in Figure 6.31. With this approach, manufacturers
simply state that for a given transistor operating under certain conditions, what you can
expect to find at the input port (port 1) and the output port (port 2), when you apply exter-
nal voltages (v1, v2) and currents (i1, i2) to it.
Two port parameters come from manufacturers in different representations. Each type
of representation can be described with names such as:
admittance – Y-parameters transfer – ABCD-parameters
hybrid – H-parameters impedance – Z-parameters
scattering – S-parameters Smith chart information
For radio frequency work, the most favoured parameters are the y-, s- and h-parameters
and information on Smith charts which is a convenient graphical display of y- and s-
Fig. 6.31 Two port network representation
A.C. equivalent circuits of transistors 297
6.4.7 Radio frequency amplifiers
Radio frequency (r.f.) amplifiers will be investigated by first considering one method of
modelling transistors at radio frequencies. This method will then be used to design an
aerial distribution amplifier.
R.F. transistor modelling
Transistor modelling serves two main purposes. First, it enables a transistor designer to
analyse what is happening within a transistor and to design the necessary modifications to
improve performance. Second, it enables a circuit designer to understand what is happen-
ing within a circuit and to carry out the necessary adjustments to achieve optimum circuit
performance from the transistor.
The hybrid p model (Figure 6.29) is particularly good for representing the properties of
a transistor but as frequencies increase, its shunt reactances cannot be neglected and its
equivalent representation becomes increasingly complex. To minimise these complica-
tions, electronic circuit designers prefer to treat a transistor as a complete unit or ‘black
box’ and to consider its performance characteristics rather than the individual components
in its equivalent circuit.
One way to do this is to use admittance parameters or y-parameters. In this
approach, the transistor is represented as a two-port network with input port (port 1) and
output port (port 2) as shown in Figure 6.32(a). In Figure 6.32(a), the details of the compo-
nents within the ‘box’ are not given. The equivalent representation shown in Figure 6.32(b)
simply tells us that if the input and output voltages v1 and v2 are changing, then the currents
i1 and i2 must also be changing in accordance with the equations
i1 = y11v1 + y12v2 (6.10)
i2 = y21v1 + y22v2 (6.11)
i1 is the a.c. current flowing into the input port (port 1)
i2 is the a.c. current flowing into the output port (port 2)
v1 is the a.c. voltage at the input port
v2 is the a.c. voltage at the output port
In practice, the values of the y-parameters, y11, y12, y21, y22, are specified at a particu-
lar frequency, in a particular configuration (common base, common emitter, or common
Fig. 6.32 (a) Admittance parameters; (b) equivalent representation
298 High frequency transistor amplifiers
collector) and with stated values of transistor operating voltages and currents. It is impor-
tant to remember that y-parameters are measured phasor quantities, obtained by measur-
ing external phasor voltages and currents for a particular transistor.
From inspection of Equations 6.10 and 6.11, we define
⎧i ⎫
y11 = ⎨ 1 ⎬ (6.12)
⎩ v1 ⎭v2 = 0
⎧i ⎫
y12 = ⎨ 1 ⎬ (6.13)
⎩ v2 ⎭v1 = 0
⎧i ⎫
y21 = ⎨ 2 ⎬ (6.14)
⎩ v1 ⎭v2 = 0
⎧i ⎫
y22 = ⎨ 2 ⎬ (6.15)
⎩ v2 ⎭v1 = 0
Figure 6.33 shows the same transistor (represented by its ‘y’-parameters) being driven
by a constant current signal source (is) with a source admittance (Ys). The transistor feeds
a load admittance (YL). By inspection of the circuit in Figure 6.33, it can be seen that
is = Ysv1 + i1
i1 = is – Ysv1 (6.16)
i2 = –YLv2 (6.17)
Note the minus sign. This is because current is flowing in the opposite direction to that
indicated in the diagram.
Circuit parameters can be calculated as follows.
Fig. 6.33 Equivalent circuit of a transistor with source (Ys) and load (YL)
A.C. equivalent circuits of transistors 299
Voltage gain. Voltage gain (Av) is defined as (v2/v1). Substituting Equation 6.17 in
Equation 6.11
i2 = y21v1 + y22v2 = –YLv2
v2(YL + y22) = –y21v1
v2 –y21
— = ———— = Av (6.18)
v1 YL + y22
Input admittance (yin). Input admittance (yin) is defined as i1/v1. Substituting Equation
6.18 in Equation 6.10
{ }
i1 = y11v1 + y12v2 = y11v1 + y12 ————
y22 + YL
and transposing
i1 y12y21
— = y11 – ——— = Yin (6.19)
v1 y22 + yL
Current gain (Ai). Current gain (Ai) is defined as i2/i1. From Equations 6.19 and 6.17
i1 –i2
v1 = — and v2 = ——
Yin YL
and substituting in Equation 6.11
[ ] [ ]
i1 i2
i2 = y21 —— – y22 ——
Yin YL
[ ][ ]
YL + Y22 Y21
i2 ———— = —— i1
YL Yin
i2 y21YL
— = —————— = Ai (6.20)
i1 Yin (y22 + YL)
Output admittance (Yout). Output admittance (Yout) is defined as [i2/v2] . Substituting
is = 0
Equation 6.16 in Equation 6.10
i1 = y11v1 + y12v2 = –Ysv1 (remember is = 0)
300 High frequency transistor amplifiers
{ }
v1 = ————
y11 + Ys
Substituting in Equation 6.11
{ }
i2 = y21 ———— + y22v2
y11 + YS
Transposing v2 results in
i2 y12y21
— = y22 – ——— = Yout (6.21)
v2 y11 + YS
Equations 6.18 to 6.21 enable us to calculate the performance of a transistor circuit.
The equations are in a general form and apply to a transistor regardless of whether it
is operating in the common emitter, common base or common collector mode. The
only stipulations are that you recognise that signal enters and leaves the transistor at
port 1 and port 2 respectively and that you use the correct set of ‘y’-parameters in the
Design case: aerial amplifier design using ‘y’-parameters
In this design study (Figure 6.34), the signal is picked up by an aerial whose source imped-
ance is 75 W. The signal is then fed into an amplifier whose load is a 300 W distribution
system which feeds signals to all the domestic VHF/FM receivers in the house. The design
was carried out in the following manner.
1 Manufacturers’ data sheets were used to find a transistor which will operate satisfac-
torily at 100 MHz; the approximate centre of the VHF broadcast band. The transistor is
assumed to be unconditionally stable.
2 A decision was made on transistor operating conditions. Guidelines are usually given in
the data sheets for operating conditions and ‘y’-parameters. Typical operating conditions
for a well known transistor with d.c. conditions (Vce = 6 V, Ic = 1 mA) operating in the
common emitter mode were found to be:
Fig. 6.34 Aerial amplifier design study
A.C. equivalent circuits of transistors 301
y11 = (13.752 + j13.946) mS y12 = (–0.146 – j1.148) mS
y21 = (1.094 – j17.511) mS y22 = (0.3 + j1.571) mS
The relevant information is summarised below.
Zs= 75 W or Ys = 13.33 mS ∠ 0° ZL = 300 W or YL = 3.33 mS ∠ 0°
y11 = (13.75 + j13.95) mS y12 = (–0.15 – j1.15) mS
y21 = (1.09 – j17.51) mS y22 = (0.3 + j1.57) mS
Required: (a) Voltage gain (Av), (b) input admittance (Yin), (c) output admittance (Yout).
Solution. In this solution, you should concentrate on the method used, rather than the
laborious arithmetic which can be easily checked by a calculator. Simpler numerical para-
meters were not used because they do not reflect realistic design problems.
(a) Use Equation 6.18 to calculate the voltage gain (Av):
–y21 1/180° × (1.09–j17.5) mS
Av = ———— = ——————————————
YL + y22 3.33 mS/0° + (0.3 + j1.57) mS
17.54 mS/93.56°
= ———————— = 4.44/70.17°
3.95 mS/23.39°
From the above answer, you should note the following.
• The phase relationship between the input and output signals is not always the usual 180°
phase reversal expected in a low frequency common emitter amplifier. This is because
of transistor feedback caused mainly by the reduced reactance of the internal collec-
tor–base capacitance at higher radio frequencies.
• The gain and phase relationship is also dependent on the magnitude and phase of the load.
Figure 6.35 shows the relationships between the input (OA) and output (OB) voltages.
It is readily seen that there is a component of the output signal which is in phase with the
Fig. 6.35 OA and OB represent input and output phasor voltages
302 High frequency transistor amplifiers
input signal. This in-phase component can cause instability problems if it is allowed to
stray back into the input port. To keep the two signals apart, good layout, short connecting
leads and shielding are essential.
If a tuned circuit is used as an amplifier load, its impedance and phase will vary with
tuning. This in turn affects the amplitude and phase relationships between amplifier input
and output voltages. These variations must be incorporated into the amplifier design other-
wise instability will occur.
Input admittance (Yin)
(b) For this calculation, we use Equation 6.19:
Yin = Y11 – ————
Y22 + YL
(–0.15 – j1.15) mS (1.09 – j17.51) mS
= (13.75 + j13.95) mS – ————————————————
(0.3 + j1.57) mS + 3.33 mS/0°
(1.16 mS2 ∠ –97.4°)(17.54 mS ∠ –86.44°)
= (13.75 + j13.95) mS – —————————————————
(3.63 + j1.57) mS
20.35 mS ∠ –183.84°
= (13.75 + j13.95) mS – ——————————
3.95 mS ∠ 23.39°
= (13.75 + j13.95) mS – 5.15 mS ∠ –207.23°
= (13.75 + j13.95) mS – (–4.58 + j2.36) mS
= (18.33 + j11.59) mS
From the answer, you should note Yin is also dependent on the phase of the load admit-
tance. This is particularly important in multi-stage amplifiers where the input admittance
of the last amplifier provides the load for the amplifier before it. In this case, altering or
tuning the load of the last stage amplifier will affect its input admittance and in turn affect
the load of the amplifier driving it. This is the reason why the procedure for tuning a multi-
stage amplifier usually requires that the last stage be adjusted before the earlier stages.
Output admittance (Yout)
(c) For this calculation, we use Equation 6.21:
Yout = y22 – ————
y11 + Ys
(–0.15 – j1.15) mS (1.09 – j17.51) mS
= (0.3 + j1.57) mS – ————————————————
(13.75 + j13.95) mS + 13.33 mS ∠ 0°)
(1.16 ∠ –97.4°) mS (17.54 ∠ –86.44°) mS
= (0.3 + j1.57) mS – ————————————————— —
(13.75 + j13.95) mS + 13.33 mS ∠ 0°
20.35 mS2 ∠ –183.84°
= (0.3 + j1.57) mS – —————————
(27.08 + j13.95) mS
A.C. equivalent circuits of transistors 303
20.35 mS2 ∠ –183.84°
= (0.3 + j1.57) mS – —————————
30.46 mS ∠ 27.25°
= (0.3 + j1.57) mS – 0.67 mS ∠ –211.09°
= (0.3 + j1.57) mS – (–0.57 + j0.35) mS
= (0.87 + j1.22) mS
From the answer, you should note Yout is also dependent on the phase of the source admit-
tance. This is particularly important in multi-stage amplifiers where the load admittance of
the first amplifier provides the source for the amplifier following it. In this case, altering
or tuning the load of the first stage will affect the source admittance of the second stage
and so on. This is why, having tuned a multi-stage high frequency amplifier once, you
usually have to repeat the tuning again to compensate for the changes in the output admit-
Theoretically, it would appear that there is no satisfactory way of tuning a multi-stage
amplifier because individual amplifiers affect each other. In practice, it is found that after
the second tuning, little improvement is obtained if a subsequent re-tune is carried out.
Therefore as a compromise between performance and labour costs, most multi-stage
amplifiers are considered to be tuned after the second tuning.
Summary of the design case. From this design study, you have learnt how to calculate the
voltage gain, input and output admittances of a simple amplifier. You should also have under-
stood why good shielding and layout practices are important in high frequency amplifiers
and the reasons for the procedures used in tuning multi-stage high frequency amplifiers.
Example 6.13
Using the parameters given in the case study, calculate the current gain (Ai) of the amplifier.
Solution. Using Equation 6.20, current gain (Ai) is defined as i2/i1. Hence
i2 y21 YL
Ai = — = —————
i1 Yin (y22 + YL)
(1.09 – j17.51) mS × 3.33 mS/0°
= ——————————————————————
(18.33 + j11.59) mS [(0.3 + j1.57) mS + 3.33 mS/0° ]
(17.54 ∠ –86.44°) mS × 3.33 × mS/0°
= —————————————————
(21.69 ∠ 32.31°) mS (3.96 mS ∠ 23.39°)
(58.41 ∠ –86.44°) mS2
= —————————
(85.89 ∠ 55.70°) mS2
= 0.68 ∠ –142.14°
Note: Current gain is less than unity. This is because the load admittance is low.
304 High frequency transistor amplifiers
6.5 General r.f. design considerations
The example given in the Design Study has been based on a design methodology, where
we have assumed that the transistor is unconditionally stable, that gain is not of paramount
importance, and that the inherent electrical noise of the amplifier is not prevalent. In real
life, ideal conditions do not exist and we must trade off some properties at the expense of
others. The designs that follow show you how these trade-offs can be carried out.
Design of linear r.f. small signal amplifiers is usually based on requirements for specific
power gain at specific frequencies. Other design considerations include stability, band-
width, input–output isolation and production reproducibility. After a basic circuit type is
selected, the applicable design equations can be solved.
Many r.f. amplifier designs fail because the incorrect transistor has been chosen for the
required purpose. Two of the most important considerations in choosing a transistor for use
in any amplifier design are its stability and its maximum available gain (MAG). Stability,
as it is used here, is a measure of the transistor’s tendency to oscillate, that is to provide an
output signal with no intended input signal. MAG is a figure-of-merit for a transistor which
indicates the maximum theoretical power gain which can be obtained from a transistor
when it is conjugately matched12 to its source and load impedances. MAG is never
achieved in practice because of resistive losses in a circuit; nevertheless MAG is extremely
useful in evaluating the initial capabilities of a transistor.
6.5.1 Stability
A major factor in the overall design is the potential stability of the transistor. A transistor
is stable if there is no output signal when there is no input signal. There are two main
stability factors that concern us in amplifier design, (i) the stability factor of the transistor
on its own, and (ii) the stability factor of an amplifier circuit.
Linvill stability factor
The Linvill stability factor is used to determine the stability of a transistor on its own, that
is when its input and output ports are open-circuited. Linvill’s stability factor (C) can be
calculated by using the following expression:
|yf yr|
C = ———————— (6.22)
2gigo – Re (yf yr)
|yf yr| = magnitude of the product in brackets
yf = forward-transfer admittance
yr = reverse-transfer admittance
gi = input conductance
go = output conductance
Re = real part of the product in parentheses
12 A signal source generator (Z ) will deliver maximum power to a load (Z ) when its source impedance (Z
g L g
= Rg + jXg) = (ZL = RL – jXL). The circuit is said to be conjugately matched because Rg = RL and Xg = –XL.
General r.f. design considerations 305
When C < 1, the transistor is unconditionally stable at the bias point and the frequency
which you have chosen. This means that you can choose any possible combination of
source and load impedance for your device and the amplifier will remain stable providing
that no external feedback paths exist between the input and output ports.
When C > 1, the transistor is potentially unstable and will oscillate for certain values
of source and load impedance. However, a C factor greater than 1 does not indicate that
the transistor cannot be used as an amplifier. It merely indicates that you must exercise
extreme care in selecting your source and load impedances otherwise oscillations may
occur. You should also be aware that a potentially unstable transistor at a particular
frequency and/or operating point may not necessarily be unstable at another frequency
and/or operating point. If for technical or economical reasons, you must use a transistor
with C > 1, then try using the transistor with a different bias point, and/or mismatch the
input and output impedances of the transistor to reduce the gain of the stage.
The Linvill stability factor (C) is useful in predicting a potential stability problem. It
does not indicate the actual impedance values between which the transistor will go un-
stable. Obviously, if a transistor is chosen for a particular design problem, and the tran-
sistor’s C factor is less than 1 (unconditionally stable), that transistor will be much easier
to work with than a transistor which is potentially unstable. Bear in mind also that if C is
less than but very close to 1 for any transistor, then any change in operating point due to
temperature variations can cause the transistor to become potentially unstable and most
likely oscillate at some frequency. This is because Y-parameters are specified at a partic-
ular operating point which varies with temperature. The important rule is: make C as
small as possible.
Example 6.14
When operated at 500 MHz with a Vce of 5 V and Ic = 2 mA, a transistor has the follow-
ing parameters:
yi = (16 + j11.78) mS yr = (1.55 ∠ 258°) mS
yf = (45 ∠ 285°) mS yo = (0.19 + j5.97) mS
Calculate its Linvill stability factor.
Solution. Using Equation 6.22
|yf yr|
C = ———————
2gigo – Re (yf yr)
|(45 mS ∠ 285° × 1.55 mS ∠ 258°|
= —————————————————————————
2 × 16 mS × 0.19 mS – Re (45 mS ∠ 285° × 1.55 mS ∠ 258°)
|(69.75 mS2 ∠ 183°| |(69.75 mS2 ∠ 183°|
= —————————————— = ——————————
6.08 mS2 – Re (69.75 mS2 ∠ 183°) 6.08 mS2 – (–69.65 mS2)
69.75 mS2
= ——— —— = 0.92
75.73 mS2
306 High frequency transistor amplifiers
Since the Linvill stability factor < 1, the transistor is unconditionally stable. However, it is
only just unconditionally stable and, in production, changes in transistor parameters might
easily cause instability. If due to costs or the desire to stock a minimum inventory of parts,
you cannot change the transistor, try another operating bias point.
The Stern stability factor (K)
The Stern stability factor (K) is used to predict the stability of an amplifier when it is oper-
ated with certain values of load and source impedances. The Stern stability factor (K) can
be calculated by:
2(gi + Gs) (go + GL)
K = ————————— (6.23)
|yf yr| + Re (yf yr)
Gs = the source conductance
GL = the load conductance
If K > 1, the circuit is stable for that value of source and load impedance. If K < 1, the
circuit is potentially unstable and will most likely oscillate at some frequency or in a
production run of the circuit.
Example 6.15
A transistor operating at VCE = 5 V, IC = 2 mA at 200 MHz with a source impedance of
(50 + j0) W and a load impedance of (1000 + j0) W has the following y-parameters:
yi = (4.8 + j4.52) mS yr = (0.90 ∠ 265°) mS
yf = (61 ∠ 325°) mS yo = (0.05 + j2.26) mS
What is the Stern stability factor of the circuit?
YS = 1/(ZS) = 1/(50 + j0) = 20 mS
YL = 1/(ZL) = 1/(1000 + j0) = 1 mS
Using Equation 6.23:
2(gi + Gs)(go + GL)
K = ————————
|yf yr| + Re (yf yr)
2 (4.8 mS + 20 mS) (0.05 mS + 1 mS)
= ——————————————————————————————
|61 mS ∠ 325° × 0.9 mS ∠ 265°| + Re (61 mS ∠ 325° × 0.9 mS ∠ 265°)
2 (24.8 mS) (1.05 mS) 52.08 mS2
= ————————————— = ——————————
54.9 mS2 + Re (54.9 mS2 ∠ 230°) 54.9 mS2 + (–35.29 mS2)
General r.f. design conditions 307
52.08 mS2
= ———— = 2.656 ≈ 2.66
19.61 mS2
Since K > 1, the circuit is stable.
Summary of the Linvill and Stern stability factors
The Linvill stability factor (C) is useful in finding stable transistors:
• if C < 1, the transistor is unconditionally stable;
• if C > 1, the transistor is potentially unstable.
The Stern stability factor (K) is useful for predicting stability problems with circuits:
• if K > 1, the circuit is stable for the chosen source and load impedance;
• if K < 1, the circuit is potentially unstable for the chosen source and load.
6.5.2 Maximum available gain
The maximum available gain (MAG) of a transistor can be found by using the following
MAG = ——— (6.24)
MAG is useful in the initial search for a transistor for a particular application. It gives a
good indication as to whether a transistor will provide sufficient gain for a task.
The maximum available gain for a transistor occurs when yr = 0, and when YL and YS
are the complex conjugates of yo and yi respectively. The condition that yr must equal zero
for maximum gain to occur is due to the fact that under normal conditions, yr acts as a
negative feedback path internal to the transistor. With yr = 0, no feedback is allowed and
the gain is at a maximum.
In practical situations, it is physically impossible to reduce yr to zero and as a result
MAG can never be truly obtained. However, it is possible to very nearly achieve the MAG
calculated in Equation 6.24 through a simultaneous conjugate match of the input and
output impedances of the transistor. Therefore, Equation 6.24 remains a valuable tool in
the search for a transistor provided you understand its limitation. For example, if your
amplifier design calls for a minimum gain of 20 dB at 500 MHz, find a transistor that will
give you a small margin of extra gain, preferably at least about 3–6 dB greater than 20 dB.
In this case, find a transistor that will give a gain of approximately 23–26 dB. This will
compensate for realistic values of yr, component losses in the matching networks, and vari-
ations in bias operating points.
Example 6.16
A transistor has the following Y-parameters:
yi = (16 + j11.78) mS yr = (1.55 ∠ 258°) mS
yf = (45 ∠ 285°) mS yo = (0.19 + j5.97) mS
308 High frequency transistor amplifiers
when it is operated at VCE = 5 V and IC = 2 mA at 500 MHz. Calculate its maximum avail-
able gain?
Solution. Using Equation 6.24
|yf|2 |45 mS|2
MAG = ——— = —————————
4gigo 4 × 16 mS × 0.19 mS
2025 mS2
= ———— = 166.53 or 22.21 dB
12.16 mS2
6.5.3 Simultaneous conjugate matching
Optimum power gain is obtained from a transistor when yi and yo are conjugately matched
to Ys and YL respectively. However the reverse-transfer admittance (yr) associated with
each transistor tends to reflect13 any immittance (impedance or admittance) changes made
at one port back to the other port, causing a change in that port’s immittance characteris-
tics. This makes it difficult to design good matching networks for a transistor while using
only its input and output admittances, and totally ignoring the contribution that yr makes
to the transistor’s immittance characteristics. Although YL affects the input admittance of
the transistor and YS affects its output admittance, it is still possible to provide the transis-
tor with a simultaneous conjugate match for maximum power transfer (from source to
load) by using the following design equations:
[2 gi go − Re( yf yr )]2 − yf yr (6.25)
Gs =
2 go
when yr = 0
Gs = gi (6.25a)
Im (yf yr)
Bs = – jbi + ———— (6.26)
[2 gi go − Re( yf yr )]2 − yf yr (6.27)
GL =
2 gi
or by using Equation 6.25 for the numerator
GL = ——— (6.27a)
13 If you have forgotten this effect, refer to Equations 6.19 and 6.21.
General r.f. design considerations 309
Im(yf yr)
BL = – jbo + ———— (6.28 )
2 gi
Gs = source conductance
Bs = source susceptance
GL = load conductance
BL = load susceptance
Im = imaginary part of the product in parenthesis
The above equations may look formidable but actually they are not because the numera-
tors in these sets of equations are similar and need not be calculated twice. A case study
of how to apply these equations is shown in Example 6.17.
Example 6.17
Design an amplifier which will provide maximum gain for conjugate matching of source
and load at 300 MHz. The transistor used has the following parameters at 300 MHz with
VCE = 5 V and IC = 2 mA:
yi = 17.37 + j11.28 mS yr = 1.17 mS ∠ –91°
yo = 0.95 + j3.11 mS yf = 130.50 mS ∠ –69°
What are the admittance values which must be provided for the transistor at (a) its input
and (b) its output?
Given: yi = 17.37 + j11.28 mS yr = 1.17 mS ∠ – 91°
yo = 0.95 + j3.11 mS yf = 130.50 mS ∠ – 69°
f = 300 MHz, VCE = 5 V and IC = 2 mA
Required: (a) Its input and (b) its output admittances for conjugate match.
1 Calculate the Linvill stability factor (C) using Equation 6.22:
|yf yr|
C = ———————
2gi go – Re(yf yr)
|(130.5 mS ∠ –69°) (1.17 mS ∠ –91°)|
= ———————————————————————————
2 (17.37 mS)(0.95 mS) – Re[(130.5 mS ∠ –69°)(1.17 mS ∠ –91°)]
152.69 mS2 152.69 mS2
= ——————————— = ————— = 0.87
33.00 mS2 – (–143.48 mS2) 176.48 mS2
Since C < 1, the device is unconditionally stable and we may proceed with the design.
If C > 1, we would have to be extremely careful in matching the transistor to the source
and load as instability could occur.
310 High frequency transistor amplifiers
2 Calculate the maximum available gain (MAG) using Equation 6.24:
yf 130.5 mS ∠ − 69°
MAG = =
4 gi go 4(17.37 mS)(0.95 mS)
17 030.15 µS2
= = 258 or 24.12 dB
66.01 µS2
The actual gain achieved will be less due to yr and component losses.
3 Determine the conjugate values to match transistor input admittance using Equation
[2 gi go − Re( yf yr )]2 − yf yr
Gs =
2 go
[33 µS2 − Re(152.69 µS2 ∠ − 160°)]2 − 152.69 µS2
1.9 mS
[33 µS2 − ( −143.48 µS2 )]2 − 152.69 µS2
1.9 mS
[176.48 µS2 )]2 − 152.69 µS2 [13 145.19 − 23 314.24] pS4
= =
1.9 mS 1.9 mS
[7830.95] pS4 88.49 µS2
= = = 46.57 mS
1.9 mS 1.9 µS
Using Equation 6.26
Im(yf yr)
Bs = – jbi + ————
–52.22 mS2
= –j11.28 mS + j ————— = – j38.76 mS
2(0.95 mS)
Therefore the source admittance for the transistor is (46.57 – j38.76) mS. The transistor
input admittance is (46.57 + j38.76) mS.
4 Determine the conjugate values to match transistor output admittance using Equation
Gsgo (46.57)(0.95) mS2
— —
GL = —— = ——————— = 2.55 mS
gi 17.37 mS
Using Equation 6.28
General r.f. design conditions 311
Im(yf yr)
BL = – jbo + ————
–52.22 mS2
= – j3.11 mS + j —————— = – j4.61 mS
2 (17.37 mS)
Therefore, the load admittance required for the transistor is (2.55 – j4.61) mS. The tran-
sistor output admittance is (2.55 + j4.61) mS.
5 Calculate the Stern stability factor (K) using Equation 6.23:
2(gi + Gs)(go + GL)
K = —————————
|yf yr| + Re(yf yr)
2(17.37 + 46.57)(0.95 + 2.55) mS2
= ————————————————
|152.69| mS2 + Re(152.69 ∠ –160°) mS2
(2)(63.94)(3.50) mS2
= ——————————— = 48.60
152.69 mS2 + (–143.48) mS2
Since K > 1, the circuit is stable.
After you have satisfied yourself with the design, you need to design networks which will
give the transistor its required source and load impedances and, within reason, also its
operating bandwidth. This can be done by using the filter and matching techniques
described in Chapter 5 and/or by using the Smith chart and transmission line techniques
explained in Chapter 3. Smith chart and transmission line techniques will be expanded in
the microwave amplifiers which will be designed in Chapter 7.
Summary. The calculated parameters of Example 6.17 are:
• C = 0.87
• MAG = 24.12 dB
• conjugate input admittance = (46.57 – j38.76) mS
• conjugate output admittance = (2.55 – j4.61) mS
• K = 4.45
6.5.4 Transducer gain (GT)
Transducer gain (GT) of an amplifier stage is the gain achieved after taking into account
the gain of the device and the actual input and output impedances used. This is the term
most often used in r.f. amplifier design work. Transducer gain includes the effects of input
and output impedance matching as well as the contribution that the transistor makes to the
overall gain of the amplifier stage. Component resistive losses are neglected. Transducer
gain (GT) can be calculated from:
GT = ——————————— (6.29)
|(yi + Ys)(yo + YL) – yf yr|2
312 High frequency transistor amplifiers
where Ys and YL are respectively the source and load admittances used to terminate the
Example 6.18
Find the gain of the circuit that was designed in Example 6.17. Disregard any component
Solution. The transducer gain for the amplifier is determined by substituting the values
given in Example 6.17 into Equation 6.29:
GT = ———————————
|(yi + Ys)(yo + YL) – yf yr|2
4(46.57)(2.55)|130.50|2 × 10–12
= ————————————————————————————
|(63.94 – j27.48)(3.50 – j1.50) × 10–6 – (152.69 ∠ –160°) × 10–6|2
8 089 607.17 × 10–12
= ———————————————————————————————
|69.60 × 10–6 ∠ –23.26° × 3.81 × 10–6 ∠ –23.20° – 152.69 × 10–6 ∠ –160°|2
8 089 607.17 × 10–12
= ————————————————————
|265 × 10–6 ∠ –46.46° – 152.69 × 10–6 ∠ –160°|2
8 089 607.17 × 10–12
= ————————————————————————————
|182.55 × 10–6 – j192.10 × 10–6 – (–143.48 × 10–6 – j52.22 × 10–6)|2
8 089 607.17 × 10–12
= ————————————
|326.03 – j139.88|2 × 10–12
8 089 607.17
= ————————— = 64.28 or 18.08 dB
|354.74 ∠ –23.22|2
The transistor gain calculated in Example 6.18 is approximately 6 dB less than the MAG
that was calculated in Example 6.17. In this particular case, the reverse-transfer admittance
(yr) of the transistor has taken an appreciable toll on gain. It is best to calculate GT imme-
diately after the transistor’s load and source admittances are determined to see if the gain
is sufficient for your purpose.
If you cannot tolerate the lower gain, the alternatives are:
• increase the operating current to increase gm and hopefully achieve more gain;
• unilaterise or neutralise the transistor to increase gain (this is explained shortly);
• find a transistor with a higher fT, to reduce the effect of yr.
If you carry out one or more of the items above, you will have to go through all the calcu-
lations in Examples 6.17 and 6.18 again.
General r.f. design considerations 313
6.5.5 Designing amplifiers with conditionally stable transistors
If the Linvill stability factor (C) calculated with Equation 6.22 is greater than 1, the tran-
sistor chosen is potentially unstable and may oscillate under certain conditions of source
and load impedance. If this is the case, there are several options available that will enable
use of the transistor in a stable amplifier configuration:
• select a new bias point for the transistor; this will alter gi and go;
• unilaterise or neutralise the transistor; this is explained shortly;
• mismatch the input and output impedance of the transistor to reduce stage gain.
Alternative bias point
The simplest solution is probably a new bias point, as any change in a transistor’s biasing
point will affect its r.f. parameters. If this approach is taken, it is absolutely critical that the
bias point be temperature-stable over the operational temperature range especially if C is
close to unity.
Unilaterisation and neutralisation
Unilaterisation consists of providing an external feedback circuit (Cn and Rn in Figure
6.36) from the output to the input. The external current is designed to be equal but oppo-
site to the internal yr current so that the net current feedback is zero. Stated mathematically,
I(R C ) = –I(y ) and the effective composite reverse-transfer admittance is zero. With this
n n r
condition, the device is unconditionally stable. This can be verified by substituting yr = 0
in Equation 6.22. The Linvill stability factor in this case becomes zero, indicating uncon-
ditional stability.
One method of applying unilaterisation is shown in Figure 6.36(a). The principle of
operation is explained in Figure 6.36(b). Referring to the latter figure, Vt is the total volt-
age across the tuned circuit, n1 and n2 form the arms of one side of the bridge while Cn and
Rn (external components) and the yr components (Rr and Cr) form the other arm of the
bridge. Cn and Rn are adjusted until the bridge is balanced for zero feedback, i.e. VBE = 0.
It follows that at balance
Fig. 6.36 (a) Transformer unilaterisation circuit Fig. 6.36 (b) Equivalent unilaterisation circuit
314 High frequency transistor amplifiers
n2 Zr
VEC = ———— Vt = ——— Vt
n1 + n2 Zr + Zn
and after cross-multiplication
Zn = Zr —— (6.30)
Often when yr is a complex admittance consisting of gr ± jbr, it becomes very difficult to
provide the correct external reverse admittance needed to totally eliminate the effect of yr.
In such cases neutralisation is often used. Neutralisation is similar to unilaterisation
except that only the imaginary component of yr is counteracted. An external feedback path
is constructed as before, from output to input such that Bf = br (n1/n2). Thus, the compos-
ite reverse-transfer susceptance is effectively zero. Neutralisation is very helpful in stabil-
ising amplifiers because in most transistors, gr is negligible when compared to br. The
effective cancellation of br very nearly cancels out yr. In practical cases, you will find that
neutralization is used instead of unilaterisation. However, be warned: the addition of exter-
nal components increases the costs and the complexity of a circuit. Also, most neutralisa-
tion circuits tend to neutralise the amplifier at the operating frequency only and may cause
problems (instability) at other frequencies.
Summing up, unilaterisation/neutralisation is an effective way of minimising the effects
of yr and increasing amplifier gain, but it costs more and is inherently a narrow-band
compensation method.
Mismatching techniques
A more economical method stabilising an amplifier is to use selective mismatching.
Another look at the Stern stability factor (K) in Equation 6.23 will reveal how this can be
done. If Gs and GL are made large enough to increase the numerator sufficiently, it is pos-
sible to make K greater than 1, and the amplifier will then become stable for those termi-
nations. This suggests selectively mismatching the transistor to achieve stability. The price
you pay is that the gain of the amplifier will be less than that which would have been pos-
sible with a simultaneous conjugate match.
Procedure for amplifier design using conditionally stable
The procedure for a design using conditionally-stable devices is as follows.
1 Choose Gs based on some other criteria such as convenience of input-network, Q factor.
Alternatively, from the transistor’s data sheet, choose Gs to be that value which gives
you transistor operation with minimal noise figure.
2 Select a value of K that will assure a stable amplifier (K > 1).
3 Substitute the above values for K and Gs into Equation 6.2. and solve for GL.
4 Now that Gs and GL are known, all that remains is to find Bs and BL. Choose a value of
BL equal to –bo of the transistor. The corresponding YL which results will then be very
close to the true YL that is theoretically needed to complete the design.
5 Next calculate the transistor input admittance (Yin) using the load chosen in step 4 and
Equation 6.19:
General r.f. design considerations 315
y12 y21
yin = y11 – ————
y22 + YL
where YL = GL ± jBL (found in steps 3 and 4).
6 Once Yin is known, set bs equal to the negative of the imaginary part of Yin or Bs = –Bin.
7 Calculate the gain of the stage using Equation 6.29.
From this point forward, it is only necessary to produce input and output admittance
networks that will present the calculated Ys and YL to the transistor. Example 6.19 shows
how the procedure outlined above can be carried out.
Example 6.19
A transistor has the following y-parameters at 200 MHz:
yi = 2.25 + j7.2 yr = 0.70 /–85.9°
yf = 44.72 /–26.6° yo = 0.4 + j1.9
All of the above parameters are in mS. Find the source and load admittances that will
assure you of a stable design. Find the gain of the amplifier.
Given: yi = 2.25 + j7.2 yr = 0.70 /–85.9°
yf = 44.72 /–26.6° yo = 0.4 + j1.9
Required: (a) Load admittance, (b) source admittance and (c) gain when the circuit is
designed for a Stern stability factor (K) of 3.
Solution. If you were to use Equation 6.22 to calculate the Linvill stability factor (C) for
the transistor, you will find C = 2.27. Therefore the device is potentially unstable and you
must exercise extreme caution in choosing source and load admittances for the transistor.
1 The data sheet for the transistor states that the source resistance for optimum noise
figure is 250 W. Choosing this value results in Gs = 1/Rs = 4 mS.
2 For an adequate safety margin choose a Stern stability factor of K = 3.
3 Substituting GS and K into Equation 6.23 and solving for GL yields
2(gi + GS)(go + GL)
K = —————————
|yf yr| + Re(yf yr)
(2)(2.25 + 4)(0.4 + GL)
3 = ———————————
|31.35| + Re(–12)
GL = 4.24 mS
4 Setting BL = –bo of the transistor
BL = –j1.9 mS
The load admittance is now defined as
YL = (4.24 – j1.9) mS
316 High frequency transistor amplifiers
5 The input admittance of the transistor is calculated using Equation 6.19:
y12 y21
yin = y11 – ————
y22 + YL
(0.701 ∠ –85.9°)(44.72 ∠ –26.6°)
= 2.25 + j7.2 – ——————————————
0.4 + j1.9 + 4.24 – j1.9
= (4.84 + j13.44) mS
6 Setting Bs equal to the negative of the imaginary part of Yin, yields
Bs = –j13.44 mS
The source admittance needed for the design is now defined as:
Ys = (4 – j13.44) mS
7 Now that Ys and YL are known, we can use Equation 6.29 to calculate the gain of the
GT = ———————————
|(yi + Ys)(yo + YL) – yf yr|2
= ————————————————
|(6.25 – j6.24)(4.64) – (–12 – j28.96)|2
135 671.7
= ————— = 80.71 or 19.1 dB
Therefore even though the transistor is not conjugately matched, you can still realise a
respectable amount of gain while maintaining a stable amplifier.
After you have satisfied yourself with the design, you need to design networks which
will give the transistor its required source and load impedances and, within reason, also its
operating bandwidth. This can be done by using the filter and matching techniques
described in Part 5 and/or by using Smith chart and transmission line techniques explained
in Part 3. Smith chart and transmission line techniques will be expanded in the microwave
amplifiers which will be designed in Part 7.
6.6 Transistor operating configurations
6.6.1 Introduction
Sometimes you will find that you want one set of y-parameters (e.g. common base para-
meters) while the manufacturer has only supplied y-parameters for the common emitter
configuration. What do you do? Well, you simply use the indefinite admittance matrix
to convert from one set of parameters to another.
Transistor operating configurations 317
6.6.2 The indefinite admittance matrix
Admittance parameters provide an easy way of changing the operating configuration of a
transistor. For example, if y-parameters for the common emitter configuration are known,
it is easy to derive the parameters for common base and common collector configurations.
These derivations are carried out using the indefinite admittance matrix method. The use
of this matrix is best shown by example but to avoid confusion in the discussions which
follow, it is best to first clarify the meaning of the suffixes attached to y-parameters.
Each y-parameter is associated with two sets of suffixes. The first set, yi, yr, yf and yo,
refer to y11, y12, y21, y22 respectively. (The symbols i, r, f, and o stand for input, reverse
transconductance, forward transconductance and output respectively.) The second set, e, b,
c, refer to the emitter, base or collector configuration respectively. For example, yie refers
to y11 in the common emitter mode, yrb refers to y12 in the common base mode, yfc refers
to y21 in the common collector mode and so on.
An admittance matrix for a transistor is made as follows.
1 Construct Table 6.2.
2. Insert the appropriate set of y-parameters into the correct places in the table. If the
common emitter parameters for a transistor are:
yie = (13.75 + j13.95) × 10–3 S yre = (–0.15 – j1.15) × 10–3 S
yfe = (1.09 – j17.51) × 10–3 S yoe = (0.30 + j1.57) × 10–3 S
This set should be inserted as shown in Table 6.3.
Note: No entries are made in the emitter row and column. Similarly, if a set of common
base parameters is used instead, no entries will be made in the base row and column.
The same applies for a set of common collector parameters; no entries are made in the
collector row and column.
3. Sum real and imaginary parts of all rows and columns to zero. See Table 6.4.
4. Extract the required set of parameters.
Table 6.2 Blank indefinite admittance matrix
Base Emitter Collector
Table 6.3 Indefinite admittance matrix with common emitter entries
Base Emitter Collector
Base (13.75 + j13.95)10–3 S (–0.15 – j1.15)10–3 S
Collector (1.09 – j1.75)10–3 S (0.30 + j1.57)10–3 S
Table 6.4 Indefinite admittance matrix with rows and columns summed to zero
Base Emitter Collector
Base (13.75 + j13.95)10–3 S (–13.60 – j12.80)10–3 S (–0.15 – j1.15)10–3 S
Emitter (–14.84 – j12.20)10–3 S (14.99 + j12.62)10–3 S (–0.15 – j0.42)10–3 S
Collector (1.09 – j1.75)10–3 S (–1.39 + j0.18)10–3 S (0.30 + j1.57)10–3 S
318 High frequency transistor amplifiers
To obtain the y-parameters for the common base configuration, ignore all data in the base
row and base column but extract the remaining four parameters. These are:
yib = (14.99 + j12.62) × 10–3 S yrb = (–0.15 – j0.42) × 10–3 S
yfb = (–1.39 + j0.18) × 10–3 S yob = (0.30 + j1.57) × 10–3 S
To obtain the y-parameters for the common collector configuration, ignore all data in
the collector row and collector column but extract the remaining four parameters. These
yic = (13.75 + j13.95) × 10–3 S yrc = (–13.60 – j12.80) × 10–3 S
yfc = (–14.84 – j12.20) × 10–3 S yoc = (14.99 + j12.62) × 10–3 S
This information can now be applied to the general Equations 6.18 to 6.21 and subsequent
The information given above has been shown for a bi-polar transistor but the method is
general and applies to other transistor types as well. For FETS, replace the words base,
emitter and collector in Tables 6.2 to 6.4 by gate, source and drain respectively.
Example 6.20
A transistor operating with the d.c. conditions of VCE = 5 V, IC = 2 mA and at a frequency
of 500 MHz is stated by the manufacturer to have the following y-parameters in the
common emitter mode.
y11 = (16 + j12) mS y12 = 1.55 mS ∠ 258°
y21 = 45 mS ∠ 285° y22 = (0.19 + j6) mS
Calculate its equivalent y-parameters for the base configuration when the transistor is oper-
ating with the same d.c. operating conditions and at the same frequency.
Solution. We will use the indefinite admittance matrix but, before doing so, the parame-
ters y12 and y21 must be converted from its polar form:
y12 = 1.55 mS ∠ 258° = (–0.32 – j1.52) mS
y21 = 45 mS ∠ 285° = (11.65 – j43.47) mS
Fill in the indefinite admittance matrix and sum all rows and columns to zero results in
Table 6.5.
Table 6.5
Base (mS) Emitter (mS) Collector (mS)
Base (16 + j12) (–15.68 – j10.48) (–0.32 – j1.52)
Emitter (–27.65 + 31.47) (27.52 – j26.99) (0.13 – j4.48)
Collector (11.65 – j43.47) (–11.84 + j37.47) (0.19 + j6)
Extracting the common base parameters yields:
yib = (27.52 – j26.99) mS yrb = (0.13 – j4.48) mS
yfb = (–11.84 + j37.47) mS yob = (0.19 + j6) mS
Summary 319
6.7 Summary
In Chapter 6, we have reviewed the operating principle of bi-polar transistors FETs and
MOSFETs. We have also reviewed some transistor biasing methods. A brief resumé of a.c.
equivalent circuits was introduced. Admittance parameters (y) were re-introduced, derived
and applied to the design of amplifiers. You were shown methods on how to design ampli-
fiers, conjugate matched amplifiers and amplifiers using conditionally-stable transistors.
Our software program, PUFF, has no facilities for employing y-parameters directly.
However, if you want, you can use Table 3.1 of Chapter 3 to convert all the y-parameters
(or at least the results) into s-parameters. You can then use the PUFF techniques of Chapter
7 for amplifier design. Details of this technique are explained in Examples 7.1 and 7.2
followed by its PUFF design results in Figure 7.4 in the next chapter.
Microwave amplifiers
7.1 Introduction
In Part 2, we introduced transmission lines. Part 3 was devoted to Smith charts and scat-
tering parameters while Part 4 covered the use of PUFF as a computing aid. In Part 5, we
discussed the behaviour of passive devices such as capacitors and inductors at radio
frequencies and investigated the use of these elements in the design of resonant tuned
circuits, filters, transformers and impedance matching networks. We showed how the
indefinite admittance matrix can be used to convert transistor parameters given in one
configuration to another configuration. In Chapter 6, we investigated biasing techniques,
the a.c. equivalent circuit of transistors, admittance parameters, and their use in high
frequency amplifier design. We will now combine all of this information and use it in the
design of microwave amplifiers.
7.1.1 Aim
The main aim of this chapter is to show how microwave amplifiers can be designed using
scattering parameters.
7.1.2 Objectives
After you have read this chapter, you should be able to:
• calculate transistor stability;
• calculate maximum available gain of an amplifying device;
• design amplifiers with conjugate matching impedances;
• design amplifiers using conditionally stable transistors;
• design amplifiers for a specific gain;
• understand and calculate stability circles;
• design amplifiers for optimum noise figure;
• understand broadband matching amplifier techniques;
• design broadband amplifiers;
• understand feedback amplifier techniques;
• design feedback amplifiers.
Transistors and S-parameters 321
7.2 Transistors and s-parameters
7.2.1 Introduction
The purpose of this section is to show you how s-parameters can be used in the design of
transistor amplifiers. It has already been shown that transistors can be characterised by
their s-parameters. Smith charts have been introduced, therefore it is now time to apply
these parameters to produce practical design amplifiers.
7.2.2 Transistor stability
Before designing a circuit, it is important to check whether the active device which we will
use is (i) unconditionally stable or (ii) conditionally stable. This is necessary because
different conditions require the appropriate design method. It is possible to calculate
potential instabilities in transistors even before an amplifier is built. This calculation serves
as a useful aid in finding a suitable transistor for a particular application.
To calculate a transistor’s stability with s-parameters, we first calculate an intermediate
quantity Ds where
Ds = s11s22 – s12s21 (7.1)
We do this because in the expressions that follow, you will find that the quantity Ds is used
many times and we can save ourselves considerable work by doing this.
The Rollett stability factor (K) is calculated as:
1 + |Ds| 2 – |s11| 2 – |s22| 2
K = —————————— (7.2)
(2)(|s21|) (|s12|)
If K is greater than 1, the transistor is unconditionally stable for any combination of
source and load impedance. If K is less than 1, the transistor is potentially unstable and
will most likely oscillate with certain combinations of source and load impedances. With
K less than 1, we must be extremely careful in choosing source and load impedances for
the transistor. It does not mean that the transistor cannot be used for a particular applica-
tion; it merely indicates that the transistor will have to be used with more care.
If K is less than 1, there are several approaches that we can take to complete the design:
• select another bias point for the transistor;
• choose a different transistor;
• design the amplifier heeding carefully detailed procedures that we will introduce shortly.
7.2.3 Maximum available gain
The maximum gain we can ever get from a transistor under conjugately matched condi-
tions is called the maximum available gain (MAG). Maximum available gain is calcu-
lated in two steps.
(1) Calculate an intermediate quantity called B1, where
B1 = 1 + |s11| 2 – |s22| 2 – |Ds| 2 (7.3)
Ds is calculated from Equation 7.1.
322 Microwave amplifiers
Note: The reason B1 has to be calculated first is because its polarity determines which sign
(+ or –) to use before the radical in Equation 7.4 which follows shortly.
(2) Calculate MAG using the result from Equation 7.2:
MAG = 10 log + 10 log K ± K 2 − 1 (7.4)
MAG = maximum available gain in dB
K = stability factor from Equation 7.2
Note: K must be greater than 1 (unconditionally stable) otherwise Equation 7.4 will be
undefined because the radical sign will become an imaginary number rendering the equa-
tion invalid. Thus, MAG is undefined for unstable transistors.
7.3 Design of amplifiers with conjugately matched impedances
This method of design is only applicable to transistors which are stable and give sufficient
gain for our design aims. For other conditions we will have to use other design methods.
This design procedure results in load and source reflection coefficients which provide a
conjugate match for the actual output and input impedances of the transistor. However,
remember that the actual output impedance of a transistor is dependent on its source
impedance and vice-versa. This dependency is caused by the reverse gain (s12) of the tran-
sistor. If s12 was zero, then of course the load and source impedances would have no effect
on the transistor’s input and output impedances.
7.3.1 Output reflection coefficient
To find the desired load reflection coefficient for a conjugate match
C2 = s22 – (Ds s11*) (7.5)
where the asterisk indicates the complex conjugate of s11 (same magnitude but opposite
angle). The quantity Ds is the quantity calculated in Equation 7.1.
Next we calculate B2:
B2 = 1 + |s22| 2 – |s11| 2 – |Ds| 2 (7.6)
The magnitude of the reflection coefficient is found from the equation
B2 ± B2 − 4 C2 (7.7)
ΓL =
2 C2
The sign preceding the radical is opposite to the sign of B2 previously calculated in
Equation 7.6. The angle of the load reflection coefficient is simply the negative of the angle
of C2 calculated in Equation 7.5.
Design of amplifiers with conjugately matched impedances 323
After the desired load reflection coefficient is found, we can either (a) plot GL on a
Smith chart to find the load impedance (Z) directly, or (b) substitute GL from the equation
ZL – Zo
GL = ———— (7.8)
ZL + Zo
You have encountered the above equation in Chapter 2 when we were looking at trans-
mission lines.
7.3.2 Input reflection coefficient
With the desired load reflection coefficient specified, the source reflection coefficient
needed to terminate the transistor’s input can now be calculated:
⎡ s s Γ ⎤
Γs = ⎢s11 + 12 21 L ⎥ (7.9)
⎣ 1 − s22 ΓL ⎦
The asterisk sign again indicates the conjugate of the quantity in brackets (same magni-
tude but opposite sign for the angle). In other words, once we complete the calculation
within brackets of Equation 7.9, we will have the correct magnitude but the incorrect angle
sign and will have to change the sign of the angle.
As before when Gs is found, it can be plotted on a Smith chart or we can again use the
Zo – Zs
Gs = – —— —— (7.10)
Zo + Zs
All the foregoing is best clarified by an example.
Example 7.1
A transistor has the following s-parameters at 150 MHz with a Vce = 12 V and Ic = 8 mA:
s11 = 0.3 ∠ 160° s12 = 0.03 ∠ 62°
s21 = 6.1 ∠ 65° s22 = 0.40 ∠ –38°
The amplifier must operate between 50 W terminations. Design (a) input and (b) output
matching networks for simultaneously conjugate matching of the transistor for maximum
Given: ƒ = 150 MHz, Vce = 12 V, Ic = 8 mA, s11 = 0.3 ∠ 160°, s12 = 0.03 ∠ 62°, s21 = 6.1
∠ 65°, s22 = 0.40 ∠ –38°.
Required: Conjugate input and output matching networks for maximum gain to 50 W
source and load impedances.
Solution. Using Equations 7.1 and 7.2, check for stability:
Ds = s11s22 – s12s21
= (0.3 ∠ 160°)(0.4 ∠ – 38°) – (0.03 ∠ 62°)(6.1 ∠ 65°)
= (0.120 ∠ 122°) – (0.183 ∠ 127°)
324 Microwave amplifiers
= (–0.064 + j0.102) – (–0.110 + j0.146)
= (0.046 – j0.044)
= (0.064 ∠ –43.73°)
Using the magnitude of Ds, calculate K:
1 + |Ds| 2 – |s11| 2 – |s22| 2
K = ——————————
1 + (0.064) 2 – (0.3)2 – (0.4)2
= ————————————
= 2.06
Since K > 1, the transistor is unconditionally stable and we may proceed with the design.
Using Equation 7.3, calculate B1:
B1 = 1 + |s11| 2 – |s22| 2 – |Ds| 2
B1 = 1 + (0.3)2 – (0.4)2 – (0.064)2
= 0.926
Using Equation 7.4, calculate maximum available gain (MAG):
MAG = 10 log + 10 log K ± K 2 − 1
Since B1 is positive, the negative sign will be used in front of the square root sign and
MAG = 10 log + 10 log 2.06 − (2.06)2 − 1
= 23.08 + ( −5.87)
= 17.21 dB
We will consider 17.21 dB to be adequate for our design gain of 16 dB minimum. If the
design had called for a minimum gain greater than 16 dB, a different transistor would be
To find the load reflection coefficient for a conjugate match, the two intermediate quan-
tities (C2 and B2) must first be found. Using Equation 7.5:
C2 = s22 – (Dss11*)
= (0.4 ∠ –38°) – [(0.064 ∠ –43.73°) (0.3 ∠ –160°) ]
= 0.315 – j0.246 – [–0.018 + j0.008]
= (0.419 ∠ –37.33°)
Using Equation 7.6
B2 = 1 + |s22| 2 – |s11| 2 – |Ds| 2
= 1 + (0.4)2 – (0.3)2 – (0.064)2
= 1.066
Therefore the magnitude of the load reflection coefficient can now be found using
Equation 7.7:
Design of amplifiers with conjugately matched impedances 325
B2 ± B2 − 4 C2
ΓL =
2 C2
1.066 − (1.066)2 − 4(0.419)2
= 0.486
The angle of the load reflection coefficient is simply equal to the negative of the angle of
C2 or +37.33°. Thus
GL = 0.486 ∠ 37.33°
GL can now be substituted in Equation 7.9 to calculate Gs:
s12s21 GL *
Gs = s11 + ————
1 – s22 GL ]
(0.03 ∠ 62°)(6.1 ∠ 65°)(0.486 ∠ 37.33°) *
[ 0.3 ∠ 160° + —————————————————
1 – (0.4 ∠ –38°)(0.486 ∠ 37.33°) ]
(0.089 ∠ 164.33°) *
= (–0.282 + j0.103) + —————————
(1 – 0.194 ∠ –0.670°) ]
(0.089 ∠ 164.33°) *
= (–0.282 + j0.103) + ————————
(0.806 ∠ 0.142°)
= [(–0.282 + j0.103) + (–0.160 + j0.030)]*
= [0.463 ∠ 163.29°]*
= 0.463 ∠ –163.29°
Once the desired Gs and GL are known, all that remains is to provide the transistor with
components which ‘look like’ Gs and GL.
(a) Input matching network. The input matching network design is shown on the Smith
chart of Figure 7.1. The object of the design is to force the 50 Ω source to present a reflec-
tion coefficient1 of 0.463 ∠ –163.29° to the transistor input. With Gs plotted as shown, the
corresponding desired and normalised impedance is read directly from the chart as Z s =
(0.36 – j0.12) Ω. Remember this is a normalised impedance because the chart has been
normalised to 50 Ω. The actual impedance represented by Gs is equal to 50(0.36 – j0.12)
Ω = (18.6 – j6) Ω.
Now we must transform the 50 Ω source to (18.6 – j6) Ω impedance. The most
common circuit used is a low pass filter configuration consisting of a shunt C and a series
1 In all examples containing reflection coefficients and Smith charts, the reflection coefficients are plotted on
the Smith chart and the resultant values are read from it. Theoretically, the Smith chart should give you an exact
answer. In practice, reading difficulties and interpolations must be made, so expect slightly different answers if
you are using mathematics to derive these values. Strictly speaking, this is not a problem because transistor char-
acteristics change, even when the same type number is used. Hence perfect match with one transistor does not
mean perfect match with another transistor. The most difficult part of amplifier design is choosing loads that will
produce the same circuit characteristics in spite of transistor changes.
326 Microwave amplifiers
Fig. 7.1 Input matching network A = 1 + j0, B = 0.43 – j0.5, C = 0.43 – j0.14
L. Remember that when using a Smith chart with shunt elements, you use it as an admit-
tance chart, and when using the chart for series elements, you use it as an impedance chart.
For ease of transformation, we use the Smith chart type ZY-10-N which was introduced in
Chapter 3. Proceeding from the source, we have:
Arc AB = shunt C = j1.33 S
Arc BC = series L = j0.34 Ω
If you have difficulty following the above construction, look first at Figure 7.3 where you
will see the schematic of the amplifier. It starts off with a 50 Ω source (point A on Figure
7.1). Across this point, we have a shunt capacitor; therefore we must use the admittance
part of the chart. This shunt capacitance moves us to point B in Figure 7.1, i.e. along arc
AB. The next element in Figure 7.1 is a series inductor. We must therefore use the imped-
ance part of the Smith chart in Figure 7.1. Our final destination is the required source
Design of amplifiers with conjugately matched impedances 327
impedance for transistor (point C in Figure 7.1) – hence the arc BC. The Smith chart values
are read according to the part of the chart being used; admittance values for the shunt
components and impedance values for the series components.
The actual component values are found using Equations 7.11 to 7.14 which are:
Cseries = ——— (7.11)
ω XN
Cshunt = —— (7.12)
L series = —— (7.13)
Lshunt = —— (7.14)
N = normalisation value
B = susceptance in Siemens
X = reactance in ohms
w = frequency in radians/second
For this example
C1 = ———————— = 28.22 pF ≈ 28 pF
2p(150 MHz)(50)
L1 = —————— = 18.04 nH ≈ 18 nH
2p(150 MHz)
This completes the input matching network.
(b) Output matching network. The load reflection coefficient is plotted in Figure 7.2
and after plotting in GL = 0.486 ∠ 37.33° the Smith chart shows a normalised load imped-
ance of (1.649 + j1.272) Ω. After re-normalisation, it represents a load impedance ZL =
50 (1.649 + j1.272) Ω or (82.430 + j63.611) Ω. The matching network is designed as
follows. Proceeding from the load:
Arc AB = series C = –j1.1 Ω
Arc BC = shunt L = –j0.8 S
If you have difficulty with the Smith chart, look at the schematic on Figure 7.3. The
final load is 50 Ω (point A in Figure 7.2). The transistor load is point C in Figure 7.2.
328 Microwave amplifiers
Fig. 7.2 Output matching network A = 1 + j0, B = 1 –j1.28, C = 0.486 ∠ 37.33° or 1.65 + j1.272
Starting from the 50 Ω point (point A in Figure 7.2), you encounter a series capacitor.
Therefore the series part of the chart must be used. This takes you to point B on the
chart, along arc AB. Next you have a shunt inductor (see Figure 7.3); therefore you
must use the admittance part of the chart to get to your destination (point C), that is
along arc BC in Figure 7.2. The Smith chart values are read according to the part of the
chart being used; admittance values for the shunt components and impedance values for
the series components.
The actual component values are found using Equations 7.11 and 7.14 which are for
this example:
C2 = ————————— = 19.29 pF ≈ 19 pF
2 p(150 MHz)(1.1)(50)
Design of amplifiers with conjugately matched impedances 329
Fig. 7.3 R.F. circuit for Example 7.1
L2 = ———————— = 66.31 nH ≈ 66 nH
2 p(150 MHz)(0.8)
The completed design (minus biasing network) is shown in Figure 7.3.
Transducer gain (GT)
The transducer gain is the actual gain of an amplifier stage including the effects of input
and output matching and device gain. It does not include losses attributed to power dissi-
pation in imperfect components. Transducer gain is calculated as follows:
|s21|2(1 – |Gs|2)(1 – |GL|2)
GT = ————————— ———— ——— (7.15)
|(1 – s11Gs)(1 – s22GL) – s12s21GLGs|2
Gs and GL are the source and load reflection coefficients respectively
Calculation of GT is a useful method of checking the power gain of an amplifier before it
is built. This is shown by Example 7.2.
Example 7.2
Calculate the transducer gain of the amplifier that was designed in Example 7.1.
Given: As in Example 7.1.
Required: Transducer gain.
Solution. Using Equation 7.15
|s21|2(1 – |Gs|2)(1 – |GL|2)
GT = ———————————————
|(1 – s11Gs)(1 – s22GL) – s12s21GLGs|2
(6.1)2(1 – (0.463)2)(1 – (0.486)2)
= ———————————————————————————————————————————
|(1 – 0.139 ∠ 3.3°)(1 – 0.194 ∠ 01.7°) – (0.03 ∠ 62°)(6.1 ∠ 65°)(0.486 ∠ 37.33°)(0.463 ∠ –163.29°)|2
22.329 22.329
≈ ——————————— = ————————————
|0.694 – (0.041 ∠ 1.04°)|2 |0.694 – (0.041 + j0.001)|2
330 Microwave amplifiers
= ———
= 52.365 or 17.19 dB
Note: The transducer gain calculates to be very close to MAG. This is due to the fact that
s12 is not equal to zero and is therefore providing some internal transistor feedback.
PUFF results
The results of Examples 7.1 and 7.2 using PUFF are shown in Figure 7.4. However, you
should be aware of how PUFF was modified for the results.
• The amplitude range and the frequency range in the rectangular plot had to be modified
as explained in Chapter 4 to obtain the required amplitude and frequency ranges.
• There is no transistor device in PUFF that meets the requirements of the transistor in the
examples. A new transistor template called ‘ed701’ was generated by copying the
FHX04.device template and modifying it to suit the requirements of the example.
• A resistor of 1 pΩ was generated within PUFF to provide a spacer for ease in laying out
the circuit.
• The input matching circuit of Figure 7.1 was plotted at port 1. At the match frequency,
the input match reflection coefficient is shown as –35.39 dB in Figure 7.4.
• The output matching circuit of Figure 7.2 was plotted at port 2. At the match frequency,
Fig. 7.4 Results of Examples 7.1 and 7.2 using PUFF
Design of amplifiers with conjugately matched impedances 331
the output match reflection coefficient is shown as –21.21 dB in Figure 7.4.
• The gain of the circuit is given by PUFF as 17.13 dB. We calculated the gain as 17.19 dB.
Note: There is a very slight discrepancy between the results but this is to be expected
because the examples were carried out graphically. Nonetheless, this should convince you
that our design methods are reliable!
Example 7.3
A MESFET has the following S-parameters at 5 GHz with Vce = 15 V and Ic = 10 mA:
s11 = 0.3 ∠ 140° s12 = 0.03 ∠ 65°
s21 = 2.1 ∠ 62° s22 = 0.40 ∠ –38°
Calculate the maximum available gain (MAG) for the transistor under these operating
Given: ƒ = 5 GHz, Vce = 15 V, Ic = 10 mA, s11 = 0.3 ∠ 140°, s12 = 0.03 ∠ 65°, s21 = 2.1
∠ 62°, s22 = 0.40 ∠ –38°.
Required: MAG for the MESFET at 5 GHz.
Solution. Using Equations 7.1 and 7.2, check for stability:
Ds = s11s22 – s12s22
= (0.3 ∠ 140°)(0.4 ∠ –38°) – (0.03 ∠ 65°)(2.1 ∠ 62°)
= (0.120 ∠ 102°) – (0.063 ∠ 127°)
= (–0.025 + j0.117) – (–0.038 + j0.050)
= (0.013 + j0.067)
= (0.068 ∠ 79.06°)
Using the magnitude of Ds, calculate K:
1 + |Ds|2 – |s11|2 – |s22|2
K = ——————————
(2)(|s21|) (|s12|)
1 + (0.068)2 – (0.3)2 – (0.4)2
= ————————————
= 5.99
Since K > 1, the transistor is unconditionally stable and we may proceed with the design.
Using Equation 7.3, calculate B1:
B1 = 1 + |s11|2 – |s22|2 – |Ds|2
= 1 + (0.3)2 – (0.4)2 – (0.068)2
= 0.925
Using Equation 7.4, calculate maximum available gain (MAG):
Since B1 is positive, the negative sign will be used in front of the square root sign:
MAG = 10 log + 10 log K ± K 2 − 1
332 Microwave amplifiers
If you wish, check the answer by using PUFF.
MAG = 10 log + 10 log 5.99 − (5.99)2 − 1
= 18.45 + ( −10.75)
≈ 7.7 dB
Example 7.4
An integrated circuit has the following S-parameters:
s11 = 0.3 ∠ 140° s12 = 0.03 ∠ 65°
s21 = 2.1 ∠ 62° s22 = 0.40 ∠ –38°
If its source reflection coefficient Gs = 0.463 ∠ –164° and its load reflection coefficient GL
= 0.486 ∠ 38°, calculate the transducer gain of the amplifier.
Given: Gs = 0.463 ∠ –140° GL = 0.486 ∠ 38°
s11 = 0.3 ∠ 140° s12 = 0.03 ∠ 65°
s21 = 2.1 ∠ 62° s22 = 0.40 ∠ –38°
Required: Amplifier transducer gain.
Solution. Using Equation 7.15
|s21|2(1 – |Gs|2)(1 – |GL|2)
GT = ———————————————
|(1 – s11Gs)(1 – s22GL) – s12s21GLGs|2
(2.1)2(1 – (0.463)2)(1 – (0.486)2)
= —————————————————————————————————
|(1 – 0.139)(1 – 0.194) – (0.03 ∠ 65°)(2.1 ∠ 62°)(0.486 ∠ 38°)(0.463 ∠ –140°)|2
2.646 2.646
= ——————— ———— = ———————————
|0.694 – (0.014 ∠ 25°)| 2 |0.694 – (0.014 – j0.002)|2
= ———
= 5.708 or 7.6 dB
If you wish, check the answer by using PUFF.
7.4 Design of amplifiers for a specific gain
In cases where a specific gain is required, it is normal practice to provide selective
mismatching so that transistor gain can be reduced to the desired gain. Selective
mismatching is a relatively inexpensive method used to decrease gain by not matching a
transistor to its conjugate load.
One of the easiest ways of selective mismatching is through the use of a constant gain
circle plotted on the Smith chart. A constant gain circle is merely a circle, the circumfer-
ence of which represents a locus of points (load impedances) that will force the amplifier
Design of amplifiers for a specific gain 333
gain to a specific value. For instance, any of the infinite number of impedances located on
the circumference of a 12 dB constant gain circle would force the amplifier stage gain to
12 dB. Once the circle is drawn on a Smith chart, you can see the load impedances that
will provide a desired gain.
7.4.1 Constant gain circles
To plot a constant gain circle on a Smith chart, it is necessary to know (i) where the centre
of the circle is located and (ii) its radius. The procedure for calculating a constant gain
circle is as follows.
1 Calculate Ds as in Equation 7.1.
2 Calculate D2:
D2 = |s22|2 – |Ds|2 (7.16)
3 Calculate C2:
C2 = s22 – Dss11* (7.17)
4 Calculate G:
|desired gain|
G = —————— (7.18)
5 Calculate centre location of constant gain circle:
6 Calculate radius of the circle:
GC2 (7.19
ro =
1 + D2 G
Equation 7.19 produces a complex number in magnitude–angle format similar to that of a
1 − 2 K s12 s21 G + s12 s21 G 2
po = (7.20
1 + D2 G
reflection coefficient. This number is plotted on the Smith chart exactly as you would plot
a value of reflection coefficient.
The radius of the circle that is calculated with Equation 7.20 is simply a fractional
number between 0 and 1 which represents the size of that circle in relation to a Smith chart.
A circle with a radius of 1 has the same radius as a Smith chart; a radius of 0.5 represents
half the radius of a Smith chart and so on.
After the load reflection coefficient (in effect the load impedance) is chosen by the
designer, the next step will be to determine the source reflection coefficient that is required
to prevent any further decrease in gain. This value is of course the conjugate of the actual
input reflection coefficient of the transistor with the specified load calculated by Equation
7.9. To clarify the procedure, we now present Example 7.5.
334 Microwave amplifiers
Example 7.5
A transistor has the following S-parameters at 1 GHz, with Vce = 15 V and Ic = 5 mA:
s11 = 0.28 ∠ –58° s12 = 0.08 ∠ 92°
s21 = 2.1 ∠ 65° s22 = 0.8 ∠ –30°
Design an amplifier to present 9 dB of gain at 1 GHz. The source impedance Zs = (35 –
j60) Ω and the load impedance ZL = (50 – j50) W. The transistor is unconditionally stable
with K = 1.168. Design the output and input networks.
Given: ƒ = 1 GHz Vce = 15 V Ic = 5 mA s11 = 0.28 ∠ –58°
s12 = 0.08 ∠ 92° s21 = 2.1 ∠ 65° s22 = 0.8 ∠ –30°
Zs = (35 – j60) Ω ZL = (50 – j50) K = 1.168
Required: 9 dB amplifier, output network, input network.
Solution. Using Equation 7.1, find Ds for substitution in Equations 7.16 and 7.17:
Ds = s11s22 – s12s21
= (0.28 ∠ –58°)(0.8 ∠ –30°) – (0.08 ∠ 92°)(2.1 ∠ 65°)
= (0.224 ∠ –88°) – (0.168 ∠ 157°)
= 0.008 – j0.224 + 0.155 – j0.066
= 0.333 ∠ –60.66°
Using Equation 7.16, find D2 for subsequent insertion in Equation 7.19:
D2 = |s22|2 – |Ds|2
= (0.8)2 – (0.333)2
= 0.529
Using Equation 7.17, find C2 for subsequent insertion in Equation 7.19:
C2 = s22 – Dss11*
= (0.8 ∠ –30°) – (0.333 ∠ –60.66°)(0.28 ∠ 58°)
= (0.693 – j0.400) – (0.093 – j0.004)
= (0.719 ∠ –33.42°)
Bearing in mind that a power ratio of 9 dB is a power ratio of 7.94, use Equation 7.18 to
find G for subsequent insertion in Equation 7.19:
|desired gain| 7.94
G = —————— = ——— = 1.80
|s21|2 (2.1)2
Using Equation 7.19, find the centre of the constant gain circle:
GC2* 1.80 (0.719 ∠ 33.42°)
ro = ———— = —————————— = 0.663 ∠ 33.42°
1 + D2G 1 + (0.529)(1.80)
Using Equation 7.20, find the radius for the 9 dB constant gain circle:
1 − 2 K s12 s21 G + s12 s21 G 2
po =
1 + D2 G
Design of amplifiers for a specific gain 335
1 − 2(1.168)(0.08)(2.1)(1.80) + (0.08 × 2.1)2 (1.80)2
1 + (0.529)(1.80)
1 − 0.706 + 0.091
1 + 0.952
= 0.318
The Smith chart construction is shown in Figure 7.5. Note that any load impedance located
on the circumference of the circle will produce an amplifier gain of 9 dB if the input
impedance of the transistor is conjugately matched. The actual load impedance we have to
Fig. 7.5 Output network using 9 dB constant gain circle A = 1 – j1, B = 1 – j3, C = 0.1 – j0.11, ro = 0.663 ∠ 33.43°,
po = 0.318. Angle between point D and E = 33.43°
336 Microwave amplifiers
work with is (50 – j50) which in its normalised form is (1 – j1) Ω on the Smith chart and
denoted by point A.
Output network. The transistor’s output network must transform the actual load imped-
ance into a value that falls on the constant gain 9 dB circle. Obviously there are many
circuit configurations which will satisfy these conditions. The configuration shown in
Figure 7.5 has been chosen for convenience. Proceeding from the load:
Arc AB = series C = –j2.0 Ω
Arc BC = shunt L = –j0.41 S
Using Equation 7.11 for a series C:
C1 = ———————— ≈ 1.6 pF
2p(1 GHz)(2)(50)
Using Equation 7.14 for a shunt L:
L1 = ———————— ≈ 19.4 nH
2p(1 GHz)(0.41)
Input network. For a conjugate match at the input to the transistor with GL = 0.82 ∠ 13°
(point C in Figure 7.5), the desired source reflection coefficient must be (using Equation
s12s21 GL *
Gs =
[(s11 + ————
1 – s22 GL ]
(0.08 ∠ 92°)(2.1 ∠ 65°)(0.82 ∠ 13°) *
[ 0.28 ∠ –58° + ———————————————
1 – (0.8 ∠ –30°)(0.82 ∠ 13°) ]
(0.138 ∠ 170.00°) *
= 0.28 ∠ –58° + ————————
1 – (0.656 ∠ –17°) ]
(0.138 ∠ 170.00°) *
= 0.148 – j0.237 + ————————
(0.420 ∠ 27.24°) ]
= [0.148 – j0.237 + (–0.262 + j0.199]*
= [0.120 ∠ –161.56°]*
= 0.120 ∠ 161.56°
The point is plotted as point D in Figure 7.6. The actual normalised source impedance is
plotted at point A (0.7 – j1.2) Ω. The input network must transform the actual impedance
at point A to the desired impedance at point D. We have used a three element matching
network this time:
Design of amplifiers for a specific gain 337
Fig. 7.6 Input network of Example 7.5 A = 0.7 – j1.2, B = 0.37 + j1.25, C = 0.2 + j0.33, D = 1.25 – j0.1
Arc AB = shunt C2 = j0.63 S
Arc BC = series L2 = j1.08 Ω
Arc CD = shunt C3 = j2.15 S
Using Equation 7.12 for a shunt capacitance:
C2 = ——————— ≈ 2 pF
2p(1 GHz)(50)
Using Equation 7.13 for a series inductance:
L2 = ————— ≈ 8.5 nH
2p(1 GHz)
338 Microwave amplifiers
Fig. 7.7 R.F. circuit for Example 7.5
Using Equation 7.12 for a shunt capacitance:
C3 = —————— ≈ 6.8 pF
2p(1 GHz)(50)
The completed design (minus biasing network) is shown in Figure 7.7.
Example 7.6
Use the information of Example 7.5 to calculate a constant gain circle of 8 dB.
Given: ƒ = 1 GHz Vce = 15 V Ic = 5 mA s11 = 0.28 ∠ –58°
s12 = 0.08 ∠ 92° s21 = 2.1 ∠ 65° s22 = 0.8 ∠ –30°
Ds = 0.333 ∠ –60.66° D2 = 0.529° C2 = 0.719 ∠ –33.42°
K = 1.168
Required: A constant gain circle of 8 dB.
Solution. Bearing in mind that a power ratio of 8 dB is a power ratio of 6.31, use equa-
tion 7.18 to find G for subsequent insertion in Equation 7.19:
|desired gain| 6.31
G = —————— = ——— = 1.43
|s21| 2 (2.1)2
Using Equation 7.19, find the centre of the constant gain circle:
GC2* 1.43(0.719 ∠ 33.42°)
ro = ———— = ————————— = 0.585 ∠ 33.42°
1 + D2G 1 + (0.529)(1.43)
Using Equation 7.20, find the radius for the 9 dB constant gain circle:
1 − 2 K s12 s21 G + s12 s21 G 2
po =
1 + D2 G
Design of amplifiers for a specific gain 339
1 − 2(1.168)(0.08)(2.1)(1.43) + (0.08 × 2.1)2 (1.43)2
1 + (0.529)(1.43)
1 − 0.561 + 0.058
1 + 0.756
= 0.401
7.4.2 Design of amplifiers with conditionally stable devices
When the Rollett stability factor (K) calculates to be less than unity, it is a certainty that
with some combinations of source and load impedances, the transistor will oscillate. To
prevent oscillation the source and load impedances must be chosen very carefully. One of
the best methods of determining those source and load impedances that will cause the tran-
sistor to go unstable is to plot stability circles on a Smith chart.
7.4.3 Stability circles
A stability circle is simply a circle on a Smith chart which represents the boundary
between those values of source or load impedance that cause instability and those that do
not. The circumference of the circle represents the locus of points which forces K = 1.
Either the inside or the outside of the circle may represent the unstable region and that
determination must be made after the circles are drawn.
The location and radii of the input and output stability circles are found as follows.
1 Calculate Ds using Equation 7.1.
2 Calculate C1:
C1 = s11 – Dss22* (7.21)
3 Calculate C2 using Equation 7.5.
4 Calculate the centre location of the input stability circle:
Gs1 = —————— (7.22)
|s11|2 – |Ds|2
5 Calculate the radius of the input stability circle:
| s12s21
ps1 = ——————
|s11|2 – |Ds|2 | (7.23)
6 Calculate the centre location of the output stability circle:
Gs2 = —————— (7.24)
|s22|2 – |Ds|2
7 Calculate the radius of the output stability circle:
| s11s21
ps2 = ——————
|s22|2 – |Ds|2 | (7.25)
340 Microwave amplifiers
Fig. 7.8 Unstable regions for a potentially unstable transistor
Once the calculations are made, circles can be plotted directly on the Smith chart. For a
potentially unstable transistor, the stability circles might resemble those shown in Figure
After the stability circles are plotted on the chart, the next step is to determine which
side of the stability circles (inside or outside) represents the stable region. This is easily
done if s11 and s22 for the transistor are less than 1. If the stability circles do not enclose
the centre of the Smith chart, then regions inside the stability circles are unstable and all
regions outside the stability circles on the Smith chart are stable regions. See Figure 7.8.
If one of the stability circles encloses the centre of the Smith chart, then the region inside
that stability circle is stable. This is because the S-parameters were measured with a 50 Ω
source and load, and since the transistor remained stable (s11 and s22 < |1|) under these
conditions, then the centre of the Smith chart must be part of the stable regions. Example
7.7 will help to clarify this.
Example 7.7
The S-parameters for a transistor at 200 MHz with Vce = 6 V and Ic = 5 mA are:
s11 = 0.4 ∠ 280° s12 = 0.048 ∠ 65°
s21 = 5.4 ∠ 103° s22 = 0.78 ∠ 345°
Design and choose stable load and source reflection coefficients that will provide a power
gain of 12 dB at 200 MHz.
Given: ƒ = 200 MHz Vce = 6 V Ic = 5 mA
s11 = 0.4 ∠ 280° s12 = 0.048 ∠ 65°
s21 = 5.4 ∠ 103° s22 = 0.78 ∠ 345°
Required: Stable load reflection coefficient, stable source reflection coefficient, and
power gain of 12 dB at 200 MHz.
Design of amplifiers for a specific gain 341
1 Using Equation 7.1, find Ds:
Ds = s11s22 – s12s21
= (0.4 ∠ 280°)(0.78 ∠ 345°) – (0.048 ∠ 65°)(5.4 ∠ 103°)
= (–0.027 – j0.311) – (–0.254 + j0.054)
= (0.429 ∠ –58.2°)
2 Using Equation 7.2, calculate Rollett’s stability factor (K):
1 + |Ds|2 – |s11|2 – |s22|2
K = ——————————
(2)(|s21|) (|s12|)
1 + (0.429)2 – (0.4)2 – (0.78)2
= —————————————
= 0.802
Since K < 1, we must exercise extreme care in choosing source and load impedances
otherwise the transistor will oscillate so stability circles must be plotted.
3 Using Equation 7.21, calculate C1:
C1 = s11 – Dss22*
= (0.4 ∠ 280°) – (0.429 ∠ –58.2°)(0.78 ∠ –345°)
= (0.069 – j0.394) – (0.244 – j0.229)
= (0.241 ∠ –136.7°)
4 Using Equation 7.5, calculate C2:
C2 = s22 – (Dss11*)
= (0.78 ∠ 345°) – (0.429 ∠ –58.18°)(0.4 ∠ –280°)
= (0.753 – j0.202) – (0.159 + j0.064)
= 0.651 ∠ –24.1°
5 Using Equation 7.22, calculate the centre location of the input stability circle:
Gs1 = ——————
|s11|2 – |Ds|2
(0.241 ∠ 136.7°)
= ————————
(0.4)2 – (0.429)2
= –10 ∠ 136.7° or 10 ∠ –43.4°
6 Using Equation 7.23, calculate the radius of the input stability circle:
| s12s21
ps1 = ——————
|s11|2 – |Ds|2 |
342 Microwave amplifiers
(0.048 ∠ 65°)(5.4 ∠ 103°)
= ————————————
(0.4)2 – (0.429)2 |
= 10.78
7 Using Equation 7.24, calculate the centre location of the output stability circle:
Gs2 = —————
|s22| 2 – |D |2
(0.651 ∠ 24.1°)
= ————————
(0.78)2 – (0.429)2
= 1.534 ∠ 24.1°
8 Using Equation 7.25, calculate the radius of the output stability circle:
| s12s21
ps2 = ——————
|s22|2 – |Ds|2 |
(0.048 ∠ 65°)(5.4 ∠ 103°)
= ———————————
(0.78)2 – (0.429)2 |
= 0.611
These circles are shown in Figure 7.9. Note that the input stability circle is so large that
it is actually drawn as a straight line on the Smith chart. Since s11 and s22 are both < 1, we
can deduce that the inside of the input stability circle represents the region of stable source
impedances, while the outside of the output stability circle represents the region of stable
load impedances for the device.
The 12 dB gain circle is also shown plotted in Figure 7.9. It is found using Equations
7.1 and 7.16 to 7.20 in a manner similar to that given in Example 7.7. The centre location
for the 12 dB gain circle is GO = 0.287 ∠ 24°. The radius for the 12 dB gain circle is po =
The only load impedances we may not select for the transistor are located inside of the
output stability circle. Any other load impedance located on the 12 dB gain circle will
provide the needed gain provided the input of the transistor is conjugately matched and as
long as the input impedance required for a conjugate match falls inside of the input stabil-
ity circle.
A convenient point on the 12 dB gain circle will be selected and for this example, we
GL = 0.89 ∠ 70°
Using Equation 7.9 to calculate the source reflection coefficient for a conjugate match and
plotting this point on the Smith chart
Gs = 0.678 ∠ 79.4°
Design of amplifiers for a specific gain 343
Fig. 7.9 Stability parameters A = 1 + j0, GL = 0.89 ∠ 70°, GS = 0.678 ∠ 79.4°, Go = 0.287 ∠ 24°
Notice that Gs falls within the stable region of the input stability circle and therefore repre-
sents a stable termination for the transistor. The input and output matching networks are
then designed in the manner detailed in Example 7.5.
Example 7.8
In Example 7.7, the centre location for the 12 dB gain circle is given as ro = 0.287 ∠ 24°
and the radius for the 12 dB circle is given as po = 0.724. Show that these values are
Given: Values of Example 7.7.
Required: ro and po.
Solution. Bearing in mind that a power gain of 12 dB represents a power ratio of 15.85
and using Equation 7.18
344 Microwave amplifiers
gain desired 15.85
G = —————— = ——— = 0.544
|s21|2 5.42
Using Equation 7.16, find D2 for subsequent insertion in Equation 7.19:
D2 = |s22|2 – |Ds|2
= |0.78|2 – |0.429|2
= 0.424
Using Equation 7.19, find the centre of the constant gain circle:
GC2* 0.544(0.651 ∠ 24.1°)
ro = ———— = —————————— = 0.288 ∠ 24.1°
1 + D2G 1 + (0.424)(0.544)
≈ 0.287 ∠ 24°
Using Equation 7.20, find the radius for the 12 dB constant gain circle:
1 − 2 K s12 s21 G + s12 s21 G 2
po =
1 + D2 G
1 − 2(0.802)(0.048)(5.4)(0.544) + (0.048 × 5.4)2 (0.544)2
1 + (0.424)(0.544)
1 − 0.266 + 0.020
1 + 0.231
= 0.724
Example 7.9
In Example 7.8, it is stated that Gs is 0.678 ∠ 79.4°. Show that this value is correct for a
GL of 0.89 ∠ 70°.
Given: Values of Example 7.8.
Required: Gs.
Solution. Using Equation 7.9
s12s21GL *
Gs = s11 + ————
1 – s22GL ]
(0.048 ∠ 65°)(5.4 ∠ 103°)(0.89 ∠ 70°) *
= 0.4 ∠ 280° + ————————————————
1 – (0.78 ∠ 345°)(0.89 ∠ 70°) ]
(0.231 ∠ 238°) *
= (0.069 – j0.394) + ———————
(1 – 0.694 ∠ 55°) ]
Design of amplifiers for optimum noise figure 345
(0.231 ∠ 238°) *
= (0.069 – j0.394) + ———————
(0.828 ∠ –43.36°) ]
= [(0.069 – j0.394) + (0.055 – j0.274)]*
= [0.679 ∠ –79.48°]*
≈ 0.678 ∠ 79.5°
7.5 Design of amplifiers for optimum noise figure
Many manufacturers specify optimum driving resistances and operating currents on their
data sheets for their transistors to operate with minimum noise figures.2 Designing ampli-
fiers for a minimum noise figure then becomes simply a matter of setting the optimum
conditions for a particular transistor. In practice, it means that the input network must be
made to transform the input source generator impedance (generally 50 Ω) to that of the
optimum driving resistance for the transistor to achieve its minimum noise operating
After providing the transistor with its optimum source impedance, the next step is to
determine the optimum load reflection coefficient needed to properly terminate the tran-
sistor’s output. This is given by:
s12s21Gs *
GL = s22 + ————
1 – s11Gs ] (7.26)
Gs is the source reflection coefficient for minimum noise figure
The rest of the design then follows the conventional design methods as you will see in
Example 7.10.
Example 7.10
The optimum source reflection coefficient (Gs) for a transistor under minimum noise
figure operating conditions is Gs = 0.68 ∠ 142°. Its s-parameters under the same condi-
tions are:
s11 = 0.35 ∠ 165° s12 = 0.035 ∠ 58°
s21 = 5.9 ∠ 66° s22 = 0.46 ∠ –31°
The d.c parameters for the transistor are Vce = 15 V, Ic = 4 mA, and the operating frequency
is 300 MHz. Design a low noise amplifier to operate between a 75 Ω source and a 100 Ω
load at 300 MHz.
2 All devices produce electrical noise. In a transistor, noise figure is defined as the ratio of (signal-to-noise
ratio at the input) to (signal-to-noise ratio at the output). It follows that if the transistor has a noise figure of 0 dB
(power ratio = 1) then the signal-to-noise ratio at both the output and input remains the same and we are said to
have a ‘noise free’ transistor. This does not happen in practice although at the time of writing noise figures of
0.5 dB are now being achieved.
346 Microwave amplifiers
Given: s11 = 0.35 ∠ 165° s12 = 0.035 ∠ 58° s21 = 5.9 ∠ 66° s22 = 0.46 ∠ –31°
ƒ = 300 MHz Vce = 12 V Ic = 4 mA hFE gain = 100
Gs = 0.68 ∠ 142° R L = 100 Ω
Required: Low noise amplifier with Zs = 75 Ω, R L = 100 Ω.
1 Using Equation 7.1
Ds = s11s22 – s12s21
= (0.35 ∠ 165°)(0.46 ∠ –31°) – (0.035 ∠ 58°)(5.9 ∠ 66°)
= (–0.122 + j0.116) – (–0.115 + j0.171)
= (0.056 ∠ –86.25°)
2 Using Equation 7.2, calculate Rollett’s stability factor (K):
1 + |Ds|2 – |s11|2 – |s22|2
K = ——————————
1 + (0.056)2 – (0.35)2 – (0.46)2
= —————————————
= 1.620
The Rollett stability factor (K) calculates to be 1.62 which indicates unconditional stabil-
ity. Therefore we may proceed with the design.
3 Input Matching Network
The design values of the matching network are shown in Figures 7.10 and 7.12. Here the
normalised 75 Ω source resistance is transformed to Gs using two components:
Arc AB = shunt C = j1.65 S
Arc BC = series L = j0.85 Ω
4 Using Equation 7.12
C1 = ———————— 17.5 pF
2p(300 MHz) (50)
5 Using Equation 7.13
L1 = ——————— ≈ 22.5 nH
2p(300 MHz)
Design of amplifiers for optimum noise figure 347
Fig. 7.10 Input matching network for Example 7.10 A = (1.5 + j0) Ω or (0.667 – j0) S, B = (0.21 – j0.52) Ω or (0.667
+ j1.653) S, C = (0.212 + j0.33) Ω or (1.433 – j2.184) S
6 Output Matching Network
The load reflection coefficient needed to properly terminate the transistor is found from
Equation 7.26:
s12s21Gs *
GL = s22 + ————
1 – s11Gs ]
(0.035 ∠ 58°)(5.9 ∠ 66°) (0.68 ∠ 142°) *
= 0.46 ∠ –31° + ————————————————
1 – (0.35 ∠ 165°)(0.68 ∠ 142°) ]
(0.140 ∠ 266°)
[ ]
= 0.46 ∠ –31° + ————————
1 – (0.143 – j0.190)
= [(0.394 – j0.237) + (–0.045 – j0.152)]*
= 0.523 ∠ 48.2°
348 Microwave amplifiers
Fig. 7.11 Output matching network for Example 7.10 A = 1.252 + j1.234, B = 2.0 + j1.2, C = 2.0 + j0, D = 0.524
∠ –48.4°
This value along with the normalised load resistance value is plotted in Figure 7.11. The
100 Ω load must be transformed into GL. One possible method is shown in Figure 7.11:
Arc AB = shunt L = –j0.72 S
Arc BC = series C = –j1.07 Ω
Using Equation 7.14, the inductor’s value is
L2 = —————————— ≈ 43 nH
2p (300 MHz) (0.62)
Using Equation 7.11, the series capacitance is
C2 = ——————————— ≈ 8.8 pF
2p (300 MHz) (1.2) (50)
The final design including a typical bias network is shown in Figure 7.12. The 0.1 mF
capacitors are used only as bypass and coupling elements. The gain of the amplifier can be
calculated with Equation 7.15.
Design of broadband amplifiers 349
43 nH
8.8 pF
22.5 nH
Fig. 7.12 Final circuit for Example 7.10
Using PUFF. Here again, you can use PUFF software to design or verify the amplifier
7.6 Design of broadband amplifiers
7.6.1 Design methods
There are many approaches to broadband amplifier design. We can use amplifier
mismatching, feedback amplifiers and distributed amplifiers. We show how amplifier
mismatching can be used in Example 7.11.
7.6.2 Broadband design using mismatch techniques
This method is explained and illustrated by Example 7.11.
Example 7.11
A broadband amplifier is to be designed to operate in the 1.5–2.5 GHz frequency range,
with a 12 dB transducer power gain, using the HP-Avantek 41410 BJT. The S-parameters
of the transistor at the operating range are shown in Table 7.1.
Solution. For the purposes of the example we will assume that s12 ≈ 0 and therefore the
unilateral case is considered. The expression for the transducer power gain in the unilat-
eral case is given as Equation 7.27:
Table 7.1 Scattering parameters of HP-AVANTEK BJT
f (GHz) s11 s12 s21 s22
1.5 0.6 169° 0.04 58° 5.21 58° 0.41 –40°
2.0 0.6 157° 0.05 55° 3.94 55° 0.41 –45°
2.5 0.61 151° 0.06 55° 3.20 50° 0.4 –49°
350 Microwave amplifiers
GTU = GSGOGL (7.27)
GTU = gain of amplifier circuit
GS = ‘gain’ of source network
GO = gain of transistor
GL = ‘gain’ of load network
1 – |GS|2
GS = ————— (7.28)
|1 – S11GS|2
GO = |S21|2 (7.29)
1 – |GL|2
GL = ————— (7.30)
|1 – S22GL|2
Table 7.1 shows that there is a considerable variation of s-parameters with frequency and
the degree of variation can be calculated by Equations 7.27 to 7.30.
The circuit gain is given by Equation 7.27 and it is dependent on GS, GO and GL. If the
individual gains are calculated for a conjugate match at the input and output ports, we will
get the results shown in Table 7.2. The maximum gain we can ever hope to achieve over
the bandwidth 1.5–2.5 GHz is limited by the minimum gain of the three frequencies, i.e.
12.88 dB at 2.5 GHz. The other two frequencies (1.5 GHz and 2.0 GHz) show gains of
17.07 dB and 14.64 dB respectively but these gains can be reduced to 12.88 dB by
mismatching of the ports. Hence, by designing for circuit losses, it is realistic to expect
gains of approximately 12 dB over the three frequencies. Inspection of Table 7.2 reveals
that little variation of GL occurs with frequency. It is therefore easier to manipulate Gs to
achieve the required controlled loss.
Table 7.3 shows the gain characteristic required for Gs to achieve an overall average
gain of 12 dB over the frequency range. The input circuit should now be designed to
produce the required response for Gs shown in Table 7.3. This is carried out by using
Table 7.2 Circuit gain vs frequency (when the input and output circuits are conjugately matched)
f (GHz) GS,max (dB) GO (dB) GL,max (dB) GTU (dB)
1.5 1.94 14.34 0.79 17.07
2.0 1.94 11.91 0.79 14.64
2.5 2.02 10.1 0.76 12.88
Table 7.3 Expected gains for an average overall gain of 12 dB
f (GHz) GS,max (dB) GO (dB) GL,max (dB) GTU (dB)
1.5 – 3.13 14.34 0.79 12
2.0 – 0.7 11.91 0.79 12
2.5 +1.14 10.1 0.76 12
Design of broadband amplifiers 351
Fig. 7.13 Gain of the broadband amplifier
constant gain circles for the three frequencies and by choosing a network that will satisfy
the response for Gs in Table 7.3. An exact response is not always possible and a compro-
mise is often the case.
The process of design is one of trial and error and as such is greatly assisted by opti-
mising software. As the necessary software is not available with this book, no attempt will
be made to do the necessary input broadband matching. Instead we will simply look at the
results to see what can be achieved after CAD matching.
In Figure 7.13, we show how the gain-frequency response of the amplifier has been
improved after optimisation. The amplifier now has a nominal gain of 12 dB è 0.25 dB
over the band 1.5–2.5 GHz, instead of the original 4 dB gain fall-off in gain as calculated
in Table 7.2.
This levelling of gain has been achieved by using a T network as the input matching
circuit. This circuit is shown in Figure 7.14. However, the penalty paid for this levelling of
gain is poor matching at the input circuit.
The return loss of the matching networks of this amplifier is shown in Figure 7.15. Note
that the return loss of the input circuit is poor at 1.5 GHz (about 2 dB) but gradually
improves towards about 11 dB at 2.5 GHz. The return loss of the output circuit is more
even, and ranges from 7.5 dB at 1.5 GHz to about 5 dB at 2.5 GHz.
Fig. 7.14 Circuit diagram of the broadband amplifier
352 Microwave amplifiers
Fig. 7.15 Return loss of the broadband amplifier
7.7 Feedback amplifiers
7.7.1 Introduction
R.F. feedback amplifiers are used in much the same way as feedback elements are intro-
duced in operational amplifier circuits to produce constant gain over a desired bandwidth.
In this section we shall show you how these amplifiers can be designed. Feedback ampli-
fiers are usually designed by first decomposing the combined circuit into individual sub-
systems. They are then re-combined into a composite amplifier and its parameters are then
calculated to yield the desired results.
7.7.2 Design of feedback amplifiers
Consider the basic feedback circuit shown in Figure 7.16. If you look at it closely, you will
find that it consists of two basic parts; the feedback circuit which comprises RFB, LFB and
its d.c. blocking capacitor CFB situated between points A and B, and the transistor circuit
and inductor LD which is also situated between the same two points. Since both circuits
are in parallel, we can draw them as shown in Figure 7.17. For this example, we will make
considerable use of Y-parameters which were originally introduced in Chapter 6.
In Figure 7.17(a), YFB now represents the feedback network, RFB, LFB and its d.c. block-
ing capacitor CFB situated between points A and B. Block YA represents the amplifier and
the inductor LD. Each network is subject to the same voltage across its terminals; therefore
it follows that the currents of each network can be added together to form a composite
network YC. This is also shown diagramatically in Figure 7.17(b). From the composite YC
Feedback amplifiers 353
Fig. 7.16 A radio frequency feedback amplifier
Fig. 7.17 Block diagram of the feedback amplifier of Figure 7.16: (a) composite sections, YA and YFB; (b) combined
network, YC
network, it is now possible to calculate the circuit gain, input and output admittance as a
single circuit in Y-parameters or if you wish you may change them into s-parameters and
carry out the calculations using s-parameters. The conversion tables for changing from one
type of parameter to another are given in Table 3.1. Hence, the parameters of the circuit
can be evaluated using the system above.
To clarify the design method, we will show you a very simple example where this tech-
nique is used. We will assume that the values for the circuit of Figure 7.18 have already
been chosen. Furthermore in order to simplify matters, we will assume that the values are
already in Y-parameters and that only resistances are used in the network. The last assump-
tion simplifies the mathematics considerably yet it does not obscure the principles which
we are trying to use.
Example 7.12
In the circuit of Figure 7.18, the open-circuit generator voltage is 200 mV. Calculate (a)
the input impedance (Zin), (b) the gain (Av) of the circuit and (c) Vout. You can assume that
the d.c. blocking capacitor in the feedback chain has negligible reactance. The transistor
Y-parameters for the given frequency of operation are:
[ 1/1200
1/40 000 ]
Solution. The Y-parameters of the transistor YA are:
YA =
[ 1/1200
1/40 000 ] [
S =
58 333.3
25 ]
354 Microwave amplifiers
Fig. 7.18 Negative feedback amplifier
From inspection of the circuit, the Y-parameters of the feedback element YF are:
YF =
[ 1/10k
1/10k ][
100 ] mS
The composite admittance matrix [YC] = [YA] + [YF]. Hence
[YC] =
[ 833.3
58 333.3
25] [
mS +
[ 933.3
58 233.3 ]
We will now obtain the answers.
(a) Defining [Dy] as [y11y22 – y12y21]
[Dy] = [y11y22 – y12y21] = [9.33 × 1.25 + 1 × 582.3] × 10–8 = 593.96 × 10–8
Using Equation 6.19:
y12y21 Dy + y11YL
yin = y11 – ———— = —————
y22 + yL y22 + YL
y22 + YL [1.25 + 10] × 10–4
Zin = ————— = ————————————
∆y + y11YL [593.96 + 9.33 × 10] × 10–8
11.25 × 10–4
= —————— = 163.9 W
687.26 × 10–8
R.F. power transistors 355
(b) First find vin:
ZinVg 163.69 × 200
vin = ———— = —————— = 15.13 mV
Zin + Zg 2163.7
Using Equation 6.18:
vout –y21 –582.3
Av = — = —— —— = ————
vin y22 + yL 1.25 + 10
= —— — = –51.76
(c) Output voltage is given by
vout = vin × Av
= 15.13 × [–51.76]
= –783 mV
PUFF results. If you wish, you can carry out this example on PUFF by converting the
transistor admittance parameters into scattering parameters and generating a transistor
device as described in Section 4.9. You can then insert your feedback components and vary
them accordingly.
7.7.3 Summary of feedback amplifiers
Example 7.12 should now convince you that the procedure used above is useful for evalu-
ating feedback amplifiers. However, this method is laborious especially without the use of
a computer program.
The disadvantage of this method is that each block, YA, YF and YC, is only applicable
for one frequency at one time. Thus, if you were designing a broadband circuit, you would
have to calculate the parameters for each frequency and then sum up the results. This
involves considerable work if hand calculators are used.
Another great disadvantage is that the component values that you may have chosen in
the first instance may not produce the desired result. Therefore you must carry out the
complete procedure again and again until the desired result is achieved. However, it is
fortunate that good computer programs, such as ‘SUPERCOMPACT, SPICE, etc.’, provide
optimisation facilities and allow you to design the circuit quickly and efficiently.
7.8 R.F. power transistors
The design of r.f. power transistors is treated differently from that of the low power linear
transistors described in the early part of this chapter. The reason for this is because r.f.
power transistors are normally operated in a non-linear mode. This means that manufac-
turers tend to only specify output power and output capacitance for a given input power
and input capacitance. A typical example is shown in Table 7.4 where values of input
356 Microwave amplifiers
Table 7.4 Typical optimum input and conjugate of load impedances for MRF658. Pout= 65 W, Vdc = 12.5 V
Frequency (MHz) Zin W Zout W
400 0.620 + j0.28 1.2 + j2.5
440 0.720 + 0j3.1 1.1 + j2.8
470 0.790 + 0j3.3 0.98 + j3.0
490 0.84 + j3.4 0.91 + j3.2
512 0.88 + j3.5 0.84 + j3.3
520 0.90 + j3.6 0.80 + j3.4
impedance and conjugate of load impedances are specified for a given output power and
operating d.c. voltage. However, once these values are known, the matching networks are
designed in a similar way. You can find a good introduction to the design of power ampli-
fiers by consulting Baeten.3
7.9 Summary
Many of the commonly used techniques in amplifier design have been covered in this
chapter. The circuit topics discussed included transistor stability, maximum available gain
and matching techniques. In addition, we produced design examples of conjugate matched
amplifiers, conditionally stable transistor amplifiers, optimum noise figure amplifiers, and
amplifiers designed for a specific gain.
The design techniques of broadband amplifiers, feedback amplifiers and power ampli-
fiers were investigated. We also showed how some designs can be carried out using the
PUFF software supplied with this book.
You should now have a good knowledge of microwave engineering principles that will
allow you to do simple amplifier designs and to understand more complicated devices and
I would like to remind you of the article ‘Practical Circuit Design’ which has been
reproduced on the disk supplied with this book. This article provides many more examples
of how PUFF can be used in practical circuit design. There are particularly interesting
sections on components, earthing techniques, biasing, passive, active and circuit layout
techniques. The article is crowned by the complete design of a 5 GHz microwave ampli-
fier from its conception as transistor data to final layout. R.F. design calculations are
carried out in Appendix A, input and output line matching and layout using PUFF are
shown. Frequency response and gain are checked with PUFF. Bias design for this ampli-
fier is also given in Appendix B. Calculations for the design of input and output matching
filters are shown in Appendix C.
3 R. Baeten, CAD of a broadband class C 65 watt UHF amplifier, RF Design, March 1993, 132–9.
Oscillators and frequency
8.1 Introduction
Prior to the invention of an amplifying device (vacuum tube, transistor, special negative-
resistance device, etc.) great difficulty was experienced in producing an undamped radio
signal. The early radio transmitters used a high frequency a.c. generator to produce a high
voltage which was increased by a step-up transformer. The output voltage was applied to
a series resonant circuit and the Q of the circuit produced a high enough voltage to jump
across a ‘capacitor gap’ to produce a spark1 which in turn produced a radio signal. This
principle is still used in a petrol engine today where the spark is used to ignite the petrol
mixture. You can frequently hear it on your car radio when the engine cover is removed.
Such a radio signal is damped, i.e. it decays exponentially and it produces many harmon-
ics which interfere with other communication systems. It is now illegal to transmit a
damped oscillation.
There are many criteria in choosing an oscillator, but the main ones are:
• frequency stability
• amplitude stability
• low noise
• low power consumption
• size.
Frequency stability is important because it enables narrow-band communication
systems to be accurately fixed within a frequency band. An unstable frequency oscilla-
tor also behaves like an unstable f.m. modulator and produces unwanted f.m. noise. An
amplitude unstable oscillator behaves like an amplitude modulated modulator because
it produces unwanted a.m. modulation noise. Even if the oscillation frequency and
amplitude can be held precisely, it is inevitable that noise will be produced in an oscil-
lator because of transistor noise which includes ‘flicker noise’, ‘shot noise’ and
‘1/frequency’ noise. In other words, oscillator noise is inevitable but it should be kept
as low as possible. Low power consumption and small size are specially important in
portable equipment.
1 This is the reason why radio operators are often called sparkies.
358 Oscillators and frequency synthesizers
8.1.1 Aim
The aims of this chapter are to explain radio frequency oscillators and frequency synthe-
sizers. Radio frequency oscillators produce radio frequency signals without an input
signal. Frequency synthesizers are used to control and vary the frequency of an oscillator
very precisely.
8.1.2 Objectives
After reading this chapter, you should be able to:
• understand the criteria for oscillation
• calculate the criteria (gain and frequency) of
• Hartley oscillators
• Colpitts oscillators
• Clapp oscillators
• crystal oscillators
• voltage controlled oscillators
• phase locked loops
• frequency synthesizers
8.2 Sine wave type oscillators
An oscillator is a device which produces an output signal without requiring an external
input signal. An amplifier can be made into an oscillator if its output signal is fed back into
its own input terminals to provide an input signal of the correct amplitude and phase.
One easy way of producing an r.f. oscillator is to use an r.f. amplifier and to feed its
output signal (with the correct amplitude and phase) back to its input. Figure 8.1 shows a
typical common emitter r.f. amplifier. The important things to note about this circuit are its
waveforms. Vin is the input sinusoidal applied to the amplifier. Vtc is the inverted voltage
appearing across the collector, and Vout is the voltage appearing across the output. The
phasor relationship between Vtc and Vout is dependent on the manner in which the
secondary winding of T1 is connected. In Figure 8.1, the output winding has been earthed
in a manner that will cause Vout to appear with a similar phase to Vin.
Fig. 8.1 An r.f. amplifier with associated waveforms
Sine wave type oscillators 359
Fig. 8.2 An r.f. oscillator constructed by feeding output to input
As Vin and Vout both have similar phases, there is no reason why Vout cannot be
connected back to the amplifier input to supply its own input voltage. This is shown
schematically in Figure 8.2 where a connection (thick line) has been made between points
A and B. Examination of these waveforms shows clearly that if, in addition to the phase
requirements, Vout > Vin, then the amplifier will supply its own input and no external signal
source will be needed. Therefore the amplifier will produce an output on its own and will
become an oscillator!
One question still remains unanswered. How do we produce Vout in the first instance
without an external Vin? Any operating amplifier produces inherent wideband noise
which contains an almost infinite number of frequencies. The collector tuned circuit
selects only its resonant frequency for amplification and rejects all other frequencies;
therefore only the resonant frequency of the tuned circuit will appear as Vout. Initially,
Vout will probably have insufficient amplitude to cause oscillation but as it is fed back
around again and again to the amplifier input terminals, Vin will increase in amplitude
and, if the circuit has been designed properly, Vin will soon be large enough to cause
Example 8.1
The tuned circuit of the oscillator circuit shown in Figure 8.2 has an effective inductance
of 630 nH and a total capacitance (CT) of 400 pF. If conditions are set so that oscillations
can take place, what is its frequency of oscillation (fosc)?
Solution. In Figure 8.2, the frequency of oscillation is determined by the resonant
frequency of the tuned circuit. For the values given
fosc = = = 10.026 MHz
2π LC 2π 630 nH × 400 pF
8.2.1 Barkhausen criteria
The introduction to oscillators above was to provide you with an elementary idea of oscil-
lator requirements. To design oscillators, we need a more systematic method. Consider an
amplifier with a positive feedback loop (Figure 8.3).
360 Oscillators and frequency synthesizers
β Vo
Fig. 8.3 An amplifier with positive feedback signal
The following terms are defined.
Voltage gain (Av) of the amplifier on its own is defined as
Av = — (8.1)
Voltage gain (Avf) of the amplifier with feedback applied is
Avf = — (8.2)
By inspection of Figure 8.3
b = output voltage fraction fed back (8.3)
v1 = vin + bvo (8.4)
vin = v1 – bvo (8.4a)
Using Equations 8.2 and 8.4a and dividing each term by v1
vo vo/v1
Avf = ——— = —————
v1 – bvo 1 – bvo/v1
Avf = ——— (8.5)
1 – Avb
If b is positive, and if Avb (defined as loop gain) = 1, then the denominator (1–Avb) = 0, or
Avb = 1 (8.6)
Substituting Equation 8.6 into Equation 8.5 yields
Avf = — = ∞ (8.7)
Wien bridge oscillator 361
which means that there is an output (vo) in spite of there being no input signal. This system
is known as an oscillator.
The two main requirements for oscillation are:
loop gain amplitude (Avb) = 1 (8.8)
loop gain phase = 0° or n360° n is any integer (8.9)
Equations 8.8 and 8.9 are known as the Barkhausen criteria for oscillation.
Note: Avb = 1 is the minimum condition for oscillation. If Avb > 1, it merely means that
the oscillation will start more easily but then, due to non-linearity in the amplifier, Avb will
revert back to 1.
For ease of understanding the above explanation, I have assumed that there is no phase
change in Av and b. In practice, Av is a phasor quantity and if it produces a phase shift of,
say, 170°, then b must produce a complementary phase shift of 190° to make the total
phase shift of the signal feedback equal to 360° (or any multiple of it). This enables the
returned feedback (input) signal to be in the correct phase to aid oscillation.
8.2.2 Summary
The Barkhausen criteria state that for an oscillator, the loop gain (Avb) must equal unity
and the loop gain phase must be 0° of any integer multiple of 360°.
It follows that if the Barkhausen criteria can be met then any amplifier may be made
into an oscillator. It is relatively easy to calculate the conditions required for oscillation,
but it is important to realise that when oscillation occurs, linear theory no longer applies
because the transistor is no longer working in its linear mode. In the discussion that
follows, we will show how (i) the conditions for oscillation and (ii) the desired frequency
of oscillation may be achieved with various circuits.
8.3 Low frequency sine wave oscillators
At frequencies less than about 2 MHz, oscillators are often made using resistances and
capacitances as the frequency determining elements instead of LC circuits. This is because
at these frequencies, LC elements are physically larger, more expensive, and more difficult
to control in production. Two main types of RC oscillators will be previewed. One is the
well known Wien bridge oscillator which is used extensively in instruments and the other
is the Phase-Shift oscillator.
8.4 Wien bridge oscillator
A block diagram of the Wien bridge oscillator is shown in Figure 8.4. The basic parts of
this oscillator consist of a non-inverting amplifier2 and an RC network which determines
its frequency of operation.
2 This non-inverting amplifier can consist of either two common emitter amplifiers in cascade (one follow-
ing the other) or a common base or operational amplifier.
362 Oscillators and frequency synthesizers
Fig. 8.4 Wien bridge oscillator
8.4.1 Operation
When power is applied to the circuit, currents (including inherent noise currents and volt-
ages) appear in the amplifier. This noise voltage is fed back through the Wien (RC
network) back to the input of the amplifier. The circuit is designed to allow sufficient feed-
back voltage to satisfy the Barkhausen criterion on loop gain (Avb = 1), but only noise
frequencies which satisfy the second Barkhausen criterion (∠ Avb = n360°) will cause the
oscillation. The circuit will oscillate at a frequency w2 = 1/(R1R2C1C2) radians per second.
8.4.2 Wien bridge oscillator analysis
In this analysis, it is assumed that the amplifier does not load the Wien bridge network
shown in Figure 8.5.
Fig. 8.5 Wien bridge network
By inspection
Z1 = R1 + 1/(jwC1)
1 1 R2
Z2 = — = ————— = —————
Y2 1/R2 + jwC2 1 + jwC2R2
Wien bridge oscillator 363
vin = ———— (vo)
Z1 + Z2
vo Z1 + Z2
— = ——— = 1 + Z1Y2
vin Z2
Substituting for Z1 and Y2
vo (R1 + 1/jwC1)(1 + jwC2R2)
— = 1 + ———————————
vin R 2
Multiplying out and sorting the real and imaginary terms
vo R1 C2
— — —
— = 1 + — + — + j(wC2R1 – 1/(wC1R2)) (8.10)
vin R2 C1
For the phase to equal zero, the quadrature or j terms = 0 which gives
wC2R1 = 1/(wC1R2)
w2 = ———— (8.11)
From Equation 8.10, the real part of the equation indicates that the gain of the amplifier
(Av) at the oscillation frequency must be
|vo| R1 C2
— —
Av = — = 1 + — + —— (8.12)
|vin| R2 C1
From Figure 8.4, the fraction of the voltage fed-back (b) = |vin|/|vout|. From Equation 8.10
|vin| 1
b = — = ————
— ———— (8.13)
|vout| 1 + R1/R2 + C2/C1
For oscillation, the Barkhausen criteria is Avb = 1. Using Equations 8.12 and 8.13
|vo| |vin|
Avb = — × —
— —
|vin| |vo|
= (1 + R1/R2 + C2/C1) × ———————— = 1
(1 + R1/R2 + C2/C1)
Therefore the Barkhausen gain criteria are satisfied and the circuit will oscillate.
364 Oscillators and frequency synthesizers
8.4.3 Practical Wien bridge oscillator circuits
In practical designs and for reasons of economy, variable frequency Wien bridge oscilla-
tors use twin-gang3 variable capacitors (C1 = C2 = C) or twin-gang resistors (R1 = R2 = R)
to vary the oscillation frequency.
For the case where C1 = C2 = C and R1 = R2 = R, Equations 8.11 and 8.12 become
w = — rads s–1
— (8.11a)
Av = — = 1 + 1 + 1 = 3 (8.12a)
Example 8.2
In the Wien bridge oscillator of Figure 8.4, R1 = 100 kΩ, R2 = 10 kΩ, C1 = 10 nF and
C2 = 100 nF. Calculate (a) the frequency of oscillation and (b) the minimum gain of the
amplifier for oscillation.
(a) Using Equation 8.11
w2 = 1/(100 kΩ × 10 kΩ × 10 nF × 100 nF) = 1 000 000
w = 1000 rads–1
fosc = 159.15 Hz
(b) Using Equation 8.12, the minimum gain of the amplifier is
|vo| R1 C2
— —
— =1+— +——
|vin| R2 C1
100 kΩ 100 nF
= 1 + —— —— + ———— = 1 + 10 + 10 = 21
10 kΩ 10 nF
8.5 Phase shift oscillators
8.5.1 Introduction
The circuit of a phase shift oscillator is shown in Figure 8.6. In this circuit, an inverting
amplifier (180° phase shift) is used. To feed the signal back in the correct phase, RC
3 Twin-gang capacitors or resistors are variable elements whose values are changed by the same rotating
Phase shift oscillators 365
Fig. 8.6 Phase shift oscillator
networks are used to produce an additional nominal 180° phase shift. The theoretical maxi-
mum phase shift for one RC section is 90° but this is not easily obtained in practice so
three RC stages are used to produce the required phase shift. The transmission (gain or
loss) analysis of a three section RC circuit can be difficult unless some simplifying meth-
ods are employed. To do this, I shall use matrix methods and assume that you are familiar
with matrix addition, subtraction, multiplication and division.
8.5.2 Analysis of the phase shift network
In the analysis that follows, it is assumed that the input and output impedances of the tran-
sistor are sufficiently large so that they do not load the phase shifting network. It can be
shown4 that the two port transmission matrix for a series impedance (Z) is:
[ ]1
It can also be shown5 that the two port transmission matrix for a shunt admittance (Y) is:
| |
V1 V1
A=—— =1 B=—— =Z
V2 I2 = 0 I2 V2 = 0
| |
I1 I1
C = —— =0‡ D=—— =1
V2 I2 = 0 I2 V2 = 0
‡ This follows from the diagram since I1 = I2 = 0
| |
V1 V1
A=—— =1 B=—— =0‡
V2 I2 = 0 I2 V2 = 0
| |
I1 I1
C=—— =Y D=— — =1
V2 I2 = 0 I2 V2 = 0
‡ This follows from the diagram since V1 = V2
366 Oscillators and frequency synthesizers
[ ]
The transmission parameters for the six element network shown in Figure 8.7 can be easily
obtained by multiplying out the matrices of the individual components. I will simplify the
arithmetic by making Z1 = Z2 = Z3 = Z and Y1 = Y2 = Y3 = Y. I have also drawn vo and vin
in the conventional manner but the analysis will show that the amplifier gain must be
inverted. By inspection of Figure 8.7
Fig. 8.7 Six element network
⎡vo ⎤ ⎡1 Z ⎤ ⎡1 0 ⎤ ⎡1 Z ⎤ ⎡ 1 0 ⎤ ⎡1 Z ⎤ ⎡ 1 0 ⎤ ⎡vin ⎤
⎢ I ⎥ = ⎢0 1 ⎥ ⎢Y 1 ⎥ ⎢0 1 ⎥ ⎢Y 1 ⎥ ⎢0 1 ⎥ ⎢Y 1 ⎥ ⎢ I2 ⎥
⎣ 1⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦⎣ ⎦⎣ ⎦⎣ ⎦
⎡vo ⎤ ⎡1 + YZ Z ⎤ ⎡1 + YZ Z ⎤ ⎡1 + YZ Z ⎤ ⎡vin ⎤
⎢I ⎥ = ⎢ Y 1⎥ ⎢ Y 1⎥ ⎢ Y 1 ⎥ ⎢ I2 ⎥
⎣ 1⎦ ⎣ ⎦⎣ ⎦⎣ ⎦⎣ ⎦
⎡vo ⎤ ⎡1 + YZ Z ⎤ ⎡1 + 3YZ + Y 2 Z 2 YZ 2 + 2 Z ⎤ ⎡vin ⎤
⎢I ⎥ = ⎢ Y ⎢
1 ⎥ ⎣ Y 2 Z + 2Y
⎥ ⎢I ⎥
⎣ 1⎦ ⎣ ⎦⎢ YZ + 1 ⎦ ⎥ ⎣ 2⎦
⎡vo ⎤ ⎡Y 3 Z 3 + 5Y 2 Z 2 + 6YZ + 1 Y 2 Z 3 + 4YZ 2 + 3Z ⎤ ⎡vin ⎤
⎢I ⎥ = ⎢ 3 2 2 ⎥ ⎢I ⎥
⎣ 1 ⎦ ⎣ Y Z + 4Y Z + 3Y
⎢ Y 2 Z 2 + 3YZ + 1 ⎦⎥ ⎣ 2⎦
Since we have assumed that the transistor impedance does not load the circuit, I2 = 0.
= A = Y 3 Z 3 + 5Y 2 Z 2 + 6YZ + 1
vin I2 = 0
Substituting for Y and Z
vo 1 5 6
= + + +1 (8.16)
vin I2 = 0 ( jωCR) 3
( jωCR) 2 ( jωCR)
Sorting out real and imaginary terms
vo 1 5 6
= + + +1
vin I2 = 0 ( jωCR) 3
( jωCR) 2 ( jωCR)
Phase shift oscillators 367
⎡ 5 ⎤ ⎡ 1 6 ⎤
= ⎢1 − 2⎥
+ j⎢ −
3 ωCR ⎥ (18.6a)
⎣ (ωCR) ⎦ ⎣ (ωCR) ⎦
[real part] [imaginary part]
The Barkhausen criterion for oscillation is that the voltage through the network
must undergo a phase change of 180°, i.e. imaginary or quadrature terms are zero.
–6 1
—— + ——— = 0
ωCR (ωCR)3
6 = ——— and w2 = ——— (8.17)
(wCR)2 6(CR)2
The real part of Equation 8.16a at resonance is:
vo –5
— = ——— + 1 (8.18)
vin (wCR)2
and using Equation 8.17 to substitute for 1/(wCR)2
— = –5 × 6 + 1 = –29 = 29 ∠ 180°
Since vo/vin = Av
Av = 29 ∠ 180° (8.19)
—— = 29 (8.19a)
From Figure 8.6, the fraction of the voltage fed back b = |vin|/|vo|. Using Equation 8.19
|vin| 1
b = —— = —— (8.20)
|vo| 29
To check for oscillation at resonance, the Barkhausen criterion is Avb = 1. Using Equations
8.19a and 8.20
|vo| |vin| 1
Avb = —— × —— = 29 × ——= 1
|vin| |vo| 29
Therefore the Barkhausen criteria are met and the circuit will oscillate.
368 Oscillators and frequency synthesizers
8.6 Radio frequency (LC) oscillators
8.6.1 Introduction
Oscillators operating at frequencies greater than 500 kHz tend to use inductors and capac-
itors as their frequency controlling elements because:
• RC values are beginning to get inconveniently small;
• LC values are beginning to assume practical and economical values.
8.6.2 General analysis of (LC) oscillators
In the analysis of the oscillators within this section, it must be realised that all calculations
to establish the conditions and frequency of oscillation are based on linear theory. When
oscillation occurs, the transistor no longer operates in a linear mode and some modifica-
tion (particularly bias) is inevitable.
The simplified solutions derived for each type of oscillator are based on the following
• The input impedance of the transistor does not load the feedback circuit.
• The output impedance of the transistor does not load the feedback circuit.
• The collector–emitter voltage (VCE) is the output voltage.
• The emitter–base voltage (VEB) is the input voltage.
• The feedback circuit is purely reactive (no resistive losses).
• If the transistor is operated in the common emitter configuration, a positive base input
voltage will result in an inverted collector voltage. If the transistor is operated in the
common base configuration, there is zero phase shift through the transistor.
• There is 0 or 2p radians (360°) shift through the loop gain circuit. In the case of a
common emitter amplifier, if p radians (180°) shift is caused by the transistor when its
collector load is resistive and/or resonant, then a further p radians shift will be required
in the feedback circuit to return the feedback signal in the correct phase. With a
common-base amplifier, if 0 radians phase shift is produced by the transistor, then zero
phase shift through the feedback network is required to return the feedback signal in the
correct phase.
• For clarity, oscillator outputs are not shown in Figures 8.8, 8.9 and 8.10. Outputs are
taken from either the collector or emitter via capacitance coupling or magnetic coupling
from the inductor.
The above assumptions are approximately true in practice and form a reasonably accurate
starting point for oscillator design.
8.7 Colpitts oscillator
A schematic diagram of the Colpitts oscillator circuit is shown in Figure 8.8. Two capaci-
tors, C1 and C2, are connected in series to provide a divider network for the voltage devel-
oped across points C and B. The tuned circuit is formed by the series equivalent
capacitance of C1 and C2 and the inductor L.
Colpitts oscillator 369
Fig. 8.8 Colpitts oscillator
The transistor is operated in the common base configuration. The base is a.c. earthed
via C4. The point B is a.c. earthed through capacitor C3. If the voltage at point C is posi-
tive with respect to earth, then point E is also positive with respect to earth. Hence the tran-
sistor supplies its own input voltage in the correct phase.
8.7.1 Frequency of oscillation
Using the assumptions of Section 8.6.2 and since we assume that there is no loading of the
tuned circuit at resonance, XL = XC, yielding
—— + —— = wL
ωC1 wC2
w2(L) = — + —
— —
C1 C2
w2 = —
L [ —
— +—
C2 ] (8.21)
8.7.2 Conditions for oscillation
From Figure 8.8 and using the assumptions of Section 8.6.2
VEB = vin and VCB = vo
By inspection of Figure 8.8
vin = ————— vo (8.22)
Xc1 + Xc2
370 Oscillators and frequency synthesizers
By definition, b = vin/vo, therefore
b = ———— (8.23)
Xc1 + Xc2
By definition, Av = vo/vin and using Equation 8.22
Xc1 + Xc2 C2 C2
Av = ———— = — + 1 or 1+—— (8.24)
Xc2 C1 C1
For oscillation, we must satisfy the Barkhausen criteria and ensure that the loop gain Avb
= 1. Using Equations 8.23 and 8.24
Xc1 + Xc2 Xc2
Avb = ———— × ———— = 1
Xc2 Xc1 + Xc2
Therefore the circuit will oscillate provided
[ ]C2
Av ≥ 1 + —— (8.25)
Summing up Equation 8.21 determines the frequency of oscillation and Equation 8.25
determines the minimum gain of the amplifier for oscillation.
Example 8.3
If in Figure 8.8 C1 = 10 pF and C2 = 100 pF and the desired oscillation frequency is 100
MHz, calculate (a) the value of the inductor and (b) the minimum voltage gain of the
amplifier. Assume that the transistor does not load the tuned circuit.
(a) From Equation 8.21
L [
w2 = — — + —
C1 C2
Transposing and substituting for C1 and C2
[ ]
L = ——————— —— + ——— H —
(2p × 100 MHz)2 10 pF 100 pF
11 × 1012 F
= ————— —————
3.948 × 1017 [ 100 ] H = 278.6 nH
(b) From Equation 8.24 the minimum voltage gain of the amplifier is
C2 100 pF
— = 1 + ——— = 11
C1 10 pF
Hartley oscillator 371
8.8 Hartley oscillator
A schematic diagram of the Hartley oscillator circuit is shown in Figure 8.9. Note that the
Hartley circuit is the dual of the Colpitts circuit where inductors and capacitors have been
interchanged. Inductor L in conjunction with capacitor C forms the tuned circuit. Inductor
L also serves as an auto-transformer. In an auto-transformer, the voltage developed6 across
points E and B is proportional to the number of turns (n2) between E and B. Similarly, the
voltage developed across points C and B is proportional to the number of turns (n1 + n2)
between points C and B.
Fig. 8.9 Hartley oscillator
The transistor is operated in the common base configuration. The base is a.c. earthed via
capacitor C4. The point B is a.c. earthed through capacitor C3. C2 is a d.c. blocking capaci-
tor. If the voltage at point C is positive with respect to earth, then point E is also positive with
respect to earth. Hence the transistor supplies its own input voltage in the correct phase.
8.8.1 Frequency of oscillation
Using the assumptions of Section 8.6.2 and since we assume that there is no loading of the
tuned circuit at resonance, XL = XC, yielding
wL = ——
w2 = —— (8.26)
8.8.2 Conditions for oscillation
From Figure 8.9 and using the assumptions of Section 8.6.2
VEB = vin and VCB = vo
6 This assumes that the same flux embraces both parts of the auto-transformer.
372 Oscillators and frequency synthesizers
From Figure 8.9, since inductor L serves as an auto-transformer
vin = ——— vo (8.27)
n1 + n2
By definition, b = vin/vo. Therefore
b = ——— (8.28)
n1 + n2
By definition, Av = vo/vin and using Equation 8.27
n1 + n2 n1 n1
Av = —— —
—— = — + 1 or 1+—— (8.29)
n2 n2 n2
For oscillation, we must satisfy the Barkhausen criteria and ensure that the loop gain Av b
= 1. Using Equations 8.28 and 8.29
n1 + n2 n2
Avb = ——— × ——— = 1
n2 n1 + n2
Therefore the circuit will oscillate provided
[ ]n1
Av ≥ 1 + —
— (8.30)
Summing up Equation 8.26 determines the frequency of oscillation and Equation 8.30
determines the minimum gain of the amplifier.
8.9 Clapp oscillator
A schematic diagram of the Clapp oscillator circuit is shown in Figure 8.10. Two capaci-
tors, C1 and C2, are connected in series to provide a divider network for the voltage devel-
oped across points C and B. The tuned circuit is formed by the equivalent series
capacitance of C1, C2 and CT and the inductor L. The Clapp oscillator is a later develop-
ment of the Colpitts oscillator except that an additional capacitance CT has been added to
improve frequency stability and facilitate design.
The transistor is operated in the common base configuration. An r.f. choke is used to
feed d.c. power to the collector. The reactance of the r.f. choke is made deliberately high
so that it does not shunt the tuned circuit. The base is a.c. earthed via capacitor C3. The
point B is earthed directly. If the voltage at point C is positive with respect to earth, then
point E is also positive with respect to earth. Hence the transistor supplies its own input
voltage in the correct phase.
8.9.1 Frequency of oscillation
Using the assumptions of Section 8.6.2 and since we assume that there is no loading of the
tuned circuit at resonance, XL = XC, yielding
Clapp oscillator 373
Fig. 8.10 Clapp oscillator
wL = — + — + —
— — —
wC1 wC2 wCT
w2L = — + — + —
— — —
C1 C2 CT
[ ]
CT CT
w2 = —— 1 + — + ——
— (8.31)
LCT C1 C2
If [CT/C1 + CT/C2] << 1, then Equation 8.31 becomes
w2 = —— (8.31a)
Equation 8.31a is the preferred mode of operation for the Clapp oscillator for the follow-
ing reasons.
• It allows CT and L to be the main contributors for determining the oscillation
frequency. This is particularly useful when the oscillation is to be set to another
frequency because only one control is needed. In many cases, CT is a varactor (capac-
itance diode) whose capacitance can be changed electronically by applying a d.c.
control voltage. This is particularly usefully in crystal oscillators which we shall be
describing shortly.
• It provides freedom for setting C1 and C2 to get the required values for easy oscillation.
C1 and C2 can be made reasonably large provided their ratio remains the same.
• Larger values of C1 and C2 help to swamp transistor inter-electrode capacitances which
change with operating bias and temperature.
374 Oscillators and frequency synthesizers
8.9.2 Conditions for oscillation
From Figure 8.10 and using the assumptions of Section 8.6.2
VEB = vin and VCB = vo
By inspection of Figure 8.10
vin = ————— vo (8.32)
Xc1 + Xc2
By definition, b = vin/vo. Therefore
b = ————— (8.33)
Xc1 + Xc2
By definition, Av = vo/vin and using Equation 8.32
Xc1 + Xc2 C2 C2
Av = ———— = —— + 1 or 1 + —— (8.34)
Xc2 C1 C1
For oscillation, we must satisfy the Barkhausen criteria and ensure that the loop gain Avb
= 1. Using Equations 8.33 and 8.34
Xc1 + Xc2 Xc2
Avb = ———— × ———— = 1
Xc2 Xc1 + Xc2
Therefore the circuit will oscillate provided
[ ]C2
Av ≥ 1 + —— (8.35)
Summing up Equation 8.31 determines the frequency of oscillation and Equation 8.35
determines the minimum gain of the amplifier for oscillation.
Example 8.4
Calculate the approximate frequency of oscillation for the Clapp oscillator circuit of
Figure 8.10 when CT = 15 pF, C1 = 47 pF, C2 = 100 pF and L = 300 nH.
Solution. Using Equation 8.31
[ ] [ ]
CT CT 1 15 15
w2 = —— 1 + — + — = ————
— — —
———— 1 + — + ——
LCT C1 C2 300 nH × 15 pF 47 100
= 2.222 × 1017 × 1.469 = 3.264 × 1017
w = 571 314 274.3 radians/s
Voltage-controlled oscillator 375
fosc = 90.93 MHz
An alternative approach is to calculate the combined series capacitance of the circuit:
Ctotal = [1/15 pF + 1/47 pF + 1/100 pF]–1 = 10.21 pF
fosc = = 90.93 MHz
300 nH × 10.21 pF
8.10 Voltage-controlled oscillator
A voltage-controlled oscillator is shown in Figure 8.11(a). If you examine this circuit, you will
find that it is almost identical to that of the Clapp oscillator (Figure 8.10). The exception is that
CT has been replaced by CV which is a variable capacitance diode or varactor. The voltage
across the varactor and hence its capacitance is controlled by varying the varactor voltage. C3
is a d.c. blocking capacitor used to isolate the varactor voltage from the collector voltage.
All the conditions relating to the Clapp oscillator apply here except that CT in all the
equations must be replaced by Cv.
8.10.1 Frequency of oscillation
The frequency of oscillation is that calculated by Equation 8.31 except that CT must be
replaced by Cv.
8.10.2 Oscillator gain
If you plotted the frequency of oscillation against the varactor voltage, you would get a
curve similar to that of Figure 8.11(b). The frequency sensitivity or frequency gain of the
oscillator is defined as
(a) (b)
Fig. 8.11 (a) Voltage-controlled oscillator; (b) oscillator frequency (fr) vs varactor (Vr)
376 Oscillators and frequency synthesizers
ko = —— (8.36)
Equation 8.36 is important because it tells us what voltage must be applied to the varactor
to alter the oscillator frequency. In this case, the sensitivity is positive because the slope
Dƒ/Dv is positive.
However, if the varactor in Figure 8.11(a) is connected in the opposite direction then a
negative voltage would be needed for control and, in this case, the sensitivity slope will be
negative. We will be returning to the question of frequency sensitivity when we discuss
phase locked loops.
Summing up Equation 8.31 determines the frequency of oscillation when you substi-
tute Cv for CT. Equation 8.35 determines the minimum gain of the amplifier for oscilla-
tion. Equation 8.36 describes the frequency sensitivity or frequency gain of the oscillator.
8.11 Comparison of the Hartley, Colpitts, Clapp and
voltage-controlled oscillators
• The Hartley oscillator is very popular and is used extensively in low powered oscillators
where the inductor value can be increased by winding the coil on ferrite cored material.
It is used extensively as the local oscillator in domestic superhet radio receivers.
• The Colpitts oscillator is used in cases where a piezo-electric crystal is used in place of
the inductor.
• The Clapp oscillator finds favour in electronically controlled circuits. It is more
frequency stable than the other two oscillators and can be easily adapted for crystal
control oscillators.
• The voltage-controlled oscillator is ideal for varying the frequency of an oscillator elec-
tronically. This is particularly true in phase locked loops and frequency synthesizers
which will be explained shortly.
8.12 Crystal control oscillators
8.12.1 Crystals
Crystals are electromechanical circuits made from thin plates of quartz crystal or
lead–zirconate–titanate. They are sometimes used in place of LCR-tuned circuits. The
electrical symbol for a crystal7 and its equivalent circuit are shown in Figure 8.12.
In this circuit, the capacitance between the connecting plates is represented by Co, while
L and C represent the electrical effect of the vibrating plate’s mass stiffness, and R repre-
sents the effect of damping. The circuit has two main resonant modes; one when L, C and
R are in series resonance and the other at a frequency slightly above the series resonant
frequency when the total series combination is inductive and resonates with Co to form a
parallel resonant circuit. Engineers can use either of these resonant modes in their designs.
7 It is common practice to abbreviate the term crystal resonator to xtal.
Crystal control oscillators 377
Fig. 8.12 Electrical equivalent circuit of a crystal resonator
Quartz and lead–zirconate–titanate are piezo-electric, i.e. they vibrate mechanically
when an electrical signal is applied to them and vice-versa. The resonant frequencies at
which they vibrate are dictated by their geometrical sizes, mainly plate thickness and angle
of crystal cut with respect to the main electrical axes of the crystal. Crystals are normally
cut to give the correct frequency in a specified oscillator circuit at a given temperature,
nominally 15°C.
Plate thickness reduces with frequency and at the higher frequencies, plate thickness
becomes so thin that the crystal is fragile. To avoid this condition and still operate at
frequencies up to 200 MHz, manufacturers often resort to overtone operation where they
cut the crystal to a lower frequency but mount it in such a manner that it operates at a
higher harmonic. Overtone crystals which operate at third, fifth and seventh overtones are
The angle of cut determines the temperature stability of the crystal. Typical types of
cuts are the AT cut, BT cut and SC cut. The frequency stabilities of these cuts against
temperature are shown in Figure 8.13. The AT cut is popular because it is reasonably easy
and cheap to make. It has a positive temperature coefficient (frequency increases) when the
ambient temperature changes outside its design (turnover) temperature. The BT cut crys-
tal has a negative temperature coefficient when the temperature is outside its turnover
Fig. 8.13 Frequency stability against temperature
378 Oscillators and frequency synthesizers
temperature. The SC cut crystal has an almost zero temperature coefficient but this crystal
is difficult to make because it is first cut at an angle with respect to one axis and then
rotated and cut again at a second angle to another axis.
From Figure 8.13, you should also note that changes of frequency for the crystal
itself against temperature are very small. To put the matter in perspective, 1 part in
108 is equivalent to about 1 Hz in 100 MHz, but remember in a practical circuit, there
are other parts (transistors, external capacitors, etc.) which will affect frequency as
Crystal resonators manifest very high Qs and values of Qs greater than 100 000 are
common in 10 MHz crystals. Metallic plates are used to make electrical connections to the
piezo material and the whole assembly is usually enclosed in a hermetically sealed can or
glass bulb to minimise oxidation and the ingress of contaminants.
8.12.2 Crystal controlled oscillator circuits
Crystal oscillators are particularly useful because of their frequency stability and low
noise properties. Figure 8.14 shows how a crystal is used in its parallel mode as an induc-
tor in a Colpitts oscillator circuit. When used in this manner, it is sometimes called a
Pierce oscillator. Such a circuit has the advantage of quick frequency changes by simply
switching in different crystals. The resonance frequency of oscillation is determined by
the equivalent circuit capacitance across the crystal and the equivalent inductance of the
crystal. To order such a crystal from the manufacturer, it is essential to tell the manufac-
turer the model of the crystal required, its mode of operation (series or parallel), operat-
ing frequency and parallel capacitance loading. Typical capacitance loads are 10 pF, 30
pF and/or 50 pF.
Figure 8.15 shows how a crystal is used in its series resonance mode as an inductor in
a Clapp type oscillator.
Crystal oscillators are used because they offer vastly superior frequency stability when
compared with LC circuits. As crystal Qs are very high, crystal oscillators tend to produce
much less noise than LC types. Typical values of stabilities and frequency ranges offered
Fig. 8.14 Colpitts xtal oscillator
Crystal control oscillators 379
Fig. 8.15 Clapp xtal oscillator
by crystal oscillators are given in Table 8.1. Explanatory terms for some abbreviations are
given below the table. Cathodeon and OSA are company names.
Table 8.1
Oscillator type VCO OCXO TCXO TCXO
(analogue) (digital)
Type no Cathodeon FS 5909 Cathodeon FS 5951 Cathodeon FS 5805 OSA DTCXO 8500
Frequency range 5–20 MHz 300kHz–40 MHz 5–15 MHz 1–20 MHz
Output TTL TTL TTL or sine Sine
Temperature range –25 to 80˚C 0–60˚C è0.1 ppm or –10 to 55˚C –40 to 85˚C
(°C) and stability è50 ppm on set –40 to 70˚C è1 ppm or –20 0.5 ppm or –20 to
frequency ±0.2 ppm to 70˚C ±2 ppm 70˚C ±0.3 ppm
Frequency adjust External volt Internal trimmer External resistor External resistor
(50 ppm)
Ageing rate 2 ppm/year 3 × 10–9 ppm day 1 ppm/year <1 ppm/year
Oscillator supply 5–15 V 5V 9V 12 V
Oven supply 9–24 V
Power oscillator 20 mA at 12 V 40/60 mA 9 mA <200 mW
Power oven 3–8 W
Package size (mm) 36.1 × 26.7 × 15 36.1 × 26.8 × 25.4 36.1 × 26.8 × 19.2 35.33 × 26.9 × 7.19
VCO = voltage-controlled oscillator usually like that of Figure 8.11(a).
OCXO = oven-controlled crystal oscillator. This type of oscillator is usually mounted in an oven operating at
75°C. The frequency stability is extremely good because, as you can see from Figure 8.13, the frequency of a
crystal is very stable. The disadvantages are additional bulk, size, weight and oven consumption of additional
electrical power. The last is particularly undesirable in battery operated equipment.
TCXO (analogue) = temperature-compensated crystal oscillator which uses compensating circuits (usually ther-
mistors) to correct frequency drifts with temperature.
TCXO (digital) = temperature-compensating oscillator which uses a microprocessor or look-up tables to correct
frequency drifts with temperature.
8.12.3 Summary for crystal oscillators
In Section 8.12 you have gained an insight into quartz crystals and their properties. You
have also seen how crystals are used in crystal oscillators, how they operate, their
frequency stability with temperature, and factors which affect the ordering of crystals for
use in oscillators.
380 Oscillators and frequency synthesizers
8.13 Phase lock loops
8.13.1 Introduction
The necessity for stable, low noise, power oscillators at very high frequencies has led to
many innovative oscillator design systems. Consider the case where a stable, low noise 3.6
GHz oscillator is required for a transmitter. Ideally we would like to use a crystal oscilla-
tor operating at 3.6 GHz. However, this is not possible because the maximum operating
range for a crystal oscillator is about 300 MHz.
Fig. 8.16 Producing a crystal controlled high frequency signal
One way of producing the 3.6 GHz signal would be to use the scheme shown in Figure
8.16 where a 300 MHz oscillator is fed into a cascade of frequency multipliers8 which
amplify and select various harmonics of 300 MHz signal to produce the final 3.6 GHz.
This method is expensive, requires a lot of circuit adjustment and is relatively inefficient.
However, it can be used and is still in use particularly at frequencies which cannot be easily
amplified. The great disadvantage of this method is that frequency multiplication increases
unwanted f.m. noise and the oscillator output is generally noisy.
Another method of producing this signal would be to use a voltage-controlled LC oscil-
lator at 3.6 GHz, divide its frequency, and compare its divided frequency against a refer-
ence crystal oscillator through a frequency or phase comparator which then emits a
controlling voltage to shift the frequency back to the desired frequency. Such an arrange-
ment is shown in Figure 8.17. This system is the basis of phase locked loop systems which
we will now discuss more extensively.
Fig. 8.17 A simple phase lock oscillator system
8 A frequency multiplier can be made by over-driving an amplifier with a large signal so that the amplifer
limits and produces a quasi-square waveform output. Fourier analysis tells us that a square waveform consists of
many harmonics and we can select the desired harmonic to give the required signal.
Phase lock loops 381
8.13.2 Elements of a phase locked loop system
The block diagram of the basic phase locked loop (PLL) is shown in Figure 8.18. The
phase detector, or phase comparator, compares the phase of the output waveform from the
voltage-controlled oscillator with the phase of the r.f. reference oscillator. Their phase
difference causes an output voltage from the phase detector, and this output voltage is fed
to a low pass filter which removes frequency components at and above the frequencies of
the r.f. input and the VCO. The filter output is a low frequency voltage which controls the
frequency of the VCO.
Fig. 8.18 The basic phase locked loop
When the loop is ‘in lock’, the phase difference has a steady value, which causes a d.c.
voltage output from the filter. This d.c. voltage is sufficient to cause the VCO output
frequency to become exactly equal to the input frequency. The two frequencies must be
synchronised, otherwise there will be a continually-changing phase difference, the VCO
input voltage will not be steady, and the loop will not be locked. Thus the loop ‘locks onto’
the reference frequency. Once the loop has locked, the reference frequency may vary, and the
VCO output will follow it, over a range of frequencies called the hold-in range. That is the
PLL will stay in lock, providing the output frequency does not fall outside the hold-in range.
In the following sections, we will look at each of the components of the PLL in turn,
followed by the closed loop frequency response and step response.
8.13.3 The phase detector
The basic principle behind phase detection is signal multiplication. Figure 8.19 shows the
principle using an ideal analogue multiplier. The VCO output voltage is represented by vv
= sin wvt where wv is its angular frequency. The reference input signal is an unmodulated
carrier vr sin (wrt + f) where wr is the input angular frequency and f is its relative phase to
vv at t = 0. The multiplier output is
vr vv = sin (wrt + f) sin wvt (8.37)
A little trigonometry9 shows that
vr vv = sin (wrt + f) sin wvt
= 0.5 {cos[(wr – wv)t + f] – cos (wr + wv)t + f}
9 cos (A – B) = cos A cos B + sin A sin B and cos (A+B) = cos A cos B – sin A sin B and subtracting the two
equations yields
cos (A – B) – cos (A + B) = 2 sin A sin B or sin A sin B = 0.5 [cos (A – B) – cos (A + B)]
382 Oscillators and frequency synthesizers
Fig. 8.19 Multiplication of two sine waves
The low pass filter removes the sum frequency (wr + wv) and the oscillator frequencies,
leaving the difference–frequency component:
vf = 0.5{cos[(wr – wv)t + f]} (8.38)
When the loop is in lock, the VCO frequency becomes equal to the reference input
frequency and wr = wv and Equation 8.38 becomes
vf = 0.5[cos f] when locked (8.39)
This is a d.c. level proportional to the cosine of the phase difference (f) between the two
signals at the input of the phase detector. Figure 8.20 shows the variation of this voltage
with phase difference.
The sensitivity of Figure 8.20 is defined as rate of filter d.c./phase difference. It is maxi-
mum at points A and B when the phase difference between the two signals is è90° and
minimum at point C where the phase difference is zero. It follows that if we want maxi-
mum sensitivity, then the reference oscillator and the VCO should be out of phase by 90°.
For the sake of clarity, let us choose point A. For this point we see:
• filtered d.c. output is zero for f = –90°
• filtered d.c. output is positive for –90° < f < 0°
• filtered d.c. output is negative for –180° < f < –90°
Fig. 8.20 Phase detector output when locked
Phase lock loops 383
Assume that the reference input frequency is constant. When the VCO drifts so that the
relative phase shift –90° < φ < 0°, a positive voltage will be generated for its frequency
correction. When the VCO drifts in the opposite direction so the phase shift is –180° < φ
< –90°, a negative voltage will be generated for frequency correction. If the VCO is
designed for the right sense of correction, it follows that the filter d.c. output voltage will
keep the phase and hence the frequency constant.
8.13.4 Types of phase detectors
In practice, analogue multipliers are seldom used as phase detectors, because there are
simpler circuits which can achieve the same overall result cheaper and faster. However, the
theory of the analogue multiplier applies to these circuits too. We will now examine two
basic phase detectors: the analogue switch, and the digital type.
Analogue switch type
One typical analogue switch phase detector is shown in Figure 8.21. In this circuit, the
principle of operation described earlier is carried out by multiplying the two signals as
described earlier. Vr remains the reference input signal, but part of the VCO signal is used
to produce a square wave which switches the diodes ON and OFF when the square wave-
form is 1 and 0 respectively. Since a square wave is composed of a series of sine waves, it
is apparent that the multiplication process is obtained and if a low pass filter is used after
vo, we will get the required d.c. term for VCO control as before.
Fig. 8.21 Analogue type phase detector
Digital phase detectors
Digital phase detectors have both inputs in the form of digital waveforms. Typically, they
use digital logic circuitry, such as TTL or CMOS.
The AND gate type
The most obvious digital equivalent of the analogue multiplier is an AND gate, as shown
in Figure 8.22(a). For the AND gate, with logical inputs A and B, the output is given by Y
= A.B. So, when the two input square waves have the same frequency and are in phase, the
average output voltage is maximum and equal to half the logic 1 output voltage (Vo).
384 Oscillators and frequency synthesizers
Fig. 8.22 The AND gate digital phase detector: (a) AND gate phase detector; (b) input and output waveforms for
different phase shifts; (c) filter d.c. output versus phase difference, with locked loop
Figure 8.22(b) shows input square waves with various amounts of phase shift and corre-
sponding output waveforms.
Figure 8.22(c) shows that, with input signals of the same frequency, the average output
voltage varies linearly with the phase (f). The filter output10 voltage (vd), when the loop is
locked, is
vd = (V1/2)(1 + f/p) for –p < f < 0 (see slope side A)
vd = (V1/2)(1 – f/p) for 0 < f < p (see slope side B)
10 The filter output voltage (v ) is defined as the output voltage from the detector after removal of the oscilla-
tor and sum frequencies.
Phase lock loops 385
This function, which peaks at f = 0, is analogous to the cosine function of the analogue
multiplier, which also peaks at f = 0.
Note: Although an AND gate has been featured in this case, you should be aware that it
is also possible to use a NAND gate or an exclusive-OR gate as a phase detector but expect
phase inversion in the output signal.
Gain sensitivity of the AND gate
The gain sensitivity or gain (kf) of a phase detector is defined as output voltage change
(vd)/phase difference (qe) change or
kf = —— (8.40)
Using the same definition for Figure 8.22 we obtain for an AND gate
kf = ——— (8.40a)
The sign of kf is dependent on the point used for the definition. If you use point A in Figure
8.22(c), you will get a positive slope and hence a positive value for kf whereas point B will
give a negative slope and a negative value for kf.
The flip-flop type phase detector
In some applications of the phase locked loop, one or the other of the digital inputs to the
phase detector may not be a square wave. For instance one input may come from the
output of a counter used as a frequency divider, whose output waveform has a mark:space
ratio which changes when the division factor is changed. Such a waveform is not suitable
for use with the AND gate or the EX-OR gate. An alternative, which avoids this disad-
vantage, is the flip-flop. This is shown in Figure 8.23(a). Here, rising edges of one input
set the output (Q) to logical 1, and rising edges of the other input reset the output (Q) to
logical 0, as shown in the waveforms of Figure 8.23(b). You should see that the average
output vd varies with f as in Figure 8.23(c) and not with the mark: space ratios of the input
Gain sensitivity of the flip-flop detector
The gain sensitivity or gain (kf) of Figure 8.23 is defined as
kf = — (8.41)
The sign of kf is dependent on the point used for the definition. If you use slope B in Figure
8.23(c), you will get a positive slope and hence a positive value for kf whereas point A will
give a negative slope and a negative value for kf.
386 Oscillators and frequency synthesizers
Fig. 8.23 A set–reset (SR) flip-flop used as a phase detector, with triggering on the rising edges of the input wave-
form: (a) set–reset flip-flop; (b) input and output waveforms; (c) average output voltage versus phase difference
8.13.5 The filter
The purpose of the filter is to remove the two oscillator frequencies ƒr and ƒVCO and the
sum frequencies so that they do not cause instabilities in the loop system. The low pass
filter is a simple RC network. In many applications it is a simple first-order single lag filter
comprising just a series resistor followed by a shunt capacitor.
In some cases a lower frequency single lag filter is used for loop compensation. In other
cases a slightly more complicated lag-lead RC network is used. Loop compensation is
discussed further in the section on closed-loop response.
8.13.6 The d.c. amplifier
In many cases, the output from the phase detector is not sufficient to control the voltage-
control led oscillator (VCO); therefore some form of d.c. amplification is necessary. Figure
8.24 shows three voltage gain amplifiers and their gain (volts out/volts in) parameters kA
= vo/vi.
Phase lock loops 387
Fig. 8.24 (a) Common emitter; (b) inverting op-amp; (c) non-inverting op-amp
For Figure 8.24(a)
kA ≈ –Rc/Re (8.42)
For Figure 8.24(b)
kA ≈ –Rf/R1 (8.43)
For Figure 8.24(c)
kA ≈ (Rf + R1)/R1 (8.44)
The bandwidth of the d.c. amplifier must be very high when compared to the loop band-
width (explained later) otherwise loop instability will occur.
8.13.7 The voltage-controlled oscillator (VCO)
The sine wave VCO described in Section 8.10 is a suitable oscillator for use in phase
locked loops, but other types such as the digital type oscillator shown in Figure 8.25 can
also be used. The sine wave type is used as the oscillator in r.f. transmitters, and as local
oscillators in radio and television receivers and as synthesised oscillators in mobile phones
and signal generators. In these applications a pure sine wave is the ideal. With varactor
diode type oscillators, the relation between the oscillator frequency and the control volt-
age is chiefly dependent on the chosen varactor and its direction of connection. Most
varactors vary their capacitance as Va where a can range from 0.5 to 3.
Fig. 8.25 Digital free-running multi-vibrator with voltage-controlled frequency
388 Oscillators and frequency synthesizers
The linear, square wave type of VCO is suitable for use in a PLL system used for
frequency demodulation. This is done at the receiver intermediate frequency, so no great
strain is made on the operating frequency of the digital circuitry. If the loop is in lock, then
the VCO frequency must be following the input frequency. In this case the VCO’s control
voltage, which is also the demodulated output from the loop, is linearly related to the
frequency shift and, hence, to the original frequency modulation.
Gain sensitivity of a VCO
The gain sensitivity of a VCO is defined as
ko = dƒ/dvc (8.45)
dƒ is the change in VCO frequency
dvc is the change in control voltage vc
Frequency of a VCO
If we define ƒFR as the free-running frequency of a VCO when there is zero correction volt-
age from the phase detector, and assume oscillator control linearity, then by using Equation
8.45 we can describe the frequency (ƒ) of the VCO as
ƒ = ƒFR + Dƒ = ƒFR + kovc (8.46)
Some VCOs are designed for a control voltage centred on 0 V, at which point they gener-
ate their ‘free-running’ frequency, ƒFR. This is shown in Figure 8.26(a). For a linear type
VCO, the output frequency is given by
ƒ = ƒFR + kovc (8.46a)
Positive control voltages may increase the frequency, and negative ones decrease it, result-
ing in a positive value for ko; or the converse may be true with ko negative.
Fig. 8.26 Transfer characteristics of linear VCOs: (a) vc = 0 for ƒFR and a positive control slope; (b) vc ≠ 0 for ƒFR and
a negative control slope
Phase lock loops 389
Fig. 8.27 Basic phase lock loop
Other VCOs have a control voltage range centred on a non-zero voltage. Figure 8.26(b)
shows an example of this type of transfer characteristics. In this case and because of the
negative slope
ƒ = ƒFR – kovc (8.46b)
Note: In each case, the slope of the characteristic ∂ƒ/∂vc gives ko.
8.13.8 Loop gain and static phase error
At this stage, it would be prudent if we consolidate what we have discussed using the
diagram shown in Figure 8.27. For clarity of understanding, we will consider the PLL
initially as locked and tracked. Later we will consider how the PLL becomes locked. In
Figure 8.27, each block has its own gain parameters. From Equation 8.40, we know that
the phase detector develops an output parameter (vd) in response to a phase difference (qe)
between the reference input (ƒi) and the VCO frequency (ƒo). The transfer gain (kf) has
units of volt/radians of phase difference.
At this locked stage, the main function of the filter is simply to remove ƒi, ƒo and the
sum of these frequencies. We will temporarily ignore the parameters of the filter in the
dashed block of Figure 8.27 because in the locked state, the filter has a parameter of unity.
The amplifier has a gain of kA and its unit of gain (Vo/Vd) is dimensionless. Thus Vo =
kAvd. The VCO free-running frequency is ƒFR. The VCO frequency (ƒo) will change in
response to an input voltage change. The transfer gain (ko) has units of Hertz/volt.
The loop gain (kL) for this system is simply the product of the individual blocks:
kL = kf kAko (8.47)
The units for kL = (v/rad) (v/v) (Hz/v) = Hz/rad.
In the diagram ƒi is the input reference frequency to the phase detector and the system
is in the locked stage. If the frequency difference before lock was Dƒ = ƒi – fFR, then a volt-
age vc = Dƒ/ko is required to keep the VCO frequency equal to ƒi. The amplifier must
supply this vc so its input must be vc/kA and this is the output voltage (Vd) required from
the phase detector.
Summing up
vd = vc/kA = (Dƒ/ko)/kA = Dƒ/(kokA) (8.48)
390 Oscillators and frequency synthesizers
To produce vd, we would need a phase difference error of qe radians between ƒi and ƒo
because kf = vd/qe or vd = kfqe and substituting this in Equation 8.48 yields
qe = Dƒ/(kokAkf)
and using Equation 8.47, we get
Dƒ Dƒ
qe = ——— = — — (8.49)
kokAkf kL
8.13.9 Hold-in range of frequencies
The hold-in range of frequency (Dƒ in Equation 8.49) is defined as the frequency range
within which the VCO can drift before it becomes out of lock and stops being controlled
by ƒi. In practical circuits, the hold-in frequency range is limited by the operating range
of the phase detector because the VCO has a much wider frequency range of operation.
In all of the types of phase detector described in Section 8.13.4, except for the flip-flop
type, the useful phase–difference range is limited to èp/2. Outside this range, the slope of
the phase detector’s characteristic changes, altering the loop feedback from negative to
positive, and causing the loop to lose lock. So the static phase error qe is limited to èp/2.
Re-arranging Equation 8.49 for qe we get
Dƒ = qekL (8.50)
For the AND gate type phase detector, where qe has its maximum possible value of èp/2,
Dƒ is the maximum possible deviation, Dƒmax = (èp/2)(kL) and its hold-in range = èp/2 ×
(d.c. loop gain). For the flip-flop type, the static phase error (qe) is limited to èp, and its
hold-in range = èp × (d.c. loop gain).
Again from Equation 8.50, you can see that if you want a wide hold-in frequency range,
then you should increase the loop gain (kL). This is true but care should be exercised in
doing this because too high a loop gain will result in loop instability. Let us now consoli-
date our thoughts by carrying out Example 8.5.
Example 8.5
Example 8.5 summarises much of the information acquired on PLL at this stage. Figure
8.28 provides enough information to calculate the static behaviour of a phase locked loop.
Calculate (a) the voltage gain (ka) for the op-amp, (b) the loop gain (kL) in units of
second–1 and in decibels (w = 1 rad/s). (c) With S1 open as shown, what is observed at vo
with an oscilloscope? (d) When the loop is closed, determine: (i) the VCO output
frequency; (ii) the static phase error (qe) at the phase comparator output; (iii) Vo. (e)
Calculate the hold-in range Dƒ (assume that the VCO and op-amp are not saturating). (f)
Determine the maximum value of vd.
(a) Using Equation 8.44
kA = (Rf/R1) + 1 = (9 kΩ/1 kΩ) + 1 = 10
Phase lock loops 391
ko = –40 kHz/v
Fig. 8.28 Closed loop system
(b) Using Equation 8.47
kL = kfkAko = (0.12 v/rad)(10)(–40 kHz/v) = –48 000 Hz/rad
48 000 cycles/sec 48 000 cycles/sec
= ———————— = ————————
rad cycles/(2p)
= [2p × 48 000] s–1 = 301 593 s–1
and in terms of dB at 1 rad, we have
kL = 20 log (301 593) ≈ 109.6 dB at 1 rad/s
(c) With S1 open, there is no phase lock. If we assume that ƒo, ƒi and the sum frequencies
have been removed by the filter, then all that will be seen is the beat or difference
frequency ƒo – ƒi = 120 kHz – 100 kHz = 20 kHz.
(d) (i) When the loop is closed and phase locked, then by definition ƒo = ƒi and since ƒi
= 100 kHz, it follows that ƒo = 100 kHz. There is no frequency error but there is
a phase error between the two signals.
(ii) The free-running frequency fFR of the VCO = 120 kHz. For the VCO output to be
100 kHz, we transpose Equation 8.45 to give
Vo = Dƒ/ko = (100 – 120) kHz/(–40 kHz/V) = 0.500 V
We want vd, the input to the amplifier whose gain = 10. Using Equation 8.48
vd = vo/kA = 0.5 V/10 = 0.050 V
Finally, using Equation 8.40 and transposing it, we obtain
qe = vd/kf = 0.050 V/0.12 V/rad = 0.417 radians
Alternatively, we could have used Equation 8.49 where
392 Oscillators and frequency synthesizers
Dƒ (100 – 120) kHz
qe = —— = ———————
—— —————— = 0.417 rad
kokAkf (–40 kHz/V)(0.12 V/rad) (10)
(iii) Vo was calculated in (ii) as 0.500 V d.c.
(e) Since the VCO and op-amp are assumed not to be saturating, then the limitation will
obviously depend on the phase detector output. Clearly vd can only increase until vd
→ vmax = A, at which point qe = p/2. Beyond this point, vd decreases for increasing
static phase error, and the phase detector simply cannot produce more output voltage
to increasing ƒo, and the loop breaks lock. The total hold-in range is èp/2 or p radi-
ans. The total hold-in frequency range between these two break-lock points can be
found by using Equation 8.50:
Dƒ = qekL = (p) (–40 kHz/V)(0.12 V/rad)(10) = 48.00 kHz
In the answer, I have dropped the minus sign because we are only interested in the
frequency range.
(f) At the frequency where qe = p/2, we have vd(max) = A. Therefore
vd = kfqe = 0.12 V/rad × p/2 rad = 0.188 V d.c.
This example shows clearly the conditions existing within a phase locked loop system
when it is in lock.
Summary. Example 8.5 has shown clearly what happens in a phase locked loop when it
is in lock. Certain facts are required to make a PLL function correctly.
• The sensitivity of the phase locked loop detector (kf) must be known. It can be obtained
from measurement or calculation.
• The amplifier gain (kA) must be known. This can be obtained by calculation or measure-
• The sensitivity of the VCO must be known. The usual way to obtain this is by measurement.
• The hold-in range (Dƒ) must be known or measured.
• If the oscillator drifts outside this range either due to noise, instability, temperature, etc.,
then the phase locked loop will be erratic and will break lock and behave like a free-
running oscillator.
The above conditions are basic to a phase locked loop when it is in a hold-in range situation.
Lock-in range
The lock-in range is defined as the range of input frequencies over which an unlocked loop
will acquire lock. If ƒh and ƒL are respectively the highest and lowest frequencies at which
the loop will attain lock, then
lock-in range = ƒh – ƒL
The lock-in range is usually smaller than the hold-in range. In practice when loop-lock is
lost, the PLL generates a saw-tooth wave to sweep the VCO in the hope that lock-in may
be re-captured.
There are many problems to be considered in acquiring frequency-lock. These include
frequency sweep range, step response time, loop bandwidth frequency response, loop
Frequency synthesizers 393
bandwidth gain, and the response of the individual components to the sweep range. The
VCO phase response is important because its phase gain falls off with an increase in
frequency sweep. The loop filter is also extremely important because it determines the step
response time and hence the settle-time of the VCO to its new frequency.
The PLL is a very complicated system and to properly design such a system, the
designer has to take into account many of the problems mentioned above. We do not
propose to do it here because we have achieved our aim of showing the principles of a
phase lock loop system. However, if you want an extensive source of material that covers
phase lock loop design techniques, consult F. M. Gardner’s Phaselock Techniques.11
8.14 Frequency synthesizers
8.14.1 Introduction
A frequency synthesizer is a variable-frequency oscillator with the frequency stability of a
crystal-controlled oscillator. Synthesizers are used in radio transmitters and receivers
because of their output signal stability which is essential in today’s narrow band commu-
nication systems. In fact, modern communication systems cannot exist without them.
Two basic approaches are used in the design of synthesizers. They are the direct method
and the indirect method. The direct method generates the output signal by combining one or
more crystal-controlled oscillator outputs with frequency dividers/comb generators, filters
and mixers. The indirect method utilises a spectrally pure VCO and programmable phase
locked circuitry. Although slower than the direct approach and susceptible to f.m. noise on
the VCO, indirect frequency synthesis using the phase lock loop principle is less expensive,
requires less filtering, and offers greater output power with lower spurious harmonics.
8.14.2 The direct type synthesizer
This type uses no PLLs or VCOs, but only harmonic multipliers, dividers and filters. It
may use only one crystal oscillator, with multiple harmonic multipliers and dividers, or it
may use several crystal oscillators. An example of the single crystal type is shown in
Figure 8.29. The 1 MHz crystal oscillator is followed by a harmonic multiplier, or comb
generator. This is a circuit which ‘squares-up’ the signal from the crystal oscillator to
generate a train of pulses, rich in harmonics of the crystal frequency. It is called a comb
generator because its frequency spectrum resembles a comb. Harmonic selector filter 1
(HSF 1) allows harmonics of the 1 MHz signal to appear at 1 MHz intervals. Hence, by
adjusting the filter, it is possible to get frequencies of 1, 2, 3, 4, etc. MHz. This selection
of 1 MHz is rather a coarse adjustment.
The second output of the 1 MHz crystal oscillator is divided down to 100 kHz, fed to
another comb generator and into harmonic selector filter 2 (HSF 2). This filter allows 100
kHz harmonics to pass throught it. The harmonic selected is dependent on the setting of
the filter.
The two outputs from harmonic selector filter 1 and harmonic selector filter 2 are then
mixed to produce sum and difference frequencies, amplified and filtered through the
11 Gardner’s book (1966) is published by John Wiley & Sons. It is old but it remains the classic on phase lock
394 Oscillators and frequency synthesizers
Fig. 8.29 An early direct frequency synthesizer
output filter. For example, if an output frequency of 6.5 MHz is required. The 6th harmonic
of 1 MHz will be selected by (HSF 1), i.e. 6 MHz, and the 5th harmonic of 100 kHz will
be selected by (HSF 2), i.e. 500 kHz. The two frequencies are fed into the mixer to produce
sum 6.5 MHz and difference frequency 5.5 MHz. The four signals, 500 kHz, 5.5 MHz, 6
MHz and 6.5 MHz, are amplified but the switchable output filter rejects all but the desired
6.5 MHz signal.
The frequency resolution can be improved still further by adding further divider, comb
generator and filter sections. For instance, a second decade divider and comb generator
would provide 10 kHz steps, selectable by a second filter. Its output would be mixed with
the output of the 100 kHz step filter in a second mixer, whose output would be filtered to
select the sum or difference frequency. This would then be mixed with the selected 1 MHz
step in the first mixer. Thus this circuit would provide frequencies up to 10.99 MHz, with
10 kHz resolution.
The biggest disadvantage of such a system is that it places stringent requirements on the
output filter. This is because, in some cases, the sum and difference frequencies can differ
by very little and selecting one frequency and rejecting the other means extremely steep
filter slopes.
Example 8.6
A synthesizer with four decade divider, comb generator and harmonic filter sections has a
crystal oscillator frequency of 10 MHz. State the frequency which could be selected by
each section to produce a final output at 75.48 MHz, assuming the sum frequency is
selected from each mixer output. State also the frequencies at the output of each mixer, and
the frequency selected by the filter following each mixer.
Solution. Assuming that the sum frequency is selected from each mixer output, the
frequencies are:
Frequency synthesizers 395
Section 1: 70 MHz
Section 2: 5 MHz
Section 3: 400 kHz
Section 4: 80 kHz
Mixer 3: (400 è 80) kHz = 480 kHz and 320 kHz
After filter 3: 480 kHz
Mixer 2: (5 è 0.48) MHz = 5.48 MHz and 4.52 MHz
After filter 2: 5.48 MHz
Mixer 1: (70 è 5.48) MHz = 75.48 MHz and 64.52 MHz
After filter 1: 75.48 MHz
8.14.3 Direct digital waveform synthesis
The system described in Section 8.14.2 is clumsy and is seldom used today. This is a more
recent technique which synthesises a sine wave digitally. The basic principle is illustrated
in Figure 8.30. Values of the sine function, at regularly-spaced angular intervals over one
complete cycle, are stored digitally in a look-up table, typically in ROM. The values are
clocked sequentially out of the look-up table to a digital-to-analogue convertor and via a
filter to the output. The filter removes clock-frequency components.
The output frequency ƒo = ƒc/S, where ƒc is the clock frequency and S is the number of
samples per cycle stored in the look-up table. So the output frequency is determined by the
clock frequency. This can be changed by changing the division factor of the program-
mable counter, which divides the crystal oscillator frequency ƒref by the factor N. Thus the
output frequency is ƒo = ƒref /S N. The value of S must be at least 4, but preferably 10 or
so, unless complicated tuneable analogue filtering is used, which would defeat the object
of the cheap digital chip.
Clearly, this type of synthesizer cannot produce frequencies much higher than, say, 100
MHz because the clock must run at S times the output frequency, and the fast test digital
Fig. 8.30 Basic direct synthesizer using sinusoidal waveform synthesis
396 Oscillators and frequency synthesizers
Fig. 8.31 Basic PLL synthesizer
circuits limit is currently about 2 GHz. In practice, a look-up table in such fast circuitry is
currently too expensive, and the digital technique’s cost advantage is realised only at lower
8.14.4 Indirect synthesizers (phase-lock types)
These synthesizers use phase lock loops to control voltage-controlled oscillators, with good
spectral purity, at frequencies locked to harmonics or sub-harmonics of crystal oscillators.
Single loop type
The simplest example is the single loop type of Figure 8.31 where a digital counter type
frequency divider, set to divide by a programmable factor N, follows the VCO. The loop keeps
the divider’s output frequency equal to the crystal frequency (ƒref) so the output frequency
from the VCO is an integer multiple (harmonic) of the crystal frequency: ƒo = Nƒref.
The output frequency can be changed by changing the division factor N of the divider.
The highest output frequency is ƒo = Nmax ƒref, where Nmax is the maximum division capac-
ity of the counter. The frequency resolution, which is the minimum output frequency step
size, is equal to the crystal frequency ƒref.
All the previous analysis of PLLs applies to this loop, but with the added complication
that the loop gain is divided by the factor N, so the loop gain changes as the output
frequency is changed. Because of this, the loop filter must be designed to maintain loop
stability in the worst case. As the output frequency is raised, N is increased, which lowers
the loop gain and the loop crossover frequency. So, if a lead-lag filter is used, its break
points must be chosen well below the lowest loop crossover frequency, which is obtained
at the highest output frequency.
The simple single loop synthesizer of Figure 8.31 cannot produce output frequencies
any higher than can he handled by the digital divider. With moderately-priced TTL-variant
programmable counters, this limits the frequency to the order of 100 MHz.
A simple modification called pre-scaling enables higher frequency VCOs to be used.
Figure 8.31 shows an example. This synthesizer is used to generate the local-oscillator
frequency for a UHF television receiver. The broadcast vision carriers12 have frequencies
12 Television channels and television channel spacing differ in different parts of the world and you should only
take this frequency as representative.
Frequency synthesizers 397
Fig. 8.32 A PLL synthesizer using pre-scaling, for generating the local-oscillator signal for a television receiver
from 471.25 to 847.25 MHz, spaced at exactly 8 MHz intervals. With a receiver vision i.f.
of 39.75 MHz, the local oscillator must be tuneable over the range 511 to 887 MHz,
assuming it works above the carrier frequency.
The pre-scaler is a high speed divider using ECL or Gallium Arsenide (GaAs) circuitry,
which can work at these frequencies. It divides by a fixed factor M. The output frequency
is now ƒo = MNƒref, and the frequency resolution is increased to Mƒref. We still have ƒo =
(N × resolution), as with the simple loop.
The poorer resolution is not a problem in this case, because the TV channels are spaced
8 MHz apart. However, 8 MHz does not divide integrally into the required local-oscillator
frequencies, so a frequency resolution must be chosen which does. The highest such
frequency is 1 MHz, so this is the choice for the resolution, although of course the chan-
nel control logic will restrict N to selecting just channel frequencies at 8 MHz intervals. A
value of 64 is chosen for M, to reduce the highest local-oscillator frequency down to less
than 20 MHz so that a cheap, low power TTL or CMOS chip can be used for the variable
divider. Since resolution = Mƒref, we have 1 MHz = 64ƒref, making ƒref = 15.625 kHz. This
is best obtained from a cheap, higher frequency crystal and divider, such as a 1 MHz crys-
tal with a divide-by-64 counter.
Example 8.7
Calculate the values of N to produce the lowest and the highest required local-oscillator
ƒo = MNƒref = resolution
ƒo = N × resolution = N × 1 MHz and N = ƒo/1 MHZ
398 Oscillators and frequency synthesizers
The lowest local-oscillator frequency is 511 MHz, so N = 511. The highest local-oscilla-
tor frequency is 887 MHz, so N = 887.
8.14.5 Translation loops: frequency offset using heterodyne
Another technique for avoiding the need for a VHF programmable divider is to translate
the output frequency down to a lower frequency by mixing with a stable ‘offset’ oscillator.
Figure 8.33 shows the principle. The balanced mixer output contains the sum and differ-
ence frequencies, ƒo + ƒosc and ƒo – ƒosc, is of the same order as ƒo then (ƒo + ƒosc) is much
greater than (ƒo – ƒosc), and a simple low pass filter following the mixer can remove the
sum easily, leaving only the difference frequency at the input to the divider. The loop
containing the frequency-translation circuitry is called a translation loop.
In the example shown in Figure 8.33, the synthesizer is used as the local oscillator for
a VHF FM receiver. The carrier signal band is 88.0–108.0 MHz, and the i.f. is 10.7 MHz
so, with the local-oscillator frequency above the received frequency, the synthesizer must
tune from 98.7 to 118.7 MHz. With ƒosc = 80 MHz, the divider input frequency (ƒo – ƒosc)
ranges from 18.7 to 38.7 MHz. The required resolution is 50 kHz, so the reference
frequency is chosen as 50 kHz. This could be obtained by dividing from a 1 MHz crystal.
In that case the 80 MHz source would be a separate crystal. Alternatively, one crystal oscil-
lator running at 16 MHz could have its frequency multiplied and the appropriate harmonic
(the fifth in this case) selected for the offset source, and its frequency divided for the refer-
ence source.
Example 8.8
Calculate the lowest and highest values of the division factor N of the divider in the
Fig. 8.33 A PLL synthesizer using an offset oscillator to generate the local oscillator for a VHF FM receiver
Summary 399
Solution. The lowest frequency at the divider input is 18.7 MHz and the divider output is
50 kHz, so in this case N = 18.7 MHz/50 kHz = 374. The highest frequency at the divider
input is 38.7 MHz, so in this case N = 774.
8.15 Summary
This part has been mainly devoted to the more popular types of oscillators. The configu-
rations discussed have included the Wien, phase shift, Hartley, Colpitts, Clapp, crystal and
the phase locked loop. Frequency synthesizers and their basic operating configurations
have also been shown. Oscillator design, particularly phased lock loop synthesizers, is a
specialist subject but the information presented in this chapter has provided sufficient
background information to allow further study.
Further topics
9.1 Aims
The primary aim of this chapter is to provide an introduction to signal flow graph analysis
so that you will be able to analyse more complicated networks and to follow more advanced
publications and papers. The secondary aim of this chapter is to offer you some comments
on the use of small software packages in high frequency and microwave engineering.
9.2 Signal flow graph analysis
9.2.1 Introduction
Occasions often arise where a network is extremely cumbersome and difficult to under-
stand, and it is hard to solve the circuit parameters by algebraic means. In such cases,
signal flow graph analysis is used to help understanding, and to reduce circuit complex-
ity until it can be handled easily by more conventional algebraic methods.
Signal flow analysis is used as a means of writing and solving linear microwave
network equations. It is direct and relatively simple; variables are represented by points
and the inter-relations between points are represented by directed lines giving a direct
picture of signal flow. The easiest way to understand signal flow manupulation is by exam-
ples and in the following sections, we will show you several methods of applying signal
flow analysis.
You are already familiar with Figure 9.1 which was used when we introduced you to
Fig. 9.1 General microwave two-port network showing incoming and outgoing waves
Signal flow graph analysis 401
Fig. 9.2 Alternative signal flow graph for a two port network
scattering parameters. Figure 9.1 gives a semi-pictorial view of s-parameters. An alterna-
tive representation of Figure 9.1 is shown in Figure 9.2. This figure is often used in signal
flow analysis because of its greater simplicity. Figure 9.2 shows a two-port network with
wave a1 entering port 1 and wave a2 entering port 2. The emerging waves from the corre-
sponding ports are represented by b1 and b2.
Figure 9.3 is a signal flow graph representation of Figure 9.2. In Figure 9.3 each port is
represented by two nodes. Node an represents the wave coming into the device from another
device at port n and node bn represents the wave leaving the device at port n. A directed
branch runs from each a node to each b node within the device. Each of these branches has
one or more scattering coefficients associated with it. The coefficient/s shows how an
incoming wave gets changed to become an outgoing wave at the node b. Scattering coeffi-
cients are complex quantities because they represent both amplitude and phase changes
associated with a branch. The value of a wave at a b node when waves are coming in at both
a nodes is the superposition of the individual waves arriving at b from each of the separate
a nodes.
As you already know the relationship of the emergent waves to incident waves is writ-
ten as the linear equations
b1 = s11a1 + s12a2 (9.1)
b2 = s21a1 + s22a2 (9.2)
These are called scattering equations and the smns are the scattering coefficients.
By comparing Equations 9.1 and 9.2 and Figure 9.3 it is seen that s11 is the reflection
coefficient looking into port 1 when port 2 is terminated in a perfect match (a2 = 0).
Similarly s22 is the reflection coefficient looking into port 2 when port 1 is terminated in
a perfect match (a1 = 0).
Fig. 9.3 Signal flow representation of Figure 9.2
402 Further topics
Fig. 9.4 Cascading of networks
The parameter s21 is the transmission coefficient from port 1 to port 2 when port 2 is
terminated in a perfect match (a2 = 0), and s12 is the transmission coefficient from port 2
to port 1 when port 1 is terminated in a perfect match (a1 = 0).
Networks are cascaded by joining their individual flow graphs as in Figure 9.4. Note
how a′2 is synonymous with b2 and how b′2 is synonymous with a2. This can be shown by
a connecting branch of unity. See Figure 9.4.
9.2.2 Signal flow representation of elements
Some examples of transmission line elements and their equivalent signal flow graphs are
shown in Figures 9.5 to 9.10.
In Figure 9.5, r represents the magnitude of the reflection coefficient. The phase change
produced by the termination is shown as ejfL where fL represents the load phase change in
Fig. 9.5 Signal flow graph representation of a termination
Figure 9.6 shows a length of lossless transmission line which has no reflection coeffi-
cient. When compared to the flow graph of Figure 9.7, the s11 and s22 branches have the
value zero or can be left out entirely. The term e–jfL represents the phase change within the
Fig. 9.6 Signal flow graph of a length of lossless transmission line
Figure 9.7 shows a detector and k denotes the scalar conversion efficiency relating the
incoming wave amplitude to a meter reading. The meter reading M is assumed calibrated
to take into account the detector law so k is independent of level.
Signal flow graph analysis 403
Fig. 9.7 Signal flow graph of a detector
Figure 9.8 shows the signal flow graph for a shunt admittance.
Fig. 9.8 Signal flow graph of a shunt admittance
Figure 9.9 shows the signal flow graph for a series impedance.
Fig. 9.9 Signal flow graph of a series impedance
Figure 9.10 shows a flow graph representation of a generator. In microwave systems, it
is generally more convenient to think of a generator as a constant source of outward trav-
elling wave with a reflection coefficient looking back into the generator output.
Fig. 9.10 Signal flow graph of a generator
9.2.3 Topological manipulation of signal flow graphs
The method of finding the value of a wave at a certain node may be arrived at with a series
of topological manipulations which reduce a flow graph to simpler and simpler forms until
the answer is apparent or until such a stage is reached when Mason’s non-touching rule
404 Further topics
can be used easily. Mason’s rule will be explained later. Four rules are given for easier
understanding of signal flow manipulations.
Rule I: Branches in series (where the common node has only one incoming and one
outgoing branch) may be combined to form a single branch whose coefficient is the prod-
uct of the coefficients of the individual branches. A typical example of this is shown in
Figure 9.11 where
E2 = S21E1 E3 = S32E2
E3 = S32S21E1 (9.3)
Fig. 9.11 Branches in series
Rule II: Two branches pointing from a common node to another common node (branches
in parallel) may be combined into a single branch whose coefficient is the sum of the indi-
vidual coefficients. A typical example of this is shown in Figure 9.12 where
Fig. 9.12 Branches in parallel
E2 = SAE1 E2 = SBE1
E2 = (SA + SB)E1 (9.4)
Fig. 9.13 Reduction of a feedback loop
Rule III: When node n possesses a self loop (a branch which begins and ends at n) of
coefficient Snn the self loop may be eliminated by dividing the coefficient of every other
branch entering node n by (1 – Snn). A typical example of this is shown in Figure 9.13
where the loop S22 at node E2 may be eliminated by dividing the coefficient (S21) entering
the node E2, by (1 – S22).
Signal flow graph analysis 405
Rule IV: A node may be duplicated, i.e. split into two nodes which may be subsequently
treated as two separate nodes, providing the resulting signal flow graph contains, once and
only once, each combination of separate (not a branch which forms a self loop) input and
output branches which connect to the original node. Any self loop attached to the original
node must also be attached to each of the nodes resulting from duplication. A typical
example is shown in Figure 9.14(a) to (e) where a complicated signal flow graph is
reduced to that of a far simpler one.
Fig. 9.14 Node duplication with a feedback loop: (a) original graph; (b) duplication of node E3 ; (c) elimination of node
E ’3 to form a self loop; (d) duplication of node E2 with self loop; (e) elimination of self loops
406 Further topics
9.2.4 Mason’s non-touching loop rule
Mason’s non-touching loop rule is extremely useful for calculating the wave parameters in
a network. At first glance, Mason’s expression appears to be very frightening and formi-
dable but it can be easily applied once the fundamentals are understood. Some of the
fundamentals relating to this rule have already been discussed in the preceding sections but
for the sake of clarity, some of the material will be repeated in the application of Mason’s
rule to the example given in Figure 9.15 which shows the flow diagram of a network
cascaded between a generator (E) and a load (GL).
Fig. 9.15 Signal flow representation of a network
When networks are cascaded it is only necessary to cascade the flow graphs since the
outgoing wave from the earlier network is the incoming wave to the next network. In
Figure 9.15, the system has only one independent variable, the generator amplitude (E).
The flow graph contains paths and loops.
A path is a series of directed lines followed in sequence and in the same direction in
such a way that no node is touched more than once. The value of the path is the product
of all coefficients encountered en route. In Figure 9.15, there is one path from E to b2. It
has the value S21. There are two paths from E to b1, namely S11 and S21GLS12.
A first-order loop is a series of directed lines coming to a closure when followed in
sequence and in the same direction with no node passed more than once. The value of
the loop is the product of all coefficients encountered en route. In Figure 9.15, there are
three first-order loops, namely GgS11, S22GL, and GgS21GLS12. A second-order loop is
the product of any two first-order loops that do not touch at any point. In Figure 9.15,
there is one second-order loop, namely GgS11S22GL. A third-order loop is the product
of any three first-order loops which do not touch. There is no third-order loop in Figure
9.15. An nth-order loop is the product of any n first-order loops which do not touch
and so on.
The solution of a flow graph is accomplished by application of Mason’s non-touching
loop rule1 which, written symbolically, is
P1[1 – ∑L(1)(1) + ∑L(2)(1) – ∑L(3)(1) + . . .] + P2[1 – ∑L(1)(2) + ∑L(2)(2) . . . ] + P3[1 – ∑L(1)(3) + . . .]
T = ——————————————————————————————————————————
1 – ∑L(1) + ∑L(2) – ∑L(3) + . . .
1 Do not panic! Detailed examples follow. If you really want to know more about this topic see Mason, S.J.,
‘Feedback theory – some properties of signal flow graphs’. Proc. IRE (41) 1144–56, Sept 1953. See also Mason,
S.J., ‘Feedback theory – further properties of signal flow graphs’. Proc. IRE (44) 920–26, July 1956.
Signal flow graph analysis 407
Here, P1, P2, P3, etc. are the values of all the various paths which can be followed from
the independent variable node to the node whose value is desired.
∑L(1) denotes the sum of all first-order loops. ∑L(2) denotes the sum of all second-
order loops and so on. ∑L(1)(1) denotes the sum of all first-order loops which do not touch
P1 at any point, and so on. ∑L(2)(1) denotes the sum of all second-order loops which do
not touch P1 at any point, the superscript (1) denoting path 1. Similarly, ∑L(1)(2) denotes
the sum of all first-order loops which do not touch P2 at any point and so on. In other
words, each path is multiplied by the factor in brackets which involves all the loops of all
orders which that path does not touch.
T is a general symbol representing the ratio between the dependent variable of interest
and the independent variable. This process is repeated for each independent variable of the
system and the results are summed.
As examples of the application of the rules in Figure 9.15, the transmission b2/E and
the reflection coefficient b1/a1 are written as follows:
b2 S21
—— = ——————————————————— (9.6)
E 1 – GgS11 – S22GL – GgS21GL S12 + Gg S11S22GL
b1 S11(1 – S22GL) + S21GLS12 S12S21GL
——= ——————————— = S11 + ———— (9.7)
a1 1 – S22GL 1 – S22GL
Note that the generator flow graph is unnecessary when solving for b1/a1, and the loops
associated with it are deleted when writing this solution. It is worth mentioning at this
point that second- and higher-order loops can quite often be neglected while writing down
the solution if one has orders of magnitude for the components in mind.
9.2.5 Signal flow applications
Signal flow graphs are best understood by some illustrative examples.
Example 9.1
Figure 9.16 illustrates a simple system where a generator is connected to a detector. The
signal flow graph for the system is illustrated in Figure 9.17. It is made up from the basic
Fig. 9.16 Generator and detector system
Fig. 9.17 Signal flow graph for system of Figure 9.16
408 Further topics
building blocks of the generator and detector illustrated in Figures 9.7 and 9.10. The phase
of the generator is not considered for ease of understanding.
By inspection, and by using Rules I and III, you can readily see that
[ ]
M1 = Eg ———— k (9.8)
1 – ρg ρ d
Example 9.2
Fig. 9.18 Generator, two-port network and detector
Fig. 9.19 Signal flow graph of system of Figure 9.18
Figure 9.18 depicts the case where a two port network is placed between the generator and
the detector. The signal flow graph of Figure 9.19 is again made up from the basic build-
ing blocks of Figures 9.3, 9.7 and 9.10. Note particularly that we have used the signal flow
graph of Figure 9.3 for the two-port network but for ease of working have replaced the S-
parameters, s11, s21, s22, s12, with r1, T, r2, T, respectively.
In Figure 9.20, nodes (2) and (4) have been duplicated into nodes (2′), (2″), (4′), (4″)
by Rule IV.
Figure 9.21 follows from the sequence of manipulations below.
Fig. 9.20 Simplification process 1
1 Eliminate node (4′) by Rule I giving a self loop at node (3) of the value ρd ρ2.
2 Eliminate this self loop by Rule III changing the value of the branch node (1) to node
(3) from T to T/(1 – ρd ρ2).
3 Eliminate node (2′) by Rule I giving a self loop at node (1) of value ρg ρ1.
4 Eliminate this self loop by Rule III changing the value of the branch leading from the
generator to 1/(1 – ρg ρ1) and also changing the branch from node (2″) to node (1) to
ρg/(1 – ρg ρ1).
Signal flow graph analysis 409
5 Eliminate nodes (2″) and (4″) giving a branch from node (3) to node (1) of value
(Tρgρd)/(1 – ρg ρ1).
Fig. 9.21 Simplification process 2
Figure 9.22 shows the duplication of node (3) into nodes (3′) and (3″).
Fig. 9.22 Simplification process 3
Figure 9.23 shows the signal flow graph which results from eliminating node (3″) by
Rule I and then eliminating the resulting self loop at node (1) by Rule III.
Fig. 9.23 Simplification process 4
In Figure 9.23, there now exists a single path from Eg to M2 and nodes (1) and (3) can
be eliminated by Rule 1 yielding:
Eg Tk
M2 =
⎡ T 2 ρg ρd ⎤
(1 − ρg ρ1 )⎢1 − ⎥(1 − ρd ρ2 )
⎢ (1 − ρd ρ2 )(1 − ρg ρ1 ) ⎥
⎣ ⎦
which simplifies to
⎡ 1 ⎤
M2 = Eg kT ⎢ 2
⎥ (9.9)
⎣1 − ρg ρ1 − ρd ρ2 − T ρg ρd + ρg ρd ρ1ρ2 ⎦
⎢ ⎥
410 Further topics
9.2.6 Summary on signal flow graphs
The chief advantages of signal flow graphs over matrix algebra in solving cascaded
networks are the convenient pictorial representations and the painless method of proceed-
ing directly to the solution with approximations being obvious in the process. As your
study in microwave engineering continues, you will come across more and more examples
on the use of S-parameters and signal flow techniques.
9.3 Small effective microwave CAD packages
9.3.1 Introduction
The subject of software is a volatile one because new programs are being constantly intro-
duced, old programs are constantly being updated, and last but not least, personal prefer-
ences come into the choice of a particular program.
Excellent radio engineering computer programs such as Hewlett Packard Design
System 85150 series, EEsof’s Libra*, Touchstone*, Super Compact*, and Academy* have
been available for many years. These programs are excellent and extremely versatile.
However, these facilities cost money and require quality computers with large RAM and
disk facilities. If you or your company can afford these systems then by all means go for
one or more of these software packages.
If you have an average personal computer with average RAM and only a little money,
consider intermediate software programs such as ARRL (American Radio Relay League)
‘ARRL Radio Designer’, or Barnard’s Microwave System’s ‘Wavemaker’ or Number One
System’s ‘Z match for Windows (Professional)’. The ARRL’s program is a subset of
‘Super Compact’ and is quite powerful. If your frequency usage is not confined to
microwave systems, then consider general purpose programs such as ‘PSpice’, ‘HSpice’
and ‘Spice Age’ which incorporate limited Smith chart facilities.
If money is nearly non-existent, then look in the Internet for free programs. Some of
these programs are excellent. You can also go to the large commercial firms and enquire
about the possibility of hiring software programs for a limited period. In some cases, it is
possible to obtain reduced rates for educational establishments. Some firms such as
Hewlett Packard offer internet sites where universities send in examples of their teaching
and research work and may offer free copies of the programs they have developed. Some
firms may offer you the free use of their small programs for a limited period.
Many people and some small radio engineering courses cannot afford even these costs.
In long distance learning situations, software expenditure becomes doubly important
because each student must be provided with programs which can be run on their home
computers. Therefore low cost, small, powerful packages requiring minimum storage and
RAM requirements are vitally important.
For personal and home use, I use inexpensive but relatively powerful programs such as
Hewlett Packard’s AppCAD, CalTech’s PUFF and Motorola’s Impedance Matching
Program (MIMP) to investigate and produce simple designs. AppCAD is useful for the
calculation of individual components, losses and gains. PUFF is valuable in calculating
two port parameters, input and output matches, gains and losses and frequency responses
and has plotting facilities for layout artwork. MIMP provides real time facilities for narrow
and broadband matching using Smith charts and rectangular plots.
Small effective microwave CAD packages 411
9.3.2 Hewlett-Packard’s computer aided design program – AppCAD
Hewlett-Packard’s Applications Computer Aided Design Program AppCAD is a collection
of software tools or modules, which aid in the design of RF (radio frequency) and
microwave circuits. AppCAD also includes a selection guide for Hewlett Packard RF and
Microwave semiconductors. The modules considered in AppCAD are listed in Figure 9.24.
Each main program is listed on the left table while the highlighted item is described on the
right table.
Fig. 9.24 Screen print-out of AppCAD
AppCAD modules
The modules considered in AppCAD are as follows.
(i) Transistor design data. This module computes various gain and stability data from
S-parameters. Input S-parameters can be read from a TOUCHSTONE formatted file or
entered manually. S-parameters input from a TOUCHSTONE formatted file may also be
modified manually via an edit function.
The program
• calculates stability circles;
• converts S-parameters to Y-parameters, Z-parameters or H-parameters;
• calculates Gms and Gml.
(ii) Mixer spurious search. Mixers will generate harmonics of the RF input signal and
the local-oscillator frequencies. This means that there exists a wide range of frequencies,
many of which may lie outside the desired input passband, which will give unwanted
responses in the IF band. This module will calculate all spurious responses for user chosen
frequencies, passbands, and harmonics of the local oscillator and signal.
(iii) Microwave calculator. This is a computerised version of the famous Hewlett
Packard ‘Reflectometer’ slide rule calculator. Calculations include reflection coefficients,
412 Further topics
standing wave ratios, mismatch loss, return loss, mismatch phase error, coupler directivity
uncertainty and maximum standing wave ratio from mismatches.
(iv) Microwave path calculations. This module calculates the signal-to-noise (S/N)
performance resulting from the following factors: receiver noise figure, antenna gain,
transmitter power, path distance, frequency and line losses. The systems covered are one-
way (communication) and two-way (radar).
(v) Transmission lines. From physical dimensions, this module calculates the following
properties: characteristic impedance, effective dielectric constant, electrical line length and
coupling factor. The structures covered include microstrip, stripline, coplanar waveguide
with and without ground plane and coaxial lines.
(vi) Two-port circuit analysis. This easy to use linear circuit analysis program can
include lumped or distributed circuit elements, which may be represented by an S-para-
meter file in TOUCHSTONE format. The calculated S-parameters can be displayed on the
screen and printed. The data can also be graphed on to either a Smith chart or a linear X–Y
plot, which can be printed to a Epson compatible printer.
(vii) Spiral inductor design. This module calculates the inductance of a circular spiral
from its number of turns, conductor width, substrate height, inner radius and dielectric
constant. Inductance is calculated for two cases, with and without a ground plane.
(viii) Impedance matching. This module determines both lumped and distributed
elements for impedance matching of a source impedance and load impedance. The lumped
or distributed elements determined are those for either an L-section, T-section, pi-section,
transmission line transformer or tandem 3/8 wavelength transformer.
(ix) Pin attenuator and switch design. Insertion loss and isolation are calculated from
PIN diode characteristics for both the series and shunt configurations. A built-in menu
automatically returns the diode parameters from a menu selection of part numbers.
Alternatively, custom diode series resistance and junction capacitance may be used. The
program calculates the required resistance values for both Pi and bridged T attenuators
from the desired attenuation in decibels (dB).
(x) Schottky detector calculations. This module calculates the effect of video amplifier
characteristics, RF bypassing, amplifier input resistance and voltage sensitivity on pulse
response, detected video bandwidth and TSS.
(xi) Transistor bias circuits. This module examines bias networks for microwave bi-
polar transistors. Bias network resistors are calculated for a given collector current and
voltage. The change in collector current with temperature is also computed for each
network. Networks covered include non-stabilised, voltage feedback and voltage feedback
constant base current.
(xii) Noise calculations. This module calculates the cascade noise figure and other
performance parameters for a sub-system block diagram such as a receiver. This type of
analysis allows system planning for the tradeoffs of important characteristics such as noise
Small effective microwave CAD packages 413
figure (sensitivity), gain distribution, dynamic range, signal levels and intermodulation
products. Provision is made for system analysis with temperature.
(xiii) Thermal analysis. A general introduction to heat transfer is presented with empha-
sis on applications to semiconductors. This module includes a tutorial section, thermal
resistance calculations for semiconductors, thermal analysis of MIC (hybrid) circuits and
a table of thermal conductivities.
Details of programs
A program listed in Figure 9.24 usually sub-divides into other programs. For example, the
transmission lines program of Figure 9.25 sub-divides into seven other types of transmission
line. See the right column of Figure 9.25 for a description of the program. Figure 9.25 further
sub-divides into another screen for the calculation of seven types of lines (Figure 9.26).
Fig. 9.25 Selection of transmission lines program
Fig. 9.26 Sub-division of the transmission lines program
414 Further topics
Fig. 9.27 Calculation of co-planar waveguide with ground plane
Figure 9.27 shows the case for the calculation of co-planar waveguide with a ground
The same is true of the microwave calculator program. The selection of this program
leads to further sub-division and four separate programs. See Figure 9.28.
AppCAD also has facilities for calculating the inductance of spiral inductors. You
merely have to state the number of turns and dimensions of your inductor in Figure 9.29
and APPCAD will give you the inductance in the result box.
AppCAD is also extremely useful for conversion from one set of parameters to another.
Figure 9.30 shows what happens when you select the ‘Transistor Design Data’ program. If
you now select the Convert to Y, Z, or H and press the return key. You will now be given
a choice of what type of parameters you require. See Figure 9.31. Ensure that the Convert
Fig. 9.28 AppCAD’s calculator program
Small effective microwave CAD packages 415
Fig. 9.29 Calculating the inductance of spiral inductors
Fig. 9.30 Selection of transistor design data
Fig. 9.31 Selection of Y-, Z- or H-parameters
416 Further topics
Fig. 9.32 Y-parameters of transistor HPMA0285.S2P
to Y-parameters line is highlighted. Press the return key and you get Figure 9.32. These
parameters have been calculated from the s-parameters supplied by the manufacturer for
transistor HPMA0285.S2P.
If you had wanted the Z-parameters, then you would have chosen the Convert to Z-
parameters line and pressed return to get Figure 9.33. These parameters have been calcu-
lated from the s-parameters supplied by the manufacturer for transistor HPMA0285.S2P.
Similarly if you had wanted H-parameters, you would have selected the Convert to H-
parameters and pressed return to get Figure 9.34.
So you can see for yourself how much time and effort can be saved by using AppCAD.
In most cases, you might be able to get the program free from Hewlett Packard who own
the copyright. The hardware requirements for the program are very modest and the
Fig. 9.33 Z-parameters of transistor HPMA0285.S2P
Small effective microwave CAD packages 417
Fig. 9.34 H-parameters of transistor HPMA0285.S2P
DOS version will even run with an 8086 processor. There is also a Windows version of
AppCAD available on the Internet.
9.3.3 PUFF Version 2.1
Fig. 9.35 Feedback amplifier response
418 Further topics
Puff Version 2.1 was chosen for this book because of: (i) its ease of use – PC format, (ii)
its versatility, (iii) its computer requirement flexibility – ≈290 KB on a floppy disk, any
processor from an 8080 to Pentium, choice of display, CGA, EGA or VGA, and (iv) choice
of printer, dot-matrix, bubble jet or laser and (v) its low costs. As you have seen for your-
self, it can be used for (i) lumped and distributed filter design, evaluation, layout and fabri-
cation, (ii) evaluation of s-parameter networks, (iii) lumped and distributed matching
techniques, layout and construction, (iv) amplifier design layout and construction and (v)
determination of input impedance and admittance of networks.
There are also many features to PUFF which have not been used in this book. For exam-
ple, PUFF can be used for the design of oscillators; it has compressed Smith chart facili-
ties. An example of this is shown in Figure 9.35.
If you want further information on PUFF, I suggest you contact PUFF Distribution,
CalTech in Pasadena, California, USA. They can supply you with the source code, a
manual for the program, and lecture notes for carrying out more advanced work with
9.3.4 Motorola’s Impedance Matching Program (MIMP)
MIMP is excellent for narrow-band and wide-band matching of impedances. It has
facilities for matching complex source and load impedances and designing lumped or
distributed circuits with the desired Q graphically. This program can be explained by
an example. Consider the case where the output impedance, ≈(20 + j0) Ω, of a trans-
mitter operating between 470 MHZ and 500 MHz is to be matched to an antenna whose
nominal impedance is 50 Ω. A return loss of ≈20 dB is required. The conditions are
entered into MIMPs as in Figure 9.36. Three frequencies are used to cover the band and
the load and source impedances are also entered into the figure. When Figure 9.36 is
completed, the ESC key is pressed to move on to Figure 9.37 where a network is
chosen. For this case, a T network has been chosen. At this stage, the exact values of
the components and the Q of the matching network are unknown so nominal values are
inserted. The exact values will be derived later. For clarity, Zin and the load have been
annotated in Figure 9.37.
After completion of Figure 9.37, the ESC key is pressed to enter Figure 9.38.
Starting from the top left line in Figure 9.37, we have SERIES CAP and up/down
arrows which allow any component to be selected. C1 is shown in this case. Capacitors
are shown in this box with its arrow keys. The next right block shows values of induc-
tors. Adjustment is provided by up/down arrow keys. The next right box with its arrow
keys is the Q selection box. Q = 3 has been selected. The next box after the logo is the
line impedance (Zo) box. Its default position is 10 Ω but it can be changed and the
Smith chart plot values will automatically change accordingly. The FREQ box allows
selection of frequency. It has been set to mid-point, i.e. 485 MHz. The remaining three
boxes are self-evident.
The middle left-hand box provides a read-out of impedance at points to the ‘right’ of a
junction. The Smith chart return loss circle size is determined by the Return loss boundary
set in the lower left box. Arcs AB, BC and CD are adjusted by components C1, C2 and L1
respectively until the desired matching is obtained.
Small effective microwave CAD packages 419
Fig. 9.36 Input data for Motorola’s MIMP
Fig. 9.37 Matching circuit for Motorola’s MIMP
420 Further topics
Fig. 9.38 Smith chart and input return loss
The great advantage of MIMP is that matching is carried out electronically and quickly.
There are no peripheral scales on it like a conventional paper Smith chart but this is unnec-
essary because each individual point on the chart can be read from the information boxes.
A good description of how this program works can be found in ‘MIMP Analyzes
Impedance Matching Network’ RF Design, Jan 1993, 30–42.
MIMP is a Motorola copyright program but it is usually available free from your
friendly Motorola agents. The program requirements are very modest with processor
80286 or higher, VGA graphics and 640k RAM.
9.4 Summary of software
The above computer aided design programs, namely AppCAD, PUFF and MIMP, are
extremely low cost and provide a very good cross-section of theoretical and practical
constructional techniques for microwave radio devices and circuits. AppCAD was used
extensively for checking the bias and matching circuits. MIMP was used extensively for
checking the Smith chart results in the book. I am sure that these programs will be useful
additions to your software library.
These references are provided as a guide to readers who want more knowledge on the main items
discussed in this book. They have been compiled into seven categories. These are circuit fundamen-
tals, transmission lines, components, computer aided design, amplifiers, oscillators, and signal flow
diagrams. The references are in alphabetical order in each section.
References soon become antiquated and to keep up with developments, it is best to read mater-
ial, such as the IEEE Transactions on various topics, IEE journals, Microwave Engineering journal,
etc. These journals are essential because they provide knowledge of the latest developments in the
high frequency and microwave world. Attending conferences is also very important and many large
firms like Hewlett Packard, Motorola, and Texas Instruments often provide free study seminars to
keep engineers up to date on amplifier, oscillator, CAD and measurement techniques.
Many large firms such as Hewlett Packard also provide education material on the Internet. For
example, many universities put their experimental work on http://www.hp.com/info/college_lab101.
Circuit Fundamentals
Avantek 1982: High frequency transistor primer, Part 1. Santa Clara CA: Avantek.
Festing, D. 1990. Realizing the theoretical harmonic attenuation of transmitter output matching and
filter circuits, RF Design, February.
Granberg, H. O. 1980. Good RF construction practices and techniques. RF Design, September/
Hewlett Packard. S parameters, circuit analysis and design. Hewlett Packard Application Note 95,
Palo Alto CA: Hewlett Packard.
Johnsen, R.J. Thermal rating of RF power transistors. Application Note AN790. Phoenix Az:
Motorola Semiconductor Products.
Jordan, E. 1979: Reference data for engineers: radio, electronics, computers and communications.
Seventh Edition. Indianapolis IN: Howard Sams & Co.
Motorola. Controlled Q RF technology – what it means, how it is done. Engineering bulletin EB19,
Phoenix AZ: Motorola Semiconductor Products Sector.
Motorola 1991. RF data book DL110, Revision 4, Phoenix AZ: Motorola Semiconductor Sector.
Saal, R. 1979: Handbook of filter design. Telefunken Aktiengesellschaft, 715 Backnang (Wurtt),
Gerberstrasse 34 PO Box 129, Germany.
Transistor manual, Technical Series SC12, RCA, Electronic Components and Devices, Harrison NJ,
Transmission Lines
Babl, I.J. and Trivedi, D.K. 1977. A designer’s guide to microstrip. Micorwaves, May.
Chipman, R.A. 1968: Transmission lines. New York NY: Schaum, McGraw-Hill.
422 References
Davidson, C.W. 1978: Transmission lines for communications. London: Macmillan.
Edwards, T.C. 1992: Foundations for microstrip circuit design. Second Edition. John Wiley & Sons.
Ho, C.Y. 1989. Design of wideband quadrature couplers for UHF/VHF. RF Design, 58–61,
November (with further useful references).
Smith, P.H. 1944. An improved line calculator. Electronics, January, 130.
Acrian Handbook 1987. Various Application Notes. The Acrian Handbook, Acrian Power Solutions,
490 Race Street, San Jose CA.
Blockmore, R.K. 1986. Practical wideband RF power transformers, combiners and splitters.
Proceedings of RF Expo West, January.
Fair-Rite. Use of ferrities for wide band transformers. Application Note. Fair-Rite Products
Haupt, D.N. 1990. Broadband-impedance matching transformers as applied to high-frequency power
amplifiers. Proceedings of RG Expo West, March.
Myer, D. 1990. Equal delay networks match impedances over wide bandwidths. Microwaves and
RF, April.
Phillips, 1969–72. On the design of HF wideband transformers, parts I and II. Electronic Application
Reports ECO69007 & ECOP7213. Phillips Discrete Semiconductor Group.
Computer aided design
CAD Roundtable 1996. Diverse views on the future of RF design. Microwave Engineering Europe
Directory, 20–26.
Da Silva, E. 1997: Low cost microwave packages. Fourth International Conference, Computer Aided
Engineering Education CAEE97, Krakow, Poland.
Da Silva, E. 1997: Low cost radio & microwave CAL packages. EAEEIE, Eighth Annual conference,
Edinburgh, Scotland.
Davis, F. Matching network designs with computer solutions. Application Note AN267. Phoenix AZ:
Motorola Semiconductor Sector.
Edwards, T.C. 1992: Foundations for microstrip circuit design. Second Edition. John Wiley &
Gillick, M., Robertson, I.D. and Aghvami, A.H. 1994. Uniplanar techniques for MMICs. Electronic
and Communications Engineering Journal, August, 187–94.
Hammerstad, E.O. 1975. Equations for microstrip circuit design. Proceedings Fifth European
Conference, Hamburg.
Kirchning, N. 1983. Measurement of computer-aided modelling of microstrip discontinuities by an
improved resonator method. IEEE Trans MIT, International Symposium Digest, 495–8.
Koster, W., Norbert, H.L. and Hanse, R.H. 1986. The microstrip discontinuity; a revised description.
IEEE MTT 34 (2), 213–23.
Matthei, G.L., Young, L. and Jones, E.M.T. 1964: Microwave filters, impedance matching networks
and coupling structures. New York NY: McGraw-Hill.
MMICAD (for IBM PCs). Optotek, 62 Steacie Drive, Kanata, Ontario, Canada, K2K2A9.
Moline, D. 1993. MIMP analyzes impedance matching networks. RF Design, January, 30–42.
Nagel, L.W. and Pederson, D.O. 1973. Simulation program with integrated circuit emphasis.
Electronics Research Lab Rep No ERL-M382, University of Calif, Berkeley.
PSpice by MicroSim Corporation, 20 Fairbands, Irvine CA 92718.
Rutledge, D. 1996: EE153 Microwave Circuits. California Institute of Technology.
SpiceAge. Those Engineers Ltd, 31 Birbeck Road, Mill Hill, London, England.
Wheeler, H.A. 1977. Transmission line properties of a strip on a dielectric sheet on a plane. IEEE
MTT 25 (8), 631–47.
References 423
Bowick, C. 1982: RF circuit design. Indianapolis IN: Howard Sams.
Carson, R.S. 1975: High frequency amplifiers. New York NY: John Wiley and Sons.
Dye, N. and Shields, M. Considerations in using the MHW801 and MHW851 series power modules.
Application Note AN-1106. Phoenix AZ: Motorola Semiconductor Sector.
Froehner, W.H. 1967. Quick amplifier design with scattering parameters. Electronics, October.
Gonzales, G. 1984: Microwave transistor amplifier analysis and design. Englewood Cliffs NJ:
Prentice Hall.
Granberg, H.O. A two stage 1 kW linear amplifier. Motorola Application Note A758. Phoenix AZ:
Motorola Semiconductor Sector.
Granberg, H.O. 1987. Building push-pull VHF power amplifiers. Microwave and RF, November.
Hejhall, R. RF small signal design using two port parameters. Motorola Application Report AN
Hewlett Packard. S parameter design. Application Note 154. Palo Alto CA: Hewlett Packard Co.
ITT Semiconductors. VHF/UHF power transistor amplifier design. Application Note AN-1-1, ITT
Liechti, C.A. and Tillman, R.L. 1974. Design and performance of microwave amplifiers with GaAs
Schottky-gate-field-effect transistors. IEEE MTT-22, May, 510–17.
Pengelly, R.S. 1987: Microwave field effect transistors theory, design and applications. Second
Edition. Chichester, England: Research Studies Press, division of John Wiley and Sons.
Rohde, U.L. 1986. Designing a matched low noise amplifier using CAD tools. Microwave Journal,
October 29, 154–60.
Vendelin, G., Pavio, A. and Rohde, U. Microwave circuit design. New York NY: John Wiley &
Vendelin, G.D., Archer, J. and Bechtel, G. 1974. A low-noise integrated s-band amplifier. Microwave
Journal, February. Also IEEE International Solid-state Circuits Conference, February 1974.
Young, G.P. and Scalan, S.O. 1981. Matching network design studies for microwave transistor
amplifiers. IEEE MTT 29, No 10, October, 1027–35.
Abe, H. A highly stabilized low-noise Ga-As FET integrated oscillator with a dielectric resonator in
the C Band. IEEE Trans MTT 20, March.
Gilmore, R.J. and Rosenbaum, F.J. 1983. An analytical approach to optimum oscillator design using
S-parameters. IEEE Trans on Microwave Theory and Techniques MTT 31, August, 663–9.
Johnson, K.M. 1980. Large signal GaAs FET oscillator design. IEEE MTT-28, No 8, August.
Khanna, A.P.S. and Obregon, J. 1981. Microwave oscillator analysis. IEEE MTT-29, June, 606–7.
Kotzebue, K.L. and Parrish, W.J. 1975. The use of large signal S-parameters in microwave oscillator
design. Proceedings of the International IEEE Microwave Symposium on circuits and systems.
Rohde, Ulrich L. 1983: Digital PLL frequency-synthesizers theory and design. Englewood Cliffs NJ:
Prentice Hall.
Vendelin, G.D. 1982: Design of amplifiers and oscillators by the S-parameter method. New York
NY: John Wiley and Sons.
Signal flow
Chow, Y. and Cassignol, E. 1962: Linear signal-flow graphs and applications. New York NY: John
Wiley and Sons.
Horizon House 1963: Microwave engineers’ handbook and buyers’ guide. Horison House Inc,
Brookline Mass, T-15.
Hunton, J.K. 1960. Analysis of microwave measurement techniques by means of signal flow graphs.
Trans IRE MTT-8, March, 206–12.
424 References
Mason, S. J. 1955. Feedback theory – some properties of signal flow graphs. Proc IRE 41, 1144–56,
Mason, S.J. 1955. Feedback theory – further properties of signal flow graphs. Proc IRE 44, 920–26,
Mason and Zimmerman. 1960: Electronic circuits, signals and systems. New York NY: John Wiley
and Sons.
Montgomery et al 1948: Principles of microwave circuits. New York NY: McGraw-Hill Book Co.
AC equivalent circuits, transistors 291, 292 Band-stop filter 222
Active bias circuit 276 design 223
Adjacent channel selectivity 35 Barkhausen criteria 359
Admittance manipulation, on Smith chart 98 Base-voltage potential divider bias circuit 272
Admittance parameters 297 Biasing:
Admittance, found using Smith charts 172 bi-polar transistors 275
Aerial amplifier design 300 depletion mode MOSFETS 290
Aerial distribution systems using amplifiers 31 MOSFETs 286
Aerials 14 n-channel FETs 278
Alternative bias point 313 Bipolar transistors 262
Amplifier design 195 biasing 275
broadband amplifiers 349 construction 291
feedback amplifiers 352 Branch line coupler 82, 170
for optimum noise figure 345 Broadband amplifiers, design 349
for specific gain 332 Broadband matching 112
s parameters 321 Broadband matching networks 260
using conditionally stable transistors 313, 314 BT cut 377
with conditionally stable devices 339 Butterworth filter 208
with conjugately matched impedances 322 Butterworth normalized values 210
Amplitude distortion 63
Amplitude modulation 4 Capacitance end effects for an open circuit 191
Analogue-type phase detector 383 Capacitive divider matching 240
AND gate phase detector 383 Capacitive matching 240
Antennas 14 Capacitive stub matching 181
distribution systems 25 Capacitors:
AppCAD 411 quality factor 242
AT cut 377 series and parallel forms 242
Attenuation 46, 62 Cascading of tuned circuits 201
Auto-transformers 235 Characteristic impedance 45
Clapp oscillator 372
Balanced antenna 25 Coaxial line 48
Balanced line 26 characteristic impedance 50
Balanced/unbalanced transformer 26 Collector current characteristics 265
Bandpass filter 152, 217 Colpitts oscillator 368
design 218 Combined modulation 9
using microstrip lines 221 Commercial cable 60
426 Index
Constant gain circles 333 Gain sensitivity:
Co-planar waveguides 47 AND gate 385
Coupled lines 50 flip-flop detector 385
Coupler evaluation, using PUFF software 170 Group velocity 63
Crystal control oscillators 376
Crystal temperature stability 377 Half-power supply principle 268
Crystals 376 Hartley oscillators 271
Current gain, transistors 266, 299 High frequency equivalent circuit of a
transistor 294
DC amplifier 386 High-pass filter 214
Depletion mode MOSFETs 288 design 215
Digital phase detectors 383 using microstrip lines 217
Dipole 16 Hybrid p equivalent circuit 294
Direct digital waveform synthesis 395
Direct type synthesizer 393 Image channel interference 39
Directivity of radiation 14 Immittance Smith chart 97
Discontinuities 189 Impedance:
in transmission systems 46 conversion to Admittance using Smith chart
Dispersion, in transmission systems 46 95
Double stub matching 122 of distributed circuits 106
Double stub tuning, verification using PUFF Impedance manipulation, on Smith chart 94
software 186 Impedance matching 110, 233, 412, 418
Double superhet receivers 40 using a l/4 transformer 111
using a stub tuner 115
Electromagnetic waves 9 using circuit configurations 241
Equivalence of the series and parallel using multiple stubs 122
representations 242 Impedance relations in transmission lines 65,
Excess capacitance of a corner 190 77
Impedance values, plotted on Smith chart 92
Feedback amplifiers, design 352 Incident waves 45
FETs 277 Indefinite admittance matrix 316
construction 292 Indirect synthesizers 396
n-channel 277 Inductive stub matching 181
biasing 278 Inductors:
properties 290 quality factor 245
Field strength 12 series and parallel forms 245
Field-effect transistors see FETs Input admittance 302
Filter design 203 Input impedance 168
Filters 203 Input impedance of line, verified using PUFF
Butterworth filter 208 software 174
normalised parameters 210 Input impedance of low loss transmission lines
specification of 206 79
Tchebyscheff filter 225 Input reflection coefficient 323
First order loop 406 Intermediate frequency amplifier with
Flip-flop-type phase detector 385 transformers 236
gain sensitivity 385 Isotropic radiation 14
Folded dipole 17
Free-space radiation 11 L matching network 247
Frequency distortion 63 Ladder network 188
Frequency modulation 6 Line impedance derivation 55
Frequency synthesisers 393 Line impedances, found using Smith charts 107
Index 427
Linvill stability factor 304 radio frequency oscillators 368
Lock in range 392 sine wave type oscillators 358
Loop gain, phase lock loops 389 voltage-controlled oscillator 375, 387
Low frequency sine wave oscillators 361 Wein bridge oscillators 361
Low frequency equivalent circuit of a Output admittance 302
transistor 294 Output impedance matching, verification using
Low pass filter: PUFF software 185
design 211 Output reflection coefficient 322
using microstrip lines 213 Oven controlled crystal oscillator 379
Mason’s non-touching loop rule 406 Parallel circuits 200
Matched transmission lines 64 Parallel wire line 49
Maximum available gain 307, 321 p-channel 281
Microstrip lines 48, 165 Phase detector 381
as band pass filters 221 types 383
as low pass filters 213 Phase lock loops 380
characteristic impedance 51 Phase modulation 8
step change in width 191 Phase shift oscillators 364
Microwave amplifiers 320 Phase velocities, 63
Microwave CAD packages 407 p network 253
Microwave calculator 412 p -equivalent circuit 293
Microwave path calculations 412 Pin attenuator and switch design 412
MIMP 418 Polar diagram 15
Mismatched loss 69 Polarisation 12
Mismatching techniques 314 Power density 13
Mixer spurious search 411 Pre-scaling 396
MOSFETs 284 Primary line constants 54
biasing 286 Propagation constant in terms of the primary
depletion mode MOSFETS 288 constants 72
n-channel enhancement mode type 284 Propagation constant of transmission lines 72
p-channel enhancement-mode type 285 Propagation delay 61
properties 290 Propagation of energy 45
Multi-loop antennas 19 Propagation time delay, in transmission systems
Neper 62 Propagation velocity, in transmission systems
Network impedances, verification using PUFF 46
software 173 PUFF 2.1 software 43, 142, 198, 330, 417,
Noise calculations 412 418
Noise figure 37 bandpass filters 152
nth order loop 406 commands 156
evaluating couplers 170
Operating point, transistors 266 installation 143
Oscillators 357 printing and fabrication of artwork 150, 154
Colpitts oscillator 368 running PUFF 144
comparison of types 376 Smith chart expansion 149
crystal control oscillators 376 s-parameter calculation 186
frequency synthesisers 393 templates 157
Hartley oscillator 371 modification of transistor templates 164
low frequency sine wave oscillators 361 verification of Smith chart applications
phase lock loops 380 172
phase shift oscillators 364 Pulse propagation 61
428 Index
Quality factor (Q): of detector 403
capacitors 242 of generator 403
inductors 245 of lossless transmission line 402
on Smith charts 93 of series impedance 403
Quarter wave transformers, verification using *of shunt admittance 403
PUFF software 175 Signal propagation on transmission lines 61
Signal to noise ratio 36
Radiating resistance 15 Simultaneous conjugate matching 308
Radio frequency amplifiers 297 Sine wave type oscillators 358
Radio frequency design conditions 304 Single loop antennas 19
Radio frequency oscillators 368 Slot line 48
Radio frequency power transistor 355 Smith charts 88
Radio frequency transistor modelling 297 admittance manipulation 98
Radio receivers 32 applications
properties 33 expansion 149
Rat race coupler 85, 170 immitance 97
impedance manipulation 94
Reactances using transmission lines 80 impedance matching 110
Reflected waves in transmission lines 46 impedance networks 104
Reflection coefficient 46, 64, 70, 105, 166 impedance of distributed networks 106
Reflections in transmission systems 46 impedance-to-admittance conversion 95
Return loss 75 plotting impedance values 92
Ring coupler 85 PUFF 2.1 software 149
Q values 94
s parameters see Scattering parameters reflection coefficients 102
SC cut 377 theory and applications 99
Scattering coefficients 401 using PUFF 172
Scattering parameters 125, 186, 321, 401 Spiral inductor design 412
calculation using PUFF 186 Stability circles 339
conversion between s-parameters and Stability of tyransistor 304
y-parameters 131 Standing wave ratio 47
examples in two-port networks 131 Static phase error, phase lock loops 389
in terms of impedances 130 Stern stability factor 306
in transistor amplifier design 321 Strip line 48
incident and reflected waves 127 Stub matching, verification using PUFF
Schottky detector calculations 412 software 176, 185
Second channel interference 40 Superheterodyne receiver 38
Second order loop 406
Selectivity 34 T network 257
Sensitivity 36 Tchebyscheff filter 225
Series circuits 197 design procedures 228
Series connected L networks for lower Q Tchebyscheff high pass filter 230
applications 260 Tchebyscheff low pass filter 228
Series elements 187 Tchebyscheff tables 227
Shunt elements 187 Temperature compensated crystal oscillator 379
Signal flow applications 407 Templates, PUFF software 157
Signal flow graph: Thermal analysis 413
analysis, 400 Third order loop 406
*topological manipulation 403 Three element matching networks 252
Signal flow representation: Topological manipulation of signal flow graphs
of termination 402 403
Index 429
Transducer gain 311, 329 Tuned circuits 196
PUFF software results 330 cascading 201
Transformer matching 233 Twin lines 49
Transistor action 263 Twin parallel wire characteristic impedance 51
Transistor bias circuits 412 Two-port circuit analysis 412
Transistor biasing 266 Two-port networks 295
Transistor design data 411 Two-way splitter 30
Transistor impedances, matching 181
Transistor operating configurations 316 Unbalanced antennas 25
Transistor stability 320, 321 Unbalanced line 26
Transistor template modification, PUFF Unilateralization and neutralisation 313
software 157 Unmatched transmission lines 64
AC equivalent circuits 291 Vertical rod antennas 18
as a two port network 295 Voltage controlled oscillator 375, 387
high-frequency transistor amplifiers 262 Voltage feedback bias circuit 267, 270
low-frequency equivalent circuit 294 Voltage gain 299
Translation loop, 398 Voltage reflection coefficient 65
Transmission coefficient 46, 64, 69 VSWR 64, 69
Transmission line couplers 82
Transmission lines 412 Wave impedance 12
as transformers 81 Waveguide 45, 47
as electrical components 77 Wein bridge oscillators 361
matched 64
unmatched 64 Yagi-Uda array 23
Transmission path, energy 45
TRF receivers 36 Z0 by measurement 59
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October 2
A teacher gave a test to a class in which 10% of the students are juniors and 90% are seniors. The average score on the test was 84. The juniors all received the same score, and the average score of
the seniors was 83. What score did each of the juniors receive on the test?
Interactive version of problem and solution.
Buy Origami, Eleusis, and the Soma Cube
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General Maths @ Coburg
The idea of correlation
In analysing the scatterplot we look for a pattern in the way the points lie. Certain patterns tell us that certain relationships exist between the two variables. This is referred to as a correlation
Bob and Jim work for the same company. Bob drives a Porsche, costing $200000, and Jim drives an Austin Allegro, costing $9000. Which man has the greater salary?
In this case, we can reasonably assume that it must be Bob who earns more, as he drives the more expensive car. As he earns a larger salary, the chances are that he can afford a more expensive car.
We can’t be absolutely certain, of course. It could be that Bob’s Porsche was a gift from a friend, or part of the divorce settlement from his wife – or he could have stolen it! However, most of the
time, an expensive car means a larger salary.
So we say that there is a correlation between someone’s salary and the cost of the car that he/she drives. This means that as one figure change, we can expect the other to change in a fairly regular
q-correlation coefficient
The q-correlation coefficient is a measure of the strength of the association between two variables. The calculation of the q-correlation coefficient aids us considerably in making that judgment.
To calculate the q-correlation coefficient:
The value of the q-correlation coefficient in the above example indicates a strong correlation. The diagram below gives a rough guide to the strength of the correlation suggested by the value of q.
Practice Questions
Question 1
Question 2
a. Calculate the q-correlation coefficient for your scatter plot of height vs arm span.
b. Write down what type of relationship exists between the pair of variables and comment on this relationship.
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Math Help
April 20th 2011, 10:26 AM #1
Nov 2010
Let f (xy) =f(x) f(y) for every x, y belongs to R.if f(x) is continuous at any one point x=a, then prove that f(x) is continuous for all x belongs to R-{0}.
F(x) = 3 if x=0
and [(1+ax+bx^3)/(x^2)]^(1/x) if x is not equal to zero
Find a, b if F(x) is continuous at 0.(looks wrong question )
Let f (xy) =f(x) f(y) for every x, y belongs to R.if f(x) is continuous at any one point x=a, then prove that f(x) is continuous for all x belongs to R-{0}.
F(x) = 3 if x=0
and [(1+ax+bx^3)/(x^2)]^(1/x) if x is not equal to zero
Find a, b if F(x) is continuous at 0.(looks wrong question )
How about showing some work?!
For Q1, you could write out what it means for f to be continuous on the given set. That's what they call "beginning with the end in mind".
Use the given information about a, have a creative (or logical) insight, and string it all together!
I assume that f(xy) means f(x,y), that is to say, the function is not of the product of x and y, but rather it is a function of x and y separately.
If that's true then this first part should be false by taking y/(x-1). It will be continuous, say, at x = 2 but discontinuous at x = 1, which is in R-{0}. Perhaps you've made a mistake typing in
the statement of the question.
I assume that f(xy) means f(x,y), that is to say, the function is not of the product of x and y, but rather it is a function of x and y separately.
If that's true then this first part should be false by taking y/(x-1). It will be continuous, say, at x = 2 but discontinuous at x = 1, which is in R-{0}. Perhaps you've made a mistake typing in
the statement of the question.
Perhaps you have made a mistake in the interpretation of the question!
Since the function is defined on R (NOT RxR), xy is a product.
NO question is correctly typed.
If f is continuous at any (non-zero) a, then, for any x and b, let x= by/a y= bx/a. The f(x)= f(by/a)= f(b/a)f(y). In particular, [tex]\lim_{x\to b}f(x)= \lim_{y\to a} f(by/a)= f(b/a)\lim_{y\to
a} f(y)= f(b/a)f(a)= f((b/a)(a))= f(b)
However, I believe that $ae 0$ is necessary for this.
April 20th 2011, 10:41 AM #2
April 20th 2011, 10:42 AM #3
Jun 2010
April 20th 2011, 10:43 AM #4
April 20th 2011, 10:44 AM #5
Nov 2010
April 20th 2011, 01:36 PM #6
MHF Contributor
Apr 2005
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[FOM] Mathematical explanation
Richard Heck rgheck at brown.edu
Fri Oct 28 16:48:35 EDT 2005
mjmurphy wrote:
>On Mon, 24 Oct 2005, I wrote:
>>... there is an example of Searle's (borrowed from Wittgenstein, but I don't quite know where). The example involves the mathematical proposition 3 + 4.
>Neil said: That's not a proposition, but let's not bother ...
>Actually, lets, as it may make Searle's point clearer. [snip] Searle then asks: are there any types of sentence which are immune from this kind of context dependence, and considers mathematical statements in this light:
>"Perhaps one might show, for example, that an arithmetical sentence such as "3+4=7" is not dependent on any contextual assumptions for the applicability of its literal meaning. Even here, however, it appears that certain assumptions about the nature of mathematical operations such as addition must be made in order to apply the literal meaning of the sentence."
What kinds of assumptions about addition are we supposed to have to
make? The example mentioned is one in which there are three nuts in one
circle and four in another, but the circles overlap. So what Searle is
suggesting is that you can't `apply' "3+4+7" in such a situation. That's
obviously true, if what that is supposed to mean is that you shouldn't
conclude that there are seven nuts in the union of the circles. But
"3+4=7" does not say anything about nuts and circles. (That was Neil's
point.) What it implies, and what one can prove logically, is that, if
there are three Fs and there are four Gs, and no F is a G, then there
are seven F-or-Gs. That has no "conditions of applicability", so far as
I can see. And if you're tempted to say it's useless if there are Fs
that are G, don't forget you can reason by modus tollens: If you find
there aren't seven F-or-Gs, and you know there are three Fs and four Gs,
then you can conclude that some F is G. (Frege was rather fond of this
sort of point.)
>So, to the question "Is mathematics necessary?" It seems to me that if an arithmetical sentence with its literal meaning can be applied under differing assumptions about the nature of mathematical operations, than we have a counter[example] to its application under any particular set of assumptions, and so mathematics is not necessary.
No such example could serve to undermine the necessity of mathematical
claims. To think that it could is to misunderstand both what
context-dependence is and what Searle is arguing: Searle is arguing that
the literal meaning of most (or all) sentences underdetermines the
propositions expressed by utterances of them, which are determined only
contextually. What Murphy says is that "the literal utterance of a
statement does not supply us with a proposition which can be used to
assess its truth value without reference to context". That comment is
ambiguous, and I'll guess that the ambiguity is part of the trouble.
What Searle is saying is not that the *assessment* is relative to
context (Searle isn't a relativist) but that what proposition is
expressed is relative to context. If so, however, then the literal
meaning of a sentence is not, in general, truth-evaluable, so of course
its literal meaning isn't necessary. But that's trivial.
Compare: The proposition expressed by the sentence "I am Richard Heck"
is necessarily true if I utter it, but necessarily false if anyone else
does. The *sentence* is neither necessary nor contingent, since it isn't
truth-evaluable, and the same goes for its literal meaning. Searle is
claiming that all sentences are kind of like "I am Richard Heck", but he
really thinks they are more like "That is Richard Heck", about which
similar things could be said.
That mathematical claims are necessary is a thesis about the
propositions those claims express. It would be silly to think that the
sentence "3+4=7" could not have expressed a falsehood, though that is
sometimes said, sloppily. It could have, and it would have had "3" meant
what "2" does. Similarly, if there were certain contexts in which
utterances of "3+4=7", that would not show that what it expressed in
certain other contexts was not necessary. So to argue is seriously confused.
Richard Heck
Richard G Heck, Jr
Professor of Philosophy
Brown University
Get my public key from http://sks.keyserver.penguin.de
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Curriculum and Program Information
Hours and Location
Mon - Fri:8:30am - 5:00pm
Grace Jacobs, 6th Floor
• Curriculum and Program Information
Curriculum and Program Information
Department of Mathematics and Computer Science
The Department of Mathematics and Computer Science offers a major and a minor in both Mathematics and Computer Science.
Within the Mathematics major the student may choose a concentration in Mathematics Secondary Education. The Mathematics major is intended to prepare students for any of the following:
• the study of mathematics on the graduate level
• employment in business, government, or industry
• teaching mathematics at the secondary level
• study in subject areas requiring a strong mathematics background, such as chemistry, economics, engineering, operations research, and actuarial science.
The Computer Science major is intended to provide students with the knowledge, aptitudes, and skills required for successful employment in computer-related fields and for the study of computer
science on the graduate level.
General Education Requirement in Mathematics
The General Education Requirements of the University include three semester credit hours in mathematics, excluding credits earned for courses with the DVMT code. Each entering student is required to
take a mathematics placement exam. The student's achievement level on this exam and high school mathematics record are used to place the student in DVMT 108, DVMT 109, or a course to satisfy the
General Education Requirement.
Usually this is one of the following courses, depending on the student's major:
MATH 103 Mathematics for Elementary Teachers I
MATH 110 College Algebra
MATH 125 Mathematics for Liberal Arts
MATH 131 College Algebra for Mathematics and Science Majors
MATH 203 Basic Statistics
The student should consult his/her academic advisor to determine which course to take to satisfy the General Education Requirement in mathematics. The following table summarizes these departmental
Mathematics General Education Requirement by Major/Department
Department Requirement
Applied Psychology MATH 125
Criminal Justice MATH 125 or MATH 203
Education MATH 103
History MATH 125
Humanities and Media MATH 125
Liberal Arts MATH 125 or MATH 203 (or any other MATH course)
Management Science MATH 131
Math/Computer Science MATH 131
Natural Science MATH 131
Nursing MATH 110
Social Sciences MATH 125
Social Work MATH 203
Sports Management MATH 125 (except MATH 110 for Sports Medicine concentration)
Urban Arts Production MATH 125 or MATH 203 or MATH 203
Course Prerequisites
For courses in mathematics and computer science, prerequisites are specified. It is department policy that these prerequisites must be completed with a grade of C or better.
Assessment of Majors
The extent to which students majoring in both Mathematics (Liberal Arts) and Computer Science have met the goals of the program will be measured before each student graduates.
Each student will be required to take a capstone course (MATH 417 or COSC 417 ) in the senior year. The course is intended to cover current and advanced topics in Mathematics (or Computer Science).
It will draw together all of the material the students have encountered in their earlier training. The assessment will involve either a project undertaken by a student or group of students and/or a
test developed by members of the Mathematics and Computer Science department to measure knowledge of topics taught in the major-requirement courses. Both the project and the test may involve
Related items
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Robert Recorde
From Wikipedia, the free encyclopedia
Robert Recorde (c. 1510 – 1558) was a Welsh physician and mathematician. He introduced the "equals" sign (=) in 1557.
A member of a respectable family of Tenby, Wales, he entered the University of Oxford in about 1525, and was elected a fellow of All Souls College in 1531. Having adopted medicine as a profession, he
went to the University of Cambridge to take the degree of M.D. in 1545. He afterwards returned to Oxford, where he publicly taught mathematics, as he had done prior to going to Cambridge. It appears
that he afterwards went to London, and acted as physician to King Edward VI and to Queen Mary, to whom some of his books are dedicated. He was also controller of the Royal Mint and served as
"Comptroller of Mines and Monies" in Ireland.^[1] After being sued for defamation by a political enemy, he was arrested for debt and died in the King's Bench Prison, Southwark, in 1558.
Recorde published several works upon mathematical subjects, chiefly in the form of dialogue between master and scholar, such as the following:
• The Grounde of Artes, teachings the Worke and Practise, of Arithmeticke, both in whole numbers and fractions (c. 1540), the first English book on algebra.
• The Pathway to Knowledge, containing the First Principles of Geometry ... bothe for the use of Instrumentes Geometricall and Astronomicall, and also for Projection of Plattes (London, 1551)
• The Castle of Knowledge, containing the Explication of the Sphere both Celestiall and Materiall, etc. (London, 1556)
• The Whetstone of Witte, whiche is the seconde parte of Arithmeteke: containing the extraction of rootes; the cossike practise, with the rule of equation; and the workes of Surde Nombers (London,
1557). This was the book in which the equals sign was introduced. With the publication of this book Recorde is credited with introducing algebra into England.^[2]
• a medical work, The Urinal of Physic (1548), frequently reprinted.
Sherburne states that Recorde also published Cosmographiae isagoge, and that he wrote a book De Arte faciendi Horologium and another De Usu Globorum et de Statu temporum. Recorde's chief
contributions to the progress of algebra were in the way of systematizing its notation.
See also
1. ^ Newman, James R. 1956, "The World Of Mathematics"
2. ^ Jourdain, Philip E. B. 1913, "The Nature Of Mathematics"
• This article incorporates text from the article "Robert Recorde" in the Encyclopædia Britannica, Eleventh Edition, a publication now in the public domain.
• Newman, James R. (1956), "The World Of Mathematics" Vol. 1 "Commentary On Robert Recorde"
• Jourdain, Philip E.B. (1913), "The Nature Of Mathematics"
External links
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Topic: Who cares the null hypotheses (two-tail test) is really TRUE?
Replies: 2 Last Post: Dec 22, 2012 7:42 PM
Messages: [ Previous | Next ]
Luis A. Re: Who cares the null hypotheses (two-tail test) is really TRUE?
Afonso Posted: Dec 22, 2012 7:42 PM
Posts: Comment about
From: ?Should Psychology abandon p-values and teach CI´s instead? Evidence-based reform in Statistics education?.
LIsbon Fiona Fidler (2006).
(Portugal) www.stat.auckland.ac.nz/~iase/publications/17/5E4_FIDL.pdf -
2/16/05 I, myself, never search if a hypothesis is true or false . . . a chimerical goal I do not pursue. We should not care, at all, such a thing: is impossible and uninteresting. What we are
only able to achieve is if there is (or not) sufficient evidence/plausibility to reject the Null Hypothesis giving the data. To bring the Aristotelian logic (true/false) to events
subjected to chance is plainly erroneous.
The A. said, all people agree: ?particularly serious misconception associated with NHST, namely that statistical non-significance is equivalent of evidence of `no effect`. But immediately
she state that ? from a low power study with a no-trivial effect size - as evidence the null hypotheses is true . . . suggest that she have the concern, about she criticise as a method?s
weakness: a sufficiently large sample will invalidate this result. But this is expected, of course, this is natural: even a little difference will be evident from data, therefore leading
to H0 rejection. A difference that could be, evidently, insignificant for practical purposes.
Luis A. Afonso
Date Subject Author
8/5/11 Who cares the null hypotheses (two-tail test) is really TRUE? Luis A. Afonso
8/6/11 You NEVER know if a coin is FAIR! Luis A. Afonso
12/22/12 Re: Who cares the null hypotheses (two-tail test) is really TRUE? Luis A. Afonso
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Crystal Growth in a Computer
U, potential energy or chemical binding energy;
k[b], Boltzmann's constant or 1.38041x10-23
T, absolute Temperature (Kelvin) where the temperature for the triple point of water is 273.15 Kelvin.
The study of the growth of solid crystals, whether in a liquid solution or by vapor deposition in a vacuum chamber, is an important and intriguing process. It is a process which has many important
uses in such industries as pharmaceuticals to semiconductors. The process for this project is simple. We start with a seed crystal or substrate with which atoms from the solution or vapor collide.
Some of these atoms become a part of the crystal and the substrate grows. Many students have studied the growth of sugar crystals on a string placed in a supersaturated solution of sugar-water. A
process of great importance today is the growth of near perfect crystals for the semiconductor industry. The computer industry is interested only in those processes which may consistently produce a
well ordered crystalline solid, free of defects or vacancies. Amorphous solids, unstructured groups of atoms, are rejected. The production of crystals are determined by factors such as the quality
and temperature of the seed crystal and the rate of growth.
We begin with a substrate of atoms, the seed crystal, and randomly drop atoms onto the substrate. Three things happen:
1. the atom sticks at that site
2. the atom bounces off the substrate completely
3. the atom may move to a neighboring position on the surface of the substrate
The outcome is determined by the temperature and the potential (chemical binding) energy U of the atom on the surface of the substrate. Boltzmanns factor e^(-U/kbT) determines the probability of
sticking. Qualitatively we would expect the crystals to develop differently at different temperatures, at low temperatures the atoms tend to stick to the substrate and at high temperatures the atoms'
thermal energy allow them to bounce off the substrate.
Figure 1. At location [i,j], there are eight neighbors. The atom at [i,j] would have two types of bonds: a bond with a neighbor in the same row or column is a perpendicular bond, while the bonds with
the four remaining neighbors are called diagonal bonds.
This procedure is accomplished for each of the four perpendicular neighbors of the site, top, right, bottom, and left. (Here, let j=1 be our present position, j=2: top, j=3:right, j=4:bottom, j=
5:1eft). The five values of the PE's are summed to calculate a total potential energy for that site, PETOTAL. Now for each of the five sites a probability factor is calculated by dividing the
individual energy values by the total: PROBj=PEj/PETOTAL. Next the simulation calculates a random number between zero and one, RAND, against which we will compare the probability of current position:
1. if ProB 1 >= RAND we record a 1 at location [i,j] and the "clock" value at time[i,j];
2. else if PROB 1 + PROB2 > RAND let [i,j+l], the top, be our current position and go tbrough the whole process again;
3. else if PROB 1 + PROB2 + PROB3 > RAND let [i+l,j], the right, be our current position and go through the whole process again;
4. else if PROB l+PROB2+PROB3+PROB4>RAND let [i,j-l ], the bottom, be our current position and go through the whole process again;
5. else let [i-1,j], the left, be our current position.
This process gives the atom the ability to jump to different sites and calculate whether the atom might stick at a site which borders the selected site. The procedures exist if:
• ST = 0 and DI = O (there are no neighbors)
• the column chosen is either the first or last colutnn, or the row selected is at the top of the grid definition, the top row.
Whether or not the atom sticks, the programs should increment the 'clock' and go on to find the next random column. The clock for this program will be an interation count through the routine. The
clock will be initialized to zero at the beginning of the program. The time required to fill up the grid with crystals depends on the Boltzmann temperature constant and tbe bonding strengths. For a
75x30 grid a thousand clock cycles should be sufficient. Due to system and time constraints only results from some of the clock cycles will be saved. The user will determine which clock cycles to
save. In order for the user to choose, they should be prompted for:
1. the name of the output file
2. the minimum clock value at which to start saving the data
3. the maximum clock value at which time the algorithm terminates
4. the number of clock cycles skipped to the next data saved, this is done by using the Pascal modulus operator, 'mod'
Main program:
initialize; {call the initialize procedure}
get_parameters; {call the procedure which asks the user for the constant values}
loop: while clock <= maximum time
{get a random column and find the row just above the
monte_carlo(thiscolumn, thisrow); {pass the column and row
of the algorithm}
if: the time checks are correct
output the data in AVS format;
increment clock;
clock = 0;
all the entries for location[i,j] will be zero except for the bottom row, location[i, 1] will
Get parameters:
prompt user for the parameters kbT, diagonal bond strength, perpendicular bond strength, minimum and maximum clock cycles, and the number of clock cycles to skip before saving the data, and the
name of the output file.
Random Column:
get a random integer value between 1 and the maximum column; drop from the top to the bottom to find the row just above the substrate below, this includes the diagonal neghbors, also;
Monte Carlo procedure:
we need the arrays of reals: BP[1..5], PE[1..5], PROB[1..5]
Sticking calculation procedure (i, j, k):
if location [i, j] = 1
BP[k] = PE[k] = 0.0;
calculate the number of diagonal neighbors and
perpendicular neighbors; calculate BP[k], PE[k] and
PETOTAL as discussed earlier;
[it is important to keep in mind that not every location
has eight neighbors]
initialize BP[ ] and PE [ ] to zero;
call sticking (this column, thisrow, neighborlocation);
if there are no neighbors
exit routine;
else if the column is the last or first column or the top row
pass the up (neighbor location = 2), right (nl=3), bottom
(nl = 4), and left (nl = 5) column, row, and "nl" values to
sticking procedure.
calculate a random value, RAND, between 0 and 1;
check the random comparisons as discussed earlier,
updating the crystal information when necessary
otherwise allow the atom to jump to its perpendicular
Output procedure:
for each time iteration saved, there are two output files - an AVS header file and the data file.
With the flexibility of the user prompted parameters we may investigate the various properties of the crystal growth process. Figures 2 & 3 show some of the different crystals which develop under
different conditions.
Figure 2. kT=28, diagonal bond = 100, perpendicular bond =10
Figure 3. kT=5, diagonal bond=100, perpendicular bond=10
SI 1998
This code was developed from a simulation developed by Berkowitz, S.J., and Haase, D.G., and was originally written in BASICA for the IBM PC.
Leslie Southern is the OSC coordinator for the Crystal Growth project. Leslie's office is in 420-3. Please contact Leslie to set up appointment(s) for consultation.
For assistance, write si-contact@osc.edu or call 614-292-0890.
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What's the Division Property of Equality?
Ever wondered what rules you're allowed to follow when you're working with inequalities? Well, one of those rules is called the division property of inequality, and it basically says that if you
divide one side of an inequality by a number, you can divide the other side of the inequality by the same number. However, you have to be very careful about the direction of the inequality! Watch the
tutorial to see how this looks in terms of algebra!
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basic linear algebra questions
April 17th 2009, 07:32 PM
Hikari Clover
basic linear algebra questions
let ABC be a triangle and let p be the midpoint of side AB
let http://i52.photobucket.com/albums/g37/mmmmms/1-7.jpg and http://i52.photobucket.com/albums/g37/mmmmms/2-6.jpg
express http://i52.photobucket.com/albums/g37/mmmmms/5-3.jpg and http://i52.photobucket.com/albums/g37/mmmmms/6-2.jpg in terms of http://i52.photobucket.com/albums/g37/mmmmms/7-1.jpg and http://
i52.photobucket.com/albums/g37/mmmmms/8-1.jpg and http://i52.photobucket.com/albums/g37/mmmmms/9-1.jpg
hence prove that the midpoint of the hypotenuse of a right-angle triangle is equidistant from the three vertices
let p be the plane with cartesian equation 2x+y-3z=9 and let A,B be points with coordinates (1,-2,3) and (-2,1,-1)
l is the line passing through the point A and B
find the coordinates of the point C where the line l intersects the plane p
i have no ideas how to find the intersection between a line and a plane
April 17th 2009, 11:43 PM
let ABC be a triangle and let p be the midpoint of side AB
let http://i52.photobucket.com/albums/g37/mmmmms/1-7.jpg and http://i52.photobucket.com/albums/g37/mmmmms/2-6.jpg
express http://i52.photobucket.com/albums/g37/mmmmms/5-3.jpg and http://i52.photobucket.com/albums/g37/mmmmms/6-2.jpg in terms of http://i52.photobucket.com/albums/g37/mmmmms/7-1.jpg and http://
i52.photobucket.com/albums/g37/mmmmms/8-1.jpg and http://i52.photobucket.com/albums/g37/mmmmms/9-1.jpg
hence prove that the midpoint of the hypotenuse of a right-angle triangle is equidistant from the three vertices
$\overrightarrow{AP}^2 = \frac12\:\overrightarrow{AB}\cdot\frac12\:\overrig htarrow{AB}\cdot = \frac14\:\overrightarrow{AB}\cdot\overrightarrow{A B}$
$\overrightarrow{AB} = \overrightarrow{AC} + \overrightarrow{CB} = -\vec{a} + \vec{b}$
$\overrightarrow{AP}^2 = \frac14\:\left(-\vec{a} + \vec{b})^2\right) = \frac14\:\left(\vec{a}^2 + \vec{b}^2 -2 \vec{a} \cdot \vec{b}\right)$
$\overrightarrow{CP} = \frac12\:\left(\overrightarrow{CA} + \overrightarrow{CB}\right) = \frac12\:\left(\vec{a} + \vec{b}\right)$
let p be the plane with cartesian equation 2x+y-3z=9 and let A,B be points with coordinates (1,-2,3) and (-2,1,-1)
l is the line passing through the point A and B
find the coordinates of the point C where the line l intersects the plane p
i have no ideas how to find the intersection between a line and a plane
AB coordinates are (-3,3,-4)
Therefore one parametric equation of l is :
x = -3t + 1
y = 3t -2
z = -4t +3
C(x,y,z) is on the plane iff 2x+y-3z=9
C is on l iff there exists t such that
x = -3t + 1
y = 3t -2
z = -4t +3
Substitute x,y,z in the Cartesian equation of the plane to get one linear equation. Solve for t and substitute in x,y,z expressions.
April 18th 2009, 04:20 AM
Hikari Clover
$\overrightarrow{AP}^2 = \frac12\:\overrightarrow{AB}\cdot\frac12\:\overrig htarrow{AB}\cdot = \frac14\:\overrightarrow{AB}\cdot\overrightarrow{A B}$
$\overrightarrow{AB} = \overrightarrow{AC} + \overrightarrow{CB} = -\vec{a} + \vec{b}$
$\overrightarrow{AP}^2 = \frac14\:\left(-\vec{a} + \vec{b})^2\right) = \frac14\:\left(\vec{a}^2 + \vec{b}^2 -2 \vec{a} \cdot \vec{b}\right)$
$\overrightarrow{CP} = \frac12\:\left(\overrightarrow{CA} + \overrightarrow{CB}\right) = \frac12\:\left(\vec{a} + \vec{b}\right)$
AB coordinates are (-3,3,-4)
Therefore one parametric equation of l is :
x = -3t + 1
y = 3t -2
z = -4t +3
C(x,y,z) is on the plane iff 2x+y-3z=9
C is on l iff there exists t such that
x = -3t + 1
y = 3t -2
z = -4t +3
Substitute x,y,z in the Cartesian equation of the plane to get one linear equation. Solve for t and substitute in x,y,z expressions.
hey , thx for ur replying(Clapping)
but for the first question,did u forget to put that absolute value sign? or it doesnot matter?
April 18th 2009, 05:58 AM
Are you talking about the modulus ?
$||\overrightarrow{AB}||^2 = \overrightarrow{AB}\cdot\overrightarrow{AB} = \overrightarrow{AB}^2 = AB^2$
April 18th 2009, 06:17 AM
Hikari Clover
oh i got it
thanks so much ^_^
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help with proof
December 1st 2009, 04:13 PM
help with proof
Prove the following:
An accumulation point of a set S is either an interior point of S or a boundary point of S.
This seems very logical to me, I just dont know how to prove it.
I know that an accumulation point may or may not be in S.
If it is in S then it would be an interior and boundary. If it is not in S then it would be a boundary point. Pretty simple I think, but how can I actually prove this?
December 1st 2009, 04:57 PM
Prove the following:
An accumulation point of a set S is either an interior point of S or a boundary point of S.
This seems very logical to me, I just dont know how to prove it.
I know that an accumulation point may or may not be in S.
If it is in S then it would be an interior and boundary. If it is not in S then it would be a boundary point. Pretty simple I think, but how can I actually prove this?
Let $E$ be a set and $\xi$ a limit point of $E$. Then for every neighborhood $N_{\varepsilon}(\xi)$ there contains a point of $E$ different from $E$. Note now that if we consider let $P$ be the
statement "there exists some neighborhood of $\xi$ such that it is contained in $E$" then "Every neigborhood of $\xi$ contains a point of $E^{c}$" is the statement $eg P$. And since we are
working in first order logic it follows that $P\vee eg P$ is a tautology. You get the idea.
December 15th 2009, 07:44 PM
I have been reading this post for awhile and the way you structured the proof does not make any sense to me.
Is there any other way to show this proof?
like with if X is in S or if x is not S sort of version then there exists a deleted neighborhood intersected with S that is nonempty?
I kind of have the jist of what accumulation points, boundary and interior points are but im having trouble forming a proof in my own words.
S is a subset of R.
I know that if x is an accumulation point there exists $N*(x, {\varepsilon} ) \cap$ S =/= $\emptyset$. So by that if x is in the deleted neighborhood and an accumulation point (meaning x is in
S'), then x is not in S.
If x is not in S then x is in R/S.
So if x is in R/S then that means that $N(x , \varepsilon) \cap (R/S)$ =/= $\emptyset.$
So x in the bd of S.
Is this kind of on the right track?
December 15th 2009, 08:25 PM
I have been reading this post for awhile and the way you structured the proof does not make any sense to me.
Is there any other way to show this proof?
like with if X is in S or if x is not S sort of version then there exists a deleted neighborhood intersected with S that is nonempty?
I kind of have the jist of what accumulation points, boundary and interior points are but im having trouble forming a proof in my own words.
S is a subset of R.
I know that if x is an accumulation point there exists $N*(x, {\varepsilon} ) \cap$ S =/= $\emptyset$. So by that if x is in the deleted neighborhood and an accumulation point (meaning x is in
S'), then x is not in S.
If x is not in S then x is in R/S.
So if x is in R/S then that means that $N(x , \varepsilon) \cap (R/S)$ =/= $\emptyset.$
So x in the bd of S.
Is this kind of on the right track?
I was a little lazy. I think if I rephrase my argument more rigorously it will make more sense.
Assume that $\xi\in S'$, then $\forall\varepsilon>0$ there exists some $x\in S,xe\xi$ such that $x\in N_{\varepsilon}(\xi)$. Now if there exists some $\varepsilon'>0$ such that $N_{\varepsilon'}
(\xi)\subseteq S$ then $\xi\in S^{\circ}$. If not, then $\forall\varepsilon>0$ it is true that $N_{\varepsilon}(\xi)subseteq S$ which means there exists some point $y\in S^{c}$ such that $y\in N_
{\varepsilon}(\xi)$. Consequently, every neighborhood of $\xi$ contains a point of $S^c$, but since $\xi\in S'$ it is true that every neighborhood also contains a point of $S$. This tells us that
$\xi\in\partial S$.
$\partial S$-boundary
$S'$-limit points
December 15th 2009, 08:41 PM
I was a little lazy. I think if I rephrase my argument more rigorously it will make more sense.
Assume that $\xi\in S'$, then $\forall\varepsilon>0$ there exists some $x\in S,xe\xi$ such that $x\in N_{\varepsilon}(\xi)$. Now if there exists some $\varepsilon'>0$ such that $N_{\varepsilon'}
(\xi)\subseteq S$ then $\xi\in S^{\circ}$. If not, then $\forall\varepsilon>0$ it is true that $N_{\varepsilon}(\xi)subseteq S$ which means there exists some point $y\in S^{c}$ such that $y\in N_
{\varepsilon}(\xi)$. Consequently, every neighborhood of $\xi$ contains a point of $S^c$, but since $\xi\in S'$ it is true that every neighborhood also contains a point of $S$. This tells us that
$\xi\in\partial S$.
$\partial S$-boundary
$S'$-limit points
Thank you, its kind of late so im really having trouble putting anything of value together but at first glance this makes ALOT more sense..
December 16th 2009, 03:01 AM
I would be inclined to use indirect proof: Assume a point, p, is NOT either an interior point or a boundary point of S. Then p is an interior point of the complemennt of S. Prove that s cannot be
an accumulation point of S.
December 16th 2009, 07:39 AM
the second part of the question says:
Prove the following:
(b)A boundary of a set S is either an accumulation point of S or an isolated point of S.
I tried doing it by contradiction but Im not sure you can do this part that way.
If I assume a point p is not an accumulation point nor an isolated point, then this point could be an interior point of S, or an interior point of R/S?
is that correct?
December 16th 2009, 07:57 AM
Suppose that $x\in\beta(S)$, the boundary.
If $x$ is an isolated point of $S$ we are done.
Suppose that it is not. For all open sets $\mathcal{O}$ if $x \in \mathcal{O} \Rightarrow \quad \left( {\exists y \in \mathcal{O} \cap S\backslash \{ x\} } \right)$.
What does that tell us?
December 16th 2009, 08:11 AM
Suppose that $x\in\beta(S)$, the boundary.
If $x$ is an isolated point of $S$ we are done.
Suppose that it is not. For all open sets $\mathcal{O}$ if $x \in \mathcal{O} \Rightarrow \quad \left( {\exists y \in \mathcal{O} \cap S\backslash \{ x\} } \right)$.
What does that tell us?
x and y are both in that intersection?
December 16th 2009, 08:44 AM
Does it mean that every open set that contains $x$ contains a point of $S$ distinct from $x$?
What is that the definition of?
December 16th 2009, 08:46 AM
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Using Stock Fundamental Analysis to Value a Company
By: InvestorGuide Staff, dated January 25th, 2013
Fundamental analysis is a method used to determine the value of a stock by analyzing the financial data that is ‘fundamental’ to the company. That means that fundamental analysis takes into
consideration only those variables that are directly related to the company itself, such as its earnings, its dividends, and its sales. Fundamental analysis does not look at the overall state of the
market nor does it include behavioral variables in its methodology. It focuses exclusively on the company’s business in order to determine whether or not the stock should be bought or sold.
Critics of fundamental analysis often charge that the practice is either irrelevant or that it is inherently flawed. The first group, made up largely of proponents of the efficient market hypothesis,
say that fundamental analysis is a useless practice since a stock’s price will always already take into account the company’s financial data . In other words, they argue that it is impossible to
learn anything new about a company by analyzing its fundamentals that the market as a whole does not already know, since everyone has access to the same financial information. The other major
argument against fundamental analysis is more practical than theoretical. These critics charge that fundamental analysis is too unscientific a process, and that it’s difficult to get a clear picture
of a company’s value when there are so many qualitative factors such as a company’s management and its competitive landscape.
However, such critics are in the minority. Most individual investors and investment professionals believe that fundamental analysis is useful, either alone or in combination with other techniques. If
you decide that fundamental analysis is the method for you, you’ll find that a company’s financial statements (its income statement, its balance sheet and its cash flow statement) will be
indispensable resources for your analysis . And even if you’re not totally sold on the idea of fundamental analysis, it’s probably a good idea for you to familiarize yourself with some of the
valuation measures it uses since they are often talked about in other types of stock valuation techniques as well.
It is often said that earnings are the “bottom line” when it comes to valuing a company’s stock, and indeed fundamental analysis places much emphasis upon a company’s earnings. Simply put, earnings
are how much profit (or loss) a company has made after subtracting expenses. During a specific period of time, all public companies are required to report their earnings on a quarterly basis through
a 10-Q Report . Earnings are important to investors because they give an indication of the company’s expected dividends and its potential for growth and capital appreciation. That does not
necessarily mean, however, that low or negative earnings always indicate a bad stock; for example, many young companies report negative earnings as they attempt to grow quickly enough to capture a
new market, at which point they’ll be even more profitable than they otherwise might have been. The key is to look at the data underlying a company’s earnings on its financial statements and to use
the following profitability ratios to determine whether or not the stock is a sound investment.
Earnings Per Share
Comparing total net earnings for various companies is usually not a good idea, since net earnings numbers don’t take into account how many shares of stock are outstanding (in other words, they don’t
take into account how many owners you have to divide the earnings among). In order to make earnings comparisons more useful across companies, fundamental analysts instead look at a company’s earnings
per share (EPS). EPS is calculated by taking a company’s net earnings and dividing by the number of outstanding shares of stock the company has. For example, if a company reports $10 million in net
earnings for the previous year and has 5 million shares of stock outstanding, then that company has an EPS of $2 per share. EPS can be calculated for the previous year (“trailing EPS”), for the
current year (“current EPS”), or for the coming year (“forward EPS”). Note that last year’s EPS would be actual, while current year and forward year EPS would be estimates.
P/E Ratio
EPS is a great way to compare earnings across companies, but it doesn’t tell you anything about how the market values the stock. That’s why fundamental analysts use the price-to-earnings ratio, more
commonly known as the P/E ratio, to figure out how much the market is willing to pay for a company’s earnings. You can calculate a stock’s P/E ratio by taking its price per share and dividing by its
EPS. For instance, if a stock is priced at $50 per share and it has an EPS of $5 per share, then it has a P/E ratio of 10. (Or equivalently, you could calculate the P/E ratio by dividing the
company’s total market cap by the company’s total earnings; this would result in the same number.) P/E can be calculated for the previous year (“trailing P/E”), for the current year (“current P/E”),
or for the coming year (“forward P/E”). The higher the P/E, the more the market is willing to pay for each dollar of annual earnings. Note that last year’s P/E would be actual, while current year and
forward year P/E would be estimates, but in each case, the “P” in the equation is the current price. Companies that are not currently profitable (that is, ones which have negative earnings) don’t
have a P/E ratio at all. For those companies you may want to calculate the price-to-sales ratio (PSR) instead.
So is a stock with a high P/E ratio always overvalued? Not necessarily. The stock could have a high P/E ratio because investors are convinced that it will have strong earnings growth in the future
and so they bid up the stock’s price now. Fortunately, there is another ratio that you can use that takes into consideration a stock’s projected earnings growth: it’s called the PEG. PEG is
calculated by taking a stock’s P/E ratio and dividing by its expected percentage earnings growth for the next year. So, a stock with a P/E ratio of 40 that is expected to grow its earnings by 20% the
next year would have a PEG of 2. In general, the lower the PEG, the better the value, because you would be paying less for each unit of earnings growth.
Dividend Yield
The dividend yield measures what percentage return a company pays out to its shareholders in the form of dividends . It is calculated by taking the amount of dividends paid per share over the course
of a year and dividing by the stock’s price. For example, if a stock pays out $2 in dividends over the course of a year and trades at $40, then it has a dividend yield of 5%. Mature, well-established
companies tend to have higher dividend yields, while young, growth-oriented companies tend to have lower ones, and most small growing companies don’t have a dividend yield at all because they don’t
pay out dividends.
Dividend Payout Ratio
The dividend payout ratio shows what percentage of a company’s earnings it is paying out to investors in the form of dividends. It is calculated by taking the company’s annual dividends per share and
dividing by its annual earnings per share (EPS). So, if a company pays out $1 per share annually in dividends and it has an EPS of $2 for the year, then that company has a dividend payout ratio of
50%; in other words, the company paid out 50% of its earnings in dividends. Companies that distribute dividends typically use about 25% to 50% of their earnings for dividend payments. The higher the
payout ratio, the less confidence the company has that it would’ve been able to find better uses for the money it earned. This is not necessarily either good or bad; companies that are still growing
will tend to have lower dividend payout ratios than very large companies, because they are more likely to have other productive uses for the earnings.
Book Value
The book value of a company is the company’s net worth, as measured by its total assets minus its total liabilities. This is how much the company would have left over in assets if it went out of
business immediately. Since companies are usually expected to grow and generate more profits in the future, most companies end up being worth far more in the marketplace than their book value would
suggest. For this reason, book value is of more interest to value investors than growth investors. In order to compare book values across companies, you should use book value per share, which is
simply the company’s last quarterly book value divided by the number of shares of stock it has outstanding.
Price / Book
A company’s price-to-book ratio (P/B ratio) is determined by taking the company’s per share stock price and dividing by the company’s book value per share. For instance, if a company currently trades
at $100 and has a book value per share of $5, then that company has a P/B ratio of 20. The higher the ratio, the higher the premium the market is willing to pay for the company above its hard assets.
Price-to-book ratio is of more interest to value investors than growth investors.
Price / Sales Ratio
As with earnings and book value, you can find out how much the market is valuing a company by comparing the company’s price to its annual sales. This measure is known as the price-to-sales ratio (P/S
or PSR). You can calculate the P/S by taking the stock’s current price and dividing by the company’s total sales per share for the past year (or equivalently, by dividing the entire company’s market
cap by its total sales). That means that a company whose stock trades at $1 per share and which had $2 per share in sales last year will have a P/S of 0.5. Low P/S ratios (below one) are usually
thought to be the better investment since their sales are priced cheaply. However, P/S, like P/E ratios and P/B ratios, are numbers that are subject to much interpretation and debate. Sales obviously
don’t reveal the whole picture: a company could be selling dollar bills for 90 cents each, and have huge sales but be terribly unprofitable. Because of the limitations, P/S ratios are usually used
only for unprofitable companies, since such companies don’t have a P/E ratio .
Return on Equity
Return on equity (ROE) shows you how much profit a company generates in comparison to its book value . The ratio is calculated by taking a company’s after-tax income (after preferred stock dividends
but before common stock dividends) and dividing by its book value (which is equal to its assets minus its liabilities). It is used as a general indication of the company’s efficiency; in other words,
how much profit it is able to generate given the resources provided by its stockholders. Investors usually look for companies with ROEs that are high and growing.
Other relevant articles you may like
One Response to “Using Stock Fundamental Analysis to Value a Company”
1. Neat summary of stock valuation ratios and measures. Thanks.
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Basic Rules of Algebra
There are basic properties in math that apply to all real numbers. When working with variables in algebra, these properties still apply. We will apply most of the following properties to solve
various Algebraic problems.
Algebraic Properties
Let a, b, and c be real numbers, variables, or algebraic expressions.
Commutative Property of Addition
We can add numbers in any order.
Commmutative Property of Multiplication
We can also multiply numbers in any order.
Associative Property of Addition
We can group numbers in a sum any way we want and get the same answer.
Associative Property of Multiplication
We can group numbers in a product any way we want and get the same answer.
Distributive Property
When we are adding and multiplying with a parenthesis, we can distribute the multiplication through the addition.
For an in depth discussion, see Distributive Property
Additive Identity Property
If we add 0 to any number, we will end up with the same number.
Multiplicative Identity Property
If we multiply 1 to any number, we will end up with the same number.
Additive Inverse Property
If we adda number by the opposite of itself, we will end up with 0.
Multiplicative Inverse Property
If we multiply a number by its reciprocal, we will end up with 1.
Keep in mind that subtraction is also considered addition, but with a negative number. Similarly, divison can be thought of as inverse multiplication, but with a restriction that the denominator
cannot be equal to 0.
Properties of Negation
We must be careful not to make arithmetic mistakes when dealing with negative signs and subtraction.
Properties of Equality
Add c to each side
Multiply both sides by c
Subtract c from both sides
Divide both sides by c
Properties of Zero
0 added or subtracted to anything equals itself
0 multiplied by anything equals 0
0 divided by anything equals 0
We cannot divide by 0
Zero Product Property
If the product of two or more things equals 0, at least one of the values must be 0
Properties and Operations of Fractions
Let a, b, c and d be real numbers, variables, or algebraic expressions such that b and d do not equal 0.
Equivalent Fractions
cross multiply
Rules of Signs
the negative can go anywhere in the fraction and two negatives equal a positive
Generate Equivalent Fractions
multiplying the top and bottom by the same thing keeps the fraction the same value
Add/Subtract with Like Denominators
if the denominators are the same, add or subtract the top of the fraction
Add/Subtract with Unlike Denominators
find a common denominator
Multiply Fractions
top times the top and bottom times the bottom
Divide Fractions
when dividing two fracitons, multiply the divisor by the reciprocal
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JEE Main
The syllabus contains two Sections - A and B. Section - A pertains to the Theory Part having 80% weightage, while Section - B contains Practical Component (Experimental Skills) having 20% weightage.
SECTION – A
Physics, technology and society, S I units, Fundamental and derived units. Least count, accuracy and precision of measuring instruments, Errors in measurement, Dimensions of Physical quantities,
dimensional analysis and its applications.
UNIT 2: KINEMATICS
Frame of reference. Motion in a straight line: Position-time graph, speed and velocity. Uniform and non-uniform motion, average speed and instantaneous velocity Uniformly accelerated motion,
velocity-time, position-time graphs, relations for uniformly accelerated motion. Scalars and Vectors, Vector addition and Subtraction, Zero Vector, Scalar and Vector products, Unit Vector, Resolution
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UNIT 3: LAWS OF MOTION
Force and Inertia, Newton’s First Law of motion; Momentum, Newton’s Second Law of motion; Impulse; Newton’s Third Law of motion. Law of conservation of linear momentum and its applications,
Equilibrium of concurrent forces. Static and Kinetic friction, laws of friction, rolling friction. Dynamics of uniform circular motion: Centripetal force and its applications.
UNIT 4: WORK, ENERGY AND POWER
Work done by a constant force and a variable force; kinetic and potential energies, workenergy theorem, power. Potential energy of a spring, conservation of mechanical energy, conservative and
nonconservative forces; Elastic and inelastic collisions in one and two dimensions.
Centre of mass of a two-particle system, Centre of mass of a rigid body; Basic concepts of rotational motion; moment of a force, torque, angular momentum, conservation of angular momentum and its
applications; moment of inertia, radius of gyration. Values of moments of inertia for simple geometrical objects, parallel and perpendicular axes theorems and their applications. Rigid body rotation,
equations of rotational motion.
UNIT 6: GRAVITATION
The universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Kepler’s laws of planetary motion. Gravitational potential energy; gravitational potential.
Escape velocity. Orbital velocity of a satellite. Geo-stationary satellites.
Elastic behaviour, Stress-strain relationship, Hooke’s Law, Young’s modulus, bulk modulus, modulus of rigidity. Pressure due to a fluid column; Pascal’s law and its applications. Viscosity, Stokes’
law, terminal velocity, streamline and turbulent flow, Reynolds number. Bernoulli’s principle and its applications. Surface energy and surface tension, angle of contact, application of surface
tension - drops, bubbles and capillary rise. Heat, temperature, thermal expansion; specific heat capacity, calorimetry; change of state, latent heat. Heat transfer-conduction, convection and
radiation, Newton’s law of cooling.
Thermal equilibrium, zeroth law of thermodynamics, concept of temperature. Heat, work and internal energy. First law of thermodynamics. Second law of thermodynamics: reversible and irreversible
processes. Carnot engine and its efficiency.
UNIT 9: KINETIC THEORY OF GASES
Equation of state of a perfect gas, work doneon compressing a gas.Kinetic theory of gases - assumptions, concept of pressure. Kinetic energy and temperature: rms speed of gas molecules; Degrees of
freedom, Law of equipartition of energy,applications to specific heat capacities of gases; Mean free path, Avogadro’s number.
UNIT 10: OSCILLATIONS AND WAVES
Periodic motion - period, frequency, displacement as a function of time. Periodic functions. Simple harmonic motion (S.H.M.) and its equation; phase; oscillations of a spring -restoring force and
force constant; energy in S.H.M. - kinetic and potential energies; Simple pendulum - derivation of expression for its time period; Free, forced and damped oscillations, resonance. Wave motion.
Longitudinal and transverse waves, speed of a wave. Displacement relation for a progressive wave. Principle of superposition of waves, reflection of waves, Standing waves in strings and organ pipes,
fundamental mode and harmonics, Beats, Doppler effect in sound
UNIT 11: ELECTROSTATICS
Electric charges: Conservation of charge, Coulomb’s law-forces between two point charges, forces between multiple charges; superposition principle and continuous charge distribution. Electric field:
Electric field due to a point charge, Electric field lines, Electric dipole, Electric field due to a dipole, Torque on a dipole in a uniform electric field. Electric flux, Gauss’s law and its
applications to find field due to infinitely long uniformly charged straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Electric potential and its
calculation for a point charge, electric dipole and system of charges; Equipotential surfaces, Electrical potential energy of a system of two point charges in an electrostatic field. Conductors and
insulators, Dielectrics and electric polarization, capacitor, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between
the plates, Energy stored in a capacitor.
UNIT 12: CURRRENT ELECTRICITY
Electric current, Drift velocity, Ohm’s law, Electrical resistance, Resistances of different materials, V-I characteristics of Ohmic and nonohmic conductors, Electrical energy and power, Electrical
resistivity, Colour code for resistors; Series and parallel combinations of resistors; Temperature dependence of resistance. Electric Cell and its Internal resistance, potential difference and emf of
a cell, combination of cells in series and in parallel. Kirchhoff’s laws and their applications. Wheatstone bridge, Metre bridge. Potentiometer - principle and its applications.
Biot - Savart law and its application to current carrying circular loop. Ampere’s law and its applications to infinitely long current carrying straight wire and solenoid. Force on a moving charge in
uniform magnetic and electric fields. Cyclotron.
Force on a current-carrying conductor in a uniform magnetic field. Force between two parallel current-carrying conductors-definition of ampere. Torque experienced by a current loop in uniform
magnetic field; Moving coil galvanometer, its current sensitivity and conversion to ammeter and voltmeter. Current loop as a magnetic dipole and its magnetic dipole moment. Bar magnet as an
equivalent solenoid, magnetic field lines; Earth’s magnetic field and magnetic elements. Para-, dia- and ferro- magnetic substances. Magnetic susceptibility and permeability, Hysteresis,
Electromagnets and permanent magnets.
Electromagnetic induction; Faraday’s law, induced emf and current; Lenz’s Law, Eddy currents. Self and mutual inductance. Alternating currents, peak and rms value of alternating current/ voltage;
reactance and impedance; LCR series circuit, resonance; Quality factor, power in AC circuits, wattless current. AC generator and transformer.
Electromagnetic waves and their characteristics. Transverse nature of electromagnetic waves. Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, Xrays, gamma rays).
Applications of e.m. waves.
UNIT 16: OPTICS
Reflection and refraction of light at plane and spherical surfaces, mirror formula, Total internal reflection and its applications, Deviation and Dispersion of light by a prism, Lens Formula,
Magnification, Power of a Lens, Combination of thin lenses in contact, Microscope and Astronomical Telescope (reflecting and refracting) and their magnifyingpowers. Wave optics: wavefront and
Huygens’ principle, Laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width. Diffraction due to a single slit, width
of central maximum. Resolving power of microscopes and astronomical telescopes, Polarisation, plane polarized light; Brewster’s law, uses of plane polarized light and Polaroids.
UNIT 17: DUAL NATURE OF MATTER ANDRADIATION
Dual nature of radiation. Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation; particle nature of light. Matter waves-wave nature of particle, de Broglie
relation. Davisson-Germer experiment.
UNIT 18: ATOMS AND NUCLEI
Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model, energy levels, hydrogen spectrum. Composition and size of nucleus, atomic masses, isotopes, isobars; isotones.
Radioactivity-alpha, beta and gamma particles/rays and their properties; radioactive decay law. Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number,
nuclear fission and fusion.
UNIT 19: ELECTRONIC DEVICES
Semiconductors; semiconductor diode: I-V characteristics in forward and reverse bias; diode as a rectifier; I-V characteristics of LED, photodiode, solar cell and Zener diode; Zener diode as a
voltage regulator. Junction transistor, transistor action, characteristics of a transistor; transistor as an amplifier (common emitter configuration) and oscillator. Logic gates (OR, AND, NOT, NAND
and NOR). Transistor as a switch.
Propagation of electromagnetic waves in the atmosphere; Sky and space wave propagation, Need for modulation, Amplitude and Frequency Modulation, Bandwidth of signals, Bandwidth of Transmission
medium, Basic Elements of a Communication System (Block Diagram only).
UNIT 21: EXPERIMENTAL SKILLS
Familiarity with the basic approach and observations of the experiments and activities:
1. Vernier callipers-its use to measure internal and external diameter and depth of a vessel.
2. Screw gauge-its use to determine thickness/diameter of thin sheet/wire.
3. Simple Pendulum-dissipation of energy by plotting a graph between square of amplitude and time.
4. Metre Scale - mass of a given object by principle of moments.
5. Young’s modulus of elasticity of the material of a metallic wire.
6. Surface tension of water by capillary rise and effect of detergents.
7. Co-efficient of Viscosity of a given viscous liquid by measuring terminal velocity of a given spherical body.
8. Plotting a cooling curve for the relationship between the temperature of a hot body and time.
9. Speed of sound in air at room temperature using a resonance tube.
10. Specific heat capacity of a given (i) solid and (ii) liquid by method of mixtures.
11. Resistivity of the material of a given wire using metre bridge.
12. Resistance of a given wire using Ohm’s law.
13. Potentiometer – (i) Comparison of emf of two primary cells. (ii) Determination of internal resistance of a cell.
14. Resistance and figure of merit of a galvanometer by half deflection method.
15. Focal length of: (i) Convex mirror (ii) Concave mirror, and (iii) Convex lens using parallax method.
16. Plot of angle of deviation vs angle of incidence for a triangular prism.
17. Refractive index of a glass slab using a travelling microscope.
18. Characteristic curves of a p-n junction diode in forward and reverse bias.
19. Characteristic curves of a Zener diode and finding reverse break down voltage.
20. Characteristic curves of a transistor and finding current gain and voltage gain.
21. Identification of Diode, LED, Transistor, IC, Resistor, Capacitor from mixed collection of such items.
22. Using multimeter to:
(i) Identify base of a transistor
(ii) Distinguish between npn and pnp type transistor
(iii) See the unidirectional flow of current in case of a diode and an LED.
(iv) Check the correctness or otherwise of a given electronic component (diode, transistor or IC).
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Agreeing with Bill James
In 1988, the Bill James Abstract included
"A Bill James Primer"
, with 15 statements expressing what he deemed to be useful knowledge. On that list was:
2. Talent in baseball is not normally distributed. It is a pyramid. For every player who is 10 percent above the average player, there are probably twenty players who are 10 percent below average. I
agree. (Others don't; for further discussion also see here.) But what is this thing called "talent"? Talent is a combination of a high level of skill and sustained, consistent performance. Skill in
baseball is measured through metrics such as ERA (earned run average) and OPS (on-base average plus slugging percentage) -- measures that turn counting stats into an efficiency or rate measure. While
this type of measure is important, they fail to account for the fact that some players have lengthy careers, while other players have a very short MLB career. Teams will sign long-term contracts with
aging superstars because the player's skill is still above average, even though they may have diminished with age. In short, career length becomes a valid proxy for talent.
The charts below plot the number of pitchers over the period 1996-2009, by both the number of games played (which favours the relief pitchers) and innings pitched (which favours the starters). During
this period a total of 2,134 individuals pitched in MLB -- but the chart shows that very few of them stuck around for any length of time. At the head of the "games" list at 898 is the still-active
Mariano Rivera, while the pitcher with the most innings over this period was Greg Maddux (2887.67 innings; and Maddux threw more than 2,100 innings before 1996, as well). These two individuals, and
other Hall of Fame calibre pitchers, are out at the far right of the long tail. Close to the origin at the left are pitchers whose entire career lasted but 1/3 of an inning -- a single out. Figure 1:
Number of Pitchers, by Career Innings Pitched (1996-2009) Figure 2: Number of Pitchers, by Career Games (1996-2009)
But what of the average skill level of those pitchers? Pitchers who get a small amount of MLB experience (fewer than 27 innings) have a higher ERA than those who get more opportunities to pitch. This
group -- 27% of all MLB pitchers -- recorded an average ERA of 8.08, compared to 5.15 for the 42% who pitched between 27 to 269 innings, and 4.45 for the 27% who threw between 270 and 1349 innings.
The elite, those who pitched 1350 innings and above, recorded the lowest ERA of all, 4.17.
In spite of the wide variance in the ERAs of the coffee drinkers, the differences in the mean scores are statistically significant.
Figure 3: MLB Pitchers, average ERA, by number of innings pitched (1996-2009)
In summary: there is an abundance of players who are less talented than the major league average, while at the same time the number of above-average talents is low. The distribution, at the major
league level, is not normal. Just like Bill James said 22 years ago.
No comments:
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Golden Ratio Dividers
This allows anyone to build a golden ratio divider of any size at all.
The method is independent of measurement units (inches, mm, cubits, ad
The attached drawing shows the design layout. Note that all angles are
either 90 or 45 degrees (It's pretty obvious which are which).
Wherever two lines meet, there's a pivot.
Decide on the longest measurement you want the dividers to handle.
Let's call it L. Either multiply L by 0.618 or divide it by 1.618. The
result will be the span between the left point and the "middle" point.
Let's call that Dmajor. Then the span between the "middle" point and
the right point will be Sminor = L - Smajor.
The only other piece of information needed is that the length of a
side of a square is (roughly) 0.707 times the length of a diagonal.
One of the things that ocurred to me as I made the drawing was that
there's no limit to the number of "middle" arms this thing can have,
and that even with just one arm, you can set up any ratio you want
(eg: divide any span into thirds).
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
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La Mirada Calculus Tutor
Find a La Mirada Calculus Tutor
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Cheyney Algebra 2 Tutor
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Ordnance Survey
By Steve Pratchett
• How do we measure and describe gradients on maps and in the field?
• How does a map indicate gradients?
• What is the relationship between gradient and land use?
One arrow indicates a gradient of 1 in 7 to 1 in 5
Two arrows indicate a gradient of 1 in 5 or steeper
Modelling the locality
To develop these ideas, children constructed a 3-D model from the Ordnance Survey map. A section of the map was enlarged in colour (to make it easier to read the contour lines). I then traced over
the contour lines with a dark brown pen so that black-and-white copies could be made. The children took the copies and cut around each contour line. They glued their cut-out onto a piece of 5 mm
Styrofoam® mounting board (a much cheaper option is to use corrugated cardboard from boxes). The boards gave a convenient arbitrary unit for the interval between each contour line. Although the
children were using 5 mm board, the original map had been enlarged by 400%, so distortion was less than x2. The children then cut around the Styrofoam mounted contour lines using keyhole saws.
When the children viewed the model from the side they, in effect, experienced a cross section of the landscape. The children deduced that the number of arrows indicated the steepness of the hill.
A field trip with map and model
The ideal situation is to take the children on the field trip to a steep hill. When I did this the benefits were clear. On the outward journey, one group of children navigated with the map and the
other with the model. They were delighted to find the model echoing even the slightest change in the incline of the road. Their fingers became the car, kinaesthetically tracing the route and
mimicking its movement as it ascended and descended. This exercise also helped with map orientation skills, as the model had to be turned to face the way we were going.
Where a field trip is impractical, good discussion results can be obtained by using photographs.
Relating gradient to land use
The children were encouraged to deduce why areas of mixed woodland appeared in particular locations on their Ordnance Survey map of the Bere Peninsula. These tended to occur along the sides of stream
and river valleys. The rest of the peninsula is covered by farmland belonging to numerous dairy and arable farms, such as Hole & Well Farm.
3D Model
The children offered several explanations –
'The trees grow near the rivers because they get more water.'
'The trees are planted to stop the valleys eroding.'
'The contour lines are close together so it's steep ...you can't farm the fields, it's too steep here.'
The latter response from one observant child stimulated a flurry of discussion as the children began to check all the patches of woodland on the map and model to see if they only occurred where the
contour lines were closely packed. This proved to be the case. To reinforce this observation, the children were given a tiny toy tractor and tried driving it on different parts of the
three-dimensional model of the peninsula. On the steeply wooded slopes, the tractor rolled over into the river or stream!
This led to a discussion of why farmers did not plough steeply sloping land for arable or cereal production and the identification of more such areas on the local map. Some humorous comments were
also made about the problems dairy cows would have 'rolling down the hill!'
Try the following suggestions on modelling gradients as ratios and gradients as percentages. Lots of lovely links with numeracy!
*Styrofoam is a registered trade mark of the Dow Chemical company
Modelling gradients as percentages using metre rulers and elastic
To develop children's skills in measuring and expressing gradients as percentages, metre rulers were used. A metre ruler laid flat on the ground gives a convenient horizontal distance of 100 units
(cms) and another ruler held upright at one end gives a selection of the same units for vertical climb. A length of shirring elastic can be used to link the two rulers for example, a triangle formed
in this way with a 100 cm base and 20 cm height has a 20% gradient. This means that a toy car would climb 20 cms vertically for every 100 cms it travels horizontally. 20 units out of every 100 = 20%.
Using the same triangle, it is also possible to translate percentages into ratios and vice versa. If the child travels 5 cms along the horizontal ruler with a toy car and then measures up vertically
from this point to the elastic slope, the height will be 1 cm. This is a 1: 5 triangle. These ratios can be multiplied e.g.:
8 cms along .......................... 1.6 cms up
12 cms along .......................... 2.4 cms up
16 cms along .......................... 3.2 cms up
100 cms along .......................... 20 cms up (20%)
Extension of work on gradients - understanding the shape of the ground
An extension to work on gradients and their corresponding contour line patterns is to model convex and concave slopes in clay. The children slice the models into horizontal layers at regular
intervals and then draw around each slice to create 'contour lines'. The models can be directly compared to the configuration of contour lines they generate, highlighting the relationship between the
change in gradients and the change in spacing between the contours. This work will lay a foundation for using contours to understand the shape of the ground when map-reading.
With thanks to Kim Wilde, Headteacher, at Bere Alston Primary School.
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Probabilty problem
October 21st 2010, 06:14 PM #1
Oct 2010
Probabilty problem
Hello forum :]
I am in need of some help on this probability problem.
The probability of being born on a Friday the 13th is 1/214. Assume the days of people's birth are independent of each other.
(a) You meet two new friends; what is the probability that both new friends were born on a Friday the 13th?
I got this one, just use the multiplication rule so: (1/214)(1/214) = 1/45796
(b) You meet 31 new friends; what is the probability that at least one of these new friends was born on a Friday the 13th?
I cant figure this out. If they were disjoint then I could just use the addition rule, but they aren't so I am very confused.
Any help will be greatly appreciated.
Hello, Phodot!
The probability of being born on a Friday the 13th is 1/214.
Assume the days of people's birth are independent of each other.
(a) You meet two new friends. What is the probability that
both new friends were born on a Friday the 13th?
I got this one, just use the multiplication rule so: $(\frac{1}{214})(\frac{1}{214}) \:=\: \frac{1}{45,\!796}$ . Right!
(b) You meet 31 new friends; what is the probability that
at least one of these new friends was born on a Friday the 13th?
The opposite of "at least one" is "none."
The probability that none were borh on Friday the 13th is: . $(\frac{213}{214})^{31}$
Therefore: . $P(\text{at least one Friday the 13th}) \;=\;1 - \left(\frac{213}{214}\right)^{31} \;\approx\;13.5\%$
Thank you very much :]
October 21st 2010, 08:42 PM #2
Super Member
May 2006
Lexington, MA (USA)
October 21st 2010, 09:04 PM #3
Oct 2010
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13th July 2012: Scott Aaronson (MIT)
Title: Quantum Computing and the Limits of the Efficiently Computable
By Scott Aaronson (Associate Professor of Electrical Engineering and Computer Science at MIT)
Abstract: I'll discuss what can and can't be feasibly computed according to physical law. I'll argue that this is a fundamental question, not only for mathematics and computer science, but also for
physics; and that the infeasibility of certain computational problems (such as NP-complete problems) could plausibly be taken as a physical principle, analogous to the Second Law or the impossibility
of superluminal signalling. I'll first explain the basics of computational complexity, including the infamous P versus NP problem and the Extended Church-Turing Thesis. Then I'll discuss quantum
computers: what they are, whether they can be scalably built, and what's known today about their capabilities and limitations. Lastly, I'll touch on speculative models of computation that would go
even beyond quantum computers, using (for example) closed timelike curves or nonlinearities in the Schrodinger equation. I'll emphasize that, even if "intractable" computations occur in a particular
description of a physical system, what really matters is whether those computations have observable consequences.
Bio: Scott Aaronson is an Associate Professor of Electrical Engineering and Computer Science at Massachusetts Institute of Technology (MIT). He received his PhD in computer science from University of
California, Berkeley and did postdocs at the Institute for Advanced Study and the University of Waterloo. Scott's research interests center around fundamental limits on what can efficiently be
computed in the physical world. This has entailed studying quantum computing, the most powerful model of computation we have based on known physical theory. He also writes a popular blog
(www.scottaaronson.com/blog), and is the creator of the Complexity Zoo (www.complexityzoo.com), an online encyclopedia of computational complexity theory. He is the recipient of NSF's Alan T.
Waterman Award for 2012.
This article was published on Nov 7, 2012
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The Black Vault Message Forums
at1with0 wrote:Infinity might be a misnomer.
We keep arguing over the same concepts every couple of years.
http://en.wikipedia.org/wiki/Infinity#A ... e_sciences
From the perspective of cognitive scientists George Lakoff, concepts of infinity in mathematics and the sciences are metaphors, based on what they term the Basic Metaphor of Infinity (BMI),
namely the ever-increasing sequence <1,2,3,...>.
Infinity is a metaphor and can be manipulated in various ways mathematically.
Indefinite succession?
"I can conceive of nothing in religion, science, or philosophy, that is anything more than the proper thing to wear, for a while." ~ Charles Fort
The only reason there is a set of natural numbers, i.e. the only reason there is an infinite set, is that it is assumed (the axiom of infinity). AFAIK there is no way to prove that there is a set N
such that x is in N iff x is a natural number.
"it is easy to grow crazy"
the definitional parameters of the set concept are not well formed.
What is truth? Baby, don't hurt me, don't hurt me... no more.
"I can conceive of nothing in religion, science, or philosophy, that is anything more than the proper thing to wear, for a while." ~ Charles Fort
Truth is the state of the status quo encapsulating both the objective and the subjective eye of the beholder.
"I can conceive of nothing in religion, science, or philosophy, that is anything more than the proper thing to wear, for a while." ~ Charles Fort
Individual self stands in contrast to the continuum of all possible selves.
Hopefully At1with0 will be able to figure out how to get his rich self from the Richie Rich universe to cross the impassible gulf between universes into this universe, in order to give him his
desperately sought after prize.
khanster wrote:Individual self stands in contrast to the continuum of all possible selves.
Hopefully At1with0 will be able to figure out how to get his rich self from the Richie Rich universe to cross the impassible gulf between universes into this universe, in order to give him his
desperately sought after prize.
The pot of gold argument just shows that not every conceivable event occurs somewhere in the multiverse.
"it is easy to grow crazy"
at1with0 wrote:
khanster wrote:Individual self stands in contrast to the continuum of all possible selves.
Hopefully At1with0 will be able to figure out how to get his rich self from the Richie Rich universe to cross the impassible gulf between universes into this universe, in order to give him
his desperately sought after prize.
The pot of gold argument just shows that not every conceivable event occurs somewhere in the multiverse.
Actually you mentioned the pot of gold argument just off the top of your head in an insincere fashion. You never really seriously meditated on the pot of gold nor did you ever work on your powers of
concentration, so why should you stand out amongst the infinite number of other at1with0's?
What makes you so special and deserving of that pot of gold?
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On multidimensional curves with Hilbert property
- NIC Series , 2002
"... this paper a parallelisable and cheap method based on space-filling curves is proposed. The partitioning is embedded into the parallel solution algorithm using multilevel iterative solvers and
adaptive grid refinement. Numerical experiments on two massively parallel computers prove the efficienc ..."
Cited by 8 (0 self)
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this paper a parallelisable and cheap method based on space-filling curves is proposed. The partitioning is embedded into the parallel solution algorithm using multilevel iterative solvers and
adaptive grid refinement. Numerical experiments on two massively parallel computers prove the efficiency of this approach
, 2007
"... Abstract. Let X = S G where G is a countable group and S is a finite set. A cellular automaton (CA) is an endomorphism T: X → X (continuous, commuting with the action of G). Shereshevsky [14]
proved that for G = Z d with d> 1 no CA can be forward expansive, raising the following conjecture: For G = ..."
Cited by 3 (1 self)
Add to MetaCart
Abstract. Let X = S G where G is a countable group and S is a finite set. A cellular automaton (CA) is an endomorphism T: X → X (continuous, commuting with the action of G). Shereshevsky [14] proved
that for G = Z d with d> 1 no CA can be forward expansive, raising the following conjecture: For G = Z d, d> 1 the topological entropy of any CA is either zero or infinite. Morris and Ward [11],
proved this for linear CA’s, leaving the original conjecture open. We show that this conjecture is false, proving that for any d there exist a d-dimensional CA with finite, nonzero topological
entropy. We also discuss a measure-theoretic counterpart of this question for measure-preserving CA’s. 1.
"... Two-dimensional R-trees are a class of spatial index structures in which objects are arranged to enable fast window queries: report all objects that intersect a given query window. One of the
most successful methods of arranging the objects in the index structure is based on sorting the objects acco ..."
Cited by 3 (2 self)
Add to MetaCart
Two-dimensional R-trees are a class of spatial index structures in which objects are arranged to enable fast window queries: report all objects that intersect a given query window. One of the most
successful methods of arranging the objects in the index structure is based on sorting the objects according to the positions of their centres along a two-dimensional Hilbert spacefilling curve.
Alternatively one may use the coordinates of the objects ’ bounding boxes to represent each object by a four-dimensional point, and sort these points along a four-dimensional Hilbert-type curve. In
experiments by Kamel and Faloutsos and by Arge et al. the first solution consistently outperformed the latter when applied to point data, while the latter solution clearly outperformed the first on
certain artificial rectangle data. These authors did not specify which four-dimensional Hilbert-type curve was used; many exist. In this paper we show that the results of the previous papers can be
explained by the choice of the fourdimensional Hilbert-type curve that was used and by the way it was rotated in four-dimensional space. By selecting a curve that has certain properties and choosing
the right rotation one can combine the strengths of the two-dimensional and the four-dimensional approach into one, while avoiding their apparent weaknesses. The effectiveness of our approach is
demonstrated with experiments on various data sets. For real data taken from VLSI design, our new curve yields R-trees with query times that are better than those of R-trees that were obtained with
previously used curves. 1
- in: ESA
"... Space-filling curves can be used to organise points in the plane into bounding-box hierarchies (such as R-trees). We develop measures of the bounding-box quality of space-filling curves that
express how effective different space-filling curves are for this purpose. We give general lower bounds on th ..."
Cited by 3 (2 self)
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Space-filling curves can be used to organise points in the plane into bounding-box hierarchies (such as R-trees). We develop measures of the bounding-box quality of space-filling curves that express
how effective different space-filling curves are for this purpose. We give general lower bounds on the bounding-box quality measures and on locality according to Gotsman and Lindenbaum for a large
class of space-filling curves. We describe a generic algorithm to approximate these and similar quality measures for any given curve. Using our algorithm we find good approximations of the locality
and the bounding-box quality of several known and new space-filling curves. Surprisingly, some curves with relatively bad locality by Gotsman and Lindenbaum’s measure, have good bounding-box quality,
while the curve with the best-known locality has relatively bad bounding-box quality. 1
"... A discrete space-filling curve provides a linear traversal/indexing of a multi-dimensional grid space. This paper presents an application of random walk to the study of inter-clustering of
space-filling curves and an analytical study on the inter-clustering performances of 2-dimensional Hilbert and ..."
Cited by 1 (1 self)
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A discrete space-filling curve provides a linear traversal/indexing of a multi-dimensional grid space. This paper presents an application of random walk to the study of inter-clustering of
space-filling curves and an analytical study on the inter-clustering performances of 2-dimensional Hilbert and z-order curve families. Two underlying measures are employed: the mean inter-cluster
distance over all inter-cluster gaps and the mean total inter-cluster distance over all subgrids. We show how approximating the mean inter-cluster distance statistics of continuous multi-dimensional
space-filling curves fits into the formalism of random walk, and derive the exact formulas for the two statistics for both curve families. The excellent agreement in the approximate and true mean
inter-cluster distance statistics suggests that the random walk may furnish an effective model to develop approximations to clustering and locality statistics for space-filling curves. Based upon the
analytical results, the asymptotic comparisons indicate that z-order curve family performs better than Hilbert curve family with respect to both statistics.
"... The geometric structural complexity of spatial objects does not render an intuitive distance metric on the data space that measures spatial proximity. However, such a metric provides a formal
basis for analytical work in transformation-based multidimensional spatial access methods, including localit ..."
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The geometric structural complexity of spatial objects does not render an intuitive distance metric on the data space that measures spatial proximity. However, such a metric provides a formal basis
for analytical work in transformation-based multidimensional spatial access methods, including locality preservation of the underlying transformation and distance-based spatial queries. We study the
Hausdorff distance metric on the space of multidimensional polytopes, and prove a tight relationship between the metric on the original space of k-dimensional hyperrectangles and the standard
p-normed metric on the transform space of 2kdimensional points under the corner transformation, which justifies the effectiveness of the transformationbased technique in preserving spatial locality.
Keywords: databases, multidimensional spatial access methods, corner transformation, locality
, 2007
"... Abstract. Let X = S G where G is a countable group and S is a finite set. A cellular automaton (CA) is an endomorphism T: X → X (continuous, commuting with the action of G). Shereshevsky [13]
proved that for G = Z d with d> 1 no CA can be forward expansive, raising the following conjecture: For G = ..."
Add to MetaCart
Abstract. Let X = S G where G is a countable group and S is a finite set. A cellular automaton (CA) is an endomorphism T: X → X (continuous, commuting with the action of G). Shereshevsky [13] proved
that for G = Z d with d> 1 no CA can be forward expansive, raising the following conjecture: For G = Z d, d> 1 the topological entropy of any CA is either zero or infinite. Morris and Ward [10],
proved this for linear CA’s, leaving the original conjecture open. We show that this conjecture is false, proving that for any d there exist a d-dimensional CA with finite, nonzero topological
entropy. We also discuss a measure-theoretic counterpart of this question for measure-preserving CA’s. 1.
, 909
"... Column-oriented indexes—such as projection or bitmap indexes—are compressed by run-length encoding to reduce storage and increase speed. Sorting the tables improves compression. On realistic
data sets, permuting the columns in the right order before sorting can reduce the number of runs by a factor ..."
Add to MetaCart
Column-oriented indexes—such as projection or bitmap indexes—are compressed by run-length encoding to reduce storage and increase speed. Sorting the tables improves compression. On realistic data
sets, permuting the columns in the right order before sorting can reduce the number of runs by a factor of two or more. For many cases, we prove that the number of runs in table columns is minimized
if we sort columns by increasing cardinality. Yet—maybe surprisingly—we must sometimes maximize the number of runs to minimize the index size. Experimentally, sorting based on Hilbert space-filling
curves is poor at minimizing the number of runs. Key words:
"... Two-dimensional R-trees are a class of spatial index structures in which objects are arranged to enable fast window queries: report all objects that intersect a given query window. One of the
most successful methods of arranging the objects in the index structure is based on sorting the objects acco ..."
Add to MetaCart
Two-dimensional R-trees are a class of spatial index structures in which objects are arranged to enable fast window queries: report all objects that intersect a given query window. One of the most
successful methods of arranging the objects in the index structure is based on sorting the objects according to the positions of their centres along a two-dimensional Hilbert space-filling curve.
Alternatively one may use the coordinates of the objects ’ bounding boxes to represent each object by a four-dimensional point, and sort these points along a four-dimensional Hilbert-type curve. In
experiments by Kamel and Faloutsos and by Arge et al. the first solution consistently outperformed the latter when applied to point data, while the latter solution clearly outperformed the first on
certain artificial rectangle data. These authors did not specify which four-dimensional Hilbert-type curve was used; many exist. In this paper we show that the results of the previous papers can be
explained by the choice of the four-dimensional Hilbert-type curve that was used and by the way it was rotated in fourdimensional space. By selecting a curve that has certain properties and choosing
the right rotation one can combine the strengths of the two-dimensional and the four-dimensional approach into one, while avoiding their apparent weaknesses. The effectiveness of our approach is
demonstrated with
|
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|
The u/v coordinates for the hemisphere x ≥ 0 are derived from the phi and theta angles, as follows:
u = sin(θ) cos(φ)
v = sin(θ) sin(φ)
In these expressions, φ and θ are the phi and theta angles, respectively.
In terms of azimuth and elevation, the u and v coordinates are
u = cos(el) sin(az)
v = sin(el)
The values of u and v satisfy the inequalities
–1 ≤ u ≤ 1
–1 ≤ v ≤ 1
u^2 + v^2 ≤ 1
Conversely, the phi and theta angles can be written in terms of u and v
tan(φ) = v/u
sin(θ) = sqrt(u^2 + v^2)
The azimuth and elevation angles can also be written in terms of u and v
sin(el) = v
tan(az) = u/sqrt(1 – u^2 – v^2)
The φ angle is the angle from the positive y-axis toward the positive z-axis, to the vector's orthogonal projection onto the yz plane. The φ angle is between 0 and 360 degrees. The θ angle is the
angle from the x-axis toward the yz plane, to the vector itself. The θ angle is between 0 and 180 degrees.
The figure illustrates φ and θ for a vector that appears as a green solid line. The coordinate system is relative to the center of a uniform linear array, whose elements appear as blue circles.
The coordinate transformations between φ/θ and az/el are described by the following equations
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R-statistics blog
Category Archives: R bloggers
(Written by Ian Fellows)
The RForge build error has been fixed. the package can now be tried with: install.packages("Deducer",,"http://www.rforge.net",type="source")
As prolific as the CRAN website is of packages, there are several packages to R that succeeds in standing out for their wide spread use (and quality), Hadley Wickhams ggplot2 and plyr are two such
And today (through twitter) Hadley has updates the rest of us with the news:
just released new versions of plyr and ggplot2. source versions available on cran, compiled will follow soon #rstats
Going to the CRAN website shows that plyr has gone through the most major update, with the last update (before the current one) taking place on 2009-06-23. And now, over a year later, we are
presented with plyr version 1, which includes New functions, New features some Bug fixes and a much anticipated Speed improvements.
ggplot2, has made a tiny leap from version 0.8.7 to 0.8.8, and was previously last updated on 2010-03-03.
Me, and I am sure many R users are very thankful for the amazing work that Hadley Wickham is doing (both on his code, and with helping other useRs on the help lists). So Hadley, thank you!
Here is the complete change-log list for both packages:
Continue reading
Update (07.07.10): The function in this post has a more mature version in the “arm” package. See at the end of this post for more details.
* * * *
Imagine you want to give a presentation or report of your latest findings running some sort of regression analysis. How would you do it?
This was exactly the question Wincent Rong-gui HUANG has recently asked on the R mailing list.
One person, Bernd Weiss, responded by linking to the chapter “Plotting Regression Coefficients” on an interesting online book (I have never heard of before) called “Using Graphs Instead of Tables” (I
should add this link to the free statistics e-books list…)
Letter in the conversation, Achim Zeileis, has surprised us (well, me) saying the following
I’ve thought about adding a plot() method for the coeftest() function in the “lmtest” package. Essentially, it relies on a coef() and a vcov() method being available – and that a central limit
theorem holds. For releasing it as a general function in the package the code is still too raw, but maybe it’s useful for someone on the list. Hence, I’ve included it below.
(I allowed myself to add some bolds in the text)
So for the convenience of all of us, I uploaded Achim’s code in a file for easy access. Here is an example of how to use it:
data("Mroz", package = "car")
fm <- glm(lfp ~ ., data = Mroz, family = binomial)
coefplot(fm, parm = -1)
I hope Achim will get around to improve the function so he might think it worthy of joining his“lmtest” package. I am glad he shared his code for the rest of us to have something to work with in the
* * *
Update (07.07.10):
Thanks to a comment by David Atkins, I found out there is a more mature version of this function (called coefplot) inside the {arm} package. This version offers many features, one of which is the
ability to easily stack several confidence intervals one on top of the other.
It works for baysglm, glm, lm, polr objects and a default method is available which takes pre-computed coefficients and associated standard errors from any suitable model.
(Notice that the Poisson model in comparison with the binomial models does not make much sense, but is enough to illustrate the use of the function)
data("Mroz", package = "car")
M1<- glm(lfp ~ ., data = Mroz, family = binomial)
M2<- bayesglm(lfp ~ ., data = Mroz, family = binomial)
M3<- glm(lfp ~ ., data = Mroz, family = binomial(probit))
coefplot(M2, xlim=c(-2, 6), intercept=TRUE)
coefplot(M1, add=TRUE, col.pts="red", intercept=TRUE)
coefplot(M3, add=TRUE, col.pts="blue", intercept=TRUE, offset=0.2)
(hat tip goes to Allan Engelhardt for help improving the code, and for Achim Zeileis in extending and improving the narration for the example)
Resulting plot
* * *
Lastly, another method worth mentioning is the Nomogram, implemented by Frank Harrell’a rms package.
About prize baring contests
Competition with prizes are an amazing thing. If you are not sure of that, I urge you to listened to Peter Diamandis talk about his experience with the X prize (start listening at minute 11:40):
At short – prizes can give up to 1 to 50 ratio of return on investment of the people giving funding to the prize. The money is spent only when results are achieved. And there is a lot of value in
terms of public opinion and publicity. And the best of all (for the promoter of the competition) – prizes encourage people to take risks (at their own expense) in order to get results done.
All of that said, I look at prize baring competition as something worth spreading, especially in cases where the results of the winning team will be shared with the public.
About the IEEE ICDM Contest
The IEEE ICDM Contest (“Road Traffic Prediction for Intelligent GPS Navigation”), seems to be one of those cases. Due to a polite request, I am republishing here the details of this new competition,
in the hope that some of my R colleagues will bring the community some pride
Continue reading
(Written by Ian Fellows)
Below is a link to the first of a weekly (or bi-weekly) screen-cast vlog of my progress building a GUI for the ggplot2 package.
comments and suggestions are more than welcome, and can e-mailed to me at: hefell@gmail.com
About Clustergrams
In 2002, Matthias Schonlau published in “The Stata Journal” an article named “The Clustergram: A graph for visualizing hierarchical and . As explained in the abstract:
In hierarchical cluster analysis dendrogram graphs are used to visualize how clusters are formed. I propose an alternative graph named “clustergram” to examine how cluster members are assigned to
clusters as the number of clusters increases.
This graph is useful in exploratory analysis for non-hierarchical clustering algorithms like k-means and for hierarchical cluster algorithms when the number of observations is large enough to
make dendrograms impractical.
A similar article was later written and was (maybe) published in “computational statistics”.
Both articles gives some nice background to known methods like k-means and methods for hierarchical clustering, and then goes on to present examples of using these methods (with the Clustergarm) to
analyse some datasets.
Personally, I understand the clustergram to be a type of parallel coordinates plot where each observation is given a vector. The vector contains the observation’s location according to how many
clusters the dataset was split into. The scale of the vector is the scale of the first principal component of the data.
Clustergram in R (a basic function)
After finding out about this method of visualization, I was hunted by the curiosity to play with it a bit. Therefore, and since I didn’t find any implementation of the graph in R, I went about
writing the code to implement it.
The code only works for kmeans, but it shows how such a plot can be produced, and could be later modified so to offer methods that will connect with different clustering algorithms.
How does the function work: The function I present here gets a data.frame/matrix with a row for each observation, and the variable dimensions present in the columns.
The function assumes the data is scaled.
The function then goes about calculating the cluster centers for our data, for varying number of clusters.
For each cluster iteration, the cluster centers are multiplied by the first loading of the principal components of the original data. Thus offering a weighted mean of the each cluster center
dimensions that might give a decent representation of that cluster (this method has the known limitations of using the first component of a PCA for dimensionality reduction, but I won’t go into that
in this post).
Finally all of our data points are ordered according to their respective cluster first component, and plotted against the number of clusters (thus creating the clustergram).
My thank goes to Hadley Wickham for offering some good tips on how to prepare the graph.
Here is the code (example follows)
The R function can be downloaded from here
Corrections and remarks can be added in the comments bellow, or on the github code page.
Example on the iris dataset
The iris data set is a favorite example of many R bloggers when writing about R accessors , Data Exporting, Data importing, and for different visualization techniques.
So it seemed only natural to experiment on it here.
source("http://www.r-statistics.com/wp-content/uploads/2012/01/source_https.r.txt") # Making sure we can source code from github
par(cex.lab = 1.5, cex.main = 1.2)
Data <- scale(iris[,-5]) # notice I am scaling the vectors)
clustergram(Data, k.range = 2:8, line.width = 0.004) # notice how I am using line.width. Play with it on your problem, according to the scale of Y.
Here is the output:
Looking at the image we can notice a few interesting things. We notice that one of the clusters formed (the lower one) stays as is no matter how many clusters we are allowing (except for one
observation that goes way and then beck).
We can also see that the second split is a solid one (in the sense that it splits the first cluster into two clusters which are not “close” to each other, and that about half the observations goes to
each of the new clusters).
And then notice how moving to 5 clusters makes almost no difference.
Lastly, notice how when going for 8 clusters, we are practically left with 4 clusters (remember – this is according the mean of cluster centers by the loading of the first component of the PCA on the
If I where to take something from this graph, I would say I have a strong tendency to use 3-4 clusters on this data.
But wait, did our clustering algorithm do a stable job?
Let’s try running the algorithm 6 more times (each run will have a different starting point for the clusters)
source("http://www.r-statistics.com/wp-content/uploads/2012/01/source_https.r.txt") # Making sure we can source code from github
Data <- scale(iris[,-5]) # notice I am scaling the vectors)
par(cex.lab = 1.2, cex.main = .7)
par(mfrow = c(3,2))
for(i in 1:6) clustergram(Data, k.range = 2:8 , line.width = .004, add.center.points = T)
Resulting with: (press the image to enlarge it)
Repeating the analysis offers even more insights.
First, it would appear that until 3 clusters, the algorithm gives rather stable results.
From 4 onwards we get various outcomes at each iteration.
At some of the cases, we got 3 clusters when we asked for 4 or even 5 clusters.
Reviewing the new plots, I would prefer to go with the 3 clusters option. Noting how the two “upper” clusters might have similar properties while the lower cluster is quite distinct from the other
By the way, the Iris data set is composed of three types of flowers. I imagine the kmeans had done a decent job in distinguishing the three.
Limitation of the method (and a possible way to overcome it?!)
It is worth noting that the current way the algorithm is built has a fundamental limitation: The plot is good for detecting a situation where there are several clusters but each of them is clearly
“bigger” then the one before it (on the first principal component of the data).
For example, let’s create a dataset with 3 clusters, each one is taken from a normal distribution with a higher mean:
source("http://www.r-statistics.com/wp-content/uploads/2012/01/source_https.r.txt") # Making sure we can source code from github
Data <- rbind(
cbind(rnorm(100,0, sd = 0.3),rnorm(100,0, sd = 0.3),rnorm(100,0, sd = 0.3)),
cbind(rnorm(100,1, sd = 0.3),rnorm(100,1, sd = 0.3),rnorm(100,1, sd = 0.3)),
cbind(rnorm(100,2, sd = 0.3),rnorm(100,2, sd = 0.3),rnorm(100,2, sd = 0.3))
clustergram(Data, k.range = 2:5 , line.width = .004, add.center.points = T)
The resulting plot for this is the following:
The image shows a clear distinction between three ranks of clusters. There is no doubt (for me) from looking at this image, that three clusters would be the correct number of clusters.
But what if the clusters where different but didn’t have an ordering to them?
For example, look at the following 4 dimensional data:
source("http://www.r-statistics.com/wp-content/uploads/2012/01/source_https.r.txt") # Making sure we can source code from github
Data <- rbind(
cbind(rnorm(100,1, sd = 0.3),rnorm(100,0, sd = 0.3),rnorm(100,0, sd = 0.3),rnorm(100,0, sd = 0.3)),
cbind(rnorm(100,0, sd = 0.3),rnorm(100,1, sd = 0.3),rnorm(100,0, sd = 0.3),rnorm(100,0, sd = 0.3)),
cbind(rnorm(100,0, sd = 0.3),rnorm(100,1, sd = 0.3),rnorm(100,1, sd = 0.3),rnorm(100,0, sd = 0.3)),
cbind(rnorm(100,0, sd = 0.3),rnorm(100,0, sd = 0.3),rnorm(100,0, sd = 0.3),rnorm(100,1, sd = 0.3))
clustergram(Data, k.range = 2:8 , line.width = .004, add.center.points = T)
In this situation, it is not clear from the location of the clusters on the Y axis that we are dealing with 4 clusters.
But what is interesting, is that through the growing number of clusters, we can notice that there are 4 “strands” of data points moving more or less together (until we reached 4 clusters, at which
point the clusters started breaking up).
Another hope for handling this might be using the color of the lines in some way, but I haven’t yet figured out how.
Clustergram with ggplot2
Hadley Wickham has kindly played with recreating the clustergram using the ggplot2 engine. You can see the result here:
And this is what he wrote about it in the comments:
I’ve broken it down into three components:
* run the clustering algorithm and get predictions (many_kmeans and all_hclust)
* produce the data for the clustergram (clustergram)
* plot it (plot.clustergram)
I don’t think I have the logic behind the y-position adjustment quite right though.
Conclusions (some rules of thumb and questions for the future)
In a first look, it would appear that the clustergram can be of use. I can imagine using this graph to quickly run various clustering algorithms and then compare them to each other and review their
stability (In the way I just demonstrated in the example above).
The three rules of thumb I have noticed by now are:
1. Look at the location of the cluster points on the Y axis. See when they remain stable, when they start flying around, and what happens to them in higher number of clusters (do they re-group
2. Observe the strands of the datapoints. Even if the clusters centers are not ordered, the lines for each item might (needs more research and thinking) tend to move together – hinting at the real
number of clusters
3. Run the plot multiple times to observe the stability of the cluster formation (and location)
Yet there is more work to be done and questions to seek answers to:
• The code needs to be extended to offer methods to various clustering algorithms.
• How can the colors of the lines be used better?
• How can this be done using other graphical engines (ggplot2/lattice?) – (Update: look at Hadley’s reply in the comments)
• What to do in case the first principal component doesn’t capture enough of the data? (maybe plot this graph to all the relevant components. but then – how do you make conclusions of it?)
• What other uses/conclusions can be made based on this graph?
I am looking forward to reading your input/ideas in the comments (or in reply posts).
useR!2010 is coming. I am going to give two talks there (I will write more of that soon), but in the meantime, please note that the online registration deadline is coming to an end.
This was published on the R-help mailing list today:
The final registration deadline for the R User Conference is June 20,
2010, one week away. Later registration will not be possible on site!
Conference webpage: http://www.R-project.org/useR-2010
Conference program: http://www.R-project.org/useR-2010/program.html
The conference is scheduled for July 21-23, 2010, and will take place at
the campus of the National Institute of Standards and Technology (NIST) in
Gaithersburg, Maryland, USA.
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Geometry proof
October 4th 2008, 06:49 AM #1
Junior Member
Aug 2008
Geometry proof
A radius drawn to bisect a chord in a circle will always meet the chord at 90 degrees.
Use a mathematical method (that does not use vectors), to prove this.
I've never been too great with geometric proofs
let the chord be segment AB. let the circle center be point O.
let M be the intersection point of the bisecting radius and chord AB.
M is the midpoint of AB ... why?
now, prove that triangle OMA is congruent to triangle OMB, then use corresponding parts of congruent triangles to show that angle OMA is congruent to angle OMB.
from this point, it should be easy to show that OM is perpendicular to AB.
formally, you can't just say M is the midpoint of AB ... the reason must be given, i.e. "definition of a segment bisector".
you do understand that every statement in a proof requires a reason, correct?
that's all I'm trying to tell you.
October 4th 2008, 07:14 AM #2
October 4th 2008, 07:22 AM #3
Junior Member
Aug 2008
October 4th 2008, 07:37 AM #4
October 4th 2008, 07:39 AM #5
Junior Member
Aug 2008
October 4th 2008, 08:10 AM #6
October 4th 2008, 08:12 AM #7
Junior Member
Aug 2008
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Fibonacci problem, proof by induction?
November 13th 2012, 09:27 PM #1
Nov 2012
alpha = (1+ sqrt5)/2 and beta = (1-sqrt5)/2
alpha^2 = 1 + alpha and beta^2 = 1+ beta
Use induction to prove that for all integers n >= 1 we have
alpha^n = f(n-1)+ f(n)*(alpha) and beta^n = f(n-1)+ f(n)*(beta)
I've did my base case and plug in k+1 to n but I can't get them equal to each other. Please help, I've been playing with these numbers for hours.
Re: Fibonacci problem, proof by induction?
Having demonstrated the base case $P_1$ is true, then state the induction hypothesis $P_k$
Multiply through by $\alpha$:
Now, since $F_{n+1}=F_{n}+F_{n-1}$, can you continue?
Re: Fibonacci problem, proof by induction?
so alpha^(k+1) = (alpha)*f(k+1) + f(k)
sorry, I still don't know how is this a proof alpha^n = f(n-1)+ f(n)*(alpha)
Re: Fibonacci problem, proof by induction?
We may write this result as:
You see, we have arrived at $P_{k+1}$ which we derived from $P_k$, completing the proof by induction.
Observe that this is the same as the induction hypothesis, except $k$ is replaced with $k+1$. This means it is true for all $n\in\mathbb{N}$.
Have you learned the analogy of climbing a ladder or falling dominoes to mathematical induction?
Re: Fibonacci problem, proof by induction?
oh okay, thanks. This problem is just very different to the kind of Fibonacci problem I've been doing. But they're just the same in the end. Thanks again : )
Re: Fibonacci problem, proof by induction?
Glad to help, and welcome to the forum!
November 13th 2012, 09:46 PM #2
November 13th 2012, 10:01 PM #3
Nov 2012
November 13th 2012, 10:19 PM #4
November 13th 2012, 10:34 PM #5
Nov 2012
November 13th 2012, 10:38 PM #6
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Solution for a particular column vector implies solution for all column vectors.
February 1st 2011, 07:31 PM
Solution for a particular column vector implies solution for all column vectors.
Let $A$ be a square matrix. Show that if the system $AX=B$ has a unique solution for some particular column vector $B$, then it has a unique solution for all $B$.
So I'm not really sure how to go about this since there's no assumption that $A$ is invertible, and I'm assuming I have to utilize row operations or the like, but I just can't make any
connections. Any help would be appreciated, thanks.
February 1st 2011, 08:56 PM
Let $A$ be a square matrix. Show that if the system $AX=B$ has a unique solution for some particular column vector $B$, then it has a unique solution for all $B$.
So I'm not really sure how to go about this since there's no assumption that $A$ is invertible, and I'm assuming I have to utilize row operations or the like, but I just can't make any
connections. Any help would be appreciated, thanks.
Here is my thoughts on the question.
Every solution to a linear system can be decomposed into two separate pieces.
$X=X_c+X_p$ where $X_c$ is the complimentary solution (the solution to the homogeneous equation) and $X_p$ is a particular solution.
Since we know the solution is unique for some particular $b$
We have
$Ax=A(x_c+x_p)=b$ Since the solution is unique The complementary solution cannot have any free parameters.
e.g the kernel of this Matrix has only one vector in it and since $\vec{0} \in x_c$ the only solution to the system
$Av=0$ is $v=0$
This will give the desired conclusion.
February 1st 2011, 09:09 PM
I'm not sure I exactly follow (we haven't covered kernel, etc).
Are you saying that since the solution is unique, it must be the case that $x_{c}$ is the zero column vector?
And I'm not sure I make the connection to how it follows that there must be a unique solution for any $B$.
February 2nd 2011, 12:34 AM
If $AX=B_0$ has unique solution, then
$E_n\cdot\ldots\cdot E_1AX=E_n\cdot\ldots\cdot E_1B_0\Leftrightarrow UX=E_n\cdot\ldots\cdot E_1B_0$
(where $U$ is a row echelon form) has unique solution. This implies that all leading coefficients of $U$ are different from $0$. Now, you can conclude.
Fernando Revilla
February 2nd 2011, 04:16 PM
Whew, I get it now. Thanks so much!
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14. How Strongly Do Physical Characteristics Of ... | Chegg.com
14. How strongly do physical characteristics of sisters and brothers correlate? Here are data on the heights (in inches) of 11 adult pairs.
Brother height (x) 71 68 66 67 70 71 70 73 72 65 66
Sister height (y) 69 64 65 63 65 62 65 64 66 59 62
Use your calculator to answer a, b and c below and answer all questions in 3 decimal places. Show your work for d ~ g for full credit.
a. The correlation between brother and sister heights
b. The mean of the brother and sister height
c. The standard deviation of brother and sister height
d. The slope of the least-square regression line, use the answers found from a and c.
e. The intercept of the least-squares regression line, use the answers found from b and d.
f. The equation of the least-square regression line
g. The residual corresponding to the example data point (67” for a brother height and 63” for his sister)
Statistics and Probability
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La Mirada Calculus Tutor
Find a La Mirada Calculus Tutor
...It describes how the world around us works, and it is the foundation of the other sciences (chemistry, biology, etc., have their roots in physics). I love talking about and teaching physics, as
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THE NATURAL COMMUNITIES OF VIRGINIA
CLASSIFICATION OF ECOLOGICAL COMMUNITY GROUPS
Second Approximation (Version 2.6)
Information current as of July, 2013
Ordination is a multivariate technique that arranges vegetation samples in relation to each other based on compositional similarity and relative species-abundances. Ordination procedures summarize
multidimensional data in a reduced coordinate system, extracting those axes that explain the most variation in the data. DCR-DNH ecologists use non-metric multidimensional scaling (NMDS) , an
ordination technique based on indirect gradient analysis that maximizes, to the extent possible, the rank-order (i.e ., non-parametric) correlation between inter-sample dissimilarity and inter-sample
distance in ordination space. The results of ordination analyses are depicted by diagrams, in which each point represents a plot and the distance between points roughly indicates the degree of
compositional similarity. Statistically significant correlations between measured environmental variables and sample coordinates on each axis may be plotted as vectors and overlain on the diagram.
The direction of a vector indicates the direction of maximum correlation through ordination space, while vector line lengths are determined by the strength of the correlation. This diagram shows a
two-dimensional ordination of the same dataset used to illustrate cluster analysis. Symbols indicate the four groups identified in the dendrogram and significant environmental gradients are plotted.
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Finding largest no
Join Date
Nov 2007
Rep Power
I have following code for finding largest no among 3 nos.
Java Code:
import java.util.*;
public class LargestNo{
public static void main (String [] arg) {
Scanner scan = new Scanner (System.in);
int [] numbers = new int [3];
int x;
int largestNumber;
System.out.print("Put in (three) numbers");
for (x=0; x<numbers.length; x++) {
numbers[x]=scan.nextInt ();
largestNumber = 0;
for (x=0; x<numbers.length; x++) {
if (x == 0) {
largestNumber = numbers[0];
if (numbers[x] > largestNumber) {
largestNumber = numbers[x];
System.out.println("The largest number is " + largestNumber);
I want to print the largest no and also its position in the array. how to do that? Please give me tips.
Join Date
Aug 2007
Rep Power
You can have a separate int variable for position.
something like this...
Java Code:
import java.util.*;
public class LargestNo{
public static void main (String [] arg) {
Scanner scan = new Scanner (System.in);
int [] numbers = new int [3];
int x;
int largestNumber;
int pos=0;//position in the array
System.out.print("Put in (three) numbers");
for (x=0; x<numbers.length; x++) {
numbers[x]=scan.nextInt ();
largestNumber = 0;
for (x=0; x<numbers.length; x++) {
if (x == 0) {
largestNumber = numbers[0];
pos = 0;
if (numbers[x] > largestNumber) {
largestNumber = numbers[x];
pos = x;//x is the position of the number in the array.
System.out.println("The largest number is " + largestNumber);
System.out.println("Position of the largest number:"+pos);
Join Date
Nov 2007
Rep Power
Ok kool. Cant I fetch the index from array just specifying the value. For example, I am assuming that array will only contain unique values and having a value I now want to get its index from
Just like same thing you have to done. Use another dummy variable. At each of the comparison you have done to check whether the number is large or small, update that dummy value.
Say first number is large, your dummy should be 0. In the next iteration the second number is small, still your dummy should be 0, because it holds the position.
I think my logic is clear.
At the same time, I think you have use additional } at last. Compile and check it.
Join Date
Nov 2007
Rep Power
Thanks Eranga. But consider the following:
For example, I am assuming that array will only contain unique values and having a value I now want to get its index from array.
I'm not get you. What you mean unique values. Between array index and array values there is no connection. Can you explain little more.
Join Date
Nov 2007
Rep Power
No there is no connection. I think you know that, array is indexing by starting with 0. That mean maximum indexing is less than one by number of element.
element 12 5 23
index 0 1 2
#of element 1 2 3
Join Date
Aug 2007
Rep Power
There is actually a way to get the index..Arrays has a method
binarySearch(array to be searched,key) which returns the position of the specified key..
Using it in your program:
Java Code:
import java.util.*;
public class LargestNo{
public static void main (String [] arg) {
Scanner scan = new Scanner (System.in);
int [] numbers = new int [3];
int x;
int largestNumber;
//int pos=0;//position in the array
int index=0;
System.out.print("Put in (three) numbers");
for (x=0; x<numbers.length; x++) {
numbers[x]=scan.nextInt ();
largestNumber = 0;
for (x=0; x<numbers.length; x++) {
if (x == 0) {
largestNumber = numbers[0];
//pos = 0;
if (numbers[x] > largestNumber) {
largestNumber = numbers[x];
//pos = x;//x is the position of the number in the array.
index = Arrays.binarySearch(numbers, largestNumber);
System.out.println("The largest number is " + largestNumber);
System.out.println("Position of the largest number:"+index);
I am not sure though if this would be a correct way to get the index of an array element...
Yep that's correct. One of the efficient way it is.
The way what I've told is the basis way to do it, comparing how to find the largest number. Because the same way should follow.
Join Date
Nov 2007
Rep Power
Its clear now. Thanks all of you.
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KS3 Maths Revision
Ronald Wayne was the third founder of the business "Apple"! However he sold his 10% stake for how much money in 1976?
Mouseover to see answer
$800 - Apple is now valued roughly at $620 Billion!
KS3 Maths Revision
(Resources for school year 7, year 8 and year 9)
Whether you love Maths or you hate it, you can be encouraged by the fact that 50% of all doctors graduate in the bottom half of their class! If you love Maths then we hope our KS3 Maths quizzes help
you love it a little more and if you hate Maths then we hope the quizzes help you hate it a little less.
Whatever else happens, you will agree that interactive Maths revision quizzes are more exciting than Maths text books!
Maths is a subject that needs to be learned step by step and because of this we categorize our quizzes using the same levels, ages and years as the National Curriculum. If you are confused by all the
numbers then we hope the table below helps you to revise the right subjects at the right time:
Level = 3/4: Age = 11/12: Year = 7
Level = 5/6: Age = 12/13: Year = 8
Level = 7/8: Age = 13/14: Year = 9
How To Play
Each quiz consists of 10 questions and each question has 4 multiple choice answers. At the top of each quiz you are given a choice of how you want to play it and this affects what happens when you
provide an incorrect answer.
You can either have the correct answer given immediately, or you can choose to have the questions presented again at the end of the quiz.
To print any of the quizzes (both questions and answers) click the “Print” link at the bottom of the quiz
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Perth Amboy Math Tutor
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and databases (SQL). I have taken various history classes as well and one of my hobbies is teaching mysel...
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23 Subjects: including algebra 2, electrical engineering, ASVAB, differential equations
...Algebra II/Trigonometry:I cover as much topics as the student needs or all the topics, especially the ratio of those most pertinent to the Regents exam. From Algebraic expressions, Functions
and Relations, Composition and Inverses of Functions, Exponential and Logarithmic Functions, Trigonometri...
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4 Subjects: including trigonometry, algebra 1, prealgebra, precalculus
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A collection of tutorials (Flash and Java) on various applications of integration including area between two curves, volumes,...
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Material Type:
Lawrence Husch
Date Added:
Jul 19, 2006
Date Modified:
Oct 22, 2013
A Calculus Tutorial developed at Oregon State University. "Contains wonderful illustrations ofcalculus concepts using stories...
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Material Type:
William Bogley and Robby Robson
Date Added:
Aug 08, 2000
Date Modified:
Feb 08, 2011
This subsite of Mathematics Tutorials and Problems (with applets) is divided into Interactive Tutorials, Calculus Problems,...
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Material Type:
kader dendane
Date Added:
Jun 28, 2008
Date Modified:
Oct 22, 2013
This site contains an electronic version of a Calculus textbook linked to a collection of interactive Java applets. The book...
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Dan Sloughter
Date Added:
Nov 14, 2005
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Oct 22, 2013
Tutorials and animations (Flash and Java) on finding volumes of solids of revolution; helpful for students who are visual...
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Lawrence Husch
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Jul 20, 2006
Date Modified:
Jan 28, 2013
The CCP includes modules that combine the flexibility and connectivity of the Web with the power of computer algebra systems...
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David Smith & Lawrence Moore, CCP Co-Directors
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Quoted from the site: [This site contains...] "Free mathematics tutorials to help you explore and gain deep understanding of...
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kader dendane
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This webpage has all lecture material, homeworks, homework solutions, and computer laboratories for a Calculus for Biology...
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Joe Mahaffy
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Collection of tutorials for the first year calculus student. Includes collections of online quizzes and drill problems.
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real derivative of the magnitude of a complex function?
October 13th 2010, 05:29 PM #1
Oct 2010
real derivative of the magnitude of a complex function?
Dear all,
I'm new to these forums and not sure if this is in the right section, but calculus I think is the main part of the problem. So I'm trying to solve an optimization problem that requires me to take
the (real) derivative of the magnitude of a complex function. Given a real vector $\mathbf{x} \in \mathbb{R}^n$ and a function $f : \mathbb{R}^n \rightarrow \mathbb{C}$, how do I compute
$abla |f(\mathbf{x})| = ?$
I know that the complex derivative of the magnitude of a complex function is undefined, but the real derivative should be defined, right?
For those interested in the entire problem, the function is of the form
$f(\mathbf{x}) = \mathbf{c}^T \mathbf{M(x) y(x)}$, where $\mathbf{M}$ is a complex matrix and $\mathbf{y}$ is a complex vector, but both are functions of a set of real inputs $\mathbf{x}$. $\
mathbf{c}$ is a constant vector. I'm fairly confident I can take the derivative of $f(\mathbf{x})$, but it's the magnitude that's giving me trouble.
Appreciate any help or insight. Thanks so much!
g(x)=|f(x)| is only an ordinary function from R^n to R. So grad(g) = ( g_1, g_2, ..., g_n), where g_i is the i-th partial derivative.
For your case, f is actually a map from R^n to R^2, g=\sqrt<f,f>. Then you can do the differential via chain rule and the Leibniz rule that D<f1,f2> = <Df1, f2> + <f1, Df2>. <,> is the inner
product of R^2.
Thanks for the prompt reply. My interpretation of your post is that I need to somehow write the complex function f(x) into f1(x) = Re(f) and f2(x) = Im(f) when I can then go ahead and use the
chain rule on sqrt(f1^2 + f2^2). Unfortunately this requires me to write down f1 and f2 explicitly which is probably possible to do but likely to be messy. For example, Re (M * y) = Re(M) Re(y) -
Im(M) Im(y), but M and y themselves are expressed in block matrix form are written in terms of multiple products of smaller complex matrices, so I'm afraid calculating Re and Im will quickly go
out of hand. Is there an alternative that can directly allow me to use the chain rule on the magnitude function, provided I can calculate grad(f)?
Thanks again!
|z|^2 = z z* where z* is the complex conjugate. So d|z|/dt = 1/(2|z|) (dz/dt z* + z dz*/dt)
= 1/(2|z|) (dz/dt z* + z (dz/dt)* ) = Re(dz/dt z*)/|z|
not sure this will reduce some of the effort or not
hmm... you know, it might, actually. I'm going to go try it and see if it's easier. Thanks for your help!
If, using the standard notation, z= x+ iy, f(z)= u(x,y)+ iv(x,y), then $|f(z)|= \sqrt{u^2(x,y)+ v^2(x,y)$ and $abla |f(z)|= \frac{\partial \sqrt{u^2+ v^2}}{\partial x}\vec{i}+ \frac{\partial \
sqrt{u^2+ v^2}}{\partial y}\vec{j}$.
More generally, if f is a function from $R^n$ to C, then $f(x)= u(x_1, x_2, ..., x_n)+ iv(x_1, x_2, ...,, x_n)$, $|f(x)|= \sqrt{u^2+ v^2}$ and
$abla |f(x)|= \sum_{i= 1}^n \frac{\partial\sqrt{u^2+ v^2}}{\partial x_i}\vec{e_i}$
where $e_i$ is the unit vector in the $x_i$ direction.
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Weights for etale cohomology: why does Deligne's definition work?
up vote 3 down vote favorite
For a field $K$ and a variety $X/K$ (whose characteristic could be $0$) I need a 'simple' explanation for the (Deligne's) method of defining weights of the $l$-adic etale cohomology of $\overline{X}$
(the base change of $X$ to the algebraic closure of $K$). Which 'complicated' statements does one need to define and study weights, and what statements here could be proved 'easily' (using basic
properties of etale cohomology)? What is the best reference for obtaining an 'understanding' of these things (I prefer reading in English and in Russian:))?
Upd. I know some references on the subject (Weil II, Kiehl-Weissauer? SGA IV3, SGAVII2); yet it is difficult to understand which parts of these books contain the information I need. Does there exist
any 'guide' to any of these texts?
On the other hand, "Weights in arithmetic geometry" by Jannsen is too short.
etale-cohomology reference-request ag.algebraic-geometry
1 There might be some useful things in the text "Cohomology of algebraic varieties" by Danilov, or rather the parts of it about étale cohomology. It appeared in English translation in an EMS volume,
the Russion original can be found here: mi.mathnet.ru/eng/intf124 – Dan Petersen Oct 21 '12 at 11:32
Thank you; this is an interesting text! – Mikhail Bondarko Oct 22 '12 at 5:47
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2 Answers
active oldest votes
Complicated (the special case $f: X \to \mathbf{F}_q$ proper smooth is Weil I!): Let $\mathcal{F}$ be mixed of weight $\leq i$. Then $R^q\pi_!\mathcal{F}$ is mixed of weight $\leq q+i$
up vote 1 (see Deligne, Weil II, Théorème 1 (3.3.1) or Kiehl-Weissauer, Theorem I.7.1, strengthened in I.9.3)
down vote
1 Thank you! Yet I also need to bound weights from below.:) Certainly, I can apply the Verdier duality to this end; yet is there an 'easier' way? – Mikhail Bondarko Oct 20 '12 at 19:59
add comment
It seems that your question is not well defined unless $K$ is finitely generated over its prime field.
See for instance
Jannsen, Uwe Weights in arithmetic geometry. Jpn. J. Math. 5 (2010), no. 1, 73–102. http://arxiv.org/abs/1003.0927 or http://www.springerlink.com/content/207j13t274004070/
up vote 1
down vote and also (this is in French)
Deligne, Pierre Poids dans la cohomologie des variétés algébriques. Proceedings of the International Congress of Mathematicians (Vancouver, B. C., 1974), Vol. 1, pp. 79–85 Canad. Math.
Congress, Montreal, Que., 1975. http://www.mathunion.org/ICM/ICM1974.1/Main/icm1974.1.0079.0086.ocr.pdf
About the first reference: do you know where I can find some details on the isomorphism $b_{\eta,s}$ in the formula (2.3)? – Mikhail Bondarko Oct 20 '12 at 21:03
You can find all details there Théorie des topos et cohomologie étale des schémas. Tome 3.Séminaire de Géométrie Algébrique du Bois-Marie 1963–1964 (SGA 4). Lecture Notes in
Mathematics, Vol. 305. springerlink.com/content/n880177446r8 – Niels Oct 21 '12 at 7:52
I'm sorry; could you give a more precise reference? I know the smooth and proper base change theorems; yet how do they help here? – Mikhail Bondarko Oct 21 '12 at 9:29
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Math - Please Help
Posted by Barb on Sunday, February 17, 2013 at 10:25pm.
A right circular cone has a volume of 140 in^3. The height of the cone is the same length as the diameter of the base. Find the radius and height.
• Math - Please Help - Jeff, Sunday, February 17, 2013 at 11:37pm
We know the formula for the Volume of a Right Circular Cone is given by
V=140 in^3
The height of the cone = diameter of the base. The diameter = 2 times the radius, so h = 2r
The formula for Volume can now be written as
which simplifies to
You plug in 140 in^3 for V and solve for r. Then you can plug the value you find for r into the equation h=2r
• Math - Please Help - Reiny, Monday, February 18, 2013 at 12:13am
V = (1/3 π r^2 h , but h = 2r
3V = π r^2 (2r) = 2π r^3
420 = 2πr^3
r^3 = 210/π
r = (210/π)^(1/3) = 4.0584
r = 4.0584
h = 8.11683
V = (1/3)π(4.0584)^2 (8.11683) = 139.9989.. , not bad
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is there an easier way than listing all possibilities then solving to find all rational roots for P(x)=0
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we list all of the possibles to satisfy the p/q theorem actually it is called the possible rational theorem but it is not necessary to list them all when solving for all zeros of a polynomial if
nothing else, it gives us a starting point of what value a zero could be
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the first resource for mathematics
New stability and stabilization for switched neutral control systems.
(English) Zbl 1198.93187
Summary: This paper concerns stability and stabilization issues for switched neutral systems and presents new classes of piecewise Lyapunov functionals and multiple Lyapunov functionals, based on
which, two new switching rules are introduced to stabilize the neutral systems. One switching rule is designed from the solution of the so-called Lyapunov-Metzler linear matrix inequalities. The
other is based on the determination of average dwell time computed from a new class of linear matrix inequalities (LMIs). And then, state-feedback control is derived for the switched neutral control
system mainly based on the state switching rules. Finally, three examples are given to demonstrate the effectiveness of the proposed method.
Editorial remark: There are doubts about a proper peer-reviewing procedure of this journal. The editor-in-chief has retired, but, according to a statement of the publisher, articles accepted under
his guidance are published without additional control.
93D15 Stabilization of systems by feedback
34K20 Stability theory of functional-differential equations
34K40 Neutral functional-differential equations
93C23 Systems governed by functional-differential equations
93C30 Control systems governed by other functional relations
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[Numpy-discussion] Best way to construct/slice 3-dimensional ndarray from multiple 2d ndarrays?
Aronne Merrelli aronne.merrelli@gmail....
Wed Aug 17 12:46:12 CDT 2011
On Wed, Aug 17, 2011 at 9:04 AM, Keith Hughitt <keith.hughitt@gmail.com>wrote:
> Also, when subclassing ndarray and calling obj = data.view(cls) for an
> ndarray "data", does this copy the data into the new object by value or
> reference? The method which extracts the 2d slice actually returns a
> subclass of ndarray created using the extracted data, so this is why I ask.
I think it should pass a reference - the following code suggests the
subclass is sharing the same fundamental array object. You can use the .base
attribute of the ndarray object to see if it is a view back to another
ndarray object:
import numpy as np
class TestClass(np.ndarray):
def __new__(cls, inp_array):
return inp_array.view(cls)
In [2]: x = np.ones(5)
In [3]: obj = TestClass(x)
In [4]: id(x), id(obj), id(obj.base)
Out[4]: (23517648, 19708080, 23517648)
In [5]: print x, obj
[ 1. 1. 1. 1. 1.] [ 1. 1. 1. 1. 1.]
In [6]: x[2] = 2
In [7]: print x, obj
[ 1. 1. 2. 1. 1.] [ 1. 1. 2. 1. 1.]
If you change the TestClass.__new__() to: "return
np.array(inp_array).view(cls)" then you will make a copy of the input array
instead, if that is needed. In that case, it looks like the .base attribute
is a new ndarray, copied from the input array.
[PS - also note that .base is set to None, if the ndarray is not a view into
another ndarray; it turns out that None has a valid object number, which
confused me at first - see id(None).]
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10.1.1 Applications and Extensions
Next: 10.1.2 The Components of Up: DIME: Portable Software Previous: DIME: Portable Software
The most efficient speed for aircraft flight is just below the speed of sound: the transonic regime. Simulations of flight at these speeds consume large quantities of computer time, and are a natural
candidate for a DIME application. In addition to the complex geometries of airfoils and turbines for which these simulations are required, the flow tends to develop singular regions or shocks in
places that cannot be predicted in advance; the adaptive refinement capability of a DIME mesh allows the mesh to be fine and detail resolved near shocks while keeping the regions of smooth flow
coarsely meshed for economy (Section 12.3).
The version of DIME developed within C10.1.7. The manifold may, however, be embedded in a higher-dimensional space. In collaboration with the Biology division at Caltech, we have simulated the
electrosensory system of the weakly electric fish Apteronotus leptorhynchus. The simulation involves creating a mesh covering the skin of the fish, and using the boundary element method to calculate
field strengths in the three-dimensional space surrounding the fish (Section 12.2).
In the same vein of embedding the mesh in higher dimensions, we have simulated a bosonic string of high-energy physics, embedding the mesh in up to 26 spatial dimensions. The problem here is to
integrate over not only all positions of the mesh nodes, but also over all triangulations of the mesh (Section 7.2).
The information available to a DIME application is certain data stored in the elements and nodes of the mesh. When doing finite-element calculations, one would like a somewhat higher level of
abstraction, which is to refer to functions defined on a domain, with certain smoothness constraints and boundary conditions. We have made a further software layer on top of DIME to facilitate this:
DIMEFEM. With this we may add, multiply, differentiate and integrate functions defined in terms of the Lagrangian finite-element family, and define linear, bilinear, and nonlinear operators acting on
these functions. When a bilinear operator is defined, a variational principle may be solved by conjugate-gradient methods. The preconditioner for the CG method may in itself involve solving a
variational principle. The DIMEFEM package has been applied to a sophisticated incompressible flow algorithm (Section 10.2).
Next: 10.1.2 The Components of Up: DIME: Portable Software Previous: DIME: Portable Software
Guy Robinson
Wed Mar 1 10:19:35 EST 1995
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When is there a deRham duality relation between the fundamental class and a top form.?
up vote 1 down vote favorite
Hi, everyone: I am reading a small expository paper on properties of CP^2, in which the intersection form is defined as an integral of the wedge of two forms $w_1$, $w_2$, and these forms $w_1$,
$w_2$ (no problem with compact support, since CP^2 is compact) seem to have been obtained from the fundamental class [z] of H[2](CP^2)-- a copy of CP^1 (embedded in CP^2), after which we integrate
$w:=w1\wedge w2$ to get the intersection number.
I am curious on whether I am reading the above correctly, i.e., that the volume form in CP^2 is obtained by using the fund. class [z] in H[2]. If not, would someone explain; if this is correct, if we
are we using some form of deRham's theorem to turn a purely topological object like [z] into an object like $w$, for which we must have a differentiable structure defined)?
Thanks in Advance.
homology cohomology dg.differential-geometry
I took the liberty of editing the formatting of your question a little - hope that's OK, if not let me know. – Yemon Choi Apr 28 '10 at 23:46
Short answer - for all compact oriented manifolds. – Somnath Basu Apr 29 '10 at 1:36
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1 Answer
active oldest votes
Consider a suitably small tubular neighbourhood $\mathcal{N}$ of $\mathbb{CP}^1$, thought of as sitting inside $\mathbb{CP}^2$. Then $\mathcal{N}$ locally looks like $\mathbb{CP}^1\times
D_2$. The volume form $\omega$ of $\mathbb{CP}^1$ is not necessarily a $2$-form in $\mathbb{CP}^2$. However, one can imagine changing it so that we have new $2$-form $\widetilde{\omega}$ in $
\mathbb{CP}^2$, supported in $\mathcal{N}$, such that $\widetilde{\omega}$ restrcited to $\{p\}\times D_2$ (for $p\in\mathbb{CP}^1$) looks like a smooth bump function which integrates to $1$.
up vote This can be taken to be $w_1$ in your case. Now assume you take your copy of $\mathbb{CP}^1$ inside $\mathbb{CP}^2$ and perturb it a bit (i.e., make it transversal to itself) to get another
1 down copy. Apply what we said before and get $w_2$ supported in a suitable tubular neighbourhood of this perturbed copy. Now integrating $w_1\wedge w_2$ over $\mathbb{CP}^2$ gives you an
vote integration over balls around points where self-intersections occur. The normalization were so chosen that it counts the intersection number of $\mathbb{CP}^1$ with itself.
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Not the answer you're looking for? Browse other questions tagged homology cohomology dg.differential-geometry or ask your own question.
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Navy Electricity and Electronics Training Series (NEETS)
Module 12—Modulation Principles
i - ix
1-1 to 1-10
1-11 to 1-20
1-21 to 1-30
1-31 to 1-40
1-41 to 1-50
1-51 to 1-60
1-61 to 1-70
1-71 to 1-75
2-1 to 2-10
2-11 to 2-20
2-21 to 2-30
2-31 to 2-40
2-41 to 2-50
2-51 to 2-60
2-61 to 2-64
3-1 to 3-10
3-11 to 3-20
3-21 to 3-30
3-31 to 3-35
AI-1 to AI-6, Index-1 to 2, Assignment 1 , 2
is referred to as the MODULATED WAVE and is the waveform that is transmitted through space. When the modulated wave is received and demodulated, the original component waves (carrier and modulating
waves) are reproduced with their respective frequencies, phases, and amplitudes unchanged.
Modulation of a carrier can be achieved by any of several methods. Generally, the methods are named for the sine-wave characteristic that is altered by the modulation process. In this module, you
will study AMPLITUDE MODULATION, which includes CONTINUOUS-WAVE MODULATION. You will also learn about two forms of ANGLE MODULATION (FREQUENCY MODULATION and PHASE MODULATION). A special type of
modulation, known as PULSE MODULATION, will also be discussed. Before we present the methods involved in developing modulation, you need to study a process that is essential to the modulation of a
carrier, known as heterodyning.
To help you understand the operation of heterodyning circuits, we will begin with a discussion of LINEAR and NONLINEAR devices. In linear devices, the output rises and falls directly with the input.
In nonlinear devices, the output does not rise and fall directly with the input.
Whether the impedance of a device is linear or nonlinear can be determined by comparing the change in current through the device to the change in voltage applied to the device. The simple circuit
shown in view (A) of figure 1-4 is used to explain this process.
Figure 1-4A.—Circuit with one linear impedance.
First, the current through the device must be measured as the voltage is varied. Then the current and voltage values can be plotted on a graph, such as the one shown in view (B), to determine the
impedance of the device. For example, assume the voltage is varied from 0 to 200 volts in 50-volt steps, as shown in view (B). At the first 50-volt point, the ammeter reads 0.5 ampere. These
ordinates are plotted as point a in view (B). With 100 volts applied, the ammeter reads 1 ampere; this value is plotted as point b. As these steps are continued, the values are plotted as points c
and d. These points are connected with a straight line to show the linear relationship between current and voltage. For every change in voltage applied to the device, a proportional change occurs in
the current through the device. When the change in current is proportional to the change in applied voltage, the impedance of the device is linear and a straight line is developed in the graph.
Figure 1-4B.—Circuit with one linear impedance.
The principle of linear impedance can be extended by connecting two impedance devices in series, as shown in figure 1-5, view (A). The characteristics of both individual impedances are determined as
explained in the preceding section. For example, assume voltmeter V1 shows 50 volts and the ammeter shows 0.5 ampere. Point a in view (B) represents this ordinate. In the same manner, increasing the
voltage in increments of 50 volts gives points b, c, and d. Lines Z1 and Z2 show the characteristics of the two impedances. The total voltage of the series combination can be determined by adding the
voltages across Z1 and Z2. For example, at 0.5 ampere, point a (50 volts) plus point e (75 volts) produces point i (125 volts). Also, at 1 ampere, point b plus point f produces point j. Line Z1 + Z2
represents the combined voltage-current characteristics of the two devices.
Figure 1-5A.—Circuit with two linear impedances.
Figure 1-5B.—Circuit with two linear impedances.
View (A) of figure 1-6 shows two impedances in parallel. View (B) plots the impedances both individually (Z1 and Z2) and combined (Z1 x Z2)/(Z1 + Z2). Note that Z1 and Z2 are not equal. At 100 volts,
Z1 has 1 ampere of current plotted at point b and Z2 has 0.5 ampere plotted at point f. The coordinates of the equivalent impedance of the parallel combination are found by adding the current through
Z1 to the current through Z2. For example, at 100 volts, point b is added to point f to determine point j (1.5 amperes).
Figure 1-6.—Circuit with parallel linear impedances.
Positive or negative voltage values can be used to plot the voltage-current graph. Figure 1-7 shows an example of this situation. First, the voltage versus current is plotted with the battery
polarity as shown in view (A). Then the battery polarity is reversed and the remaining voltage versus current points are plotted. As a result, the line shown in view (C) is obtained.
Figure 1-7A.—Linear impedance circuit.
Figure 1-7B.—Linear impedance circuit.
Figure 1-7C.—Linear impedance circuit
The battery in view (A) could be replaced with an ac generator, as shown in view (B), to plot the characteristic chart. The same linear voltage-current chart would result. Current flow in either
direction is directly proportional to the change in voltage.
In conclusion, when dc or sine-wave voltages are applied to a linear impedance, the current through the impedance will vary directly with a change in the voltage. The device could be a resistor, an
air-core inductor, a capacitor, or any other linear device. In other words, if a sine-wave generator output is applied to a combination of linear impedances, the resultant current will be a sine wave
which is directly proportional to the change in voltage of the generator. The linear impedances do not alter the waveform of the sine wave. The amplitude of the voltage developed across each linear
component may vary, or the phase of the wave may shift, but the shape of the wave will remain the same.
You have studied that a linear impedance is one in which the resulting current is directly proportional to a change in the applied voltage. A nonlinear impedance is one in which the resulting current
is not directly proportional to the change in the applied voltage. View (A) of figure 1-8 illustrates a circuit which contains a nonlinear impedance (Z), and view (B) shows its voltage-current curve.
Figure 1-8A.—Nonlinear impedance circuit.
Figure 1-8B.—Nonlinear impedance circuit.
As the applied voltage is varied, ammeter readings which correspond with the various voltages can be recorded. For example, assume that 50 volts yields 0.4 milliampere (point a), 100 volts produces 1
milliampere (point b), and 150 volts causes 2.2 milliamperes (point c). Current through the nonlinear impedance does not vary proportionally with the voltage; the chart is not a straight line.
Therefore, Z is a nonlinear impedance; that is, the current through the impedance does not faithfully follow the change in voltage. Various combinations of voltage and current for this particular
nonlinear impedance may be obtained by use of this voltage-current curve.
The series combination of a linear and a nonlinear impedance is illustrated in view (A) of figure 1-9. The voltage-current charts of Z1 and Z2 are shown in view (B). A chart of the combined impedance
can be plotted by adding the amount of voltage required to produce a particular current through linear
impedance Z1 to the amount of voltage required to produce the same amount of current through nonlinear impedance Z2. The total will be the amount of voltage required to produce that particular
current through the series combination. For example, point a (25 volts) is added to point c (50 volts) which yields point e (75 volts); and point b (50 volts) is added to point d (100 volts) which
yields point f (150 volts). Intermediate points may be determined in the same manner and the resultant characteristic curve (Z1 + Z2) is obtained for the series combination.
Figure 1-9A.—Combined linear and nonlinear impedances.
Figure 1-9B.—Combined linear and nonlinear impedances.
You should see from this graphic analysis that when a linear impedance is combined with a nonlinear impedance, the resulting characteristic curve is nonlinear. Some examples of nonlinear impedances
are crystal diodes, transistors, iron-core transformers, and electron tubes.
Figure 1-10 illustrates an ac sine-wave generator applied to a circuit containing several linear impedances. A sine-wave voltage applied to linear impedances will cause a sine wave of current through
them. The wave shape across each linear impedance will be identical to the applied waveform.
Figure 1-10.—Sine wave generator applied to several impedances.
The amplitude, on the other hand, may differ from the amplitude of the applied voltage. Furthermore, the phase of the voltage developed by any of the impedances may not be identical to the phase of
the voltage across any of the other impedances or the phase of the applied voltage. If an impedance is a reactive component (coil or capacitor), voltage or current may lead or lag, but the wave shape
will remain the same. In a linear circuit, the output of the generator is not distorted. The frequency remains the same throughout the entire circuit and no new frequencies are generated.
View (A) of figure 1-11 illustrates a circuit that contains a combination of linear and nonlinear impedances with a sine wave of voltage applied. Impedances Z2, Z3, and Z4 are linear; and Z1 is
nonlinear. The result of a linear and nonlinear combination of impedances is a nonlinear waveform. The curve Z, shown in view (B), is the nonlinear curve for the circuit of view (A). Because of the
nonlinear impedance, current can flow in the circuit only during the positive alternation of the sine-wave generator. If an oscilloscope is connected, as shown in view (A), the waveform across Z3
will not be a sine wave. Figure 1-12, view (A), illustrates the sine wave from the generator and view (B) shows the waveform across the linear impedance Z3. Notice that the nonlinear impedance Z1 has
eliminated the negative half cycles.
Figure 1-11A.—Circuit with nonlinear impedances.
Figure 1-11B.—Circuit with nonlinear impedances.
Figure 1-12A.—Waveform in a circuit with nonlinear impedances.
Figure 1-12B.—Waveform in a circuit with nonlinear impedances.
The waveform in view (B) is no longer identical to that of view (A) and the nonlinear impedance network has generated HARMONIC FREQUENCIES. The waveform now consists of the fundamental frequency and
its harmonics. (Harmonics were discussed in NEETS, Module 9, Introduction to
Wave-Generation and Wave-Shaping Circuits.)
A circuit composed of two sine-wave generators, G1 and G2, and two linear impedances, Z1 and Z2, is shown in figure 1-13. The voltage applied to Z1 and Z2 will be the vector sum of the generator
voltages. The sum of the individual instantaneous voltages across each impedance will equal the applied voltages.
Figure 1-13.—Two sine-wave generators with linear impedances.
If the two generator outputs are of the same frequency, then the waveform across Z1 and Z2 will be a sine wave, as shown in figure 1-14, views (A) and (B). No new frequencies will be created.
Relative amplitude and phase will be determined by the relative values and types of the impedances.
Introduction to Matter, Energy, and Direct Current,
Introduction to Alternating Current and Transformers,
Introduction to Circuit Protection, Control, and Measurement
Introduction to Electrical Conductors, Wiring Techniques, and Schematic Reading
Introduction to Generators and Motors
Introduction to Electronic Emission, Tubes, and Power Supplies, Introduction to Solid-State Devices and Power Supplies
Introduction to Amplifiers, Introduction to Wave-Generation and Wave-Shaping Circuits
Introduction to Wave Propagation, Transmission Lines, and Antennas
Microwave Principles, Modulation Principles
, Introduction to Number Systems and Logic Circuits, Introduction to Microelectronics,
Principles of Synchros, Servos, and Gyros
Introduction to Test Equipment
Radio-Frequency Communications Principles
Radar Principles,
The Technician's Handbook, Master Glossary,
Test Methods and Practices,
Introduction to Digital Computers, Magnetic Recording, Introduction to Fiber Optics
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For potential Ph.D. students
Over the next few years, I may take on a few additional Ph.D. students, although times may come when I'll be too full (e.g. a time that ended recently). This page is intended for those considering
working with me, although it also contains some tips for graduate students in general, as well as an idea of what I expect.
Algebraic geometry (or at least my take on it) is a technical subject that also requires a good deal of background in other subjects, as well as geometric intuition. So before I take you on as a new
student, you should be comfortable with the foundations of the subject, which means having done the majority of the exercises in Hartshorne or my course notes, and being able to explain them on
demand. (You shouldn't do this on your own; I'm happy talking with you through this process.) You should also be actively interested in learning about nearby subjects that interest you. Which
subjects they are is up to you. If you're not interested in regularly attending talks, and being broadly interested in mathematics outside of your thesis topic, or if you don't feel like getting
technical in a rather serious way, I'm probably not a good fit for you.
If you are interested in some of the ideas of algebraic geometry, you should also consider a number of other advisors. In this department there are a good number of people interested either directly
or indirectly in algebro-geometric ideas. You can read about them here. I will of course be happy to talk with you no matter whom you are working with.
My personal style as an advisor
I'll suggest problems to think about, starting from small toy problems (which have a habit of growing into interesting serious research). You'll have to pick what to work on, and find your own thesis
problem. Mathematics isn't just about answering questions; even more so, it is about asking the right questions, and that skill is a difficult one to master.
I like to meet my students every week (except for exceptional weeks, of which there are many). You may prefer not to meet in a given week if you have nothing much to report, but those weeks are
particularly important to meet.
The disadvantage of being a student of a young parent is that you'll have to be prepared to be more independent.
I will be a demanding advisor, more demanding than most.
I have pretty broad interests in and near algebraic geometry. To get an idea of the things I think about, see some of the things I've written. However, some of those subjects may not be ideal for a
Ph.D. student for a number of reasons. I'm interested in lots of things. I may however not be the ideal person to supervise lots of things. For example, I will not supervise a thesis in a nearby
field. But I definitely do not require that you work on problems directly related to my own research.
General advice (which would apply particularly to my own students)
Think actively about the creative process. A subtle leap is required from undergraduate thinking to active research (even if you have done undergraduate research). Think explicitly about the process,
and talk about it (with me, and with others). For example, in an undergraduate class any Ph.D. student at Stanford will have tried to learn absolutely all the material flawlessly. But in order to
know everything needed to tackle an important problem on the frontier of human knowledge, one would have to spend years reading many books and articles. So you'll have to learn differently. But how?
Don't be narrow and concentrate only on your particular problem. Learn things from all over the field, and beyond. The facts, methods, and insights from elsewhere will be much more useful than you
might realize, possibly in your thesis, and most definitely afterwards. Being broad is a good way of learning to develop interesting questions.
When you learn the theory, you should try to calculate some toy cases, and think of some explicit basic examples.
Talk to other graduate students. A lot. Organize reading groups. Also talk to post-docs, faculty, visitors, and people you run into on the street. I learn the most from talking with other people.
Maybe that's true for you too.
On seminars:
• Older graduate students will verify that there is a high correlation between those students who are doing the broadest and deepest work and those who are regularly attending seminars. Many people
erroneously conclude that those who are the strongest students therefore go to seminars, while in fact the causation goes very much in the opposite direction.
• Go to research seminars earlier than you think you should. Do not just go to seminars that you think are directly related to what you do (or more precisely, what you currently think you currently
do). You should certainly go to every single seminar related to algebraic geometry that you can, and likely drop by other seminars occasionally too. Learning to get information out of research
seminars is an acquired skill, usually acquired much later than the skill of reading mathematics. You may think it isn't helpful to go to a seminar where you understand just 5% of what the
speaker says, and may want to wait until you are closer to 100%; but no one is anywhere near 100% (even the speaker!), so you should go anyway.
• Try to follow the thread of the talk, and when you get thrown, try to get back on again. (This isn't always possible, and admittedly often the fault lies with the speaker.)
• At the end of the talk, you should try to answer the questions: What question(s) is the speaker trying to answer? Why should we care about them? What flavor of results has the speaker proved? Do
I have a small example of the phenonenon under discussion? You can even scribble down these questions at the start of the talk, and jot down answers to them during the talk.
• Try to extract three words from the talk (no matter how tangentially related to the subject at hand) that you want to know the definition of. Then after the talk, ask me what they mean. (In
general, feel free to touch base with me after every seminar. I might tell you something interesting related to the talk.)
• New version of the previous jot: try the "three things" exercise.
• See if you can get one lesson from the talk (broadly interpreted). If you manage to get one lesson from each talk you go to, you'll learn a huge amount over time, although you'll only realize
this after quite a while. (If you are unable to learn even one thing about mathematics from a talk, think about what the speaker could have done differently so that you could have learned
something. You can learn a lot about giving good talks by thinking about what makes bad talks bad.)
• Try to ask one question at as many seminars as possible, either during the talk, or privately afterwards. The act of trying to formulating an interesting question (for you, not the speaker!) is a
worthwhile exercise, and can focus the mind.
• Here's a phenomenon I was surprised to find: you'll go to talks, and hear various words, whose definitions you're not so sure about. At some point you'll be able to make a sentence using those
words; you won't know what the words mean, but you'll know the sentence is correct. You'll also be able to ask a question using those words. You still won't know what the words mean, but you'll
know the question is interesting, and you'll want to know the answer. Then later on, you'll learn what the words mean more precisely, and your sense of how they fit together will make that
learning much easier. The reason for this phenomenon is that mathematics is so rich and infinite that it is impossible to learn it systematically, and if you wait to master one topic before
moving on to the next, you'll never get anywhere. Instead, you'll have tendrils of knowledge extending far from your comfort zone. Then you can later backfill from these tendrils, and extend your
comfort zone; this is much easier to do than learning "forwards". (Caution: this backfilling is necessary. There can be a temptation to learn lots of fancy words and to use them in fancy
sentences without being able to say precisely what you mean. You should feel free to do that, but you should always feel a pang of guilt when you do.)
• Your thesis problem may well come out of an idea you have while sitting in a seminar.
• Go to seminar dinners when at all possible, even though it is scary, and no one else is going.
• Go to colloquia fairly often, so you have a reasonable idea of what is happening in other parts of mathematics. It is amazing what can become relevant to your research. You won't believe it until
it happens to you. And it won't happen to you unless you go to colloquia. Ditto for seminars in other fields.
On giving talks
Here's a great story from Mark Meckes that simultaneously illustrates a number of points. By chance, I recently saw a PhD thesis whose acknowledgements ended with the sentence "Finally, I would like
to thank Dr. Mark Meckes, whose talk in Marseille in May of this year [2008] provided the final insight I needed to completely answer Kuperberg's Conjecture." What is interesting about this is that
not only had I never heard of Kuperberg's Conjecture, but my talk was completely unrelated to the subject of the thesis, and even after reading the relevant section of the thesis I still couldn't see
the connection. So one truly never knows where useful insights will come from. One of the many things I love about this story is that I don't find it at all surprising! So go to talks --- and give
talks --- and talk to people!
On writing:
• For people writing research papers for the first time (or not for the first time), here is a lecture by Serre, one of the best mathematical writers of all time, with some opinions on good (and
bad) writing.
• Terry Tao on writing.
When thinking about advisors, talk to past and current graduate students. (My former and current students: Eric Katz 2004, Rob Easton 2007, Andy Schultz 2007, Jarod Alper 2008, Joe Rabinoff 2009,
Nikola Penev 2009, Jack Hall 2010, Dung Nguyen 2010, Atoshi Chowdhury, Yuncheng Lin, Daniel Litt. I also collaborated with Kirsten Wickelgren 2009, who worked with Gunnar Carlsson.)
Advice from others:
Specific advice about algebraic geometry at Stanford
Sign up for the algebraic geometry mailing list.
Go to the Western Algebraic Geometry Seminar, a twice-yearly conference.
Occasionally go to Berkeley when you hear about something particularly interesting.
When you are up to it, subscribe to the daily mailing of abstracts of algebraic geometry papers posted to the arXiv. Then most days, just delete them, but when you have some time, browse through
them, and read the abstracts that catch your eye. You'll gradually get a sense of what is going on in the field. Caution: this can be psychologically damaging, as you'll feel "here I am stuck on this
simple problem, and thousands of papers are coming out...". So only do this if and when you're ready. I might delete this paragraph at some point if I realize it is counterproductive.
(Thanks to many people for advice about this page, including Yvonne Lai, Daniel Erman, and Mark Meckes.)
Return to my homepage
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Compactness of the class of connected sets with perimetre smaller than 1?
up vote 8 down vote favorite
It is my first post, so please be indulgent!
Here is the problem: I am in the class S of closed subsets of [0,1]^2 that are connected and have perimeter less or equal to 1.
I endow this space with the Hausdorff metric (or equivalently Fell topology), that says that for two compacts A, B, d(A,B)<=r iff every point of A is at distance less than r of B, and the other way
Question: Is this space compact?
Equivalent question: Is the function "perimeter" lower semi continuous on this set? (It is equivalent because the Hausdorff metric is compact for the class of all compact sets, and the class of
connected sets is closed).
In other words, given a sequence of connected sets with perimeter at most 1, is it possible to see suddenly some perimeter appear at the limit? (which would be an answer "no" to the question).
The perimeter here is the 1-dimensional measure of the boundary, i.e. the infimum over all coverings of the boundary by balls of the sum of the diameters of the balls.
Remark: If one drops the assumption of connectedness, it is not true anymore (consider for instance a set of points -thus with zero perimeter- that becomes dense in the square). So this assumption is
very important!
Thak you in advance!
Hi Remark: I don't know the answer to your question, but here's a potential strategy. As Tom Leinster has been explaining for some time now (see eg his posts on n-Category Cafe), another good
definition of perimeter of a closed set $X$ is $\frac{\partial}{\partial\epsilon} \bigr|_{\epsilon=0} \mathrm{Vol}(X_\epsilon)$, where $X_\epsilon$ is the set of all points within distance ϵ of X.
Perhaps this is the same as your definition. Anyway, then perhaps you can successfully swap a limit somewhere when computing a limit in Hausdorff topology. – Theo Johnson-Freyd Aug 3 '11 at 19:06
add comment
1 Answer
active oldest votes
No. Let $B$ be a closed ball of radius 1/2 and $I$ a diameter of $B$. Construct a Cantor-like set $K\subset I$ of lengths $\mathcal H^1(K)=0.9$ (where $\mathcal H^1$ denotes the
1-dimensional Hausdorff measure). We have $I\setminus K=\bigcup I_i$ where $I_1,I_2,\dots$ are disjoint open subintervals of $I$ and $\sum_{i=1}^\infty\mathcal H_1(I_i)=0.1$. Let $B_i$
be the open ball for which $I_i$ is a diameter, and let $X_k=B\setminus\bigcup_{i=1}^k B_i$. Then each $X_k$ is a closed connected set and the sequence $\{X_k\}$ Hausdorff converges to
$X=\bigcap_{k=1}^\infty X_k=B\setminus\bigcup_{i=1}^\infty B_i$.
up vote 12 The boundary of $X_k$ is a union of $k+1$ disjoint circles $\partial B$ and $\partial B_i$, $1\le i\le k$, hence $$ \mathcal H^1(\partial X_k) = \mathcal H^1(\partial B)+\sum_{i=1}^k \
down vote mathcal H^1(\partial B_i) = \pi+\sum_{i=1}^k\pi\mathcal H^1(I_i) \le \pi+0.1\pi =: C. $$ However the boundary of $X$ contains $K$, hence $\mathcal H^1(\partial X)\ge\mathcal H^1(\
accepted partial B)+\mathcal H^1(K)=\pi+0.9> C$.
Thus the space of connected sets of perimeter $\le C$ is not closed.
Nice example. It might be interesting to notice that when you consider a subset of the boundary with the Cantor set is removed, the result is true. See R. Cerf, The Hausdorff Lower
Semicontinuous Envelope Of The Length In The Plane Ann. Scuola Norm. Sup. Pisa Cl. Sci. (5) I (2002), 33$-$71. archive.numdam.org/ARCHIVE/ASNSP/ASNSP_2002_5_1_1/… – Tapio Rajala Aug 4
'11 at 9:51
Very nice example (I actually believed the answer would be yes), Cantor sets are even more vicious than I thought. Thank you very much! – Raphael L Aug 5 '11 at 8:25
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Not the answer you're looking for? Browse other questions tagged gt.geometric-topology or ask your own question.
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Image self consistency from xkcd
I love xkcd. A comic combining fun and math by definition has to be good and geeky and the author, Randall Munroe, is a real genius on this. The latest comic is pretty interesting
The image is self-descriptive, meaning that each graph represents information about the image itself. For example, the first panel contains a pie chart which says how many pixels are either white or
black on the image. Clearly, the relative amount of black pixels in the image depends on the size of the slice of that piechart representing the amount of black pixels, a “chicken-egg” kind of
problem. It is apparently difficult to obtain such image, because the plotted data must be consistent with themselves via the graphical representation. This kind of problems, where the solution
depends on itself, is quite common in many scientific problems, and it’s solved through self-consistency.
The trick is as follows: we start with a first, approximate solution, called a guess, and we apply a method that gives us a result depending on this guess. Then, we take this newly obtained result,
and reapply the method again, to obtain a new result, and then again, and again, until, hopefully, the input and the output of the method are the same. When this occurs, we solved our problem via
self-consistency. Of course, this convergence is not guaranteed to occur, but if it occurs, we found a solution (there could be more than one).
Let’s see it in action in a simplified form. I wrote two small python programs. They use matplotlib and the Python Image Library. The first (called piechart.py) creates a pie chart from a given data
import sys
from matplotlib import pyplot
white = int(sys.argv[1])
black = int(sys.argv[2])
pyplot.pie([white, black], colors=('w', 'k'))
pyplot.savefig(sys.argv[3], format="pdf")
If we call this program specifying two values (the absolute values are not important, as the pie chart shows relative amount), it draws the pie chart accordingly:
python piechart.py 100 400 piechart_100w_400b.pdf
convert -geometry 210x158 piechart_100w_400b.pdf piechart_100w_400b.png
This creates a pie chart where white is 1/5 of the pie chart area and black is 4/5. Please note that due to a setup problem of my matplotlib I can only create pdf, so I convert the pdf into png of
defined size, in our case, 210×158, using the convert program. The total size of the image is of course important, having an influence on the total number of pixels. I chose a good value for
presentation purposes which guarantees quick convergence.
The second program is called imagedata.py and extracts size and number of white and black pixels from an image.
import sys
from PIL import Image
im = Image.open(sys.argv[1])
white = 0
black = 0
for i in im.getdata():
if i == (255,255,255):
white += 1
# we assume black everything that is not white:
black += 1
print im.size[0],im.size[1],white,black
If we run this program on the png image, it will tell us how many pixels are white, and how many are black.
$ python imagedata.py piechart_100w_400b.png
Of the 33.180 pixels defining the full image above (border included, not only the pie chart circle), 23988 are white (72%), and 9192 are black (28%). Hence the image is not representing itself: the
plot represents our initial values of 20 % white and 80 % black.
Now we create a new image, in agreement with the iterative procedure, passing the most recently obtained values
python piechart.py 23988 9192 piechart_23988w_9192b.pdf
convert -geometry 210x158 piechart_23988w_9192b.pdf piechart_23988w_9192b.png
and repeat the process. This becomes tedious very soon, so I wrote a driver (driver.sh) to perform the process for me
# generates the starting guess
python piechart.py 100 400 iter_0.pdf
convert -geometry 210x158 iter_0.pdf iter_0.png
# iterative process
echo "step w h white black"
while true;
data=`python imagedata.py iter_$(($step-1)).png`
echo "$step - $data"
python piechart.py `echo $data|awk '{print $3}'` `echo $data|awk '{print $4}'` iter_$step.pdf
convert -geometry 210x158 iter_$step.pdf iter_$step.png
If we run it, we immediately see a very interesting result:
step w h white black
1 - 210 158 23988 9192
2 - 210 158 29075 4105
3 - 210 158 30551 2629
4 - 210 158 30977 2203
5 - 210 158 31108 2072
6 - 210 158 31158 2022
7 - 210 158 31164 2016
8 - 210 158 31169 2011
9 - 210 158 31172 2008
10 - 210 158 31172 2008
11 - 210 158 31172 2008
12 - 210 158 31172 2008
The number of black pixels decreases, and the number of white ones increases. At every step, the image slightly changes, until it reaches a point where it does not change anymore: it achieved
self-consistency, and it is representing itself. This is a movie of the various steps until convergence
What if we started from the other direction, namely, with a guess containing zero as the number of black pixels? The result would have been the same
1 - 210 158 31750 1430
2 - 210 158 31320 1860
3 - 210 158 31221 1959
4 - 210 158 31184 1996
5 - 210 158 31178 2002
6 - 210 158 31174 2006
7 - 210 158 31172 2008
8 - 210 158 31172 2008
9 - 210 158 31172 2008
Again, even with a different starting guess, we obtain the same result, here depicted as a movie
I hope this gave a brief explanation on how Randall achieved the self-consistent image. His case was more complex, having three plots. Also, the comic is scribbled, so either he drew it by hand,
approximating the computed result, or he performed some scribble-like transformation preserving the pixel count. I assume it is the former.
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Calculus Tutors
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MathGroup Archive: April 2010 [00069]
[Date Index] [Thread Index] [Author Index]
Re: Combining elements of a list
• To: mathgroup at smc.vnet.net
• Subject: [mg108850] Re: Combining elements of a list
• From: Murray Eisenberg <murray at math.umass.edu>
• Date: Sun, 4 Apr 2010 07:44:11 -0400 (EDT)
I'm going to assume you don't really have symbols (t, h, etc.) but
rather one-letter strings ("t", "h", etc.) here.
The basic function you want is StringJoin. Thus:
StringJoin[{"t", "h", "i", "s"}]
Next, you have your list of lists:
lis = {{"t", "h", "i", "s"}, {"i", "s"}, {"a"}, {"t", "e", "s", "t"}};
(I left out the last word.) So Map the function StringJoin onto that at
level 1:
Map[StringJoin, lis, 1]
{"this", "is", "a", "test"}
Now you need to append a blank to each word (except the last). For a
single string str StringInsert[str," ",-1] does it, as in:
trial=StringInsert["this"," ",-1]
(* you can't see the trailing blank here, so... *)
(You could also use StringJoin instead of StringInsert.)
Now Map that function of appending a blank onto the list of words:
appended = Map[StringInsert[#, " ", -1] &, words]
{this ,is ,a ,test }
Same thing, abbreviated:
StringInsert[#, " ", -1] & /@ words
Finally, join all the individual (blank-trailed) words into a single
string and delete the final trailing blank:
Obviously, all those steps could be combined, encapsulated into a single
sentenceFromLetters[lis_] :=
StringJoin[StringInsert[#, " ", -1] & /@ Map[StringJoin, lis, 1]]
Doubtless there are terser, or at least alternative ways, to do this
using pattern-matching for at least some of the steps.
If your original list really consisted of lists of symbols rather than
lists of one-character strings, you could first convert that to a list
of lists of one-character strings by using ToString.
On 4/3/2010 7:09 AM, ebaugh at illinois.edu wrote:
> How could I go from:
> {h,e,l,l,o}
> to
> {hello}?
> That is the first step of what I am really looking to do. Which is
> to go from:
> {{t,h,i,s},{i,s},{a},{t,e,s,t},{m,e,s,s,a,g,e}}
> to
> "this is a test message"
> Thanks all,
> Jason Ebaugh
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
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Dominoes program... HELP PLEASE!
04-27-2010, 11:51 PM
Dominoes program... HELP PLEASE!
I have attached the assignment given by my professor. I'm just a little confused. I can't develop and algorithm for this. I know I need a 2-D array for the board, but I don't know how to fill it
and then rearrange the Dominoes to find the number of different ways to fill the board. If I'm not clear just tell me and I will clarify.
04-27-2010, 11:55 PM
my file didn't attach so here it is:
CSCE 1040.001/.002
Program # 4, 100 points
Due: April 30, 2010 at 10:00pm.
You are to write a Java method which will count the number of different ways that a 3xN board can be completely filled with dominos (a domino is 1x2). Completely, in this case, means there are no
unfilled cells in the board. The best way to fill the board is to fill columns left-to-right using recursion (see Chapter 11 of the text). The value of N (N>1) will be passed to your program via
Dominoes may be placed horizontally or vertical. At any position, both possible domino placements should be tried. I.e – placing a domino horizontally.
Your program (main) is to call the method defined above and produce one number, the number of possible ways to completely fill a 3xN board, with appropriate formatting and description, of course.
Also, your program should contain appropriate comments.
Submit a single java file via Blackboard.
There are three ways to completely fill a 3 x 2.
There are zero ways to completely fill a 3 x 3.
The board can be a 2-dimentional integer array. Your program must work for any value of N (N>1). Therefore you must allocate memory for any arrays dynamically, either in main or inside of a
I strongly recommend that you write a printBoard method to help you with debugging. A good way to keep track of dominoes (mark them on the board) is to number them sequentially and to print them
out mod 10. Thus, the first solution of the 3x3 would be printed as
One possible way to split up the work would be to write the two methods.
1. tryHorizontal(row, col) which would try to place a domino horizontally at board[row, col] and board[row, col+1], checking, of course, for conditions like board position already filled, placing
part of a domino outside of the board, etc.
2. and tryVertical(row, col) which would try to place a domino vertically.
These two methods would be used inside of a method tryRowCol(row, col).
public tryRowCol( int row, int col){
if (.…) tryHorizontal(row, col);
if (…..) tryVertical(row, col);
} // end tryRowCol
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Smooth Differential Graded Algebra?
Posted by Urs Schreiber
A while ago I had a discussion with Todd Trimble about how to define “generalized smooth” differential graded-commutative algebras (DGCAs), generalizing the “generalized smooth”-algebras, called $C^\
infty$-algebras, discussed in the book by Moerdijk & Reyes. I think back then we fell short of arriving at a satisfactory conclusion.
As I mentioned, I would like to pick up that thread again and chat about some ideas.
This here is the general motivation:
given a category $S$ of “test objects”, generalized spaces are presheaves on $S$, namely things that can be probed by throwing objects of $S$ into them, and generalized quantities (to be thought of
as numbers, functions, sections, etc. as discussed in more detail below) are co-presheaves on $S$.
If $S := CartesianSpaces$ is the full subcategory of Manifolds on the manifolds $\mathbb{R}^n$, for all $n \in \mathbb{N}$, then the spaces in question are something like “smooth spaces” (in
particular, if their underlying presheaves happen to be concrete sheaves, these these are diffeological or Chen-smooth spaces) and then the quantities are something like “smooth quantities” (in
particular, if the underlying co-presheaves happen to be monoidal, these are those $C^\infty$-algebras).
Given such a notion of spaces, there is an obvious notion of higher spaces: pick your favorite definition of $\infty$-groupoid. Then a higher degree space, an $\infty$-space, should be an $\infty$
-groupoid internal to the above spaces.
What is the analog of this on the side of “quantities”? What is an $\infty$-quantity? There are several possible answers one could come up with, I suppose, such as the answer by David Spivak, who
replaces co-presheaves by simplicial co-presheaves and hence essentially follows the above $\infty$-ization of spaces.
But here I am interested in a different kind of answer which supposes that $\infty$-quantities corresponding to $\infty$-spaces in the above sense are something like “quasi-free differential graded
commutative $C^\infty$-algebras” , $C^\infty$qDGCAs – to be be determined.
There is a reason for this assumption, namely $\infty$-Lie theory, but that is not of concern right now. Here I just want to talk about possible definitions of $C^\infty$-qDGCAs, an interesting
question in its own right.
The idea I want to propose is simple. The goal is to have it “as simple as possible but no simpler”. Maybe you can help me check if that’s achieved, especially concerning the “no simpler”-part (i.e.
the mistakes).
Here goes:
Write $Quantities := Set^{CartesianSpaces}$ for the category of co-presheaves on CartesianSpaces. As a co-presheaf category, this is a monoidal category with tensor product of $A,B \in Quantities$
given by $A \times B : \mathbb{R}^k \mapsto A(\mathbb{R}^k) \times B(\mathbb{R}^k) \,.$
In any monoidal category we can consider monoids:
Write $Algebras := Monoids(Quantities)$
for the category of monoids internal to Quantities. Every $C^\infty$-algebra of Moerdijk-Reyes canonically becomes an object of Algebras by using postcomposition with the maps $\cdot^k : \mathbb{R}^k
\times \mathbb{R}^k \to \mathbb{R}^k$ of componentwise multiplication in $\mathbb{R}$.
Write $Spaces := Sheaves(CartesianSpaces)$ for the category of sheaves on CartesianSpaces. For every $X \in Spaces$ we get an object $C^\infty(X) \in Algebras$, the algebra of functions whose
underlying co-presheaf is $C^\infty(X) : \mathbb{R}^k \mapsto Hom_{Spaces}(X, \mathbb{R}^k)$ and whose monoidal structure comes from postcomposition with the $\cdot^k$ from above: $Hom(X, \mathbb{R}^
k) \times Hom(X, \mathbb{R}^k) \stackrel{\simeq}{\to} Hom(X, \mathbb{R}^k \times \mathbb{R}^k) \stackrel{Hom(-,\cdot^k)}{\to} Hom(X, \mathbb{R}^k) \,.$ For $X$ an ordinary manifold, $C^\infty(X)$ is
its ordinary algebra of smooth functions.
Now comes the point: Since the $Algebras$ above are monoid objects, we can consider modules internal to $Quantities$ of objects in $Algebras$. (In fact, there should be a monoidal bicategory $Bimod
Let $E \to X$ be a vector bundle internal to $Spaces$ and consider the set $Sections(E) \in Sets$ of its sections. The assignment $\Gamma(E) : \mathbb{R}^k \mapsto Sections(E \otimes \mathbb{R}^k)$
extends naturally to a co-presheaf on $CartesianSpaces$, hence to an object in $Quantities$. This naturally comes with an action of $C^\infty(X) \in Algebras$, where in components the action is given
by postcomposition with $\cdot^k$ acting on the trivial bundle part: $\array{ Hom(X,\mathbb{R}^k) \times Sections(E \otimes \mathb{R}^k) \\ \downarrow^{\subset} \\ Hom(X,\mathbb{R}^k) \times Hom(X,E
\otimes \mathbb{R}^k) \\ \downarrow^\simeq \\ Hom(X,( E \otimes \mathbb{R}^k) \times \mathbb{R}^k ) \\ \downarrow \\ Hom(X,E \otimes \mathbb{R}^k) }$
For $C \in Quantities$ an $(A \in Algebras)$-module, There is the obvious notion of dual-over-$A$ module $C^* := Hom_{A-Modules}(C,A)$. Using all this, the standard definition of qDGCAs should now
straightforwardly generalize to the generalized smooth context:
Definition: A quasi-free differential graded-commutative algebra (qDGCA) over $A \in Algebras$ is a non-positively graded cochain complex $V$ of $A$-modules internal to $Quantities$ together with a
degree +1 differential $d : \wedge^\bullet_A V^* \to \wedge^\bullet_A V^*$.
I am thinking that all the ingredients I glossed over here have the obvious straightforward definition. But maybe I should check this in more detail.
(One might want to add to the above definition the condition that $V$ is degree-wise projective.)
Posted at September 9, 2008 9:33 AM UTC
Re: Smooth Differential Graded Algebra?
How does Lawvere’s treatment of intensive and extensive quantities fit with what you’re saying?
Posted by: David Corfield on September 9, 2008 12:07 PM | Permalink | Reply to this
Re: Smooth Differential Graded Algebra?
Lawvere’s treatment of intensive and extensive quantities
So, from p. 14, we have
“extensive quantity” = homology cycles
“intensive quantitiy” = cohomology cocycle
On page 15 Grassman’s “extensive quantities” are mentioned. Now, that’s interesting, since these are of course the elemens of a Grassman algbra $\wedge^\bullet V$ which play a central role in the “$C
^\infty qDGCAs$” mentioned above.
Still reading…
Posted by: Urs Schreiber on September 9, 2008 12:29 PM | Permalink | Reply to this
Re: Smooth Differential Graded Algebra?
Perhaps a candidate for a tac reprint. It would be nice to know what happens on pp. 18-19 and pp. 25-26.
Posted by: David Corfield on September 9, 2008 12:33 PM | Permalink | Reply to this
Re: Smooth Differential Graded Algebra?
I did some searching. If I understand correctly then an “intensive quantity” is taken to be essentially one that depends contravariantly on Spaces, while an “extensive quantity” is essentially one
that depends covariantly.
So function algebras, $C^\infty : Spaces \to Algebras$ is an intensive quantity, while linear duals of such (distributions) are “extensive”.
I must admit that I haven’t figured out yet in which sense the words “intensive” and “extensive” are supposed to be suggestive here.
But in any case I suppose I can now answer your question: the quantities I was talking about in the above entry are “intensive”, in this sense (essentially being functions and sections on spaces).
Posted by: Urs Schreiber on September 9, 2008 12:50 PM | Permalink | Reply to this
Re: Smooth Differential Graded Algebra?
A classic example of the difference between intensive and extensive properties is mass (extensive) and density (intensive).
Intensive quantities work well with products via projection.
Extensive quantities work well with coproducts via sum.
Posted by: David Corfield on September 9, 2008 1:20 PM | Permalink | Reply to this
Re: Smooth Differential Graded Algebra?
Is it an issue that normally one thinks of multiplying an intensive quantity by an extensive quantity to yield an extensive quantity (density $\times$ volume = mass or integrating a density against a
measure), where you are multiplying intensive quantities?
Posted by: David Corfield on September 10, 2008 2:25 PM | Permalink | Reply to this
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Rebuilding the Tower of Hanoi
The Tower of Hanoi is a puzzle that consists of three pegs and a set of disks. Each disk has a different diameter and a hole in the middle so that the disk can fit onto any of the pegs. The initial
puzzle setup has two of the pegs empty and all the disks on the third peg (source) in monotonically decreasing order of diameter from bottom to top, which forms a structure that is reminiscent of a
tower. The goal of the puzzle is to move the tower from the source peg to a specified peg (destination) using the other peg to temporarily hold disks. The two rules of the Tower of Hanoi puzzle are
that only one disk at a time can be moved from the top of a stack of disks on a peg to some other peg, and disks with larger diameter cannot be placed on top of smaller diameter disks.
French mathematician Edouard Lucas developed the puzzle in 1883. He also created the myth for his puzzle that claimed monks have been working non-stop on a 64-disk version from the beginning of time.
Once they have solved this problem, the tower will crumble and the world will end. It's a good story and, I'm sure, it helped to promote the puzzle, but everyone knows that the world ends at 03:14:07
on Tuesday, January 19, 2038.
A programmed solution can be implemented using recursion in a most elegant way. (Again with the recursion? Just one more to finish up, I swear. I'll start my next post with something completely
different.) Here's one way to code up a solution.
void tower(char source, char dest, char temp, int level)
if (level > 0) {
tower(source, temp, dest, level-1);
printf("Move from %c to %c\n", source, dest);
tower(temp, dest, source, level-1);
printf("Move from %c to %c\n", source, dest);
. . .
tower('A', ‘C', 'B', numDisks-1);
. . .
I've used single characters to denote the pegs ('A' is the initial source peg, 'C' is the final destination, and 'B' is the temporary peg). The recursion is halted when the level parameter reaches
zero. This corresponds to the source peg having one disk remaining on it, so the code simply prints the move of a disk from the source to the destination peg without further recursive calls. As you
can see, the purpose of the code is to print out the set of instructions that can be mechanically followed by someone trying to solve the puzzle in the fewest moves.
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Math Forum Discussions
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Patent application title: SORTING A DATASET OF INCREMENTALLY RECEIVED DATA
Inventors: Jeremy Eric Elson (Seattle, WA, US) Edmund Bernard Nightingale (Redmond, WA, US) Owen Sebastian Hofmann (Austin, TX, US)
Assignees: Microsoft Corporation
IPC8 Class: AG06F1730FI
USPC Class: 707752
Class name:
Publication date: 2012-12-27
Patent application number: 20120330979
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A method of sorting a dataset includes incrementally receiving data from the dataset, and incrementally storing the received data as individual input data subsets as the data is received, thereby
sequentially generating a plurality of filled data subsets of unsorted data. The method includes individually sorting each filled data subset of unsorted data concurrently with receiving data for a
next one of the individual input data subsets, thereby sequentially generating a plurality of sorted input data subsets, and performing a merge sort on the plurality of sorted input data subsets,
thereby incrementally generating a sorted version of the dataset.
A method of sorting a dataset, comprising: incrementally receiving data from the dataset; incrementally storing the received data as individual input data subsets as the data is received, thereby
sequentially generating a plurality of filled data subsets of unsorted data; individually sorting each filled data subset of unsorted data concurrently with receiving data for a next one of the
individual input data subsets, thereby sequentially generating a plurality of sorted input data subsets; and performing a merge sort on the plurality of sorted input data subsets, thereby
incrementally generating a sorted version of the dataset.
The method of claim 1, wherein the sorted version of the dataset includes a plurality of sequentially generated sorted output data subsets, and wherein the method further comprises: outputting each
of the sorted output data subsets concurrently with generating a next one of the sorted output data subsets.
The method of claim 2, wherein the sorted output data subsets each have a same size as the individual input data subsets.
The method of claim 2, wherein the outputting each of the sorted output data subsets comprises outputting each of the sorted output data subsets to a storage medium.
The method of claim 2, wherein the outputting each of the sorted output data subsets comprises outputting each of the sorted output data subsets to a network file system.
The method of claim 1, and further comprising: varying a size of the individual input data subsets based on a size of the dataset.
The method of claim 1, wherein the individual input data subsets each have a size that is a predetermined fraction of a size of the dataset.
The method of claim 1, wherein the dataset is stored as a plurality of portions on a plurality of computing devices, and wherein the data from the dataset is incrementally received from the plurality
of computing devices.
The method of claim 1, wherein the individually sorting each filled data subset of unsorted data is performed using a quick-sort algorithm.
The method of claim 1, wherein the data incrementally received from the dataset is received from a storage medium.
The method of claim 1, wherein the data incrementally received from the dataset is received from a network file system.
A computer-readable storage medium storing computer-executable instructions that when executed by at least one processor cause the at least one processor to perform a method of sorting a dataset, the
method comprising: incrementally receiving data from the dataset; sequentially generating a plurality of filled data subsets by incrementally storing the received data as individual input data
subsets as the data is received; sequentially generating a plurality of sorted input data subsets by individually sorting each filled data subsets concurrently with receiving data for a next one of
the individual data subsets; and incrementally generating a sorted version of the dataset by performing a merge sort on the plurality of sorted input data subsets.
The computer-readable storage medium of claim 12, wherein the sorted version of the dataset includes a plurality of sequentially generated sorted output data subsets, and wherein the method further
comprises: outputting each of the sorted output data subsets concurrently with generating a next one of the sorted output data subsets.
The computer-readable storage medium of claim 13, wherein the sorted output data subsets each have a same size as the individual input data subsets.
The computer-readable storage medium of claim 12, wherein the method further comprises: varying a size of the individual input data subsets based on a size of the dataset.
The computer-readable storage medium of claim 12, wherein the individual input data subsets each have a size that is a predetermined fraction of a size of the dataset.
The computer-readable storage medium of claim 12, wherein the dataset is stored as a plurality of portions on a plurality of computing devices, and wherein the data from the dataset is incrementally
received from the plurality of computing devices.
The computer-readable storage medium of claim 12, wherein the individually sorting each filled data subsets is performed using a quick-sort algorithm.
The computer-readable storage medium of claim 12, wherein the filled data subsets have a non-uniform size.
A method of sorting a dataset, comprising: incrementally receiving data from the dataset; sequentially generating a plurality of filled data subsets by incrementally storing the received data as
individual input data subsets as the data is received; sequentially generating a plurality of sorted input data subsets by individually sorting each filled data subset concurrently with receiving
data for a subsequent one of the individual input data subsets; incrementally generating a sorted version of the dataset by performing a merge sort on the plurality of sorted input data subsets,
wherein the sorted version of the dataset includes a plurality of sequentially generated sorted output data subsets; and outputting each of the sorted output data subsets concurrently with generating
a subsequent one of the sorted output data subsets.
BACKGROUND [0001]
Sorting a large dataset is a problem commonly found in many applications. The total time required to sort a large dataset can be split into two parts: first, the input/output (I/O) delay in reading
all the unsorted data from stable storage (e.g., disk) and writing the sorted data back. Second, there are CPU requirements for comparing enough of the data elements sufficiently to sort them.
The I/O portion of the sorting process is typically much slower than computation, particularly if the amount of computation done per unit of data is small. The time to sort data tends to be dominated
by the time it takes to read or write the data from or to either the network or the storage medium (e.g. disk). This has changed in some recent storage systems, where I/O is dramatically faster than
in previous systems--often by an order of magnitude. When sorting is implemented on such systems, the time required for computation becomes more significant, and it becomes more significant to
optimize this portion of the sorting process.
SUMMARY [0003]
This summary is provided to introduce a selection of concepts in a simplified form that are further described below in the Detailed Description. This summary is not intended to identify key features
or essential features of the claimed subject matter, nor is it intended to be used to limit the scope of the claimed subject matter.
One embodiment is directed to system that splits unsorted input data into smaller subsets as it arrives, and sorts each input subset while the subsequent input subset is being read (or received, in
the case of a network file system). The system according to one embodiment performs a merge sort on the sorted subsets once the output stage begins, and performs a merge to produce an output subset
while the previous output subset is being written (or transmitted, in the case of a network file system).
One embodiment is directed to a method of sorting a dataset, which includes incrementally receiving data from the dataset, and incrementally storing the received data as individual input data subsets
as the data is received, thereby sequentially generating a plurality of filled data subsets of unsorted data. The method includes individually sorting each filled data subset of unsorted data
concurrently with receiving data for a next one of the individual input data subsets, thereby sequentially generating a plurality of sorted input data subsets, and performing a merge sort on the
plurality of sorted input data subsets, thereby incrementally generating a sorted version of the dataset.
BRIEF DESCRIPTION OF THE DRAWINGS [0006]
The accompanying drawings are included to provide a further understanding of embodiments and are incorporated in and constitute a part of this specification. The drawings illustrate embodiments and
together with the description serve to explain principles of embodiments. Other embodiments and many of the intended advantages of embodiments will be readily appreciated, as they become better
understood by reference to the following detailed description. The elements of the drawings are not necessarily to scale relative to each other. Like reference numerals designate corresponding
similar parts.
FIG. 1 is a block diagram illustrating a computing environment suitable for implementing aspects of a system for sorting a dataset according to one embodiment.
FIG. 2 is a block diagram illustrating a system for sorting a dataset according to one embodiment.
FIG. 3 is a flow diagram illustrating a method of sorting a dataset according to one embodiment.
DETAILED DESCRIPTION [0010]
In the following Detailed Description, reference is made to the accompanying drawings, which form a part hereof, and in which is shown by way of illustration specific embodiments in which the
invention may be practiced. It is to be understood that other embodiments may be utilized and structural or logical changes may be made without departing from the scope of the present invention. The
following detailed description, therefore, is not to be taken in a limiting sense, and the scope of the present invention is defined by the appended claims.
It is to be understood that features of the various exemplary embodiments described herein may be combined with each other, unless specifically noted otherwise.
In a naive implementation, a program might be split the sorting process into three stages: (1) read unsorted data; (2) sort; (3) write sorted data. One embodiment of the system disclosed herein
overlaps almost 100% of the compute time (step 2) with the time for reading (step 1) and the time for writing (step 3), reducing the total time for the second step to almost zero. Thus, the system
hides the majority of the compute time for sorting by overlapping it with the time for I/O.
One embodiment is directed to system that splits unsorted input data into smaller subsets as it arrives, and sorts each input subset while the subsequent input subset is being read (or received, in
the case of a network file system). The system according to one embodiment performs a merge sort on the sorted subsets once the output stage begins, and performs a merge to produce an output subset
while the previous output subset is being written (or transmitted, in the case of a network file system).
One potential method for sorting is to use an incremental sorting mechanism like heap sort. Each time a datum arrives, it can be added to the heap. In this way, in theory at least, all data can be
incrementally sorted as it arrives, and as soon as the last piece of data arrives the heap is entirely sorted and ready for output. However, it has been found that, in practice, this method is slow,
because it does not exploit the locality of reference required for good performance in the CPU's memory cache. Thus, one embodiment incrementally sorts data using a quick sort, which is more
FIG. 1 is a diagram illustrating a computing environment 10 suitable for implementing aspects of a system for sorting a dataset according to one embodiment. In the illustrated embodiment, the
computing system or computing device 10 includes one or more processing units 12 and system memory 14. Depending on the exact configuration and type of computing device, memory 14 may be volatile
(such as RAM), non-volatile (such as ROM, flash memory, etc.), or some combination of the two.
Computing device 10 may also have additional features/functionality. For example, computing device 10 may also include additional storage (removable and/or non-removable) including, but not limited
to, magnetic or optical disks or tape. Such additional storage is illustrated in FIG. 1 by removable storage 16 and non-removable storage 18. Computer storage media includes volatile and nonvolatile,
removable and non-removable media implemented in any suitable method or technology for storage of information such as computer readable instructions, data structures, program modules or other data.
Memory 14, removable storage 16 and non-removable storage 18 are all examples of computer storage media (e.g., computer-readable storage media storing computer-executable instructions that when
executed by at least one processor cause the at least one processor to perform a method). Computer storage media includes, but is not limited to, RAM, ROM, EEPROM, flash memory or other memory
technology, CD-ROM, digital versatile disks (DVD) or other optical storage, magnetic cassettes, magnetic tape, magnetic disk storage or other magnetic storage devices, or any other medium that can be
used to store the desired information and that can be accessed by computing device 10. Any such computer storage media may be part of computing device 10.
The various elements of computing device 10 are communicatively coupled together via one or more communication links 15. Computing device 10 also includes one or more communication connections 24
that allow computing device 10 to communicate with other computers/applications 26. Computing device 10 may also include input device(s) 22, such as keyboard, pointing device (e.g., mouse), pen,
voice input device, touch input device, etc. Computing device 10 may also include output device(s) 20, such as a display, speakers, printer, etc.
FIG. 1 and the above discussion are intended to provide a brief general description of a suitable computing environment in which one or more embodiments may be implemented. It should be understood,
however, that handheld, portable, and other computing devices of all kinds are contemplated for use. FIG. 1 thus illustrates an example of a suitable computing system environment 10 in which the
embodiments may be implemented, although as made clear above, the computing system environment 10 is only one example of a suitable computing environment and is not intended to suggest any limitation
as to the scope of use or functionality of the embodiments. Neither should the computing environment 10 be interpreted as having any dependency or requirement relating to any one or combination of
components illustrated in the exemplary operating environment 10.
FIG. 2 is a block diagram illustrating a system 200 for sorting a dataset according to one embodiment. System 200 includes a plurality of computing devices 204(1)-204(N) (collectively referred to as
computing devices 204), and a sorting device 208, where N is an integer greater than one. In one embodiment, computing devices 204 and sorting device 208 are each implemented as computers, such as
that shown in FIG. 1. Sorting device 208 is configured to sort dataset 202. In the illustrated embodiment, dataset 202 is divided into a plurality of data portions 206(1)-206(N) (collectively
referred to as data portions 206), which are stored on the plurality of computing devices 204(1)-204(N), respectively. In other embodiments, dataset 202 may be stored on a single computing device.
Sorting device 208 incrementally reads or receives unsorted data from data portions 206 stored on the computing devices 204. As unsorted data is being received, it is separated into independent input
data subsets 210(1)-210(X) (collectively referred to as input data subsets 210) by sorting device 208, where X is an integer greater than one. As unsorted data arrives at sorting device 208, it is
added to a current input data subset 210, and once the current input data subset 210 fills, it is closed, and future unsorted data that arrives goes into the next input data subset 210. Each input
data subset 210 according to one embodiment has a finite capacity (e.g., 1/100
or 1/1000
of the total size of the dataset 202 to be sorted). As each subset 210 is filled, it is sorted by sorting device 208 (referred to as a "subset-sort"), thereby generating respective sorted input data
subsets 211(1)-211(X) (collectively referred to as sorted input data subsets 211). In one embodiment, all of the subset-sorts, except for the last subset-sort, are overlapped with the read of the
data for the subsequent subset 210. Thus, the subset-sort for each current subset is performed while the subsequent subset is being filled. In one embodiment, each of the subset-sorts is performed
using a quick-sort algorithm.
After the last subset 210(X) is closed, its data is subset-sorted, and then a merge-sort is performed on all of the sorted input data subsets 211 to produce a sorted dataset 212 in total sorted
order. The time for performing this last subset-sort is not overlapped with I/O in one embodiment, but the amount of data in the last subset 210(X) is only a small fraction of the entire data set
202, so the subset-sort can be performed relatively quickly. The merge-sort incrementally generates (completely) sorted data from the (partially) sorted input data subsets 211. The merge-sort
according to one embodiment involves repeatedly picking the smallest data element from the entire set of sorted input data subsets 211. In one embodiment, the sorted dataset 212 is divided into a
plurality of sorted output data subsets 214(1)-214(Y), where Y is an integer greater than one. In one embodiment, the total number, X, of input data subsets 210 equals the total number, Y, of sorted
output data subsets 214, and the input data subsets 210 have the same size (e.g., same number of data elements) as the sorted output data subsets 214. In other embodiments, the number and size of the
input data subsets 210 may vary from that of the sorted output data subsets 214. In one embodiment, sorting device 208 adjusts the size of the input data subsets 210 and/or the sorted output data
subsets 214 based on the size of the data set 202 (e.g., making these elements to be, for example, 1/100
or 1/1000
of the total size of the data set 202, so that these elements will be larger (i.e., contain a greater number of data elements) for a larger data set 202, and will be smaller (i.e., contain a smaller
number of data elements) for a smaller data set 202.
In one embodiment, the input data subsets 210 have a uniform size, and in another embodiment have a non-uniform size. In one embodiment, the sorted output data subsets 214 have a uniform size, and in
another embodiment have a non-uniform size. In one embodiment, sorting device 208 is configured to dynamically size the input data subsets 210 and the sorted output data subsets 214 during the
sorting process.
After the first sorted output data subset 214(1) has been generated (e.g., after the first 1/100
or 1/1000
of the data in the sorted input data subsets 211 has been merge-sorted), the output or writing phase begins. In one embodiment, each subsequent portion of the merge-sort is done in the background
while the results of the previous merge-sort are being output (e.g., written to disk or output to a network). Thus, sorted output data subset 214(1) is output from sorting device 208 while sorted
output data subset 214(2) is being generated by sorting device 208, and sorted output data subset 214(2) is output from sorting device 208 while the next sorted output data subset 214 is being
generated by sorting device 208, and this process continues until the last sorted output data subset 214(Y) is output by sorting device 208. In one embodiment, the sorted data that is being generated
for each current output data subset 214 is stored in a memory cache as it is generated, and is output from the memory cache while the next output data subset 214 is being generated.
In this way, by splitting the data into X shards or subsets 210, the only CPU time that is not overlapped with I/O is the time involved in subset-sorting 1/Xth of the data, followed by the time to
merge-sort 1/Xth of the data. This makes virtually all of the CPU time for sorting disappear into the I/O time, even in systems where the I/O time is not much more than the compute time. For example,
for subsets 210 that are each 1/100
of the total size of the input dataset 202, the only CPU time that is not overlapped with an I/O operation is the time for subset-sorting 1/100
of the total data plus the time to merge-sort 1/100
of the data.
FIG. 3 is a flow diagram illustrating a method 300 of sorting a dataset according to one embodiment. In one embodiment, sorting device 208 (FIG. 2) is configured to perform method 300. At 302 in
method 300, data from a dataset is incrementally received. At 304, the received data is incrementally stored as individual input data subsets as the data is received, thereby sequentially generating
a plurality of filled data subsets of unsorted data. At 306, each filled data subset of unsorted data is individually sorted concurrently with receiving data for a next one of the individual input
data subsets, thereby sequentially generating a plurality of sorted input data subsets. At 308, a merge sort is performed on the plurality of sorted input data subsets, thereby incrementally
generating a sorted version of the dataset, wherein the sorted version of the dataset includes a plurality of sequentially generated sorted output data subsets. At 310, each of the sorted output data
subsets is output concurrently with generating a next one of the sorted output data subsets.
In one embodiment, the sorted output data subsets in method 300 each have a same size as the individual input data subsets. The outputting each of the sorted output data subsets in method 300
according to one embodiment comprises outputting each of the sorted output data subsets to a storage medium. In another embodiment, the outputting each of the sorted output data subsets comprises
outputting each of the sorted output data subsets to a network file system. In one embodiment, a size of the individual input data subsets in method 300 is varied based on a size of the dataset. The
individual input data subsets according to one embodiment each have a size that is a predetermined fraction of a size of the dataset. In one embodiment of method 300, the dataset is stored as a
plurality of portions on a plurality of computing devices, and the data from the dataset is incrementally received from the plurality of computing devices. The individually sorting each filled data
subset of unsorted data in method 300 according to one embodiment is performed using a quick-sort algorithm. In one embodiment, the data incrementally received from the dataset is received from a
storage medium, and in another embodiment the data is received from a network file system.
Although specific embodiments have been illustrated and described herein, it will be appreciated by those of ordinary skill in the art that a variety of alternate and/or equivalent implementations
may be substituted for the specific embodiments shown and described without departing from the scope of the present invention. This application is intended to cover any adaptations or variations of
the specific embodiments discussed herein. Therefore, it is intended that this invention be limited only by the claims and the equivalents thereof.
Patent applications by Microsoft Corporation
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A Hölder continuous function which does not belong to any Sobolev space
up vote 12 down vote favorite
I'm seeking a function which is Hölder continuous but does not belong to any Sobolev space.
Question: More precisely, I'm searching for a function $u$ which is in $C^{0,\gamma}(\Omega)$ for $\gamma \in (0,1)$ and $\Omega$ a bounded set such that $u \notin W_{loc}^{1,p}(\Omega)$ for any $1 \
leq p \leq \infty$. Take $\Omega$ to be bounded, open.
My first guess is to do a construction with a Weierstrass function. I know this is differentiable 'nowhere' but that doesn't convince me it isn't weakly differentiable in some bizarre way. Hopefully
someone knows of an explicit example.
ca.analysis-and-odes ap.analysis-of-pdes
Function on what domain (1-dim or higher)? What exponent in your Sobolev space (L^p-flavoured or just L^2)? – Yemon Choi Sep 15 '10 at 1:07
I clarified the question. – Dorian Sep 15 '10 at 1:25
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3 Answers
active oldest votes
Your guess is indeed right. Following a similar idea gives you the Takagi or blancmange function. It is even quasi-Lipschitz (it has a modulus of continuity $\omega(t)=ct(|\log(t)|+1)$ for
a suitable constant $c>0$), thus it's Hoelder of any positive exponent less than 1. It is not even BV in any open interval, thus $W^{1,p} _ {loc}$ for no $p\geq1$.
Rmk 1. The above example is for dimension 1: but of course it holds in any dimension a fortiori.
Rmk 2. To get an example with a more classical flavor, actually a Weierstrass function, replace $s(x)$ with $\cos(x)$. I'd say that the resulting Fourier series defines a function with the
same features, by the same reasons (the function $\cos(x)$ works better than $\sin(x),$ in view of point 2 below.)
Rmk 3. Once you know that the Weierstrass function $f(x):=\sum_{k=0}^\infty 2^{-k}\cos(2^k x)$ is nowhere differentiable, you also have that it is BV on no open interval, for BV on an
interval would imply differentiability a.e. there. However, for your needs it seems more direct just showing it has infinite variation on any interval.
1. To prove that the Takagi function $f(x)$ admits the above modulus of continuity, recall that that $f$ is characterized as the fixed point of the affine contraction $T:C_b(\mathbb{R})\
to C_b(\mathbb{R})$ such that $(Tf)(x)=\frac{1}{2}f(2x)+s(x),$ for all $x\in\mathbb{R}$, where $s(x)$ is the distance function from the integers (a zig-zag piecewise 1-periodic
up vote 20 function). Just find a $c$ such that the subset of $C_b(\mathbb{R})$ of functions that admit $\omega$ as modulus of continuity is a $T$-invariant set. The latter subset is obviously
down vote closed and non-empty, so the fixed point is there. (The above illustrated a standard general technique to prove properties of objects found by means of the contraction principle).
2. Proving that $f$ is not of bounded variation on $[0,1]$ (hence in no open interval, due to the self-similarity encoded in the fixed point equation), requires a small computation on the
partial sum $f_n$ of the series defining $f$. Let $$f_n(x):=\sum_{k=0}^{n-1}\, 2^{-k} s(2^k x).$$ First note that the derivative of $f_n$ only takes integer values, which of course
come as a result of the sum of $n$ terms $\pm 1$ (with all the $2^n$ possible signs). In particular, for any $n\in\mathbb{N}$ the function $f_{2n}$ has ${2n \choose n} $ flat intervals
of lenght $2^{-2n}$ within the unit interval $I$, and has derivative larger than $2$ in absolute value elsewhere in $I$. Thus, for the subsequent odd index $2n+1,$ the function $f_
{2n+1}$ has ${2n \choose n}$ local maxima in $I$ (located in the mid-points of the above intervals). Moreover, passing to $f_{2n+1}$ each maximum point contributes to the increment of
the total variation with $2^{-2n}$, while the total variation remains unchanged passing from $2n+1$ to the next even index $2n+2$. The conclusion is that, for any $n$, the total
variation of $f_n$ on $I$ is $$V(f_n;I)=\sum_{0\leq k < n/2}{2k\choose k}2^{-2k} =O\big(\sqrt{n}\big),$$
since by the classical asymptotics for the central binomial coefficient, ${2k \choose k}=\frac{4^k}{\sqrt{\pi k}}(1+o(1)),\, k\to\infty.$ So actually $V(f_n;I)$ diverges. Yet this
would not be sufficient to conclude that $V(f,I)=\infty,$ as the total variation is only lower semicontinuous with respect to the uniform convergence. However, the discrete variation
on a given subdivision $P:=\{t_0 < \dots < t_r \}$ $$V(f_n; P\, )=\sum_{i=0}^{r-1}\, \big|f_n(t_{i+1})-f(t_i)\big|$$ does of course pass to the limit under even pointwise convergence.
Now the point is that, for the binary subdivision $P_m:=\{ k2^{-m} \, : \, 0 \le k \le 2^m \},$ we have $V(f_n;I)=V(f_n;P_m)$ as soon as $n \geq m$. So for all $m$ letting $n\to\infty$
$$V(f;P_m)=\lim_{n\to\infty }V(f_n;P_m)=V(f_m;P_m)$$ and $$V(f;I)=\sup_{m\in\mathbb{N}}V(f;P_m)=\infty,$$ as we wished to show.
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Another simple approach is to take the lacunary series $f(x)=\sum_k a_k e^{i(m_k,x)}$ where $m_k\in \mathbb Z^d$ and $|m_{k+1}\gg |m_k|$. For any modulus of continuity $\omega$ such that $\
Omega(t)=\omega(t)/t\to \infty$ as $t\to 0$, the condition that $f$ has this modulus of continuity is equivalent to the condition that $|a_k|\le C\omega(|m_k|^{-1})$ if the spectrum is
sparce enough to ensure that $\Omega(|m_{k+1}|^{-1})\ge 2\Omega(|m_k|^{-1})$ and $\omega(|m_k|^{-1})\ge\sum_{\ell>k}\omega(|m_\ell|^{-1})$. If $f\in W^{1.p}$, then $ \left|\int f\nabla\psi\
right|\le C\|\psi\|_{L^q}$ for smooth $\psi$. Plugging $e^{-i(m_k,x)}$, we see that unless $a_k=O(|m_k|^{-1})$, we have no chance. Thus, nothing short of Lipschitzness will force $f$ to be
up vote in $f\in W^{1.p}$.
6 down
vote This formally works only on the torus but you can take any smooth partition of unity $g_j$ on the torus and notice that one of the functions $g_j f$ is also bad. But any of them can be
replanted to $\mathbb R^d$ if the supports are small enough.
Nice argument... so if I got it this way you can make a counterexample with any modulus of continuity satisfying $t=o(\omega(t).$ – Pietro Majer Sep 15 '10 at 13:34
1 A doubt. If we choose $a_k:=2^{-k}$ and $m_k:=\pi2^k$ the resulting $f$ is a Weierstrass function (or better its real part); which in this case is nowhere differentiable (with these
parameters I think its a result by Hardy), hence certainly not locally lipschitz. But the condition you wrote holds with $\omega(t)=t$. – Pietro Majer Sep 15 '10 at 14:39
You are right, I was a little bit sloppy in how exactly I stated it. I'll ix it now. – fedja Sep 15 '10 at 14:46
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I believe that the Cantor ternary function (aka devil's staircase) is a simple example.
up vote 1 down vote
For Holder exponent $< \log_3 2$. – Willie Wong Dec 8 '10 at 0:23
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Estimating the reproductive number in the presence of spatial heterogeneity of transmission patterns
Estimates of parameters for disease transmission in large-scale infectious disease outbreaks are often obtained to represent large groups of people, providing an average over a potentially very
diverse area. For control measures to be more effective, a measure of the heterogeneity of the parameters is desirable.
We propose a novel extension of a network-based approach to estimating the reproductive number. With this we can incorporate spatial and/or demographic information through a similarity matrix. We
apply this to the 2009 Influenza pandemic in South Africa to understand the spatial variability across provinces. We explore the use of five similarity matrices to illustrate their impact on the
subsequent epidemic parameter estimates.
When treating South Africa as a single entity with homogeneous transmission characteristics across the country, the basic reproductive number, R[0], (and imputation range) is 1.33 (1.31, 1.36). When
fitting a new model for each province with no inter-province connections this estimate varies little (1.23-1.37). Using the proposed method with any of the four similarity measures yields an overall
R[0] that varies little across the four new models (1.33 to 1.34). However, when allowed to vary across provinces, the estimated R[0] is greater than one consistently in only two of the nine
provinces, the most densely populated provinces of Gauteng and Western Cape.
Our results suggest that the spatial heterogeneity of influenza transmission was compelling in South Africa during the 2009 pandemic. This variability makes a qualitative difference in our
understanding of the epidemic. While the cause of this fluctuation might be partially due to reporting differences, there is substantial evidence to warrant further investigation.
Influenza; Reproductive number; Infectious disease outbreak
In an emerging outbreak of an infectious disease, such as influenza, there is great interest in determining, amongst other things, its transmissibility, which is typically quantified by its
reproductive number. Initially, there is interest in estimation of the idealized basic reproductive number, R[0], which measures the average number of cases generated by an infected individual in an
entirely susceptible population. Post hoc analyses of an outbreak may include estimation of the time-varying effective reproductive number, R[t], which represents the impact on R[0] of acquired
immunity and public health interventions that typically lead to decreased transmission and a decrease in the growth of the outbreak [1]. Several methods exist for estimating these quantities either
in real time as the epidemic progresses, or after the epidemic is over [2,3]. Frequently, these estimators assume a homogenously mixing population, even though often this simplifying assumption may
not be realistic. One possible source of heterogeneity is the lack of spatial uniformity in transmission across the population in question. In such an instance, the question arises of how to modify
our inference—on the reproductive number, for example—and the impact this might have on our understanding of transmission dynamics.
Typically epidemiological studies of influenza report reproductive numbers over large geographic regions and populations, often estimating a reproductive number for an entire country [4,5]. This may
be useful as a metric to compare to previous measures of the same quantity in an effort to determine relative transmissibility of a disease. However, this overall measure confounds information that
may impact on transmissibility, or its measure, and it is likely not an insufficiently informative representation of the reproductive number.
Many studies have attempted to measure the spatial dynamics of influenza spread, often in an effort to create better control strategies and predict the occurrence of specific strains in coming
influenza seasons [6-8]. Implicit in this work is the reality that influenza transmission dynamics are not spatially uniform, though details of how transmission may vary spatially are lacking.
Spatial considerations are clearly important, as those who live great distances from each other are much less likely to infect one another than those who live in closer proximity to each other.
Further, spatial heterogeneity is important to consider since differences may exist in behavioral patterns, demographics, control measures, climate, and other factors that may affect transmission
differently in different geographical regions. The issue is made even more complex because reporting issues and healthcare seeking behaviors can vary geographically and also influence data quality.
Thus identifying heterogeneity does not necessarily indicate its cause simply.
In this work we introduce a simple modification of the method originally proposed by Wallinga and Teunis [1] to estimate the effective reproductive number. This modification allows for the estimation
of the reproductive number(s) in the presence of greater heterogeneity in transmission. We apply this method to data from the 2009 pandemic influenza outbreak in South Africa and estimate the
reproductive number for each province. We discuss the potential implications of the results we obtain on future research and surveillance activities.
Wallinga and Teunis [1] (denoted WT method hereafter) proposed a network-based method for the estimation of the effective reproductive number by making use of the epidemic curve, N={N[1],…, N[T]},
where N[t] is the number of cases at time point t, and an estimate of the serial interval, p[1],…, p[k], where p[i] describes the probability of a serial interval of length i and the maximum serial
interval length is k. The estimator for R[t] is a function of the relative probability that case t[i] was infected by the j^th case on day t′, denoted q t i , t ′ j and is given by
R t j ' = ∑ s = t ′ + 1 min T , t ′ + k ∑ i = 1 n s q s i , t j ' = ∑ s = t ′ + 1 min T , t ′ + k n s q s , t j ' ,
where n[s] denotes the number with symptom onset on day s. The relative probability that case t[i] was infected by the j^th case on day t ′ , q t i , t ′ j , is a function of the probability that
case t[i] was infected by case t[j]′, and is entirely a function of the serial interval, such that P(t[j]′ →t[i]) = p t i − t ′ j .
Spatial transmission data
We propose the use of additional structure to describe the probability of an infection event occurring between two cases. We modify the probability of an infectious event between two cases, P(t[j]′
→t[i]), to incorporate spatial information:
P t ′ j → t i = p t i − t ′ j d t i ′ t j ,
where d[ti′ ,tj] is a measure of similarity between the cases t[j]′ and t[i]. Note we assume independence between space and the serial interval. By making use of information on conditional
probabilities, if known, one could relax this assumption. Since this method only modifies the method of constructing the probabilities that connect individuals in the network, the properties of the
estimator originally proposed by Wallinga and Teunis still apply.
This measure can be calculated in a number of ways. The simplest being
d t i ′ t j = 0 , if the cases are far apart , 1 , if the cases are close to each other .
The similarity measure can depend on geographical proximity and/or demographic proximity, and can also be informed by observed data on travel or contact patterns. The similarity matrix does not
necessarily contain probabilities, but represents the relative similarity between two locations and/or demographic features. Additionally by similarity we are describing the potential for an
infective event. Alternatively, we can also entertain a Bayesian formulation and attach a prior to the parameters of the distance measure. Below, we investigate some different measures.
We use data from South Africa describing the pandemic influenza H1N1 outbreak in 2009. The data contains information on 12,543 reported laboratory confirmed cases, including specimen collection date,
gender, age, province of residence, and symptom onset date [9]. There are nine provinces in South Africa that vary substantially in size, population density, climate, and accessibility to healthcare
[10]. The symptom onset date was available in 758 cases (6% of the cases). We impute the missing onset dates using a multiple imputation method. To this end we first fit a Poisson model to the lag
between onset date and collection date for those who had both dates recorded (715 cases) incorporating statistically significant predictors: (i) the province where the report originated, and (ii) an
indicator of weekend versus weekday for the day of collection. The Poisson model was used to randomly generate missing onset times. This process was repeated 500 times creating 500 imputed data sets.
All analyses are performed on each of the 500 imputed datasets and results are combined across the individual dataset results and these summaries are the ones we report [11].
Our model requires a similarity matrix. As we are uncertain of which similarity matrix would be most appropriate for influenza in South Africa in 2009, we investigate a variety of matrices in the
model and comment on the variability resulting from each similarity matrix. To this end, we investigate five different similarity matrices to describe, what would seem to us to be, plausible
transmission patterns between the provinces. The matrices are shown in Tables 1, 2, 3, 4, 5 and are, respectively:
Table 1. All transmission occurs within provinces
Table 2. Matrix based on reported travel patterns in South Africa
Table 3. Uniform probability of transmission between different provinces
Table 4. Increased probability of transmission for neighboring provinces
Table 5. Increased transmission for most densely populated provinces
a. Diagonal matrix. This model assumes that all transmissions occur within each province and there is no transmission between individuals in different provinces. This is comparable to applying the
original WT method to each province separately.
b. Travel patterns. Using data on travel patterns reported by the Department of Environmental Affairs and Tourism in South Africa [12] we construct a second transmission matrix. This one assumes that
transmission probabilities mirror the probability of travel between provinces.
c. Increased transmission between those in the same province, i.e. d[ij]=2 if i and j are in the same province and d[ij]=1 otherwise. This is an attempt at giving more weight to infection between
individuals within a state, but allowing for infection from an individual from another state; a less extreme isolation model than in a., above. Of course, we could entertain values other than 1 and 2
for elements of this matrix.
d. Neighboring provinces up-weighted. We define a similarity metric such that if i and j are in the same province the similarity is 2, if they are in neighboring provinces the similarity is 1,
otherwise the similarity is 0.5. This model can be thought of as between the model described in a. and the one described in c.
e. Higher transmission between densely populated provinces. This matrix allows the three provinces that are the most populous and likely experience the greatest rates of travel to have a greater
chance of cross infection. This is done by using the same arrangement as described in c., but allowing the provinces of Gauteng, West Cape, and KwaZulu-Natal to have a similarity measure of 1.5 to
each other.
This approach requires an estimate of the serial interval. We make use of the SI distribution between primary cases and suspected plus laboratory-confirmed secondary cases (30%, 17%, 20%, 23%, 7% and
3% for days 1 to 6, respectively) [11].
Sensitivity analysis
We also perform a sensitivity analysis to assess the robustness of our results to potential errors in the data. Our general approach is to allow the onset date of 10% of the individuals to shift
randomly within a 30-day window. We choose one imputed dataset and create fifty “sensitivity” data sets. All analyses are performed on these 50 datasets and compared to the results for the imputed
dataset used in the original analysis. Complete results are reported in the appendix.
All analyses were performed in R 2.13.0 (http://www.r-project.org webcite). Programs are available upon request to the corresponding author.
Figures 1 and 2 show a sample of the imputed epidemic curves first overall (Figure 1) and then for each of the nine provinces (Figure 2). The first case had symptom onset on June 12, 2009 and the
final specimen was collected on November 23, 2009. Gauteng Province had the largest number of cases with 5541 confirmed cases during the epidemic.
Figure 1. Imputed epidemic curves. Gray shading indicates the variability in the imputed data. The dashed line indicates the observed onset data.
Figure 2. Epidemic curves for each of the nine provinces. The shaded area indicates the variability in the imputed values.
Figure 3 shows the overall estimates of R[t] when the different transmission matrices are used, as well as when the original WT method (Matrix a, ignoring spatial transmission patterns) is used. We
note that there are only very minor differences between the estimates. The only noticeable differences exist during the initial and final phases of the epidemic; however, these differences are
minimal. Figure 4 shows the results by province, and again there are only minor differences between the estimates obtained with the five different transmission matrices.
Figure 3. Estimates of R[t ]using the transmission matrices, as described in the text. The estimates shown represent the average of the R[t] estimates obtained across the 500 imputed epidemics. Days
when no cases were reported have a R[t] of 0, though we smooth through this for the purpose of visual presentation.
Figure 4. Estimated R[t ]by province. The line types for each plot are the same as those used in the previous figure. Days when no cases were reported have a R[t] of 0, though we smooth through this
for the purpose of visual presentation.
Table 6 shows an estimate of R[0] obtained by averaging the R[t] estimates obtained over the period of exponential growth in the epidemic (days 10 through 70 reported, though other ranges were used
with similar results). We note sizable differences in the estimate of R[0] between provinces and the transmission matrices used. The biggest differences are between the original WT method (Matrix a)
and the methods using a nonhomogeneous transmission matrix (Matrix b-e). Those using a similarity matrix implying heterogeneity are almost identical to each other, but quite different from the value
obtained by the original WT method that assumes homogeneity. We note that when nonhomogeneous transmission is assumed, R[0] is only above 1 for Gauteng and the Western Cape, where Johannesburg and
Cape Town are located. One possible explanation is that a certain population density or degree of travel in/out of an area is required to sustain a local epidemic of the flu. The sensitivity analysis
yielded results that are consistent with these findings (Table 1).
Table 6. The R[0 ]estimates overall and by region
We further explore this result in Figure 5, where the estimates of R[0] are plotted against the population size, land area and population density with the least squares regression estimates shown.
The estimate of R[0] increases with an increase in population size and density (p=0.0004 for matrix b; p=0.04 for matrix c) and decreases with land area. The proportion infected in each province also
appears to be related to the population size in the province, though this is not significant (p=0.16). When the WT estimator is used, the model that ignores geography, these relationships disappear
(p=0.53 for population density; similar results hold for the other plots), as one might expect.
Figure 5. The relationship between characteristics of each of the provinces and the outbreak. Lines drawn reflect the least squares regression line for the relationship between the two variables. The
first panel shows the relationship between the population size and the size of the outbreak in each province. The second panel describes the relationship between the population size and R[0]. The
third panel illustrates the relationship between the land area and R[0] obtained for each of the transmission matrices. The final panel plots the relationship between population density and R[0] for
each transmission matrix. Line types follow the legend in Figure 3.
The results in this paper argue that disease transmission is a function of more than just biology, as is well known, but often ignored. The impact of adjusting the assumption of homogeneous mixing in
this South African outbreak, is that apart from the densely populated, urban areas, the pandemic would likely not have been sustained just in the rural, sparsely populated provinces. This finding
reinforces the obvious: if individuals have very limited contact with each other, then the outbreaks would probably be small in numbers, limited to small groups, and would likely not propagate to
become a larger and more noticeable outbreak. Our estimates of the reproductive number for the more populous provinces are consistent with results reported elsewhere, but the results we obtain for
more rural provinces are notably lower [5,13,14].
In our analysis, there are other important issues to consider. Throughout we have assumed that the reporting of cases is uniform throughout the country, and this was the basis for our sizable
imputation of the number of symptom onset dates. Even if this reporting is less than 100%, but still spatially uniform, the results we observe will hold [15]. However, if reporting is not uniform
between provinces and some provinces have much better reporting of cases than others, we can expect the results to change. For instance, if reporting was lower in the more rural provinces, then it is
likely that the estimated reproductive numbers would increase in these provinces if some adjustment for this underreporting were made. Without a more detailed study, this is difficult to quantify and
explain. Clearly, there is a certain amount of confounding present, and data reporting issues can be part of an explanation for the results we obtain.
Another factor that can, at least partly, explain the results are the choice of transmission matrices used. We show results for various transmission matrices in order to quantify the degree to which
transmission occurs between provinces. Four of these matrices are somewhat arbitrary and not based on actual data. One matrix is based on actual travel patterns in South Africa. But the results are
reasonably consistent for the four matrices that assume some degree of transmission between provinces, even when the amount is very small, as in the travel-based matrix. This argues that the results
are influenced more by the fact that such a matrix is used and less by the form that such a matrix takes. In all of these cases, despite the substantial differences in the matrices, the result is the
same: transmission is maintained in more urban areas and rural areas fail to sustain transmission.
We note dramatic differences between the results when transmission between provinces is incorporated into the estimation (matrices b-e) and the results that assume that no transmission occurs between
provinces (matrix a). This reflects the impact of using such matrices and the importance of performing sensitivity analyses to determine the impact of the matrix on the results. Possibly why such
matrices have not been used in the past, even though they have a qualitative impact on the results, is that these matrices are difficult to come by, and in practice, they are likely to be estimated
in a somewhat ad hoc manner. In some cases, there may be little or no data to inform a transmission matrix. In this case, a wide variety of matrices can be used to determine the plausible range of
values that the estimates can assume. Ultimately deriving a method for estimating these matrices, ideally using Bayesian tools, would mitigate this challenge. The framework we provide here lends
itself to such an approach, although we have not carried out such an analysis.
We further note the coarseness of the spatial resolution of our data. Our implicit assumption is that individuals within a province are homogenous. While assuming homogeneity within a province is
more general than assuming homogeneity over the entire country, it is still a substantial assumption that ignores potentially important variations within a province. As with any analysis, we are
limited by the available data, and acknowledge that data on a finer spatial scale would be desirable.
Our method also makes a strong assumption of independence between space and time. That is, we assume that the probability of a particular infector-infectee pair is influenced independently by the
temporal and spatial distance between the two individuals. Clearly violations of this assumption are feasible and could impact our results. Without further information on the potential correlations
that exist between space and time, any adjustment would be arbitrary and potentially misleading.
While the results we obtain might partially be explained by data quality issues and care-seeking behaviors in rural versus urban populations, there are other potential explanations. A recent cross
sectional, serum study reports differential exposure to influenza strains in China across five communities [16], with the most urban community reporting the highest exposure to influenza strains.
This provocative result begs further study as it is likely to be attributable to a number of factors and is consistent with the results we have obtained here.
The marked difference in estimates of transmission in rural versus urban areas in our study is also consistent with recent work on social contacts [17,18]. In a study in Japan, there was a
significant relationship between the number of social contacts and urbanicity amongst the elderly. Additionally, they similarly found that those in more urban areas have a greater chance of having
more supportive interactions [17]. Influenza transmission requires proximity between individuals and social contacts could be one surrogate measure of this proximity.
Additionally, there has been an observed influence of climate and relative humidity on influenza transmission [19-22]. The climate across South Africa is variable with some of the more rural
provinces being characterized by a drier climate; the country’s climate is mostly semi-arid, but subtropical along the east coast. So this is not an ideal country to test the transmission theory, but
KwaZulu-Natal is the only relatively humid province, so it does not appear that the humidity hypothesis is borne out by these data.
Travel patterns have been correlated with the movement of influenza on a large scale [6,7]. Indeed, Viboud et al show that an outbreak that starts in a rural area will spread slowly until it reaches
an urban center, at which point it will spread much more quickly [6]. We attempt to incorporate travel patterns in South Africa in our analysis. Travel between the more rural provinces and other
areas is much more limited and in general individuals tend to travel to larger provinces, rather than individuals in more urban provinces coming to rural provinces. Thus, the lack of movement between
these provinces and areas where transmission is occurring could lead to later onset of sustained transmission locally and lower levels of transmission in the absence of more individuals entering the
province and interacting with the local population.
Another explanation is that what we see here could be similar to that observed in the Netherlands in the early phases of the pandemic where the reproductive number was estimated to be below one,
indicating that sustained transmission was not occurring and the cases were being generated through imported infections [4]. In South Africa, this would imply that individuals traveled to larger
urban centers and became infected there. Their case was reported upon returning home so that the case is not attributed to the location where the transmission event actually took place, at least for
the initial cases. We did not account for the possibility of this taking place.
There has been significant work pointing to the great spatial heterogeneity that exists in influenza; however little work has been done to directly estimate the influence of local transmission in
propagating this trend. Intensive network models have the capability of investigating these dynamics, but are challenging to implement without extensive resources. Our study introduces a novel and
simple approach for doing this by making use of the epidemic curve, information on the serial interval distribution and some prior knowledge of transmission dynamics. We have shown results for
estimation over the entire outbreak period, but the modification proposed by Cauchemez et al [2] allowing for real-time estimation of R[t] could be implemented straightforwardly with this
modification, as well.
Our results are suggestive of substantial spatial heterogeneity in transmission dynamics, however further study is warranted due to the limitations of the data at hand and uncertainties on reporting
dynamics. At a minimum, these results should argue for modifications in the data that is collected from surveillance and other data collection systems to better understand reporting patterns and the
dynamics of interaction between individuals that would lead to substantial heterogeneity in transmission. An improved understanding of heterogeneity will aid in targeting limited interventions in the
most effective way possible.
Authors’ contributions
LFW and MP conceived the study and developed the methods used. LFW performed the analyses and wrote the manuscript. BA collected the data. All authors reviewed the manuscript. All authors read and
approved the final manuscript.
LFW and MP were supported by Award Number U54GM088558 from the National Institute of General Medical Sciences. The content is solely the responsibility of the authors and does not necessarily
represent the official views of the National Institute Of General Medical Sciences or the National Institutes of Health.
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Efficient Light-Scattering Calculations for Aggregates of Large Spheres
Calculation of the scattering pattern from aggregates of spheres through the T-matrix approach yields high-precision results but at a high-computational cost, especially when the aggregate concerned
is large or is composed of large-size spheres. With reference to a specific but representative aggregate, we discuss how and to what extent the computational effort can be reduced but still preserve
the qualitative features of the signature of the aggregate concerned.
© 2003 Optical Society of America
OCIS Codes
(190.5890) Nonlinear optics : Scattering, stimulated
(290.1990) Scattering : Diffusion
(290.4210) Scattering : Multiple scattering
(290.5850) Scattering : Scattering, particles
Rosalba Saija, Maria Antonia Iatì, Paolo Denti, Ferdinando Borghese, Arianna Giusto, and Orazio I. Sindoni, "Efficient Light-Scattering Calculations for Aggregates of Large Spheres," Appl. Opt. 42,
2785-2793 (2003)
Sort: Year | Journal | Reset
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2. P. J. Wyatt, “Scattering of electromagnetic plane waves from inhomogeneous spherically symmetric objects,” Phys. Rev. B 127, 1837–1843 (1962).
3. M. I. Mishchenko, W. J. Wiscombe, J. H. Hovenier, and L. D. Travis, “Overview of scattering by nonspherical particles,” in Light Scattering by Nonspherical Particles, M. I. Mishchenko, J. W.
Hovenier, and L. D. Travis, eds. (Academic, New York, 2000), pp. 30–59.
4. S. Holler, J.-C. Auger, B. Stout, Y. Pan, J. R. Bottiger, R. K. Chang, and G. Videen, “Observations and calculations of light scattering from clusters of spheres,” Appl. Opt. 39, 6873–6887
5. M. I. Mishchenko, J. W. Hovenier, and L. D. Travis, “Concepts, terms, notation,” in Light Scattering by Nonspherical Particles, M. I. Mishchenko, J. W. Hovenier, and L. D. Travis, eds. (Academic,
New York, 2000), pp. 3–25.
6. J. D. Jackson, Classical Electrodynamics (Wiley, New York, 1975).
7. F. Borghese, P. Denti, R. Saija, G. Toscano, and O. I. Sindoni, “Multiple electromagnetic scattering from a cluster of spheres. I. Theory,” Aerosol Sci. Technol. 3, 227–235 (1984).
8. F. Borghese, P. Denti, G. Toscano, and O. I. Sindoni, “An addition theorem for vector Helmholtz harmonics,” J. Math. Phys. 21, 2754–2755 (1980).
9. F. Borghese, P. Denti, and R. Saija, “Optical properties of spheres containing several spherical inclusions,” Appl. Opt. 33, 484–491 (1994).
10. P. C. Waterman, “Symmetry, unitarity and geometry in electromagnetic scattering,” Phys. Rev. D 3, 825–839 (1971).
11. E. Fucile, F. Borghese, P. Denti, R. Saija, and O. I. Sindoni, “General reflection rule for electromagnetic multipole fields on a plane interface,” IEEE Trans. Antennas Propag. 45, 868–875
12. E. Fucile, P. Denti, F. Borghese, R. Saija, and O. I. Sindoni, “Optical properties of a sphere in the vicinity of a plane surface,” J. Opt. Soc. Am. A 14, 1505–1514 (1997).
13. Y.-L. Xu, “Electromagnetic scattering by an aggregate of spheres: far field,” Appl. Opt. 36, 9496–9508 (1997).
14. F. Borghese, P. Denti, R. Saija, and O. I. Sindoni, “Reliability of the theoretical description of electromagnetic scattering from non-spherical particles,” J. Aerosol Sci. 20, 1079–1081 (1989).
15. R. T. Wang, J. M. Greenberg, and D. W. Schuerman, “Experimental results of dependent light scattering by two spheres,” Opt. Lett. 11, 543–545 (1981).
16. D. W. Schuerman and R. T. Wang, “Experimental results of multiple scattering,” Contractor Rep. ARCSL-CR-81003 (U.S. Army Chemical Systems Laboratory, Aberdeen Proving Grounds, Md., July 1980).
17. B. Stout, J.-C. Auger, and J. Lafait, “Individual and aggregate scattering matrices and cross sections: conservation laws and reciprocity,” J. Mod. Opt. 48, 2105–2128 (2001).
18. D. W. Mackowski and M. I. Mishchenko, “Calculation of the T matrix and the scattering matrix for ensembles of spheres,” J. Opt. Soc. Am. A 13, 2266–2278 (1996).
19. M. I. Mishchenko, L. D. Travis, and A. Macke, “T-Matrix method and its applications,” in Light Scattering by Nonspherical Particles, M. I. Mishchenko, J. W. Hovenier, and L. D. Travis, eds.
(Academic, New York, 2000), pp. 147–172.
20. F. Borghese, P. Denti, R. Saija, M. A. Iatì, and O. I. Sindoni, “Optical properties of a dispersion of anisotropic particles with nonrandomly distributed orientations. The case of atmospheric ice
crystals,” J. Quant. Spectrosc. Radiat. Transfer 70, 237–251 (2001).
21. W. C. Chew, Waves and Fields in Inhomogeneous Media, IEEE Press Series on Electromagnetic Waves (Institute of Electrical and Electronic Engineers, Piscataway, N.J., 1990).
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Schuerman, ed. (Plenum, New York, 1980), pp. 283–290.
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Brooklawn, NJ Precalculus Tutor
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Copyright © University of Cambridge. All rights reserved.
'Right Angles' printed from http://nrich.maths.org/
Samantha and Shummus both realised that in order to create a triangle with a right angle, the band had to go through the centre of the circle. Shummus writes:
I noticed that the bands had to be started in the centre.
Xianglong Ni notes that:
If we have 9 points on the circle then you can't create a right-angle using the points. This is so because a right angle is inscribed in a semicircle; It is facing a diameter. But you can only create
a diameter when there is an even amount of points on the circle. If the number of points on the circle is even then yes. If the number is odd then no.
Rachel from Newstead sent us a few diagrams to illustrate examples of right-angled triangles in circles with an even number of points.
Indika of Helena Romanes 6th Form College sent us her explanation for why the band must go through the centre of the circle:
The only way that a right angle triangle can be created between 3 points round the edge is when the angle subtended at the centre by two of the points is 180 degrees, this therefore proves that two
of the points have to be opposite each other (this means having an equal number of pegs).
This is because the angle subtended at the centre by two points are exactly double the angle subtended at the edge by the same points. This rule will apply to all circles, i.e. there will be a right
angled triangle if two pegs are placed opposite each other.
If you haven't met this idea before, you may want to look at another problem from August 2005, Subtended angles Here are another couple of examples of right-angled triangles using the same
eight-point and ten-point circles that Rachel used:
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second order variable coeff homogeneous ode
May 6th 2010, 01:31 PM #1
May 2010
second order variable coeff homogeneous ode
Could someone help me solve the following:
y'' +(1/3x)y' - (1/3x)y = 0
I go through the steps I would normally to solve the DE, but when attempting to find the roots I end up with a function of x which I'm not exactly sure what to do with. It appears that there must
be different solutions for different ranges of x. How would I go about solving this?
Could someone help me solve the following:
y'' +(1/3x)y' - (1/3x)y = 0
I go through the steps I would normally to solve the DE, but when attempting to find the roots I end up with a function of x which I'm not exactly sure what to do with. It appears that there must
be different solutions for different ranges of x. How would I go about solving this?
Are you sure the ODE is $<br /> y'' + \frac{1}{3x} \, y' - \frac{1}{3x} \, y= 0?<br />$
yes, confusing... isn't it?
just so everyone knows, I'm pretty sure this is solved using the Frobenius method
You are correct, we just need to switch our variables around to obtain the required equation for the Frobenius method: Frobenius method - Wikipedia, the free encyclopedia
$<br /> x^2 y'' + \frac{1}{3} xy' - \frac{1}{3} x y= 0<br />$
$p(x) = \frac{1}{3}$
$q(x) = \frac{1}{3} x$
Then continue with the Frobenius method. There is an excellent example on that wikipedia page (first example) that will walk you through all the steps (so I won't repeat them here out of
May 6th 2010, 01:47 PM #2
May 6th 2010, 02:05 PM #3
May 2010
May 6th 2010, 05:04 PM #4
May 2010
May 7th 2010, 08:26 AM #5
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Div, Grad, Curl and All That
Cosmic Variance had three interesting &ldquot;greatest&rdquot; discussion threads:
In the comments to the Greatest Physics Textbook, Clifford (the original poster) joked that no self-respecting mathematician ever read Schey’s Div, Grad, Curl and All That. I don’t know about anyone
else, but that’s the book I learned the subject from. The book gives incredibly hand-wavy proofs, and if I remember right it trumpets its lack of rigor, but it does a good job of giving the intuition
behind the Green, Gauss, and Stokes theorems. Reading it made reading something like Spivak’s Calculus on Manifolds much easier.
7 thoughts on “Div, Grad, Curl and All That”
1. Is it just me, or was the first time you saw this stuff slightly mystifying, but when you see the material in the Manifolds context, the results seem almost obvious?
(I mean aside from the being new to the material vs not being new to the material)
2. Actually, I think I was the opposite. I tried to learn it from the manifolds point of view, and didn’t get any feel for what it was about. What finally allowed to learn it was hearing the
sentence “the determinant of a matrix is a volume”, hearing the sentence “the derivative of volume is surface area”, and then reading Div, Grad, Curl and All That which explains all of the vector
calculus theorems from the “derivative of volume is surface area” point of view. Then I was able to make sense of it from the manifolds point of view.
(Unless by “manifolds point of view”, you just mean that differential forms are much easier to understand than curl. Then I agree. In fact, I’ve completely forgotten everything I’ve ever known
about curl.”
3. I guess I mean the latter. I too have forgotten everything I ever knew about curl except for how to spell it.
4. For good intuition about calculus on manifolds I recommend Misner, Thorne and Wheeler’s ‘Gravitation’ which has nice pictures of forms. When I think about differential forms I actually use a
‘dual’ form of MWT’s diagrams that I’ve described (very handwavingly) here.
5. My link vanished. ‘here’ should point to: http://homepage.mac.com/sigfpe/Mathematics/forms.pdf
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Introducing myself
My name is
David Corfield
and I'm very grateful to have been invited to join this blog as a contributor. With five years of blogging behind me at the
-Category Café, I relish the opportunity to talk with a new audience. My rate of blogging may have slowed, especially as my adminstrative load has increased - I'm now Head of
at the University of Kent - but I'm looking forward to writing some posts here.
I have interests in a variety of approaches to mathematical philosophy, including the statistical learning theory I picked up from my time at the Max Planck Institute for Biological Cybernetics in
Tübingen, but the main idea I would like to promote to the audience here is that category theory is worth exploring as a resource for the mathematical philosopher.
I have recently published a couple of articles which examines the light category theory can throw on familiar infinite structures. In
Understanding the Infinite I: Niceness, Robustness, and Realism
, I look at the phenomenon where an infinite entity is defined by a universal property, and through this inherits 'for free' a range of other nice properties. In
Understanding the Infinite II: Coalgebra
, I look at the duality between minimally and maximally defined entities in the context of the duality between 'algebra' and 'coalgebra'.
Perhaps had I known of Shaughan Lavine (1994)
Understanding the Infinite
, Harvard University Press, I might have opted for a different title.
There's much to do to understand the relationship between category theory and the traditional foundational branches, which have drawn most philosophical attention. Recently, I posed a
on MathOverflow concerning category theory and Joel Hamkins' set theoretic multiverse. The answer by Joel there shows just the sort of joint investigation needed. A few years ago at the Café, we had
on the relationship between category theory and model theory.
Category theory also has an interface with
proof theory
, but I know less about this. Something to look out for in the future is the new
Homotopy type theory
, and associated
Univalent foundations
8 comments:
1. Cool! Glad to have you as a contributor, David
- Jeff
2. Welcome, David! So this is at least one fruitful outcome of our meeting in Gent last month :) (Hopefully, there will be others...) I look forward to your M-Phi posts!
3. Could this post be tagged "PlanetMO" so that it can be found at mathblogging.org/planetmo ?
4. The views of my MO question have risen to 997, only 3 more for a badge. Sad how enjoyable these pointless rewards are.
Peter, I'm not sure what you're asking for.
5. David, I think on blogspot tags are called "labels", they can be added in the "edit post" page.
http://www.mathblogging.org/planetmo came out of a discussion on meta.MO, cf. http://meta.mathoverflow.net/discussion/1002/should-there-be-a-corner-for-discussion-close-to-mo/
6. The Introduction "Introducing Myself" does not have any name.
How does one know who is being introduced here?
7. Anonymous, the answer was only one click away, but why should you have to do that, so I've added some words at the start of the post so you know who I am.
8. David, thanks for adding the label!
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Projectile Motion with vector functions (3D Motion in space velocity)
February 19th 2009, 01:53 PM
Projectile Motion with vector functions (3D Motion in space velocity)
A gun is fired with angle of elevation 30°. What is the muzzle speed if the maximum height of the shell is 484 m? g=gravity (9.8)
What is the initial velocity?
My instructor gave us a hint:
v_0 is initial velocity.
At max height vertical velocity = 0. This gives v_0 sin(alpha) - gt = 0.
Vertical distance = -0.5 gt^2 + v_0 sin(alpha).
Find v_0 from the above two equation.
So I set those two equations equal to each other since they both equal zero. I'm not sure if I can do that. So my initial velocities cancelled out, I solved for t and got sqrt(2). I plugged that
back into the vertical velocity and got v_0 = 28. I then plugged it back into the other equation to check it, but didn't get zero, I got 4.06.
I'm pretty lost.
February 19th 2009, 02:30 PM
$v_0\sin{\alpha} - gt = 0$
$t = \frac{v_0\sin{\alpha}}{g}$
$\Delta y = v_0\sin{\alpha} \cdot t - \frac{1}{2}gt^2$
$\Delta y = v_0\sin{\alpha} \cdot \frac{v_0\sin{\alpha}}{g} - \frac{1}{2}g\left(\frac{v_0\sin{\alpha}}{g}\right) ^2$
$\Delta y = \frac{v_0^2 \sin^2{\alpha}}{g} - \frac{v_0^2 \sin^2{\alpha}}{2g}<br />$
$\Delta y = \frac{v_0^2 \sin^2{\alpha}}{2g}$
$v_0 = \sqrt{\frac{2g \Delta y}{\sin^2{\alpha}}}$
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Torelli theorems for kähler k3 surfaces
Results 1 - 10 of 23
, 1996
"... Abstract. We consider derived categories of coherent sheaves on smooth projective varieties. We prove that any equivalence between them can be represented by an object on the product. Using
this, we give a necessary and sufficient condition for equivalence of derived categories of two K3 surfaces. ..."
Cited by 91 (6 self)
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Abstract. We consider derived categories of coherent sheaves on smooth projective varieties. We prove that any equivalence between them can be represented by an object on the product. Using this, we
give a necessary and sufficient condition for equivalence of derived categories of two K3 surfaces.
- I, Topology , 1993
"... (i) Topology of embedded surfaces. Let X be a smooth, simply-connected 4-manifold, and ξ a 2-dimensional homology class in X. One of the features of topology in dimension 4 is the fact that,
although one may always represent ξ as the fundamental class of some smoothly ..."
Cited by 68 (6 self)
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(i) Topology of embedded surfaces. Let X be a smooth, simply-connected 4-manifold, and ξ a 2-dimensional homology class in X. One of the features of topology in dimension 4 is the fact that, although
one may always represent ξ as the fundamental class of some smoothly
"... In this paper we propose a way to construct an analytic space over a non-archimedean field, starting with a real manifold with an affine structure which has integral monodromy. Our construction
is motivated by the junction of Homological Mirror conjecture and geometric Strominger-Yau-Zaslow conjectu ..."
Cited by 35 (3 self)
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In this paper we propose a way to construct an analytic space over a non-archimedean field, starting with a real manifold with an affine structure which has integral monodromy. Our construction is
motivated by the junction of Homological Mirror conjecture and geometric Strominger-Yau-Zaslow conjecture. In particular, we glue from “flat pieces ” an analytic K3 surface. As a byproduct of our
approach we obtain an action of an arithmetic subgroup of the group SO(1,18) by piecewise-linear transformations on the 2-dimensional sphere S 2 equipped with naturally defined singular affine
- Geom. Funct. Anal , 2001
"... One main problem in the theory of irreducible holomorphic symplectic manifolds is the description of the ample cone in the Picard group. The goal of ..."
Cited by 18 (7 self)
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One main problem in the theory of irreducible holomorphic symplectic manifolds is the description of the ample cone in the Picard group. The goal of
- Nucl. Phys , 1999
"... hep-th/9810210 utfa-98/26 spin-98/4 ..."
"... Given a variety over a number field, are its rational points potentially dense, i.e., does there exist a finite extension over which rational points are Zariski dense? We study the question of
potential density for symmetric products of surfaces. Contrary to the situation for curves, rational points ..."
Cited by 17 (5 self)
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Given a variety over a number field, are its rational points potentially dense, i.e., does there exist a finite extension over which rational points are Zariski dense? We study the question of
potential density for symmetric products of surfaces. Contrary to the situation for curves, rational points are not necessarily potentially dense on a sufficiently high symmetric product. Our main
result is that rational points are potentially dense for the Nth symmetric product of a K3 surface, where N is explicitly determined by the geometry of the surface. The basic construction is that for
some N, the Nth symmetric power of a K3 surface is birational to an abelian fibration over P N. It is an interesting geometric problem to find the smallest N with this property. 1
- on T 6 ,” JHEP 0803 (2008) 022 arXiv:0712.0043 [hep-th
"... For heterotic string theory compactified on T 6, we derive the complete set of T-duality invariants which characterize a pair of charge vectors (Q, P) labelling the electric and magnetic charges
of the dyon. Using this we can identify the complete set of dyons to which the previously derived degener ..."
Cited by 16 (9 self)
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For heterotic string theory compactified on T 6, we derive the complete set of T-duality invariants which characterize a pair of charge vectors (Q, P) labelling the electric and magnetic charges of
the dyon. Using this we can identify the complete set of dyons to which the previously derived degeneracy formula can be extended. By going near special points in the moduli space of the theory we
derive the spectrum of quarter BPS dyons in N = 4 supersymmetric gauge theory with simply laced gauge groups. The results are in agreement with those derived from
, 2007
"... Abstract. We analyze the ample and moving cones of holomorphic symplectic manifolds, in light of recent advances in the minimal model program. As an application, we establish a numerical
criterion for ampleness of divisors on fourfolds deformationequivalent to punctual Hilbert schemes of K3 surfaces ..."
Cited by 9 (5 self)
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Abstract. We analyze the ample and moving cones of holomorphic symplectic manifolds, in light of recent advances in the minimal model program. As an application, we establish a numerical criterion
for ampleness of divisors on fourfolds deformationequivalent to punctual Hilbert schemes of K3 surfaces. 1.
, 909
"... Suppose X is a smooth projective complex variety. Let N1(X, Z) ⊂ H2(X, Z) and N 1 (X, Z) ⊂ H 2 (X, Z) denote the group of curve classes modulo homological equivalence and the Néron-Severi group
respectively. The monoids of effective classes in each group generate cones ..."
Cited by 9 (3 self)
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Suppose X is a smooth projective complex variety. Let N1(X, Z) ⊂ H2(X, Z) and N 1 (X, Z) ⊂ H 2 (X, Z) denote the group of curve classes modulo homological equivalence and the Néron-Severi group
respectively. The monoids of effective classes in each group generate cones
- Manuscripta Math
"... Abstract. A rational Lagrangian fibration f on an irreducible symplecitc variety V is a rational map which is birationally equivalent to a regular surjective morphism with Lagrangian fibers. By
analogy with K3 surfaces, it is natural to expect that a rational Lagrangian fibration exists if and only ..."
Cited by 7 (1 self)
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Abstract. A rational Lagrangian fibration f on an irreducible symplecitc variety V is a rational map which is birationally equivalent to a regular surjective morphism with Lagrangian fibers. By
analogy with K3 surfaces, it is natural to expect that a rational Lagrangian fibration exists if and only if V has a divisor D with Bogomolov–Beauville square 0. This conjecture is proved in the case
when V is the punctual Hilbert scheme of a generic algebraic K3 surface S. The construction of f uses a twisted Fourier–Mukai transform which induces an isomorphism of V with a certain moduli space
of twisted sheaves on another K3 surface M, obtained from S as its Fourier–Mukai partner. According to Beauville [Beau-1], [Beau-2], the d-th symmetric power S (d) of a K3 surface S has a natural
resolution of singularities, the punctual Hilbert scheme S [d] = Hilb d S, which is a 2d-dimensional irreducible
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Enhanced Privacy ID
Sketch of EPID Scheme
We have developed two EPID schemes, one from the strong RSA assumption [7] and the other from bilinear maps [6]. In this article, we briefly sketch the EPID scheme from bilinear maps. (The full
scheme can be found in [6]).
Let us first review some background on bilinear maps. Let G1 and G2 be two multiplicative cyclic groups of prime order p. Let g1 be a generator of G1, and g[2] be a generator of G2. We say e: G1 x G2
→ GT is an admissible bilinear map function, if it satisfies the following properties:
For all u ∈ G1, v ∈ G2, and for all integers a, b, equation e(u^a, v^b) = e(u, v)^ab holds. The result of e(g[1]g[2] ) is a generator of GT. There exists an efficient algorithm for computing e(u, v)
for any u ∈ G1, v ∈ G2.
Our EPID scheme is derived from Boneh, Boyen, and Shacham's group signatures scheme [2] and has the following operations:
Setup: The issuer does the following:
1. Chooses G1 and G2 of prime order p and a bilinear map function e : G1 x G2 → GT.
2. Chooses a group G3 of prime order p with generator g[3].
3. Chooses at random g[1] h[1], h[2] ∈ G1 and g[2] ∈ G2.
4. Chooses a random r ∈ [1, p-1] and computes w = g[2]^r. The public key is (g[1], g[2], g[3], h[1], h[2], w) and the issuing private key is r.
Join: The join protocol is an interactive protocol between the issuer and a member as follows:
1. The member chooses at random f and y' from [0, p-1] and computes
2. The member sends T to the issuer and performs the following proof of knowledge to the issuer:
The issuer chooses at random x and y" from [0, p-1] and computes
3. The issuer sends (A, x, y" ) to the member.
4. The member computes y = y' + y"(mod p). The member's private key is (A, x, y, f ).
Note that given a valid private key (A, x, y, f ), the following equation satisfies:
Sign: Let (A, x, y, f ) be the member's private key. The member does the following:
1. If the random base option is used, the member chooses B at random from G3.
2. If the name base option is used, the member computes B = Hash (verifier's basename).
Computes K = B^f
Computes the following zero-knowledge proof
This essentially proves that the member has a valid EPID private key issued by the issuer.
3. Computes the following zero-knowledge proof
for each (B', K') pair in SIG-RL. This step proves that the member has not been revoked in SIR-RL; that is, the member did not create those (B', K') pairs in SIG-RL
4. Converts all the above zero-knowledge proofs into a signature by using the Fiat-Shamir heuristic [9].
Verify: Given the public key, PRIV-RL, SIG-RL, and an EPID signature, the verifier does the following:
1. If the random base option is used, the verifier verifies that B is an element in G3.
2. If the name base option is used, the verifier verifies that B = Hash (verifier's basename).
3. Verifies that K is an element in G3.
4. Verifies the following proof
This step verifies that the member has a valid EPID private key.
5. Verifies that K ≠ B^f' for each f' in PRIV-RL. This step verifies that the member has not been revoked in PRIV-RL.
6. Verifies the following zero-knowledge proof
for each (B', K') pair in SIG-RL. This step verifies that the member has not been revoked in SIG-RL.
Comparison with Other Techniques
There are other techniques to remotely authenticate hardware, and in this section we review these techniques and compare them with our EPID scheme.
Public Key Infrastructure (PKI). Each hardware device has a unique public and private key pair as well as a device certificate. To authenticate hardware by using PKI, the device simply shows its
certificate to the verifier along with a signature created by using the device's private key. As mentioned previously, this PKI approach does not satisfy the privacy requirement.
Direct Anonymous Attestation (DAA). DAA was designed for anonymous attestation of TPM [4, 5]. DAA satisfies all the design requirements of remote hardware authentication; however, it has limited
revocation capabilities compared to those of EPID. In the DAA scheme, there are two options for a balance between linkability and revocation. If the random base option is used, that is, a
different base is used every time a DAA signature is performed, then any two signatures by a device are unlinkable, but revocation only works if the corrupted device private key has been revealed
to the public. If a device has been compromised, but its private key has not been distributed to the verifiers (for example, if the corrupted device's private key is still under the control of
the adversary), the corrupted TPM cannot be revoked. If the name base option is used, then any two signatures produced by a device, using the same base, are linkable. Thus, if the verifier
determines that a device private key, used in a signature, has been compromised, that verifier can revoke that key locally; that is, the verifier can reject all future signatures generated by
that private key, without knowledge of the compromised private key. However, the verifier cannot tell if a different verifier uses a different name base to revoke that private key, because when a
different name is used, the revoked key cannot be identified. Furthermore, the name-based option does not safeguard privacy, because the verifier can link the transactions.
Group Signatures (GS). A group signature scheme [1, 2] has similar properties to those of the EPID scheme. In a group signature scheme, an issuer creates a group public key and issues unique
private keys to each group member. Each group member can use the private key to sign a message, and the resulting signature is called a group signature. The verifier can verify a group signature
by using the group public key. Unlike EPID, group signature schemes have an additional property called traceability. This property enables the issuer to open any group signature and identify the
actual group member who created the signature. In other words, a group signature is anonymous to the verifiers but not to the issuer. Again, as compared to this scheme, EPID keeps the identity of
the group member from the issuer.
Pseudonym System (PS). The pseudonym system [3], designed by Brands, can also be used for remote hardware authentication. In the pseudonym system, the display of a credential is anonymous by
virtue of the fact that efficient zero-knowledge proof techniques are used for proving relations among committed values. To use the pseudonym system for hardware authentication, each hardware
device obtains a credential from the issuer and uses the pseudonym credential for proof of membership. However, a credential in that system is linkable for multiple displays. To be unlinkable, a
hardware device has to get multiple credentials from the issuer and use one credential at a time. This approach has limited application for hardware authentication, as the hardware device may
never be able connect back to the issuer (the device manufacturer) once it has been produced. Thus, it cannot maintain the unlinkable property by continuing to get new credentials from the
In Table 1, we summarize a comparison between different approaches to the remote hardware authentication problem. The EPID scheme is the only scheme that satisfies all the design requirements
mentioned earlier.
[1] G. Ateniese, J. Camenisch, M. Joye, and G. Tsudik. “A practical and provably secure coalition-resistant group signature scheme.” In Advances in Cryptology -- Crypto, Volume 1880 of Lecture
Notes in Computer Science, pages 255–270, 2000.
[2] D. Boneh, X. Boyen, and H. Shacham. “Short group signatures.” In Advances in Cryptology -- Crypto, Volume 3152 of Lecture Notes in Computer Science, pages 41–55, 2004.
[3] S. Brands. Rethinking Public Key Infrastructures and Digital Certificates: Building in Privacy. MIT Press, Cambridge, MA, 2000.
[4] E. Brickell, J. Camenisch, and L. Chen. “Direct Anonymous Attestation.” In Proceedings of the 11th ACM Conference on Computer and Communications Security, pages 132–145, 2004.
[5] E. Brickell, L. Chen, and J. Li. “A New Direct Anonymous Attestation Scheme from Bilinear Maps.” In Proceedings of 1st International Conference on Trusted Computing, Volume 4968 of Lecture
Notes in Computer Science, pages 166–178, 2008.
[6] E. Brickell and J. Li. “Enhanced Privacy ID from Bilinear Pairing.” Cryptology ePrint Archive, Report 2009/095, 2009.
[7] E. Brickell and J. Li. “Enhanced Privacy ID: a Direct Anonymous Attestation Scheme with Enhanced Revocation Capabilities.” In Proceedings of the 6th ACM Workshop on Privacy in the Electronic
Society, pages 21–30, 2007.
[8] J. Camenisch and V. Shoup. “Practical Verifiable Encryption and Decryption of Discrete Logarithms.” In Advances in Cryptology -- Crypto, Volume 2729 of Lecture Notes in Computer Science,
pages 126–144, 2003.
[9] A. Fiat and A. Shamir. “How to Prove Yourself: Practical Solutions to Identification and Signature Problems.” In Advances in Cryptology -- Crypto, Volume 263 of Lecture Notes in Computer
Science, pages 186–194, 1987.
[10] O. Goldreich, S. Micali, and A. Wigderson. “Proofs that Yield Nothing but their Validity.” Journal of the ACM, Volume 38(3), pages 690-728, 1991.
[11] Trusted Computing Group. “TCG TPM Specification 1.2,” 2003, http://www.trustedcomputinggroup.org
This article and more on similar subjects may be found in the Intel Technology Journal, June 2009 Edition, "Advances in Internet Security.
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Infinitely Divisible Matrices, Kernels, and Functions
Infinitely Divisible Matrices, Kernels, and Functions
Roger Alan Horn
Department of Mathematics, Stanford University., 1967 - Matrices - 244 pages
From inside the book
10 pages matching Schur product theorem in this book
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00 Theorem Chapter choice of arguments completely monotonic function completely monotonic sequence conformal mapping consistent choice continuous function continuous kernel Conversely Corollary cp(x
Cq(S define diagonal difference quotient direct product divisible characteristic function divisible completely monotonic divisible positive definite dM(t dM(x dp(t dv(s dv(t equivalent finite
function f graph Green's function Hermitian Hilbert space ij i,j=l implies incidence matrix inequality infinitely divisible characteristic infinitely divisible completely infinitely divisible
kernels infinitely divisible positive integer interpolation problem irreducible component kernel K(x,y Krein-Milman theorem Lemma Let K(P,Q Let K(x,y limiting argument Loewner Lq(S n x n matrix
nodes nonnegative quadratic form Pick's theorem polynomial principally infinitely divisible probability measure dp Proof prove real valued representation formula Schur product theorem semigroup
stochastic process symmetric matrix teristic function Theorem 2.2 three point property unit disc univalent analytic function
Bibliographic information
Infinitely Divisible Matrices, Kernels, and Functions
Roger Alan Horn
Department of Mathematics, Stanford University., 1967 - Matrices - 244 pages
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Understanding RF power amplifiers | EE Times
Design How-To
Understanding RF power amplifiers
The following is excerpted from Chapter 7 from a new edition of the book, RF Circuit Design, 2e by Christopher Bowick. (If you order a copy of this book before March 30, 2008 you can receive
additional 20% off. Visit www.newnespress.com or call 1-800-545-2522 and use code 91603. )
Class-A Amplifiers and Linearity
A class-A amplifier is defined as an amplifier that is biased so that the output current flows at all times. Thus, the input signal-drive level to the amplifier is kept small enough to avoid driving
the transistor into cutoff. Another way of stating this is to say that the conduction angle of the transistor is 360deg., meaning that the transistor conducts for the full cycle of the input signal.
The class-A amplifier is the most linear of all amplifier types. Linearity is simply a measure of how closely the output signal of the amplifier resembles the input signal. A linear amplifier is one
in which the output signal is proportional to the input signal, as shown in Fig. 7-4. Notice that, in this case, the output signal level is equal to twice the input signal level, and the transfer
function from input to output is a straight line.
7-4. Transfer characteristic for a linear amplifier.
No transistor is perfectly linear, however, and, therefore, the output signal of an amplifier is never an exact replica of the input signal. There are always spurious components added to a signal in
the form of harmonic generation or intermodulation distortion (IMD). These types of nonlinearities in transistors produce amplifier transfer functions that no longer resemble straight lines.
7-5. Nonlinear amplifier characteristics.
Instead, a curved characteristic appears, as shown in Fig. 7-5A. The distortion caused to an input signal of such an amplifier is shown in Fig. 7-5B. Notice the flat topping of the output signal that
occurs due to the second-harmonic content generated by the amplifier. This type of distortion is called harmonic distortion and is expressed by the equation:
The second term of Equation 7-1 is known as the second harmonic or second-order distortion. The third term is called the third harmonic or third-order distortion. Of course, a perfectly linear
amplifier will produce no second, third, or higher order products to distort the signal.
Notice in Fig. 7-5, where the amplifier's transfer function is given as Vout =5V[in] +2V^2[in], that the second-order distortion component increases as the square of the input signal. Thus, with
increasing input-signal levels, the second-order component will increase much faster than the fundamental component in the output signal. Eventually, the second-order content in the output signal
will equal the amplitude of the fundamental. This effect is shown graphically in Fig. 7-6.
7-6. Second-order intercept point.
The point at which the second-order and first-order content of the output signal are equal is called the second-order intercept point. A similar graph may be drawn for an amplifier which exhibits a
third-order distortion characteristic. In this case, the third-order term is plotted along with the fundamental gain term of the amplifier. In this manner, the third-order intercept may be
determined. The second- and third-order intercept of an amplifier are often used as figures of merit. The higher the intercept point, the better the amplifier is at amplifying large signals.
When two or more signals are input to an amplifier simultaneously, the second-, third-, and higher-order intermodulation components are caused by the sum and difference products of each of the
fundamental input signals and their associated harmonics. For example, when two perfect sinusoidal signals, at frequencies f1 and f2, are input to any nonlinear amplifier, the following output
components will result:
fundamental: f[1], f[2]
second order: 2f[1], 2f[2], f[1] +f[2], f[1] - f[2]
third order: 3f[1], 3f[2], 2f[1] ±f[2], 2f[2] ±f[1] +higher order terms
Under normal circuit operation, the second-, third-, and higher-order terms are usually at a much smaller signal level than the fundamental component and, in the time domain, this is seen as
distortion. Note that, if f[1] and f[2] are very close in frequency, the 2 f[1] - f[2] and 2[2] -f[1] terms fall very close to the two fundamental terms. Third-order distortion products are,
therefore, much more difficult to eliminate through filtering once they are generated within an amplifier.
The bias requirements for a class-A power amplifier are the same as those for small-signal amplifiers. In fact, the distinction between a class-A power amplifier and its small-signal counterpart is a
hazy one at best. For all practical purposes, they are equivalent except for input and output signal levels.
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From Gerris
Revision as of 14:13, 5 July 2012; view current revision
←Older revision
| Newer revision→
The Hypre module makes available some features of the high-performance preconditioning library Hypre, in particular its algebraic multigrid for use instead of Gerris' native multigrid solver for
Poisson equations.
There are some special notes on installing Hypre for use with Gerris.
The parameters of the Hypre module can be tuned using an optional parameter block. For example
GModule hypre {
solver_type = boomer_amg
precond_type = none
relax_type = gs-j
coarsening_type = cgc_e
cycle_type = 1
nlevel = 0
verbose = 0
where the values are set to their default. The parameters are
the type of solver to be used,
the type of relaxation to use for the AMG solver,
the coarsening algorithm to use for the AMG solver,
the type of cycle to use for the AMG solver,
the maximum number of multigrid levels for the AMG solver (setting this to zero will use the depth of the quad/octree),
selects whether Hypre should print its own statistics.
So far, only two types of solvers are available:
the AMG solver,
a preconjugate gradient solver.
For the AMG solver various types of cycles can be used and several algorithms are available for relaxation and coarsening. Please refer to the documentation page of the Hypre library for more details
on the different algorithms. The keywords for the different relaxation algorithms are:
Jacobi or CF-Jacobi,
Gauss-Seidel, sequential (very slow!),
Hybrid: SOR-J mix off-processor, SOR on-processor with outer relaxation parameters (forward solve),
Hybrid: SOR-J mix off-processor, SOR on-processor with outer relaxation parameters (backward solve),
Hybrid: SSOR-J mix off-processor, SSOR on-processor with outer relaxation parameters,
Hybrid: GS-J mix off-processor, chaotic GS on-node,
Jacobi (uses Matvec), only needed in CGNR.
The keywords for the different coarsening algorithms are:
CLJP-coarsening - a parallel coarsening algorithm using independent sets,
Classical Ruge-Stueben coarsening on each processor, followed by a third pass, which adds coarse points on the boundaries,
Falgout coarsening (uses 1 first, followed by CLJP using the interior coarse points generated by 1 as its first independent set) ,
PMIS-coarsening (a parallel coarsening algorithm using independent sets, generating lower complexities than CLJP, might also lead to slower convergence),
HMIS-coarsening (uses one pass Ruge-Stueben on each processor independently, followed by PMIS using the interior C-points generated as its first independent set),
CGC coarsening by M. Griebel, B. Metsch and A. Schweitzer,
CGC-E coarsening by M. Griebel, B. Metsch and A. Schweitzer.
For compatibility with Gerris' default solver, the Hypre module also uses the following parameters, set using GfsApproxProjectionsParams and GfsProjectionParams:
the convergence threshold,
the minimum number of iterations,
the maximum number of iterations,
the number of relaxation sweeps on each level.
Note that Hypre uses the RMS-norm of the residual to define its convergence tolerance criteria (rather than the maximum for Gerris). It also calculates the RMS-norm in a different way than Gerris.
This is partly related to the fact that the Hypre module solves a linear probem for which each stencil has been non-dimensionalised by the size of the diagonal cell, whereas the residual norm in
Gerris takes into account the volume of all the cells of the domain. Also, whereas Gerris calculates the norm of the residual only over all the cells of the domain, Hypre takes into accounts the
ghost cells at the boundaries of the domain. To account partially for the difference, a correction factor defined as <math> \sqrt \right( \frac{ \sum_1^n cell_volume }{n} \left) </math> where n is
the number of cells in the domain is used to rescale the norm residual calculated by Hypre.
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Pine Hill, NJ Math Tutor
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My name is Mrs. W. I have 14 years' experience teaching in New Jersey.
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...During my time in college, I took one 3-credit course in Differential Equations. While I was studying, I worked in the Math Center at my college. This gave me the opportunity to tutor students
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Kew Gardens Algebra 2 Tutor
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a "homological dimension" for embedding of manifolds
up vote 7 down vote favorite
Let $A\to B$ be a surjective map of commutative $k$-algebras, and suppose $C\to B$ is a free resolution of $B$ as an $A$-algebra, meaning that $C$ is a free non-negatively graded commutative
$A$-algebra with a differential decreasing degree by 1, and the map to $B$ is a quasi-isomorphism (e.g. $C$ might be a Koszul-Tate resolution). Let $d(C)$ denote the highest degree of a generator of
$C$ over $A$, and let $d(B,A)$ be the smallest possible value of $d(C)$ over all such $C$; it is a kind of non-linear version of homological dimension.
I am particularly interested in the case when $A=C^\infty(M)$ and $B=C^\infty(S)$ for a closed embedding $S\hookrightarrow M$ of manifolds, in which case I denote the quantity above by $d(S,M)$. For
instance, if $S$ is the zero locus of a generic section of a trivial vector bundle $E$ on $M$, $d(S,M)=1$ (just take the Koszul resolution), whereas if the bundle is not trivial but stably trivial,
$d(S,M)=2$. In general, the value of $d$ is the higher the farther away $E$ is from being trivial, in a certain sense that can be made precise.
My question is, has this invariant been studied before, and if so, is there a formula expressing it in terms of some known topological invariants? I'd be happy with the case when $M$ is itself a
vector bundle and $S$ the zero section.
N.B. Asking for $C$ to be merely projective over $A$ would yield a different value of $d$: e.g. one would have $d=1$ for the zero locus of a generic section of any vector bundle. In this case, higher
values of $d$ would detect singularities.
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Fairchilds, TX Science Tutor
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...Usually when I tutor my focus has been in Science and Mathematics, but my minor in Creative Writing and passion for writing have also resulted in helping to proof read countless essays and
papers. If there's a topic you're interested in but it isn't listed below contact me and there's a good cha...
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takeWhile, applied to a predicate p and a list xs, returns the longest prefix (possibly empty) of xs of elements that satisfy p: > takeWhile (< 3) [1,2,3,4,1,2,3,4] == [1,2] > takeWhile (< 9) [1,2,3]
== [1,2,3] > takeWhile (< 0) [1,2,3] == []
takeWhile, applied to a predicate p and a ByteString xs, returns the longest prefix (possibly empty) of xs of elements that satisfy p.
O(n) takeWhile, applied to a predicate p and a Text, returns the longest prefix (possibly empty) of elements that satisfy p. Subject to fusion.
takeWhile, applied to a predicate p and a ByteString xs, returns the longest prefix (possibly empty) of xs of elements that satisfy p.
O(i) applied to a predicate p and a sequence xs, returns the longest prefix (possibly empty) of xs of elements that satisfy p.
O(i) applied to a predicate p and a sequence xs, returns the longest suffix (possibly empty) of xs of elements that satisfy p. takeWhileR p xs is equivalent to reverse (takeWhileL p (reverse xs)).
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Posts by
Total # Posts: 696
For the piecewise function, find the values g(-8), 8(3), and g(8). g(x)={X+6,FOR X<=3 7-x, for x>3
For the piecewise function, find the values g(-8), 8(3), and g(8). g(x)={X+6,FOR X<=3 7-x, for x>3
The researcher structures the study by addressing three topics which include the research question, the researcher s perspective and also the sample selection. On the above sentence, do I need to put
a colon after include, or is the sentence fine the way it is?
Do I use a colon or a period at the end of the following sentence? According to the author of this article, the hypothesis is as follows
How do you solve this problem 1 1/5 minus 3/8?
i will like to make a word with these letter aaiemctmsh
Ratios Math
i need help with ratios in math
Marine Biology
Any ideas where I can read online articles about marine species and the different types of habitats along coastal beaches?
whats limestone used for?
What do algae, kelp, and seaweeds have in common? How are they different?
CIS Microcomputers Application
It could be a cookie................
I dont know what to make for a physics experiment?
RE: Water
If I boil water at home, when and how do I know I've boiled the water long enough to make distilled water?
What is the difference between drinking water and distilled water? Can we make our own distilled water?
Could you tell me which words in column A have the same vowel sound as the words in clumn B. Column A Column B Path Butter Here Weight Saw Trouble Wide Put Hit-------------- File Look Weird Stool
Laugh Luck Taught Heat Cheese Table Prune Pronunciation Eye Pan Matter Thanks.
Where can I download a bowling score sheet?
how do I compute sales tax? $1,367.85. sales tax rate is 5.75%. also, how do I add in city tax?
Where can I download free score-keeping sheets for bowling?
math please help
12x^2y+18xy^2 and the answer is 6xy(2x+3y) how do you get that answer?
I go to the Middletown High School and I lost the combination to my locker which has my Global Histoory textbook in it. I was hoping somebody would be kind enough to tell me what the questions are on
page 26 numbers 3-7 and page 31 numbers 3-7. I realize that we might not have...
science (graphing data)
on the axis of my graph i want to have my scale begin with 100 and then count by tens to 300. can I just start with 100 or do i have to put in a symbol or something first? If so, what's the symbol?
math 117
A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft2, what is the width of the path? using the quadratic formula I did
the following: Let x = distance between the inner rectangle and the outer bound...
Life Science
How did Mendel novel approach contribute to his success in describing how traits are inherited Thank you for using the Jiskha Homework Help Forum. Here are some sites for you: 1. http://
en.wikipedia.org/wiki/Heredity (at the bottom, be sure to click for additional information ...
17 simplify 3/(x-5) + 1/ 1 - 4/(x-5) The non-use of grouping symbols makes this sentence without meaning. The non-use of grouping symbols makes this sentence without meaning. 3/x-5 +1 / 1 - 4/x-5 = I
never know how to write the problems so someone can understnad what I mean, s...
20 Divide: x^2 -49y^2/6x^2+42y divided by (x^2-7xy) = factor the first term. Multiply the first term by the inverse of the second.Combine terms. I will be happy to critique your work. (x-3)(x+5)/2(2x
^2) * 2x-10 /x^2 - 25 = x-3/2x^2 ? or am I missing it somewhere :You are not w...
22. solve x/($a) - x/($b) = c what is this? The world wonders. believe it or not this was actually a problem that I have been given to solve. Any suggestions as to how I accomplish this?
subtract. express in simplest form 13x/30 - 4x/ 15 = factor the first denominator. Get a common denominator. Combine terms. I will be happy to critique your work. ok 13x/30 - 4x/ 15 = This is where I
get lost, I know that I have to fins the LCD but when I do that it looks like...
add:Express in simplest form 4x/(x^2 -18x +72) + 4/(x-6) factor the first denominator. Get a common denominator. Combine terms. I will be happy to critique your work. ok hers goes 4x/(x-6)(x-12) + 4/
(x-6)= 4x + 4/(x-6)(x -12) or am I completely lost? you factored correctly. Yo...
subtract: express your answer in simplest form. 5x-3/6 - (x+3)/6 answer choices a) 4x-1 b) 2x-3/3 c) 2x/3 -1 d) 2x/3 show me hou to arrive at the correct answer get a common denominator. get a common
denominator. I thought 6 was the common denominator, why can't I use it s...
divide: x^2 + 2x -15/4x^2 divided by x^2-25/2x-10 answer choices a) x-3/2x^2 b) x + 3/2x^2 c) 2x^2/x + 3 d) 2x^2/x - 3 any your work is? any your work is? (x-3)(x+5)/4x^2 * 2x-1/x^2-25= x-3/2x^2
Subtract: express your answer in simplest form. 2/5 - y/y-2 answers a)-3y-2/y-2 b) -3y -2/5(4-2) c) -3y-4/5(y-2) d) -3y-4/y-2 The way you have written the problem, is 2/5 minus y^3. I dont think that
is what you meant. Please use grouping symbols on fractions () or [] The way ...
Simplify: 3/4/4/5 answer choices are: a) 3/5 b) 5/3 C) 15/16 d)16/15 How do I go about solving this fraction, please show me how to work this. 3/4 divided by 4/5 1. Invert the second fraction. 3/4 /
5/4 2. Multiply the numerators and denominators. What answer do you get? 15/16...
During rush hour, Fernando can drive 20 miles using the side roads in the same time that it takes to travel 15 miles on the freeway. If Fernando's rate on the side roads is 9 mi/h faster than his
rate on the freeway, find his rate on the side roads. What is the formula and...
What values for x must be exclued in the following fraction? x + 5/ -1 a) 5 b)none c) 0 d)-5 and how do I find the answer? if this is (x+5)/-1, no values. if you had a term like 1/(x-s), hten when x
was equal to s, that is not allowed (rule: exclude all zero solutions in the d...
math 117
x/6- x/8=1 Find a common denominator for the two x-terms on the left. Then do the subraction to get x/? = 1 You do the rest. These are not hard questions
math 117
x/(x-2)- (x+1)/x= 8/(x^2-2x) notice that x^2 - 2x factors to x(x-2), which are found as the denominators of the first two fractions. Multiply each term by x(x-2), the rest is easy.
math 117
5/(x+6)+ 2/(x^2+7x+6)= 3/(x+1) Rewrite x^2 +7x +1 as (x+6)(x+1) Multiply both sides of the equation by that expression. You get 5(x+1) + 2 = 3(x+6) Multiply out the parentheses and combine similar
terms. You will be left with a simple linear equation for x.
math 117
2/(5 )=(x-2)/20
math 117
Kevin earned $165 interest for 1 year on an investment of $1500. At the same rate, what amount of interest would be earned by an investment of $2500? At the same rate of interest, and period, the
interest earned will be proprtional to the original principle. Therefore, you can...
math 117
A plane flies 720 mi against a steady 30 mi/h headwind and then returns to the same point with the wind. If the entire trip takes 10 h, what is the plane s speed in still air?
math 117
Ariana took 2 h longer to drive 360 mi on the first day of a trip than she took to drive 270 mi on the second day. If her speed was the same on both days, what was the driving time each day? Use the
same method indicated in my response to your previous post. Thanks for asking.
math 117
A passenger train can travel 325 mi in the same time a freight train takes to travel 200 mi. If the speed of the passenger train is 25 mi/h faster than the speed of the freight train, find the speed
of each. Speed = distance/time Therefore time = distance/speed. Assume the tim...
If one half of one integer is subtracted from three fifths of the next consecutive integer, the difference is 3. What are the two integers? see other post.
math 117
If one half of one integer is subtracted from three fifths of the next consecutive integer, the difference is 3. What are the two integers? i got 24 and 25. (3/5 of 25) - (1/2 of 24) = 3 (n+1)*3/5 -
1/2 n =3 Right? I will be happy to critique your thinking. I suggest start by ...
math 117/algebra
simplify comples fractions. w+3/4w-----w-3/2w it is w-3 over 4w over w-3 over 2w q/f divided by t/u is the same as q/f multiplied by u/t, or qu/ft Division is the reciprocal of multiplication. q/f
divided by t/u is the same as q/f multiplied by u/t, or qu/ft Division is the re...
math 117/ algebra
Fungicides account for 1/10 of the pesticides used in the US. Insecticides account for 1/4 of all pesticides used in the US. The two ratio of herbicides to insecticides used in the US can be written
1/10 ÷ 1/4 . Write this ratio in simplest form. see the other post abou...
math 117/algebra
The combined resistance of two resistors R_(1 amd R_2 )in a parallel circuit is given by the formula R_(r= 1/(1/R_1 + 1/R_2 )) Simplify the formula. Help what is this talking about and how do I work
it? find R as a function fo r`1 and r2. 1/R= 1/r1 + `1/R2 R= ??? Get a common ...
mat 117/algebra
2/(5w+10 )- 3/(2w-4) please show me how to work this problem. I assume this is some sort of combine the fractions, as there is nothing equal to the sentence above. factor out 1/((5w+10)(2w-4)) 1/
((5w+10)(2w-4)) * ( 2*(2w-4) - 3(5w+10) ) do the multiplication in the paren.. 1/(...
1)Refer to the triangle in the figure. Find an expression that represents its perimeter. lft side 3/4x, right side 5/x^2 third side 1/x^2. 2)Find the perimeter of the given figure. x/2x - 5 and 8/2x
- 3. 3)The volume of the box is represented by (x^2 + 5x +6)(x + 5) Find the p...
Factor completely. ax - ay + x^2 - xy please show me how to factor this problem a(x-y) + x(x-y) = (a+x)(x-y)
find the value of the polynomial 8x - 6 when x = 1 and when x = -1 8(1) - 6= 2 8(-1) - 6= -8 -6 = is the answer -14 or -2 -8 -6 = -14
mat 117/ algebra
My question is about scientific notations. Here is the problem. What I need clarification on is the power signs. When to add, subtract or multiply? (2.4 x 10^-5)(4 x 10^-4) is the anwer 9.6 x 10^-9
or 9.6 x 10^20 The answer is NOT 9.6 x 10^20. Rember your rules of exponents: w...
If the sides of a square are increased by 3 cm, the area is decreased by 36 cm^2. What were the dimensions of the original square? Please show me the formula and how to work. That doesn't sound
possible. If the sides of a square are increased, the area must also be increas...
factor each polynomial completely. to begin, state which method should be applied to the first step, given the guidelines of this section. Then continue the exercise and factor each polynomial
completely. 2p - 6q + pq - 3q^2 Please help, I don't know how or where to begin....
Rewrite the middle term as the sumof two terms and then factor completely. Please show me the formula and how to use it correctly. 12w^2 + 19w + 4 19w can be written as 16w + 3w. Those are the cross
product terms when multiplying 4w times 4 and 3w times 1. That is a clue that ...
Factor each expression a^2(b-c)-16b^2(b-c) Help show me how (b-c) appears in both terms and can be factored out, giving you (b-c)(a^2-16b^2) Now note that the second term can also be factored since
it is the difference of two perfect squares. (a^2-16b^2) = (a+4b)(a-4b) Make th...
Determine whetere each of the following trinomials is a perfect square. If it is, factor the trinomial. x^2 - 24x + 48 I cannot get the 24x I've tried (x-24)(x - 2) What am I doing wrong? It is not a
square. THe factors are 21.797 and 2.20, approximately.
mat 117/algebra
Find a value for k so the 9m^2-kn^2 will have the factors 3m + 7n and 3m -7n. Please show me how to get this and show the work please. Use that (a-b)(a+b) = a^2 - b^2 So, you can read-off that k =
mat 117/algebra
Find all positive values for k which each for each of the following can be factored. (1) x^2 + x + k) k<= 1/4
Which species of Penguins is the largest? Check this site for the penguin species that's the largest. http://www.enchantedlearning.com/subjects/birds/printouts/penguins.shtml
The area of a rectangle of length x is given by 3x^2 + 5x. Find the width of the rectangle. Show me the formula and how to use it please. area= length times width 3x^2 + 5x = x * width divide both
sides by x, then you have width.
mat 117/algebra
Find the GCF of each product. Please show the steps so I can work the problem by myself. it is 6y squared - 3y (6y^2-3y)(y+7) 3y(2y-1)(y+7) 3y will be the greatest common factor, as each term when
you multiply it out will have a 3y. In case you want to work it out.. 3y(2y^2+13...
The price of an item is given by p=2xsquared - 100. Find the polymonial that represents the revenue generated from the sale of x items. Multiply the price term, 2x^2 - 100, by the number of sales, x.
That should be the revenue from selling x units. The p(x) relationship you ha...
math 117/algebra
The US population 1990 was approximately 250 million, and the average growth rate for the past 30 years gives a doubling time of 66 years. The above formula for the US then becomes P(in millions) =
250 x 1(y - 1990)/66 (1) What will the population of the US be in 2025 if this ...
math 117 algebra
The cost i dollarsof manufacturing w wing nuts is given by the expression 0.07w + 13.3. Find the cost when 375 wing nuts are made. What is the average cost to manufacture one wing nut? Put in 375 for
w, and compute the cost. Average cost= costabove/w I'm sorry this still i...
1) The length of one of the equal legs of an isisceles triangle is 8cm less than 4 time the length of the base. If the perimeter is 29 cm, find the length of one of the equal legs. a) 4 cm B) 5 cm C)
11 cm d) 12 cm 2) The perimeter of a rectangle is to be no greater than 300 i...
The sum of tow numbers is 71. The second is 7 more than 3 time the first. What are the two numbers? let x = the smaller number then you know that 7+3x +x =71 solve for x
Sally bought three chocolate bars and a pack of gum and paid $1.75. jake bought two chocolate bars and four packs of gum and paid $2.00. Find the cost of a chocolate bar nad the cost of a pack of
gum. 3C + 1G= 1.75 2C + 4G=2.00 solve. I know that i have to eliminate the c'...
Rewrite the equation -x -6y = -6 as a function of x. add x to both sides, add six to both sides.
solve each of the following systems by substitution. 16) 5x -2y = -5 y - 5x = 3 20) 8x -4y = 16 y = 2x - 4 28) 4x -12y = 5 -x + 3y = -1 Step one: Solve on of the equations for one of the variables. y
=2x-4 Step two:Substitute the expression for the variable found in step one in...
if f(x= 5x - 1 find f(a-2) Please show me the work so I can understand how to arrive at the solution. Thank you. f(a-2)=5(a-2)-1
the inventor charges $4.00 per unit, then her profit for producing and selling x units is given by the function P(x) = 2.25x - 7000 (a) What is her profit if she sells 2000 units? (b) What is per
profit if she sells 5000 units? (c) What is the break-even point for sales? (a) S...
I have a graph that shows a ---- kind of 3 line above the solid line for x & y. My assignment is as follows. Whe have graphed the boundary line for the liear inequality. Determine the correct
half-plane. y >3 First how do I solve and 2 how do I graph it? If the allowed area...
How do you factor 3x^2-x-4? thanks in advance How do you factor 3x^2-x-4? thanks in advance 3 times what = 3? 3 of course. giving you (3x+/-a)(1x+/-b). a and b can only be 1 and 4 or 2 and 2. Can you
take it from here? (3x-4)(x+1)?
If you have steel and wood at 0 Celsius, which is colder Both are the same temperature.
What are Active X controls? Are we supposed to "enable" or "disable" them? does not matter http://www.webopedia.com/TERM/A/ActiveX_control.html You have to use your judgment, depending on which
program seems to be requiring it.
What are the main points to consider when choosing a monitor? What is the footprint? The user's eye sight. Cost. There may be othrs.
computer science
Is there a website that explains computer terms and definitions? http://www.computeruser.com/resources/dictionary/ http://www.webopedia.com/ Another site from your first post is below.
computer science
Is there a website that explains computer terms and definitions? http://whatis.techtarget.com/
I don't know why you said it was perfectly inelastic when is not. Is relative elastic
Which one is correct: As president Bush walked to rhe podium,... or As President Bush walked to the podium,.... Since that is his official title, it would be treated the same way as Mr., or Gen. Cap
on Pres
My thesis is: Our nation needs a health care plan capable of addressing the needs of each individual. My Question is: How can I make this into a powerful thesis staement? You could start out by
indicating that there is a lack of adequate health care for many individuals in an ...
Can you tell me what type of essay I would be writing in response to this statement: "What, if anything, is significant about an issue--expressed in a claim?" My instructor told us to write an essay
to this question. A "claim" is usually the term for the th...
can you tell me what direction to take with this? My essay: During the last century, the average life span of Americans nearly doubled, from 49 years in 1900 to nearly 80 years in 2000. Increased
life expectancy and advances in U.S. healthcare means that Americans now live lon...
Based on what I have so far, what direction do you think I should go--what do you feel the thesis is about? This will eventually end up 10 pages long. I am concerned that I will not find enough to
write about. The assignment is, "What, if anything, is significant about an...
Can you explain the differences in these terms, and the function they serve when writing an essay? thesis and thesis statement central reasoning and central relationship The thesis is your central
idea, your main idea, the point you are trying to prove in your paper. The thesi...
social science
Is iot possible to find out the number of physicians in a given practice--per state? example: how many doctors have a practice in geriatrics in the state of Utah? http://www.ama-assn.org/cgi-bin/
sserver/datalist.cgi I found it by state and county, but not just by state.
how can I send material that is only intended for teachers, instructors, etc..and not the world? i don't think you can. you have to copy and paste your material onto this forum and teachers will help
you with it. I believe that's the only way. "anon" is right...
May I submit my draft for an essay for your comments? Please put your draft on the board and one of the teachers will be glad to comment.
Please don't post my question or your answer because I do not want others to steal my idea! The only way we have to communicate with you is by posting on our message boards. Sorry. =( I need to learn
to type on A computer. http://www.bbc.co.uk/schools/typing/flash/stage1.s...
Do you think it is truly possible to get all of the information I would need to write an essay about the sudden "halt" or end of production of electric cars? And, what about the Hybreds? I am
apprehensive about this topic because I may not be able to find all there i...
Is there a method of determining if you debit or credit an account? How do you know if you should debit or credit a transaction? If it's a balance sheet accout you need to learn (and mrmorize) which
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which is saltier the gulf of mexico or pacific ocean http://www.windows.ucar.edu/tour/link=/earth/Water/images/salinity_big_gif_image.html&edu=high The more orange, the higher salinity
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The length of a rectangle is 5 in. more than twice its width.? How do I find the width of the rectangle. duplicate post Divide the length by 2 it's going to be a decimal
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such as the area or the perimeter. I don't think there is enough information ...
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prove M is complete
December 3rd 2011, 05:15 PM #1
Senior Member
Jan 2010
prove M is complete
Suppose that every countable, closed subset of M is complete. Prove that M is complete.
my idea is to show that M is closed first, then let (Xn) be a cauchy sequence in M, so it's cauchy in all of its subset and then converge, so M is complete.
Re: prove M is complete
But, $M$ may not be countable, so this doesn't work. Try proving that if $(x_n)$ is a Cauchy sequence then $X=\overline{\{x_n:n\in\mathbb{N}\}}$ contains at most one more element and thus is also
countable. But, then $(x_n)$ is a Cauchy sequence in the countable closed subset $X\subseteq M$.
Re: prove M is complete
if A is subset of M, and (Xn) is cauchy in M, does it imply (Xn) is cauchy in A?
Re: prove M is complete
December 3rd 2011, 05:23 PM #2
December 3rd 2011, 05:45 PM #3
Senior Member
Jan 2010
December 3rd 2011, 05:46 PM #4
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I’ve often heard it argued that learning maths is like learning another language. There is a whole vocabulary and a way of speaking that is alien to people who don’t live in the land of maths.
Abstract concepts are understood by saying things that only maths people understand. The conjunction is the equals sign; verbs are operators; a degree-level literature essay is a second-order
differential equation.
If that’s true, then Fraction County is the kind of place where the banjo stops playing when you walk into a bar. The talking stops. The locals all put down their home-made moonshine and all that can
be heard is the faint rustle of tumbleweed blowing along the street outside. And you realise that the language they were talking is a completely different dialect from one that you’ve understood
It is no wonder that many children panic when they hear the word “fraction”.
Think about this.
The children walk into a room and see ¼ written on the board. The teacher asks “how do you say this?”
A brave child sticks their hand up and says “one line four”. Another child, emboldened by the first contribution, suggests “one point four”. Then someone asks “is it a fourth?”
“That’s not exactly how we say it,” corrects the teacher, obliquely referring to some shadowy group of people the children have never heard of. A group of people that obviously can already speak
‘Fraction’. “We say ‘quarter’” The teacher smiles reassuringly, but inside is concerned. She knows that the children should already be able to read and say a quarter and she utters a silent curse
at the children’s previous teacher.
The lesson continues. The children learn that fractions are something to do with pizza (or if you listen to Sal Khan, pie). Then, after seeing that ¼ of a pizza is one piece out of 4, the teacher
holds up 4 multilink cubes that are all joined together in a small tower. She asks the children how many cubes are in the tower. The children say “four”. The teacher breaks off a cube. She asks
how many cubes she broke off. The children say “one”. “Ah, but what fraction did I break off?” asks the teacher, with an air of mystery. “Half?” asks a child. “A third?” asks another.
Ever patient, the teacher persists. “How many cubes were in the tower?”
“So what is the ‘out of’ number?”
“So this cube is one out of four,” declares the teacher triumphantly, writing ¼ on the board again. “How do we say that?”
“One line four” says a child.
“One four” says another.
“Quarter” says a third.
“Yes,” says the teacher, pouncing on the learning. She vigorously shakes the child in sheer joy that someone has got it. “And we write a quarter, one over four.”
The problem is in the language. The children have already learned that division is one word that means two different things – sharing and grouping. Now there’s the whole same thing going on with
fractions. They’re sharing pizzas and calling each piece a fraction. Then they’re grouping sets of objects into equal subsets and calling each subset a fraction. Then despite the fraction being
called “a quarter”, the teacher describes it as being “one out of four” whilst explaining that you write it “one over four.” The concepts behind these aren’t impossible to grasp, but the language we
use to describe them is just so inefficient.
This is one of the reasons that my favourite thing to come out of the old National Numeracy Strategy was the book on maths vocabulary – describing the kind of words that children should be taught in
each subsequent year.
But knowing the words is only part of the problem. I know some French words and some Spanish words but (to my shame) I find it hard to put them in the right order. The language of ‘Fraction’ is
similar. It takes practice and good teaching to put them in order. If your teacher is woolly in their teaching and you don’t practice enough, you won’t learn the language. Worse, I know plenty of
people whose maths teacher lost patience with them during some maths lesson or the other and shouted at them for not getting it quickly enough. This is often a reflection on that teacher’s subject
knowledge, not the maths ability of the student. It is a reason why I recommend Derek Haylock’s excellent book on teaching maths.
So next time you’re on the road to Fraction County, make sure you’ve rehearsed some of your lines – you may just teach your child to know their denominators from their numerators.
Good for the fractions learning; bad for the coffee mug
Sometimes children hear the word 'fractions' and they turn off.
I saw it on Wednesday when I started my lesson on comparing and ordering fractions. I had barely uttered the words when I saw a few heads drop. A few children joined in when I asked them what they
knew about fractions – one knew the word 'third'; someone else knew 'part'; yet another one knew they have something to do with division. But quite a few heads with dropped.
So while the keen had their hands up, and others were looking to avoid eye contact, I slid an empty coffee mug into an empty plastic bag. Then, for security, whilst the conversation continued, I
placed the first plastic bag into a second one.
Then I smacked it against the wall. Really hard.
All the children looked – some jumped.
I proceeded to pull pieces out of the bag and estimate how much of the mug each piece had been, from the large chunks (1/3 or 1/5) to the tiny chips that were only 1/1000 or maybe even smaller.
The children were engaged and by the end of the lesson all of them had made some progress about ordering and comparing fractions. Even the special needs group children who, according to their data,
struggle to order numbers 1-100.
As a bonus, we even specified that the bottom of the fraction was called the denominator and the top number the numerator – I love it when children learn proper maths words, although it was amusing
to hear one child call the top number the nominator and the bottom number the dominator.
So, if you're stuck with teaching fractions – break something. At least you'll stop the heads from dropping…
Valuing misconceptions on the way to explaining fractions
I filmed this about 6 months ago, following an excellent session about fractions on the Mathematics Specialist Teacher Programme. The challenge that we were given, and then I in turn gave to the
children, was given a 4-pint bottle of milk that gets 3/5 of a pint drunk each day, how many days does the milk bottle last for? Those of us with a formal background in maths would say:
4 ÷ 3/5
= 4 ÷ 3 x 5
= 4 x 5 ÷ 3
= 20 ÷ 3
= 6 r 2.
So the milk lasts for 6 and a bit days. If we wanted to be really fancy we would say the milk lasts for 6 and 2/3 days. And isn't it more practical to say the milk lasts for 6 days and there's 2/5 of
a pint left over? Does our understanding of the algorithms let us say that?
Also can children, who are without the drilled-in knowledge that when you divide a divisor you actually multiply, do this question?
That's what the video explores – and there's some interesting misconceptions on the way.
Fractions: learning something new
Yesterday was a complete surprise to me. I learnt something new about maths. And I enjoyed it.
Without trying to show off, I do know a lot of maths. I won’t bore you with too many of the details, but I am both interested in maths and quite good at it. I recognise that there are a lot of people
who are much better than me – without some of those people I would never have got through my ‘A’ level maths (thanks Greg, thanks Yao) nor my Engineering degree (thanks, Jim, thanks Dan). However, in
primary teaching I haven’t met too many of those people. Most of my colleagues are good at teaching maths, but would say that it is not their main interest. Some would demonstrate an enthusiasm for a
particular branch of maths, whilst a few would express some negativity about areas of maths, particularly at the higher levels of the primary age range. Yesterday’s topic at the local area meeting of
the MAST programme was ‘fractions’ – an area of maths which usually generates the word ‘Hmph’ from children, parents and teachers alike. I was so excited by some of the fractions problems we
attempted I took them straight back to school the next day and filmed my Year 6 children trying to solve them. Here’s the video: http://www.youtube.com/get_playerHopefully you can see how the
children progressed in the lesson. Many of the children, despite being the most able in the school, had quite a negative attitude to solving problems involving fractions. Through using models and
images the children now have a better conceptual understanding of fractions – they have linked the visual to the concrete – and are now ready to move on to using the abstract: numerical fractions
themselves. It struck me that as teachers we often move too quickly from the concrete to the abstract. If the highest ability children needed this level of input to begin to ‘get it’, then younger
children and less able will need far more input at the concrete and visual stage before they move on to the abstract. This makes complete common sense, but in our overly prescriptive curriculum, how
often do we rush children on to using and failing with the numbers when they don’t get the concept? So if two and half men take two and half days to dig two and half trenches, how many trenches can
one man dig in one day? My answer was one trench and I was completely wrong. The feeling was exquisite – some maths that I didn’t get. My table group had to work hard to try and solve the problem and
we still didn’t get it. Finally when someone provided a solution and the concept started to sink in it was marvellous to realise that I had been challenged with something and learnt something new as
a consequence.
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just an integral...
Hi everyone! Could someone help me with evaluating the following integral?
[itex]\int_{2 \pi /L}^{\pi/l_0} \int_{2 \pi /L}^{\pi/l_0} \frac{\cos(k_x \Delta x)}{k_x^2 + k_y^2} dk_x dk_y [/itex]
I have a good reason to believe that it will end up with some
[itex] \frac{1}{2} \ln (\frac{L}{\Delta x})[/itex]
though this might just be some approximation of it, since
[itex] l_0 \ll L, \Delta x \ll L[/itex] .
Any help would be appreciated! Thank you!
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The plausibility of alternative placements for theropod taxa
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The plausibility of alternative placements for theropod taxa
• To: dinosaur@usc.edu
• Subject: The plausibility of alternative placements for theropod taxa
• From: Michael Mortimer <mickey_mortimer111@msn.com>
• Date: Fri, 21 Mar 2008 03:55:38 -0700
• Reply-to: mickey_mortimer111@msn.com
• Sender: owner-DINOSAUR@usc.edu
• Sun-java-system-smtp-warning: Lines longer than SMTP allows found and wrapped.
David Marjanovic wrote-
>> I should try adding the holotype to a non-maniraptoriform supermatrix I've
>> been working on (including all codings from Smith et al., 2007; Tykoski,
>> 2005; Carrano and Sampson, 2007; Rauhut, 2003; Azuma and Currie, 2000;
>> Allain, 2002; Ezcurra and Novas, 2006; etc.).
> Wow. Is that supermatrix your thesis?
Nah. As you know, ensuring a preexisting matrix is coded correctly and has
objective character states takes as much time as creating your own matrix. So
I would never use such a supermatrix for more than just seeing where the
currently published matrices suggest taxa belong. I made it because I was
curious about ceratosaur and coelophysoid monophyly. So far it contains all
the characters, taxa and codings from-
Smith et al.'s (2007) large theropod matrix in the Cryolophosaurus paper. (347
Tykoski's (2005) matrix focusing on coelophysoids and ceratosaurs from his
thesis. (95 additional characters)
Carrano and Sampson's (2007) ceratosaur matrix. (39 additional characters)
Excurra and Novas' (2006) matrix focusing on coelophysoids from their
Zupaysaurus paper. (61 additional characters)
Rauhut's (2003) theropod matrix, including Xu et al.'s (2006) codings for
Guanlong and Dilong, Lamanna's (2004) codings for Megaraptor, and Yates' (2005)
codings for Dracovenator and several other taxa. (20 additional characters)
Allain's (2002) matrix focusing on spinosauroids and carnosaurs from his
Dubreuillosaurus paper (7 additional characters)
Azuma and Currie's (2000) matrix from their Fukuiraptor paper (31 additional
characters), which is basically the same as Currie and Carpenter's (2000)
Acrocanthosaurus paper.
Langer and Benton's (2006) matrix of basal dinosaur relationships (25
additional characters).
Yates' (2006) sauropodomorph matrix (134 additional characters so far), in
order to get good representation of the theropod sister group.
I only checked character accuracy when two matrices disagreed, and generally
didn't code taxa for characters nobody else had. Thus while superior to the
studies listed above, it's by no means a finished product and represents more
of a compilation of prior work than my own work. Coding 131 taxa for 763
characters the proper way is too large a project for me at the moment- that's
what my coelurosaur analysis is and it's taking years. :) Obvious future steps
include adding Holtz et al.'s (2004), Carrano et al.'s (2005), Sereno et al.'s
new carcharodontosaurid stuff, and the info from the Lophostropheus,
Tyrannotitan and Mapusaurus papers. Important taxa to add would be
Scleromochlus, Lewisuchus, Sacisaurus, Eucoelophysis, Agnostiphys,
Chindesaurus, Sarcosaurus, Chuandongocoelurus, Berberosaurus, Brontoraptor,
Spinosaurus, Marshosaurus, Erectopus, Gasosaurus, Lourinhanosaurus,
Yangchuanosaurus, Mapusaurus, Saurophaganax, Huaxiagnathus, Sinocalliopteryx,
Tanycolagreus, Eotyrannus,
Aniksosaurus, Juravenator, Nedcolbertia, Nqwebasaurus, Santanaraptor and
Scipionyx. Also, coding Falcarius, Protarchaeopteryx, Incisivosaurus,
Sinovenator, Mei, Buitreraptor and/or Rahonavis in addition to the two birds
and two derived dromaeosaurids would quite probably affect the placement of
Ornitholestes and Bagaraatan.
Current results are-
`--+--Ornithischia (5 taxa)
|--Sauropodomorpha (4 taxa)
| `--Herrerasaurus
| | |--Dilophosaurus wetherilli
| | |--"Dilophosaurus" sinensis
| | `--Cryolophosaurus
| `--+*-Gojirasaurus
| |--Lophostropheus
| |--Liliensternus
| |--Zupaysaurus
| `--+*-Shake'n'Bake taxon
| |--+--Segisaurus
| | `--"Syntarsus" kayentakatae
| `--+--Coelophysis
| `--+--Megapnosaurus
| `--Procompsognathus
| | `--Spinostropheus
| `--+--Ceratosaurus
| `--+--+--Noasaurus
| | |--Genusaurus
| | |--Masiakasaurus
| | `--Laevisuchus
| `--+--Ekrixinatosaurus
| |--Rugops
| `--+--Abelisaurus
| `--+--+--Rajasaurus
| | |--Majungasaurus
| | `--Indosuchus
| `--+--Ilokelesia
| `--+--Aucasaurus
| `--Carnotaurus
| `--+--+--+--Eustreptospondylus
| | `--+--Piveteausaurus
| | |--Dubreuillosaurus
| | `--Afrovenator
| `--+--Torvosaurus
| `--+--Chilantaisaurus
| |--Irritator
| |--Baryonyx
| `--Suchomimus
| |--Megaraptor
| |--Carcharodontosaurus
| `--Giganotosaurus
| `--+--Acrocanthosaurus
| `--+--Allosaurus
| `--Sinraptor
| `--+--Stokesosaurus
| `--Tyrannosaurus
| |--Mirischia
| `--Aristosuchus
| `--Coelurus
| `--+--Compsognathus
| `--Siamotyrannus
`--+--Ornithomimosauria (2
`--Paraves (4 taxa)
I forced various constraint trees to estimate the rough liklihood that
alternative hypotheses are correct. I tried to test as many possibilities as I
could recall were ever suggested in the last thirty years, with some older ones
in there for good measure. Of course, with near certain coding errors, not
every taxon coded for every character possible, and missing characters and
taxa, these numbers shouldn't be taken as the final word. But a topology that
needs 5 more steps is far more likely to be true than one that needs 20 more
steps. So, starting with the most likely alternatives and working to the the
least likely ones, with my subjective divisions of how liklihood correlates to
extra steps in the trees...
Extremely possible-
The current tree, with the following non-consensus aspects-
Silesaurus as a basal saurischian. Never suggested before, but all the data
from Langer and Benton (2006) which kept it out of Dinosauria was included.
Eoraptor and herrerasaurids as successively closer outgroups to Avepoda, as
suggested by Sereno et al. (1993). This despite the fact Langer and Benton's
data were included.
Dilophosaurids as coelophysoids, as in Paul (1988) and most cladistic analyses.
With Rauhut (2003), Yates (2005) and Smith et al. (2007) all contributing to
the data, it certainly seems to support coelophysoid dilophosaurids.
"Syntarsus" kayentakatae being closer to Segisaurus than to Megapnosaurus, as
suggested by Tykoski (2005).
Spinostropheus clading with Elaphrosaurus to the exclusion of other
ceratosaurs, as in Lapparent (1960) and Carrano and Sampson (2007).
Ceratosaurus closer to abelisauroids than Elaphrosaurus is, as in Holtz (2000).
Ilokelesia being a carnotaurine, as in Carrano and Sampson (2007).
Streptospondylus being a basal tetanurine outside the
Spinosauroidea+Avetheropoda clade, which hasn't been suggested before.
Poekilopleuron being a basal spinosauroid, which has not been suggested before.
Torvosaurus being closer to spinosaurids than to eustreptospondylids, as in
Rauhut (2003).
The generally basalmost tetaurines Piatnitzkysaurus, Condorraptor,
Xuanhanosaurus and "Szechuanoraptor" being closer to avetheropods than
spinosauroids are. This has been suggested previously for Piatnitzkysaurus
(Novas, 1992) and "Szechuanraptor" (Chure, 2000).
Monolophosaurus being outside Avetheropoda, as in Smith et al. (2007).
Carcharodontosaurids being outside Avetheropoda, as in Coria and Salgado (1995).
Sinraptor being closer to Allosaurus than to Neovenator or Acrocanthosaurus,
which I don't think has been suggested before.
Neovenator being closer to allosaurids than carcharodontosaurids, as in Hutt et
al. (1996).
Acrocanthosaurus being closer to allosaurids than carcharodontosaurids, as in
Stovall and Langston (1950).
Fukuiraptor being a coelurosaur, as in Longrich (2001).
Deltadromeus being a tyrannosauroid, which hasn't been suggested yet. Though
the possibly synonymous Bahariasaurus has (Paul, 1988; Chure, 2000).
Dilong being closer to birds than tyrannosaurids, as in Turner et al. (2007).
This leaves no reason to postulate secondarily featherless tyrannosaurids.
Mirischia and Aristosuchus being closer to Dilong than Compsognathus (as in
Niaish, online 2006; though he had Dilong as a tyrannosauroid).
Guanlong being closer to birds than tyrannosaurids (and Dilong!), which hasn't
been suggested before and is frankly the opposite of what I expected.
Siamotyrannus being a compsognathid, which hasn't been suggested before.
Bagaraatan being a maniraptoran, as in Rauhut (2003).
Forcing Elaphrosaurus to be closer to abelisaurids than Ceratosaurus is (as in
Holtz, 1994) adds only one step.
Forcing Streptospondylus to be sister to Eustreptospondylus (as in Allain,
2001) adds only one step. Streptospondylus moves into Spinosauroidea, whose
topology remains the same.
Forcing Piatnitzkysaurus to be outside the spinosauroid-avetheropod clade (as
in Rauhut, 2003 and Smith et al., 2007) adds only one step. Condorraptor is
its sister taxon in this case, as in Smith et al. (2007), and Xuanhanosaurus
and "Szechuanoraptor" move along with it (as in Rauhut, 2003).
Forcing Neovenator to be a carcharodontosaurid (as in Rauhut, 2003 and others)
adds one step. The (Acrocanthosaurus(Sinraptor+Allosaurus)) clade remains,
while Neovenator is the sister to Megaraptor in basal Carcharodontosauridae
outside Avetheropoda.
Forcing Siamotyrannus to be a carnosaur (Pharris, 1997) adds one step. The
topology among paraphyletic carnosaurs is unchanged and it is a sinraptorid
(which is where Pharris predicted it would be).
Forcing Mirischia to be a compsognathid (as in Martill et al., 2000) adds one
Forcing Coelophysis and Megapnosaurus rhodesiensis to be sister taxa (as in
Tykoski, 2005 and others) adds 2 steps.
Forcing Eustreptospondylus to be closer to spinosaurids than Torvosaurus (as in
Smith et al., 2007) adds 2 steps. The other eustreptospondylids move with
Forcing Poekilopleuron to be a torvosaurid (as in Galton and Jensen, 1979) adds
2 steps. The newly formed Torvosauridae are still sister to Spinosauridae.
Forcing Chilantaisaurus to be a tyrannosauroid (as in Paul, 1988) adds 2 steps.
Interestingly, this is rather close to where "Chilantaisaurus" maortuensis
ends up.
Forcing Allosauroidea (allosaurids, carcharodontosaurids and sinraptorids) (as
in Sereno et al., 1996) to be monophyletic adds 2 steps. The topology within
it stays the same, so that carcharodontosaurids are most basal and Sinraptor is
nested within allosaurids. Monolophosaurus is still basal to Avetheropoda,
while Fukuiraptor and Siamotyrannus are coelurosaurs.
Forcing Monolophosaurus to be a spinosauroid (as suggested by Headden on the
DML, 2002) adds 2 steps. It is the most basal spinosauroid in these trees.
Forcing Fukuiraptor to be sister to Siamotyrannus (as in Holtz et al., 2004)
adds 2 steps. The two are basal coelurosaurs outside Tyrannoraptora.
Forcing Procompsognathinae sensu Sereno (Procompsognathus + Segisaurus) (as in
Sereno and Wild, 1992) adds 3 steps.
Forcing Dracovenator to be closer to Ceratosauria + Tetanurae than to
Coelophysoidea (as in Yates, 2005) adds 3 more steps. Oddly, it ends up as a
spinosaurid, while Cryolophosaurus is sister to Ceratosauria+Tetanurae and
"Dilophosaurus" sinensis is a basal tetanurine. Dilophosaurus wetherilli stays
in Coelophysoidea.
Forcing Deltadromeus to be a ceratosaur (as suggested by Sereno et al., 2002)
adds 3 more steps. It falls in the Elaphrosaurus + Spinostropheus clade.
Forcing Carnotaurus and Majungasaurus to clade to the exclusion of Rajasaurus
(as in Wilson et al., 2003) adds 3 steps.
Forcing Sinraptoridae to be basal to Avetheropoda (as in Paul, 1988) adds 3
steps. In this tree, carcharodontosaurids are coelurosaurs.
Forcing Fukuiraptor to be a carnosaur (as in Azuma and Currie, 2000 and Holtz
et al., 2004) adds 3 more steps. It is placed sister to Allosaurus, while
carcharodontosaurids and Monolophosaurus are still outside Avetheropoda.
Forcing Aristosuchus to be a coelurid (as in Lydekker, 1889) adds 3 steps. In
this tree, Guanlong is also a coelurid.
Forcing Coelurus to be a compsognathid (as in Lull, 1911; technically,
Compsognathus would be a coelurid...) adds 3 steps.
Forcing Eusaurischia to exclude herrerasaurids and Eoraptor (as in Langer and
Benton, 2006) adds only 4 more steps. These trees have the same topology as
Langer and Benton's paper, with Silesaurus outside Dinosauria, Eoraptor closer
to eusaurischians than herrerasaurids are, and Guaibasaurus as a basal theropod.
Forcing Procompsognathus to be the most basal avepod (as in Paul, 1988) adds 4
Forcing Cryolophosaurus to be a basal tetanurine (as in Smith et al., 2005)
adds 4 more steps. In these trees, "Dilophosaurus" sinensis is its sister
Forcing Piatnitzkysaurus to be a spinosauroid (as in Holtz et al., 2004) adds 4
steps. It falls out (often with Condorraptor) as the basalmost spinosauroid,
even more basal than Poekilopleuron.
Forcing Neovenator to be the sister taxon to Allosaurus (as in Hutt et al.,
1996) adds 4 steps. Acrocanthosaurus, Sinraptor and Megaraptor form successive
outgroups, while carcharodontosaurids are coelurosaurs.
Forcing Coelurus to be a maniraptoran (as in Gauthier, 1986) adds 4 steps. It
is directly basal to Ornitholestes in this tree.
Forcing Proceratosaurus to be sister to Ornitholestes (as in Paul, 1988) adds 4
steps. This pairing ends up in basal Maniraptora.
Forcing Megapnosaurus rhodesiensis and "Syntarsus" kayentakatae to be sister
taxa (as in Rowe, 1989) adds 5 more steps. In these trees, Procompsognathus is
closest to Megapnosaurus sensu lato, with Coelophysis one step further out.
Segisaurus is now in the Zupaysaurus-Liliensternus mess.
Forcing Deltadromeus to be sister to ornithomimosaurs (as in Rauhut, 2003) adds
5 more steps.
Forcing Spinostropheus to be sister to abelisaurs (but keeping Elaphrosaurus as
the basalmost ceratosaur; as in Sereno et al., 2004) adds 5 more steps.
Forcing Megaraptor to be a spinosauroid (as suggested by Calvo et al., 2004)
adds 5 steps. It ends up sister to Torvosaurus+Spinosauridae.
Forcing Siamotyrannus to be a tyrannosauroid (as in Buffetaut et al., 1996)
adds 5 steps. Guanlong and Dilong are still closer to birds than to
Forcing Bagaraatan to be a tyrannosauroid (as in my analyses from 2003 onward)
adds 5 steps.
Forcing Torvosaurus to be closer to eustreptospondylids than to spinosaurids
(as in Sereno et al., 1994) adds 6 steps. In these trees, the four 'basalmost
tetanurines' (Piatnitzkysaurus, Xuanhanosaurus, Condorraptor and
"Szechuanoraptor") all move outside the spinosauroid-avetheropod clade.
Forcing Afrovenator to be outside the Eustreptospondylus + Torvosaurus +
Spinosauridae clade (as in Sereno et al., 1994) adds 6 steps. Dubreuillosaurus
and Piveteausaurus stay with Afrovenator, and Torvosaurus is still closer to
spinosaurids than Eustreptospondylus.
Forcing Monolophosaurus to be a carnosaur (as suggested by Zhao and Currie,
1993) adds 6 steps. It is the basalmost carnosaur, and the topology of
Allosauroidea is the same as when its monophyly is forced, with Fukuiraptor and
Siamotyrannus still coelurosaurs.
Forcing Ornitholestes to be a coelurid (as in Matthew and Brown, 1922) adds 6
Forcing Ornitholestes to be outside Maniraptoriformes (as in Paul, 1988) adds 6
Forcing Proceratosaurus to be a tyrannosauroid (as suggested by me on the DML)
adds 6 steps.
Forcing Dilophosaurus to be sister to Ceratosauria + Tetanurae (as in Rauhut,
2003) adds 7 more steps. In these trees, Dracovenator is sister to
Dilophosaurus, while "Dilophosaurus" sinensis and Cryolophosaurus form a clade
one node closer to ceratosaurs+tetanurines.
Forcing Deltadromeus to be a noasaurid (as in Wilson et al., 2003) adds 7 more
Forcing Xuanhanosaurus to be a torvosaurid adds 7 steps.
Forcing Acrocanthosaurus to be a carcharodontosaurid (as in Sereno et al.,
1996) adds 7 steps. In these trees, Neovenator is a more basal
carcharodontosaurid, while Sinraptor and Allosaurus form a clade sister to
Carcharodontosauridae. Monolophosaurus and Megaraptor are outside
Avetheropoda, Fukuiraptor and Siamotyrannus are coelurosaurs, and Tyrannotitan
is basal to Carcharodontosaurus+Giganotosaurus.
Forcing Guanlong (as in Xu et al., 2006), Dilong (as in Xu et al., 2004) and/or
Mirischia (as in Naish, online 2006) to be tyrannosauroids adds 7 steps. They
form a clade with Aristosuchus to the exclusion of Tyrannosauridae, with
"Alashansaurus" as the basalmost tyrannosauroid.
Forcing Compsognathus to be outside Tyrannoraptora (as in Holtz, 1994) adds 7
Forcing Proceratosaurus to be basal to Coelurus and tyrannosauroids (as in
Holtz, 2000) adds 7 steps.
Possible but not well supported-
Forcing Liliensternus to be more closely related to Dilophosaurus than to
Coelophysis (as in Paul, 1988) adds 8 steps. Dracovenator (and sometimes
Lophostropheus) joins this 'halticosaur' clade, while Zupaysaurus and
Gojirasaurus stay closer top Coelophysis. Cryolophosaurus and "Dilophosaurus"
sinensis are then closer to Ceratosauria+Tetanurae.
Forcing "Szechuanoraptor" to be an allosaurid (as in Molnar et al., 1990) adds
8 steps.
Forcing Afrovenator to be closer to Avetheropoda than Piatnitzkysaurus,
Torvosaurus, Eustreptospondylus and Spinosauridae (as in Holtz, 2000) adds 8
steps. Dubreuillosaurus and Piveteausaurus stay in Spinosauroidea.
Forcing spinosauroids to be carnosaurs (as in Rauhut, 2003 and older sources
with more extensive Carnosauria's) adds 8 steps. Fukuiraptor also falls out in
this clade, though Siamotyrannus remains a coelurosaur.
Forcing Zupaysaurus to be closer to ceratosaurs+tetanurines than to
coelophysoids (as in Arcucci and Coria, 1997) adds 9 steps. As in that study,
dilophosaurids end up even closer to birds.
Forcing Piatnitzkysaurus to be a basal carnosaur (as in Harris, 1998) adds 9
Forcing Coelurus to be a tyrannosauroid (as in Senter, 2007) adds 9 steps. The
topology is similar to when Guanlong is forced to be a tyrannosauroid, with
Coelurus outside the Dilong+Guanlong+Aristosuchus+Mirischia clade.
Forcing Coelurus to be basal to tyrannoraptorans (as in Makovicky, 1995) adds 9
Forcing Bagaraatan to be outside Tyrannoraptora (as in Holtz, 2000) adds 9
Forcing Proceratosaurus to be a ceratosaurid (as in Huene, 1926) adds 9 steps.
Forcing "Szechuanoraptor" to be a sinraptorid (as in Paul, 1988) adds 10 steps.
Forcing Piatnitzkysaurus to be a eustreptospondylid (as in Paul, 1988) adds 10
steps. It ends up sister to other eustreptospondylids.
Forcing Sinraptor to be basal to allosaurids and carcharodontosaurids within
Allosauroidea (as in Harris, 1998) adds 10 steps. Oddly, this forces
spinosauroids to be carnosaurs, in addition to Monolophosaurus. While
Neovenator and Acrocanthosaurus are carcharodontosaurids, Fukuiraptor is an
Forcing Dilong to be a compsognathid (as suggested by Olshevsky I believe,
online) adds 10 steps.
Forcing Eoraptor to be a basal theropod, but herrerasaurids to be outside
Eusaurischia (as in Ezcurra, 2006) adds 11 steps.
Forcing Indosaurus to be a tyrannosaurid (as in Chatterjee, 1978) adds 11 steps.
Forcing Coelurus to be basal to avetheropods (as in Paul, 1988) adds 11 steps.
Oddly, instead of being just outside Avetheropoda, it moves to the base of the
Tetanurae with Tugulusaurus.
Forcing Aristosuchus to be a compsognathid (as in Naish, 2002) adds 12 steps.
Forcing Tyrannosaurus to be a maniraptoran (as in Sereno, 1997) adds 12 steps.
Forcing coelophysoids to be ceratosaurs (as in Gauthier, 1986) adds 13 more
steps. In these trees, Cryolophosaurus and "Dilophosaurus" sinensis are basal
Forcing Elaphrosaurus to be a coelophysoid (as in Paul, 1988) adds 13 more
steps. It falls out with Spinostropheus between Liliensternus+Lophostropheus
and Zupaysaurus+derived coelophysids. Amusingly right where Paul put it.
Forcing spinosaurids to be more basal than Piatnitzkysaurus,
Eustreptospondylus, Afrovenator and Torvosaurus (as in Holtz, 2000) adds 13
steps. Spinosauroidea becomes paraphyletic to Avetheropoda, with Torvosaurus,
Eustreptospondylidae, Poekilopleuron and Piatnitzkysaurus successively closer
to it.
Forcing Sinraptor to be closer to carcharodontosaurids than to Allosaurus (as
in Coria and Currie, 2002) adds 13 steps. I these trees, Neovenator and
Acrocanthosaurus are carcharodontosaurids, while Allosaurus is the basal
Forcing Compsognathus to be a tyrannosauroid (as in Olshevsky, 1991) adds 13
steps. This also makes Sinosauropteryx, Guanlong, Coelurus, Dilong, Mirischia
and Aristosuchus tyrannosauroids.
Forcing Ilokelesia to be outside Noasauridae+Abelisauridae (as in Coria and
Salgado, 2000) adds 14 steps.
Forcing Afrovenator to be sister to Allosauroidea (as in Rauhut, 2003) adds 14
steps. Dubreuillosaurus and Piveteausaurus stay in Spinosauroidea.
Forcing Ornitholestes outside Tyrannoraptora (as in Holtz, 1994 and others)
adds 14 steps.
Forcing Procompsognathus to be outside Dinosauria (as in Allen, 2004) adds 15
Forcing Deltadromeus to be closer to birds than Ornitholestes (as in Sereno et
al., 1996) adds 15 more steps, mainly from moving Ornitholestes down the tree.
Forcing Piatnitzkysaurus to be an allosaurid (as in Molnar et al., 1981) adds
15 steps.
Near certainly untrue-
Forcing Ceratosaurus to be closer to tetanurines than abelisaurids are (as in
Carrano and Sampson, 1999) adds 18 steps. Elaphrosaurs and noasaurids stay
with abelisaurids.
Forcing Compsognathus outside Avetheropoda (as in Paul, 1988) adds 18 steps.
Forcing Piatnitzkysaurus to be an abelisaur (as in Currie and Zhao, 1993) adds
21 steps.
Forcing Monolophosaurus to be sister to Guanlong (as suggested by Carr, 2006)
adds 22 steps. The pair are placed as coelurosaurs just outside Tyrannoraptora.
Forcing Cryolophosaurus to be a carnosaur (as in Sereno et al., 1996) adds 23
more steps. It changes the tree a lot, with most basal tetanurines moved into
Forcing Monolophosaurus to be a tyrannosauroid (as suggested by me on the DML)
adds 24 steps. In these trees, Guanlong and Dilong are still closer to birds
than to tyrannosaurids, though Siamotyrannus is a tyrannosauroid.
Forcing Guanlong to be a carnosaur (as in Carr, 2006) adds 24 steps.
Siamotyrannus, Fukuiraptor, Dilong and Aristosuchus are also carnosaurs in
these trees.
Forcing Tyrannosaurus to clade with ornithomimosaurs (as in Huene, 1923) adds
24 steps.
Forcing Torvosaurus to be a ceratosaur (as in Britt, 1991) adds 26 steps. It
emerges outside the elaphrosaur-ceratosaurid-abelisauroid clade, but other
spinosauroids stay in Tetanurae.
Forcing Guanlong to be a carnosaur sister taxon to Monolophosaurus (as in Carr,
2006) adds 27 steps.
Forcing Acrocanthosaurus to be a spinosaurid (as in Walker, 1964)adds 30 steps.
Spinosaurids are moved into Carnosauria.
Forcing Elaphrosaurus to be an ornithomimosaur (as in Galton, 1982) adds 33
Forcing abelisaurids to be sister to carcharodontosaurids (as in Novas, 1997)
adds 35 steps. The clade ends up in Ceratosauria, and Acrocanthosaurus and
Neovenator remain in Carnosauria.
Forcing Zupaysaurus to be a tetanurine (as in Arcucci and Coria, 2003) adds 37
Forcing Megaraptor to be a dromaeosaurid (as suggested in Rauhut, 2003) adds 38
Forcing Abelisaurus to be a carcharodontosaurid (as suggested by Lamanna et
al., 2002) adds 39 steps.
Forcing Monolophosaurus to be an ornitholestiid (as in Paul, 2002) adds 40
steps. Ornitholestes and Proceratosaurus are moved outside Avetheropoda with
Forcing Ornitholestes to be an allosaurid (as in Paul, 1988) adds 41 steps.
Forcing Sinosauropteryx outside Coelurosauria (as in Longrich, DML 2000) adds
42 steps.
Forcing the less extensive Carnosauria of Molnar et al., 1990
(Piatnitzkysaurus, Allosaurus, carcharodontosaurids and tyrannosaurids) adds 44
steps. Spinosauroids, Monolophosaurus and Sinraptor form successively closer
outgroups to Carnosauria. The Allosauridae of Molnar is paraphyletic to
tyrannosaurids, with Piatnitzkysaurus most basal and carcharodontosaurids most
closely related.
Russell and Dong's (1993) crazy topology of
adds 45 steps, mainly from moving ornithomimosaurs so basally.
Bakker et al.'s (1988) topology of
(Cerato(Allo(Dromaeo(Acrocantho(Ornithimimidae,Tyrannosauridae))))) adds 53
Forcing abelisaurids to be tetanurines closer to birds than Torvosaurus (as in
Forster, 1999) adds 55 steps. Noasaurids and Deltadreomeus clade with
abelisaurids here.
Forcing the Allosauria of Paul (Ornitholestes, Proceratosaurus, Allosaurus,
carcharodontosaurids and tyrannosaurids) adds 56 more steps. Though
constrained to be outside Paul's Allosauria, Piatnitzkysaurus, spinosauroids,
Monolophosaurus and sinraptorids end up forming successively closer outgroups
to it. Paul's ornitholestiines end up closer to tyrannosaurids than to
allosaurids, and include Dilong and Guanlong as well.
Forcing abelisauroids to be torvosaurids within Tetanurae (as in Paul, 1988)
adds 57 steps. Spinosaurids and Monolophosaurus join the torvosaur part of
this clade, while elaphrosaurs hang back with Ceratosaurus.
Forcing spinosaurids to be coelophysoids closest to dilophosaurs (as in Paul,
1988) adds 60 steps.
Forcing a traditional Carnosauria of sinraptorids, Torvosaurus, spinosaurids,
abelisaurids, eustreptospondylids, carcharodontosaurids, allosaurids and
tyrannosaurids (as in Kurzanov, 1989) adds 82 steps. Unlike Kurzanov's
topology, abelisaurids emerge as most basal, Spinosauroidea is intact, and
sinraptorids are sister to allosaurids.
Forcing paravians outside of Dinosauria (as Feduccia, Martin, etc. suggest)
adds 87 steps. Even though I specified Allosaurus as a dinosaur, all
coelurosaurs are more parsimoniously "birds" when this happens.
Forcing spinosaurids and birds to be sister taxa (as in Elzanowski and
Wellnhofer, 1992) adds 93 steps. The clade ends up in a very basal coelurosaur
position (ignoring for the moment that birds aren't with the rest of what's
normally Paraves).
Forcing Huene's original Carnosauria-Coelurosauria phylogeny of 1923, where
coelophysids, ornithomimids, tyrannosaurids and Ceratosaurus are coelurosaurs,
while Allosaurus and spinosauroids are carnosaurs adds 97 steps.
Forcing Torvosaurus and Poekilopleuron to be more closely related to
Plateosaurus than to coelophysoids and coelurosaurs (as in Galton and Jensen,
1979) adds 101 steps. The rest of Spinosauroidea, as well as Ceratosauria and
Carnosauria end up joining the two 'megalosaurs' to form a large-bodied
theropod clade sister to Sauropodomorpha. Within Saurischia, coelophysoids are
more closely related to this clade than coelurosaurs.
Mickey Mortimer
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[SciPy-dev] Implementing a distance matrix between two sets of vectors concept
David Cournapeau david@ar.media.kyoto-u.ac...
Wed Jul 4 22:38:11 CDT 2007
Peter Skomoroch wrote:
> You're right, I was thinking the sparse data structures would help
> with storing the input vectors themselves during the computation
> rather than the final matrix (which will need to be 1/2 M*N if the
> distance is symmetric)...this comes up a lot in collaborative
> filtering where the dimensionality of the vectors is high, but most of
> the vector entries are missing.
Ok, that this basically means supporting sparse input, right ? I have to
say that I don't know anything about sparse implementations issues in
numpy (or any other language for that matter). I guess that performances
mainly depend on the flexibility between matrix representation and data
storage. Are sparse arrays directly supported in numpy ?
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[Haskell-beginners] Re: [Haskell-cafe] try, seq, and IO
Daniel Fischer daniel.is.fischer at web.de
Fri Sep 17 09:18:21 EDT 2010
On Friday 17 September 2010 13:17:30, Jeroen van Maanen wrote:
> What I don't understand is the difference between:
> try $ return $ seq theCheckSum maybeUpdatedModel
> or even
> try $! return $! seq theCheckSum maybeUpdatedModel
> and
> try $ evaluate $ seq theCheckSum maybeUpdatedModel
> How is it possible that the exception escapes the former two
> expressions, but gets caught by the third try?
> Cheers, Jeroen
It's quite devilish :)
Well, the first and the third are rather straightforward.
Let's start with the third.
What that does is, evaluate (seq theCheckSum maybeUpdatedModel) to WHNF, if
that throws an exception (of appropriate type, here SomeException), return
(Left exception) else return (Right result). To evaluate `seq theCheckSum
maybeUpdatedModel' to WHNF, theCheckSum has to be evaluated to WHNF (hence
completely, since it's an Integer or something like), which in turn
requires the complete evaluation of maybeUpdatedModel.
The last throws an exception, that gets caught and wrapped in try, as
expected and desired.
The first one is `try (return thunk)' where thunk is "if needed, calculate
`seq theCheckSum maybeUpdatedModel'". The return succeeds immediately, try
wraps it in a Right and returns (Right thunk), as expected but not desired.
The exception is thrown when you demand the evaluation of the thunk, after
try has been left. Too lazy.
Now the second one.
try $! return $! seq t m
=== let z = return $! seq t m in z `seq` try z
=== let z = let v = seq t m in v `seq` return v in z `seq` try z
=== let { v = seq t m; z = v `seq` return v } in z `seq` try z
=== let v = seq t m in (v `seq` return v) `seq` try (return v)
=== ((t `seq` m) `seq` (return (t `seq` m)) `seq` try (return (t `seq` m))
So before try is even called, t has to be evaluated to WHNF, which throws
an exception. Since it's thrown before try has been entered, try can't
catch it. Too strict.
So, first gives an uncaught exception after try has been left, second gives
an uncaught exception before try has been entered.
How do we get the exception to be thrown inside the try?
That's easy. We mustn't allow try to return an exception-throwing thunk, so
we need (return $! seq t m).
But we mustn't cause the exception before try has been entered, so we need
try $ return $! seq theCheckSum maybeUpdatedModel
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