diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/099_8.11 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/099_8.11 Applications.md
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-# Section 8.11 Applications
-
-**8.78** An automobile airbag igniter is modeled by the circuit in Fig. 8.122. Determine the time it takes the voltage across the igniter to reach its first peak after switching from *A* to *B*. Let *R* = 3 Ω, *C* = 1∕30 F, and *L* = 60 mH.
-
-**8.79** A load is modeled as a 100-mH inductor in parallel with a 12-Ω resistor. A capacitor is needed to be connected to the load so that the network is critically damped at 60 Hz. Calculate the size of the capacitor.
-
-8.83 Figure 8.124 shows a typical tunnel-diode oscillator circuit. The diode is modeled as a nonlinear resistor with *iD* = *f* (*vD*), i.e., the diode current is a nonlinear function of the voltage across the diode. Derive the differential equation for the circuit in
-
-terms of *v* and *iD*.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/100_Comprehensive Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/100_Comprehensive Problems.md
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-# **PART TWO**
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/101_PART 2 - AC Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/101_PART 2 - AC Circuits.md
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-# AC Circuits
-
-# OUTLINE
-
-- 9 Sinusoids and Phasors
-- 10 Sinusoidal Steady-State Analysis
-- 11 AC Power Analysis
-- 12 Three-Phase Circuits
-- 13 Magnetically Coupled Circuits
-- 14 Frequency Response
-
-# **chapter**
-
-9
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/102_Chapter 9 - Sinusoids and Phasors.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/102_Chapter 9 - Sinusoids and Phasors.md
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-# Sinusoids and Phasors
-
-*He who knows not, and knows not that he knows not, is a fool—shun him. He who knows not, and knows that he knows not, is a c hild—teach him. He who knows, and knows not that he knows, is asleep—wake him up. He who knows, and knows that he knows, is wise—follow him.*
-
-—Persian Proverb
-
-# Enhancing Your Skills and Your Career
-
-# **ABET EC 2000 criteria (3.d), "an ability to function on multi‑disciplinary teams."**
-
-The "ability to function on multidisciplinary teams" is inherently critical for the working engineer. Engineers rarely, if ever, work by themselves. Engineers will always be part of some team. One of the things I lik e to remind students is that you do not ha ve to like everyone on a team; you just have to be a successful part of that team.
-
-Most frequently, these teams include indi viduals from a v ariety of engineering disciplines, as well as individuals from nonengineering disciplines such as marketing and finance.
-
-Students can easily de velop and enhance this skill by w orking in study groups in every course they take. Clearly, working in study groups in nonengineering courses, as well as engineering courses outside your discipline, will also give you experience with multidisciplinary teams.
-
-Photo by Charles Alexander
-
-# Historical
-
-George Westinghouse. Photo © Bettmann/Corbis
-
-**Nikola Tesla** (1856–1943) and **George Westinghouse** (1846–1914) helped establish alternating current as the primary mode of electricity transmission and distribution.
-
-Today it is obvious that ac generation is well established as the form of electric power that makes widespread distribution of electric power efficient and economical. However, at the end of the 19th century, which was the better—ac or dc—w as hotly debated and had e xtremely outspoken supporters on both sides. The dc side was led by Thomas Edison, who had earned a lot of respect for his many contributions. Power generation using ac really be gan to b uild after the successful contrib utions of Tesla. The real commercial success in ac came from Geor ge Westinghouse and the outstanding team, including Tesla, he assembled. In addition, tw o other big names were C. F. Scott and B. G. Lamme.
-
-The most significant contribution to the early success of ac w as the patenting of the polyphase ac motor by Tesla in 1888. The induction motor and polyphase generation and distrib ution systems doomed the use of dc as the prime energy source.
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Better understand sinusoids.
-- 2. Understand phasors.
-- 3. Understand the phasor relationships for circuit elements.
-- 4. Know and understand the concepts of impedance and admittance.
-- 5. Understand Kirchhoff's laws in the frequency domain.
-- 6. Comprehend the concept of phase-shift.
-- 7. Understand the concept of AC bridges.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/103_9.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/103_9.1 Introduction.md
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-# **9.1** Introduction
-
-Thus far our analysis has been limited for the most part to dc circuits: those circuits e xcited by constant or time-in variant sources. We ha ve restricted the forcing function to dc sources for the sake of simplicity, for pedagogic reasons, and also for historic reasons. Historically, dc sources were the main means of providing electric power up until the late 1800s. At the end of that century, the battle of direct current versus alternating current began. Both had their advocates among the electrical engineers of the time. Because ac is more efficient and economical to transmit over long distances, ac systems ended up the winner . Thus, it is in k eeping with the historical sequence of events that we considered dc sources first.
-
-We now begin the analysis of circuits in which the source v oltage or current is time-v arying. In this chapter , we are particularly interested in sinusoidally time-varying excitation, or simply, excitation by a *sinusoid*.
-
-## A sinusoid is a signal that has the form of the sine or cosine function.
-
-A sinusoidal current is usually referred to as *alternating current* (*ac*). Such a current reverses at regular time intervals and has alternately positive and negative values. Circuits driven by sinusoidal current or voltage sources are called *ac circuits*.
-
-We are interested in sinusoids for a number of reasons. First, nature itself is characteristically sinusoidal. We e xperience sinusoidal v ariation in the motion of a pendulum, the vibration of a string, the ripples on the ocean surface, and the natural response of underdamped secondorder systems, to mention but a few. Second, a sinusoidal signal is easy to generate and transmit. It is the form of v oltage generated throughout the world and supplied to homes, factories, laboratories, and so on. It is the dominant form of signal in the communications and electric power industries. Third, through Fourier analysis, any practical periodic signal can be represented by a sum of sinusoids. Sinusoids, therefore, play an important role in the analysis of periodic signals. Lastly , a sinu soid is easy to handle mathematically . The derivative and integral of a sinusoid are themselve s sinusoids. For these and other reasons, the sinusoid is an extremely important function in circuit analysis.
-
-A sinusoidal forcing function produces both a transient response and a steady-state response, much like the step function, which we studied in Chapters 7 and 8. The transient response dies out with time so that only the steady-state response remains. When the transient response has become negligibly small compared with the steady-state response, we say that the circuit is operating at sinusoidal steady state. It is this *sinusoidal steady-state response* that is of main interest to us in this chapter.
-
- We begin with a basic discussion of sinusoids and phasors. We then introduce the concepts of impedance and admittance. The basic circuit laws, Kirchhoff's and Ohm's, introduced for dc circuits, will be applied to ac circuits. Finally , we consider applications of ac circuits in phaseshifters and bridges.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/104_9.2 Sinusoids.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/104_9.2 Sinusoids.md
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-# **9.2** Sinusoids
-
-Consider the sinusoidal voltage
-
-$$
-v(t) = V_m \sin \omega t \tag{9.1}
-$$
-
-where
-
-*Vm* = the *amplitude* of the sinusoid
-
-*ω* = the *angular frequency* in radians/s
-
-*ωt* = the *argument* of the sinusoid
-
-The sinusoid is shown in Fig. 9.1(a) as a function of its argument and in Fig. 9.1(b) as a function of time. It is e vident that the sinusoid repeats itself every *T* seconds; thus, *T* is called the *period* of the sinusoid. From the two plots in Fig. 9.1, we observe that *ωT* = 2*π*,
-
-$$
-T = \frac{2\pi}{\omega} \tag{9.2}
-$$
-
-# Historical
-
-© Hulton Archives/Getty Images
-
-**Heinrich Rudorf Hertz** (1857–1894), a German experimental physicist, demonstrated that electromagnetic waves obey the same fundamental laws as light. His work confirmed James Clerk Maxwell's celebrated 1864 theory and prediction that such waves existed.
-
-Hertz w as born into a prosperous f amily in Hamb urg, German y. He attended the Uni versity of Berlin and did his doctorate under the prominent physicist Hermann von Helmholtz. He became a professor at Karlsruhe, where he be gan his quest for electromagnetic w aves. Hertz successfully generated and detected electromagnetic w aves; he was the first to show that light is electromagnetic ener gy. In 1887, Hertz noted for the first time the photoelectric effect of electrons in a molecular structure. Although Hertz only lived to the age of 37, his discovery of electromagnetic waves paved the way for the practical use of such waves in radio, television, and other communication systems. The unit of frequency, the hertz, bears his name.
-
-The fact that *v*(*t*) repeats itself every *T* seconds is shown by replacing *t* by *t* + *T* in Eq. (9.1). We get
-
-$$
-v(t+T) = V_m \sin \omega(t+T) = V_m \sin \omega \left(t + \frac{2\pi}{\omega}\right)
-$$
-
-= $V_m \sin (\omega t + 2\pi) = V_m \sin \omega t = v(t)$ (9.3)
-
-Hence,
-
-$$
-v(t+T) = v(t) \tag{9.4}
-$$
-
-that is, *v* has the same value at *t* + *T* as it does at *t* and *v*(*t*) is said to be *periodic*. In general,
-
-A periodic function is one that satisfies f (t) = f (t + nT ), for all t and for all integers n.
-
-As mentioned, the *period T* of the periodic function is the time of one complete cycle or the number of seconds per c ycle. The reciprocal of this quantity is the number of c ycles per second, kno wn as the *cyclic frequency f* of the sinusoid. Thus,
-
-$$
-f = \frac{1}{T}
-$$
- (9.5)
-
-From Eqs. (9.2) and (9.5), it is clear that
-
-$$
-\omega = 2 \pi f \tag{9.6}
-$$
-
-While *ω* is in radians per second (rad/s), *f* is in hertz (Hz).
-
-Let us now consider a more general expression for the sinusoid,
-
-$$
-v(t) = V_m \sin(\omega t + \phi)
-$$
-\n(9.7)
-
-where (*ωt* + *ϕ*) is the argument and *ϕ* is the *phase*. Both argument and phase can be in radians or degrees.
-
-Let us examine the two sinusoids
-
-$$
-v_1(t) = V_m \sin \omega t \qquad \text{and} \qquad v_2(t) = V_m \sin(\omega t + \phi) \tag{9.8}
-$$
-
-shown in Fig. 9.2. The starting point of *v*2 in Fig. 9.2 occurs first in time. Therefore, we say that *v*2 *leads v*1 by *ϕ* or that *v*1 *lags v*2 by *ϕ*. If *ϕ* ≠ 0, we also say that *v*1 and *v*2 are *out of phase*. If *ϕ* = 0, then *v*1 and *v*2 are said to be *in phase;* they reach their minima and maxima at e xactly the same time. We can compare *v*1 and *v*2 in this manner because they operate at the same frequency; they do not need to have the same amplitude.
-
-A sinusoid can be e xpressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes. This is achieved by using the following trigonometric identities:
-
-$$
-sin(A \pm B) = sin A cos B \pm cos A sin B
-$$
-
-\n
-$$
-cos(A \pm B) = cos A cos B \mp sin A sin B
-$$
-\n(9.9)
-
-With these identities, it is easy to show that
-
-$$
-sin(\omega t \pm 180^\circ) = -sin \omega t
-$$
-
-\n
-$$
-cos(\omega t \pm 180^\circ) = -cos \omega t
-$$
-
-\n
-$$
-sin(\omega t \pm 90^\circ) = \pm cos \omega t
-$$
-
-\n
-$$
-cos(\omega t \pm 90^\circ) = \mp sin \omega t
-$$
-\n(9.10)
-
-Using these relationships, we can transform a sinusoid from sine form to cosine form or vice versa.
-
-The unit of f is named after the German physicist Heinrich R. Hertz (1857–1894).
-
-# **Figure 9.3**
-
-A graphical means of relating cosine and sine: (a) cos(*ωt* − 90°) = sin *ωt*, (b) sin(*ωt* + 180°) = −sin *ωt*.
-
-A graphical approach may be used to relate or compare sinusoids as an alternati ve to using the trigonometric identities in Eqs. (9.9) and (9.10). Consider the set of axes shown in Fig. 9.3(a). The horizontal axis represents the magnitude of cosine, while the vertical axis (pointing down) denotes the magnitude of sine. Angles are measured positi vely counterclockwise from the horizontal, as usual in polar coordinates. This graphical technique can be used to relate tw o sinusoids. F or example, we see in Fig. 9.3(a) that subtracting 90° from the ar gument of cos *ωt* gives sin *ωt*, o r cos(*ωt* − 90°) = sin *ωt*. Similarly, adding 1 80° to the argument of sin*ωt* gives −sin *ωt*, or sin(*ωt* + 180°) = −sin *ωt*, as shown in Fig. 9.3(b).
-
-The graphical technique can also be used to add tw o sinusoids of the same frequency when one is in sine form and the other is in cosine form. To add *A* cos *ωt* and *B* sin *ωt*, we note that *A* is the magnitude of cos *ωt* while *B* is the magnitude of sin*ωt*, as shown in Fig. 9.4(a). The magnitude and argument of the resultant sinusoid in cosine form is readily obtained from the triangle. Thus,
-
-$$
-A\cos\omega t + B\sin\omega t = C\cos(\omega t - \theta)
-$$
-\n(9.11)
-
-where
-
-$$
-C = \sqrt{A^2 + B^2}
-$$
-, $\theta = \tan^{-1} \frac{B}{A}$ (9.12)
-
-For example, we may add 3 cos *ωt* and −4 sin*ωt* as shown in Fig. 9.4(b) and obtain
-
-$$
-3\cos \omega t - 4\sin \omega t = 5\cos(\omega t + 53.1^{\circ})
-$$
- (9.13)
-
-Compared with the trigonometric identities in Eqs. (9.9) and (9.10), the graphical approach eliminates memorization. Ho wever, we must not confuse the sine and cosine ax es with the ax es for comple x numbers to be discussed in the ne xt section. Something else to note in Figs. 9.3 and 9.4 is that although the natural tendenc y is to ha ve the vertical axis point up, the positive direction of the sine function is down in the present case.
-
-(a) Adding *A* cos *ωt* and *B* sin *ωt*, (b) adding 3 cos *ωt* and −4 sin *ωt*.
-
-*v*(*t*) = 12 cos(50*t* + 10°) V.
-
-# **Solution:**
-
-The amplitude is *Vm* = 12 V. The phase is *ϕ* = 10°. The angular frequency is *ω* = 50 rad/s. The period *T* = \_\_\_ 2*π ω* = \_\_\_ 2*π* 50 = 0.1257 s. The frequency is *f* = \_\_1 *T* = 7.958 Hz.
-
-Given the sinusoid 45 cos(5 *πt* + 36°), calculate its amplitude, phase, angular frequency, period, and frequency.
-
-**Answer:** 45, 36°, 15.708 rad/s, 400 ms, 2.5 Hz.
-
-Calculate the phase angle between *v*1 = −10 cos(*ωt* + 50°) and *v*2 = Example 9.2 12 sin(*ωt* − 10°). State which sinusoid is leading.
-
-# **Solution:**
-
-Let us calculate the phase in three ways. The first two methods use trigonometric identities, while the third method uses the graphical approach.
-
-■ **METHOD 1** In order to compare *v*1 and *v*2, we must e xpress them in the same form. If we e xpress them in cosine form with posi tive amplitudes,
-
-*v*1 = −10 cos(*ωt* + 50°) = 10 cos(*ωt* + 50° − 180°) *v*1 = 10 cos(*ωt* − 130°) or *v*1 = 10 cos(*ωt* + 230°) **(9.2.1)** and *v*2 = 12 sin(*ωt* − 10°) = 12 cos(*ωt* − 10° − 90°) *v*2 = 12 cos(*ωt* − 100°) **(9.2.2)**
-
-It can be deduced from Eqs. (9.2.1) and (9.2.2) that the phase difference between *v*1 and *v*2 is 30°. We can write *v*2 as
-
-*v*2 = 12 cos(*ωt* − 130° + 30°) or *v*2 = 12 cos(*ωt* + 260°) **(9.2.3)**
-
-Comparing Eqs. (9.2.1) and (9.2.3) shows clearly that *v*2 leads *v*1 by 30°.
-
-■ **METHOD 2** Alternatively, we may express *v*1 in sine form:
-
-*v*1 = −10 cos(*ωt* + 50°) = 10 sin(*ωt* + 50° − 90°) = 10 sin(*ωt* − 40°) = 10 sin(*ωt* − 10° − 30°)
-
-Practice Problem 9.1
-
-But *v*2 = 12 sin(*ωt* − 10°). Comparing the tw o shows that *v*1 lags *v*2 by 30°. This is the same as saying that *v*2 leads *v*1 by 30°.
-
-■ **METHOD 3** We may regard *v*1 as simply −10 cos*ωt* with a phase shift of +50°. Hence, *v*1 is as shown in Fig. 9.5. Similarly, *v*2 is 12 sin*ωt* with a phase shift of −10°, as shown in Fig. 9.5. It is easy to see from Fig. 9.5 that *v*2 leads *v*1 by 30°, that is, 90° − 50° − 10°.
-
-Find the phase angle between
-
-*i*1 = −4 sin(377*t* + 55°) and *i*2 = 5 cos(377*t* − 65°)
-
-Does *i*1 lead or lag *i*2?
-
-**Answer:** 210°, *i*1 leads *i*2.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/105_9.3 Phasors.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/105_9.3 Phasors.md
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-# **9.3** Phasors
-
-Sinusoids are easily expressed in terms of phasors, which are more convenient to work with than sine and cosine functions.
-
-A phasor is a complex number that represents the amplitude and phase of a sinusoid.
-
-Phasors provide a simple means of analyzing linear circuits e xcited by sinusoidal sources; solutions of such circuits would be intractable otherwise. The notion of solving ac circuits using phasors was first introduced by Charles Steinmetz in 1893. Before we completely define phasors and apply them to circuit analysis, we need to be thoroughly f amiliar with complex numbers.
-
-A complex number *z* can be written in rectangular form as
-
-$$
-z = x + jy \tag{9.14a}
-$$
-
-where *j* = √ \_\_\_ −1 ; *x* is the real part of *z*; *y* is the imaginary part of *z*. In this context, the variables *x* and *y* do not represent a location as in tw odimensional vector analysis but rather the real and imaginary parts of *z* in the complex plane. Nevertheless, we note that there are some resemblances between manipulating complex numbers and manipulating twodimensional vectors.
-
-The complex number *z* can also be written in polar or e xponential form as
-
-$$
-z = r/\phi = re^{j\phi} \tag{9.14b}
-$$
-
-Charles Proteus Steinmetz (1865–1923) was a German-Austrian mathematician and electrical engineer.
-
-Appendix B presents a short tutorial on complex numbers.
-
-# Historical
-
-**Charles Proteus Steinmetz** (1865–1923), a German-Austrian mathematician and engineer, introduced the phasor method (covered in this chapter) in ac circuit analysis. He is also noted for his w ork on the theory of hysteresis.
-
-Steinmetz was born in Breslau, Germany, and lost his mother at the age of one. As a youth, he was forced to leave Germany because of his political activities just as he w as about to complete his doctoral dis sertation in mathematics at the Uni versity of Breslau. He migrated to Switzerland and later to the United States, where he w as employed by General Electric in 1893. That same year, he published a paper in which complex numbers were used to analyze ac circuits for the first time. This led to one of his man y textbooks, *Theory and Calculation of ac Phenomena,* published by McGra w-Hill in 1897. In 1901, he became the president of the American Institute of Electrical Engineers, which later became the IEEE.
-
-where *r* is the magnitude of *z*, and *ϕ* is the phase of *z*. We notice that *z* can be represented in three ways:
-
-| z = x + jy | Rectangular form | |
-|------------|------------------|--------|
-| z = r⧸ϕ | Polar form | (9.15) |
-| z = rejϕ | Exponential form | |
-
-The relationship between the rectangular form and the polar form is shown in Fig. 9.6, where the *x* axis represents the real part and the *y* axis represents the imaginary part of a comple x number. Given *x* and *y*, we can get *r* and *ϕ* as
-
-$$
-r = \sqrt{x^2 + y^2}
-$$
-, $\phi = \tan^{-1} \frac{y}{x}$ (9.16a)
-
-On the other hand, if we know *r* and *ϕ*, we can obtain *x* and *y* as
-
-$$
-x = r \cos \phi, \qquad y = r \sin \phi \tag{9.16b}
-$$
-
-Thus, *z* may be written as
-
-$$
-z = x + jy = r/\underline{\phi} = r(\cos\phi + j\sin\phi)
-$$
- (9.17)
-
-Addition and subtraction of complex numbers are better performed in rectangular form; multiplication and division are better done in polar form. Given the complex numbers
-
-$$
-z = x + jy = r/\underline{\phi}, \qquad z_1 = x_1 + jy_1 = r_1/\underline{\phi}_1
-$$
-$$
-z_2 = x_2 + jy_2 = r_2/\underline{\phi}_2
-$$
-
-the following operations are important. **Addition:**
-
-$$
-z_1 + z_2 = (x_1 + x_2) + j(y_1 + y_2)
-$$
-\n(9.18a)
-
-# **Subtraction:**
-
-$$
-z_1 - z_2 = (x_1 - x_2) + j(y_1 - y_2)
-$$
-\n(9.18b)
-
-**Multiplication:**
-
-$$
-z_1 z_2 = r_1 r_2 / \phi_1 + \phi_2 \tag{9.18c}
-$$
-
-**Division:**
-
-$$
-\frac{z_1}{z_2} = \frac{r_1}{r_2} / \phi_1 - \phi_2 \tag{9.18d}
-$$
-
-**Reciprocal:**
-
-$$
-\frac{1}{z} = \frac{1}{r} \angle -\phi \tag{9.18e}
-$$
-
-**Square Root:**
-
-$$
-\sqrt{z} = \sqrt{r} \sqrt{\phi/2}
-$$
- (9.18f)
-
-# **Complex Conjugate:**
-
-$$
-z^* = x - jy = r'_\text{p} = re^{-j\phi} \tag{9.18g}
-$$
-
-Note that from Eq. (9.18e),
-
-$$
-\frac{1}{j} = -j \tag{9.18h}
-$$
-
-These are the basic properties of complex numbers we need. Other properties of complex numbers can be found in Appendix B.
-
-The idea of phasor representation is based on Euler' s identity. In general,
-
-$$
-e^{\pm i\phi} = \cos\phi \pm j\sin\phi \qquad (9.19)
-$$
-
-which shows that we may re gard cos *ϕ* and sin *ϕ* as the real and imagi nary parts of *e jϕ* ; we may write
-
-$$
-\cos \phi = \text{Re}(e^{j\phi}) \tag{9.20a}
-$$
-
-$$
-\sin \phi = \text{Im}(e^{j\phi})\tag{9.20b}
-$$
-
-where Re and Im stand for the *real part of* and the *imaginary part of*. Given a sinusoid *v*(*t*) = *Vm* cos(*ωt* + *ϕ*), we use Eq. (9.20a) to e xpress *v*(*t*) as
-
-$$
-v(t) = V_m \cos(\omega t + \phi) = \text{Re}(V_m e^{j(\omega t + \phi)})
-$$
- (9.21)
-
-or
-
-$$
-v(t) = \text{Re}(V_m e^{j\phi} e^{j\omega t})
-$$
-\n(9.22)
-
-Thus,
-
-$$
-v(t) = \text{Re}(\mathbf{V}e^{j\omega t})
-$$
- (9.23)
-
-where
-
-$$
-\mathbf{V} = V_m e^{j\phi} = V_m / \phi \tag{9.24}
-$$
-
-**V** is thus the *phasor representation* of the sinusoid *v*(*t*), as we saidearlier. In other words, a phasor is a complex representation of the magnitude and phase of a sinusoid. Either Eq. (9.20a) or Eq. (9.20b) can be used to develop the phasor, but the standard convention is to use Eq. (9.20a).
-
-One way of looking at Eqs. (9.23) and (9.24) is to consider the plot of the *sinor* **V***e jωt* = *Vme j*(*ωt*+*ϕ*) on the comple x plane. As time increases, the sinor rotates on a circle of radius *Vm* at an angular velocity *ω* in the counterclockwise direction, as shown in Fig. 9.7(a). We may regard *v*(*t*) as the projection of the sinor **V***e jωt* on the real axis, as shown in Fig. 9.7(b). The value of the sinor at time *t* = 0 is the phasor **V** of the sinusoid *v*(*t*). The sinor may be regarded as a rotating phasor. Thus, whenever a sinusoid is expressed as a phasor, the term *e jωt* is implicitly present. It is therefore important, when dealing with phasors, to keep in mind the frequency *ω* of the phasor; otherwise we can make serious mistakes.
-
- A phasor may be regarded as a mathematical equivalent of a sinusoid with the time dependence dropped.
-
- If we use sine for the phasor instead of cosine, then v (t) = V m sin(*ω*t + *ϕ*) = Im(Vmej(*ω*t+*ϕ*) ) and the corresponding phasor is the same as that in Eq. (9.24).
-
-Equation (9.23) states that to obtain the sinusoid corresponding to a given phasor **V**, multiply the phasor by the time f actor *ejωt* and tak e the real part. As a complex quantity, a phasor may be expressed in rectangular form, polar form, or exponential form. Because a phasor has magnitude and phase ("direction"), it behaves as a vector and is printed in boldface. For example, phasors **V** = *Vm*⧸*ϕ* and **I** = *Im*⧸−*θ* are graphically represented in Fig. 9.8. Such a graphical representation of phasors is known as a *phasor diagram*.
-
-Equations (9.21) through (9.23) re veal that to get the phasor cor responding to a sinusoid, we first express the sinusoid in the cosine form so that the sinusoid can be written as the real part of a complex number. Then we tak e out the time f actor *ejωt* , and whate ver is left is the pha sor corresponding to the sinusoid. By suppressing the time f actor, we transform the sinusoid from the time domain to the phasor domain. This transformation is summarized as follows:
-
-$$
-v(t) = V_m \cos(\omega t + \phi) \qquad \Leftrightarrow \qquad \mathbf{V} = V_m / \underline{\phi}
-$$
- (9.25)
-\n(Time-domain representation) (Phasor-domain representation)
-
-We use lightface italic letters such as z to represent complex numbers but boldface letters such as **V** to represent phasors, because phasors are vectorlike quantities.
-
-A phasor diagram showing **V** = *Vm*⧸*ϕ* and **I** = *Im*⧸−*θ*.
-
-Given a sinusoid *v*(*t*) = *Vm* cos(*ωt* + *ϕ*), we obtain the corresponding phasor as **V** = *Vm* ⧸*ϕ*. Equation (9.25) is also demonstrated in Table 9.1, where the sine function is considered in addition to the cosine function. From Eq. (9.25), we see that to get the phasor representation of a sinu soid, we e xpress it in cosine form and tak e the magnitude and phase. Given a phasor, we obtain the time domain representation as the cosine function with the same magnitude as the phasor and the ar gument as *ωt* plus the phase of the phasor. The idea of expressing information in alternate domains is fundamental to all areas of engineering.
-
-## **TABLE 9.1**
-
-Sinusoid-phasor transformation.
-
-| Phasor domain representation |
-|------------------------------|
-| Vm⧸ϕ |
-| Vm⧸ϕ − 90° |
-| Im⧸θ |
-| Im⧸θ − 90° |
-| |
-
-Note that in Eq. (9.25) the frequenc y (or time) f actor *ejωt* is sup pressed, and the frequency is not explicitly shown in the phasor domain representation because *ω* is constant. However, the response depends on *ω*. For this reason, the phasor domain is also known as the *frequency domain.*
-
-From Eqs. (9.23) and (9.24), *v*(*t*) = Re(**V***e jωt* ) = *Vm* cos(*ωt* + *ϕ*), so that
-
-$$
-\frac{dv}{dt} = -\omega V_m \sin(\omega t + \phi) = \omega V_m \cos(\omega t + \phi + 90^\circ)
-$$
-
-= Re( $\omega V_m e^{j\omega t} e^{j\phi} e^{j90^\circ}$ ) = Re( $j\omega V e^{j\omega t}$ ) (9.26)
-
-This shows that the deri vative *v*(*t*) is transformed to the phasor domain as *jω***V**
-
-$$
-\frac{dv}{dt} \qquad \Leftrightarrow \qquad j\omega V \qquad (9.27)
-$$
-\n(Time domain)
-
-\n(Phasor domain)
-
-Similarly, the inte gral of *v*(*t*) is transformed to the phasor domain as **V**∕*jω*
-
-$$
-\int v \, dt \qquad \Leftrightarrow \qquad \frac{V}{j\omega} \qquad (9.28)
-$$
-\n(Time domain)
-
-\n(Phasor domain)
-
-Equation (9.27) allows the replacement of a derivative with respect to time with multiplication of *jω* in the phasor domain, whereas Eq. (9.28) allows the replacement of an inte gral with respect to time with di vision by *jω* in the phasor domain. Equations (9.27) and (9.28) are useful in finding the steady-state solution, which does not require knowing the initial values of the variable involved. This is one of the important applications of phasors.
-
-Besides time differentiation and integration, another important use of phasors is found in summing sinusoids of the same frequency. This is best illustrated with an example, and Example 9.6 provides one.
-
-The differences between *v*(*t*) and **V** should be emphasized:
-
-- 1. *v*(*t*) is the *instantaneous or time domain* representation, while **V** is the *frequency or phasor domain* representation.
-- 2. *v*(*t*) is time dependent, while **V** is not. (This f act is often for gotten by students.)
-- 3. *v*(*t*) is al ways real with no comple x term, while **V** is generally complex.
-
-Finally, we should bear in mind that phasor analysis applies only when frequency is constant; it applies in manipulating two or more sinusoidal signals only if they are of the same frequency.
-
- Differentiating a sinusoid is equivalent to multiplying its corresponding phasor by j*ω*.
-
- Integrating a sinusoid is equivalent to dividing its corresponding phasor by j*ω*.
-
- Adding sinusoids of the same frequency is equivalent to adding their corresponding phasors.
-
-Evaluate these complex numbers: Example 9.3
-
-(a)
-$$
-(40/50^{\circ} + 20/-30^{\circ})^{1/2}
-$$
-
-\n(b)
-$$
-\frac{10/-30^{\circ} + (3-j4)}{(2+j4)(3-j5)^{*}}
-$$
-
-# **Solution:**
-
-(a) Using polar to rectangular transformation,
-
-$$
-40/50^{\circ} = 40(\cos 50^{\circ} + j \sin 50^{\circ}) = 25.71 + j30.64
-$$
-
-$$
-20\angle -30^{\circ} = 20[\cos(-30^{\circ}) + j\sin(-30^{\circ})] = 17.32 - j10
-$$
-
-Adding them up gives
-
-$$
-40/50^{\circ} + 20/-30^{\circ} = 43.03 + j20.64 = 47.72/25.63^{\circ}
-$$
-
-Taking the square root of this,
-
-$$
-(40/50^{\circ} + 20/-30^{\circ})^{1/2} = 6.91/12.81^{\circ}
-$$
-
-(b) Using polar-rectangular transformation, addition, multiplication, and division,
-
-(b) Using polar-rectangular transformation, a
-division,
-$$
-\frac{10(-30^\circ + (3 - j4))}{(2 + j4)(3 - j5)^*} = \frac{8.66 - j5 + (3 - j4)}{(2 + j4)(3 + j5)}
-$$
-$$
-= \frac{11.66 - j9}{-14 + j22} = \frac{14.73(-37.66^\circ)}{26.08(122.47^\circ)}
-$$
-$$
-= 0.565(-160.13^\circ)
-$$
-
-| Practice Problem 9.3 | Evaluate the following complex numbers: |
-|----------------------|------------------------------------------------------------------------|
-| | (a) [(5 + j2)(−1 + j4) − 5⧸ 60°]* |
-| | 10 + j5 + 3⧸ 40°
______________
(b)
+ 10⧸ 30° + j5
−3 + j4 |
-| | Answer: (a) −15.5 −
j13.67, (b) 8.293 + j7.2. |
-| | |
-
-Example 9.4 Transform these sinusoids to phasors:
-
-(a) *i* = 6 cos(50*t* − 40°) A (b) *v* = −4 sin(30*t* + 50°) V
-
-# **Solution:**
-
-(a) *i* = 6 cos(50*t* − 40°) has the phasor
-
-$$
-I = 6/–40^{\circ} A
-$$
-
-(b) Since
-$$
--\sin A = \cos(A + 90^\circ)
-$$
-,
- $v = -4 \sin(30t + 50^\circ) = 4 \cos(30t + 50^\circ + 90^\circ)$
-
-$$
-= 4\cos(30t + 140^\circ) \text{ V}
-$$
-
-The phasor form of *v* is
-
-$$
-V = 4/140^{\circ} V
-$$
-
-Practice Problem 9.4 Express these sinusoids as phasors:
-
-(a) *v* = −14 sin(5*t* − 22°) V (b) *i* = −8 cos(16*t* + 15°) A
-
-**Answer:** (a) **V** = 14⧸ 68° V, (b) **I** = 8⧸−165° A.
-
-Find the sinusoids represented by these phasors: Example 9.5
-
-(a)
-$$
-\mathbf{I} = -3 + j4 \text{ A}
-$$
-
-\n(b) $\mathbf{V} = j8e^{-j20^{\circ}} \text{ V}$
-
-# **Solution:**
-
-(a) **I** = −3 + *j*4 = 5⧸ 126.87°. Transforming this to the time domain gives
-
-$$
-i(t) = 5 \cos(\omega t + 126.87^{\circ})
-$$
- A
-
-(b) Because *j* = 1⧸ 90°,
-
-$$
-\mathbf{V} = j8 \underline{/ -20^{\circ}} = (1 \underline{/ 90^{\circ}})(8 \underline{/ -20^{\circ}})
-$$
-$$
-= 8 \underline{/ 90^{\circ} - 20^{\circ}} = 8 \underline{/ 70^{\circ}} \text{ V}
-$$
-
-Converting this to the time domain gives
-
-$$
-v(t) = 8\cos(\omega t + 70^\circ)\text{V}
-$$
-
-Find the sinusoids corresponding to these phasors: Practice Problem 9.5
-
-(a) **V** = −25⧸ 40° V (b) **I** = *j*(12 − *j*5) A **Answer:** (a) *v*(*t*) = 25 cos(*ωt* − 140°) V or 25 cos(*ωt* + 220°) V, (b) *i*(*t*) = 13 cos(*ωt* + 67.38°) A.
-
-Given *i*1(*t*) = 4 cos(*ωt* + 30°) A and *i*2(*t*) = 5 sin(*ωt* − 20°) A, find Example 9.6 their sum.
-
-# **Solution:**
-
-Here is an important use of phasors—for summing sinusoids of the same frequency. Current *i*1(*t*) is in the standard form. Its phasor is
-
-$$
-\mathbf{I}_1 = 4/30^\circ
-$$
-
-We need to express *i*2(*t*) in cosine form. The rule for converting sine to cosine is to subtract 90°. Hence,
-
-$$
-i_2 = 5\cos(\omega t - 20^\circ - 90^\circ) = 5\cos(\omega t - 110^\circ)
-$$
-
-and its phasor is
-
-**I**2 = 5⧸−110°
-
-If we let *i* = *i*1 + *i*2, then
-
-$$
-\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 = 4/30^\circ + 5/-110^\circ
-$$
-
-= 3.464 + j2 - 1.71 - j4.698 = 1.754 - j2.698
-= 3.218/-56.97° A
-
-Transforming this to the time domain, we get
-
-$$
-i(t) = 3.218 \cos(\omega t - 56.97^{\circ}) \text{ A}
-$$
-
-Of course, we can find *i*1 + *i*2 using Eq. (9.9), but that is the hard way.
-
-Practice Problem 9.6 If *v*1 = −10 sin(*ωt* − 30°) V and *v*2 = 20 cos(*ωt* + 45°) V, find *v*= *v*1 + *v*2.
-
-**Answer:** *v*(*t*) = 29.77 cos(*ωt* + 49.98°) V.
-
-Example 9.7 Using the phasor approach, determine the current *i*(*t*) in a circuit described by the integrodifferential equation
-
-$$
-4i + 8\int i\,dt - 3\frac{di}{dt} = 50\cos(2t + 75^\circ)
-$$
-
-# **Solution:**
-
-We transform each term in the equation from time domain to phasor domain. Keeping Eqs. (9.27) and (9.28) in mind, we obtain the phasor form of the given equation as
-
-$$
-4\mathbf{I} + \frac{8\mathbf{I}}{j\omega} - 3j\omega\mathbf{I} = 50/75^{\circ}
-$$
-
-But *ω* = 2, so
-
-$$
-I(4 - j4 - j6) = 50 / 75^{\circ}
-$$
-
-$$
-I = \frac{50/75^{\circ}}{4 - j10} = \frac{50/75^{\circ}}{10.77/-68.2^{\circ}} = 4.642/143.2^{\circ}
-$$
- A
-
-Converting this to the time domain,
-
-$$
-i(t) = 4.642 \cos(2t + 143.2^{\circ}) \text{ A}
-$$
-
-Keep in mind that this is only the steady-state solution, and it does not require knowing the initial values.
-
-Practice Problem 9.7 Find the v oltage *v*(*t*) in a circuit described by the inte grodifferential equation
-
-$$
-2\frac{dv}{dt} + 5v + 10 \int v \, dt = 50 \cos(5t - 30^{\circ})
-$$
-
-using the phasor approach.
-
-**Answer:**
-$$
-v(t) = 5.3 \cos(5t - 88^\circ)
-$$
- V.
-
-# **9.4** Phasor Relationships for Circuit Elements
-
-Now that we know how to represent a voltage or current in the phasor or frequency domain, one may le gitimately ask ho w we apply this to cir cuits involving the passive elements *R*, *L*, and *C*. What we need to do is to transform the voltage-current relationship from the time domain to the frequency domain for each element. Again, we will assume the passi ve sign convention.
-
-We be gin with the resistor . If the current through a resistor *R* is *i* = *Im* cos(*ωt* + *ϕ*), the voltage across it is given by Ohm's law as
-
-$$
-v = iR = RIm \cos(\omega t + \phi)
-$$
- (9.29)
-
-The phasor form of this voltage is
-
-$$
-\mathbf{V} = R I_m / \underline{\phi} \tag{9.30}
-$$
-
-But the phasor representation of the current is **I** = I*m*⧸*ϕ*. Hence,
-
-$$
-V = RI \tag{9.31}
-$$
-
-showing that the v oltage-current relation for the resistor in the phasor domain continues to be Ohm's law, as in the time domain. Figure 9.9 illustrates the voltage-current relations of a resistor. We should note from Eq. (9.31) that voltage and current are in phase, as illustrated in the phasor diagram in Fig. 9.10.
-
-For the inductor *L*, assume the current through it is *i* = *Im* cos(*ωt* + *ϕ*). The voltage across the inductor is
-
-$$
-v = L\frac{di}{dt} = -\omega L I_m \sin(\omega t + \phi)
-$$
-\n(9.32)
-
-Recall from Eq. (9.10) that −sin *A* = cos(*A* + 90°). We can write the voltage as
-
-$$
-v = \omega L I_m \cos(\omega t + \phi + 90^\circ) \tag{9.33}
-$$
-
-which transforms to the phasor
-
-$$
-\mathbf{V} = \omega L I_m e^{j(\phi + 90^\circ)} = \omega L I_m e^{j\phi} e^{j90^\circ} = \omega L I_m / \phi + 90^\circ \tag{9.34}
-$$
-
-But *Im*⧸*ϕ* = **I**, and from Eq. (9.19), *ej*90° = *j*. Thus,
-
-$$
-V = j\omega L I \tag{9.35}
-$$
-
-showing that the v oltage has a magnitude of *ωLIm* and a phase of *ϕ* + 90°. The voltage and current are 90° out of phase. Specifically, the current lags the v oltage by 90°. Figure 9.11 sho ws the v oltage-current relations for the inductor. Figure 9.12 shows the phasor diagram.
-
-For the capacitor *C*, assume the voltage across it is *v* = *Vm* cos(*ωt* + *ϕ*). The current through the capacitor is
-
-$$
-i = C \frac{dv}{dt}
-$$
- (9.36)
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/106_9.4 Phasor Relationships for Circuit Elements.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/106_9.4 Phasor Relationships for Circuit Elements.md
deleted file mode 100644
index 7e5eec7c60020292c5b79b91492a85b69561c582..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/106_9.4 Phasor Relationships for Circuit Elements.md
+++ /dev/null
@@ -1,117 +0,0 @@
-By following the same steps as we took for the inductor or by applying Eq. (9.27) on Eq. (9.36), we obtain
-
-$$
-\mathbf{I} = j\omega C \mathbf{V} \qquad \Rightarrow \qquad \mathbf{V} = \frac{\mathbf{I}}{j\omega C} \tag{9.37}
-$$
-
-# **Figure 9.9**
-
-Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain.
-
-Phasor diagram for the resistor.
-
-# **Figure 9.11**
-
-Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain.
-
-**Figure 9.12** Phasor diagram for the inductor; **I** lags **V**.
-
-Although it is equally correct to say that the inductor voltage leads the current by 90°, convention gives the current phase relative to the voltage.
-
-## **384** Chapter 9 Sinusoids and Phasors
-
-showing that the current and voltage are 90° out of phase. To be specific, the current leads the voltage by 90°. Figure 9.13 shows the voltage-current relations for the capacitor; Fig. 9.14 gives the phasor diagram. Table 9.2 summarizes the time domain and phasor domain representations of the circuit elements.
-
-# **TABLE 9.2**
-
-Summary of voltage-current relationships.
-
-| Element | Time domain | Frequency domain |
-|---------|--------------------|-------------------|
-| R | v = Ri | V = RI |
-| L | v = L __di
dt | V = jωLI |
-| C | i = C ___ dv
dt | V = ____ I
jωC |
-
-Example 9.8 The voltage *v* = 12cos(60*t* + 45°) is applied to a 0.1-H inductor. Find the steady-state current through the inductor.
-
-# **Solution:**
-
-For the inductor , **V** = *jωL***I**, where *ω* = 60 rad/s and **V** = 12 ⧸45° V. Hence,
-
-$$
-I = \frac{V}{j\omega L} = \frac{12/45^{\circ}}{j60 \times 0.1} = \frac{12/45^{\circ}}{6/90^{\circ}} = 2/45^{\circ}
-$$
- A
-
-Converting this to the time domain,
-
-$$
-i(t) = 2\cos(60t - 45^\circ) \,\mathrm{A}
-$$
-
-Practice Problem 9.8 If voltage *v* = 25 sin(100*t* − 15°) *V* is applied to a 50 *μ*F capacitor, calculate the current through the capacitor.
-
-# **Answer:** 125 sin(100*t* + 75°) mA.
-
-$$
-0.125 \sim
-$$
-
-# **9.5** Impedance and Admittance
-
-In the preceding section, we obtained the v oltage-current relations for the three passive elements as
-
-$$
-V = RI, \t V = j\omega LI, \t V = \frac{I}{j\omega C}
-$$
-(9.38)
-
-These equations may be written in terms of the ratio of the phasor v oltage to the phasor current as
-
-$$
-\frac{V}{I} = R, \qquad \frac{V}{I} = j\omega L, \qquad \frac{V}{I} = \frac{1}{j\omega C}
-$$
- (9.39)
-
-From these three e xpressions, we obtain Ohm's law in phasor form for any type of element as
-
-$$
-Z = \frac{V}{I} \qquad \text{or} \qquad V = ZI \tag{9.40}
-$$
-
-where **Z** is a frequenc y-dependent quantity known as *impedance*, measured in ohms.
-
-The impedance **Z** of a circuit is the ratio of the phasor voltage **V** to the phasor current **I**, measured in ohms (Ω).
-
-The impedance represents the opposition that the circuit e xhibits to the flow of sinusoidal current. Although the impedance is the ratio of two phasors, it is not a phasor, because it does not correspond to a sinusoidally varying quantity.
-
-The impedances of resistors, inductors, and capacitors can be readily obtained from Eq. (9.39). Table 9.3 summarizes their impedances. From the table we notice that **Z***L* = *jωL* and **Z***C* = −*j*∕*ωC*. Consider two extreme cases of angular frequenc y. When *ω* = 0 (i.e., for dc sources), **Z***L* = 0 and **Z***C* → ∞, confirming what we already know—that the inductor acts like a short circuit, while the capacitor acts like an open circuit. When *ω* → ∞ (i.e., for high frequencies), **Z***L* → ∞ and **Z***C* = 0, indicating that the inductor is an open circuit to high frequencies, while the capacitor is a short circuit. Figure 9.15 illustrates this.
-
-As a complex quantity, the impedence may be e xpressed in rectangular form as
-
-$$
-Z = R \pm jX \tag{9.41}
-$$
-
-where *R* = Re **Z** is the *resistance* and *X* = Im **Z** is the *reactance*. The reactance, *X,* is just a magnitude, a positi ve value, but when used as a vector, a *j* is associated with inductance and a −*j* is associated with capacitance. Thus, impedance **Z** = *R* + *jX* is said to be *inductive* o r lagging since current lags v oltage, while impedance **Z** = *R* − *jX* i s capacitive or leading because current leads v oltage. The impedance, resistance, and reactance are all measured in ohms. The impedance may also be expressed in polar form as
-
-$$
-\mathbf{Z} = |\mathbf{Z}| \; / \theta \tag{9.42}
-$$
-
-**TABLE 9.3**
-
-Impedances and admittances of passive elements.
-
-| Impedance | Admittance |
-|--------------------|------------------------------------------------------------|
-| Z = R | Y = __1
R |
-| Z = jωL | Y = ____ 1
jωL |
-| Z = ____ 1
jω C | Y = jωC |
-| | Short circuit at dc
Open circuit at
high frequencies |
-| (a) | |
-| | Open circuit at dc |
-| (b) | Short circuit at
high frequencies |
-| | |
-
-# **Figure 9.15**
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/107_9.5 Impedance and Admittance.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/107_9.5 Impedance and Admittance.md
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+++ /dev/null
@@ -1,126 +0,0 @@
-Equivalent circuits at dc and high frequencies: (a) inductor, (b) capacitor. Comparing Eqs. (9.41) and (9.42), we infer that
-
-$$
-Z = R \pm jX = |Z|/\theta \tag{9.43}
-$$
-
-where
-
-$$
-|Z| = \sqrt{R^2 + X^2}, \qquad \theta = \tan^{-1} \frac{\pm X}{R}
-$$
- (9.44)
-
-and
-
-$$
-R = |\mathbf{Z}| \cos \theta, \qquad X = |\mathbf{Z}| \sin \theta \tag{9.45}
-$$
-
-It is sometimes con venient to w ork with the reciprocal of imped ance, known as *admittance*.
-
-The admittance **Y** is the reciprocal of impedance, measured in siemens (S).
-
-The admittance **Y** of an element (or a circuit) is the ratio of the phasor current through it to the phasor voltage across it, or
-
-$$
-Y = \frac{1}{Z} = \frac{I}{V}
-$$
- (9.46)
-
-The admittances of resistors, inductors, and capacitors can be obtained from Eq. (9.39). They are also summarized in Table 9.3.
-
-As a complex quantity, we may write **Y** as
-
-$$
-Y = G + jB \tag{9.47}
-$$
-
-where *G* = Re **Y** is called the *conductance* and *B* = Im **Y** is called the *susceptance*. Admittance, conductance, and susceptance are all expressed in the unit of siemens (or mhos). From Eqs. (9.41) and (9.47),
-
-$$
-G + jB = \frac{1}{R + jX} \tag{9.48}
-$$
-
-By rationalization,
-
-$$
-G + jB = \frac{1}{R + jX} \cdot \frac{R - jX}{R - jX} = \frac{R - jX}{R^2 + X^2}
-$$
-(9.49)
-
-Equating the real and imaginary parts gives
-
-$$
-G = \frac{R}{R^2 + X^2}, \qquad B = -\frac{X}{R^2 + X^2}
-$$
-(9.50)
-
-showing that *G* ≠ 1∕*R* as it is in resisti ve circuits. Of course, if *X* = 0, then *G* = 1∕*R*.
-
-Find *v*(*t*) and *i*(*t*) in the circuit shown in Fig. 9.16. Example 9.9
-
-# **Solution:**
-
-From the voltage source 10 cos 4*t*, *ω* = 4,
-
-$$
-V_s = 10/0^{\circ} V
-$$
-
-The impedance is
-
-$$
-\mathbf{Z} = 5 + \frac{1}{j\omega C} = 5 + \frac{1}{j4 \times 0.1} = 5 - j2.5 \ \Omega
-$$
-
-Hence the current
-
-$$
-\mathbf{I} = \frac{\mathbf{V}_s}{\mathbf{Z}} = \frac{10/0^{\circ}}{5 - j2.5} = \frac{10(5 + j2.5)}{5^2 + 2.5^2}
-$$
-(9.9.1)
-= 1.6 + j0.8 = 1.789/26.57° A
-
-The voltage across the capacitor is
-
-The voltage across the capacitor is
-\n
-$$
-\mathbf{V} = IZ_C = \frac{I}{j\omega C} = \frac{1.789/26.57^\circ}{j4 \times 0.1} = \frac{1.789/26.57^\circ}{0.4/90^\circ} = 4.47/-63.43^\circ \text{ V}
-$$
-\n(9.9.2)
-
-Converting **I** and **V** in Eqs. (9.9.1) and (9.9.2) to the time domain, we get
-
-$$
-i(t) = 1.789 \cos(4t + 26.57^{\circ}) \text{ A}
-$$
-
-$$
-v(t) = 4.47 \cos(4t - 63.43^{\circ}) \text{ V}
-$$
-
-Notice that *i*(*t*) leads *v*(*t*) by 90° as expected.
-
-Refer to Fig. 9.17. Determine *v*(*t*) and *i*(*t*). Practice Problem 9.9
-
-**Answer:** 8.944 sin (10*t* + 93.43°) V, 4.472 sin(10*t* + 3.43°) A.
-
-**9.6** Kirchhoff's Laws in the Frequency Domain
-
-We cannot do circuit analysis in the frequenc y domain without Kirch hoff's current and v oltage laws. Therefore, we need to e xpress them in the frequency domain.
-
-For KVL, let *v*1,*v*2, … , *vn* be the voltages around a closed loop. Then
-
-$$
-v_1 + v_2 + \dots + v_n = 0 \tag{9.51}
-$$
-
-In the sinusoidal steady state, each v oltage may be written in cosine form, so that Eq. (9.51) becomes
-
-$$
-V_{m1}\cos(\omega t + \theta_1) + V_{m2}\cos(\omega t + \theta_2)
-$$
-
-+ ... + $V_{mn}\cos(\omega t + \theta_n) = 0$ (9.52)
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/108_9.6 Kirchhoff's Laws in the Frequency Domain.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/108_9.6 Kirchhoff's Laws in the Frequency Domain.md
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/108_9.6 Kirchhoff's Laws in the Frequency Domain.md
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@@ -1,46 +0,0 @@
-# **Figure 9.17**
-
-For Practice Prob. 9.9.
-
-This can be written as
-
-$$
-\text{Re}(V_{m1}e^{j\theta_1}e^{j\omega t}) + \text{Re}(V_{m2}e^{j\theta_2}e^{j\omega t}) + \dots + \text{Re}(V_{mn}e^{j\theta_n}e^{j\omega t}) = 0
-$$
-
-or
-
-$$
-\text{Re}[(V_{m1}e^{j\theta_1} + V_{m2}e^{j\theta_2} + \dots + V_{mn}e^{j\theta_n})e^{j\omega t}] = 0 \tag{9.53}
-$$
-
-If we let **V***k* = *Vmkejθk*, then
-
-$$
-Re[(V_1 + V_2 + \dots + V_n)e^{j\omega t}] = 0
-$$
-\n(9.54)
-
-Because *ejωt* ≠ 0,
-
-$$
-V_1 + V_2 + \dots + V_n = 0 \tag{9.55}
-$$
-
-indicating that Kirchhoff's voltage law holds for phasors.
-
-By following a similar procedure, we can show that Kirchhoff's current law holds for phasors. If we let *i*1, *i*2, … , *in* be the current leaving or entering a closed surface in a network at time *t*, then
-
-$$
-i_1 + i_2 + \dots + i_n = 0 \tag{9.56}
-$$
-
-If **I**1, **I**2, … , **I***n* are the phasor forms of the sinusoids *i*1,*i*2, … ,*in*, then
-
-$$
-\mathbf{I}_1 + \mathbf{I}_2 + \dots + \mathbf{I}_n = 0 \tag{9.57}
-$$
-
-which is Kirchhoff's current law in the frequency domain.
-
-Once we have shown that both KVL and KCL hold in the frequency domain, it is easy to do man y things, such as impedance combination, nodal and mesh analyses, superposition, and source transformation.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/109_9.7 Impedance Combinations.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/109_9.7 Impedance Combinations.md
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@@ -1,259 +0,0 @@
-# **9.7** Impedance Combinations
-
-Consider the *N* series-connected impedances sho wn in Fig. 9.18. The same current **I** flows through the impedances. Applying KVL around the loop gives
-
-$$
-V = V_1 + V_2 + \dots + V_N = I(Z_1 + Z_2 + \dots + Z_N)
-$$
-(9.58)
-
-*N* impedances in series.
-
-The equivalent impedance at the input terminals is
-
-$$
-\mathbf{Z}_{\text{eq}} = \frac{\mathbf{V}}{\mathbf{I}} = \mathbf{Z}_1 + \mathbf{Z}_2 + \dots + \mathbf{Z}_N
-$$
-
-$$
-Z_{eq} = Z_1 + Z_2 + \dots + Z_N \tag{9.59}
-$$
-
-or
-
-showing that the total or equi valent impedance of series-connected impedances is the sum of the indi vidual impedances. This is similar to the series connection of resistances.
-
-If *N* = 2, as shown in Fig. 9.19, the current through the imped ances is
-
-$$
-\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}_1 + \mathbf{Z}_2} \tag{9.60}
-$$
-
-Because **V**1 = **Z**1**I** and **V**2 = **Z**2**I**, then
-
-$$
-V_1 = \frac{Z_1}{Z_1 + Z_2} V, \qquad V_2 = \frac{Z_2}{Z_1 + Z_2} V
-$$
- (9.61)
-
-which is the *voltage-division* relationship.
-
-In the same manner , we can obtain the equi valent impedance or admittance of the *N* parallel-connected impedances shown in Fig. 9.20. The voltage across each impedance is the same. Applying KCL at the top node,
-
-$$
-\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 + \dots + \mathbf{I}_N = \mathbf{V} \left( \frac{1}{\mathbf{Z}_1} + \frac{1}{\mathbf{Z}_2} + \dots + \frac{1}{\mathbf{Z}_N} \right) \tag{9.62}
-$$
-
-*N* impedances in parallel.
-
-The equivalent impedance is
-
-$$
-\frac{1}{Z_{\text{eq}}} = \frac{I}{V} = \frac{1}{Z_1} + \frac{1}{Z_2} + \dots + \frac{1}{Z_N}
-$$
-(9.63)
-
-and the equivalent admittance is
-
-$$
-\mathbf{Y}_{\text{eq}} = \mathbf{Y}_1 + \mathbf{Y}_2 + \dots + \mathbf{Y}_N \tag{9.64}
-$$
-
-This indicates that the equivalent admittance of a parallel connection of admittances is the sum of the individual admittances.
-
-When *N* = 2, as sho wn in Fig. 9.21, the equi valent impedance becomes
-
-$$
-Z_{\text{eq}} = \frac{1}{Y_{\text{eq}}} = \frac{1}{Y_1 + Y_2} = \frac{1}{1/Z_1 + 1/Z_2} = \frac{Z_1 Z_2}{Z_1 + Z_2}
-$$
-(9.65)
-
-**Figure 9.21** Current division.
-
-Voltage division.
-
-Also, since
-
-$$
-\mathbf{V} = \mathbf{I}\mathbf{Z}_{\text{eq}} = \mathbf{I}_1\mathbf{Z}_1 = \mathbf{I}_2\mathbf{Z}_2
-$$
-
-the currents in the impedances are
-
-$$
-\mathbf{I}_1 = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{I}, \qquad \mathbf{I}_2 = \frac{\mathbf{Z}_1}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{I}
-$$
- (9.66)
-
-which is the *current-division* principle.
-
-The delta-to-wye and wye-to-delta transformations that we applied to resisti ve circuits are also v alid for impedances. With reference to Fig. 9.22, the conversion formulas are as follows.
-
-**Figure 9.22** Superimposed *Y* and ∆ networks.
-
-*Y*-∆ *Conversion:*
-
-$$
-Z_a = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_1}
-$$
-
-\n
-$$
-Z_b = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_2}
-$$
-
-\n
-$$
-Z_c = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_3}
-$$
-
-\n(9.67)
-
-∆-*Y Conversion:*
-
-$$
-\mathbf{Z}_{1} = \frac{\mathbf{Z}_{b}\mathbf{Z}_{c}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}}
-$$
-\n
-$$
-\mathbf{Z}_{2} = \frac{\mathbf{Z}_{c}\mathbf{Z}_{a}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}}
-$$
-\n
-$$
-\mathbf{Z}_{3} = \frac{\mathbf{Z}_{a}\mathbf{Z}_{b}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}}
-$$
-\n(9.68)
-
-A delta or wye circuit is said to be balanced if it has equal impedances in all three branches.
-
-When a ∆-*Y* circuit is balanced, Eqs. (9.67) and (9.68) become
-
-$$
-\mathbf{Z}_{\Delta} = 3\mathbf{Z}_{Y} \qquad \text{or} \qquad \mathbf{Z}_{Y} = \frac{1}{3}\mathbf{Z}_{\Delta} \qquad (9.69)
-$$
-
-where **Z***Y* = **Z**1 = **Z**2 = **Z**3 and **Z**∆ = **Z***a* = **Z***b* = **Z***c*.
-
-As you see in this section, the principles of voltage division, current division, circuit reduction, impedance equi valence, and *Y*-∆ transformation all apply to ac circuits. Chapter 10 will sho w that other c ircuit techniques—such as superposition, nodal analysis, mesh analysis, source transformation, the Thevenin theorem, and the Norton theorem are all applied to ac circuits in a manner similar to their application in dc circuits.
-
-Find the input impedance of the circuit in Fig. 9.23. Assume that the Example 9.10 circuit operates at *ω* = 50 rad/s.
-
-# **Solution:**
-
-Let
-
-- **Z**1 = Impedance of the 2-mF capacitor
-- **Z**2 = Impedance of the 3-Ω resistor in series with the10-mF capacitor
-- **Z**3 = Impedance of the 0.2-H inductor in series with the 8-Ω resistor
-
-Then
-
-$$
-\mathbf{Z}_1 = \frac{1}{j\omega C} = \frac{1}{j50 \times 2 \times 10^{-3}} = -j10 \text{ }\Omega
-$$
-
-$$
-\mathbf{Z}_2 = 3 + \frac{1}{j\omega C} = 3 + \frac{1}{j50 \times 10 \times 10^{-3}} = (3 - j2) \text{ }\Omega
-$$
-
-$$
-\mathbf{Z}_3 = 8 + j\omega L = 8 + j50 \times 0.2 = (8 + j10) \text{ }\Omega
-$$
-
-The input impedance is
-
-at impedance is
-\n
-$$
-\mathbf{Z}_{in} = \mathbf{Z}_1 + \mathbf{Z}_2 \parallel \mathbf{Z}_3 = -j10 + \frac{(3 - j2)(8 + j10)}{11 + j8}
-$$
-\n
-$$
-= -j10 + \frac{(44 + j14)(11 - j8)}{11^2 + 8^2} = -j10 + 3.22 - j1.07 \Omega
-$$
-
-Thus,
-
-$$
-\mathbf{Z}_{\text{in}} = 3.22 - j11.07 \ \Omega
-$$
-
-For Example 9.10.
-
-**Figure 9.25**
-
-**Figure 9.26**
-
-circuit in Fig. 9.25.
-
-For Example 9.11.
-
-The frequency domain equivalent of the
-
-*vs* = 20 cos(4*t* − 15°) ⇒ **V***s* = 20⧸−15° V, *ω* = 4 10 mF ⇒ \_\_\_\_ 1 *jωC* = \_\_\_\_\_\_\_\_\_\_\_\_ 1 *j*4 × 10 × 10−3 = −*j*25 Ω 5 H ⇒ *jωL* = *j*4 × 5 = *j*20 Ω
-
- **Z**1 = Impedance of the 60-Ω resistor
-
- **Z**2 = Impedance of the parallel combination of the 10-mF capacitor and the 5-H inductor
-
-Then **Z**1 = 60 Ω and
-
-$$
-\mathbf{Z}_2 = -j25 \parallel j20 = \frac{-j25 \times j20}{-j25 + j20} = j100 \text{ }\Omega
-$$
-
-By the voltage-division principle,
-
-$$
-\mathbf{V}_o = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{V}_s = \frac{j100}{60 + j100} (20/15^\circ)
-$$
-
-= (0.8575/30.96°)(20/15°) = 17.15/15.96° V
-
-We convert this to the time domain and obtain
-
-*vo*(*t*) = 17.15 cos(4*t* + 15.96°) V
-
-For Practice Prob. 9.11.
-
-# Find current **I** in the circuit of Fig. 9.28. Example 9.12
-
-# **Solution:**
-
-The delta netw ork connected to nodes *a*, *b*, and *c* can be con verted to the *Y* network of Fig. 9.29. We obtain the *Y* impedances as follows using Eq. (9.68):
-
-68):
-\n
-$$
-\mathbf{Z}_{an} = \frac{j4(2-j4)}{j4+2-j4+8} = \frac{4(4+j2)}{10} = (1.6+j0.8) \ \Omega
-$$
-\n
-$$
-\mathbf{Z}_{bn} = \frac{j4(8)}{10} = j3.2 \ \Omega, \qquad \mathbf{Z}_{cn} = \frac{8(2-j4)}{10} = (1.6-j3.2) \ \Omega
-$$
-
-The total impedance at the source terminals is
-
-$$
-\mathbf{Z} = 12 + \mathbf{Z}_{an} + (\mathbf{Z}_{bn} - j3) \parallel (\mathbf{Z}_{cn} + j6 + 8)
-$$
-
-= 12 + 1.6 + j0.8 + (j0.2) || (9.6 + j2.8)
-= 13.6 + j0.8 + $\frac{j0.2(9.6 + j2.8)}{9.6 + j3}$
-= 13.6 + j1 = 13.64/4.204° Ω
-
-The desired current is
-
-$$
-I = \frac{V}{Z} = \frac{50/0^{\circ}}{13.64/4.204^{\circ}} = 3.666/-4.204^{\circ} A
-$$
-
-# Practice Problem 9.12 Find **I** in the circuit of Fig. 9.30.
-
-**Figure 9.30** For Practice Prob. 9.12.
-
-# **Figure 9.31**
-
-Series *RC* shift circuits: (a) leading output, (b) lagging output.
-
-**Answer:** 12.728 ⧸ 63.8° A.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/110_9.8 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/110_9.8 Applications.md
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-# **9.8** Applications
-
-In Chapters 7 and 8, we sa w certain uses of *RC*, *RL*, and *RLC* circuits in dc applications. These circuits also have ac applications; among them are coupling circuits, phase-shifting circuits, filters, resonant circuits, ac bridge circuits, and transformers. This list of applications is ine xhaustive. We will consider some of them later. It will suffice here to observe two simple ones: *RC* phase-shifting circuits, and ac bridge circuits.
-
-# **9.8.1** Phase-Shifters
-
-A phase-shifting circuit is often emplo yed to correct an undesirable phase shift already present in a circuit or to produce special desired effects. An *RC* circuit is suitable for this purpose because its capacitor causes the circuit current to lead the applied v oltage. Two commonly used *RC* circuits are shown in Fig. 9.31. (*RL* circuits or any reactive circuits could also serve the same purpose.)
-
-In Fig. 9.31(a), the circuit current **I** leads the applied voltage **V***i* by some phase angle *θ*, where 0 < *θ*< 90°, depending on the values of *R* and *C*. If *XC* = −1∕*ωC*, then the total impedance is **Z** = *R* + *jXC*, and the phase shift is given by
-
-$$
-\theta = \tan^{-1} \frac{X_C}{R}
-$$
- (9.70)
-
-This shows that the amount of phase shift depends on the v alues of *R*, *C*, and the operating frequenc y. Since the output v oltage **V***o* across the resistor is in phase with the current, **V***o* leads (positive phase shift) **V***i* as shown in Fig. 9.32(a).
-
-In Fig. 9.31(b), the output is taken across the capacitor. The current **I** leads the input v oltage **V***i* by *θ*, but the output voltage *vo*(*t*) across the capacitor lags (negative phase shift) the input v oltage *vi*(*t*) as illustrated in Fig. 9.32(b).
-
-**Figure 9.32** Phase shift in *RC* circuits: (a) leading output, (b) lagging output.
-
-We should keep in mind that the simple *RC* circuits in Fig. 9.31 also act as voltage dividers. Therefore, as the phase shift *θ* approaches 90°, the output vo ltage **V***o* approaches zero. F or this reason, these simple *RC* circuits are used only when small amounts of phase shift are required. If it is desired to ha ve phase shifts greater than 60°, simple *RC* networks are cascaded, thereby providing a total phase shift equal to the sum of the indi vidual phase shifts. In practice, the phase shifts due to the stages are not equal, because the succeeding stages load down the earlier stages unless op amps are used to sepa rate the stages.
-
-# **Solution:**
-
-If we select circuit components of equal ohmic v alue, say *R* = ∣*XC*∣ = 20 Ω, at a particular frequenc y, according to Eq. (9.70), the phase shift is exactly 45°. By cascading two similar *RC* circuits in Fig. 9.31(a), we obtain the circuit in Fig. 9.33, providing a positive or leading phase shift of 90°, as we shall soon show. Using the series-parallel combination technique, **Z** in Fig. 9.33 is obtained as
-
-$$
-\mathbf{Z} = 20 \| (20 - j20) = \frac{20(20 - j20)}{40 - j20} = 12 - j4 \ \Omega
-$$
- (9.13.1)
-
-Using voltage division,
-
-$$
-\mathbf{V}_1 = \frac{\mathbf{Z}}{\mathbf{Z} - j20} \mathbf{V}_i = \frac{12 - j4}{12 - j24} \mathbf{V}_i = \frac{\sqrt{2}}{3} \angle 45^\circ \mathbf{V}_i \quad (9.13.2)
-$$
-
-and
-
-$$
-\mathbf{V}_o = \frac{20}{20 - j20} \mathbf{V}_1 = \frac{\sqrt{2}}{2} \underline{45^\circ} \mathbf{V}_1
-$$
- (9.13.3)
-
-Substituting Eq. (9.13.2) into Eq. (9.13.3) yields
-
-$$
-\mathbf{V}_o = \left(\frac{\sqrt{2}}{2} \angle 45^\circ\right) \left(\frac{\sqrt{2}}{3} \angle 45^\circ \mathbf{V}_i\right) = \frac{1}{3} \angle 90^\circ \mathbf{V}_i
-$$
-
-Thus, the output leads the input by 90° b ut its magnitude is only about 33 percent of the input.
-
-Design an *RC* circuit to pro vide a 90° lagging phase shift of the out - Practice Problem 9.13 put voltage relative to the input v oltage. If an ac v oltage of 60 V rms is applied, what is the output voltage?
-
-**Answer:** Figure 9.34 shows a typical design; 20 V rms.
-
-# **Figure 9.33**
-
-An *RC* phase shift circuit with 90° leading phase shift; for Example 9.13.
-
-**Figure 9.35** For Example 9.14.
-
-Example 9.14 For the *RL* circuit shown in Fig. 9.35(a), calculate the amount of phase shift produced at 2 kHz.
-
-# **Solution:**
-
-At 2 kHz, we transform the 10- and 5-mH inductances to the corresponding impedances.
-
-$$
-10 \text{ mH} \Rightarrow X_L = \omega L = 2\pi \times 2 \times 10^3 \times 10 \times 10^{-3}
-$$
-$$
-= 40\pi = 125.7 \text{ }\Omega
-$$
-$$
-5 \text{ mH} \Rightarrow X_L = \omega L = 2\pi \times 2 \times 10^3 \times 5 \times 10^{-3}
-$$
-$$
-= 20\pi = 62.83 \text{ }\Omega
-$$
-
-Consider the circuit in Fig. 9.35(b). The impedance **Z** is the parallel combination of *j*125.7 Ω and 100 + *j*62.83 Ω. Hence,
-
-$$
-\mathbf{Z} = j125.7 \parallel (100 + j62.83)
-$$
-
-\n
-$$
-\mathbf{Z} = j125.7 \parallel (100 + j62.83)
-$$
-
-\n
-$$
-= \frac{j125.7(100 + j62.83)}{100 + j188.5} = 69.56 / 60.1^{\circ} \ \Omega
-$$
- (9.14.1)
-
-Using voltage division,
-
-$$
-\mathbf{V}_1 = \frac{\mathbf{Z}}{\mathbf{Z} + 150} \mathbf{V}_i = \frac{69.56 / 60.1^{\circ}}{184.7 + j60.3} \mathbf{V}_i
-$$
-
-= 0.3582 / 42.02° $\mathbf{V}_i$ (9.14.2)
-
-and
-
-$$
-\mathbf{V}_o = \frac{j62.832}{100 + j62.832} \mathbf{V}_1 = 0.532 \, \text{/} 57.86^{\circ} \, \mathbf{V}_1 \tag{9.14.3}
-$$
-
-Combining Eqs. (9.14.2) and (9.14.3),
-
-**V***o* = (0.532 ⧸ 57.86°)(0.3582 ⧸ 42.02°)**V***i* = 0.1906 ⧸ 100° **V***i*
-
-showing that the output is about 19 percent of the input in magnitude but leading the input by 100°. If the circuit is terminated by a load, the load will affect the phase shift.
-
-Practice Problem 9.14 Refer to the *RL* circuit in Fig. 9.36. If 10 V is applied to the input, find the magnitude and the phase shift produced at 5 kHz. Specify whether the phase shift is leading or lagging.
-
-**Answer:** 1.7161 V, 120.39°, lagging.
-
-# **9.8.2** AC Bridges
-
-An ac bridge circuit is used in measuring the inductance *L* of an inductor or the capacitance *C* of a capacitor. It is similar in form to the Wheatstone bridge for measuring an unknown resistance (discussed in Section 4.10) and follows the same principle. To measure *L* and *C*, however, an ac
-
-source is needed as well as an ac meter instead of the galvanometer. The ac meter may be a sensitive ac ammeter or voltmeter.
-
-Consider the general ac bridge circuit displayed in Fig. 9.37. The bridge is *balanced* when no current flows through the meter. This means that **V**1 = **V**2. Applying the voltage division principle,
-
-$$
-\mathbf{V}_1 = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{V}_s = \mathbf{V}_2 = \frac{\mathbf{Z}_x}{\mathbf{Z}_3 + \mathbf{Z}_x} \mathbf{V}_s
-$$
-(9.71)
-
-Thus,
-
-$$
-\frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} = \frac{\mathbf{Z}_x}{\mathbf{Z}_3 + \mathbf{Z}_x} \qquad \Rightarrow \qquad \mathbf{Z}_2 \mathbf{Z}_3 = \mathbf{Z}_1 \mathbf{Z}_x \tag{9.72}
-$$
-
-or
-
-$$
-Z_x = \frac{Z_3}{Z_1} Z_2 \tag{9.73}
-$$
-
-This is the balanced equation for the ac bridge and is similar to Eq. (4.30) for the resistance bridge except that the *R*'s are replaced by **Z**'s.
-
-Specific ac bridges for measuring *L* and *C* are shown in Fig. 9.38, where *Lx* and *Cx* are the unknown inductance and capacitance to be measured while *Ls* and *Cs* are a standard inductance and capacitance (the values of which are kno wn to great precision). In each case, tw o resistors, *R*1 and *R*2, are varied until the ac meter reads zero. Then the bridge is balanced. From Eq. (9.73), we obtain
-
-$$
-L_x = \frac{R_2}{R_1} L_s \tag{9.74}
-$$
-
-and
-
-$$
-C_x = \frac{R_1}{R_2} C_s \tag{9.75}
-$$
-
-Notice that the balancing of the ac bridges in Fig. 9.38 does not depend on the frequency *f* of the ac source, since *f* does not appear in the relationships in Eqs. (9.74) and (9.75).
-
-Specific ac bridges: (a) for measuring *L*, (b) for measuring *C*.
-
-**Figure 9.37**
-
-A general ac bridge.
-
-Example 9.15 The ac bridge circuit of Fig. 9.37 balances when **Z**1 is a 1-kΩ resistor, **Z**2 is a 4.2-kΩ resistor, **Z**3 is a parallel combination of a 1.5-MΩ resistor and a 12-pF capacitor, and *f* = 2 kHz. Find: (a) the series com ponents that make up **Z***x*, and (b) the parallel components that make up **Z***x*.
-
-# **Solution:**
-
-- 1. **Define.** The problem is clearly stated.
-- 2. **Present.** We are to determine the unknown components subject to the fact that they balance the given quantities. Given that a parallel and series equivalent exists for this circuit, we need to find both.
-- 3. **Alternative.** Although there are alternative techniques that can be used to find the unknown values, a straightforward equality works best. Once we have answers, we can check them by using hand techniques such as nodal analysis or just using *PSpice*.
-- 4. **Attempt.** From Eq. (9.73),
-
-$$
-Z_{x} = \frac{Z_{3}}{Z_{1}} Z_{2}
-$$
- (9.15.1)
-
-where **Z***x* = *Rx* + *jXx*,
-
-$$
-\mathbf{Z}_1 = 1000 \,\Omega, \qquad \mathbf{Z}_2 = 4200 \,\Omega \tag{9.15.2}
-$$
-
-and
-
-$$
-\mathbf{Z}_3 = R_3 \parallel \frac{1}{j\omega C_3} = \frac{\frac{R_3}{j\omega C_3}}{R_3 + 1/j\omega C_3} = \frac{R_3}{1 + j\omega R_3 C_3}
-$$
-
-Since *R*3 = 1.5 MΩ and *C*3 = 12 pF,
-
-Since
-$$
-R_3 = 1.5 \text{ M}\Omega
-$$
- and $C_3 = 12 \text{ pF}$ ,
-\n
-$$
-\mathbf{Z}_3 = \frac{1.5 \times 10^6}{1 + j2\pi \times 2 \times 10^3 \times 1.5 \times 10^6 \times 12 \times 10^{-12}} = \frac{1.5 \times 10^6}{1 + j0.2262}
-$$
-
-or
-
-$$
-Z_3 = 1.427 - j0.3228 M\Omega
-$$
- (9.15.3)
-
-(a) Assuming that **Z***x* is made up of series components, we substitute Eqs. (9.15.2) and (9.15.3) in Eq. (9.15.1) and obtain
-
-$$
-R_x + jX_x = \frac{4200}{1000}(1.427 - j0.3228) \times 10^6
-$$
-
-= (5.993 - j1.356) MΩ (9.15.4)
-
-Equating the real and imaginary parts yields *Rx* = 5.993 MΩ and a capacitive reactance
-
-$$
-X_x = \frac{1}{\omega C} = 1.356 \times 10^6
-$$
-
-or
-
-$$
-C = \frac{1}{\omega X_x} = \frac{1}{2\pi \times 2 \times 10^3 \times 1.356 \times 10^6} = 58.69 \text{ pF}
-$$
-
-(b) *Zx* remains the same as in Eq. (9.15.4) but *Rx* and *Xx* are in parallel. Assuming an *RC* parallel combination,
-
-$$
-\mathbf{Z}_x = (5.993 - j1.356) \text{ M}\Omega
-$$
-$$
-= R_x \parallel \frac{1}{j\omega C_x} = \frac{R_x}{1 + j\omega R_x C_x}
-$$
-
-By equating the real and imaginary parts, we obtain
-
-By equating the real and imaginary parts, we obtain
-\n
-$$
-R_x = \frac{\text{Real}(\mathbf{Z}_x)^2 + \text{Imag}(\mathbf{Z}_x)^2}{\text{Real}(\mathbf{Z}_x)} = \frac{5.993^2 + 1.356^2}{5.993} = 6.3 \text{ M}\Omega
-$$
-\n
-$$
-C_x = -\frac{\text{Imag}(\mathbf{Z}_x)}{\omega[\text{Real}(\mathbf{Z}_x)^2 + \text{Imag}(\mathbf{Z}_x)^2]}
-$$
-\n
-$$
-= -\frac{-1.356}{2\pi(2000)(5.917^2 + 1.356^2)} = 2.852 \mu\text{F}
-$$
-
-We have assumed a parallel *RC* combination which works in this case.
-
-5. **Evaluate.** Let us now use *PSpice* to see if we indeed have the correct equalities. Running *PSpice* with the equivalent circuits, an open circuit between the "bridge" portion of the circuit, and a 10-volt input voltage yields the following voltages at the ends of the "bridge" relative to a reference at the bottom of the circuit:
-
-| FREQ | VM(\$N_0002) | VP(\$N_0002) |
-|-------------|--------------|--------------|
-| 2.000E + 03 | 9.993E + 00 | -8.634E - 03 |
-| 2.000E + 03 | 9.993E + 00 | -8.637E - 03 |
-
-Because the voltages are essentially the same, then no measurable current can flow through the "bridge" portion of the circuit for any element that connects the two points together and we have a balanced bridge, which is to be expected. This indicates we have properly determined the unknowns.
-
-There is a very important problem with what we have done! Do you know what that is? We have what can be called an ideal, "theoretical" answer, but one that really is not very good in the real world. The difference between the magnitudes of the upper impedances and the lower impedances is much too large and would never be accepted in a real bridge circuit. For greatest accuracy, the overall magnitude of the impedances must at least be within the same relative order. To increase the accuracy of the solution of this problem, I would recommend increasing the magnitude of the top impedances to be in the range of 500 kΩ to 1.5 MΩ. One additional real-world comment: The size of these impedances also creates serious problems in making actual measurements, so the appropriate instruments must be used in order to minimize their loading (which would change the actual voltage readings) on the circuit.
-
-6. **Satisfactory?** Because we solved for the unknown terms and then tested to see if they worked, we validated the results. They can now be presented as a solution to the problem.
-
-Practice Problem 9.15 In the ac bridge circuit of Fig. 9.37, suppose that balance is achie ved when **Z**1 is a 4.8-k Ω resistor , **Z**2 is a 10- Ω resistor in series with a 0.25-*μ*H inductor, **Z**3 is a 12-kΩ resistor, and *f* = 6 MHz. Determine the series components that make up **Z***x*.
-
-**Answer:** A 25-Ω resistor in series with a 0.625-*μ*H inductor.
-
-# **9.9** Summary
-
-1. A sinusoid is a signal in the form of the sine or cosine function. It has the general form
-
-$$
-v(t) = V_m \cos(\omega t + \phi)
-$$
-
-where *Vm* is the amplitude, *ω* = 2*πf* is the angular frequency, (*ωt* + *ϕ*) is the argument, and *ϕ* is the phase.
-
-2. A phasor is a complex quantity that represents both the magnitude and the phase of a sinusoid. Given the sinusoid *v*(*t*) = *Vm* cos(*ωt* + *ϕ*), its phasor **V** is
-
-$$
-\mathbf{V}=V_m\big/\phi
-$$
-
-- 3. In ac circuits, v oltage and current phasors al ways ha ve a fixed relation to one another at an y moment of time. If *v*(*t*) = *Vm* cos(*ωt* + *ϕv*) represents the v oltage through an element and *i*(*t*) = *Im* cos(*ωt* + *ϕi*) represents the current through the element, then *ϕi* = *ϕv* if the element is a resistor , *ϕi* leads *ϕv* by 90° if the element is a capacitor , and *ϕi* lags *ϕv* by 90° if the element is an inductor.
-- 4. The impedance **Z** of a circuit is the ratio of the phasor voltage across it to the phasor current through it:
-
-$$
-\mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = R(\omega) + jX(\omega)
-$$
-
-The admittance **Y** is the reciprocal of impedance:
-
-$$
-\mathbf{Y} = \frac{1}{\mathbf{Z}} = G(\omega) + jB(\omega)
-$$
-
-Impedances are combined in series or in parallel the same w ay as resistances in series or parallel; that is, impedances in series add while admittances in parallel add.
-
-- 5. For a resistor **Z** = *R*, for an inductor **Z** = *jX* = *jωL*, and for a capacitor **Z** = −*jX* = 1∕*jωC*.
-- 6. Basic circuit la ws (Ohm's and Kirchhof f's) apply to ac circuits in the same manner as they do for dc circuits; that is,
-
-$$
-\mathbf{V} = \mathbf{Z}\mathbf{I}
-$$
-$$
-\Sigma \mathbf{I}_k = 0 \quad \text{(KCL)}
-$$
-$$
-\Sigma \mathbf{V}_k = 0 \quad \text{(KVL)}
-$$
-
-- 7. The techniques of voltage/current division, series/parallel combination of impedance/admittance, circuit reduction, and *Y*-∆ transformation all apply to ac circuit analysis.
-- 8. AC circuits are applied in phase-shifters and bridges.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/111_9.9 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/111_9.9 Summary.md
deleted file mode 100644
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+++ /dev/null
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-# Review Questions
-
-**9.1** Which of the following is *not* a right way to express the sinusoid *A* cos *ωt*?
-
-> (a) *A* cos 2*π ft* (b) *A* cos(2*πt*∕*T*) (c) *A* cos *ω*(*t* − *T*) (d) *A* sin(*ωt* − 90°)
-
-**9.2** A function that repeats itself after fixed intervals is said to be:
-
-| (a) a phasor | (b) harmonic |
-|--------------|--------------|
-| (c) periodic | (d) reactive |
-
-**9.3** Which of these frequencies has the shorter period?
-
-(a) 1 krad/s (b) 1 kHz
-
-- **9.4** If *v*1 = 30 sin(*ωt* + 10°) and *v*2 = 20 sin(*ωt* + 50°), which of these statements are true?
- - (a) *v*1 leads *v*2 (b) *v*2 leads *v*1 (c) *v*2 lags *v*1 (d) *v*1 lags *v*2
- - (e) *v*1 and *v*2 are in phase
-- **9.5** The voltage across an inductor leads the current through it by 90°.
-
-(a) True (b) False
-
-- **9.6** The imaginary part of impedance is called:
- - (a) resistance (b) admittance (c) susceptance (d) conductance
- - (e) reactance
-- **9.7** The impedance of a capacitor increases with increasing frequency.
-
-(a) True (b) False
-
-**9.8** At what frequency will the output voltage *vo*(*t*) in Fig. 9.39 be equal to the input voltage *v*(*t*) ?
-
-| (a) 0 rad/s | (b) 1 rad/s | (c) 4 rad/s |
-|-------------|-----------------------|-------------|
-| (d) ∞ rad/s | (e) none of the above | |
-
-# **Figure 9.39**
-
-For Review Question 9.8.
-
-**9.9** A series *RC* circuit has ∣*VR*∣ = 12 V and ∣*VC*∣ = 5 V. The magnitude of the supply voltage is:
-
-(a) −7 V (b) 7 V (c) 13 V (d) 17 V
-
-**9.10** A series *RCL* circuit has *R* = 30 Ω, *XC* = 50 Ω, and *XL* = 90 Ω. The impedance of the circuit is:
-
-> (a) 30 + *j*140 Ω (b) 30 + *j*40 Ω (c) 30 − *j*40 Ω (d) −30 − *j*40 Ω (e) −30 + *j*40 Ω
-
-*Answers: 9.1d, 9.2c, 9.3b, 9.4b,d, 9.5a, 9.6e, 9.7b, 9.8d, 9.9c, 9.10b.*
-
-# Problems
-
-# Section 9.2 Sinusoids
-
-- **9.1** Given the sinusoidal voltage *v*(*t*) = 50 cos (30*t* + 10°) V, find: (a) the amplitude *Vm*, (b) the period *T*, (c) the frequency *f*, and (d) *v*(*t*) at *t* = 10 ms.
-- **9.2** A current source in a linear circuit has
-
-*is* = 15 cos (25 *π t* + 25°) A
-
-- (a) What is the amplitude of the current?
-- (b) What is the angular frequency?
-- (c) Find the frequency of the current.
-- (d) Calculate *is* at *t* = 2 ms.
-- **9.3** Express the following functions in cosine form:
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/112_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/112_Review Questions.md
deleted file mode 100644
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-(a) 10 sin(*ωt* + 30°) (b) −9 sin (8*t*) (c) −20 sin(*ωt* + 45°)
-
-- **9.4** Design a problem to help other students better understand sinusoids.
-- **9.5** Given *v*1 = 45 sin(*ωt* + 30°) V and *v*2 = 50 cos(*ωt* − 30°) V, determine the phase angle between the two sinusoids and which one lags the other.
-- **9.6** For the following pairs of sinusoids, determine which one leads and by how much.
-
-(a)
-$$
-v(t) = 10 \cos(4t - 60^{\circ})
-$$
- and
- $i(t) = 4 \sin(4t + 50^{\circ})$
-
-- (b) *v*1(*t*) = 4 cos(377*t* + 10°) and *v*2(*t*) = −20 cos 377*t*
-- (c) *x*(*t*) = 13 cos 2*t* + 5 sin 2*t* and *y*(*t*) = 15 cos(2*t* − 11.8°)
-
-# Section 9.3 Phasors
-
-- **9.7** If *f*(*ϕ*) = cos *ϕ* + *j* sin *ϕ*, show that *f*(*ϕ*) = *e jϕ* .
-- **9.8** Calculate these complex numbers and express your results in rectangular form:
-
-(a)
-$$
-\frac{60/45^{\circ}}{7.5 - j10} + j2
-$$
-
-\n(b)
-$$
-\frac{32/20^{\circ}}{(6 - j8)(4 + j2)} + \frac{20}{-10 + j24}
-$$
-
-\n(c)
-$$
-20 + (16/-50^{\circ})(5 + j12)
-$$
-
-**9.9** Evaluate the following complex numbers and leave your results in polar form:
-
-(a)
-$$
-5/30^{\circ}
-$$
- $\left(6 - j8 + \frac{3/60^{\circ}}{2 + j}\right)$
-(b) $\frac{(10/60^{\circ})(35/ -50^{\circ})}{(2 + j6) - (5 + j)}$
-
-**9.10** Design a problem to help other students better understand phasors.
-
-**9.11** Find the phasors corresponding to the following signals:
-
-(a)
-$$
-v(t) = 21 \cos(4t - 15^\circ) \text{V}
-$$
-
-(b)
-$$
-i(t) = -8 \sin(10t + 70^{\circ})
-$$
- mA
-
-(c)
-$$
-v(t) = 120 \sin(10t - 50^{\circ})
-$$
- V
-
-(d)
-$$
-i(t) = -60 \cos(30t + 10^{\circ})
-$$
- mA
-
-**9.12** Let **X** = 4⧸ 40° and **Y** = 20⧸−30°. Evaluate the following quantities and express your results in polar form:
-
-$$
-\text{(a)}\,(X+Y)X^*
-$$
-
-$$
-(b) (X - Y)^*
-$$
-
-(c) (**X** + **Y**)∕**X**
-
-**9.13** Evaluate the following complex numbers:
-
-(a)
-$$
-\frac{2+j3}{1-j6} + \frac{7-j8}{-5+j11}
-$$
-
-\n(b)
-$$
-\frac{(5/10°)(10/-40°)}{(4/-80°)(-6/50°)}
-$$
-
-\n(c)
-$$
-\begin{vmatrix} 2+j3 & -j2 \\ -j2 & 8-j5 \end{vmatrix}
-$$
-
-**9.14** Simplify the following expressions:
-
-Simplify the following expressions:
-\n(a)
-$$
-\frac{(5 - j6) - (2 + j8)}{(-3 + j4)(5 - j) + (4 - j6)}
-$$
-\n(b)
-$$
-\frac{(240/75^\circ + 160/–30^\circ)(60 - j80)}{(67 + j84)(20/32^\circ)}
-$$
-\n(c)
-$$
-\left(\frac{10 + j20}{3 + j4}\right)^2 \sqrt{(10 + j5)(16 - j20)}
-$$
-
-**9.15** Evaluate these determinants:
-
-(a)
-$$
-\begin{vmatrix} 10 + j6 & 2 - j3 \ -5 & -1 + j \end{vmatrix}
-$$
-
-\n(b)
-$$
-\begin{vmatrix} 20 \underline{/-30^{\circ}} & -4 \underline{/-10^{\circ}} \\ 16 \underline{/-0^{\circ}} & 3 \underline{/-40^{\circ}} \end{vmatrix}
-$$
-
-\n(c)
-$$
-\begin{vmatrix} 1 - j & -j & 0 \\ j & 1 & -j \\ 1 & j & 1 + j \end{vmatrix}
-$$
-
-**9.16** Transform the following sinusoids to phasors:
-
-(a) −20 cos(4*t* + 135°) (b) 8 sin(20*t* + 30°) (c) 20 cos (2*t*) + 15 sin (2*t*)
-
-- **9.17** Two voltages *v*1 and *v*2 appear in series so that their sum is *v* = *v*1 + *v*2. If *v*1 = 10 cos(50*t* − *π*∕3) V and *v*2 = 12 cos(50*t* + 30°) V, find *v*.
-- **9.18** Obtain the sinusoids corresponding to each of the following phasors:
-
-(a)
-$$
-V_1 = 60/15^{\circ}
-$$
- V, $\omega = 1$
-\n(b) $V_2 = 6 + j8$ V, $\omega = 40$
-\n(c) $I_1 = 2.8e^{-j\pi/3}$ A, $\omega = 377$
-\n(d) $I_2 = -0.5 - j1.2$ A, $\omega = 10^3$
-
-**9.19** Using phasors, find:
-
-(a) 3 cos(20*t* + 10°) − 5 cos(20*t* − 30°)
-
-- (b) 40 sin 50*t* + 30 cos(50*t* − 45°)
-- (c) 20 sin 400*t* + 10 cos(400*t* + 60°)
-
-$$
--5\sin(400t-20^\circ)
-$$
-
-**9.20** A linear network has a current input 7.5 cos(10*t* + 30°) A and a voltage output 120 cos(10*t* + 75°) V. Determine the associated impedance.
-
-**9.21** Simplify the following:
-
-(a)
-$$
-f(t) = 5 \cos(2t + 15^\circ) - 4 \sin(2t - 30^\circ)
-$$
-
-(b) $g(t) = 8 \sin t + 4 \cos(t + 50^\circ)$
-
-- (c) *h*(*t*) = ∫ 0 (10 cos 40*t* + 50 sin 40*t*) *dt*
-- **9.22** An alternating voltage is given by *v*(*t*) = 55 cos(5*t* + 45°) V. Use phasors to find
-
-$$
-10v(t) + 4\frac{dv}{dt} - 2\int_{-\infty}^{t} v(t) dt
-$$
-
-Assume that the v alue of the inte gral is zero at *t* = −∞.
-
-**9.23** Apply phasor analysis to evaluate the following:
-
-(a)
-$$
-v = [110 \sin(20t + 30^\circ) + 220 \cos(20t - 90^\circ)]
-$$
- V
-(b) $i = [30 \cos(5t + 60^\circ) - 20 \sin(5t + 60^\circ)]$ A
-
-**9.24** Find *v*(*t*) in the following integrodifferential equations using the phasor approach:
-
-(a)
-$$
-v(t) + \int v dt = 10 \cos t
-$$
-
-\n(b) $\frac{dv}{dt} + 5v(t) + 4 \int v dt = 20 \sin(4t + 10^{\circ})$
-
-**9.25** Using phasors, determine *i*(*t*) in the following equations:
-
-(a)
-$$
-2\frac{di}{dt} + 3i(t) = 4\cos(2t - 45^{\circ})
-$$
-
-\n(b) $10 \int i \, dt + \frac{di}{dt} + 6i(t) = 5\cos(5t + 22^{\circ})$ A
-
-**9.26** The loop equation for a series *RLC* circuit gives
-
-$$
-\frac{di}{dt} + 2i + \int_{-\infty}^{t} i \, dt = \cos 2t \, A
-$$
-
-Assuming that the v alue of the inte gral at *t* = −∞ is zero, find *i*(*t*) using the phasor method.
-
-**9.27** A parallel *RLC* circuit has the node equation
-
-$$
-\frac{dv}{dt} + 50v + 100 \int v \, dt = 110 \cos(377t - 10^{\circ}) \, \text{V}
-$$
-
-Determine *v*(*t*) using the phasor method. You may assume that the value of the integral at *t* = −∞ is zero.
-
-# Section 9.4 Phasor Relationships for Circuit Elements
-
-- **9.28** Determine the current that flows through an 20-Ω resistor connected to a voltage source *v*s = 120 cos (377*t* + 37°) V.
-- **9.29** Given that *vc*(0) = 2 cos(155°) V, what is the instantaneous voltage across a 2-*μ*F capacitor when the current through it is *i* = 4 sin(106 *t* + 25°) A?
-
-- **9.30** A voltage *v*(*t*) = 100 cos(60*t* + 20°) V is applied to a parallel combination of a 40-kΩ resistor and a 50-*μ*F capacitor. Find the steady-state currents through the resistor and the capacitor.
-- **9.31** A series *RLC* circuit has *R* = 80 Ω, *L* = 240 mH, and *C* = 5 mF. If the input voltage is *v*(*t*) = 115 cos 2*t*, find the current flowing through the circuit.
-- **9.32** Using Fig. 9.40, design a problem to help other students better understand phasor relationships for circuit elements.
-
-- **9.33** A series *RL* circuit is connected to a 220-V ac source. If the voltage across the resistor is 170 V, find the voltage across the inductor.
-- **9.34** What value of *ω* will cause the forced response, *vo*, in Fig. 9.41 to be zero?
-
-**Figure 9.41** For Prob. 9.34.
-
-# Section 9.5 Impedance and Admittance
-
-**9.35** Find the steady-state current *i* in the circuit of Fig. 9.42, when *vs*(*t*) = 115 cos 200*t* V.
-
-**Figure 9.42** For Prob. 9.35.
-
-# **9.36** Using Fig. 9.43, design a problem to help other students better understand impedance.
-
-**9.37** Determine the admittance **Y** for the circuit in Fig. 9.44.
-
-# **Figure 9.45**
-
-For Prob. 9.38.
-
-**9.39** For the circuit shown in Fig. 9.46, find *Z*eq and use that to find current **I**. Let *ω* = 10 rad/s.
-
-**Figure 9.46** For Prob. 9.39.
-
-**9.40** In the circuit of Fig. 9.47, find *io* when:
-
-(a)
-$$
-\omega = 1
-$$
- rad/s (b) $\omega = 5$ rad/s
-(c) $\omega = 10$ rad/s
-
-# **Figure 9.47** For Prob. 9.40.
-
-**9.41** Find *v*(*t*) in the *RLC* circuit of Fig. 9.48.
-
-# **Figure 9.48**
-
-For Prob. 9.41.
-
-**9.42** Calculate *vo*(*t*) in the circuit of Fig. 9.49.
-
-# **Figure 9.49**
-
-For Prob. 9.42.
-
-**9.43** Find current **I***o* in the circuit shown in Fig. 9.50.
-
-**Figure 9.50** For Prob. 9.43.
-
-**9.44** Calculate *i*(*t*) in the circuit of Fig. 9.51.
-
-**Figure 9.51** For prob. 9.44.
-
-**Figure 9.52** For Prob. 9.45.
-
-200 mH 100 mF 2 Ω 2Ω i o *v*s + ‒
-
-# **Figure 9.53**
-
-For Prob. 9.46.
-
-**9.47** In the circuit of Fig. 9.54, determine the value of *is*(*t*).
-
-# **Figure 9.54**
-
-For Prob. 9.47.
-
-**9.48** Given that *vs*(*t*) = 20 sin(100*t* − 40°) in Fig. 9.55, determine *ix*(*t*).
-
-# **Figure 9.55**
-
-For Prob. 9.48.
-
-**9.49** Find *v s* (*t*) in the circuit of Fig. 9.56 if the current *ix* through the 1-Ω resistor is 8 sin 200*t* A.
-
-**Figure 9.56** For Prob. 9.49.
-
-**9.50** Determine *vx* in the circuit of Fig. 9.57. Let *is*(*t*) = 5 cos(100*t* + 40°) A.
-
-# **Figure 9.57**
-
-For Prob. 9.50.
-
-**9.51** If the voltage *vo* across the 2-Ω resistor in the circuit of Fig. 9.58 is 90 cos 2*t* V, obtain *is*.
-
-# **Figure 9.58**
-
-For Prob. 9.51.
-
-**9.52** If **V***o* = 8⧸ 30° V in the circuit of Fig. 9.59, find **I***s*.
-
-# **Figure 9.59**
-
-For Prob. 9.52.
-
-**Figure 9.60**
-
-For Prob. 9.53.
-
-**9.54** In the circuit of Fig. 9.61, find **V***s* if **I***o* = 30⧸ 0° A.
-
-**Figure 9.61** For Prob. 9.54.
-
-# **Figure 9.62**
-
-For Prob. 9.55.
-
-# Section 9.7 Impedance Combinations
-
-**9.56** At *ω* = 377 rad/s, find the input impedance of the circuit shown in Fig. 9.63.
-
-**Figure 9.63** For Prob. 9.56.
-
-**9.57** At *ω* = 1 rad/s, obtain the input admittance in the circuit of Fig. 9.64.
-
-# **Figure 9.64**
-
-For Prob. 9.57.
-
-**9.58** Using Fig. 9.65, design a problem to help other students better understand impedance combinations.
-
-\* An asterisk indicates a challenging problem.
-
-**\* 9.59** For the network in Fig. 9.66, find **Z**in. Let *ω* = 100 rad/s.
-
-**Figure 9.66**
-
-For Prob. 9.59.
-
-**9.60** Obtain **Z**in for the circuit in Fig. 9.67.
-
-**Figure 9.67**
-
-For Prob. 9.60.
-
-**9.61** Find **Z**eq in the circuit of Fig. 9.68.
-
-# **Figure 9.68**
-
-For Prob. 9.61.
-
-**9.62** For the circuit in Fig. 9.69, find the input impedance **Z**in at 10 krad/s.
-
-**Figure 9.69** For Prob. 9.62.
-
-**9.63** For the circuit in Fig. 9.70, find the value of **Z***T*.
-
-**Figure 9.70** For Prob. 9.63.
-
-**9.64** Find **Z***T* and Vo in the circuit in Fig. 9.71. Let the value of the inductance equal *j*20 Ω.
-
-**9.65** Determine **Z***T* and **I** for the circuit in Fig. 9.72.
-
-# **Figure 9.72**
-
-For Prob. 9.65.
-
-**9.66** For the circuit in Fig. 9.73, calculate **Z***T* and **V***ab*.
-
-**Figure 9.73** For Prob. 9.66.
-
-**9.67** At *ω* = 103 rad/s, find the input admittance of each of the circuits in Fig. 9.74.
-
-# **Figure 9.74**
-
-For Prob. 9.67.
-
-**9.68** Determine **Y**eq for the circuit in Fig. 9.75.
-
-# **Figure 9.75**
-
-For Prob. 9.68.
-
-**9.69** Find the equivalent admittance **Y**eq of the circuit in Fig. 9.76.
-
-# **Figure 9.76**
-
-For Prob. 9.69.
-
-**9.70** Find the equivalent impedance of the circuit in Fig. 9.77.
-
-**Figure 9.77** For Prob. 9.70.
-
-**9.71** Obtain the equivalent impedance of the circuit in Fig. 9.78.
-
-# **Figure 9.78**
-
-For Prob. 9.71.
-
-**Figure 9.79**
-
-For Prob. 9.72.
-
-**9.73** Determine the equivalent impedance of the circuit in Fig. 9.80.
-
-# Section 9.8 Applications
-
-- **9.74** Design an *RL* circuit to provide a 90° leading phase shift.
-- **9.75** Design a circuit that will transform a sinusoidal
-- voltage input to a cosinusoidal voltage output.
-- **9.76** For the following pairs of signals, determine if *v*1 leads or lags *v*2 and by how much.
-
-(a) *v*1 = 10 cos(5*t* − 20°), *v*2 = 8 sin 5*t*
-
-(b)
-$$
-v_1 = 19 \cos(2t + 90^\circ)
-$$
-, $v_2 = 6 \sin 2t$
-
-(c)
-$$
-v_1 = -4 \cos 10t
-$$
-, $v_2 = 15 \sin 10t$
-
-- **9.77** Refer to the *RC* circuit in Fig. 9.81.
- - (a) Calculate the phase shift at 2 MHz.
- - (b) Find the frequency where the phase shift is 45°.
-
-# **Figure 9.81**
-
-For Prob. 9.77.
-
-- **9.78** A coil with impedance 8 + *j*6 Ω is connected in series with a capacitive reactance *X*. The series combination is connected in parallel with a resistor *R*. Given that the equivalent impedance of the resulting circuit is 5⧸ 0° Ω, find the value of *R* and *X*.
-- **9.79** (a) Calculate the phase shift of the circuit in Fig. 9.82. (b) State whether the phase shift is leading or lagging (output with respect to input).
- - (c) Determine the magnitude of the output when the input is 120 V.
-
-# **Figure 9.82**
-
-For Prob. 9.79.
-
-- **9.80** Consider the phase-shifting circuit in Fig. 9.83. Let **V***i* = 120 V operating at 60 Hz. Find:
- - (a) **V***o* when *R* is maximum
- - (b) **V***o* when *R* is minimum
- - (c) the value of *R* that will produce a phase shift of 45°
-
-# **Figure 9.83**
-
-For Prob. 9.80.
-
-- **9.81** The ac bridge in Fig. 9.37 is balanced when *R*1 = 400 Ω, *R*2 = 600 Ω, *R*3 = 1.2 kΩ, and *C*2 = 0.3 *μ*F. Find *Rx* and *Cx*. Assume *R*2 and *C*2 are in series.
-- **9.82** A capacitance bridge balances when *R*1 = 100 Ω, *R*2 = 2 kΩ, and *Cs* = 40 *μ*F. What is *Cx*, the capacitance of the capacitor under test?
-- **9.83** An inductive bridge balances when *R*1 = 1.2 kΩ, *R*2 = 500 Ω, and *Ls* = 250 mH. What is the value of *Lx*, the inductance of the inductor under test?
-
-**9.84** The ac bridge shown in Fig. 9.84 is known as a *Maxwell bridge* and is used for accurate measurement of inductance and resistance of a coil in terms of a standard capacitance *Cs*. Show that when the bridge is balanced,
-
-$$
-L_x = R_2 R_3 C_s \qquad \text{and} \qquad R_x = \frac{R_2}{R_1} R_3
-$$
-
-Find *Lx* and *Rx* for *R*1 = 40 k Ω, *R*2 = 1.6 k Ω, *R*3 = 4 kΩ, and *Cs* = 0.45 *μ*F.
-
-**Figure 9.84** Maxwell bridge; For Prob. 9.84.
-
-# *f* = \_\_\_\_\_\_\_\_\_\_\_\_ 1 2*π*√
-
-**9.86** The circuit shown in Fig. 9.86 is used in a television receiver. What is the total impedance of this circuit?
-
-Comprehensive Problems
-
-For Prob. 9.86.
-
-**9.87** The network in Fig. 9.87 is part of the schematic describing an industrial electronic sensing device. What is the total impedance of the circuit at 4 kHz?
-
-**Figure 9.87** For Prob. 9.87.
-
-- (a) What is the impedance of the circuit?
-- (b) If the frequency were halved, what would be the impedance of the circuit?
-
-# **Figure 9.88**
-
-For Prob. 9.88.
-
-**9.89** An industrial load is modeled as a series combination of an inductor and a resistance as shown in Fig. 9.89. Calculate the value of a capacitor *C* across the series combination so that the net impedance is resistive at a frequency of 2 kHz.
-
-For Prob. 9.89.
-
-**9.90** An industrial coil is modeled as a series combination of an inductance *L* and resistance *R*, as shown in Fig. 9.90. Since an ac voltmeter measures only the magnitude of a sinusoid, the following
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/113_Comprehensive Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/113_Comprehensive Problems.md
deleted file mode 100644
index 82324d2941c801a99d33dc9c08993bd123766492..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/113_Comprehensive Problems.md
+++ /dev/null
@@ -1,34 +0,0 @@
-**9.85** The ac bridge circuit of Fig. 9.85 is called a *Wien bridge*. It is used for measuring the frequency of a source. Show that when the bridge is balanced,
-
-> *\_\_\_\_\_\_\_\_ R*2*R*4*C*2*C*4
-
-measurements are taken at 60 Hz when the circuit operates in the steady state:
-
-$$
-|\mathbf{V}_s| = 145 \text{ V}, \qquad |\mathbf{V}_1| = 50 \text{ V}, \qquad |\mathbf{V}_o| = 110 \text{ V}
-$$
-
-Use these measurements to determine the values of *L* and *R*.
-
-# **Figure 9.90**
-
-For Prob. 9.90.
-
-**9.91** Figure 9.91 shows a series combination of an inductance and a resistance. If it is desired to connect a capacitor in parallel with the series combination such that the net impedance is resistive at 10 kHz, what is the required value of *C*?
-
-**Figure 9.91** For Prob. 9.91.
-
-- **9.92** A transmission line has a series impedance of **Z** = 100⧸ 75° Ω and a shunt admittance of **Y** = 450⧸ 48° *μ*S. Find: (a) the characteristic impedance **Z***o* = √ \_\_\_\_\_ **Z**∕**Y** , (b) the propagation constant *γ* = √ \_\_\_ **ZY** .
-- **9.93** A power transmission system is modeled as shown in Fig. 9.92. Given the source voltage and circuit elements
-
-| Vs = 115⧸ 0° V, | source impedance |
-|------------------------|---------------------------|
-| Zs = (1 + j0.5) Ω, | line impedance |
-| Zt = (0.4 + j0.3) Ω, | and load impedance |
-| ZL = (23.2 + j18.9) Ω, | find the load current IL. |
-
-**Figure 9.92** For Prob. 9.93.
-
-# **chapter**
-
-# 10
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/114_Chapter 10 - Sinusoidal Steady-State Analysis.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/114_Chapter 10 - Sinusoidal Steady-State Analysis.md
deleted file mode 100644
index dd9d1a5b443a544fc2dc8a4d44e446aac797c038..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/114_Chapter 10 - Sinusoidal Steady-State Analysis.md
+++ /dev/null
@@ -1,68 +0,0 @@
-# Sinusoidal Steady-State Analysis
-
-*Three men are my friends—he that loves me, he that hates me, he that is indifferent to me. Who loves me, teaches me tenderness; who hates me, teaches me caution; who is indifferent to me, teaches me self-reliance.* —J. E. Dinger
-
-# Enhancing Your Career
-
-# **Career in Software Engineering**
-
-Software engineering is that aspect of engineering that deals with the practical application of scientific knowledge in the design, construction, and v alidation of computer programs and the associated documenta ‑ tion required to de velop, operate, and maintain them. It is a branch of electrical engineering that is becoming increasingly important as more and more disciplines require one form of software package or another to perform routine tasks and as programmable microelectronic systems are used in more and more applications.
-
-The role of a softw are engineer should not be confused with that of a computer scientist; the softw are engineer is a practitioner , not a theoretician. A softw are engineer should ha ve good computer ‑ pro gramming skills and be familiar with programming languages, in par‑ ticular C++, which is becoming increasingly popular. Because hardware and software are closely interlinked, it is essential that a softw are engi‑ neer have a thorough understanding of hardware design. Most important, the software engineer should ha ve some specialized kno wledge of the area in which the software development skill is to be applied.
-
-All in all, the field of software engineering offers a great career to those who enjo y programming and de veloping softw are packages. The higher rewards will go to those having the best preparation, with the most interesting and challenging opportunities going to those with graduate education.
-
-A three‑dimensional printing of the output of an AutoCAD model of a NASA flywheel. © Ansoft Corporation
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Analyze electrical circuits in the frequency domain using nodal analysis.
-- 2. Analyze electrical circuits in the frequency domain using mesh analysis.
-- 3. Apply the superposition principle to frequency domain electrical circuits.
-- 4. Apply source transformation in frequency domain circuits.
-- 5. Understand how Thevenin and Norton equivalent circuits can be used in the frequency domain.
-- 6. Analyze electrical circuits with op amps.
-
-# **10.1** Introduction
-
-In Chapter 9, we learned that the forced or steady ‑state response of cir‑ cuits to sinusoidal inputs can be obtained by using phasors. We also know that Ohm's and Kirchhoff's laws are applicable to ac circuits. In this chapter, we want to see ho w nodal analysis, mesh analysis, Thevenin's theorem, Norton's theorem, superposition, and source transformations are applied in analyzing ac circuits. Since these techniques were already introduced for dc circuits, our major effort here will be to illustrate with examples.
-
-Analyzing ac circuits usually requires three steps.
-
-# Steps to Analyze AC Circuits:
-
-- 1. Transform the circuit to the phasor or frequency domain.
-- 2. Solve the problem using circuit techniques (nodal analysis, mesh analysis, superposition, etc.).
-- 3. Transform the resulting phasor to the time domain.
-
-Step 1 is not necessary if the problem is specified in the frequency domain. In step 2, the analysis is performed in the same manner as dc circuit analysis except that complex numbers are involved. Having read Chapter 9, we are adept at handling step 3.
-
-Toward the end of the chapter , we learn ho w to apply *PSpice* in solving ac circuit problems. We finally apply ac circuit analysis to two practical ac circuits: oscillators and ac transistor circuits.
-
-# **10.2** Nodal Analysis
-
-The basis of nodal analysis is Kirchhof f's current la w. Since KCL is valid for phasors, as demonstrated in Section 9.6, we can analyze ac cir‑ cuits by nodal analysis. The following examples illustrate this.
-
-Frequency domain analysis of an ac circuit via phasors is much easier than analysis of the circuit in the time domain.
-
-Find *ix* in the circuit of Fig. 10.1 using nodal analysis. Example 10.1
-
-For Example 10.1.
-
-# **Solution:**
-
-We first convert the circuit to the frequency domain:
-
-$$
-20 \cos 4t \Rightarrow 20 \underline{/0^{\circ}}, \qquad \Omega = 4 \text{ rad/s}
-$$
-
-\n
-$$
-1 \text{ H} \Rightarrow j \Omega L = j4
-$$
-
-\n
-$$
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/115_10.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/115_10.1 Introduction.md
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+++ /dev/null
@@ -1,138 +0,0 @@
-0.5 \text{ H} \Rightarrow j \Omega L = j2
-$$
-
-\n
-$$
-0.1 \text{ F} \Rightarrow \frac{1}{j \Omega C} = -j2.5
-$$
-
-Thus, the frequency domain equivalent circuit is as shown in Fig. 10.2.
-
-# **Figure 10.2**
-
-Frequency domain equivalent of the circuit in Fig. 10.1.
-
-Applying KCL at node 1,
-
-$$
-\frac{20 - V_1}{10} = \frac{V_1}{-j2.5} + \frac{V_1 - V_2}{j4}
-$$
-
-or
-
-(1 + *j*1.5)**V**1 + *j*2.5**V**2 = 20 **(10.1.1)**
-
-At node 2,
-
-$$
-2\mathbf{I}_x + \frac{\mathbf{V}_1 - \mathbf{V}_2}{j4} = \frac{\mathbf{V}_2}{j2}
-$$
-
-But **I***x* = **V**1∕−*j*2.5. Substituting this gives
-
-$$
-\frac{2V_1}{-j2.5} + \frac{V_1 - V_2}{j4} = \frac{V_2}{j2}
-$$
-
-By simplifying, we get
-
-$$
-11V_1 + 15V_2 = 0 \tag{10.1.2}
-$$
-
-Equations (10.1.1) and (10.1.2) can be put in matrix form as
-
-$$
-\begin{bmatrix} 1+j1.5 & j2.5 \\ 11 & 15 \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} 20 \\ 0 \end{bmatrix}
-$$
-
-We obtain the determinants as
-
-$$
-\Delta = \begin{vmatrix} 1+j1.5 & j2.5 \\ 11 & 15 \end{vmatrix} = 15 - j5
-$$
-
-$$
-\Delta_1 = \begin{vmatrix} 20 & j2.5 \\ 0 & 15 \end{vmatrix} = 300, \qquad \Delta_2 = \begin{vmatrix} 1+j1.5 & 20 \\ 11 & 0 \end{vmatrix} = -220
-$$
-$$
-\mathbf{V}_1 = \frac{\Delta_1}{\Delta} = \frac{300}{15 - j5} = 18.97 \underline{/18.43^\circ} \text{ V}
-$$
-$$
-\mathbf{V}_2 = \frac{\Delta_2}{\Delta} = \frac{-220}{15 - j5} = 13.91 \underline{/198.3^\circ} \text{ V}
-$$
-
-The current **I***x* is given by
-
-$$
-\mathbf{I}_x \text{ is given by}
-$$
-\n
-$$
-\mathbf{I}_x = \frac{\mathbf{V}_1}{-j2.5} = \frac{18.97 \; / 18.43^\circ}{2.5 \; / -90^\circ} = 7.59 \; / 108.4^\circ \; \text{A}
-$$
-
-Transforming this to the time domain,
-
-$$
-i_x = 7.59 \cos(4t + 108.4^\circ)
-$$
- A
-
-Using nodal analysis, find *v*1 and *v*2 Practice Problem 10.1 in the circuit of Fig. 10.3.
-
-**Figure 10.3** For Practice Prob. 10.1.
-
-**Answer:** *v*1(*t*) = 28.31 cos(2*t* + 60.01°) V, *v*2(*t*) = 82.56 cos(2*t* + 57.12°) V.
-
-Compute **V**1 and **V**2 Example 10.2 in the circuit of Fig. 10.4.
-
-For Example 10.2.
-
-# **Solution:**
-
-Nodes 1 and 2 form a supernode as shown in Fig. 10.5. Applying KCL at the supernode gives
-
-$$
-3 = \frac{\mathbf{V}_1}{-j3} + \frac{\mathbf{V}_2}{j6} + \frac{\mathbf{V}_2}{12}
-$$
-
-or
-
-$$
-36 = j4V_1 + (1 - j2)V_2 \tag{10.2.1}
-$$
-
-
-
-# **Figure 10.5**
-
-A supernode in the circuit of Fig. 10.4.
-
-But a voltage source is connected between nodes 1 and 2, so that
-
-$$
-V_1 = V_2 + 10 \frac{\angle 45^{\circ}}{\angle 0.2.2}
-$$
-
-Substituting Eq. (10.2.2) in Eq. (10.2.1) results in
-
-$$
-36 - 40 \underline{1135^\circ} = (1 + i2) \mathbf{V}_2 \implies \mathbf{V}_2 = 31.41 \underline{18^\circ} \text{ V}
-$$
-
-From Eq. (10.2.2),
-
-$$
-V_1 = V_2 + 10 \underline{745^\circ} = 25.78 \underline{770.48^\circ} \text{ V}
-$$
-
-Calculate
-$$
-V_1
-$$
- and $V_2$ in the circuit shown in Fig. 10.6. **Practice Problem 10.2**
-
-For Practice Prob. 10.2.
-
-**Answer: V**1 = 96.8 ∕69.66° V, **V**2 = 16.88∕165.72° V.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/116_10.3 Mesh Analysis.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/116_10.3 Mesh Analysis.md
deleted file mode 100644
index 5ac19638c3efa81be32ec5290b2726b4d35197e1..0000000000000000000000000000000000000000
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+++ /dev/null
@@ -1,238 +0,0 @@
-# **10.3** Mesh Analysis
-
-Kirchhoff's voltage law (KVL) forms the basis of mesh analysis. The validity of KVL for ac circuits was shown in Section 9.6 and is illustrated in the follo wing examples. Keep in mind that the v ery nature of using mesh analysis is that it is to be applied to planar circuits.
-
-Determine current **I***o* in the circuit of Fig. 10.7 using mesh analysis. Example 10.3
-
-# **Solution:**
-
-Applying KVL to mesh 1, we obtain
-
-$$
-(8+j10-j2)\mathbf{I}_1 - (-j2)\mathbf{I}_2 - j10\mathbf{I}_3 = 0 \tag{10.3.1}
-$$
-
-**Figure 10.7** For Example 10.3.
-
-For mesh 2,
-
-$$
-(4 - j2 - j2)I2 - (-j2)I1 - (-j2)I3 + 20 / 90o = 0
-$$
- (10.3.2)
-
-For mesh 3, **I**3 = 5. Substituting this in Eqs. (10.3.1) and (10.3.2), we get
-
-$$
-(8+j8)\mathbf{I}_1 + j2\mathbf{I}_2 = j50 \tag{10.3.3}
-$$
-
-$$
-j2\mathbf{I}_1 + (4 - j4)\mathbf{I}_2 = -j20 - j10
-$$
- (10.3.4)
-
-Equations (10.3.3) and (10.3.4) can be put in matrix form as
-
-[ 8 + *j*8 *j*2 *j*2 4 − *j*4 ] [ **I**1 **I**2 ] = [ *j*50 −*j*30]
-
-from which we obtain the determinants
-
-$$
-\Delta = \begin{vmatrix} 8 + j8 & j2 \\ j2 & 4 - j4 \end{vmatrix} = 32(1 + j)(1 - j) + 4 = 68
-$$
-
-$$
-\Delta_2 = \begin{vmatrix} 8 + j8 & j50 \\ j2 & -j30 \end{vmatrix} = 340 - j240 = 416.17 \underline{/-35.22^{\circ}}
-$$
-
-$$
-\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{416.17 \underline{/-35.22^{\circ}}}{68} = 6.12 \underline{/-35.22^{\circ}} A
-$$
-
-The desired current is
-
-$$
-I_o = -I_2 = 6.12 \underline{144.78}^{\circ} A
-$$
-
-For Practice Prob. 10.3.
-
-# **Solution:**
-
-As shown in Fig. 10.10, meshes 3 and 4 form a supermesh due to the current source between the meshes. For mesh 1, KVL gives
-
-$$
--10 + (8 - j2)\mathbf{I}_1 - (-j2)\mathbf{I}_2 - 8\mathbf{I}_3 = 0
-$$
-
-or
-
-$$
-(8 - j2)\mathbf{I}_1 + j2\mathbf{I}_2 - 8\mathbf{I}_3 = 10 \tag{10.4.1}
-$$
-
-For mesh 2,
-
-$$
-I_2 = -3 \tag{10.4.2}
-$$
-
-For the supermesh,
-
-$$
-(8 - j4)\mathbf{I}_3 - 8\mathbf{I}_1 + (6 + j5)\mathbf{I}_4 - j5\mathbf{I}_2 = 0 \tag{10.4.3}
-$$
-
-Due to the current source between meshes 3 and 4, at node A,
-
-$$
-\mathbf{I}_4 = \mathbf{I}_3 + 4 \tag{10.4.4}
-$$
-
-■ **METHOD 1** Instead of solving the above four equations, we re ‑ duce them to two by elimination.
-
-Combining Eqs. (10.4.1) and (10.4.2),
-
-$$
-(8 - j2)\mathbf{I}_1 - 8\mathbf{I}_3 = 10 + j6 \tag{10.4.5}
-$$
-
-Combining Eqs. (10.4.2) to (10.4.4),
-
-$$
--8I1 + (14 + j)I3 = -24 - j35
-$$
- (10.4.6)
-
-**Figure 10.10** Analysis of the circuit in Fig. 10.9.
-
-From Eqs. (10.4.5) and (10.4.6), we obtain the matrix equation
-
-$$
-\begin{bmatrix} 8-j2 & -8 \ -8 & 14+j \end{bmatrix} \begin{bmatrix} I_1 \ I_3 \end{bmatrix} = \begin{bmatrix} 10+j6 \ -24-j35 \end{bmatrix}
-$$
-
-We obtain the following determinants
-
-$$
-\Delta = \begin{vmatrix} 8 - j2 & -8 \\ -8 & 14 + j \end{vmatrix} = 112 + j8 - j28 + 2 - 64 = 50 - j20
-$$
-
-$$
-\Delta_1 = \begin{vmatrix} 10 + j6 & -8 \\ -24 - j35 & 14 + j \end{vmatrix} = 140 + j10 + j84 - 6 - 192 - j280
-$$
-
-$$
-= -58 - j186
-$$
-
-Current **I**1 is obtained as
-
-$$
-\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{-58 - j186}{50 - j20} = 3.618 \; \underline{/274.5^\circ} \, \text{A}
-$$
-
-The required voltage **V***o* is
-
-$$
-\mathbf{V}_o = -j2(\mathbf{I}_1 - \mathbf{I}_2) = -j2(3.618 \angle 274.5^\circ + 3)
-$$
-
-= -7.2134 - j6.568 = 9.756 \angle 222.32^\circ \text{V}
-
-■ **METHOD 2** We can use *MATLAB* to solve Eqs. (10.4.1) to (10.4.4). We first cast the equations as
-
-$$
-\begin{bmatrix} 8-j2 & j2 & -8 & 0 \ 0 & 1 & 0 & 0 \ -8 & -j5 & 8-j4 & 6+j5 \ 0 & 0 & -1 & 1 \ \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \\ I_4 \end{bmatrix} = \begin{bmatrix} 10 \\ -3 \\ 0 \\ 4 \end{bmatrix}
-$$
- (10.4.7a)
-
-or
-
-$$
-AI = B
-$$
-
-By inverting **A**, we can obtain **I** as
-
-$$
-\mathbf{I} = \mathbf{A}^{-1} \mathbf{B} \tag{10.4.7b}
-$$
-
-We now apply *MATLAB* as follows:
-
-```
->> A = [(8-j*2) j*2 -8 0;
- 0 1 0 0;
- -8 -j*5 (8-j*4) (6+j*5);
- 0 0 -1 1];
->> B = [10 -3 0 4]';
->> I = inv(A)*B
-I =
- 0.2828 - 3.6069i
- -3.0000
- -1.8690 - 4.4276i
- 2.1310 - 4.4276i
->> Vo = -2*j*(I(1) - I(2))
-Vo =
- -7.2138 - 6.5655i
-```
-
-as obtained previously.
-
-Calculate current **I***o* in the circuit of Fig. 10.11.
-
-**Answer:** 6.089∕5.94° A.
-
-# **10.4** Superposition Theorem
-
-Since ac circuits are linear, the superposition theorem applies to ac circuits the same way it applies to dc circuits. The theorem becomes important if the circuit has sources operating at *different* frequencies. In this case, since the impedances depend on frequency, we must have a different frequency domain circuit for each frequency. The total re ‑ sponse must be obtained by adding the individual responses in the *time* domain. It is incorrect to try to add the responses in the phasor or fre‑ quency domain. Why? Because the exponential factor *ejωt* is implicit in sinusoidal analysis, and that factor would change for every angular frequency *ω*. It would therefore not make sense to add responses at different frequencies in the phasor domain. Thus, when a circuit has sources operating at different frequencies, one must add the responses due to the individual frequencies in the time domain.
-
-Use the superposition theorem to find **I***o* in the circuit in Fig. 10.7.
-
-# **Solution:**
-
-Let
-
-$$
-\mathbf{I}_o = \mathbf{I}_o' + \mathbf{I}_o'' \tag{10.5.1}
-$$
-
-where **I**′ *o* and **I**″ *o* are due to the voltage and current sources, respectively. To find **I**′ *o*, consider the circuit in Fig. 10.12(a). If we let **Z** be the parallel combination of −*j*2 and 8 + *j*10, then
-
-2 and 8 + j10, then
-\n
-$$
-\mathbf{Z} = \frac{-j2(8+j10)}{-2j+8+j10} = 0.25 - j2.25
-$$
-
-and current **I**′ *o* is
-
-$$
-\mathbf{I}'_o = \frac{j20}{4 - j2 + \mathbf{Z}} = \frac{j20}{4.25 - j4.25}
-$$
-
-or
-
-$$
-\mathbf{I}'_o = -2.353 + j2.353\tag{10.5.2}
-$$
-
-To get **I**″ *o*, consider the circuit in Fig. 10.12(b). For mesh 1,
-
-$$
-(8 + j8)\mathbf{I}_1 - j10\mathbf{I}_3 + j2\mathbf{I}_2 = 0 \tag{10.5.3}
-$$
-
-For mesh 2,
-
-$$
-(4 - j4)\mathbf{I}_2 + j2\mathbf{I}_1 + j2\mathbf{I}_3 = 0 \tag{10.5.4}
-$$
-
-For mesh 3,
-
-**I**3 = 5 **(10.5.5)**
-
-**Figure 10.12** Solution of Example 10.5.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/117_10.4 Superposition Theorem.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/117_10.4 Superposition Theorem.md
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-# Example 10.5
-
-(a)
-
-From Eqs. (10.5.4) and (10.5.5),
-
-$$
-(4 - j4)\mathbf{I}_2 + j2\mathbf{I}_1 + j10 = 0
-$$
-
-Expressing **I**1 in terms of **I**2 gives
-
-$$
-\mathbf{I}_1 = (2 + j2)\mathbf{I}_2 - 5\tag{10.5.6}
-$$
-
-Substituting Eqs. (10.5.5) and (10.5.6) into Eq. (10.5.3), we get
-
-$$
-(8+j8)[(2+j2)I2 - 5] - j50 + j2I2 = 0
-$$
-
-or
-
-$$
-I_2 = \frac{90 - j40}{34} = 2.647 - j1.176
-$$
-
-Current **I**″ *o* is obtained as
-
-$$
-\mathbf{I}_{o}'' = -\mathbf{I}_{2} = -2.647 + j1.176
-$$
- (10.5.7)
-
-From Eqs. (10.5.2) and (10.5.7), we write
-
-$$
-I_o = I'_o + I''_o = -5 + j3.529 = 6.12 \underline{144.78^\circ} A
-$$
-
-which agrees with what we got in Example 10.3. It should be noted that applying the superposition theorem is not the best way to solve this prob‑ lem. It seems that we have made the problem twice as hard as the origi ‑ nal one by using superposition. However, in Example 10.6, superposi ‑ tion is clearly the easiest approach.
-
-| Practice Problem 10.5 | Find current | Io in the circuit of Fig. 10.8 using the superposition |
-|-----------------------|--------------|--------------------------------------------------------|
-| | theorem. | |
-| | | |
-
-**Answer:** 5.97∕65.45° A.
-
-Because the circuit operates at three different frequencies (*ω* = 0 for the dc voltage source), one way to obtain a solution is to use superposition, which breaks the problem into single‑frequency problems. So we let
-
-$$
-v_o = v_1 + v_2 + v_3 \tag{10.6.1}
-$$
-
-where *v*1 is due to the 5‑V dc voltage source, *v*2 is due to the 10 cos 2*t* V voltage source, and *v*3 is due to the 2 sin 5*t* A current source.
-
-To find *v*1, we set to zero all sources except the 5‑V dc source. We recall that at steady state, a capacitor is an open circuit to dc while an inductor is a short circuit to dc. There is an alternative way of looking at this. Because *ω* = 0, *jωL* = 0, 1∕*jωC* = ∞. Either way, the equivalent circuit is as shown in Fig. 10.14(a). By voltage division,
-
-$$
--v_1 = \frac{1}{1+4} (5) = 1 \text{ V}
-$$
- (10.6.2)
-
-To find *v*2, we set to zero both the 5‑V source and the 2 sin 5*t* current source and transform the circuit to the frequency domain.
-
-10 cos 2t
-$$
-\Rightarrow
-$$
- $10\underline{/0}^{\circ}$ , $\omega = 2 \text{ rad/s}$
-2 H $\Rightarrow$ $j\omega L = j4 \Omega$
-0.1 F $\Rightarrow$ $\frac{1}{j\omega C} = -j5 \Omega$
-
-The equivalent circuit is now as shown in Fig. 10.14(b). Let
-
-$$
-\mathbf{Z} = -j5 \| 4 = \frac{-j5 \times 4}{4 - j5} = 2.439 - j1.951
-$$
-
-**Figure 10.14**
-
-Solution of Example 10.6: (a) setting all sources to zero except the 5‑V dc source, (b) setting all sources to zero except the ac voltage source, (c) setting all sources to zero except the ac current source.
-
-By voltage division,
-
-oltage division,
-\n
-$$
-\mathbf{V}_2 = \frac{1}{1 + j4 + \mathbf{Z}} (10/0^\circ) = \frac{10}{3.439 + j2.049} = 2.498 \underline{\text{ } 20.79^\circ}
-$$
-
-In the time domain,
-
-$$
-v_2 = 2.498 \cos(2t - 30.79^\circ) \tag{10.6.3}
-$$
-
-To obtain *v*3, we set the voltage sources to zero and transform what is left to the frequency domain.
-
-$$
-2 \sin 5t \Rightarrow 2(-90^\circ, \omega = 5 \text{ rad/s})
-$$
-
-$$
-2 \text{ H} \Rightarrow j\omega L = j10 \Omega
-$$
-
-$$
-0.1 \text{ F} \Rightarrow \frac{1}{j\omega C} = -j2 \Omega
-$$
-
-The equivalent circuit is in Fig. 10.14(c). Let
-
-$$
-\mathbf{Z}_1 = -j2 \parallel 4 = \frac{-j2 \times 4}{4 - j2} = 0.8 - j1.6 \ \Omega
-$$
-
-By current division,
-
-$$
-\mathbf{I}_1 = \frac{j10}{j10 + 1 + \mathbf{Z}_1} (2 \angle -90^\circ) \text{ A}
-$$
-$$
-\mathbf{V}_3 = \mathbf{I}_1 \times 1 = \frac{j10}{1.8 + j8.4} (-j2) = 2.328 \angle -80^\circ \text{ V}
-$$
-
-In the time domain,
-
-$$
-v_3 = 2.33 \cos(5t - 80^\circ) = 2.33 \sin(5t + 10^\circ) \text{ V} \qquad (10.6.4)
-$$
-
-Substituting Eqs. (10.6.2) to (10.6.4) into Eq. (10.6.1), we have
-
-$$
-v_o(t) = -1 + 2.498 \cos(2t - 30.79^\circ) + 2.33 \sin(5t + 10^\circ) \text{ V}
-$$
-
-Practice Problem 10.6 Calculate *vo* in the circuit of Fig. 10.15 using the superposition theorem.
-
-**Answer:** 11.577 sin(5*t* − 81.12°) + 3.154 cos(10*t* − 86.24°) V.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/118_10.5 Source Transformation.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/118_10.5 Source Transformation.md
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-# **10.5** Source Transformation
-
-As Fig. 10.16 shows, source transformation in the frequency domain involves transforming a voltage source in series with an impedance to a current source in parallel with an impedance, or vice versa. As we go from one source type to another, we must keep the following relationship in mind:
-
-$$
-\mathbf{V}_s = \mathbf{Z}_s \mathbf{I}_s \qquad \Leftrightarrow \qquad \mathbf{I}_s = \frac{\mathbf{V}_s}{\mathbf{Z}_s} \qquad (10.1)
-$$
-
-Calculate **V***x* in the circuit of Fig. 10.17 using the method of source transformation.
-
-For Example 10.7.
-
-# **Solution:**
-
-We transform the voltage source to a current source and obtain the cir ‑ cuit in Fig. 10.18(a), where
-
-$$
-I_s = \frac{20/-90^{\circ}}{5} = 4/-90^{\circ} = -j4 \text{ A}
-$$
-
-The parallel combination of 5‑Ω resistance and (3 + *j*4) impedance gives
-
-$$
-\mathbf{Z}_1 = \frac{5(3+j4)}{8+j4} = 2.5 + j1.25 \ \Omega
-$$
-
-Converting the current source to a voltage source yields the circuit in Fig. 10.18(b), where
-
-$$
-\mathbf{V}_s = \mathbf{I}_s \mathbf{Z}_1 = -j4(2.5 + j1.25) = 5 - j10 \text{ V}
-$$
-
-**Figure 10.18** Solution of the circuit in Fig. 10.17.
-
-By voltage division,
-
-oltage division,
-$$
-\mathbf{V}_x = \frac{10}{10 + 2.5 + j1.25 + 4 - j13} (5 - j10) = 5.519 \underline{/-28}^\circ \text{ V}
-$$
-
-Example 10.7
-
-Practice Problem 10.7 Find **I***o* in the circuit of Fig. 10.19 using the concept of source transformation.
-
-**Answer:** 1.9727∕99.46° A.
-
-**10.6** Thevenin and Norton Equivalent Circuits
-
-Thevenin's and Norton's theorems are applied to ac circuits in the same way as they are to dc circuits. The only additional effort arises from the need to manipulate complex numbers. The frequency domain version of a Thevenin equivalent circuit is depicted in Fig. 10.20, where a linear circuit is replaced by a voltage source in series with an impedance. The Norton equivalent circuit is illustrated in Fig. 10.21, where a linear cir ‑ cuit is replaced by a current source in parallel with an impedance. Keep in mind that the two equivalent circuits are related as
-
-$$
-\mathbf{V}_{\mathrm{Th}} = \mathbf{Z}_N \mathbf{I}_N, \qquad \mathbf{Z}_{\mathrm{Th}} = \mathbf{Z}_N \tag{10.2}
-$$
-
-just as in source transformation. **V**Th is the open‑circuit voltage while **I***N* is the short‑circuit current.
-
-If the circuit has sources operating at dif ferent frequencies (see Example 10.6, for example), the Thevenin or Norton equivalent circuit must be determined at each frequenc y. This leads to entirely dif ferent equivalent circuits, one for each frequenc y, not one equi valent circuit with equivalent sources and equivalent impedances.
-
-Example 10.8 Obtain the Thevenin equivalent at terminals *a*‑*b* of the circuit in Fig. 10.22.
-
-**Figure 10.22** For Example 10.8.
-
-**Figure 10.21** Norton equivalent.
-
-# **Solution:**
-
-We find **Z**Th by setting the voltage source to zero. As shown in Fig. 10.23(a), the 8 ‑Ω resistance is now in parallel with the −*j*6 reac‑ tance, so that their combination gives
-
-$$
-\mathbf{Z}_1 = -j6 \| 8 = \frac{-j6 \times 8}{8 - j6} = 2.88 - j3.84 \ \Omega
-$$
-
-Similarly, the 4 ‑Ω resistance is in parallel with the *j*12 reactance, and their combination gives
-
-$$
-\mathbf{Z}_2 = 4 || j12 = \frac{j12 \times 4}{4 + j12} = 3.6 + j1.2 \ \Omega
-$$
-
-The Thevenin impedance is the series combination of **Z**1 and **Z**2; that is,
-
-$$
-\mathbf{Z}_{\text{Th}} = \mathbf{Z}_1 + \mathbf{Z}_2 = 6.48 - j2.64 \ \Omega
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/119_10.6 Thevenin and Norton Equivalent Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/119_10.6 Thevenin and Norton Equivalent Circuits.md
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-$$
-
-To find **V**Th, consider the circuit in Fig. 10.23(b). Currents **I**1 and **I**2 are obtained as
-
-$$
-\mathbf{I}_1 = \frac{120/75^{\circ}}{8 - j6} \text{ A}, \qquad \mathbf{I}_2 = \frac{120/75^{\circ}}{4 + j12} \text{ A}
-$$
-
-Applying KVL around loop *bcdeab* in Fig. 10.23(b) gives
-
-$$
-\mathbf{V}_{\mathrm{Th}} - 4\mathbf{I}_2 + (-j6)\mathbf{I}_1 = 0
-$$
-
-or
-
-$$
-\mathbf{V}_{\text{Th}} = 4\mathbf{I}_2 + j6\mathbf{I}_1 = \frac{480/75^\circ}{4 + j12} + \frac{720/75^\circ + 90^\circ}{8 - j6}
-$$
-$$
-= 37.95/3.43^\circ + 72/201.87^\circ
-$$
-$$
-= -28.936 - j24.55 = 37.95/220.31^\circ \text{ V}
-$$
-
-Practice Problem 10.8 Find the Thevenin equivalent at terminals *a*‑*b* of the circuit in Fig. 10.24.
-
-For Practice Prob. 10.8.
-
-**Answer: Z**Th = 12.4 − *j*3.2 Ω, **V**Th = 47.43∕−51.57° V.
-
-# Example 10.9
-
-Find the Thevenin equivalent of the circuit in Fig. 10.25 as seen from terminals *a*‑*b*.
-
-# **Figure 10.25**
-
-For Example 10.9.
-
-# **Solution:**
-
-To find **V**Th, we apply KCL at node 1 in Fig. 10.26(a).
-
-$$
-15 = Io + 0.5Io \Rightarrow Io = 10 A
-$$
-
-Applying KVL to the loop on the right ‑hand side in Fig. 10.26(a), we obtain
-
-$$
--I_o(2-j4) + 0.5I_o(4+j3) + V_{Th} = 0
-$$
-
-or
-
-$$
-\mathbf{V}_{\text{Th}} = 10(2 - j4) - 5(4 + j3) = -j55
-$$
-
-Thus, the Thevenin voltage is
-
-$$
-V_{\text{Th}} = 55 \angle -90^{\circ} \text{ V}
-$$
-
-**Figure 10.26** Solution of the problem in Fig. 10.25: (a) finding **V**Th, (b) finding **Z**Th.
-
-To obtain **Z**Th, we remove the independent source. Due to the presence of the dependent current source, we connect a 3-A current source (3 is an arbitrary value chosen for convenience here, a number divisible by the sum of currents leaving the node) to terminals *a*-*b* as shown in Fig. 10.26(b). At the node, KCL gives
-
-$$
-3 = I_o + 0.5I_o \qquad \Rightarrow \qquad I_o = 2A
-$$
-
-Applying KVL to the outer loop in Fig. 10.26(b) gives
-
-$$
-\mathbf{V}_s = \mathbf{I}_o(4 + j3 + 2 - j4) = 2(6 - j)
-$$
-
-The Thevenin impedance is
-
-$$
-\mathbf{Z}_{\text{Th}} = \frac{\mathbf{V}_s}{\mathbf{I}_s} = \frac{2(6-j)}{3} = 4 - j0.6667 \ \Omega
-$$
-
-Determine the Thevenin equivalent of the circuit in Fig. 10.27 as seen from the terminals *a*-*b*.
-
-**Answer:**
-$$
-\mathbb{Z}_{\text{Th}} = 4.473 \underline{\smash{\big)}\, -7.64^{\circ}} \,\Omega, \, \mathbf{V}_{\text{Th}} = 11.763 \underline{\smash{\big)}\, 72.9^{\circ}} \text{ volts.}
-$$
-
-**Figure 10.27** For Practice Prob. 10.9.
-
-Obtain current **I***o* in Fig. 10.28 using Norton's theorem. Example 10.10
-
-For Example 10.10.
-
-# **Solution:**
-
-Our first objective is to find the Norton equivalent at terminals *a*-*b*. **Z***N* is found in the same way as **Z**Th. We set the sources to zero as shown in Fig. 10.29(a). As evident from the figure, the (8 − *j*2) and (10 + *j*4) impedances are short-circuited, so that
-
-$$
-\mathbf{Z}_N = 5 \ \Omega
-$$
-
-To get **I***N*, we short-circuit terminals *a*-*b* as in Fig. 10.29(b) and apply mesh analysis. Notice that meshes 2 and 3 form a supermesh because of the current source linking them. For mesh 1,
-
-$$
--j40 + (18 + j2)\mathbf{I}_1 - (8 - j2)\mathbf{I}_2 - (10 + j4)\mathbf{I}_3 = 0 \tag{10.10.1}
-$$
-
-Solution of the circuit in Fig. 10.28: (a) finding **Z***N*, (b) finding **V***N*, (c) calculating **I***o*.
-
-For the supermesh,
-
-$$
-(13 - j2)I2 + (10 + j4)I3 - (18 + j2)I1 = 0 \t(10.10.2)
-$$
-
-At node *a*, due to the current source between meshes 2 and 3,
-
-$$
-I_3 = I_2 + 3 \tag{10.10.3}
-$$
-
-Adding Eqs. (10.10.1) and (10.10.2) gives
-
-$$
--j40 + 5\mathbf{I}_2 = 0 \qquad \Rightarrow \qquad \mathbf{I}_2 = j8
-$$
-
-From Eq. (10.10.3),
-
-$$
-\mathbf{I}_3 = \mathbf{I}_2 + 3 = 3 + j8
-$$
-
-The Norton current is
-
-$$
-\mathbf{I}_N = \mathbf{I}_3 = (3 + j8) \text{ A}
-$$
-
-Figure 10.29(c) shows the Norton equivalent circuit along with the im ‑ pedance at terminals *a*‑*b*. By current division,
-
-$$
-\mathbf{I}_o = \frac{5}{5 + 20 + j15} \mathbf{I}_N = \frac{3 + j8}{5 + j3} = 1.465 / 38.48^{\circ} \text{ A}
-$$
-
-# Practice Problem 10.10
-
-Determine the Norton equivalent of the circuit in Fig. 10.30 as seen from terminals *a*‑*b*. Use the equivalent to find **I***o*.
-
-# **Figure 10.30**
-
-For Practice Prob. 10.10 and Prob. 10.35.
-
-**Answer: Z***N* = 3.176 + *j*0.706 Ω, **I***N* = 8.396⧸−32.68° A, **I***o* = 1.9714⧸−2.10° A.
-
-# **10.7** Op Amp AC Circuits
-
-The three steps stated in Section 10.1 also apply to op amp circuits, as long as the op amp is operating in the linear region. As usual, we will assume ideal op amps. (See Section 5.2.) As discussed in Chapter 5, the key to analyzing op amp circuits is to keep two important properties of an ideal op amp in mind:
-
-- 1. No current enters either of its input terminals.
-- 2. The voltage across its input terminals is zero.
-
-The following examples will illustrate these ideas.
-
-# **Figure 10.31**
-
-For Example 10.11: (a) the original circuit in the time domain, (b) its frequency domain equivalent.
-
-# **Solution:**
-
-We first transform the circuit to the frequency domain, as shown in Fig. 10.31(b), where **V***s* = 3⧸ 0°, *ω* = 1000 rad/s. Applying KCL at node 1, we obtain
-
-$$
-\frac{3/0^{\circ} - V_1}{10} = \frac{V_1}{-j5} + \frac{V_1 - 0}{10} + \frac{V_1 - V_o}{20}
-$$
-
-or
-
-$$
-6 = (5 + j4)\mathbf{V}_1 - \mathbf{V}_o \tag{10.11.1}
-$$
-
-At node 2, KCL gives
-
-$$
-\frac{\mathbf{V}_1 - 0}{10} = \frac{0 - \mathbf{V}_o}{-j10}
-$$
-
-which leads to
-
-$$
-\mathbf{V}_1 = -j\mathbf{V}_o \tag{10.11.2}
-$$
-
-Substituting Eq. (10.11.2) into Eq. (10.11.1) yields
-
-$$
-6 = -j(5 + j4)\mathbf{V}_o - \mathbf{V}_o = (3 - j5)\mathbf{V}_o
-$$
-$$
-\mathbf{V}_o = \frac{6}{3 - j5} = 1.029 \angle 59.04^\circ
-$$
-
-Hence,
-
-$$
-v_o(t) = 1.029 \cos(1000t + 59.04^{\circ}) \text{ V}
-$$
-
-Practice Problem 10.11 Find *vo* and *io* in the op amp circuit of Fig. 10.32. Let *vs*= 12 cos 5000*t* V.
-
-**Answer:** 4 sin 5,000*t* V, 400 sin 5,000*t μ*A.
-
-**Figure 10.33** For Example 10.12.
-
-Example 10.12 Compute the closed ‑loop g ain and phase shift for the circuit in Fig. 10.33. Assume that *R*1 = *R*2 = 10 kΩ, *C*1 = 2 *μ*F, *C*2 = 1 *μ*F, and *ω* = 200 rad/s.
-
-# **Solution:**
-
-The feedback and input impedances are calculated as
-
-$$
-\mathbf{Z}_f = R_2 \left\| \frac{1}{j\omega C_2} = \frac{R_2}{1 + j\omega R_2 C_2}
-$$
-$$
-\mathbf{Z}_i = R_1 + \frac{1}{j\omega C_1} = \frac{1 + j\omega R_1 C_1}{j\omega C_1}
-$$
-
-Since the circuit in Fig. 10.33 is an inverting amplifier, the closed‑loop gain is given by
-
-$$
-G = \frac{V_o}{V_s} = \frac{Z_f}{Z_i} = \frac{-j\omega C_1 R_2}{(1 + j\omega R_1 C_1)(1 + j\omega R_2 C_2)}
-$$
-
-Substituting the given values of *R*1, *R*2, *C*1, *C*2, and *ω*, we obtain
-
-given values of
-$$
-R_1
-$$
-, $R_2$ , $C_1$ , $C_2$ , and $\omega$ ,
-\n
-$$
-G = \frac{-j4}{(1+j4)(1+j2)} = 0.434 / 130.6^{\circ}
-$$
-
-Thus, the closed‑loop gain is 0.434 and the phase shift is 130.6°.
-
-Practice Problem 10.12 Obtain the closed‑loop gain and phase shift for the circuit in Fig. 10.34. Let *R* = 10 kΩ, *C* = 1 *μ*F, and *ω* = 1000 rad/s.
-
-**Answer:** 1.0147, −5.6°.
-
-# **10.8** AC Analysis Using PSpice
-
-*PSpice* affords a big relief from the tedious task of manipulating com‑ plex numbers in ac circuit analysis. The procedure for using *PSpice* for ac analysis is quite similar to that required for dc analysis. The reader should read Section D.5 in Appendix D for a review of *PSpice* concepts for ac analysis. AC circuit analysis is done in the phasor or frequency domain, and all sources must have the same frequency. Although ac analysis with *PSpice* involves using AC Sweep, our analysis in this chapter requires a single frequency *f* = *ω*∕2*π*. The out‑ put file of *PSpice* contains voltage and current phasors. If necessary, the impedances can be calculated using the voltages and currents in the output file.
-
-Obtain *vo* and *io* in the circuit of Fig. 10.35 using *PSpice*. Example 10.13
-
-# **Solution:**
-
-We first convert the sine function to cosine.
-
-$$
-8 \sin(1000t + 50^{\circ}) = 8 \cos(1000t + 50^{\circ} - 90^{\circ})
-$$
-$$
-= 8 \cos(1000t - 40^{\circ})
-$$
-
-The frequency *f* is obtained from *ω* as
-
-$$
-f = \frac{\omega}{2\pi} = \frac{1000}{2\pi} = 159.155
-$$
- Hz
-
-The schematic for the circuit is shown in Fig. 10.36. Notice that the current‑controlled current source F1 is connected such that its current flows from node 0 to node 3 in conformity with the original circuit in Fig. 10.35. Since we only want the magnitude and phase of *vo* and *io*, we set the attributes of IPRINT and VPRINT1 each to *AC* = *yes*, *MAG* = *yes*, *PHASE* = *yes*. As a single ‑frequency analysis, we select **Analysis/ Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 159.155, and *Final Freq* = 159.155. After saving the schematic, we simulate it by selecting **Analysis/Simulate.** The output file includes the source fre‑ quency in addition to the attributes checked for the pseudocomponents IPRINT and VPRINT1,
-
-| FREQ | | IM(V_PRINT3) IP(V_PRINT3) |
-|-----------|-----------|---------------------------|
-| 1.592E+02 | 3.264E–03 | –3.743E+01 |
-| FREQ | VM(3) | VP(3) |
-| 1.592E+02 | 1.550E+00 | –9.518E+01 |
-
-**Figure 10.36**
-
-The schematic of the circuit in Fig. 10.35.
-
-From this output file, we obtain
-
-**V***o* = 1.55⧸−95.18° V, **I***o* = 3.264⧸−37.43° mA
-
-which are the phasors for
-
-*vo* = 1.55 cos(1000*t* − 95.18°) = 1.55 sin(1000*t* − 5.18°) V
-
-and
-
-*io* = 3.264 cos(1000*t* − 37.43°) mA
-
-For Practice Prob. 10.13.
-
-**Answer:** 3.219 cos(3,000*t* − 154.6°) V, 6.527 cos(3,000*t* − 55.12°) mA.
-
-Example 10.14
-
-Find **V**1 and **V**2 in the circuit of Fig. 10.38.
-
-# **Solution:**
-
-1. **Define.** In its present form, the problem is clearly stated. Again, we must emphasize that time spent here will save lots of time and expense later on! One thing that might have created a problem for you is that, if the reference was missing for this problem, you would then need to ask the individual assigning the problem where
-
-it is to be located. If you could not do that, then you would need to assume where it should be and then clearly state what you did and why you did it.
-
-- 2. **Present.** The given circuit is a frequency domain circuit and the unknown node voltages **V**1 and **V**2 are also frequency domain values. Clearly, we need a process to solve for these unknowns in the frequency domain.
-- 3. **Alternative.** We have two direct alternative solution techniques that we can easily use. We can do a straightforward nodal analysis approach or use *PSpice*. Since this example is in a section dedicated to using *PSpice* to solve problems, we will use *PSpice* to find **V**1 and **V**2. We can then use nodal analysis to check the answer.
-- 4. **Attempt.** The circuit in Fig. 10.35 is in the time domain, whereas the one in Fig. 10.38 is in the frequency domain. Since we are not given a particular frequency and *PSpice* requires one, we select any frequency consistent with the given impedances. For example, if we select *ω* = 1 rad/s, the corresponding frequency is *f* = *ω*∕2*π* = 0.15916 Hz. We obtain the values of the capacitance (*C* = 1∕*ωXC*) and inductances (*L* = *XL*∕*ω*). Making these changes results in the schematic in Fig. 10.39. To ease wiring, we have exchanged the positions of the voltage‑controlled current source
-
-# **Figure 10.39** Schematic for the circuit in the Fig. 10.38.
-
-G1 and the 2 + *j*2 Ω impedance. Notice that the current of G1 flows from node 1 to node 3, while the controlling voltage is across the capacitor C2, as required in Fig. 10.38. The attributes of pseudo components VPRINT1 are set as shown. As a single‑ frequency analysis, we select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 0.15916, and *Final Freq* = 0.15916. After saving the schematic, we select **Analysis/Simulate** to simulate the circuit. When this is done, the output file includes
-
-| FREQ | VM(1) | VP(1) |
-|-----------|-----------|------------|
-| 1.592E–01 | 2.708E+00 | –5.673E+01 |
-| | | |
-| FREQ | VM(3) | VP(3) |
-| 1.592E-01 | 4.468E+00 | –1.026E+02 |
-
-from which we obtain,
-
-**V**1 = **2.708**⧸**−56.74° V** and **V**2 = **6.911**⧸**−80.72° V**
-
-5. **Evaluate.** One of the most important lessons to be learned is that when using programs such as *PSpice* you still need to validate the answer. There are many opportunities for making a mistake, including coming across an unknown "bug" in *PSpice* that yields incorrect results.
-
-So, how can we validate this solution? Obviously, we can rework the entire problem with nodal analysis, and perhaps using *MATLAB*, to see if we obtain the same results. There is another way we will use here: Write the nodal equations and substitute the answers obtained in the *PSpice* solution, and see if the nodal equations are satisfied.
-
-The nodal equations for this circuit are given below. Note we have substituted **V**1 = **V***x* into the dependent source.
-
-$$
--3 + \frac{\mathbf{V}_1 - 0}{1} + \frac{\mathbf{V}_1 - 0}{-j1} + \frac{\mathbf{V}_1 - \mathbf{V}_2}{2 + j2} + 0.2\mathbf{V}_1 + \frac{\mathbf{V}_1 - \mathbf{V}_2}{-j2} = 0
-$$
-
-(1 + j + 0.25 - j0.25 + 0.2 + j0.5) $\mathbf{V}_1$
--(0.25 - j0.25 + j0.5) $\mathbf{V}_2$ = 3
-(1.45 + j1.25) $\mathbf{V}_1$ - (0.25 + j0.25) $\mathbf{V}_2$ = 3
-1.9144/40.76° $\mathbf{V}_1$ - 0.3536/45° $\mathbf{V}_2$ = 3
-
-Now, to check the answer, we substitute the *PSpice* answers into this.
-
-$$
-1.9144 \underline{/40.76^{\circ}} \times 2.708 \underline{/-56.74^{\circ}} - 0.3536 \underline{/45^{\circ}} \times 6.911 \underline{/-80.72^{\circ}}
-$$
-
-= 5.184 \underline{/-15.98^{\circ}} - 2.444 \underline{/-35.72^{\circ}}
-= 4.984 - j1.4272 - 1.9842 + j1.4269
-= 3 - j0.0003 [Answer checks]
-
-6. **Satisfactory?** Although we used only the equation from node 1 to check the answer, this is more than satisfactory to validate the answer from the *PSpice* solution. We can now present our work as a solution to the problem.
-
-Obtain **V***x* and **I***x* in the circuit depicted in Fig. 10.40. Practice Problem 10.14
-
-For Practice Prob. 10.14.
-
-**Answer:** 39.37⧸ 44.78° V, 10.336⧸158° A.
-
-# **10.9** Applications
-
-The concepts learned in this chapter will be applied in later chapters to calculate electric power and determine frequency response. The con ‑ cepts are also used in analyzing coupled circuits, three‑phase circuits, ac transistor circuits, filters, oscillators, and other ac circuits. In this section, we apply the concepts to develop two practical ac circuits: the capaci ‑ tance multiplier and the sine wave oscillators.
-
-# **10.9.1** Capacitance Multiplier
-
-The op amp circuit in Fig. 10.41 is known as a *capacitance multiplier*, for reasons that will become obvious. Such a circuit is used in integrated‑ circuit technology to produce a multiple of a small physical capacitance *C* when a large capacitance is needed. The circuit in Fig. 10.41 can be used to multiply capacitance values by a factor up to 1,000. For exam‑ ple, a 10‑pF capacitor can be made to behave like a 100‑nF capacitor.
-
-In Fig. 10.41, the first op amp operates as a voltage follower, while the second one is an inverting amplifier. The voltage follower iso‑ lates the capacitance formed by the circuit from the loading imposed by the inverting amplifier. Since no current enters the input terminals of the op amp, the input current **I***i* flows through the feedback capaci‑ tor. Hence, at node 1,
-
-$$
-\mathbf{I}_{i} = \frac{\mathbf{V}_{i} - \mathbf{V}_{o}}{1/j\omega C} = j\omega C(\mathbf{V}_{i} - \mathbf{V}_{o})
-$$
-\n(10.3)
-
-Applying KCL at node 2 gives
-
-$$
-\frac{\mathbf{V}_i - \mathbf{0}}{R_1} = \frac{\mathbf{0} - \mathbf{V}_o}{R_2}
-$$
-
-or
-
-$$
-\mathbf{V}_o = -\frac{R_2}{R_1} \mathbf{V}_i \tag{10.4}
-$$
-
-Substituting Eq. (10.4) into (10.3) gives
-
-$$
-\mathbf{I}_i = j\omega C \bigg( 1 + \frac{R_2}{R_1} \bigg) \mathbf{V}_i
-$$
-
-or
-
-$$
-\frac{\mathbf{I}_i}{\mathbf{V}_i} = j\omega \left( 1 + \frac{R_2}{R_1} \right) C \tag{10.5}
-$$
-
-The input impedance is
-
-$$
-\mathbf{Z}_{i} = \frac{\mathbf{V}_{i}}{\mathbf{I}_{i}} = \frac{1}{j\omega C_{\text{eq}}}
-$$
-(10.6)
-
-where
-
-$$
-C_{\text{eq}} = \left(1 + \frac{R_2}{R_1}\right)C\tag{10.7}
-$$
-
-Thus, by a proper selection of the values of *R*1 and *R*2, the op amp circuit in Fig. 10.41 can be made to produce an effective capacitance between the input terminal and ground, which is a multiple of the physical capaci‑ tance *C*. The size of the effective capacitance is practically limited by the inverted output voltage limitation. Thus, the larger the capacitance multiplication, the smaller is the allowable input voltage to prevent the op amps from reaching saturation.
-
-A similar op amp circuit can be designed to simulate inductance. (See Prob. 10.89.) There is also an op amp circuit configuration to create a resistance multiplier.
-
-Example 10.15 Calculate *C*eq in Fig. 10.41 when *R*1 = 10 kΩ, *R*2 = 1 MΩ, and *C* = 1 nF.
-
-# **Solution:**
-
-From Eq. (10.7) *C*eq = (1 + \_\_\_ *R*2 *R*1 )*C* = (1 + 1 × 106 \_\_\_\_\_\_\_\_ 10 × 103 ) 1 nF = 101 nF Determine the equivalent capacitance of the op amp circuit in Fig. 10.41 if *R*1 = 10 kΩ, *R*2 = 10 MΩ, and *C* = 10 nF.
-
-**Answer:** 10 *μ*F.
-
-# **10.9.2** Oscillators
-
-We know that dc is produced by batteries. But how do we produce ac? One way is using *oscillators,* which are circuits that convert dc to ac.
-
-An oscillator is a circuit that produces an ac waveform as output when powered by a dc input.
-
-The only external source an oscillator needs is the dc power supply. Ironically, the dc power supply is usually obtained by con verting the ac supplied by the electric utility company to dc. Having gone through the trouble of conversion, one may wonder why we need to use the oscillator to convert the dc to ac again. The problem is that the ac supplied by the utility company operates at a preset frequenc y of 60 Hz in the United States (50 Hz in some other nations), whereas man y applications such as electronic circuits, communication systems, and micro wave devices require internally generated frequencies that range from 0 to 10 GHz or higher. Oscillators are used for generating these frequencies.
-
-In order for sine w ave oscillators to sustain oscillations, the y must meet the *Barkhausen criteria*:
-
-- 1. The overall gain of the oscillator must be unity or greater. Therefore, losses must be compensated for by an amplifying device.
-- 2. The overall phase shift (from input to output and back to the input) must be zero.
-
-Three common types of sine wave oscillators are phase ‑shift, twin *T*, and Wien ‑bridge oscillators. Here we consider only the Wien ‑bridge oscillator.
-
-The *Wien-bridge oscillator* is widely used for generating sinusoids in the frequency range below 1 MHz. It is an *RC* op amp circuit with only a few components, easily tunable and easy to design. As shown in Fig. 10.42, the oscillator essentially consists of a noninverting amplifier with two feedback paths: The positive feedback path to the noninverting input creates oscillations, while the ne gative feedback path to the in verting input controls the gain. If we define the impedances of the *RC* series and parallel combinations as **Z***s* and **Z***p*, then
-
-$$
-Z_s = R_1 + \frac{1}{j\omega C_1} = R_1 - \frac{j}{\omega C_1}
-$$
-(10.8)
-$$
-Z_p = R_2 \Big| \frac{1}{j\omega C_2} = \frac{R_2}{1 + j\omega R_2 C_2}
-$$
-(10.9)
-
-The feedback ratio is
-
-$$
-\frac{\mathbf{V}_2}{\mathbf{V}_o} = \frac{\mathbf{Z}_p}{\mathbf{Z}_s + \mathbf{Z}_p}
-$$
-
-Practice Problem 10.15
-
-Substituting Eqs. (10.8) and (10.9) into Eq. (10.10) gives
-
-Substituting Eqs. (10.8) and (10.9) into Eq. (10.10) gives
-\n
-$$
-\frac{\mathbf{V}_2}{\mathbf{V}_o} = \frac{R_2}{R_2 + \left(R_1 - \frac{j}{\omega C_1}\right)(1 + j\omega R_2 C_2)}
-$$
-\n
-$$
-= \frac{\omega R_2 C_1}{\omega (R_2 C_1 + R_1 C_1 + R_2 C_2) + j(\omega^2 R_1 C_1 R_2 C_2 - 1)}
-$$
-\n(10.11)
-
-To satisfy the second Barkhausen criterion, **V**2 must be in phase with **V***o*, which implies that the ratio in Eq. (10.11) must be purely real. Hence, the imaginary part must be zero. Setting the imaginary part equal to zero gives the oscillation frequency *ωo* as
-
-$$
-\omega_o^2 R_1 C_1 R_2 C_2 - 1 = 0
-$$
-
-or
-
-$$
-\omega_o = \frac{1}{\sqrt{R_1 R_2 C_1 C_2}}\tag{10.12}
-$$
-
-In most practical applications, *R*1 = *R*2 = *R* and *C*1 = *C*2 = *C*, so that
-
-$$
-\omega_o = \frac{1}{RC} = 2\pi f_o \tag{10.13}
-$$
-
-or
-
-$$
-f_o = \frac{1}{2\pi RC}
-$$
- (10.14)
-
-Substituting Eq. (10.13) and *R*1 = *R*2 = *R*, *C*1 = *C*2 = *C* into Eq. (10.11) yields
-
-$$
-\frac{V_2}{V_o} = \frac{1}{3}
-$$
- (10.15)
-
-Thus, in order to satisfy the first Barkhausen criterion, the op amp must compensate by providing a gain of 3 or greater so that the overall gain is at least 1 or unity. We recall that for a noninverting amplifier,
-
-$$
-\frac{\mathbf{V}_o}{\mathbf{V}_2} = 1 + \frac{R_f}{R_g} = 3\tag{10.16}
-$$
-
-or
-
-$$
-R_f = 2R_g \tag{10.17}
-$$
-
-Due to the inherent delay caused by the op amp, Wien‑bridge oscil‑ lators are limited to operating in the frequency range of 1 MHz or less.
-
-# Example 10.16
-
-Design a Wien‑bridge circuit to oscillate at 100 kHz.
-
-# **Solution:**
-
-Using Eq. (10.14), we obtain the time constant of the circuit as
-
-14), we obtain the time constant of the circuit as
-$$
-RC = \frac{1}{2\pi f_o} = \frac{1}{2\pi \times 100 \times 10^3} = 1.59 \times 10^{-6}
-$$
- (10.16.1)
-
-If we select *R* = 10 k Ω, then we can select *C* = 159 pF to satisfy Eq. (10.16.1). Since the gain must be 3, *Rf*∕*Rg* = 2. We could select *Rf* = 20 kΩ while *Rg* = 10 kΩ.
-
-In the Wien‑bridge oscillator circuit in Fig. 10.42, let *R*1 = *R*2 = 2.5 kΩ, *C*1 = *C*2 = 1 nF. Determine the frequency *fo* of the oscillator. Practice Problem 10.16
-
-**Answer:** 63.66 kHz.
-
-# **10.10** Summary
-
-- 1. We apply nodal and mesh analysis to ac circuits by applying KCL and KVL to the phasor form of the circuits.
-- 2. In solving for the steady ‑state response of a circuit that has inde ‑ pendent sources with different frequencies, each independent source *must* be considered separately. The most natural approach to analyz‑ ing such circuits is to apply the superposition theorem. A separate phasor circuit for each frequency *must* be solved independently, and the corresponding response should be obtained in the time domain. The overall response is the sum of the time domain responses of all the individual phasor circuits.
-- 3. The concept of source transformation is also applicable in the fre ‑ quency domain.
-- 4. The Thevenin equivalent of an ac circuit consists of a voltage source **V**Th in series with the Thevenin impedance **Z**Th.
-- 5. The Norton equivalent of an ac circuit consists of a current source **I***N* in parallel with the Norton impedance **Z***N* (=**Z**Th).
-- 6. *PSpice* is a simple and powerful tool for solving ac circuit problems. It relieves us of the tedious task of working with the complex num‑ bers involved in steady‑state analysis.
-- 7. The capacitance multiplier and the ac oscillator provide two typical applications for the concepts presented in this chapter . A capaci‑ tance multiplier is an op amp circuit used in producing a multiple of a physical capacitance. An oscillator is a device that uses a dc input to generate an ac output.
-
-‒
-
-# Review Questions
-
-10 0° V ‒j1 Ω **V**o +
-
-**10.1** The voltage **V***o* across the capacitor in Fig. 10.43 is:
-
-‒
-
-For Review Question 10.1.
-
-**10.2** The value of the current **I***o* in the circuit of Fig. 10.44 is:
-
-(a)
-$$
-4\angle 0^{\circ}
-$$
- A
-(b) $2.4\angle -90^{\circ}$ A
-(c) $0.6\angle 0^{\circ}$ A
-(d) $-1$ A
-
-**Figure 10.44** For Review Question 10.2.
-
-**10.3** Using nodal analysis, the value of **V***o* in the circuit of Fig. 10.45 is:
-
-(a)
-$$
--24 \text{ V}
-$$
- (b) $-8 \text{ V}$
-
-$$
-(c) 8 V \t\t (d) 24 V
-$$
-
-# **10.6** For the circuit in Fig. 10.48, the Thevenin impedance at terminals *a*‑*b* is:
-
-| (a) 1 Ω | (b) 0.5 − j0.5 Ω |
-|------------------|------------------|
-| (c) 0.5 + j0.5 Ω | (d) 1 + j2 Ω |
-| (e) 1 − j2 Ω | |
-
-# **Figure 10.48**
-
-For Review Questions 10.6 and 10.7.
-
-- - (a) 10 cos *t* A (b) 10 sin *t* A (c) 5 cos *t* A
-
-**10.4** In the circuit of Fig. 10.46, current *i*(*t*) is:
-
-(d) 5 sin *t* A (e) 4.472 cos(*t* − 63.43°) A
-
-**Figure 10.46**
-
-**Figure 10.45**
-
-For Review Question 10.3.
-
-For Review Question 10.4.
-
-- **10.5** Refer to the circuit in Fig. 10.47 and observe that the two sources do not have the same frequency. The current *ix*(*t*) can be obtained by:
- - (a) source transformation
- - (b) the superposition theorem
- - (c) *PSpice*
-
-**Figure 10.47** For Review Question 10.5. **10.7** In the circuit of Fig. 10.48, the Thevenin voltage at terminals *a*‑*b* is:
-
-(a)
-$$
-3.535 \angle -45^{\circ} \text{ V}
-$$
- (b) $3.535 \angle 45^{\circ} \text{ V}$
-(c) $7.071 \angle -45^{\circ} \text{ V}$ (d) $7.071 \angle 45^{\circ} \text{ V}$
-
-**10.8** Refer to the circuit in Fig. 10.49. The Norton equivalent impedance at terminals *a*‑*b* is:
-
-| (a) −j4 Ω | (b) −j2 Ω |
-|-----------|-----------|
-| (c) j2 Ω | (d) j4 Ω |
-
-# **Figure 10.49**
-
-For Review Questions 10.8 and 10.9.
-
-**10.9** The Norton current at terminals *a*‑*b* in the circuit of Fig. 10.49 is:
-
-(a)
-$$
-1/\underline{0^{\circ}}
-$$
- A
-(b) $1.5/\underline{-90^{\circ}}$ A
-(c) $1.5/90^{\circ}$ A
-(d) $3/90^{\circ}$ A
-
-- **10.10** *PSpice* can handle a circuit with two independent sources of different frequencies.
- - (a) True (b) False
-
-*Answers: 10.1c, 10.2a, 10.3d, 10.4a, 10.5b, 10.6c, 10.7a, 10.8a, 10.9d, 10.10b.*
-
-# Problems
-
-# Section 10.2 Nodal Analysis
-
-**10.1** Determine *i* in the circuit of Fig. 10.50.
-
-# **Figure 10.50**
-
-For Prob. 10.1.
-
-# **Figure 10.51**
-
-For Prob. 10.2.
-
-**10.3** Determine *vo* in the circuit of Fig. 10.52.
-
-# **Figure 10.52** For Prob. 10.3.
-
-# **Figure 10.53**
-
-For Prob. 10.4.
-
-For Prob. 10.5.
-
-**10.6** Determine **V***x* in Fig. 10.55.
-
-# **Figure 10.55** For Prob. 10.6.
-
-**10.7** Use nodal analysis to find **V** in the circuit of Fig. 10.56.
-
-# **Figure 10.56** For Prob. 10.7.
-
-**10.8** Use nodal analysis to find current *io* in the circuit of Fig. 10.57. Let *is* = 6 cos(200*t* + 15°) A.
-
-# **Figure 10.57**
-
-For Prob. 10.8.
-
-For Prob. 10.9.
-
-‒
-
-**Figure 10.59**
-
-**10.11** Using nodal analysis, find *io*(*t*) in the circuit in Fig. 10.60.
-
-For Prob. 10.11.
-
-**10.12** Using Fig. 10.61, design a problem to help other students better understand nodal analysis.
-
-**Figure 10.61** For Prob. 10.12.
-
-**10.13** Determine **V***x* in the circuit of Fig. 10.62 using any method of your choice.
-
-For Prob. 10.13.
-
-**10.14** Calculate the voltage at nodes 1 and 2 in the circuit of Fig. 10.63 using nodal analysis.
-
-# **Figure 10.63**
-
-For Prob. 10.14.
-
-**10.15** Solve for the current **I** in the circuit of Fig. 10.64 using nodal analysis.
-
-**Figure 10.64**
-
-For Prob. 10.15.
-
-# **Figure 10.65**
-
-For Prob. 10.16.
-
-**10.17** By nodal analysis, obtain current **I***o* in the circuit of Fig. 10.66.
-
-**Figure 10.66** For Prob. 10.17.
-
-**10.19** Obtain **V***o* in Fig. 10.68 using nodal analysis.
-
-**10.20** Refer to Fig. 10.69. If *vs*(*t*) = *Vm* sin *ωt* and *vo*(*t*) = *A* sin(*ωt* + *ϕ*), derive the expressions for *A* and *ϕ*.
-
-**Figure 10.69** For Prob. 10.20.
-
-**10.21** For each of the circuits in Fig. 10.70, find **V***o*∕**V***i* for *ω* = 0, *ω* → ∞, and *ω*2 = 1∕*LC*.
-
-**Figure 10.70** For Prob. 10.21.
-
-**10.22** For the circuit in Fig. 10.71, determine **V***o*∕**V***s*.
-
-**Figure 10.71** For Prob. 10.22.
-
-**10.23** Using nodal analysis obtain **V** in the circuit of Fig. 10.72.
-
-**Figure 10.72**
-
-For Prob. 10.23.
-
-# Section 10.3 Mesh Analysis
-
-**10.24** Design a problem to help other students better understand mesh analysis.
-
-**10.25** Solve for *io* in Fig. 10.73 using mesh analysis.
-
-For Prob. 10.25.
-
-**10.26** Use mesh analysis to find current *io* in the circuit of Fig. 10.74.
-
-**10.29** Using Fig. 10.77, design a problem to help other students better understand mesh analysis.
-
-R3
-
-For Prob. 10.26.
-
-**10.27** Using mesh analysis, find **I**1 and **I**2 in the circuit of Fig. 10.75.
-
-jXL1
-
-**Figure 10.77** For Prob. 10.29.
-
-For Prob. 10.30.
-
-**Figure 10.76** For Prob. 10.28.
-
-**10.32** Determine **V***o* and **I***o* in the circuit of Fig. 10.80 using mesh analysis.
-
-**Figure 10.80** For Prob. 10.32.
-
-**10.33** Compute **I** in Prob. 10.15 using mesh analysis.
-
-**10.34** Use mesh analysis to find **I***o* in Fig. 10.28 (for Example 10.10).
-
-**10.35** Calculate **I***o* in Fig. 10.30 (for Practice Prob. 10.10) using mesh analysis.
-
-**10.36** Compute **V***o* in the circuit of Fig. 10.81 using mesh analysis.
-
-**Figure 10.81** For Prob. 10.36.
-
-**10.37** Use mesh analysis to find currents **I**1, **I**2, and **I**3 in the circuit of Fig. 10.82.
-
-**10.38** Using mesh analysis, obtain **I***o* in the circuit shown in Fig. 10.83.
-
-For Prob. 10.38.
-
-**10.39** Find **I**1, **I**2, **I**3, and **I***x* in the circuit of Fig. 10.84.
-
-**Figure 10.84** For Prob. 10.39.
-
-# Section 10.4 Superposition Theorem
-
-**10.40** Find *io* in the circuit shown in Fig. 10.85 using superposition.
-
-# **Figure 10.85**
-
-For Prob. 10.40.
-
-**10.41** Find *vo* for the circuit in Fig. 10.86, assuming that *is*(*t*) = 2 sin (2*t*) + 3 cos (4*t*) A.
-
-**Figure 10.86** For Prob. 10.41.
-
-**10.42** Using Fig. 10.87, design a problem to help other students better understand the superposition theorem.
-
-**Figure 10.87** For Prob. 10.42.
-
-For Prob. 10.43.
-
-**10.43** Using the superposition principle, find *ix* in the circuit of Fig. 10.88.
-
-For Prob. 10.46.
-
-**10.47** Determine *io* in the circuit of Fig. 10.92, using the superposition principle.
-
-For Prob. 10.47.
-
-**10.44** Use the superposition principle to obtain *vx* in the circuit of Fig. 10.89. Let *vs* = 50 sin 2*t* V and *is* = 12 cos(6*t* + 10°) A.
-
-**10.45** Use superposition to find *i*(*t*) in the circuit of Fig. 10.90.
-
-**Figure 10.90** For Prob. 10.45.
-
-**10.48** Find *io* in the circuit of Fig. 10.93 using superposition.
-
-**Figure 10.93**
-
-For Prob. 10.48.
-
-# Section 10.5 Source Transformation
-
-**10.49** Using source transformation, find *i* in the circuit of Fig. 10.94.
-
-For Prob. 10.49.
-
-**Figure 10.95**
-
-For Prob. 10.50.
-
-- **10.51** Use source transformation to find **I***o* in the circuit of Prob. 10.42.
-- **10.52** Use the method of source transformation to find **I***x* in the circuit of Fig. 10.96.
-
-**Figure 10.96** For Prob. 10.52.
-
-**10.53** Use the concept of source transformation to find **V***o* in the circuit of Fig. 10.97.
-
-For Prob. 10.53.
-
-**10.54** Rework Prob. 10.7 using source transformation.
-
-# Section 10.6 Thevenin and Norton Equivalent Circuits
-
-**10.55** Find the Thevenin and Norton equivalent circuits at terminals *a*‑*b* for each of the circuits in Fig. 10.98.
-
-**Figure 10.98** For Prob. 10.55.
-
-**10.56** For each of the circuits in Fig. 10.99, obtain Thevenin and Norton equivalent circuits at terminals *a*‑*b*.
-
-# **Figure 10.99**
-
-For Prob. 10.56.
-
-**10.57** Using Fig. 10.100, design a problem to help other students better understand Thevenin and Norton equivalent circuits.
-
-# **Figure 10.100** For Prob. 10.57.
-
-**10.58** For the circuit depicted in Fig. 10.101, find the Thevenin equivalent circuit at terminals *a*‑*b*.
-
-**Figure 10.101** For Prob. 10.58.
-
-**10.59** Calculate the output impedance of the circuit shown in Fig. 10.102.
-
-For Prob. 10.59.
-
-**10.60** Find the Thevenin equivalent of the circuit in Fig. 10.103 as seen from:
-
-**10.61** Find the Thevenin equivalent at terminals *a*-*b* of the circuit in Fig. 10.104.
-
-**Figure 10.104** For Prob. 10.61.
-
-**10.62** Using Thevenin's theorem, find *vo* in the circuit of Fig. 10.105.
-
-For Prob. 10.62.
-
-**10.63** Obtain the Norton equivalent of the circuit depicted in Fig. 10.106 at terminals *a*-*b*.
-
-For Prob. 10.63.
-
-**10.64** For the circuit shown in Fig. 10.107, find the Norton equivalent circuit at terminals *a*-*b*.
-
-# **Figure 10.107**
-
-For Prob. 10.64.
-
-**10.65** Using Fig. 10.108, design a problem to help other students better understand Norton's theorem.
-
-Problems **449**
-
-For Prob. 10.70.
-
-**10.71** Find *vo* in the op amp circuit of Fig. 10.114.
-
-**Figure 10.114**
-
-For Prob. 10.71.
-
-**10.72** Compute *io*(*t*) in the op amp circuit in Fig. 10.115 if *vs* = 4 cos(104 *t*) V.
-
-# **Figure 10.115**
-
-For Prob. 10.72.
-
-**10.73** If the input impedance is defined as **Z**in = **V***s*∕**I***s*, find the input impedance of the op amp circuit in Fig. 10.116 when *R*1 = 10 kΩ, *R*2 = 20 kΩ, *C*1 = 10 nF, *C*2 = 20 nF, and *ω* = 5000 rad/s.
-
-**Figure 10.116** For Prob. 10.73.
-
-**Figure 10.111** For Prob. 10.68.
-
-**Figure 10.110** For Prob. 10.67.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/120_10.7 Op Amp AC Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/120_10.7 Op Amp AC Circuits.md
deleted file mode 100644
index 7a172eecd4dcc755a3598ddae8d764db02e333d8..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/120_10.7 Op Amp AC Circuits.md
+++ /dev/null
@@ -1,109 +0,0 @@
-# Section 10.7 Op Amp AC Circuits
-
-**10.69** For the integrator shown in Fig. 10.112, obtain **V***o*∕**V***s*. Find *vo*(*t*) when *vs*(*t*) = **V***m* sin *ωt* and *ω* = 1∕*RC*.
-
-**Figure 10.112** For Prob. 10.69.
-
-**10.70** Using Fig. 10.113, design a problem to help other students better understand op amps in AC circuits. **10.74** Evaluate the voltage gain **A***v* = **V***o*∕**V***s* in the op amp circuit of Fig. 10.117. Find **A***v* at *ω* = 0, *ω* → ∞, *ω* = 1∕*R*1*C*1, and *ω* = 1∕*R*2*C*2.
-
-10 kΩ
-
-**V**o +
-
-20 kΩ
-
-i o
-
-‒ +
-
-+
-
-‒
-
-‒
-
-6 30° V ‒j2 kΩ
-
-‒j4 kΩ
-
-**10.75** In the op amp circuit of Fig. 10.118, find the closed‑ loop gain and phase shift of the output voltage with respect to the input voltage if *C*1 = *C*2 = 1 nF, *R*1 = *R*2 = 100 kΩ, *R*3 = 20 kΩ, *R*4 = 40 kΩ, and *ω* = 2000 rad/s.
-
-*v*s *v*o
-
-R2
-
-C2
-
-R1
-
-R4
-
-+
-
-‒
-
-R3
-
-‒ +
-
-C1
-
-+ ‒
-
-**Figure 10.118** For Prob. 10.75.
-
-**10.77** Compute the closed‑loop gain **V***o*∕**V***s* for the op amp circuit of Fig. 10.120.
-
-**10.78** Determine *vo*(*t*) in the op amp circuit in Fig. 10.121 below.
-
-**10.79** For the op amp circuit in Fig. 10.122, obtain **V***o*.
-
-For Prob. 10.79.
-
-**10.80** Obtain *vo*(*t*) for the op amp circuit in Fig. 10.123 if *vs* = 12 cos(1000*t* − 60°) V.
-
-**Figure 10.123** For Prob. 10.80.
-
-**10.81** Use *PSpice or MultiSim* to determine **V***o* in the circuit of Fig. 10.124. Assume *ω* = 1 rad/s.
-
-**Figure 10.124** For Prob. 10.81.
-
-**10.82** Solve Prob. 10.19 using *PSpice or MultiSim*.
-
-**10.83** Use *PSpice or MultiSim* to find *vo*(*t*) in the circuit of Fig. 10.125. Let *is* = 2 cos(103 *t*) A.
-
-# **Figure 10.125**
-
-For Prob. 10.83.
-
-i
-
-**10.84** Obtain **V***o* in the circuit of Fig. 10.126 using *PSpice or MultiSim*.
-
-# **Figure 10.126**
-
-For Prob. 10.84.
-
-**10.85** Using Fig. 10.127, design a problem to help other students better understand performing AC analysis with *PSpice or MultiSim*.
-
-# **Figure 10.127** For Prob. 10.85.
-
-**10.86** Use *PSpice or MultiSim* to find **V**1, **V**2, and **V**3 in the network of Fig. 10.128.
-
-**Figure 10.128** For Prob. 10.86.
-
-C
-
-+ + **V**o ‒ ‒
-
-R R1
-
-R2 R
-
-C
-
-**10.88** Use *PSpice or MultiSim* to find *vo* and *io* in the circuit of Fig. 10.130 below.
-
-**Figure 10.132** For Prob. 10.90.
-
-**V**i
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/121_10.9 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/121_10.9 Applications.md
deleted file mode 100644
index c03db99c5bd9d7385f87f2697e9723b44bf865a4..0000000000000000000000000000000000000000
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+++ /dev/null
@@ -1,72 +0,0 @@
-# Section 10.9 Applications
-
-**10.89** The op amp circuit in Fig. 10.131 is called an *inductance simulator*. Show that the input impedance is given by
-
-$$
-\mathbf{Z}_{in} = \frac{\mathbf{V}_{in}}{\mathbf{I}_{in}} = j\omega L_{eq}
-$$
-
-where
-
-$$
-L_{\text{eq}} = \frac{R_1 R_3 R_4}{R_2 C}
-$$
-
-**Figure 10.131** For Prob. 10.89.
-
-- **10.91** Consider the oscillator in Fig. 10.133.
- - (a) Determine the oscillation frequency.
- - (b) Obtain the minimum value of *R* for which oscillation takes place.
-
-**Figure 10.133** For Prob. 10.91.
-
-**10.87** Determine **V**1, **V**2, and **V**3 in the circuit of Fig. 10.129 using *PSpice or MultiSim*.
-
-- **10.92** The oscillator circuit in Fig. 10.134 uses an ideal op amp.
- - (a) Calculate the minimum value of *Ro* that will cause oscillation to occur.
- - (b) Find the frequency of oscillation.
-
-# **Figure 10.134**
-
-**10.93** Figure 10.135 shows a *Colpitts oscillator*. Show that the oscillation frequency is
-
-$$
-f_o = \frac{1}{2\pi\sqrt{LC_T}}
-$$
-
-where *CT* = *C*1*C*2∕(*C*1 + C2). Assume *Ri* ≫ *XC*2 .
-
-# **Figure 10.135**
-
-A Colpitts oscillator; for Prob. 10.93.
-
-(*Hint:* Set the imaginary part of the impedance in the feedback circuit equal to zero.)
-
-**10.94** Design a Colpitts oscillator that will operate at 50 kHz.
-
-**10.95** Figure 10.136 shows a *Hartley oscillator*. Show that the frequency of oscillation is
-
-# **Figure 10.136** A Hartley oscillator; for Prob. 10.95.
-
-**10.96** Refer to the oscillator in Fig. 10.137.
-
-(a) Show that
-
-$$
-\mathbf{v} \text{ that}
-$$
-\n
-$$
-\frac{\mathbf{V}_2}{\mathbf{V}_o} = \frac{1}{3 + j(oL/R - R/oL)}
-$$
-
-- (b) Determine the oscillation frequency *fo*.
-- (c) Obtain the relationship between *R*1 and *R*2 in order for oscillation to occur.
-
-**Figure 10.137** For Prob. 10.96.
-
-*This page intentionally left blank*
-
-# **chapter**
-
-11
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/122_Chapter 11 - AC Power Analysis.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/122_Chapter 11 - AC Power Analysis.md
deleted file mode 100644
index e6957e7fbfedbb997852a4cfbbdb671cfb2f1dca..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/122_Chapter 11 - AC Power Analysis.md
+++ /dev/null
@@ -1,71 +0,0 @@
-# AC Power Analysis
-
-*Four things come not back: the spoken word; the sped arrow; time past; the neglected opportunity.*
-
-—Al Halif Omar Ibn
-
-# Enhancing Your Career
-
-# **Career in Power Systems**
-
-The discovery of the principle of an ac generator by Michael Faraday in 1831 was a major breakthrough in engineering; it provided a convenient way of generating the electric po wer that is needed in e very electronic, electrical, or electromechanical device we use now.
-
-Electric power is obtained by converting energy from sources such as fossil fuels (gas, oil, and coal), nuclear fuel (uranium), h ydro energy (water falling through a head), geothermal energy (hot water, steam), wind energy, tidal ener gy, and biomass ener gy (wastes). These various ways of generating electric po wer are studied in detail in the field of power engineering, which has become an indispensable subdiscipline of electrical engineering. An electrical engineer should be f amiliar with the analysis, generation, transmission, distrib ution, and cost of electric power.
-
-The electric po wer industry is a v ery large employer of electrical engineers. The industry includes thousands of electric utility systems ranging from large, interconnected systems serving large regional areas to small power companies serving indi vidual communities or f actories. Due to the comple xity of the po wer industry, there are numerous elec trical engineering jobs in dif ferent areas of the industry: po wer plant (generation), transmission and distribution, maintenance, research, data acquisition and flow control, and management. Since electric po wer is used e verywhere, electric utility companies are e verywhere, of fering exciting training and steady emplo yment for men and w omen in thou sands of communities throughout the world.
-
-A pole-type transformer with a lowvoltage, three-wire distribution system. © Dennis Wise/Getty Images RF
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Fully understand instantaneous and average power.
-- 2. Understand the basics of maximum average power.
-- 3. Understand effective or rms values and how to calculate them and to understand their importance.
-- 4. Understand apparent power (complex power), power, and reactive power and power factor.
-- 5. Understand power factor correction and the importance of its use.
-
-# **11.1** Introduction
-
-Our effort in ac circuit analysis so f ar has been focused mainly on cal culating voltage and current. Our major concern in this chapter is power analysis.
-
-Power analysis is of paramount importance. Po wer is the most important quantity in electric utilities, electronic, and communication systems, because such systems in volve transmission of po wer from one point to another . Also, e very industrial and household electrical device—every fan, motor, lamp, pressing iron, TV, personal computer has a po wer rating that indicates ho w much po wer the equipment re quires; e xceeding the po wer rating can do permanent damage to an appliance. The most common form of electric po wer is 50- or 60-Hz ac power. The choice of ac o ver dc allowed high-voltage power transmission from the power generating plant to the consumer.
-
-We will be gin by defining and deriving *instantaneous power* and *average power*. We will then introduce other power concepts. As practical applications of these concepts, we will discuss how power is measured and reconsider how electric utility companies charge their customers.
-
-# **11.2** Instantaneous and Average Power
-
-As mentioned in Chapter 2, the *instantaneous power p*(*t*) absorbed by an element is the product of the instantaneous v oltage *v*(*t*) across the ele ment and the instantaneous current *i*(*t*) through it. Assuming the passive sign convention,
-
-$$
-p(t) = v(t)i(t)
-$$
- (11.1)
-
-The instantaneous power (in watts) is the power at any instant of time.
-
-It is the rate at which an element absorbs energy.
-
-Consider the general case of instantaneous po wer absorbed by an arbitrary combination of circuit elements under sinusoidal excitation, as
-
-We can also think of the instantaneous power as the power absorbed by the element at a specific instant of time. Instantaneous quantities are denoted by lowercase letters.
-
-shown in Fig. 11.1. Let the v oltage and current at the terminals of the circuit be
-
-$$
-v(t) = V_m \cos(\omega t + \theta_v)
-$$
- (11.2a)
-
-$$
-i(t) = I_m \cos(\omega t + \theta_i)
-$$
- (11.2b)
-
-where *Vm* and *Im* are the amplitudes (or peak values), and *θv* and *θi* are the phase angles of the voltage and current, respectively. The instantaneous power absorbed by the circuit is
-
-$$
-p(t) = v(t)i(t) = V_m I_m \cos(\omega t + \theta_v) \cos(\omega t + \theta_i)
-$$
- (11.3)
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/123_11.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/123_11.1 Introduction.md
deleted file mode 100644
index d2b7b2e79a57426df0522610c91766c7fe55a337..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/123_11.1 Introduction.md
+++ /dev/null
@@ -1,295 +0,0 @@
-We apply the trigonometric identity
-
-$$
-\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)] \tag{11.4}
-$$
-
-and express Eq. (11.3) as
-
-$$
-p(t) = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) + \frac{1}{2}V_m I_m \cos(2\omega t + \theta_v + \theta_i)
-$$
- (11.5)
-
-This shows us that the instantaneous power has two parts. The first part is constant or time independent. Its value depends on the phase dif ference between the voltage and the current. The second part is a sinusoidal function whose frequency is 2*ω*, which is twice the angular frequency of the voltage or current.
-
-A sketch of *p*(*t*) in Eq. (11.5) is shown in Fig. 11.2, where *T* = 2*π*∕*ω* is the period of v oltage or current. We observ e that *p*(*t*) is periodic, *p*(*t*) = *p*(*t* + *T*0), and has a period of *T*0 = *T*∕2, since its frequency is twice that of voltage or current. We also observe that *p*(*t*) is positive for some part of each c ycle and ne gative for the rest of the c ycle. When *p*(*t*) is positive, power is absorbed by the circuit. When *p*(*t*) is negative, power is absorbed by the source; that is, power is transferred from the circuit to the source. This is possible because of the storage elements (capacitors and inductors) in the circuit.
-
-**Figure 11.2** The instantaneous power *p*(*t*) entering a circuit.
-
-The instantaneous power changes with time and is therefore difficult to measure. The *average* power is more convenient to measure. In f act, the wattmeter, the instrument for measuring power, responds to average power.
-
-The average power, in watts, is the average of the instantaneous power over one period.
-
-# **Figure 11.1**
-
-Sinusoidal source and passive linear circuit.
-
-Thus, the average power is given by
-
-$$
-P = \frac{1}{T} \int_0^T p(t) \, dt \tag{11.6}
-$$
-
-Although Eq. (11.6) shows the averaging done over *T*, we would get the same result if we performed the integration over the actual period of *p* ( *t*) which is *T*0 = *T*∕2.
-
- Substituting *p* ( *t*) in Eq. (11.5) into Eq. (11.6) gives
-
-$$
-P = \frac{1}{T} \int_0^T \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) dt
-$$
-
-+ $\frac{1}{T} \int_0^T \frac{1}{2} V_m I_m \cos(2\omega t + \theta_v + \theta_i) dt$
-= $\frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) \frac{1}{T} \int_0^T dt$
-+ $\frac{1}{2} V_m I_m \frac{1}{T} \int_0^T \cos(2\omega t + \theta_v + \theta_i) dt$ (11.7)
-
-The first integrand is constant, and the average of a constant is the same constant. The second integrand is a sinusoid. We know that the average of a sinusoid over its period is zero because the area under the sinusoid during a positi ve half-cycle is canceled by the area under it during the following negative half-cycle. Thus, the second term in Eq. (11.7) v an ishes and the average power becomes
-
-$$
-P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i)
-$$
- (11.8)
-
-Since cos( *θv* − *θ i* ) = cos( *θi* − *θv*), what is important is the diference in the phases of the voltage and current.
-
-Note that *p* ( *t*) is time-varying while *P* does not depend on time. To find the instantaneous power, we must necessarily have *v* ( *t*) and *i* ( *t*) in the time domain. But we can find the average power when voltage and cur rent are expressed in the time domain, as in Eq. (11.8), or when they are expressed in the frequency domain. The phasor forms of *v* ( *t*) and *i* ( *t*) in Eq. (11.2) are **V** = *Vm* ⧸ *θv* and **I** = *Im* ⧸ *θ i*, respectively. *P* is calculated using Eq. (11.8) or using phasors **V** and **I**. To use phasors, we notice that
-
-$$
-\frac{1}{2}\mathbf{V}\mathbf{I}^* = \frac{1}{2}V_m I_m / \theta_v - \theta_i
-$$
-
-=
-$$
-\frac{1}{2}V_m I_m [\cos(\theta_v - \theta_i) + j \sin(\theta_v - \theta_i)]
-$$
- (11.9)
-
-We recognize the real part of this e xpression as the a verage power *P* according to Eq. (11.8). Thus,
-
-$$
-P = \frac{1}{2} \text{Re}[\mathbf{V} \mathbf{I}^*] = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i)
-$$
- (11.10)
-
-Consider two special cases of Eq. (11.10). When *θ v* = *θ i*, the voltage and current are in phase. This implies a purely resistive circuit or resis tive load *R*, and
-
-$$
-P = \frac{1}{2} V_m I_m = \frac{1}{2} I_m^2 R = \frac{1}{2} |\mathbf{I}|^2 R
-$$
- (11.11)
-
-where ∣**I**∣ 2 = **I** × **I**\*. Equation (11.11) shows that a purely resistive circuit absorbs power at all times. When *θv* − *θi* = ±90°, we have a purely reactive circuit, and
-
-$$
-P = \frac{1}{2} V_m I_m \cos 90^\circ = 0 \tag{11.12}
-$$
-
-showing that a purely reacti ve circuit absorbs no a verage po wer. In summary,
-
-A resistive load (R) absorbs power at all times, while a reactive load (L or C ) absorbs zero average power.
-
-$$
-v(t) = 120 \cos(377t + 45^{\circ})
-$$
- V and $i(t) = 10 \cos(377t - 10^{\circ})$ A
-
-find the instantaneous power and the a verage po wer absorbed by the passive linear network of Fig. 11.1.
-
-# **Solution:**
-
-The instantaneous power is given by
-
-$$
-p = vi = 1200 \cos(377t + 45^{\circ}) \cos(377t - 10^{\circ})
-$$
-
-Applying the trigonometric identity
-
-$$
-\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]
-$$
-
-gives
-
-$$
-p = 600[\cos(754t + 35^\circ) + \cos 55^\circ]
-$$
-
-or
-
-$$
-p(t) = 344.2 + 600 \cos(754t + 35^\circ)
-$$
- W
-
-The average power is
-
-$$
-P = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) = \frac{1}{2}120(10) \cos[45^\circ - (-10^\circ)]
-$$
-
-= 600 cos 55° = 344.2 W
-
-which is the constant part of *p*(*t*) above.
-
-Calculate the instantaneous power and average power absorbed by the passive linear network of Fig. 11.1 if Practice Problem 11.1
-
-*v*(*t*) = 330 cos(10*t* + 20°) V and *i*(*t*) = 33 sin(10*t* + 60°) A
-
-**Answer:** 3.5 + 5.445 cos(20*t* − 10°) kW, 3.5 kW.
-
-Calculate the average power absorbed by an impedance **Z** = 30 − *j*70 Ω Example 11.2 when a voltage **V** = 120⧸ 0**°** is applied across it.
-
-**Solution:**
-
-The current through the impedance is
-
-ent through the impedance is
-\n
-$$
-I = \frac{V}{Z} = \frac{120/0^{\circ}}{30 - j70} = \frac{120/0^{\circ}}{76.16/-66.8^{\circ}} = 1.576/66.8^{\circ} A
-$$
-
-Given that Example 11.1
-
-The average power is
-
-$$
-P = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) = \frac{1}{2}(120)(1.576)\cos(0 - 66.8^\circ) = 37.24 \text{ W}
-$$
-
-# Practice Problem 11.2
-
-A current **I** = 33⧸ 30**°** A flows through an impedance **Z** = 40⧸ −22° Ω. Find the average power delivered to the impedance.
-
-**Answer:** 20.19 kW.
-
-Example 11.3 For the circuit shown in Fig. 11.3, find the average power supplied by the source and the average power absorbed by the resistor.
-
-# **Solution:**
-
-The current **I** is given by
-
-1 is given by
-\n
-$$
-I = \frac{5/30^{\circ}}{4 - j2} = \frac{5/30^{\circ}}{4.472/-26.57^{\circ}} = 1.118/56.57^{\circ} A
-$$
-
-The average power supplied by the voltage source is
-
-$$
-P = \frac{1}{2}(5)(1.118)\cos(30^\circ - 56.57^\circ) = 2.5 \text{ W}
-$$
-
-The current through the resistor is
-
-$$
-I_R = I = 1.118 / 56.57^{\circ}
-$$
- A
-
-and the voltage across it is
-
-$$
-V_R = 4I_R = 4.472/56.57^\circ
-$$
- V
-
-The average power absorbed by the resistor is
-
-$$
-P = \frac{1}{2}(4.472)(1.118) = 2.5
-$$
- W
-
-which is the same as the average power supplied. Zero average power is absorbed by the capacitor.
-
-Practice Problem 11.3
-
-For Practice Prob. 11.3.
-
-In the circuit of Fig. 11.4, calculate the average power absorbed by the resistor and inductor. Find the average power supplied by the voltage source.
-
-**Answer:** 29.04 kW, 0 W, 29.04 kW.
-
-Determine the average power generated by each source and the average Example 11.4 power absorbed by each passive element in the circuit of Fig. 11.5(a).
-
-For Example 11.4.
-
-# **Solution:**
-
-We apply mesh analysis as shown in Fig. 11.5(b). For mesh 1,
-
-$$
-\mathbf{I}_1 = 4 \text{ A}
-$$
-
-For mesh 2,
-
-$$
-(j10 - j5)
-$$
-**I**2 - $j10$ **I**1 + 60/ $\cancel{30^{\circ}}$ = 0, **I**1 = 4 A
-
-or
-
-$$
-j5I_2 = -60/30^{\circ} + j40
-$$
- $\Rightarrow$ $I_2 = -12/-60^{\circ} + 8$
-= 10.58/79.1° A
-
-For the voltage source, the current flowing from it is **I**2 = 10.58⧸ 79.1**°** A and the voltage across it is 60⧸ 30**°** V, so that the average power is
-
-$$
-P_5 = \frac{1}{2} (60)(10.58) \cos(30^\circ - 79.1^\circ) = 207.8 \text{ W}
-$$
-
-Following the passive sign convention (see Fig. 1.8), this average power is absorbed by the source, in view of the direction of **I**2 and the polarity of the voltage source. That is, the circuit is delivering average power to the voltage source.
-
-For the current source, the current through it is **I**1 = 4⧸ 0**°** and the voltage across it is
-
-$$
-\mathbf{V}_1 = 20\mathbf{I}_1 + j10(\mathbf{I}_1 - \mathbf{I}_2) = 80 + j10(4 - 2 - j10.39)
-$$
-
-= 183.9 + j20 = 184.984/(6.21° V)
-
-The average power supplied by the current source is
-
-$$
-P_1 = \frac{1}{2} (184.984)(4) \cos(6.21^\circ - 0) = -367.8 \text{ W}
-$$
-
-It is negative according to the passive sign convention, meaning that the current source is supplying power to the circuit.
-
-For the resistor, the current through it is **I**1 = 4⧸ 0**°** and the voltage across it is 20**I**1 = 80⧸ 0**°**, so that the power absorbed by the resistor is
-
-$$
-P_2 = \frac{1}{2} (80)(4 \pm 160 \text{ W})
-$$
-
-For the capacitor, the current through it is **I**2 = 10.58⧸ 79.1**°** and the volt age across it is −*j*5**I**2 = (5⧸ −90**°**)(10.58⧸ 79.1**°**) = 52.9⧸ 79.1**°**− 90°. The average power absorbed by the capacitor is
-
-$$
-P_4 = \frac{1}{2} (52.9)(10.58)\cos(-90^\circ) = 0
-$$
-
-For the inductor, the current through it is **I**1 − **I**2 = 2 − *j*10.39 = 10.58⧸ −79.1**°**. The voltage across it is *j*10(**I**1 − **I**2) = 105.8⧸ −79.1**°** + 90°. Hence, the average power absorbed by the inductor is
-
-$$
-P_3 = \frac{1}{2} (105.8)(10.58) \text{ as } 90^\circ = 0
-$$
-
-Notice that the inductor and the capacitor absorb zero average power and that the total power supplied by the current source equals the power absorbed by the resistor and the voltage source, or
-
-*P*1 + *P*2 + *P*3 + *P*4 + *P*5 = −367.8 + 160 + 0 + 0 + 207.8 = 0
-
-indicating that power is conserved.
-
-Calculate the average power absorbed by each of the five elements in the circuit of Fig. 11.6. Practice Problem 11.4
-
-For Practice Prob. 11.4.
-
-**Answer:** 40-V Voltage source: −60 W; *j*20-V Voltage source: −40 W; resistor: 100 W; others: 0 W.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/124_11.3 Maximum Average Power Transfer.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/124_11.3 Maximum Average Power Transfer.md
deleted file mode 100644
index 56d9effb6b6d21aacfe9c2d070957dc92bd1a0cd..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/124_11.3 Maximum Average Power Transfer.md
+++ /dev/null
@@ -1,192 +0,0 @@
-# **11.3** Maximum Average Power Transfer
-
-In Section 4.8 we solv ed the problem of maximizing the po wer delivered by a power-supplying resistive network to a load *RL*. Representing the circuit by its Thevenin equi valent, we pro ved that the maximum power would be delivered to the load if the load resistance is equal to the Thevenin resistance *RL* = *R*Th. We now extend that result to ac circuits.
-
-Consider the circuit in Fig. 11.7, where an ac circuit is connected to a load **Z***L* and is represented by its Thevenin equivalent. The load is usually represented by an impedance, which may model an electric motor, an antenna, a TV, and so forth. In rectangular form, the Thevenin impedance **Z**Th and the load impedance **Z***L* are
-
-$$
-\mathbf{Z}_{\mathrm{Th}} = R_{\mathrm{Th}} + jX_{\mathrm{Th}} \tag{11.13a}
-$$
-
-$$
-\mathbf{Z}_L = R_L + jX_L \tag{11.13b}
-$$
-
-The current through the load is
-
-ough the load is
-\n
-$$
-I = \frac{V_{\text{Th}}}{Z_{\text{Th}} + Z_L} = \frac{V_{\text{Th}}}{(R_{\text{Th}} + jX_{\text{Th}}) + (R_L + jX_L)}
-$$
-\n(11.14)
-
-From Eq. (11.11), the average power delivered to the load is
-
-1), the average power delivered to the load is
-\n
-$$
-P = \frac{1}{2} |\mathbf{I}|^2 R_L = \frac{|\mathbf{V}_{\text{Th}}|^2 R_L/2}{(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2}
-$$
-\n(11.15)
-
-Our objective is to adjust the load parameters *RL* and *XL* so that *P* is maximum. To do this we set *∂P*∕*∂RL* and *∂P*∕*∂XL* equal to zero. From Eq. (11.15), we obtain
-
-num. To do this we set
-$$
-\partial P/\partial R_L
-$$
- and $\partial P/\partial X_L$ equal to zero. From
-1.15), we obtain
-$$
-\frac{\partial P}{\partial X_L} = -\frac{|\mathbf{V}_{\text{Th}}|^2 R_L (X_{\text{Th}} + X_L)}{[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2}
-$$
-(11.16a)
-
-$$
-\frac{\partial P}{\partial X_L} = -\frac{|\mathbf{V}_{\text{Th}}| \mathbf{r}_L(X_{\text{Th}} + X_L)}{[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2}
-$$
-(11.16a)
-$$
-\frac{\partial P}{\partial R_L} = \frac{|\mathbf{V}_{\text{Th}}|^2 [(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2 - 2R_L(R_{\text{Th}} + R_L)]}{2[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2}
-$$
-(11.16b)
-
-Setting *∂P*∕*∂XL* to zero gives
-
-$$
-X_L = -X_{\text{Th}} \tag{11.17}
-$$
-
-and setting *∂P*∕*∂RL* to zero results in
-
-gives us the maximum average power as
-
-ero results in
-\n
-$$
-R_L = \sqrt{R_{\text{Th}}^2 + (X_{\text{Th}} + X_L)^2}
-$$
-\n(11.18)
-
-Combining Eqs. (11.17) and (11.18) leads to the conclusion that for maximum average power transfer, **Z***L* must be selected so that *XL* = −*X*Th and *RL* = *R*Th, i.e.,
-
-$$
-Z_L = R_L + jX_L = R_{Th} - jX_{Th} = Z_{Th}^*
-$$
- (11.19)
-
-For maximum average power transfer, the load impedance **Z**L must be equal to the complex conjugate of the Thevenin impedance **Z**Th.
-
-This result is known as the *maximum average power transfer theorem* for the sinusoidal steady state. Setting *RL* = *R*Th and *XL* = −*X*Th in Eq. (11.15) When **Z**L = **Z**\*Th, we say that the load is matched to the source.
-
-In a situation in which the load is purely real, the condition for maximum power transfer is obtained from Eq. (11.18) by setting *XL* = 0; that is,
-
-*P*max = ∣**V**Th∣
-
-2 \_\_\_\_\_ 8*R*Th
-
-$$
-R_L = \sqrt{R_{\text{Th}}^2 + X_{\text{Th}}^2} = |\mathbf{Z}_{\text{Th}}|
-$$
- (11.21)
-
-**(11.20)**
-
-# **Figure 11.7** Finding the maximum average power
-
-transfer: (a) circuit with a load, (b) the Thevenin equivalent.
-
-This means that for maximum a verage power transfer to a purely resis tive load, the load impedance (or resistance) is equal to the magnitude of the Thevenin impedance.
-
-**Figure 11.8** For Example 11.5.
-
-Example 11.5 Determine the load impedance **Z***L* that maximizes the a verage po wer drawn from the circuit of Fig. 11.8. What is the maximum a verage power?
-
-# **Solution:**
-
-First we obtain the Thevenin equivalent at the load terminals. To get **Z**Th, consider the circuit shown in Fig. 11.9(a). We find
-
-$$
-\mathbf{Z}_{\text{Th}} = j5 + 4 || (8 - j6) = j5 + \frac{4(8 - j6)}{4 + 8 - j6} = 2.933 + j4.467 \ \Omega
-$$
-
-# **Figure 11.9**
-
-Finding the Thevenin equivalent of the circuit in Fig. 11.8.
-
-To find **V**Th, consider the circuit in Fig. 11.8(b). By voltage division,
-
-$$
-\mathbf{V}_{\text{Th}} = \frac{8 - j6}{4 + 8 - j6} (10) = 7.454 \underline{\text{/}} - 10.3^{\circ} \text{ V}
-$$
-
-The load impedance draws the maximum power from the circuit when
-
-$$
-\mathbf{Z}_L = \mathbf{Z}_{\text{Th}}^* = 2.933 - j4.467 \ \Omega
-$$
-
-According to Eq. (11.20), the maximum average power is
-
-$$
-P_{\text{max}} = \frac{|\mathbf{V}_{\text{Th}}|^2}{8R_{\text{Th}}} = \frac{(7.454)^2}{8(2.933)} = 2.368 \text{ W}
-$$
-
-For the circuit shown in Fig. 11.10, find the load impedance **Z***L* that absorbs the maximum average power. Calculate that maximum average power.
-
-8 Ω 5 Ω ‒j4 Ω j10 Ω **Z**L 12 A **Figure 11.10**
-
-Practice Problem 11.5
-
-For Practice Prob. 11.5.
-
-**Answer:** 3.415 − *j*0.7317 Ω, 51.47 W.
-
-In the circuit in Fig. 11.11, find the value of *RL* that will absorb the Example 11.6 maximum average power. Calculate that power.
-
-# **Solution:**
-
-We first find the Thevenin equivalent at the terminals of *RL*.
-
-rst find the Thevenin equivalent at the terminals of *RL*.
-**Z**Th = (40 − *j*30)
-$$
-||j20 = \frac{j20(40 - j30)}{j20 + 40 - j30} = 9.412 + j22.35 Ω
-$$
-
-By voltage division,
-
-division,
-\n
-$$
-\mathbf{V}_{\text{Th}} = \frac{j20}{j20 + 40 - j30} (150/30^{\circ}) = 72.76/134^{\circ} \text{ V}
-$$
-
-The value of *RL* that will absorb the maximum average power is
-
-$$
-V_L \text{ that will absorb the maximum average po}
-$$
-\n
-$$
-R_L = |\mathbf{Z}_{\text{Th}}| = \sqrt{9.412^2 + 22.35^2} = 24.25 \ \Omega
-$$
-
-The current through the load is
-
-t through the load is
-\n
-$$
-I = \frac{V_{\text{Th}}}{Z_{\text{Th}} + R_L} = \frac{72.76/134^{\circ}}{33.66 + j22.35} = 1.8/100.42^{\circ} \text{ A}
-$$
-
-The maximum average power absorbed by *RL* is
-
-$$
-P_{\text{max}} = \frac{1}{2} |\mathbf{I}|^2 R_L = \frac{1}{2} (1.8)^2 (24.25) = 39.29 \text{ W}
-$$
-
-In Fig. 11.12, the resistor *RL* is adjusted until it absorbs the maximum average power. Calculate *RL* and the maximum average power absorbed by it.
-
-**Answer:** 30 Ω, 23.06 W.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/125_11.4 Effective or RMS Value.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/125_11.4 Effective or RMS Value.md
deleted file mode 100644
index d4a552e5edd5bb560314c7171deeeb0c3fbd6a2c..0000000000000000000000000000000000000000
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@@ -1,164 +0,0 @@
-# **11.4** Effective or RMS Value
-
-The idea of *effective value* arises from the need to measure the efectiveness of a voltage or current source in delivering power to a resistive load.
-
-The effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.
-
-Practice Problem 11.6
-
-For Example 11.6.
-
-Finding the effective current: (a) ac circuit, (b) dc circuit.
-
-In Fig. 11.13, the circuit in (a) is ac while that of (b) is dc. Our objecti ve is to find *I*eff that will transfer the same po wer to resistor *R* as the sinusoid *i*. The average power absorbed by the resistor in the ac circuit is
-
-$$
-P = \frac{1}{T} \int_0^T i^2 R \, dt = \frac{R}{T} \int_0^T i^2 \, dt \tag{11.22}
-$$
-
-while the power absorbed by the resistor in the dc circuit is
-
-$$
-P = I_{\text{eff}}^2 R \tag{11.23}
-$$
-
-Equating the expressions in Eqs. (11.22) and (11.23) and solving for *I*eff, we obtain
-
-$$
-I_{\text{eff}} = \sqrt{\frac{1}{T} \int_0^T i^2 dt}
-$$
- (11.24)
-
-The effective value of the v oltage is found in the same w ay as current; that is,
-
-$$
-V_{\text{eff}} = \sqrt{\frac{1}{T} \int_0^T v^2 dt}
-$$
- (11.25)
-
-This indicates that the effective value is the (square) *root* of the *mean* (or average) of the *square* of the periodic signal. Thus, the effective value is often known as the *root-mean-square* value, or *rms* value for short; and we write
-
-$$
-I_{\rm eff} = I_{\rm rms}, \qquad V_{\rm eff} = V_{\rm rms} \tag{11.26}
-$$
-
-For any periodic function *x*(*t*) in general, the rms value is given by
-
-$$
-X_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T x^2 dt}
-$$
- (11.27)
-
-The effective value of a periodic signal is its root mean square (rms) value.
-
-Equation 11.27 states that to find the rms value of *x*(*t*), we first find its *square x*2 and then find the *mean* of that, or
-
-$$
-\frac{1}{T} \int_0^T x^2 dt
-$$
-
-and then the square *root* ( √ \_\_\_\_\_\_ ) of that mean. The rms value of a constant is the constant itself. For the sinusoid *i*(*t*) = *Im* cos *ωt*, the effective or rms value is
-
-$$
-I_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T I_m^2 \cos^2 \omega t \, dt}
-$$
-$$
-= \sqrt{\frac{I_m^2}{T} \int 0 \, T \, \frac{1}{2} (1 + \cos 2\omega t) \, dt} = \frac{I_m}{\sqrt{2}} \tag{11.28}
-$$
-
-Similarly, for *v*(*t*) = *Vm* cos *ωt*,
-
-$$
-V_{\rm rms} = \frac{V_m}{\sqrt{2}}\tag{11.29}
-$$
-
-Keep in mind that Eqs. (11.28) and (11.29) are only valid for sinusoidal signals.
-
-The average power in Eq. (11.8) can be written in terms of the rms values.
-
-$$
-P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) = \frac{V_m}{\sqrt{2}} \frac{I_m}{\sqrt{2}} \cos(\theta_v - \theta_i)
-$$
-
-= $V_{\text{rms}} I_{\text{rms}} \cos(\theta_v - \theta_i)$ (11.30)
-
-Similarly, the average power absorbed by a resistor *R* in Eq. (11.11) can be written as
-
-$$
-P = I_{\rm rms}^2 R = \frac{V_{\rm rms}^2}{R}
-$$
- (11.31)
-
-When a sinusoidal voltage or current is specified, it is often in terms of its maximum (or peak) v alue or its rms v alue, since its a verage value is zero. The power industries specify phasor magnitudes in terms of their rms values rather than peak v alues. For instance, the 110 V available at every household is the rms value of the voltage from the power company. It is convenient in power analysis to express voltage and current in their rms values. Also, analog v oltmeters and ammeters are designed to read directly the rms value of voltage and current, respectively.
-
-Determine the rms v alue of the current w aveform in Fig. 11.14. If the Example 11.7 current is passed through a 2- Ω resistor, find the average power absorbed by the resistor.
-
-# **Solution:**
-
-The period of the waveform is *T* = 4. Over a period, we can write the current waveform as
-
-$$
-i(t) = \begin{cases} 5t, & 0 < t < 2 \\ -10, & 2 < t < 4 \end{cases}
-$$
-
-The rms value is
-
-$$
-I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T i^2 dt} = \sqrt{\frac{1}{4} \left[ \int_0^2 (5t)^2 dt + \int_2^4 (-10)^2 dt \right]}
-$$
-$$
-= \sqrt{\frac{1}{4} \left[ 25 \frac{t^3}{3} \right]_0^2 + 100t \Big|_2^4} = \sqrt{\frac{1}{4} \left( \frac{200}{3} + 200 \right)} = 8.165 \text{ A}
-$$
-
-The power absorbed by a 2-Ω resistor is
-
-$$
-P = I_{\text{rms}}^2 R = (8.165)^2 (2) = 133.3 \text{ W}
-$$
-
-Find the rms value of the current waveform of Fig. 11.15. If the current flows through a 9-Ω resistor, calculate the average power absorbed by the resistor.
-
-**Answer:** 9.238 A, 768 W.
-
-**Figure 11.14** For Example 11.7.
-
-For Example 11.8.
-
-Example 11.8 The waveform shown in Fig. 11.16 is a half-w ave rectified sine wave. Find the rms value and the amount of average power dissipated in a 10-Ω resistor.
-
-# **Solution:**
-
-The period of the voltage waveform is *T* = 2*π*, and
-
-$$
-v(t) = \begin{cases} 10 \sin t, & 0 < t < \pi \\ 0, & \pi < t < 2\pi \end{cases}
-$$
-
-The rms value is obtained as
-
-$$
-V_{\text{rms}}^2 = \frac{1}{T} \int_0^T v^2(t) \, dt = \frac{1}{2\pi} \left[ \int_0^{\pi} (10 \sin t)^2 \, dt + \int_{\pi}^{2\pi} 0^2 \, dt \right]
-$$
-
-But sin2 *t* = \_\_1 2 (1 − cos 2*t*). Hence,
-
-$$
-V_{\text{rms}}^2 = \frac{1}{2\pi} \int_0^{\pi} \frac{100}{2} (1 - \cos 2t) dt = \frac{50}{2\pi} \left( t - \frac{\sin 2t}{2} \right) \Big|_0^{\pi}
-$$
-$$
-= \frac{50}{2\pi} \left( \pi - \frac{1}{2} \frac{\sin 2\pi - 0}{2} \right) = 25, \qquad V_{\text{rms}} = 5 \text{ V}
-$$
-The average power absorbed is
-$$
-\sqrt{\pi} \alpha \int_0^{\pi} \alpha \sqrt{1 - \frac{1}{2} \sin 2\pi} dt
-$$
-
-$$
-P = \frac{V_{\text{rms}}^2}{R} = \frac{5^2}{10} = 2.5 \text{ W}
-$$
-
-# Practice Problem 11.8
-
-Find the rms value of the full-wave rectified sine wave in Fig. 11.17. Calculate the average power dissipated in a 6-Ω resistor.
-
-**Answer:** 70.71 V, 833.3 W.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/126_11.5 Apparent Power and Power Factor.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/126_11.5 Apparent Power and Power Factor.md
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-# For Practice Prob. 11.8. **11.5** Apparent Power and Power Factor
-
-In Section 11.2 we saw that if the voltage and current at the terminals of a circuit are
-
-$$
-v(t) = V_m \cos(\omega t + \theta_v)
-$$
- and $i(t) = I_m \cos(\omega t + \theta_i)$ (11.32)
-
-or, in phasor form, **V** = *Vm*⧸*θv* and **I** = *Im*⧸*θi* , the average power is
-
-$$
-P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i)
-$$
- (11.33)
-
-In Section 11.4, we saw that
-
-$$
-P = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}) = S \cos(\theta_{\nu} - \theta_{i})
-$$
- (11.34)
-
-We have added a new term to the equation:
-
-$$
-S = V_{\rm rms} I_{\rm rms}
-$$
- (11.35)
-
-The average power is a product of two terms. The product *V*rms*I*rms is known as the *apparent power S* . The factor cos( *θv* − *θi*) is called the *power factor* (pf).
-
-The apparent power (in VA) is the product of the rms values of voltage and current.
-
-The apparent power is so called because it seems apparent that the power should be the v oltage-current product, by analogy with dc resisti ve circuits. It is measured in volt-amperes or VA to distinguish it from the average or real power, which is measured in watts. The power factor is dimensionless, since it is the ratio of the average power to the apparent power,
-
-$$
-pf = \frac{P}{S} = \cos(\theta_v - \theta_i)
-$$
- (11.36)
-
-The angle *θv* − *θi* is called the *power factor angle,* because it is the angle whose cosine is the power factor. The power factor angle is equal to the angle of the load impedance if **V** is the voltage across the load and **I** is the current through it. This is evident from the fact that
-
-$$
-Z = \frac{V}{I} = \frac{V_m/\theta_v}{I_m/\theta_i} = \frac{V_m}{I_m}/\theta_v - \theta_i
-$$
- (11.37)
-
-Alternatively, since
-
-$$
-\mathbf{V}_{\rm rms} = \frac{\mathbf{V}}{\sqrt{2}} = V_{\rm rms} \underline{\theta_v} \tag{11.38a}
-$$
-
-and
-
-$$
-\mathbf{I}_{\rm rms} = \frac{\mathbf{I}}{\sqrt{2}} = I_{\rm rms} / \theta_i \tag{11.38b}
-$$
-
-the impedance is
-
-$$
-Z = \frac{V}{I} = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{V_{\text{rms}}}{I_{\text{rms}}} \underbrace{\beta_{\nu} - \theta_{i}} \tag{11.39}
-$$
-
-The power factor is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of the load impedance.
-
-From Eq. (11.36), the power factor may be seen as that f actor by which the apparent power must be multiplied to obtain the real or average power. The value of pf ranges between zero and unity . For a purely resisti ve load, the voltage and current are in phase, so that *θv* − *θi* = 0 and pf = 1. This implies that the apparent po wer is equal to the a verage power. For a purely reactive load, *θv* − *θi* = ±90° and pf = 0. In this case the a verage power is zero. In between these tw o extreme cases, pf is said to be *leading* or *lagging*. Leading power factor means that current leads v oltage, which implies a capaciti ve load. Lagging po wer factor means that current lags voltage, implying an inductive load. Power factor affects the From Eq. (11.36), the power factor may also be regarded as the ratio of the real power dissipated in the load to the apparent power of the load.
-
-electric bills consumers pay the electric utility companies, as we will see in Section 11.9.2.
-
-Example 11.9 A series-connected load dra ws a current *i*(*t*) = 4 cos(100 *πt* + 10°) A when the applied v oltage is *v*(*t*) = 120 cos(100*πt* − 20°) V. Find the apparent power and the power factor of the load. Determine the element values that form the series-connected load.
-
-# **Solution:**
-
-The apparent power is
-
-$$
-S = V_{\text{rms}} I_{\text{rms}} = \frac{120}{\sqrt{2}} \frac{4}{\sqrt{2}} = 240 \text{ VA}
-$$
-
-The power factor is
-
-$$
-pf = \cos(\theta_v - \theta_i) = \cos(-20^\circ - 10^\circ) = 0.866 \quad \text{(leading)}
-$$
-
-The pf is leading because the current leads the voltage. The pf may also be obtained from the load impedance.
-
-$$
-\mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = \frac{120/-20^{\circ}}{4/10^{\circ}} = 30/-30^{\circ} = 25.98 - j15 \ \Omega
-$$
-\n
-$$
-\text{pf} = \cos(-30^{\circ}) = 0.866 \qquad \text{(leading)}
-$$
-
-The load impedance **Z** can be modeled by a 25.98-Ω resistor in series with a capacitor with
-
-$$
-X_C = -15 = -\frac{1}{\omega C}
-$$
-
-or
-
-$$
-C = \frac{1}{15\omega} = \frac{1}{15 \times 100\pi} = 212.2 \,\mu\text{F}
-$$
-
-| Practice Problem 11.9 | Obtain the power factor and the apparent power of a load whose |
-|-----------------------|------------------------------------------------------------------------------------------------|
-| | impedance is Z = 60 + j40 Ω when the applied voltage
is v(t) =
155.56 cos(377t + 10°) V. |
-| | |
-
-**Answer:** 0.8321 lagging, 167.69⧸ 33.69° VA.
-
-# **Solution:**
-
-The total impedance is
-
-$$
-\mathbf{Z} = 6 + 4 \left( (-j2) \right) = 6 + \frac{-j2 \times 4}{4 - j2} = 6.8 - j1.6 = 7 \underline{\text{/} -13.24^{\circ}} \,\Omega
-$$
-
-The power factor is
-
-$$
-pf = \cos(-13.24) = 0.9734 \text{ (leading)}
-$$
-
-since the impedance is capacitive. The rms value of the current is
-
-$$
-\mathbf{I}_{\rm rms} = \frac{\mathbf{V}_{\rm rms}}{\mathbf{Z}} = \frac{30/0^{\circ}}{7/-13.24^{\circ}} = 4.286/13.24^{\circ} \text{A}
-$$
-
-The average power supplied by the source is
-
-*P* = *V*rms*I*rmspf = (30)(4.286)0.9734 = 125 W
-
-or
-
-$$
-P = I_{\text{rms}}^2 R = (4.286)^2 (6.8) = 125 \text{ W}
-$$
-
-where *R* is the resistive part of **Z**.
-
-Calculate the power factor of the entire circuit of Fig. 11.19 as seen by the source. What is the average power supplied by the source?
-
-**Answer:** 0.936 lagging, 2.008 kW.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/127_11.6 Complex Power.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/127_11.6 Complex Power.md
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-# For Practice Prob. 11.10. **11.6** Complex Power
-
-Considerable effort has been expended over the years to express power relations as simply as possible. Po wer engineers have coined the term *complex power,* which they use to find the total effect of parallel loads. Complex power is important in po wer analysis because it contains *all* the information pertaining to the power absorbed by a given load.
-
-Consider the ac load in Fig. 11.20. Gi ven the phasor form **V** = *Vm*⧸*θv* and **I** = *Im*⧸*θi* of voltage *v*(*t*) and current *i*(*t*), the *complex power* **S** absorbed by the ac load is the product of the v oltage and the comple x conjugate of the current, or
-
-$$
-S = \frac{1}{2}VI^*
-$$
-\n(11.40)
-
-assuming the passi ve sign con vention (see Fig. 11.20). In terms of the rms values,
-
-$$
-S = V_{\rm rms}I_{\rm rms}^* \tag{11.41}
-$$
-
-where
-
-$$
-\mathbf{V}_{\rm rms} = \frac{\mathbf{V}}{\sqrt{2}} = V_{\rm rms} \underline{\theta_v} \tag{11.42}
-$$
-
-and
-
-$$
-\mathbf{I}_{\rm rms} = \frac{\mathbf{I}}{\sqrt{2}} = I_{\rm rms} / \theta_i \tag{11.43}
-$$
-
-10 Ω 8 Ω 165 0° V rms + j4 Ω ‒j6 Ω ‒ Practice Problem 11.10
-
-# **Figure 11.19**
-
-# **Figure 11.20**
-
-The voltage and current phasors associated with a load.
-
-When working with the rms values of currents or voltages, we may drop the subscript rms if no confusion will be caused by doing so.
-
-Thus, we may write Eq. (11.41) as
-
-$$
-\mathbf{S} = V_{\text{rms}} I_{\text{rms}} \underline{\beta_v - \theta_i}
-$$
-
-= $V_{\text{rms}} I_{\text{rms}} \cos(\theta_v - \theta_i) + j V_{\text{rms}} I_{\text{rms}} \sin(\theta_v - \theta_i)$ (11.44)
-
-This equation can also be obtained from Eq. (11.9). We notice from Eq. (11.44) that the magnitude of the complex power is the apparent power; hence, the comple x power is measured in v olt-amperes (VA). Also, we notice that the angle of the complex power is the power factor angle.
-
-The complex power may be expressed in terms of the load impedance **Z**. From Eq. (11.37), the load impedance **Z** may be written as
-
-$$
-Z = \frac{V}{I} = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{V_{\text{rms}}}{I_{\text{rms}}} \frac{\beta_v - \theta_i}{\beta}
-$$
-(11.45)
-
-Thus, **V**rms = **ZI**rms. Substituting this into Eq. (11.41) gives
-
-$$
-S = I_{\rm rms}^2 Z = \frac{V_{\rm rms}^2}{Z^*} = V_{\rm rms}I_{\rm rms}^*
-$$
- (11.46)
-
-Since **Z** = *R* + *jX*, Eq. (11.46) becomes
-
-$$
-S = I_{\text{rms}}^2(R + jX) = P + jQ \tag{11.47}
-$$
-
-where *P* and *Q* are the real and imaginary parts of the complex power; that is,
-
-$$
-P = \text{Re}(\mathbf{S}) = I_{\text{rms}}^2 R \tag{11.48}
-$$
-
-$$
-Q = \text{Im}(\mathbf{S}) = I_{\text{rms}}^2 X \tag{11.49}
-$$
-
-*P* is the a verage or real po wer and it depends on the load' s resistance *R*. *Q* depends on the load' s reactance *X* and is called the *reactive* (or quadrature) power.
-
-Comparing Eq. (11.44) with Eq. (11.47), we notice that
-
-$$
-P = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}), \qquad Q = V_{\text{rms}} I_{\text{rms}} \sin(\theta_{\nu} - \theta_{i}) \tag{11.50}
-$$
-
-The real power *P* is the a verage power in watts delivered to a load; it is the only useful po wer. It is the actual po wer dissipated by the load. The reactive power *Q* is a measure of the energy exchange between the source and the reactive part of the load. The unit of *Q* is the *volt-ampere reactive* (VAR) to distinguish it from the real po wer, whose unit is the w att. We know from Chapter 6 that ener gy storage elements neither dissipate nor supply power, but exchange power back and forth with the rest of the network. In the same w ay, the reactive power is being transferred back and forth between the load and the source. It represents a lossless interchange between the load and the source. Notice that:
-
-- 1. *Q* = 0 for resistive loads (unity pf).
-- 2. *Q* < 0 for capacitive loads (leading pf).
-- 3. *Q* > 0 for inductive loads (lagging pf).
-
-Thus,
-
-Complex power (in VA) is the product of the rms voltage phasor and the complex conjugate of the rms current phasor. As a complex quantity, its real part is real power P and its imaginary part is reactive power Q.
-
-Introducing the complex power enables us to obtain the real and reactive powers directly from voltage and current phasors.
-
-Complex Power =
-$$
-\mathbf{S} = P + jQ = \mathbf{V}_{\text{rms}}(\mathbf{I}_{\text{rms}})^*
-$$
-
-\n= $|\mathbf{V}_{\text{rms}}| |\mathbf{I}_{\text{rms}}| / \theta_v - \theta_i$
-\nApparent Power = $S = |\mathbf{S}| = |\mathbf{V}_{\text{rms}}| |\mathbf{I}_{\text{rms}}| = \sqrt{P^2 + Q^2}$
-\nReal Power = $P = \text{Re}(\mathbf{S}) = S \cos(\theta_v - \theta_i)$
-\nReactive Power = $Q = \text{Im}(\mathbf{S}) = S \sin(\theta_v - \theta_i)$
-\nPower Factor = $\frac{P}{S} = \cos(\theta_v - \theta_i)$
-
-This sho ws ho w the comple x po wer contains *all* the rele vant po wer information in a given load.
-
-It is a standard practice to represent **S**, *P*, and *Q* in the form of a triangle, known as the *power triangle,* shown in Fig. 11.21(a). This is similar to the impedance triangle sho wing the relationship between **Z**, *R*, and *X*, illustrated in Fig. 11.21(b). The power triangle has four items—the apparent/complex power, real power, reactive power, and the power factor angle. Given two of these items, the other two can easily be obtained from the triangle. As shown in Fig. 11.22, when **S** lies in the first quadrant, we have an inductive load and a lagging pf. When **S** lies in the fourth quadrant, the load is capacitive and the pf is leading. It is also possible for the comple x power to lie in the second or third quadrant. This requires that the load impedance ha ve a negative resistance, which is possible with active circuits.
-
-**S** contains all power information of a load. The real part of **S** is the real power P; its imaginary part is the reactive power Q; its magnitude is the apparent power S; and the cosine of its phase angle is the power factor pf.
-
-The voltage across a load is *v*(*t*) = 60 cos(*ωt* − 10°) V and the cur - Example 11.11 rent through the element in the direction of the v oltage drop is *i*(*t*) = 1.5 cos( *ωt* + 50°) A. Find: (a) the comple x and apparent po wers, (b) the real and reacti ve powers, and (c) the po wer factor and the load impedance.
-
-# **Solution:**
-
-(a) For the rms values of the voltage and current, we write
-
-$$
-\mathbf{V}_{\rm rms} = \frac{60}{\sqrt{2}} \angle 10^{\circ}, \qquad \mathbf{I}_{\rm rms} = \frac{1.5}{\sqrt{2}} \angle 50^{\circ}
-$$
-
-The complex power is
-
-$$
-\mathbf{S} = \mathbf{V}_{\rm rms} \mathbf{I}_{\rm rms}^* = \left(\frac{60}{\sqrt{2}} \angle 10^\circ \right) \left(\frac{1.5}{\sqrt{2}} \angle 50^\circ \right) = 45 \angle 60^\circ \text{ VA}
-$$
-
-The apparent power is
-
-$$
-S = |\mathbf{S}| = 45 \text{ VA}
-$$
-
-(b) We can express the complex power in rectangular form as
-
-$$
-S = 45 \underline{/-60^{\circ}} = 45 [\cos(-60^{\circ}) + j \sin(-60^{\circ})] = 22.5 - j38.97
-$$
-
-Since **S** = *P* + *jQ*, the real power is
-
-$$
-P = 22.5 \, \mathrm{W}
-$$
-
-while the reactive power is
-
-$$
-Q = -38.97
-$$
- **VAR**
-
-(c) The power factor is
-
-$$
-pf = \cos(-60^\circ) = 0.5 \text{ (leading)}
-$$
-
-It is leading, because the reactive power is negative. The load impedance is
-
-$$
-Z = \frac{V}{I} = \frac{60/-10^{\circ}}{1.5/+50^{\circ}} = 40/-60^{\circ} \,\Omega
-$$
-
-which is a capacitive impedance.
-
-For a load, **V**rms = 110⧸ 85° V, **I**rms = 3⧸ 15° A. Determine: (a) the complex and apparent powers, (b) the real and reactive powers, and (c) the power factor and the load impedance. Practice Problem 11.11
-
-> **Answer:** (a) 330 ⧸ 70° VA, 44 VA, (b) 112.87 W, 310.1 VAR, (c) 0.342 lagging, (12.541 + *j*34.46) Ω.
-
-Example 11.12 A load **Z** dra ws 12 kV A at a po wer f actor of 0.856 lagging from a 120-V rms sinusoidal source. Calculate: (a) the average and reactive powers delivered to the load, (b) the peak current, and (c) the load impedance.
-
-# **Solution:**
-
-(a) Given that pf = cos *θ* = 0.856, we obtain the power angle as *θ* = cos−1 0.856 = 31.13°. If the apparent power is *S* = 12,000 VA, then the average or real power is
-
-*P* = *S* cos *θ* = 12,000 × 0.856 = 10.272 kW
-
-while the reactive power is
-
-$$
-Q = S \sin \theta = 12,000 \times 0.517 = 6.204
-$$
-kVA
-
-(b) Since the pf is lagging, the complex power is
-
-$$
-S = P + jQ = 10.272 + j6.204
-$$
- kVA
-
-From **S** = **V**rms**I**\*rms, we obtain
-
-$$
-\text{Im } \mathbf{S} = \mathbf{V}_{\text{rms}} \mathbf{I}_{\text{rms}}^* \text{, we obtain}
-$$
-\n
-$$
-\mathbf{I}_{\text{rms}}^* = \frac{\mathbf{S}}{\mathbf{V}_{\text{rms}}} = \frac{10,272 + j6204}{120/0^{\circ}} = 85.6 + j51.7 \text{ A} = 100/31.13^{\circ} \text{ A}
-$$
-
-Thus **I**rms = 100⧸ −31.13° and the peak current is
-
-$$
-I_m = \sqrt{2}I_{\text{rms}} = \sqrt{2}(100) = 141.4 \text{ A}
-$$
-
-(c) The load impedance
-
-update
-
-\n
-$$
-\mathbf{Z} = \frac{\mathbf{V}_{\text{rms}}}{\mathbf{I}_{\text{rms}}} = \frac{120/0^{\circ}}{100/-31.13^{\circ}} = 1.2/31.13^{\circ} \ \Omega
-$$
-
-which is an inductive impedance.
-
-A sinusoidal source supplies 100 kVAR reactive power to load **Z** = 250⧸ −75° Ω. Determine: (a) the power factor, (b) the apparent power delivered to the load, and (c) the rms voltage.
-
-**Answer:** (a) 0.2588 leading, (b) 103.53 kVA, (c) 5.087 kV.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/128_11.7 Conservation of AC Power.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/128_11.7 Conservation of AC Power.md
deleted file mode 100644
index 5f2e257410ba9c117879c71a0ba80d2fc24c6d0c..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/128_11.7 Conservation of AC Power.md
+++ /dev/null
@@ -1,207 +0,0 @@
-# **11.7** Conservation of AC Power
-
-The principle of conservation of power applies to ac circuits as well as to dc circuits (see Section 1.5).
-
-To see this, consider the circuit in Fig. 11.23(a), where two load impedances **Z**1 and **Z**2 are connected in parallel across an ac source **V**. KCL gives
-
-$$
-\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 \tag{11.52}
-$$
-
-The complex power supplied by the source is (from now on, unless otherwise specified, all values of voltages and currents will be assumed to be rms values)
-
-$$
-S = VI^* = V(I_1^* + I_2^*) = VI_1^* + VI_2^* = S_1 + S_2 \qquad (11.53)
-$$
-
-11.3 and 11.4 that average power is conserved in ac circuits.
-
-In fact, we already saw in Examples
-
-Practice Problem 11.12
-
-An ac voltage source supplied loads connected in: (a) parallel, (b) series.
-
-where **S**1 and **S**2 denote the comple x powers delivered to loads **Z**1 and **Z**2, respectively.
-
-If the loads are connected in series with the voltage source, as shown in Fig. 11.23(b), KVL yields
-
-$$
-\mathbf{V} = \mathbf{V}_1 + \mathbf{V}_2 \tag{11.54}
-$$
-
-The complex power supplied by the source is
-
-$$
-S = VI^* = (V_1 + V_2)I^* = V_1I^* + V_2I^* = S_1 + S_2 \quad (11.55)
-$$
-
-where **S**1 and **S**2 denote the comple x powers delivered to loads **Z**1 and **Z**2, respectively.
-
-We conclude from Eqs. (11.53) and (11.55) that whether the loads are connected in series or in parallel (or in general), the total po wer *supplied* by the source equals the total power *delivered* to the load. Thus, in general, for a source connected to *N* loads,
-
-$$
-S = S_1 + S_2 + \dots + S_N \tag{11.56}
-$$
-
-This means that the total comple x power in a network is the sum of the complex powers of the individual components. (This is also true of real power and reactive power, but not true of apparent power.) This expresses the principle of conservation of ac power:
-
-The complex, real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads.
-
-From this we imply that the real (or reactive) power flow from sources in a network equals the real (or reactive) power flow into the other elements in the network.
-
-Example 11.13 Figure 11.24 sho ws a load being fed by a v oltage source through a transmission line. The impedance of the line is represented by the (4 + *j*2) Ω impedance and a return path. Find the real power and reactive power absorbed by: (a) the source, (b) the line, and (c) the load.
-
-# **Solution:**
-
-The total impedance is
-
-$$
-\mathbf{Z} = (4+j2) + (15-j10) = 19 - j8 = 20.62 \underline{\smash{\big)}\,22.83^\circ}
-$$
- $\Omega$
-
-In fact, all forms of ac power are conserved: instantaneous, real, reactive, and complex.
-
-The current through the circuit is
-
-through the circuit is
-\n
-$$
-\mathbf{I} = \frac{\mathbf{V}_s}{\mathbf{Z}} = \frac{220/0^{\circ}}{20.62/-22.83^{\circ}} = 10.67/22.83^{\circ} \text{ A rms}
-$$
-
-(a) For the source, the complex power is
-
-$$
-S_s = V_s I^* = (220/0^\circ)(10.67/-22.83^\circ)
-$$
-
-= 2347.4/-22.83° = (2163.5 - j910.8) VA
-
-From this, we obtain the real power as 2163.5 W and the reactive power as 910.8 VAR (leading).
-
-(b) For the line, the voltage is
-
-$$
-\mathbf{V}_{\text{line}} = (4 + j2)\mathbf{I} = (4.472 \underline{/ 26.57^{\circ}})(10.67 \underline{/ 22.83^{\circ}})
-$$
-$$
-= 47.72 \underline{/ 49.4^{\circ}} \text{ V rms}
-$$
-
-The complex power absorbed by the line is
-
-$$
-S_{line} = V_{line}I^* = (47.72/49.4^{\circ})(10.67/-22.83^{\circ})
-$$
-
-= 509.2/26.57° = 455.4 + j227.7 VA
-
-or
-
-$$
-S_{\text{line}} = |I|^2 Z_{\text{line}} = (10.67)^2 (4 + j2) = 455.4 + j227.7 VA
-$$
-
-That is, the real power is 455.4 W and the reactive power is 227.76 VAR (lagging).
-
-(c) For the load, the voltage is
-
-$$
-\mathbf{V}_L = (15 - j10)\mathbf{I} = (18.03 \text{/} - 33.7^{\circ})(10.67 \text{/} 22.83^{\circ})
-$$
-
-= 192.38 \text{/} - 10.87° V rms
-
-The complex power absorbed by the load is
-
-$$
-\mathbf{S}_L = \mathbf{V}_L \mathbf{I}^* = (192.38 \text{/} - 10.87^\circ)(10.67 \text{/} - 22.83^\circ)
-$$
-
-= 2053 \text{/} - 33.7^\circ = (1708 - j1139) VA
-
-The real power is 1708 W and the reactive power is 1139 VAR (leading). Note that **S***s* = **S**line + **S***L*, as expected. We have used the rms values of voltages and currents.
-
-In the circuit in Fig. 11.25, the 60- Ω resistor absorbs an average power of 240 W. Find **V** and the complex power of each branch of the circuit. What is the overall complex power of the circuit? (Assume the current through the 60-Ω resistor has no phase shift.)
-
-**Answer:** 240.7 ⧸ 21.45° V (rms); the 20- Ω resistor: 656 VA; the (30 − *j*10) Ω impedance: 480 − *j*160 VA; the (60 + *j*20) Ω impedance: 240 + *j*80 VA; overall: 1376 − *j*80 VA.
-
-Practice Problem 11.13
-
-For Practice Prob. 11.13.
-
-For Example 11.14.
-
-# **Solution:**
-
-The current through **Z**1 is
-
-$$
-\mathbf{I}_1 = \frac{\mathbf{V}}{\mathbf{Z}_1} = \frac{120/10^{\circ}}{60/-30^{\circ}} = 2/40^{\circ} \text{ A rms}
-$$
-
-while the current through **Z**2 is
-
-$$
-I_2 = \frac{V}{Z_2} = \frac{120/10^{\circ}}{40/45^{\circ}} = 3/-35^{\circ}
-$$
- A rms
-
-The complex powers absorbed by the impedances are
-
-$$
-\mathbf{S}_1 = \frac{V_{\text{rms}}^2}{\mathbf{Z}_1^*} = \frac{(120)^2}{60/30^\circ} = 240/-30^\circ = 207.85 - j120 \text{ VA}
-$$
-\n
-$$
-\mathbf{S}_2 = \frac{V_{\text{rms}}^2}{\mathbf{Z}_2^*} = \frac{(120)^2}{40/-45^\circ} = 360/45^\circ = 254.6 + j254.6 \text{ VA}
-$$
-
-The total complex power is
-
-$$
-S_t = S_1 + S_2 = 462.4 + j134.6 VA
-$$
-
-(a) The total apparent power is
-
-arent power is
-\n
-$$
-|\mathbf{S}_t| = \sqrt{462.4^2 + 134.6^2} = 481.6 \text{ VA}.
-$$
-
-(b) The total real power is
-
-$$
-P_t = \text{Re}(S_t) = 462.4 \text{ W or } P_t = P_1 + P_2.
-$$
-
-(c) The total reactive power is
-
-$$
-Q_t = \text{Im}(S_t) = 134.6 \text{ VAR or } Q_t = Q_1 + Q_2.
-$$
-
-(d) The pf = *Pt*∕∣**S***t*∣ = 462.4∕481.6 = 0.96 (lagging).
-
-We may cross check the result by finding the complex power **S***s* supplied by the source.
-
-$$
-\mathbf{I}_t = \mathbf{I}_1 + \mathbf{I}_2 = (1.532 + j1.286) + (2.457 - j1.721)
-$$
-
-= 4 - j0.435 = 4.024 $\underline{/$ -6.21° A rms
-$$
-\mathbf{S}_s = \mathbf{V}\mathbf{I}_t^* = (120/10°)(4.024/6.21°)
-$$
-
-= 482.88/16.21° = 463 + j135 VA
-
-which is the same as before.
-
-Two loads connected in parallel are respectively 3 kW at a pf of 0.75 leading and 6 kW at a pf of 0.95 lagging. Calculate the pf of the com bined two loads. Find the complex power supplied by the source. Practice Problem 11.14
-
-**Answer:** 0.9972 (leading), 9 − *j*0.6742 kVA.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/129_11.8 Power Factor Correction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/129_11.8 Power Factor Correction.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/129_11.8 Power Factor Correction.md
+++ /dev/null
@@ -1,109 +0,0 @@
-# **11.8** Power Factor Correction
-
-Most domestic loads (such as w ashing machines, air conditioners, and refrigerators) and industrial loads (such as induction motors) are induc tive and operate at a lo w lagging po wer factor. Although the inducti ve nature of the load cannot be changed, we can increase its power factor.
-
-The process of increasing the power factor without altering the voltage or current to the original load is known as power factor correction.
-
-Since most loads are inducti ve, as shown in Fig. 11.27(a), a load' s power factor is improved or corrected by deliberately installing a capacitor in parallel with the load, as shown in Fig. 11.27(b). The effect of adding the capacitor can be illustrated using either the power triangle or the phasor diagram of the currents in volved. Figure 11.28 sho ws the latter, where it is assumed that the circuit in Fig. 11.27(a) has a power factor of cos *θ*1, while the one in Fig. 11.27(b) has a power factor of cos *θ*2. It is evident from Fig. 11.28 that adding the capacitor has caused the phase angle between the supplied v oltage and current to reduce from *θ*1 to *θ*2, thereby increasing the power factor. We also notice from the magnitudes of the vectors in Fig. 11.28 that with the same supplied v oltage, the circuit in Fig. 11.27(a) dra ws larger current *IL* than the current *I* drawn by the circuit in Fig. 11.27(b). Power companies charge more for larger currents, because they result in increased power losses (by a squared factor, since *P* = *IL* 2 *R*). Therefore, it is beneficial to both the power company and the consumer that every effort is made to minimize current level or keep the power factor as close to unity as possible. By choosing a suitable size for the capacitor, the current can be made to be completely in phase with the voltage, implying unity power factor.
-
-Alternatively, power factor correction may be viewed as the addition of a reactive element (usually a capacitor) in parallel with the load in order to make the power factor closer to unity.
-
- An inductive load is modeled as a series combination of an inductor and a resistor.
-
-**Figure 11.27** Power factor correction: (a) original inductive load, (b) inductive load with improved power factor.
-
-We can look at the power factor correction from another perspective. Consider the power triangle in Fig. 11.29. If the original inducti ve load has apparent power *S*1, then
-
-$$
-P = S_1 \cos \theta_1
-$$
-, $Q_1 = S_1 \sin \theta_1 = P \tan \theta_1$ (11.57)
-
-**Figure 11.29** Power triangle illustrating power factor correction.
-
-If we desire to increase the po wer factor from cos *θ*1 to cos *θ*2 without altering the real power (i.e., *P* = *S*2 cos *θ*2), then the new reactive power is
-
-$$
-Q_2 = P \tan \theta_2 \tag{11.58}
-$$
-
-The reduction in the reacti ve power is caused by the shunt capacitor; that is,
-
-$$
-Q_C = Q_1 - Q_2 = P(\tan \theta_1 - \tan \theta_2)
-$$
- (11.59)
-
-But from Eq. (11.46), *QC* = *V*2 rms∕*XC* = *ωCV* 2 rms. The value of the required shunt capacitance *C* is determined as
-
-$$
-C = \frac{Q_C}{\omega V_{\text{rms}}^2} = \frac{P(\tan \theta_1 - \tan \theta_2)}{\omega V_{\text{rms}}^2}
-$$
- (11.60)
-
-Note that the real po wer *P* dissipated by the load is not af fected by the power factor correction because the a verage power due to the capaci tance is zero.
-
-Although the most common situation in practice is that of an inductive load, it is also possible that the load is capaciti ve; that is, the load is operating at a leading power factor. In this case, an inductor should be connected across the load for po wer factor correction. The required shunt inductance *L* can be calculated from
-
-$$
-Q_L = \frac{V_{\text{rms}}^2}{X_L} = \frac{V_{\text{rms}}^2}{\omega L} \qquad \Rightarrow \qquad L = \frac{V_{\text{rms}}^2}{\omega Q_L} \tag{11.61}
-$$
-
-where *QL* = *Q*1 − *Q*2, the dif ference between the ne w and old reacti ve powers.
-
-Example 11.15 When connected to a 120-V (rms), 60-Hz po wer line, a load absorbs 4 kW at a lagging po wer factor of 0.8. Find the v alue of capacitance necessary to raise the pf to 0.95.
-
-# **Solution:**
-
-If the pf = 0.8, then
-
-cos *θ*1 = 0.8 ⇒ *θ*1 = 36.87°
-
-where *θ*1 is the phase difference between voltage and current. We obtain the apparent power from the real power and the pf as
-
-$$
-S_1 = \frac{P}{\cos \theta_1} = \frac{4000}{0.8} = 5000 \text{ VA}
-$$
-
-The reactive power is
-
-$$
-Q_1 = S_1 \sin \theta = 5000 \sin 36.87 = 3000
-$$
- VAR
-
-When the pf is raised to 0.95,
-
-$$
-\cos \theta_2 = 0.95 \qquad \Rightarrow \qquad \theta_2 = 18.19^\circ
-$$
-
-The real power *P* has not changed. But the apparent power has changed; its new value is
-
-$$
-S_2 = \frac{P}{\cos \theta_2} = \frac{4000}{0.95} = 4210.5 \text{ VA}
-$$
-
-The new reactive power is
-
-$$
-Q_2 = S_2 \sin \theta_2 = 1314.4 \text{ VAR}
-$$
-
-The difference between the new and old reactive powers is due to the parallel addition of the capacitor to the load. The reactive power due to the capacitor is
-
-$$
-Q_C = Q_1 - Q_2 = 3000 - 1314.4 = 1685.6 \text{ VAR}
-$$
-
-and
-
-$$
-C = \frac{Q_C}{\omega V_{\text{rms}}^2} = \frac{1685.6}{2\pi \times 60 \times 120^2} = 310.5 \,\mu\text{F}
-$$
-
-*Note:* Capacitors are normally purchased for voltages they expect to see. In this case, the maximum voltage this capacitor will see is about 170 V peak. We would suggest purchasing a capacitor with a voltage rating equal to, say, 200 V.
-
-Find the value of parallel capacitance needed to correct a load of 140 kVAR at 0.85 lagging pf to unity pf. Assume that the load is sup plied by a 220-V (rms), 60-Hz line.
-
-**Answer:** 7.673 mF.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/130_11.9 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/130_11.9 Applications.md
deleted file mode 100644
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+++ /dev/null
@@ -1,176 +0,0 @@
-# **11.9** Applications
-
-In this section, we consider two important application areas: how power is measured and how electric utility companies determine the cost of electricity consumption.
-
-# **11.9.1** Power Measurement
-
-The average power absorbed by a load is measured by an instrument called the *wattmeter*.
-
-The wattmeter is the instrument used for measuring the average power.
-
-Figure 11.30 sho ws a w attmeter that consists essentially of tw o coils: the current coil and the voltage coil. A current coil with very low impedance (ideally zero) is connected in series with the load (Fig. 11.31) and responds to the load current. The voltage coil with very high impedance (ideally infinite) is connected in parallel with the load as shown in Fig. 11.31 and responds to the load v oltage. The current coil acts lik e a short circuit because of its low impedance; the voltage coil behaves like
-
-Reactive power is measured by an instrument called the varmeter. The varmeter is often connected to the load in the same way as the wattmeter.
-
- Some wattmeters do not have coils; the wattmeter considered here is the electromagnetic type.
-
-Practice Problem 11.15
-
-A wattmeter.
-
-**Figure 11.31** The wattmeter connected to the load.
-
-an open circuit because of its high impedance. As a result, the presence of the w attmeter does not disturb the circuit or ha ve an ef fect on the power measurement.
-
-When the two coils are energized, the mechanical inertia of the moving system produces a deflection angle that is proportional to the average value of the product *v*(*t*)*i*(*t*). If the current and voltage of the load are *v*(*t*) = *Vm* cos(*ωt* + *θv*) and *i*(*t*) = *Im* cos(*ωt* + *θi*), their corresponding rms phasors are
-
-$$
-\mathbf{V}_{\rm rms} = \frac{V_m}{\sqrt{2}} \underline{\theta_v} \quad \text{and} \quad \mathbf{I}_{\rm rms} = \frac{I_m}{\sqrt{2}} \underline{\theta_i} \quad (11.62)
-$$
-
-and the wattmeter measures the average power given by
-
-$$
-P = |\mathbf{V}_{\text{rms}}||\mathbf{I}_{\text{rms}}| \cos(\theta_{\nu} - \theta_{i}) = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}) \qquad (11.63)
-$$
-
-As shown in Fig. 11.31, each wattmeter coil has two terminals with one marked ±. To ensure upscale deflection, the ± terminal of the current coil is toward the source, while the ± terminal of the voltage coil is connected to the same line as the current coil. Re versing both coil connections still results in upscale deflection. However, reversing one coil and not the other results in downscale deflection and no wattmeter reading.
-
-# **Solution:**
-
-1. **Define.** The problem is clearly defined. Interestingly, this is a problem where the student could actually v alidate the results by doing the problem in the laboratory with a real wattmeter.
-
-- 2. **Present.** This problem consists of finding the average power delivered to a load by an external source with a series impedance.
-- 3. **Alternative.** This is a straightforward circuit problem where all we need to do is find the magnitude and phase of the current through the load and the magnitude and the phase of the voltage across the load. These quantities could also be found by using *PSpice*, which we will use as a check.
-- 4. **Attempt.** In Fig. 11.32, the w attmeter reads the average power absorbed by the (8 − *j*6) Ω impedance because the current coil is in series with the impedance while the v oltage coil is in parallel with it. The current through the circuit is
-
-The impedance while the voltage coil is in
-int through the circuit is
-$$
-I_{\rm rms} = \frac{150/0^{\circ}}{(12 + j10) + (8 - j6)} = \frac{150}{20 + j4} A
-$$
-
-The voltage across the (8 − *j*6) Ω impedance is
-
-$$
-\mathbf{V}_{\rm rms} = \mathbf{I}_{\rm rms}(8 - j6) = \frac{150(8 - j6)}{20 + j4} \text{ V}
-$$
-
-The complex power is
-
-$$
-\mathbf{S} = \mathbf{V}_{\text{rms}} \mathbf{I}_{\text{rms}}^* = \frac{150(8 - j6)}{20 + j4} \cdot \frac{150}{20 - j4} = \frac{150^2(8 - j6)}{20^2 + 4^2}
-$$
-$$
-= 423.7 - j324.6 \text{ VA}
-$$
-
-The wattmeter reads
-
-$$
-P = \text{Re}(S) = 432.7 \text{ W}
-$$
-
-5. **Evaluate.** We can check our results by using *PSpice*.
-
-To check our answer , all we need is the magnitude of the current (7.354 A) flowing through the load resistor:
-
-$$
-P = (I_L)^2 R = (7.354)^2 8 = 432.7 \text{ W}
-$$
-
-As expected, the answer does check!
-
-6. **Satisfactory?** We ha ve satisf actorily solv ed the problem and the results can now be presented as a solution to the problem.
-
-For Practice Prob. 11.16.
-
-**Answer:** 1.437 kW.
-
-# **11.9.2** Electricity Consumption Cost
-
-In Section 1.7, we considered a simplified model of the way the cost of electricity consumption is determined. But the concept of po wer factor was not included in the calculations. Now we consider the importance of power factor in electricity consumption cost.
-
-Loads with low power factors are costly to serve because they require large currents, as explained in Section 11.8. The ideal situation would be to draw minimum current from a supply so that *S* = *P*, *Q* = 0, and pf = 1. A load with nonzero *Q* means that energy flows back and forth between the load and the source, gi ving rise to additional po wer losses. In vie w of this, power companies often encourage their customers to have power factors as close to unity as possible and penalize some customers who do not improve their load power factors.
-
-Utility companies divide their customers into categories: as residential (domestic), commercial, and industrial, or as small po wer, medium power, and large power. They have different rate structures for each category. The amount of energy consumed in units of kilowatt-hours (kWh) is measured using a kilo watt-hour meter installed at the customer' s premises.
-
-Although utility companies use different methods for charging customers, the tarif f or char ge to a consumer is often tw o-part. The first part is fixed and corresponds to the cost of generation, transmission, and distribution of electricity to meet the load requirements of the con sumers. This part of the tarif f is generally e xpressed as a certain price
-
-per kW of maximum demand. Or it may be based on kVA of maximum demand, to account for the power factor (pf) of the consumer. A pf penalty charge may be imposed on the consumer whereby a certain percentage of kW or kVA maximum demand is charged for every 0.01 fall in pf below a prescribed value, say 0.85 or 0.9. On the other hand, a pf credit may be given for every 0.01 that the pf exceeds the prescribed value.
-
-The second part is proportional to the ener gy consumed in kWh; i t may be in graded form, for example, the first 100 kWh at 16 cents/kWh, the next 200 kWh at 10 cents/kWh and so forth. Thus, the bill is determined based on the following equation:
-
-Total Cost = Fixed Cost + Cost of Energy **(11.64)**
-
-A manufacturing industry consumes 200 MWh in one month. If the Example 11.17 maximum demand is 1,600 kW, calculate the electricity bill based on the following two-part rate:
-
-Demand charge: \$5.00 per month per kW of billing demand. Energy charge: 8 cents per kWh for the first 50,000 kWh, 5 cents per kWh for the remaining energy.
-
-# **Solution:**
-
-The demand charge is
-
-\$5.00 × 1,600 = \$8,000 **(11.17.1)**
-
-The energy charge for the first 50,000 kWh is
-
-$$
-$0.08 \times 50,000 = $4,000 \tag{11.17.2}
-$$
-
-The remaining energy is 200,000 kWh− 50,000 kWh = 150,000 kWh, and the corresponding energy charge is
-
-\$0.05 × 150,000 = \$7,500 **(11.17.3)**
-
-Adding the results of Eqs. (11.17.1) to (11.17.3) gives
-
-Total bill for the month = \$8,000 + \$4,000 + \$7,500 = \$19,500
-
-It may appear that the cost of electricity is too high. But this is often a small fraction of the overall cost of production of the goods manufactured or the selling price of the finished product.
-
-The monthly reading of a paper mill's meter is as follows:
-
-Maximum demand: 48,000 kW Energy consumed: 750 MWh
-
-Using the two-part rate in Example 11.17, calculate the monthly bill for the paper mill.
-
-**Answer:** \$279,000.
-
-Practice Problem 11.17
-
-Example 11.18 A 300-kW load supplied at 13 kV (rms) operates 520 hours a month at 80 percent power factor. Calculate the average cost per month based on this simplified tariff:
-
-Energy charge: 6 cents per kWh
-
-Power-factor penalty: 0.1 percent of energy charge for every 0.01 that pf falls below 0.85.
-
-Power-factor credit: 0.1 percent of energy charge for every 0.01 that pf exceeds 0.85.
-
-# **Solution:**
-
-The energy consumed is
-
-$$
-W = 300 \text{ kW} \times 520 \text{ h} = 156,000 \text{ kWh}
-$$
-
-The operating power factor pf = 80% = 0.8 is 5 × 0.01 below the prescribed power factor of 0.85. Since there is 0.1 percent energy charge for every 0.01, there is a power-factor penalty charge of 0.5 percent. This amounts to an energy charge of
-
-$$
-\Delta W = 156,000 \times \frac{5 \times 0.1}{100} = 780 \text{ kWh}
-$$
-
-The total energy is
-
-$$
-W_t = W + \Delta W = 156,000 + 780 = 156,780 \text{ kWh}
-$$
-
-The cost per month is given by
-
-Cost = 6 cents × *Wt* = \$0.06 × 156,780 = \$9,406.80
-
-An 500-kW induction furnace at 0.88 power factor operates 20 hours per day for 26 days in a month. Determine the electricity bill per month based on the tariff in Example 11.18. Practice Problem 11.18
-
-**Answer:** \$15,553.20.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/131_11.10 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/131_11.10 Summary.md
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+++ /dev/null
@@ -1,115 +0,0 @@
-# **11.10** Summary
-
-1. The instantaneous power absorbed by an element is the product of the element's terminal voltage and the current through the element:
-
-$$
-p = vi.
-$$
-
-2. Average or real po wer *P* (in w atts) is the a verage of instantaneous power *p*:
-
-$$
-P = \frac{1}{T} \int_0^T p \, dt
-$$
-
- If *v*(*t*) = *Vm*cos(*ωt* + *θv*) and *i*(*t*) = *Im* cos (*ωt* + *θi* ), then *V*rms = *Vm*∕√ 2 , *I*rms = *Im*∕ √ \_\_ 2 , and
-
-$$
-P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) = V_{\text{rms}} I_{\text{rms}} \cos(\theta_v - \theta_i)
-$$
-
- Inductors and capacitors absorb no a verage power, while the a verage power absorbed by a resistor is (1∕2)*Im* 2*R* = *I*rms 2 *R*.
-
-- 3. Maximum average power is transferred to a load when the load impedance is the comple x conjug ate of the Thevenin impedance as seen from the load terminals, **Z***L* = *Z*Th \* .
-- 4. The effective value of a periodic signal x(*t*) is its root-mean-square (rms) value.
-
-$$
-X_{\rm eff} = X_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T x^2 dt}
-$$
-
- For a sinusoid, the ef fective or rms v alue is its amplitude di vided by √ \_\_ 2 .
-
-5. The power factor is the cosine of the phase difference between voltage and current:
-
-$$
-\mathrm{pf} = \cos(\theta_v - \theta_i)
-$$
-
- It is also the cosine of the angle of the load impedance or the ratio of real power to apparent power. The pf is lagging if the current lags voltage (inductive load) and is leading when the current leads v oltage (capacitive load).
-
-6. Apparent power *S* (in VA) is the product of the rms values of voltage and current:
-
-$$
-S = V_{\rm rms} I_{\rm rms}
-$$
-
- It is also given by *S* = ∣**S**∣ = √ \_\_\_\_\_\_\_ *P*2 + *Q*2 , where *P* is the real power and *Q* is reactive power.
-
-7. Reactive power (in VAR) is:
-
-$$
-Q = \frac{1}{2} V_m I_m \sin(\theta_v - \theta_i) = V_{\text{rms}} I_{\text{rms}} \sin(\theta_v - \theta_i)
-$$
-
-8. Complex power **S** (in VA) is the product of the rms v oltage phasor and the complex conjugate of the rms current phasor . It is also the complex sum of real power *P* and reactive power *Q*.
-
-$$
-\mathbf{S} = \mathbf{V}_{\rm rms} \, \mathbf{I}_{\rm rms}^* = V_{\rm rms} I_{\rm rms} / \theta_{\rm v} - \theta_{\rm i} = P + jQ
-$$
-
-Also,
-
-$$
-\mathbf{S} = I_{\text{rms}}^2 \, \mathbf{Z} = \frac{V_{\text{rms}}^2}{\mathbf{Z}^*}
-$$
-
-- 9. The total comple x power in a netw ork is the sum of the comple x powers of the individual components. Total real power and reactive power are also, respectively, the sums of the individual real powers and the reactive powers, but the total apparent po wer is not calcu lated by the process.
-- 10. Power factor correction is necessary for economic reasons; it is the process of impro ving the po wer f actor of a load by reducing the overall reactive power.
-- 11. The wattmeter is the instrument for measuring the average power. Energy consumed is measured with a kilowatt-hour meter.
-
-# Review Questions
-
-**11.1** The average power absorbed by an inductor is zero.
-
-(a) True (b) False
-
-**11.2** The Thevenin impedance of a network seen from the load terminals is 80 + *j*55 Ω. For maximum power transfer, the load impedance must be:
-
-| (a) −80 + j55 Ω | (b) −80 − j55 Ω | | |
-|-----------------|-----------------|--|--|
-| (c) 80 − j55 Ω | (d) 80 + j55 Ω | | |
-
-**11.3** The amplitude of the voltage available in the 60-Hz, 120-V power outlet in your home is:
-
-| (a) 110 V | (b) 120 V | |
-|-----------|-----------|--|
-| (c) 170 V | (d) 210 V | |
-
-- **11.4** If the load impedance is 20 − *j*20, the power factor is
- - (a) ⧸−45*°* (b) 0 (c) 1
- - (d) 0.7071 (e) none of these
-- **11.5** A quantity that contains all the power information in a given load is the
- - (a) power factor (b) apparent power (c) average power (d) reactive power
- - (e) complex power
-- **11.6** Reactive power is measured in:
- - (a) watts (b) VA
- - (c) VAR (d) none of these
-- **11.7** In the power triangle shown in Fig. 11.34(a), the reactive power is:
- - (a) 1000 VAR leading (b) 1000 VAR lagging (c) 866 VAR leading (d) 866 VAR lagging
-
-# **Figure 11.34**
-
-For Review Questions 11.7 and 11.8.
-
-**11.8** For the power triangle in Fig. 11.34(b), the apparent power is: (a) 2000 VA (b) 1000 VAR
-
-| (c) 866 VAR | (d) 500 VAR |
-|-------------|-------------|
-
-**11.9** A source is connected to three loads **Z**1, **Z**2, and **Z**3 in parallel. Which of these is not true?
-
-| (a) P = P1 + P2 + P3 | (b) Q = Q1 + Q2 + Q3 |
-|----------------------|----------------------|
-| (c) S = S1 + S2 + S3 | (d) S = S1 + S2 + S3 |
-
-**11.10** The instrument for measuring average power is the:
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/132_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/132_Review Questions.md
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-| (a) voltmeter | (b) ammeter |
-|-------------------------|--------------|
-| (c) wattmeter | (d) varmeter |
-| (e) kilowatt-hour meter | |
-
-*Answers: 11.1a, 11.2c, 11.3c, 11.4d, 11.5e, 11.6c, 11.7d, 11.8a, 11.9c, 11.10c.*
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/133_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/133_Problems.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/133_Problems.md
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-# Problems1
-
-# Section 11.2 Instantaneous and Average Power
-
-- **11.1** If *v*(*t*) = 160 cos 50*t* V and *i*(*t*) = −33 sin (50*t* − 30°)A, calculate the instantaneous power and the average power.
-- **11.2** Given the circuit in Fig. 11.35, find the average power supplied or absorbed by each element.
-
-- **11.3** A load consists of a 60-Ω resistor in parallel with a 90-*μ*F capacitor. If the load is connected to a voltage source *vs*(*t*) = 160 cos 2000*t*, find the average power delivered to the load.
-- **11.4** Using Fig. 11.36, design a problem to help other students better understand instantaneous and average power.
-
-1Starting with problem 11.22, unless otherwise specified, assume that all values of currents and voltages are rms.
-
-## Problems **489**
-
-**11.5** ssuming that *vs* = 8 cos(2*t* − 40°) V in the circuit of Fig. 11.37, find the average power delivered to each of the passive elements.
-
-**Figure 11.37** For Prob. 11.5.
-
-**11.6** For the circuit in Fig. 11.38, *is* = 6 cos 103 *t* A. Find the average power absorbed by the 50-Ω resistor.
-
-**Figure 11.38** For Prob. 11.6.
-
-**11.7** Given the circuit of Fig. 11.39, find the average power absorbed by the 10-Ω resistor.
-
-For Prob. 11.7.
-
-**11.8** In the circuit of Fig. 11.40, determine the average power absorbed by the 40-Ω resistor.
-
-**11.9** For the op amp circuit in Fig. 11.41, **V***s* = 2⧸30*° V*. Find the average power absorbed by the 20-kΩ resistor.
-
-- For Prob. 11.9.
-- **11.10** In the op amp circuit in Fig. 11.42, find the total average power absorbed by the resistors.
-
-**Figure 11.42** For Prob. 11.10.
-
-**11.11** For the network in Fig. 11.43, assume that the port impedance is
-
-$$
-\mathbf{Z}_{ab} = \frac{R}{\sqrt{1 + \omega^2 R^2 C^2}} \sqrt{-\tan^{-1} \omega RC}
-$$
-
- Find the average power consumed by the network when *R* = 10 kΩ, *C* = 200 nF, and *i* = 33 sin(377*t* + 22°) mA.
-
-# **Figure 11.43** For Prob. 11.11.
-
-# Section 11.3 Maximum Average Power Transfer
-
-**11.12** For the circuit shown in Fig. 11.44, determine the load impedance *ZL* for maximum power transfer (to *ZL*). Calculate the maximum power absorbed by the load.
-
-# **Figure 11.44**
-
-For Prob. 11.12.
-
-- **11.13** The Thevenin impedance of a source is **Z**Th = 120 + *j*60 Ω, while the peak Thevenin voltage is **V**Th = 165 + *j*0 V. Determine the maximum available average power from the source.
-- **11.14** Using Fig. 11.45, design a problem to help other students better understand maximum average power transfer to a load *Z*.
-
-**Figure 11.45** For Prob. 11.14.
-
-**11.15** In the circuit of Fig. 11.46, find the value of **Z***L* that will absorb the maximum power and the value of the maximum power.
-
-For Prob. 11.15.
-
-**11.16** For the circuit in Fig. 11.47, find the value of **Z***L* that will receive the maximum power from the circuit. Then calculate the power delivered to the load **Z***L*.
-
-**11.17** Calculate the value of **Z***L* in the circuit of Fig. 11.48 in order for **Z***L* to receive maximum average power. ‒j3 Ω What is the maximum average power received by **Z***L*? 4Ω
-
-**Figure 11.48**
-
-For Prob. 11.17.
-
-**11.18** Find the value of **Z***L* in the circuit of Fig. 11.49 for maximum power transfer.
-
-**Figure 11.49**
-
-For Prob. 11.18.
-
-**11.19** The variable resistor *R* in the circuit of Fig. 11.50 is adjusted until it absorbs the maximum average power. Find *R* and the maximum average power absorbed.
-
-**Figure 11.50** For Prob. 11.19.
-
-**11.20** The load resistance *RL* in Fig. 11.51 is adjusted until it absorbs the maximum average power. Calculate the value of *RL* and the maximum average power.
-
-For Prob. 11.20.
-
-**11.21** Assuming that the load impedance is to be purely resistive, what load should be connected to terminals *a*-*b* of the circuits in Fig. 11.52 so that the maximum power is transferred to the load?
-
-**Figure 11.52** For Prob. 11.21.
-
-# Section 11.4 Effective or RMS Value
-
-**11.22** Find the rms value of the offset sine wave shown in Fig. 11.53.
-
-For Prob. 11.22.
-
-**11.23** Using Fig. 11.54, design a problem to help other students better understand how to find the rms value of a waveshape.
-
-- For Prob. 11.23.
-- **11.24** Determine the rms value of the waveform in Fig. 11.55.
-
-**Figure 11.55** For Prob. 11.24.
-
-**Figure 11.56** For Prob. 11.25.
-
-**11.26** Find the effective value of the voltage waveform in Fig. 11.57.
-
-For Prob. 11.26.
-
-**11.27** Calculate the rms value of the current waveform of Fig. 11.58.
-
-**11.28** Find the rms value of the voltage waveform of Fig. 11.59 as well as the average power absorbed by a 2-Ω resistor when the voltage is applied across the resistor.
-
-**11.29** Calculate the effective value of the current waveform in Fig. 11.60 and the average power delivered to a 12-Ω resistor when the current runs through the resistor.
-
-**Figure 11.60** For Prob. 11.29.
-
-**11.30** Compute the rms value of the waveform depicted in Fig. 11.61.
-
-**Figure 11.61**
-
-For Prob. 11.30.
-
-**11.31** Find the rms value of the signal shown in Fig. 11.62.
-
-**Figure 11.62** For Prob. 11.31.
-
-**11.32** Obtain the rms value of the current waveform shown in Fig. 11.63.
-
-**Figure 11.63** For Prob. 11.32.
-
-**11.33** Determine the rms value for the waveform in Fig. 11.64.
-
-**11.34** Find the effective value of *f*(*t*) defined in Fig. 11.65.
-
-**Figure 11.65** For Prob. 11.34.
-
-**11.35** One cycle of a periodic voltage waveform is depicted in Fig. 11.66. Find the effective value of the voltage. Note that the cycle starts at *t* = 0 and ends at *t* = 6 s.
-
-**Figure 11.66** For Prob. 11.35.
-
-**11.36** Calculate the rms value for each of the following functions:
-
-(a) *i*(*t*) = 10 A (b) *v*(*t*) = 4 + 3 cos 5*t* V (c) *i*(*t*) = 8 − 6 sin 2*t* A (d) *v*(*t*) = 5 sin *t* + 4 cos*t* V
-
-**11.37** Design a problem to help other students better understand how to determine the rms value of the sum of multiple currents.
-
-# Section 11.5 Apparent Power and Power Factor
-
-**11.38** For the power system in Fig. 11.67, find: (a) the average power, (b) the reactive power, (c) the power factor. Note that 440 V is an rms value.
-
-**11.39** An ac motor with impedance **Z***L* = 2 + *j*1.2 Ω is supplied by a 220-V, 60-Hz source. (a) Find pf, *P*, and *Q*. (b) Determine the capacitor required to be connected in parallel with the motor so that the power factor is corrected to unity.
-
-**11.40** Design a problem to help other students better understand apparent power and power factor.
-
-**11.41** Obtain the power factor for each of the circuits in Fig. 11.68. Specify each power factor as leading or lagging.
-
-# **Figure 11.68**
-
-For Prob. 11.41.
-
-# Section 11.6 Complex Power
-
-- **11.42** A 110-V rms, 60-Hz source is applied to a load impedance **Z**. The apparent power entering the load is 120 VA at a power factor of 0.707 lagging.
- - (a) Calculate the complex power.
- - (b) Find the rms current supplied to the load.
- - (c) Determine **Z**. (d) Assuming that **Z** = *R* + *jωL*, find the values of *R* and *L*.
-
-**11.43** Design a problem to help other students understand complex power.
-
-**11.44** Find the complex power delivered by *vs* to the network in Fig. 11.69. Let *vs* = 100 cos 2000*t* V.
-
-# **Figure 11.69** For Prob. 11.44.
-
-**11.45** The voltage across a load and the current through it are given by
-
-$$
-v(t) = 20 + 60 \cos 100t
-$$
-
-$$
-i(t) = 1 - 0.5 \sin 100t
-$$
- A
-
-Find:
-
-(a) the rms values of the voltage and of the current (b) the average power dissipated in the load
-
-**11.46** For the following voltage and current phasors, calculate the complex power, apparent power, real power, and reactive power. Specify whether the pf is leading or lagging.
-
-(a)
-$$
-V = 220/30^{\circ}
-$$
- V rms, $I = 0.5/60^{\circ}$ A rms
-
-(b)
-$$
-V = 250 \div 10^{\circ} \text{ V rms}
-$$
-,
-
-$$
-I = 6.2 \angle -25^{\circ}
-$$
- A rms
-
-(c)
-$$
-V = 120/0
-$$
-° V rms, $I = 2.4/15$ ° A rms
-
-- (d) **V** = 160⧸45*°* V rms, **I** = 8.5⧸90*°* A rms
-- **11.47** For each of the following cases, find the complex power, the average power, and the reactive power:
-
-(a)
-$$
-v(t) = 169.7 \sin(377t + 45^\circ)
-$$
- V,
- $i(t) = 5.657 \sin(377t)$ A
-
-(b)
-$$
-v(t) = 339.4 \sin (377t + 90^\circ)
-$$
- V,
-
-$$
-i(t) = 5.657 \sin (377t + 45^{\circ}) \text{ A}
-$$
-
-(c) V =
-$$
-900/90^{\circ}
-$$
- V rms, Z = $75/45^{\circ}$ $\Omega$
-
-(d)
-$$
-I = 100/60^{\circ}
-$$
- A rms, $Z = 50/60^{\circ}$ $\Omega$
-
-- **11.48** Determine the complex power for the following cases:
- - (a) *P* = 269 W, *Q* = 150 VAR (capacitive)
- - (b) *Q* = 2000 VAR, pf = 0.9 (leading)
-
-(c)
-$$
-S = 600
-$$
- VA, $Q = 450$ VAR (inductive)
-
-(d) *V*rms = 220 V, *P* = 1 kW,
-
-∣**Z**∣ = 40 Ω (inductive)
-
-**11.49** Find the complex power for the following cases:
-
-- (a) *P* = 4 kW, pf = 0.86 (lagging) (b) *S* = 2 kVA, *P* = 1.6 kW (capacitive) (c) **V**rms = 208⧸20*°* V, **I**rms = 6.5⧸−50*°* A (d) **V**rms = 120⧸30*°* V, **Z** = 40 + *j*60 Ω
-- **11.50** Obtain the overall impedance for the following cases:
- - (a) *P* = 1000 W, pf = 0.8 (leading), *V*rms = 220 V
- - (b) *P* = 1500 W, *Q* = 2000 VAR (inductive), *I*rms = 12 A
-
-(c)
-$$
-S = 4500/60^{\circ}
-$$
- VA, $V = 120/45^{\circ}$ V
-
-- **11.51** For the entire circuit in Fig. 11.70, calculate:
- - (a) the power factor
- - (b) the average power delivered by the source
- - (c) the reactive power
- - (d) the apparent power
- - (e) the complex power
-
-# **Figure 11.70**
-
-For Prob. 11.51.
-
-- **11.52** In the circuit of Fig. 11.71, device *A* receives 2 kW at 0.8 pf lagging, device *B* receives 3 kVA at 0.4 pf leading, while device *C* is inductive and consumes 1 kW and receives 500 VAR.
- - (a) Determine the power factor of the entire system.
- - (b) Find **I** given that **V***s* = 120⧸45*°* V rms.
-
-**Figure 11.71** For Prob. 11.52.
-
-- **11.53** In the circuit of Fig. 11.72, load *A* receives 4 kVA at 0.8 pf leading. Load *B* receives 2.4 kVA at 0.6 pf lagging. Box *C* is an inductive load that consumes 1 kW and receives 500 VAR.
- - (a) Determine **I**.
- - (b) Calculate the power factor of the combination.
-
-**Figure 11.72** For Prob. 11.53.
-
-# Section 11.7 Conservation of AC Power
-
-**11.54** For the network in Fig. 11.73, find the complex power absorbed by each element.
-
-# **Figure 11.73**
-
-For Prob. 11.54.
-
-**Figure 11.74**
-
-For Prob. 11.55.
-
-**11.56** Obtain the complex power delivered by the source in the circuit of Fig. 11.75.
-
-**Figure 11.75**
-
-For Prob. 11.56.
-
-For Prob. 11.57.
-
-**11.58** Obtain the complex power delivered to the 10-kΩ resistor in Fig. 11.77 below.
-
-- **11.59** Calculate the reactive power in the inductor and capacitor in the circuit of Fig. 11.78.
-- **Figure 11.78** 100 Ω 100 Ω j100 Ω 100 0° mA ‒j200 Ω 20 0°V + ‒
-
-For Prob. 11.59.
-
-**11.60** For the circuit in Fig. 11.79, find **V***o* and the input power factor.
-
-**Figure 11.79** For Prob. 11.60.
-
-**11.61** Given the circuit in Fig. 11.80, find *Io* and the overall
-
-**Figure 11.80** For Prob. 11.61.
-
-**11.62** For the circuit in Fig. 11.81, find **V***s*.
-
-complex power supplied.
-
-**11.63** Find **I***o* in the circuit of Fig. 11.82.
-
-**11.64** Determine **I***s* in the circuit of Fig. 11.83, if the voltage source supplies 6 kW and 1.2 kVAR (leading).
-
-# **Figure 11.83**
-
-For Prob. 11.64.
-
-**11.65** In the op amp circuit of Fig. 11.84, *vs* = 4 cos 104 *t* V. Find the average power delivered to the 50-kΩ resistor.
-
-For Prob. 11.65.
-
-**11.66** Obtain the average power absorbed by the 10-Ω resistor in the op amp circuit in Fig. 11.85.
-
-**Figure 11.85**
-
-For Prob. 11.66.
-
-**11.67** For the op amp circuit in Fig. 11.86, calculate:
-
-- (a) the complex power delivered by the voltage source
-- (b) the average power dissipated in the 10-Ω resistor
-
-**11.68** Compute the complex power supplied by the current source in the series *RLC* circuit in Fig. 11.87.
-
-# **Figure 11.87**
-
-For Prob. 11.68.
-
-# Section 11.8 Power Factor Correction
-
-**11.69** Refer to the circuit shown in Fig. 11.88.
-
-- (a) What is the power factor?
-- (b) What is the average power dissipated?
-- (c) What is the value of the capacitance that will give a unity power factor when connected to the load?
-
-# **Figure 11.88**
-
-For Prob. 11.69.
-
-- **11.70** Design a problem to help other students better understand power factor correction.
- - **11.71** Three loads are connected in parallel to a 120⧸0*°* V rms source. Load 1 absorbs 60 kVAR at pf = 0.85 lagging, load 2 absorbs 90 kW and 50 kVAR leading, and load 3 absorbs 100 kW at pf = 1. (a) Find the equivalent impedance. (b) Calculate the power factor of the parallel combination. (c) Determine the current supplied by the source.
- - **11.72** Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW at a 0.707 pf lagging. Determine: (a) the pf of the second load, (b) the parallel element required to correct the pf to 0.9 lagging for the two loads.
- - **11.73** A 240-V rms 60-Hz supply serves a load that is 10 kW (resistive), 15 kVAR (capacitive), and 22 kVAR (inductive). Find:
- - (a) the apparent power
- - (b) the current drawn from the supply
- - (c) the kVAR rating and capacitance required to improve the power factor to 0.96 lagging
- - (d) the current drawn from the supply under the new power-factor conditions
-
-For Prob. 11.67.
-
-- **11.74** A 120-V rms 60-Hz source supplies two loads connected in parallel, as shown in Fig. 11.89.
- - (a) Find the power factor of the parallel combination.
- - (b) Calculate the value of the capacitance connected in parallel that will raise the power factor to unity.
-
-**Figure 11.89** For Prob. 11.74.
-
-- **11.75** Consider the power system shown in Fig. 11.90. Calculate:
- - (a) the total complex power
- - (b) the power factor
- - (c) the parallel capacitance necessary to establish a unity power factor
-
-For Prob. 11.77.
-
-**11.78** Find the wattmeter reading of the circuit shown in Fig. 11.93.
-
-**11.79** Determine the wattmeter reading of the circuit in
-
-# **Figure 11.94**
-
-For Prob. 11.79.
-
-For Prob. 11.80.
-
-- **11.80** The circuit of Fig. 11.95 portrays a wattmeter connected into an ac network.
- - (a) Find the magnitude of the load current.
- - (b) Calculate the wattmeter reading.
-
-For Prob. 11.78.
-
-Fig. 11.94.
-
-# Section 11.9 Applications
-
-**11.76** Obtain the wattmeter reading of the circuit in Fig. 11.91.
-
-For Prob. 11.76.
-
-- **11.81** Design a problem to help other students better understand how to correct power factor to values other than unity.
-- **11.82** A 240-V rms 60-Hz source supplies a parallel combination of a 5-kW heater and a 30-kVA induction motor whose power factor is 0.82. Determine:
- - (a) the system apparent power
- - (b) the system reactive power
- - (c) the kVA rating of a capacitor required to adjust the system power factor to 0.9 lagging
- - (d) the value of the capacitor required
-- **11.83** Oscilloscope measurements indicate that the peak voltage across a load and the peak current through it are, respectively, 210⧸ 60*°* V and 8⧸25*°* A. Determine:
- - (a) the real power
- - (b) the apparent power
- - (c) the reactive power
- - (d) the power factor
-
-**11.84** A consumer has an annual consumption of 1200 MWh with a maximum demand of 2.4 MVA. The maximum demand charge is \$30 per kVA per annum, and the energy charge per kWh is 4 cents.
-
-(a) Determine the annual cost of energy.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/134_Comprehensive Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/134_Comprehensive Problems.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/134_Comprehensive Problems.md
+++ /dev/null
@@ -1,71 +0,0 @@
-# Comprehensive Problems
-
-- **11.86** A transmitter delivers maximum power to an antenna when the antenna is adjusted to represent a load of 75-Ω resistance in series with an inductance of 4 *μ*H. If the transmitter operates at 4.12 MHz, find its internal impedance.
-- **11.87** In a TV transmitter, a series circuit has an impedance of 3 kΩ and a total current of 50 mA. If the voltage across the resistor is 80 V, what is the power factor of the circuit?
-- **11.88** A certain electronic circuit is connected to a 110-V ac line. The root-mean-square value of the current drawn is 2 A, with a phase angle of 55°.
- - (a) Find the true power drawn by the circuit.
- - (b) Calculate the apparent power.
-
-**11.89** An industrial heater has a nameplate that reads:
-
-- 210 V 60 Hz 12 kVA 0.78 pf lagging Determine:
- - (a) the apparent and the complex power
- - (b) the impedance of the heater
-- **11.90** A 2000-kW turbine-generator of 0.85 power factor operates at the rated load. An additional load of 300 kW at 0.8 power factor is added.What kVAR \*
-
-- (b) Calculate the charge per kWh with a flat-rate tariff if the revenue to the utility company is to remain the same as for the two-part tariff.
-- **11.85** A regular household system of a single-phase threewire circuit allows the operation of both 120-V and 240-V, 60-Hz appliances. The household circuit is modeled as shown in Fig. 11.96. Calculate:
- - (a) the currents **I**1, **I**2, and I*n*
- - (b) the total complex power supplied
- - (c) the overall power factor of the circuit
-
-# **Figure 11.96**
-
-For Prob. 11.85.
-
-of capacitors is required to operate the turbinegenerator but keep it from being overloaded?
-
-**11.91** The nameplate of an electric motor has the following information:
-
-> Line voltage: 220 V rms Line current: 15 A rms Line frequency: 60 Hz Power: 2700 W
-
- Determine the power factor (lagging) of the motor. Find the value of the capacitance *C* that must be connected across the motor to raise the pf to unity.
-
-- **11.92** As shown in Fig. 11.97, a 550-V feeder line supplies an industrial plant consisting of a motor drawing 90 kW at 0.8 pf (inductive), a capacitor with a rating of 20 kVAR, and lighting drawing 10 kW.
- - (a) Calculate the total reactive power and apparent power absorbed by the plant.
- - (b) Determine the overall pf.
- - (c) Find the magnitude of the current in the feeder line.
-
-\* An asterisk indicates a challenging problem.
-
-- **11.93** A factory has the following four major loads:
- - A motor rated at 5 hp, 0.8 pf lagging (1hp = 0.7457 kW).
- - A heater rated at 1.2 kW, 1.0 pf.
- - Ten 120-W lightbulbs.
- - A synchronous motor rated at 1.6 kVAR, 0.6 pf leading.
- - (a) Calculate the total real and reactive power.
- - (b) Find the overall power factor.
-
-- (a) Calculate the cost of capacitors needed.
-- (b) Find the savings in substation capacity released.
-- (c) Are capacitors economical for releasing the amount of substation capacity?
-
-- (a) At what frequency is maximum power transferred to the speaker?
-- (b) If *Vs* = 4.6 V rms, how much power is delivered to the speaker at that frequency?
-
-# **Figure 11.98**
-
-For Prob. 11.95.
-
-- **1.96** A power amplifier has an output impedance of 40 + *j*8 Ω. It produces a no-load output voltage of 146 V at 300 Hz.
- - (a) Determine the impedance of the load that achieves maximum power transfer.
- - (b) Calculate the load power under this matching condition.
-- **1.97** A power transmission system is modeled as shown in Fig. 11.99. If **V***s* = 440⧸0*°* rms, find the average power absorbed by the load.
-
-**Figure 11.99** For Prob. 11.97.
-
-*This page intentionally left blank*
-
-# **chapter**
-
-# 12
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/135_Chapter 12 - Three-Phase Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/135_Chapter 12 - Three-Phase Circuits.md
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/135_Chapter 12 - Three-Phase Circuits.md
+++ /dev/null
@@ -1,31 +0,0 @@
-# Three-Phase Circuits
-
-*He who cannot forgive others breaks the bridge over which he must pass himself.*
-
-—G. Herbert
-
-# Enhancing Your Skills and Your Career
-
-# **ABET EC 2000 criteria (3.e), "an ability to identify, formulate, and solve engineering problems."**
-
-Developing and enhancing your "ability to identify , formulate, and solve engineering problems" is a primary focus of te xtbook. Following our six-step problem-solving process is the best w ay to practice this skill. Our recommendation is that you use this process whene ver possible. You may be pleased to learn that this process w orks well for nonengineering courses.
-
-# **ABET EC 2000 criteria (f), "an understanding of professional and ethical responsibility."**
-
-"An understanding of professional and ethical responsibility" is required of every engineer. To some e xtent, this understanding is v ery personal for each of us. Let us identify some pointers to help you de velop this understanding. One of my f avorite examples is that an engineer has the responsibility to answer what I call the "unasked question." For instance, assume that you own a car that has a problem with the transmission. In the process of selling that car, the prospective buyer asks you if there is a problem in the right-front wheel bearing. You answer no. However, as an engineer, you are required to inform the buyer that there is a problem with the transmission without being asked.
-
-Your responsibility both professionally and ethically is to perform in a manner that does not harm those around you and to whom you are responsible. Clearly, developing this capability will tak e time and ma turity on your part. I recommend practicing this by looking for profes sional and ethical components in your day-to-day activities.
-
-Photo by Charles Alexander
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Understand balanced three-phase voltages.
-- 2. Analyze balanced wye-wye circuits.
-- 3. Understand and analyze balanced wye-delta circuits.
-- 4. Analyze balanced delta-delta circuits.
-- 5. Understand and analyze balanced delta-wye circuits.
-- 6. Explain and analyze power in balanced three-phase circuits.
-- 7. Analyze unbalanced three-phase circuits.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/136_12.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/136_12.1 Introduction.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/136_12.1 Introduction.md
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@@ -1,23 +0,0 @@
-# **12.1** Introduction
-
-So far in this text, we have dealt with single-phase circuits. A singlephase ac power system consists of a generator connected through a pair of wires (a transmission line) to a load. Figure 12.1(a) depicts a singlephase two-wire system, where *Vp* is the rms magnitude of the source voltage and *ϕ* is the phase. What is more common in practice is a singlephase three-wire system, shown in Fig. 12.1(b). It contains two identical sources (equal magnitude and the same phase) that are connected to two loads by two outer wires and the neutral. For example, the normal household system is a single-phase three-wire system because the terminal voltages have the same magnitude and the same phase. Such a system allows the connection of both 120- and 240-V appliances.
-
-Historical note: Thomas Edison invented a three-wire system, using three wires instead of four.
-
-Circuits or systems in which the ac sources operate at the same fre quency but different phases are known as *polyphase*. Figure 12.2 shows a two-phase three-wire system, and Fig. 12.3 sho ws a three-phase fourwire system. As distinct from a single-phase system, a two-phase system is produced by a generator consisting of tw o coils placed perpendicular to each other so that the voltage generated by one lags the other by 90°. By the same token, a three-phase system is produced by a generator consisting of three sources having the same amplitude and frequency but out of phase with each other by 120°. Because the three-phase system is by far the most pre valent and most economical polyphase system, discus sion in this chapter is mainly on three-phase systems.
-
-**Figure 12.2** Two-phase three-wire system.
-
-Three-phase systems are important for at least three reasons. First, nearly all electric po wer is generated and distrib uted in three-phase,
-
-# Historical
-
-**Nikola Tesla** (1856–1943) was a Croatian-American engineer whose inventions—among them the induction motor and the first polyphase ac power system—greatly influenced the settlement of the ac versus dc debate in favor of ac. He was also responsible for the adoption of 60 Hz as the standard for ac power systems in the United States.
-
-Born in Austria-Hungary (now Croatia), to a clergyman, Tesla had an incredible memory and a keen affinity for mathematics. He moved to the United States in 1884 and first worked for Thomas Edison. At that time, the country was in the "battle of the currents" with George Westinghouse (1846–1914) promoting ac and Thomas Edison rigidly leading the dc forces. Tesla left Edison and joined Westinghouse be cause of his interest in ac. Through Westinghouse, Tesla gained the reputation and acceptance of his polyphase ac generation, transmission, and distribution system. He held 700 patents in his lifetime. His other inventions include high-voltage apparatus (the tesla coil) and a wireless transmission system. The unit of magnetic flux density, the tesla, was named in honor of him.
-
-Library of Congress [LC-USZ62-61761]
-
-at the operating frequenc y of 60 Hz (or *ω* = 377 rad/s) in the United States or 50 Hz (or *ω* = 314 rad/s) in some other parts of the w orld. When one-phase or tw o-phase inputs are required, the y are taken from the three-phase system rather than generated independently. Even when more than three phases are needed—such as in the aluminum industry , where 48 phases are required for melting purposes—the y can be pro vided by manipulating the three phases supplied. Second, the instanta neous power in a three-phase system can be constant (not pulsating), as we will see in Section 12.7. This results in uniform power transmission and less vibration of three-phase machines. Third, for the same amount of power, the three-phase system is more economical than the singlephase. The amount of wire required for a three-phase system is less than that required for an equivalent single-phase system.
-
-We begin with a discussion of balanced three-phase v oltages. Then we analyze each of the four possible configurations of balanced threephase systems. We also discuss the analysis of unbalanced three-phase systems. We learn how to use *PSpice for Windows* to analyze a balanced or unbalanced three-phase system. Finally, we apply the concepts developed in this chapter to three-phase po wer measurement and residential electrical wiring.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/137_12.2 Balanced Three-Phase Voltages.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/137_12.2 Balanced Three-Phase Voltages.md
deleted file mode 100644
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+++ /dev/null
@@ -1,301 +0,0 @@
-# **12.2** Balanced Three-Phase Voltages
-
-Three-phase voltages are often produced with a three-phase ac genera tor (or alternator) whose cross-sectional view is shown in Fig. 12.4. The generator basically consists of a rotating magnet (called the *rotor*) surrounded by a stationary winding (called the *stator*). Three separate windings or coils with terminals *a*-*a*′, *b*-*b*′, and *c*-*c*′ are physically placed 120° apart around the stator. Terminals *a* and *a*′, for example, stand for one of the ends of coils going into and the other end coming out of the
-
-**Figure 12.3** Three-phase four-wire system.
-
-A three-phase generator.
-
-**Figure 12.5** The generated voltages are 120° apart from each other.
-
-page. As the rotor rotates, its magnetic field "cuts" the flux from the three coils and induces voltages in the coils. Because the coils are placed 120° apart, the induced voltages in the coils are equal in magnitude but out of phase by 120° (Fig. 12.5). Since each coil can be regarded as a singlephase generator by itself, the three-phase generator can supply power to both single-phase and three-phase loads.
-
-A typical three-phase system consists of three voltage sources connected to loads by three or four wires (or transmission lines). (Threephase current sources are very scarce.) A three-phase system is equivalent to three single-phase circuits. The voltage sources can be either wyeconnected as shown in Fig. 12.6(a) or delta-connected as in Fig. 12.6(b).
-
-# **Figure 12.6**
-
-Three-phase voltage sources: (a) Y-connected source, (b) ∆-connected source.
-
-> Let us consider the wye-connected voltages in Fig. 12.6(a) for now. The voltages **V***an*, **V***bn*, and **V***cn* are respectively between lines *a*, *b*, and *c*, and the neutral line *n*. These voltages are called *phase voltages*. If the voltage sources have the same amplitude and frequency *ω* and are out of phase with each other by 120°, the voltages are said to be *balanced*. This implies that
-
-$$
-\mathbf{V}_{an} + \mathbf{V}_{bn} + \mathbf{V}_{cn} = 0 \tag{12.1}
-$$
-
-$$
-|\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}| \tag{12.2}
-$$
-
-Thus,
-
-As a common tradition in power systems, voltage and current in this chapter are in rms values unless otherwise stated.
-
-Balanced phase voltages are equal in magnitude and are out of phase with each other by 120°.
-
-Because the three-phase voltages are 120° out of phase with each other, there are tw o possible combinations. One possibility is sho wn in Fig. 12.7(a) and expressed mathematically as
-
-$$
-\mathbf{V}_{an} = V_p / \mathbf{0}^{\circ}
-$$
-\n
-$$
-\mathbf{V}_{bn} = V_p / -120^{\circ}
-$$
-\n
-$$
-\mathbf{V}_{cn} = V_p / -240^{\circ} = V_p / +120^{\circ}
-$$
-\n(12.3)
-
-where *Vp* is the effective or rms value of the phase voltages. This is known as the *abc sequence* or *positive sequence*. In this phase sequence, **V***an* leads **V***bn*, which in turn leads **V***cn*. This sequence is produced when the rotor in Fig. 12.4 rotates counterclockwise. The other possibility is shown in Fig. 12.7(b) and is given by
-
-$$
-\mathbf{V}_{an} = V_p \underline{/0^{\circ}}
-$$
-
-\n
-$$
-\mathbf{V}_{cn} = V_p \underline{/ -120^{\circ}}
-$$
-
-\n
-$$
-\mathbf{V}_{bn} = V_p \underline{/ -240^{\circ}} = V_p \underline{/ +120^{\circ}}
-$$
- (12.4)
-
-This is called the *acb sequence* or *negative sequence*. For this phase sequence, **V***an* leads **V***cn*, which in turn leads **V***bn*. The *acb* sequence is produced when the rotor in Fig. 12.4 rotates in the clockwise direction. It is easy to show that the voltages in Eqs. (12.3) or (12.4) satisfy Eqs. (12.1) and (12.2). For example, from Eq. (12.3),
-
-$$
-\mathbf{V}_{an} + \mathbf{V}_{bn} + \mathbf{V}_{cn} = V_p \underline{\hspace{0.3cm}} \left( 0^{\circ} + V_p \underline{\hspace{0.3cm}} \right) + V_p \underline{\hspace{0.3cm}} \left( 120^{\circ} + V_p \underline{\hspace{0.3cm}} \right)
-$$
-\n
-$$
-= V_p (1.0 - 0.5 - j0.866 - 0.5 + j0.866) \quad (12.5)
-$$
-\n
-$$
-= 0
-$$
-
-The phase sequence is the time order in which the voltages pass through their respective maximum values.
-
-The phase sequence is determined by the order in which the phasors pass through a fixed point in the phase diagram.
-
-In Fig. 12.7(a), as the phasors rotate in the counterclockwise direction with frequenc y *ω*, the y pass through the horizontal axis in a se quence *abcabca* . . . . Thus, the sequence is *abc* or *bca* or *cab*. Similarly, for the phasors in Fig. 12.7(b), as the y rotate in the counterclockwise direction, they pass the horizontal axis in a sequence *acbacba* . . . . This describes the *acb* sequence. The phase sequence is important in threephase power distribution. It determines the direction of the rotation of a motor connected to the power source, for example.
-
-Like the generator connections, a three-phase load can be either wye-connected or delta-connected, depending on the end application. Figure 12.8(a) sho ws a wye-connected load, and Fig. 12.8(b) sho ws a delta-connected load. The neutral line in Fig. 12.8(a) may or may not be there, depending on whether the system is four - or three-wire. (And, of course, a neutral connection is topologically impossible for a delta con nection.) A wye- or delta-connected load is said to be *unbalanced* if the phase impedances are not equal in magnitude or phase.
-
-**Figure 12.7** Phase sequences: (a) *abc* or positive sequence, (b) *acb* or negative sequence.
-
- The phase sequence may also be regarded as the order in which the phase voltages reach their peak (or maximum) values with respect to time.
-
-Reminder: As time increases, each phasor (or sinor) rotates at an angular velocity *ω*.
-
-# **Figure 12.8**
-
-Two possible three-phase load configurations: (a) a Y-connected load, (b) a ∆-connected load.
-
-A balanced load is one in which the phase impedances are equal in magnitude and in phase.
-
-For a *balanced* wye-connected load,
-
-$$
-\mathbf{Z}_1 = \mathbf{Z}_2 = \mathbf{Z}_3 = \mathbf{Z}_Y \tag{12.6}
-$$
-
-where **Z***Y* is the load impedance per phase. For a *balanced* delta-connected
-
-$$
-\mathbf{Z}_a = \mathbf{Z}_b = \mathbf{Z}_c = \mathbf{Z}_{\Delta} \tag{12.7}
-$$
-
-$$
-\mathbf{Z}_{\Delta} = 3\mathbf{Z}_{Y} \qquad \text{or} \qquad \mathbf{Z}_{Y} = \frac{1}{3}\mathbf{Z}_{\Delta} \qquad (12.8)
-$$
-
-Because both the three-phase source and the three-phase load can be either wye- or delta-connected, we have four possible connections:
-
-- Y-Y connection (i.e., Y-connected source with a Y-connected load).
-- Y-∆ connection.
-
-load,
-
-Eq. (9.69) that
-
-- ∆-∆ connection.
-- ∆-Y connection.
-
-In subsequent sections, we will consider each of these possible con figurations.
-
-It is appropriate to mention here that a balanced delta-connected load is more common than a balanced wye-connected load. This is due to the ease with which loads may be added or remo ved from each phase of a deltaconnected load. This is very difficult with a wye-connected load because the neutral may not be accessible. On the other hand, delta-connected sources are not common in practice because of the circulating current that will result in the delta-mesh if the three-phase voltages are slightly unbalanced.
-
-Example 12.1 Determine the phase sequence of the set of voltages
-
-$$
-v_{an} = 200 \cos(\omega t + 10^{\circ})
-$$
-
-*vbn* = 200 cos(*ωt* − 230°), *vcn* = 200 cos(*ωt* − 110°)
-
-# **Solution:**
-
-The voltages can be expressed in phasor form as
-
-$$
-\mathbf{V}_{an} = 200 \underline{10^{\circ}} \text{ V}, \qquad \mathbf{V}_{bn} = 200 \underline{100^{\circ}} \text{ V}, \qquad \mathbf{V}_{cn} = 200 \underline{110^{\circ}} \text{ V}
-$$
-
-We notice that **V***an* leads **V***cn* by 120° and **V***cn* in turn leads **V***bn* by 120°. Hence, we have an *acb* sequence.
-
-Practice Problem 12.1 Given that **V***bn* = 220⧸ 30° V, find **V***an* and **V***cn*, assuming a positive (*abc*) sequence.
-
-**Answer:** 220⧸150° V, 220⧸−90° V.
-
-# **12.3** Balanced Wye-Wye Connection
-
-We begin with the Y-Y system, because any balanced three-phase sys tem can be reduced to an equivalent Y-Y system. Therefore, analysis of this system should be regarded as the key to solving all balanced threephase systems.
-
-A balanced Y-Y system is a three-phase system with a balanced Y-connected source and a balanced Y-connected load.
-
-Consider the balanced four-wire Y-Y system of Fig. 12.9, where a Y-connected load is connected to a Y-connected source. We assume a balanced load so that load impedances are equal. Although the impedance **Z***Y* is the total load impedance per phase, it may also be re garded as the sum of the source impedance **Z***s*, line impedance **Z***ℓ*, and load impedance **Z***L* for each phase, since these impedances are in series. As illustrated in Fig. 12.9, **Z***s* denotes the internal impedance of the phase winding of the generator; **Z***ℓ* is the impedance of the line join ing a phase of the source with a phase of the load; **Z***L* is the impedance of each phase of the load; and **Z***n* is the impedance of the neutral line. Thus, in general
-
-$$
-\mathbf{Z}_{Y} = \mathbf{Z}_{s} + \mathbf{Z}_{\ell} + \mathbf{Z}_{L} \tag{12.9}
-$$
-
-A balanced Y-Y system, showing the source, line, and load impedances.
-
-**Z***s* and **Z***ℓ* are often very small compared with **Z***L*, so one can assume that **Z***Y* = **Z***L* if no source or line impedance is given. In any event, by lump ing the impedances together, the Y-Y system in Fig. 12.9 can be simplified to that shown in Fig. 12.10.
-
-Assuming the positi ve sequence, the *phase* v oltages (or line-toneutral voltages) are
-
-$$
-\mathbf{V}_{an} = V_p / \underline{\mathbf{0}^{\circ}}
-$$
-
-$$
-\mathbf{V}_{bn} = V_p / \underline{-120^{\circ}}, \qquad \mathbf{V}_{cn} = V_p / \underline{+120^{\circ}}
-$$
-(12.10)
-
-**Figure 12.10** Balanced Y-Y connection.
-
-The *line-to-line* voltages or simply *line* voltages **V***ab*, **V***bc*, and **V***ca* are related to the phase voltages. For example,
-
-$$
-\mathbf{V}_{ab} = \mathbf{V}_{an} + \mathbf{V}_{nb} = \mathbf{V}_{an} - \mathbf{V}_{bn} = V_p \left( \frac{0^\circ}{\rho} - V_p \right) - 120^\circ
-$$
-
-= $V_p \left( 1 + \frac{1}{2} + j \frac{\sqrt{3}}{2} \right) = \sqrt{3} V_p / 30^\circ$ (12.11a)
-
-Similarly, we can obtain
-
-$$
-\mathbf{V}_{bc} = \mathbf{V}_{bn} - \mathbf{V}_{cn} = \sqrt{3} V_p \sqrt{-90^\circ}
-$$
- (12.11b)
-
-$$
-\mathbf{V}_{ca} = \mathbf{V}_{cn} - \mathbf{V}_{an} = \sqrt{3} V_p \underline{/-210^\circ}
-$$
- (12.11c)
-
-Thus, the magnitude of the line voltages *VL* is √ 3 times the magnitude of the phase voltages *Vp*, or
-
-$$
-V_L = \sqrt{3} V_P \tag{12.12}
-$$
-
-where
-
-$$
-V_p = |\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}| \tag{12.13}
-$$
-
-and
-
-$$
-V_L = |\mathbf{V}_{ab}| = |\mathbf{V}_{bc}| = |\mathbf{V}_{ca}| \tag{12.14}
-$$
-
-Also the line voltages lead their corresponding phase voltages by 30°. Figure 12.11(a) illustrates this. Figure 12.11(a) also shows how to determine **V***ab* from the phase voltages, while Fig. 12.11(b) shows the same for the three line voltages. Notice that **V***ab* leads **V***bc* by 120°, and **V***bc* leads **V***ca* by 120°, so that the line voltages sum up to zero as do the phase voltages.
-
-Applying KVL to each phase in Fig. 12.10, we obtain the line cur rents as
-
-$$
-\mathbf{I}_{a} = \frac{\mathbf{V}_{an}}{\mathbf{Z}_{Y}}, \qquad \mathbf{I}_{b} = \frac{\mathbf{V}_{bn}}{\mathbf{Z}_{Y}} = \frac{\mathbf{V}_{an} / -120^{\circ}}{\mathbf{Z}_{Y}} = \mathbf{I}_{a} / -120^{\circ} \tag{12.15}
-$$
-\n
-$$
-\mathbf{I}_{c} = \frac{\mathbf{V}_{cn}}{\mathbf{Z}_{Y}} = \frac{\mathbf{V}_{an} / -240^{\circ}}{\mathbf{Z}_{Y}} = \mathbf{I}_{a} / -240^{\circ}
-$$
-
-We can readily infer that the line currents add up to zero,
-
-$$
-\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c = 0 \tag{12.16}
-$$
-
-so that
-
-$$
-\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) = 0 \tag{12.17a}
-$$
-
-or
-
-$$
-\mathbf{V}_{nN} = \mathbf{Z}_n \mathbf{I}_n = 0 \tag{12.17b}
-$$
-
-that is, the voltage across the neutral wire is zero. The neutral line can thus be removed without affecting the system. In fact, in long distance power transmission, conductors in multiples of three are used with the earth itself acting as the neutral conductor. Power systems designed in this way are well grounded at all critical points to ensure safety.
-
-While the *line* current is the current in each line, the *phase* current is the current in each phase of the source or load. In the Y-Y system, the line current is the same as the phase current. We will use single subscripts
-
-**V**nb **V**ab = **V**an + **V**nb
-
-**V**cn
-
-**V**bn
-
-Phasor diagrams illustrating the relationship between line voltages and phase voltages.
-
-(b)
-
-**V**bc
-
-From **I***a*, we use the phase sequence to obtain other line currents. Thus, as long as the system is balanced, we need only analyze one phase. We may do this even if the neutral line is absent, as in the three-wire system.
-
-Calculate the line currents in the three-wire Y-Y system of Fig. 12.13. Example 12.2
-
-**Figure 12.13**
-
-Three-wire Y-Y system; for Example 12.2.
-
-# **Solution:**
-
-The three-phase circuit in Fig. 12.13 is balanced; we may replace it with its single-phase equivalent circuit such as in Fig. 12.12. We obtain **I***a* from the single-phase analysis as
-
-$$
-\mathbf{I}_a = \frac{\mathbf{V}_{an}}{\mathbf{Z}_Y}
-$$
-
-where **Z***Y* = (5 − *j*2) + (10 + *j*8) = 15 + *j*6 = 16.155⧸21.8°. Hence,
-
-$$
-\mathbf{I}_a = \frac{110/0^{\circ}}{16.155/21.8^{\circ}} = 6.81/-21.8^{\circ} \text{ A}
-$$
-
-In as much as the source voltages in Fig. 12.13 are in positive sequence, the line currents are also in positive sequence:
-
-$$
-\mathbf{I}_b = \mathbf{I}_a \underline{/-120^\circ} = 6.81 \underline{/-141.8^\circ} \text{ A}
-$$
-$$
-\mathbf{I}_c = \mathbf{I}_a \underline{/-240^\circ} = 6.81 \underline{/-261.8^\circ} \text{ A} = 6.81 \underline{/-98.2^\circ} \text{ A}
-$$
-
-# A Y-connected balanced three-phase generator with an impedance of 0.4 + *j*0.3 Ω per phase is connected to a Y-connected balanced load with an impedance of 24 + *j*19 Ω per phase. The line joining the generator and the load has an impedance of 0.6 + *j*0.7 Ω per phase. Assuming a positive sequence for the source voltages and that **V***an* = 120⧸30° V, find: (a) the line voltages, (b) the line currents. Practice Problem 12.2
-
-**Answer:** (a) 207.8⧸60° V, 207.8⧸−60° V, 207.8⧸−180° V, (b) 3.75⧸−8.66° A, 3.75⧸−128.66° A, 3.75⧸ 111.34° A.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/138_12.3 Balanced Wye-Wye Connection.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/138_12.3 Balanced Wye-Wye Connection.md
deleted file mode 100644
index ddeb1acb69c5a56608c6fe41a267b2b0057c9c14..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/138_12.3 Balanced Wye-Wye Connection.md
+++ /dev/null
@@ -1,246 +0,0 @@
-# **12.4** Balanced Wye-Delta Connection
-
-A balanced Y-∆ system consists of a balanced Y-connected source feeding a balanced ∆-connected load.
-
-This is perhaps the most practical three-phase system, as the three-phase sources are usually Y-connected while the three-phase loads are usually ∆-connected.
-
-The balanced Y-delta system is sho wn in Fig. 12.14, where the source is Y-connected and the load is ∆-connected. There is, of course, no neutral connection from source to load for this case. Assuming the positive sequence, the phase voltages are again
-
-$$
-\mathbf{V}_{an} = V_p \underline{/0^{\circ}} \n\mathbf{V}_{bn} = V_p \underline{/ -120^{\circ}}, \qquad \mathbf{V}_{cn} = V_p \underline{/ +120^{\circ}}.
-$$
-\n(12.19)
-
-As shown in Section 12.3, the line voltages are
-
-$$
-\mathbf{V}_{ab} = \sqrt{3} V_p \frac{\Delta 30^\circ}{\mathbf{V}_{ca}} = \mathbf{V}_{AB}, \qquad \mathbf{V}_{bc} = \sqrt{3} V_p \frac{\Delta 90^\circ}{\Delta 10^\circ} = \mathbf{V}_{BC}
-$$
-\n
-$$
-\mathbf{V}_{ca} = \sqrt{3} V_p \frac{\Delta 150^\circ}{\Delta 10^\circ} = \mathbf{V}_{CA}
-$$
-\n(12.20)
-
-showing that the line voltages are equal to the voltages across the load impedances for this system configuration. From these voltages, we can obtain the phase currents as
-
-$$
-\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}}, \qquad \mathbf{I}_{BC} = \frac{\mathbf{V}_{BC}}{\mathbf{Z}_{\Delta}}, \qquad \mathbf{I}_{CA} = \frac{\mathbf{V}_{CA}}{\mathbf{Z}_{\Delta}}
-$$
-(12.21)
-
-These currents have the same magnitude but are out of phase with each other by 120°.
-
-**Figure 12.14** Balanced Y-∆ connection.
-
-Another way to get these phase currents is to apply KVL. For example, applying KVL around loop *aABbna* gives
-
-$$
--\mathbf{V}_{an} + \mathbf{Z}_{\Delta} \mathbf{I}_{AB} + \mathbf{V}_{bn} = 0
-$$
-
-or
-
-$$
-\mathbf{I}_{AB} = \frac{\mathbf{V}_{an} - \mathbf{V}_{bn}}{\mathbf{Z}_{\Delta}} = \frac{\mathbf{V}_{ab}}{\mathbf{Z}_{\Delta}} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}}
-$$
-(12.22)
-
-which is the same as Eq. (12.21). This is the more general way of finding the phase currents.
-
-The line currents are obtained from the phase currents by applying KCL at nodes *A*, *B*, and *C*. Thus,
-
-$$
-\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA}, \qquad \mathbf{I}_b = \mathbf{I}_{BC} - \mathbf{I}_{AB}, \qquad \mathbf{I}_c = \mathbf{I}_{CA} - \mathbf{I}_{BC} \quad (12.23)
-$$
-
-Since **I***CA* = **I***AB*⧸−240°,
-
-$$
-\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA} = \mathbf{I}_{AB} (1 - 1/240^\circ)
-$$
-
-= $\mathbf{I}_{AB} (1 + 0.5 - j0.866) = \mathbf{I}_{AB} \sqrt{3}/-30^\circ$ (12.24)
-
-showing that the magnitude *IL* of the line current is √ \_\_ 3 times the magnitude *Ip* of the phase current, or
-
-$$
-I^L = \sqrt{3}I_p \tag{12.25}
-$$
-
-where
-
-$$
-I_L = |\mathbf{I}_a| = |\mathbf{I}_b| = |\mathbf{I}_c| \tag{12.26}
-$$
-
-and
-
-$$
-I_p = |\mathbf{I}_{AB}| = |\mathbf{I}_{BC}| = |\mathbf{I}_{CA}| \tag{12.27}
-$$
-
-Also, the line currents lag the corresponding phase currents by 30°, assuming the positive sequence. Figure 12.15 is a phasor diagram illustrating the relationship between the phase and line currents.
-
-An alternative way of analyzing the Y-∆ circuit is to transform the ∆-connected load to an equi valent Y-connected load. Using the ∆-Y transformation formula in Eq. (12.8),
-
-$$
-Z_Y = \frac{Z_{\Delta}}{3}
-$$
- (12.28)
-
-After this transformation, we now have a Y-Y system as in Fig. 12.10. The three-phase Y-∆ system in Fig. 12.14 can be replaced by the singlephase equivalent circuit in Fig. 12.16. This allows us to calculate only the line currents. The phase currents are obtained using Eq. (12.25) and utilizing the fact that each of the phase currents leads the corresponding line current by 30°.
-
-A balanced *abc*-sequence Y-connected source with **V***an* = 100⧸ 10° V is Example 12.3 connected to a ∆-connected balanced load (8 + *j*4) Ω per phase. Calculate the phase and line currents.
-
-Phasor diagram illustrating the relationship between phase and line currents.
-
-**Figure 12.16** A single-phase equivalent circuit of a balanced Y-∆ circuit.
-
-# **Solution:**
-
-This can be solved in two ways.
-
-■ **METHOD 1** The load impedance is
-
-$$
-\mathbf{Z}_{\Delta} = 8 + j4 = 8.944 / 26.57^{\circ} \,\Omega
-$$
-
-If the phase voltage **V***an* = 100⧸ 10°, then the line voltage is
-
-$$
-\mathbf{V}_{ab} = \mathbf{V}_{an} \sqrt{3} / 30^{\circ} = 100 \sqrt{3} / 10^{\circ} + 30^{\circ} = \mathbf{V}_{AB}
-$$
-
-or
-
-$$
-V_{AB} = 173.2 \angle 40^{\circ}
-$$
- V
-
-The phase currents are
-
-$$
-\begin{aligned}\n\text{I}_{AB} &= \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}} = \frac{173.2/40^{\circ}}{8.944/26.57^{\circ}} = 19.36/13.43^{\circ} \text{ A} \\
-\text{I}_{BC} &= \mathbf{I}_{AB} / -120^{\circ} = 19.36 / -106.57^{\circ} \text{ A} \\
-\text{I}_{CA} &= \mathbf{I}_{AB} / +120^{\circ} = 19.36 / 133.43^{\circ} \text{ A}\n\end{aligned}
-$$
-
-The line currents are
-
-$$
-\mathbf{I}_a = \mathbf{I}_{AB} \sqrt{3} \underline{/-30^\circ} = \sqrt{3} (19.36) \underline{/13.43^\circ - 30^\circ}
-$$
-
-= 33.53} \underline{/-16.57^\circ} A
-$$
-\mathbf{I}_b = \mathbf{I}_a \underline{/-120^\circ} = 33.53 \underline{/-136.57^\circ} A
-$$
-$$
-\mathbf{I}_c = \mathbf{I}_a \underline{/+120^\circ} = 33.53 \underline{/-103.43^\circ} A
-$$
-
-■ **METHOD 2** Alternatively, using single-phase analysis,
-
-$$
-I_a = \frac{V_{an}}{Z_{\Delta}/3} = \frac{100/10^{\circ}}{2.981/26.57^{\circ}} = 33.54/-16.57^{\circ}
-$$
- A
-
-as above. Other line currents are obtained using the *abc* phase sequence.
-
-Practice Problem 12.3 One line voltage of a balanced Y-connected source is **V***AB* = 120⧸−20° V. If the source is connected to a ∆-connected load of 20 ⧸40° Ω, find the phase and line currents. Assume the *abc* sequence.
-
-> **Answer:** 6⧸−60° A, 6⧸−180° A, 6⧸60° A, 10.392⧸−90° A, 10.392⧸ 150° A, 10.392⧸30° A.
-
-# **12.5** Balanced Delta-Delta Connection
-
-A balanced ∆-∆ system is one in which both the balanced source and balanced load are ∆-connected.
-
-The source as well as the load may be delta-connected as sho wn in Fig. 12.17. Our goal is to obtain the phase and line currents as usual.
-
-Assuming a positive sequence, the phase voltages for a delta-connected source are
-
-$$
-\mathbf{V}_{ab} = V_p / \underline{\mathbf{0}^{\circ}}
-$$
-
-$$
-\mathbf{V}_{bc} = V_p / \underline{\mathbf{-120}^{\circ}}, \qquad \mathbf{V}_{ca} = V_p / \underline{\mathbf{+120}^{\circ}}
-$$
-(12.29)
-
-The line voltages are the same as the phase voltages. From Fig. 12.17, assuming there is no line impedances, the phase voltages of the de ltaconnected source are equal to the voltages across the impedances; that is,
-
-$$
-\mathbf{V}_{ab} = \mathbf{V}_{AB}, \qquad \mathbf{V}_{bc} = \mathbf{V}_{BC}, \qquad \mathbf{V}_{ca} = \mathbf{V}_{CA} \tag{12.30}
-$$
-
-Hence, the phase currents are
-
-$$
-\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{Z_{\Delta}} = \frac{\mathbf{V}_{ab}}{Z_{\Delta}}, \qquad \mathbf{I}_{BC} = \frac{\mathbf{V}_{BC}}{Z_{\Delta}} = \frac{\mathbf{V}_{bc}}{Z_{\Delta}}
-$$
-\n
-$$
-\mathbf{I}_{CA} = \frac{\mathbf{V}_{CA}}{Z_{\Delta}} = \frac{\mathbf{V}_{ca}}{Z_{\Delta}}
-$$
-\n(12.31)
-
-Because the load is delta-connected just as in the previous section, some of the formulas derived there apply here. The line currents are obtained from the phase currents by applying KCL at nodes *A*, *B*, and *C*, as we did in the previous section:
-
-$$
-\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA}, \qquad \mathbf{I}_b = \mathbf{I}_{BC} - \mathbf{I}_{AB}, \qquad \mathbf{I}_c = \mathbf{I}_{CA} - \mathbf{I}_{BC} \tag{12.32}
-$$
-
-Also, as shown in the last section, each line current lags the corre sponding phase current by 30°; the magnitude *IL* of the line current is √ \_\_ 3 times the magnitude *Ip* of the phase current,
-
-$$
-I_L = \sqrt{3}I_p \tag{12.33}
-$$
-
-An alternative way of analyzing the ∆-∆ circuit is to convert both the source and the load to their Y equivalents. We already kno w that **Z***Y* = **Z**∆∕3. To convert a ∆-connected source to a Y-connected source, see the next section.
-
-A balanced ∆-connected load ha ving an impedance 20 − *j*15 Ω is Example 12.4 connected to a ∆-connected, positi ve-sequence generator ha ving **V***ab* = 330⧸ 0° V. Calculate the phase currents of the load and the line currents.
-
-# **Solution:**
-
-The load impedance per phase is
-
-$$
-Z_{\Delta} = 20 - j15 = 25 \sqrt{-36.57^{\circ}} \,\Omega
-$$
-
-Since **V***AB* = **V***ab*, the phase currents are
-
-$$
-\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}} = \frac{330/0^{\circ}}{25/-36.87^{\circ}} = 13.2/36.87^{\circ} \text{ A}
-$$
-$$
-\mathbf{I}_{BC} = \mathbf{I}_{AB}/-120^{\circ} = 13.2/-83.13^{\circ} \text{ A}
-$$
-$$
-\mathbf{I}_{CA} = \mathbf{I}_{AB}/+120^{\circ} = 13.2/156.87^{\circ} \text{ A}
-$$
-
-For a delta load, the line current always lags the corresponding phase current by 30° and has a magnitude √ \_\_ 3 times that of the phase current. Hence, the line currents are
-
-$$
-\mathbf{I}_a = \mathbf{I}_{AB}\sqrt{3}/-30^\circ = (13.2/36.87^\circ)(\sqrt{3}/-30^\circ)
-$$
-
-= 22.86/6.87° A
-$$
-\mathbf{I}_b = \mathbf{I}_a/-120^\circ = 22.86/-113.13^\circ
-$$
- A
-$$
-\mathbf{I}_c = \mathbf{I}_a/+120^\circ = 22.86/126.87^\circ
-$$
- A
-
-A positive-sequence, balanced ∆-connected source supplies a balanced ∆-connected load. If the impedance per phase of the load is 18 + *j*12 Ω and **I***a* = 9.609⧸35° A, find **I***AB* and **V***AB*. Practice Problem 12.4
-
-**Answer:** 5.548⧸65° A, 120⧸98.69° V.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/139_12.4 Balanced Wye-Delta Connection.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/139_12.4 Balanced Wye-Delta Connection.md
deleted file mode 100644
index 03bb9be2da72b9b2369932ae6bc947a201cbf3cc..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/139_12.4 Balanced Wye-Delta Connection.md
+++ /dev/null
@@ -1,158 +0,0 @@
-# **12.6** Balanced Delta-Wye Connection
-
-A balanced ∆-Y system consists of a balanced ∆-connected source feeding a balanced Y-connected load.
-
-Consider the ∆-Y circuit in Fig. 12.18. Again, assuming the *abc* sequence, the phase voltages of a delta-connected source are
-
-$$
-\mathbf{V}_{ab} = V_p \underline{\hat{\mathbf{O}}^{\circ}}, \qquad \mathbf{V}_{bc} = V_p \underline{\hat{\mathbf{O}}^{\circ}} = V_p \underline{\hat{\mathbf{O}}^{\circ}}
-$$
-\n
-$$
-\mathbf{V}_{ca} = V_p \underline{\hat{\mathbf{O}}^{\circ}} \tag{12.34}
-$$
-
-These are also the line voltages as well as the phase voltages.
-
-We can obtain the line currents in man y ways. One way is to apply KVL to loop *aANBba* in Fig. 12.18, writing
-
-$$
--\mathbf{V}_{ab} + \mathbf{Z}_{Y}\mathbf{I}_{a} - \mathbf{Z}_{Y}\mathbf{I}_{b} = 0
-$$
-
-or
-
-Thus,
-
-$$
-\mathbf{Z}_{Y}(\mathbf{I}_{a}-\mathbf{I}_{b})=\mathbf{V}_{ab}=V_{p}\underline{\int_{0}^{\circ}}
-$$
-
-**I***a* − **I***b* = *Vp*⧸ 0° \_\_\_\_\_ **Z***Y* **(12.35)**
-
-**Figure 12.18** A balanced ∆-Y connection.
-
-But **I***b* lags **I***a* by 120°, since we assumed the *abc* sequence; that is, **I***b* = **I***a*⧸−120°. Hence,
-
-$$
-\mathbf{I}_a - \mathbf{I}_b = \mathbf{I}_a (1 - 1/ -120^\circ)
-$$
-
-= $\mathbf{I}_a \left( 1 + \frac{1}{2} + j\frac{\sqrt{3}}{2} \right) = \mathbf{I}_a \sqrt{3}/30^\circ$ (12.36)
-
-Substituting Eq. (12.36) into Eq. (12.35) gives
-
-$$
-I_a = \frac{\left(V_p / \sqrt{3}\right) / -30^{\circ}}{Z_Y}
-$$
- (12.37)
-
-From this, we obtain the other line currents **I***b* and **I***c* using the positive phase sequence, i.e., **I***b* = **I***a*⧸−120°, **I***c* = **I***a*⧸+120°. The phase currents are equal to the line currents.
-
-Another w ay to obtain the line currents is to replace the deltaconnected source with its equi valent wye-connected source, as sho wn
-
-Transforming a ∆-connected source to an equivalent Y-connected source.
-
-per phase, according to Eq. (9.69).
-
-Once the source is transformed to wye, the circuit becomes a wyewye system. Therefore, we can use the equi valent single-phase circuit shown in Fig. 12.20, from which the line current for phase *a* is
-
-$$
-I_a = \frac{V_p / \sqrt{3} / -30^{\circ}}{Z_Y}
-$$
- (12.39)
-
-**Z**Y **I**a **V**p ‒30° √3 + ‒
-
-which is the same as Eq. (12.37).
-
-Alternatively, we may transform the wye-connected load to an equivalent delta-connected load. This results in a delta-delta system, which can be analyzed as in Section 12.5. Note that
-
-$$
-\mathbf{V}_{AN} = \mathbf{I}_a \mathbf{Z}_Y = \frac{V_p}{\sqrt{3}} \frac{1 - 30^\circ}{\sqrt{3}} \tag{12.40}
-$$
-\n
-$$
-\mathbf{V}_{BN} = \mathbf{V}_{AN} \frac{1 - 120^\circ}{\sqrt{3}} \qquad \mathbf{V}_{CN} = \mathbf{V}_{AN} \frac{1 + 120^\circ}{\sqrt{3}} \tag{12.40}
-$$
-
-As stated earlier , the delta-connected load is more desirable than the wye-connected load. It is easier to alter the loads in any one phase of the delta-connected loads, as the individual loads are connected directly across the lines. Ho wever, the delta-connected source is hardly used in practice because any slight imbalance in the phase voltages will result in unwanted circulating currents.
-
-Table 12.1 presents a summary of the formulas for phase currents and voltages and line currents and voltages for the four connections. Students are advised not to memorize the formulas but to understand ho w
-
-## **TABLE 12.1**
-
-| | Summary of phase and line voltages/currents for | |
-|-------------------------------|-------------------------------------------------|--|
-| balanced three-phase systems. | 1 | |
-
-| Connection | Phase voltages/currents | Line voltages/currents |
-|-------------|-----------------------------|--------------------------------------------------|
-| Y-Y | Van =
Vp
⧸ 0° | __
√
Vab =
3
Vp
⧸ 30° |
-| | Vbn =
Vp
−120°
⧸ | Vbc =
Vab
−120°
⧸ |
-| | Vcn =
Vp
+120°
⧸ | Vca =
Vab
+120°
⧸ |
-| | Same as line currents | I
Van
Z
a =
∕
Y |
-| | | I
I
−120°
b =
⧸
a |
-| | | I
I
+120°
c =
⧸
a |
-| Y-
∆ | Van =
Vp
⧸ 0° | __
√
Vab =
VAB =
3
Vp
30°
⧸ |
-| | Vbn =
Vp
−120°
⧸ | Vbc =
VBC =
Vab
−120°
⧸ |
-| | Vcn =
Vp
+120°
⧸ | Vca =
VCA =
Vab
+120°
⧸
__ |
-| | IAB =
VAB
Z
∕
∆ | √
I
IAB
3
−30°
a =
⧸ |
-| | IBC =
VBC
Z
∕
∆ | I
I
−120°
b =
⧸
a |
-| | ICA =
VCA
Z
∕
∆ | I
I
+120°
c =
⧸
a |
-| -
∆
∆ | Vab =
Vp
⧸ 0° | Same as phase voltages |
-| | Vbc =
Vp
−120°
⧸ | |
-| | Vca =
Vp
+120°
⧸ | __ |
-| | IAB =
Vab
Z
∕
∆ | √
I
a =
IAB
3
−30°
⧸ |
-| | IBC =
Vbc
Z
∕
∆ | I
b =
I
−120°
⧸
a |
-| | ICA =
Vca
Z
∕
∆ | I
c =
I
+120°
⧸
a |
-| ∆-Y | Vab =
Vp
⧸ 0° | Same as phase voltages |
-| | Vbc =
Vp
−120°
⧸ | |
-| | Vca =
Vp
+120°
⧸ | −30° ________
Vp
⧸ |
-| | Same as line currents | I
a =
__
√
3
Z
Y |
-| | | I
I
−120°
b =
⧸
a |
-| | | I
I
+120°
c =
⧸
a |
-
-1 Positive or *abc* sequence is assumed. they are derived. The formulas can always be obtained by directly applying KCL and KVL to the appropriate three-phase circuits.
-
-A balanced Y-connected load with a phase impedance of 40 + *j*25 Ω is supplied by a balanced, positive sequence ∆-connected source with a line voltage of 210 V. Calculate the phase currents. Use **V***ab* as a reference.
-
-# **Solution:**
-
-The load impedance is
-
-$$
-\mathbf{Z}_Y = 40 + j25 = 47.17 / 32^{\circ} \,\Omega
-$$
-
-and the source voltage is
-
-$$
-\mathbf{V}_{ab} = 210 \underline{\big/ 0^{\circ}} \, \mathrm{V}
-$$
-
-When the ∆-connected source is transformed to a Y-connected source,
-
-$$
-\mathbf{V}_{an} = \frac{\mathbf{V}_{ab}}{\sqrt{3}} \underline{/-30^{\circ}} = 121.2 \underline{/-30^{\circ}} \text{ V}
-$$
-
-The line currents are
-
-$$
-\mathbf{I}_a = \frac{\mathbf{V}_{an}}{\mathbf{Z}_Y} = \frac{121.2 \div 30^\circ}{47.12 \div 32^\circ} = 2.57 \div 62^\circ \text{ A}
-$$
-\n
-$$
-\mathbf{I}_b = \mathbf{I}_a \div 120^\circ = 2.57 \div 178^\circ \text{ A}
-$$
-\n
-$$
-\mathbf{I}_c = \mathbf{I}_a \div 120^\circ = 2.57 \div 58^\circ \text{ A}
-$$
-
-which are the same as the phase currents.
-
-In a balanced ∆-Y circuit, **V***ab* = 440⧸ 15° and **Z***Y* = (12 + *j*15) Ω. Practice Problem 12.5 Calculate the line currents.
-
-**Answer:** 13.224⧸−66.34° A, 13.224⧸+173.66° A, 13.224⧸ 53.66° A.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/140_12.7 Power in a Balanced System.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/140_12.7 Power in a Balanced System.md
deleted file mode 100644
index 72253b32292fc6e560e36c4183e0e8ab24ac9ff1..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/140_12.7 Power in a Balanced System.md
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@@ -1,369 +0,0 @@
-# **12.7** Power in a Balanced System
-
-Let us now consider the power in a balanced three-phase system. We begin by examining the instantaneous power absorbed by the load. This requires that the analysis be done in the time domain. For a Y-connected load, the phase voltages are
-
-$$
-v_{AN} = \sqrt{2} V_p \cos \omega t, \qquad v_{BN} = \sqrt{2} V_p \cos(\omega t - 120^\circ)
-$$
-
-$$
-v_{CN} = \sqrt{2} V_p \cos(\omega t + 120^\circ)
-$$
- (12.41)
-
-where the factor √ 2is necessary because *Vp* has been defined as the rms value of the phase voltage. If **Z***Y* = *Z*⧸*θ*, the phase currents lag behind their corresponding phase voltages by *θ*. Thus,
-
-$$
-i_a = \sqrt{2} I_p \cos(\omega t - \theta), \qquad i_b = \sqrt{2} I_p \cos(\omega t - \theta - 120^\circ) \quad (12.42)
-$$
-$$
-i_c = \sqrt{2} I_p \cos(\omega t - \theta + 120^\circ)
-$$
-
-Example 12.5
-
-where *Ip* is the rms value of the phase current. The total instantaneous power in the load is the sum of the instantaneous powers in the three phases; that is,
-
-$$
-p = p_a + p_b + p_c = v_{AN}i_a + v_{BN}i_b + v_{CN}i_c
-$$
-
-= $2V_pI_p[\cos \omega t \cos(\omega t - \theta)$
-+ $\cos(\omega t - 120^\circ) \cos(\omega t - \theta - 120^\circ)$
-+ $\cos(\omega t + 120^\circ) \cos(\omega t - \theta + 120^\circ)$ (12.43)
-
-Applying the trigonometric identity
-
-$$
-\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \quad (12.44)
-$$
-
-gives
-
-$$
-p = V_p I_p [3 \cos \theta + \cos(2\omega t - \theta) + \cos(2\omega t - \theta - 240^\circ) + \cos(2\omega t - \theta + 240^\circ)]
-$$
-
-= $V_p I_p [3 \cos \theta + \cos \alpha + \cos \alpha \cos 240^\circ + \sin \alpha \sin 240^\circ + \cos \alpha \cos 240^\circ - \sin \alpha \sin 240^\circ]$ (12.45)
-where $\alpha = 2\omega t - \theta$
-= $V_p I_p [3 \cos \theta + \cos \alpha + 2(-\frac{1}{2}) \cos \alpha] = 3V_p I_p \cos \theta$
-
-Thus the total instantaneous power in a balanced three-phase system is constant—it does not change with time as the instantaneous power of each phase does. This result is true whether the load is Y- or ∆- connected. This is one important reason for using a three-phase system to generate and distribute power. We will look into another reason a little later.
-
-Since the total instantaneous po wer is independent of time, the average po wer per phase *Pp* for either the ∆-connected load or the Y-connected load is *p*∕3, or
-
-$$
-P_p = V_p I_p \cos \theta \tag{12.46}
-$$
-
-and the reactive power per phase is
-
-$$
-Q_p = V_p I_p \sin \theta \tag{12.47}
-$$
-
-The apparent power per phase is
-
-$$
-S_p = V_p I_p \tag{12.48}
-$$
-
-The complex power per phase is
-
-$$
-S_p = P_p + jQ_p = V_p I_p^* \tag{12.49}
-$$
-
-\_\_
-
-where **V***p* and **I***p* are the phase voltage and phase current with magnitudes *Vp* and *Ip*, respectively. The total average power is the sum of the average powers in the phases:
-
-$$
-P = P_a + P_b + P_c = 3P_p = 3V_p I_p \cos \theta = \sqrt{3} V_L I_L \cos \theta \qquad (12.50)
-$$
-
-For a Y-connected load, *IL* = *Ip* but *VL* = √ 3 *Vp*, whereas for a ∆-connected load, *IL* = √ 3 *Ip* but *VL* = *Vp*. Thus, Eq. (12.50) applies for both Y-connected and ∆-connected loads. Similarly, the total reactive power is
-
-$$
-Q = 3V_p I_p \sin \theta = 3Q_p = \sqrt{3} V_L I_L \sin \theta \qquad (12.51)
-$$
-
-and the total complex power is
-
-$$
-\mathbf{S} = 3\mathbf{S}_p = 3\mathbf{V}_p \mathbf{I}_p^* = 3I_p^2 \mathbf{Z}_p = \frac{3V_p^2}{\mathbf{Z}_p^*}
-$$
- (12.52)
-
-where **Z***p* = *Zp*⧸*θ* is the load impedance per phase. (**Z***p* could be **Z***Y* or **Z**∆.) Alternatively, we may write Eq. (12.52) as
-
-$$
-\mathbf{S} = P + jQ = \sqrt{3} V_L I_L \underline{\theta} \tag{12.53}
-$$
-
-Remember that *Vp*, *Ip*, *VL*, and *IL* are all rms values and that *θ* is the angle of the load impedance or the angle between the phase voltage and the phase current.
-
-A second major adv antage of three-phase systems for po wer distribution is that the three-phase system uses a lesser amount of wire than the single-phase system for the same line voltage *VL* and the same absorbed power *PL*. We will compare these cases and assume in both that the wires are of the same material (e.g., copper with resisti vity *ρ*), of the same length *ℓ*, and that the loads are resisti ve (i.e., unity power factor). For the tw o-wire single-phase system in Fig. 12.21(a), *IL* = *PL*∕*VL*, so the power loss in the two wires is
-
-$$
-P_{\text{loss}} = 2I_L^2 R = 2R \frac{P_L^2}{V_L^2}
-$$
- (12.54)
-
-# **Figure 12.21**
-
-Comparing the power loss in (a) a single-phase system, and (b) a three-phase system.
-
-For the three-wire three-phase system in Fig. 12.21(b), *IL*′ = |**I***a*| = |**I***b*| = |**I***c*| = *PL*∕ √ \_\_ 3 *VL* from Eq. (12.50). The power loss in the three wires is
-
-$$
-P'_{\text{loss}} = 3(I'_L)^2 R' = 3R' \frac{P_L^2}{3V_L^2} = R' \frac{P_L^2}{V_L^2}
-$$
- (12.55)
-
-Equations (12.54) and (12.55) show that for the same total power delivered *PL* and same line voltage *VL*,
-
-$$
-\frac{P_{\text{loss}}}{P'_{\text{loss}}} = \frac{2R}{R'}
-$$
-\n(12.56)
-
-But from Chapter 2, *R* = *ρℓ*∕*πr* 2 and *R*′ = *ρℓ*∕*πr*′ 2 , where *r* and *r*′ are the radii of the wires. Thus,
-
-$$
-\frac{P_{\text{loss}}}{P'_{\text{loss}}} = \frac{2r'^2}{r^2}
-$$
-\n(12.57)
-
-If the same power loss is tolerated in both systems, then *r* 2 = 2r′ 2 . The ratio of material required is determined by the number of wires and their volumes, so \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ Material for three-phase = 2(π*r*
-
-$$
-\frac{\text{Material for single-phase}}{\text{Material for three-phase}} = \frac{2(\pi r^2 \ell)}{3(\pi r^2 \ell)} = \frac{2r^2}{3r^2}
-$$
-\n
-$$
-= \frac{2}{3}(2) = 1.333
-$$
-\n(12.58)
-
-since *r* 2 = 2*r*′ 2 . Equation (12.58) shows that the single-phase system uses 33 percent more material than the three-phase system or that the threephase system uses only 75 percent of the material used in the equivalent single-phase system. In other words, considerably less material is needed to deliver the same power with a three-phase system than is required for a single-phase system.
-
-Refer to the circuit in Fig. 12.13 (in Example 12.2). Determine the total average power, reactive power, and complex power at the source and at the load.
-
-# **Solution:**
-
-It is sufficient to consider one phase, as the system is balanced. For phase *a*,
-
-$$
-V_p = 110\frac{0}{0} \text{ V}
-$$
- and $I_p = 6.81\frac{1}{21.8^{\circ}} \text{ A}$
-
-Thus, at the source, the complex power absorbed is
-
-$$
-\mathbf{S}_s = -3\mathbf{V}_p \mathbf{I}_p^* = -3(110/0^\circ)(6.81/21.8^\circ)
-$$
-
-= -2247/21.8° = -(2087 + j834.6) VA
-
-The real or average power absorbed is −2087 W and the reactive power is −834.6 VAR.
-
-At the load, the complex power absorbed is
-
-$$
-\mathbf{S}_{L} = 3|\mathbf{I}_{p}|^{2}\mathbf{Z}_{p}
-$$
-
-where $\mathbf{Z}_{p} = 10 + j8 = 12.81 \underline{ / 38.66^{\circ}}$ and $\mathbf{I}_{p} = \mathbf{I}_{a} = 6.81 \underline{ / -21.8^{\circ}}$ . Hence,
-$$
-\mathbf{S}_{L} = 3(6.81)^{2} 12.81 \underline{ / 38.66^{\circ}} = 1782 \underline{ / 38.66^{\circ}}
-$$
-$$
-= (1392 + j1113) \text{ VA}
-$$
-
-The real power absorbed is 1391.7 W and the reactive power absorbed is 1113.3 VAR. The difference between the two complex powers is ab sorbed by the line impedance (5 − *j*2) Ω. To show that this is the case, we find the complex power absorbed by the line as
-
-$$
-\mathbf{S}_{\ell} = 3|\mathbf{I}_{p}|^{2}\mathbf{Z}_{\ell} = 3(6.81)^{2}(5 - j2) = 695.6 - j278.3 \text{ VA}
-$$
-
-which is the difference between **S***s* and **S***L*; that is, **S***s* + **S***ℓ* + **S***L* = 0, as expected.
-
-Example 12.6
-
-For the Y-Y circuit in Practice Prob. 12.2, calculate the complex power at the source and at the load.
-
-**Answer:** −(1054.2 + *j*843.3) VA, (1012 + *j*801.6) VA.
-
-A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the motor.
-
-# **Solution:**
-
-The apparent power is
-
-$$
-S = \sqrt{3}V_L I_L = \sqrt{3}(220)(18.2) = 6935.13 \text{ VA}
-$$
-
-Since the real power is
-
-$$
-P = S \cos \theta = 5600 \text{ W}
-$$
-
-the power factor is
-
-pf =
-$$
-\cos \theta = \frac{P}{S} = \frac{5600}{6935.13} = 0.8075
-$$
-
-Calculate the line current required for a 30-kW three-phase motor having a power factor of 0.85 lagging if it is connected to a balanced source with a line voltage of 550 V.
-
-**Answer:** 37.05 A.
-
-Two balanced loads are connected to a 240-kV rms 60-Hz line, as shown in Fig. 12.22(a). Load 1 dra ws 30 kW at a po wer factor of 0.6 lagging, while load 2 draws 45 kVAR at a power factor of 0.8 lagging. Assuming the *abc* sequence, determine: (a) the complex, real, and reactive powers absorbed by the combined load, (b) the line currents, and (c) the kVAR rating of the three capacitors ∆-connected in parallel with the load that will raise the power factor to 0.9 lagging and the capacitance of each capacitor.
-
-# **Solution:**
-
-(a) For load 1, given that *P*1 = 30 kW and cos *θ*1 = 0.6, then sin *θ*1 = 0.8. Hence,
-
-$$
-S_1 = \frac{P_1}{\cos \theta_1} = \frac{30 \text{ kW}}{0.6} = 50 \text{ kVA}
-$$
-
-and *Q*1 = *S*1 sin *θ*1 = 50(0.8) = 40 kVAR. Thus, the complex power due to load 1 is
-
-$$
-S_1 = P_1 + jQ_1 = 30 + j40 \text{ kVA}
-$$
- (12.8.1)
-
-Practice Problem 12.7
-
-Example 12.8
-
-Practice Problem 12.6
-
-Example 12.7
-
-# **Figure 12.22**
-
-For Example 12.8: (a) The original balanced loads, (b) the combined load with improved power factor.
-
-For load 2, if
-$$
-Q_2 = 45
-$$
- kVAR and $\cos \theta_2 = 0.8$ , then $\sin \theta_2 = 0.6$ . We find
-
-$$
-S_2 = \frac{Q_2}{\sin \theta_2} = \frac{45 \text{ kVA}}{0.6} = 75 \text{ kVA}
-$$
-
-and *P*2 = *S*2 cos *θ*2 75(0.8) = 60 kW. Therefore the complex power due to load 2 is
-
-$$
-S_2 = P_2 + jQ_2 = 60 + j45 \text{ kVA}
-$$
- (12.8.2)
-
-From Eqs. (12.8.1) and (12.8.2), the total complex power absorbed by the load is
-
-$$
-S = S_1 + S_2 = 90 + j85 \text{ kVA} = 123.8 \underline{43.36^{\circ}} \text{ kVA} \quad (12.8.3)
-$$
-
-which has a power factor of cos 43.36° = 0.727 lagging. The real power is then 90 kW, while the reactive power is 85 kVAR.
-
-It will help with the calculations to assume that the loads are wye connected and then to work with the phase voltages, i.e. the magnitude of *VAN* = (240∕√ \_\_ 3 ) kV.
-
-(b) Since
-$$
-S = 3((240 \text{ kV}/\sqrt{3})I_L) = \sqrt{3}(240 \text{ kV})I_L
-$$
-
-the magnitude of the line current is
-
-$$
-I_L = \frac{S}{\sqrt{3}(240,000)}
-$$
-(12.8.4)
-
-We apply this to each load keeping in mind that the magnitude of the phase voltages is equal to (240∕√ \_\_ 3 ) kV. For load 1,
-
-$$
-I_{L1} = \frac{50,000}{\sqrt{3} \, 240,000} = 120.28 \, \text{mA}
-$$
-
-Since the power factor is lagging, the line current lags the line voltage by *θ*1 = cos−1 0.6 = 53.13°. Thus,
-
-$$
-I_{a1} = 120.28 \sqrt{-53.13^{\circ}}
-$$
-
-For load 2,
-
-$$
-I_{L2} = \frac{75,000}{\sqrt{3} \, 240,000} = 180.42 \, \text{mA}
-$$
-
-and the line current lags the line voltage by *θ*2 = cos−1 0.8 = 36.87°. Hence,
-
-$$
-I_{a2} = 180.42 \underline{/-36.87^{\circ}}
-$$
-
-The total line current is
-
-$$
-\mathbf{I}_a = \mathbf{I}_{a1} + \mathbf{I}_{a2} = 120.28 \underline{/ -53.13^\circ} + 180.42 \underline{/ -36.87^\circ}
-$$
-
-= (72.168 - j96.224) + (144.336 - j108.252)
-= 216.5 - j204.472 = 297.8 \underline{/ -43.36^\circ} mA
-
-Alternatively, we could obtain the current from the total complex power using Eq. (12.8.4) as
-
-$$
-I_L = \frac{123,800}{\sqrt{3} \, 240,000} = 297.82 \, \text{mA}
-$$
-
-and
-
-$$
-I_a = 297.82 \div 43.36^\circ \text{ mA}
-$$
-
-which is the same as before. The other line currents, **I***b*2 and **I***ca*, can be obtained according to the *abc* sequence (i.e., **I***b* = 297.82⧸−163.36° mA and **I***c* = 297.82⧸76.64° mA).
-
-(c) We can find the reactive power needed to bring the power factor to 0.9 lagging using Eq. (11.59),
-
-$$
-QC = P(\tan \theta_{\text{old}} - \tan \theta_{\text{new}})
-$$
-
-where *P* = 90 kW, *θ*old = 43.36°, and *θ*new = cos−1 0.9 = 25.84°. Hence,
-
-$$
-Q_C = 90,000 \text{(tan } 43.36^\circ - \text{tan } 25.84^\circ) = 41.4 \text{ kVAR}
-$$
-
-This reactive power is for the three capacitors. For each capacitor, the rating *QC*′ = 13.8 kVAR. From Eq. (11.60), the required capacitance is
-
-$$
-C = \frac{Q'_C}{\omega V_{\text{rms}}^2}
-$$
-
-Since the capacitors are ∆-connected as shown in Fig. 12.22(b), *V*rms in the above formula is the line-to-line or line voltage, which is 240 kV. Thus,
-
-is the line-to-ine of the voltage,
-$$
-C = \frac{13,800}{(2\pi60)(240,000)^2} = 635.5 \text{ pF}
-$$
-
-Assume that the two balanced loads in Fig. 12.22(a) are supplied by an 840-V rms 60-Hz line. Load 1 is Y-connected with 30 + *j*40 Ω per phase, while load 2 is a balanced three-phase motor drawing 48 kW at a power factor of 0.8 lagging. Assuming the *abc* sequence, calculate: (a) the complex power absorbed by the combined load, (b) the kVAR rating of each of the three capacitors ∆-connected in parallel with the load to raise the power factor to unity, and (c) the current drawn from the supply at unity power factor condition.
-
-**Answer:** (a) 56.47 + *j*47.29 kVA, (b) 15.76 kVAR, (c) 38.81 A.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/141_12.8 Unbalanced Three-Phase Systems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/141_12.8 Unbalanced Three-Phase Systems.md
deleted file mode 100644
index df0649cc47a930d4ca91ea7cc2cd48cdd6c0d238..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/141_12.8 Unbalanced Three-Phase Systems.md
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@@ -1,216 +0,0 @@
-# **12.8** Unbalanced Three-Phase Systems
-
-This chapter would be incomplete without mentioning unbalanced threephase systems. An unbalanced system is caused by two possible situa tions: (1) The source voltages are not equal in magnitude and/or differ in phase by angles that are unequal, or (2) load impedances are unequal. Thus,
-
-An unbalanced system is due to unbalanced voltage sources or an unbalanced load.
-
-To simplify analysis, we will assume balanced source voltages, but an unbalanced load.
-
-Unbalanced three-phase systems are solved by direct application of mesh and nodal analysis. Figure 12.23 sho ws an e xample of an unbal anced three-phase system that consists of balanced source v oltages (not shown in the figure) and an unbalanced Y-connected load (shown in the figure). Since the load is unbalanced, **Z***A*, **Z***B*, and **Z***C* are not equal. The line currents are determined by Ohm's law as
-
-$$
-\mathbf{I}_a = \frac{\mathbf{V}_{AN}}{\mathbf{Z}_A}, \qquad \mathbf{I}_b = \frac{\mathbf{V}_{BN}}{\mathbf{Z}_B}, \qquad \mathbf{I}_c = \frac{\mathbf{V}_{CN}}{\mathbf{Z}_C}
-$$
-(12.59)
-
-# **Figure 12.23** Unbalanced three-phase Y-connected load.
-
-A special technique for handling unbalanced three-phase systems is the method of symmetrical components, which is beyond the scope of this text.
-
-**I**a
-
-Practice Problem 12.8
-
-This set of unbalanced line currents produces current in the neutral line, which is not zero as in a balanced system. Applying KCL at node *N* gives the neutral line current as
-
-$$
-\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) \tag{12.60}
-$$
-
-In a three-wire system where the neutral line is absent, we can still find the line currents **I***a*, **I***b*, and **I***c* using mesh analysis. At node *N*, KCL must be satisfied so that **I***a* + **I***b* + **I***c* = 0 in this case. The same could be done for an unbalanced ∆-Y, Y-∆, or ∆-∆ three-wire system. As mentioned earlier, in long distance po wer transmission, conductors in mul tiples of three (multiple three-wire systems) are used, with the earth itself acting as the neutral conductor.
-
-To calculate po wer in an unbalanced three-phase system requires that we find the power in each phase using Eqs. (12.46) to (12.49). The total power is not simply three times the power in one phase but the sum of the powers in the three phases.
-
-The unbalanced Y-load of Fig. 12.23 has balanced voltages of 100 V and the *acb* sequence. Calculate the line currents and the neutral current. Take **Z***A* = 15 Ω, **Z***B* = 10 + *j*5 Ω, **Z***C* = 6 − *j*8 Ω.
-
-# **Solution:**
-
-Using Eq. (12.59), the line currents are
-
-$$
-\mathbf{I}_a = \frac{100/0^{\circ}}{15} = 6.67/0^{\circ} \text{ A}
-$$
-$$
-\mathbf{I}_b = \frac{100/120^{\circ}}{10 + j5} = \frac{100/120^{\circ}}{11.18/26.56^{\circ}} = 8.94/93.44^{\circ} \text{ A}
-$$
-$$
-\mathbf{I}_c = \frac{100/-120^{\circ}}{6 - j8} = \frac{100/-120^{\circ}}{10/-53.13^{\circ}} = 10/-66.87^{\circ} \text{ A}
-$$
-
-Using Eq. (12.60), the current in the neutral line is
-
-$$
-\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) = -(6.67 - 0.54 + j8.92 + 3.93 - j9.2)
-$$
-
-= -10.06 + j0.28 = 10.06/178.4° A
-
-The unbalanced ∆-load of Fig. 12.24 is supplied by balanced line-to-line voltages of 440 V in the positive sequence. Find the line currents. Take **V***ab* as reference.
-
-**Answer:**
-$$
-39.71 \underline{/-41.06^{\circ}}
-$$
- A, $64.12 \underline{/-139.8^{\circ}}$ A, $70.13 \underline{/74.27^{\circ}}$ A.
-
-Example 12.9
-
-Practice Problem 12.9
-
-**Figure 12.24** Unbalanced ∆-load; for Practice Prob. 12.9. For the unbalanced circuit in Fig. 12.25, find: (a) the line currents, (b) the total complex power absorbed by the load, and (c) the total complex power absorbed by the source.
-
-For Example 12.10.
-
-# **Solution:**
-
-(a) We use mesh analysis to find the required currents. For mesh 1,
-
-$$
-120 \underline{/- 120^{\circ}} - 120 \underline{/0^{\circ}} + (10 + j5)I_1 - 10I_2 = 0
-$$
-
-or
-
-$$
-(10+j5)\mathbf{I}_1 - 10\mathbf{I}_2 = 120\sqrt{3}/30^{\circ}
-$$
- (12.10.1)
-
-For mesh 2,
-
-$$
-120\underline{\bigg/120^{\circ}} - 120\underline{\bigg/}-120^{\circ} + (10 - j10)\mathbf{I}_2 - 10\mathbf{I}_1 = 0
-$$
-
-or
-
-$$
--10I1 + (10 - j10)I2 = 120\sqrt{3} / -90^{\circ}
-$$
- (12.10.2)
-
-Equations (12.10.1) and (12.10.2) form a matrix equation:
-
-2.10.1) and (12.10.2) form a matrix equation:
-\n
-$$
-\begin{bmatrix}\n10 + j5 & -10 \\
--10 & 10 - j10\n\end{bmatrix}\n\begin{bmatrix}\n\mathbf{I}_1 \\
-\mathbf{I}_2\n\end{bmatrix} = \n\begin{bmatrix}\n120\sqrt{3}/30^\circ \\
-120\sqrt{3}/-90^\circ\n\end{bmatrix}
-$$
-
-The determinants are
-
-erminants are
-\n
-$$
-\Delta = \begin{vmatrix} 10 + j5 & -10 \\ -10 & 10 - j10 \end{vmatrix} = 50 - j50 = 70.71 \underline{/-45^{\circ}}
-$$
-\n
-$$
-\Delta_1 = \begin{vmatrix} 120\sqrt{3}/30^{\circ} & -10 \\ 120\sqrt{3}/-90^{\circ} & 10 - j10 \end{vmatrix} = 207.85(13.66 - j13.66)
-$$
-\n
-$$
-= 4015 \underline{/-45^{\circ}}
-$$
-\n
-$$
-\Delta_2 = \begin{vmatrix} 10 + j5 & 120\sqrt{3}/30^{\circ} \\ -10 & 120\sqrt{3}/-90^{\circ} \end{vmatrix} = 207.85(13.66 - j5)
-$$
-\n
-$$
-= 3023.4 \underline{/-20.1^{\circ}}
-$$
-
-The mesh currents are
-
-rents are
-\n
-$$
-\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{4015.23/-45^{\circ}}{70.71/-45^{\circ}} = 56.78 \text{ A}
-$$
-\n
-$$
-\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{3023.4/-20.1^{\circ}}{70.71/-45^{\circ}} = 42.75/24.9^{\circ} \text{ A}
-$$
-
-The line currents are
-
-$$
-\mathbf{I}_a = \mathbf{I}_1 = 56.78 \text{ A}, \qquad \mathbf{I}_c = -\mathbf{I}_2 = 42.75 \underline{\smash{\big)}\, - 155.1^\circ} \text{ A}
-$$
-\n
-$$
-\mathbf{I}_b = \mathbf{I}_2 - \mathbf{I}_1 = 38.78 + j18 - 56.78 = 25.46 \underline{\smash{\big)}\, 135^\circ} \text{ A}
-$$
-
-(b) We can now calculate the complex power absorbed by the load. For phase A,
-
-$$
-\mathbf{S}_A = \mathbf{I} \mathbf{I}_a \mathbf{I}^2 \mathbf{Z}_A = (56.78)^2 (j5) = j16,120 \text{ VA}
-$$
-
-For phase B,
-
-$$
-\mathbf{S}_B = |\mathbf{I}_b|^2 \mathbf{Z}_B = (25.46)^2 (10) = 6480 \text{ VA}
-$$
-
-For phase C,
-
-$$
-S_C = I I_c^2 Z_C = (42.75)^2(-j10) = -j18,276 \text{ VA}
-$$
-
-The total complex power absorbed by the load is
-
-$$
-S_L = S_A + S_B + S_C = 6480 - j2156
-$$
- VA
-
-(c) We check the result above by finding the power absorbed by the source. For the voltage source in phase *a,*
-
-$$
-S_a = -V_{an}I_a^* = -(120/0^\circ)(56.78) = -6813.6
-$$
- VA
-
-For the source in phase *b,*
-
-$$
-\mathbf{S}_b = -\mathbf{V}_{bn}\mathbf{I}_b^* = -(120/-120°)(25.46/-135°)
-$$
-
-= -3055.2/105° = 790 - j2951.1 VA
-
-For the source in phase *c*,
-
-$$
-\mathbf{S}_c = -\mathbf{V}_{bn}\mathbf{I}_c^* = -(120/120^\circ)(42.75/155.1^\circ)
-$$
-
-= -5130/275.1° = -456.03 + j5109.7 VA
-
-The total complex power absorbed by the three-phase source is
-
-$$
-S_s = S_a + S_b + S_c = -6480 + j2156
-$$
- VA
-
-showing that **S***s* + **S***L* = 0 and confirming the conservation principle of ac power.
-
-**Answer:** 128.01⧸ 80.1° A, 76.21⧸−60° A, 85⧸−135° A, 19.36 kW.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/142_12.9 PSpice for Three-Phase Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/142_12.9 PSpice for Three-Phase Circuits.md
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/142_12.9 PSpice for Three-Phase Circuits.md
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@@ -1,882 +0,0 @@
-# **12.9** PSpice for Three-Phase Circuits
-
-*PSpice* can be used to analyze three-phase balanced or unbalanced cir cuits in the same way it is used to analyze single-phase ac circuits. However, a delta-connected source presents two major problems to *PSpice*. First, a delta-connected source is a loop of voltage sources—which *PSpice* does not like. To avoid this problem, we insert a resistor of neg ligible resistance (say, 1 *μ*Ω per phase) into each phase of the deltaconnected source. Second, the delta-connected source does not provide a convenient node for the ground node, which is necessary to run *PSpice*. This problem can be eliminated by inserting balanced wye-connected large resistors (say, 1 MΩ per phase) in the delta-connected source so that the neutral node of the wye-connected resistors serves as the ground node 0. Example 12.12 will illustrate this.
-
-For the balanced Y-∆ circuit in Fig. 12.27, use *PSpice* to find the line current **I***aA*, the phase voltage **V***AB*, and the phase current **I***AC*. Assume that the source frequency is 60 Hz.
-
-# **Solution:**
-
-The schematic is shown in Fig. 12.28. The pseudocomponents IPRINT are inserted in the appropriate lines to obtain **I***aA* and **I***AC*, while VPRINT2 is inserted between nodes A and B to print differential voltage **V***AB*. We set the attributes of IPRINT and VPRINT2 each to *AC* = *yes*, *MAG* = *yes*, *PHASE* = *yes*, to print only the magnitude and phase of the currents and voltages. As a single-frequency analysis, we select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 60, and *Final Freq* = 60. Once the circuit is saved, it is simulated by selecting **Analysis/Simulate**. The output file includes the following:
-
-| FREQ | V(A,B) | VP(A,B) |
-|-----------|--------------|--------------|
-| 6.000E+01 | 1.699E+02 | 3.081E+01 |
-| FREQ | IM(V_PRINT2) | IP(V_PRINT2) |
-| 6.000E+01 | 2.350E+00 | -3.620E+01 |
-| FREQ | IM(V_PRINT3) | IP(V_PRINT3) |
-| 6.000E+01 | 1.357E+00 | -6.620E+01 |
-
-# Example 12.11
-
-Schematic for the circuit in Fig. 12.27.
-
-From this, we obtain
-
-$$
-I_{aA} = 2.35 \underline{/-36.2^{\circ}} A
-$$
-
-$$
-V_{AB} = 169.9 \underline{/30.81^{\circ}} V, \quad I_{AC} = 1.357 \underline{/-66.2^{\circ}} A
-$$
-
-# Practice Problem 12.11
-
-Refer to the balanced Y-Y circuit of Fig. 12.29. Use *PSpice* to find the line current **I***bB* and the phase voltage **V***AN*. Take *f* = 100 Hz.
-
-**Answer:** 100.9⧸ 60.87° V, 8.547⧸−91.27° A.
-
-Consider the unbalanced ∆-∆ circuit in Fig. 12.30. Use *PSpice* t o find the generator current **I***ab*, the line current **I***bB*, and the phase current **I***BC*.
-
-Example 12.12
-
-# **Solution:**
-
-- 1. **Define.** The problem and solution process are clearly defined.
-- 2. **Present.** We are to find the generator current flowing from *a* to *b,* the line current flowing from *b* to *B,* and the phase current flowing from *B* to *C*.
-- 3. **Alternative.** Although there are different approaches to solving this problem, the use of *PSpice* is mandated. Therefore, we will not use another approach.
-- 4. **Attempt.** As mentioned above, we avoid the loop of voltage sources by inserting a 1-*μ*Ω series resistor in the delta-connected source. To provide a ground node 0, we insert balanced wyeconnected resistors (1 MΩ per phase) in the delta-connected source, as shown in the schematic in Fig. 12.31. Three IPRINT pseudocomponents with their attributes are inserted to be able
-
-**Figure 12.31** Schematic for the circuit in Fig. 12.30.
-
-to get the required currents **I***ab*, **I***bB*, and **I***BC*. Since the operating frequency is not given and the inductances and capacitances should be specified instead of impedances, we assume *ω* = 1 rad/s so that *f* = 1∕2*π* = 0.159155 Hz. Thus,
-
-$$
-L = \frac{X_L}{\omega} \quad \text{and} \quad C = \frac{1}{\omega X_C}
-$$
-
- We select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 0.159155, and *Final Freq* = 0.159155. Once the schematic is saved, we select **Analysis/Simulate** to simulate the circuit. The output file includes:
-
-| FREQ | IM(V_PRINT1) | IP(V_PRINT1) |
-|-----------|--------------|--------------|
-| 1.592E-01 | 9.106E+00 | 1.685E+02 |
-| FREQ | IM(V_PRINT2) | IP(V_PRINT2) |
-| 1.592E-01 | 5.959E+00 | -1.772E+02 |
-| FREQ | IM(V_PRINT3) | IP(V_PRINT3) |
-| 1.592E-01 | 5.500E+00 | 1.725E+02 |
-
-which yields
-
-$$
-I_{ab} = 5.595 \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \
-$$
-
-5. **Evaluate.** We can check our results by using mesh analysis. Let the loop *aABb* be loop 1, the loop *bBCc* be loop 2, and the loop *ACB* be loop 3, with the three loop currents all flowing in the clockwise direction. We then end up with the following loop equations:
-
-Loop 1
-
-(54 + *j*10)*I*1 − (2 + *j*5)*I*2 − (50)*I*3 = 208⧸ 10° = 204.8 + *j*36.12
-
-Loop 2
-
-$$
--(2+j5)I1 + (4+j40)I2 - (j30)I3 = 208/ -110o
-$$
-
-= -71.14 - j195.46
-
-Loop 3
-
-$$
--(50)I_1 - (j30)I_2 + (50 - j10)I_3 = 0
-$$
-
-Using MATLAB to solve this we get,
-
-```
->>Z = [(54+10i),(-2-5i),-50;(-2-5i),(4+40i),
--30i;-50,-30i,(50-10i)]
-```
-
-Z =
-
-```
-54.0000+10.0000i-2.0000-5.0000i-50.0000
--2.0000-5.0000i 4.0000 + 40.0000i 0-30.0000i
--50.0000 0-30.0000i 50.0000-10.0000i
-```
-
-```
->>V = [(204.8+36.12i);(-71.14-195.46i);0]
-V =
-1.0e+002*
-2.0480+0.3612i
--0.7114-1.9546i
- 0
->>I = inv(Z)*V
-I =
-8.9317+2.6983i
-0.0096+4.5175i
-5.4619+3.7964i
- IbB = −I1 + I2 = −(8.932 + j2.698) + (0.0096 + j4.518)
- = −8.922 + j1.82 = 9.106⧸168.47° A Answer checks
- IBC = I2 − I3 = (0.0096 + j4.518) − (5.462 + j3.796)
- = −5.452 + j0.722 = 5.5⧸172.46° A Answer checks
-```
-
-Now to solve for *Iab*. If we assume a small internal impedance for each source, we can obtain a reasonably good estimate for *Iab*. Adding in internal resistors of 0.01 Ω, and adding a fourth loop around the source circuit, we now get
-
-Loop 1
-
-$$
-(54.01 + j10)I1 - (2 + j5)I2 - (50)I3 - 0.01I4 = 208/10o
-$$
-
-= 204.8 + j36.12
-
-Loop 2
-
-$$
--(2+j5)I1 + (4.01+j40)I2 - (j30)I3 - 0.01I4
-$$
-
-= 208/ $-110^{\circ}$ = -71.14 - j195.46
-
-Loop 3
-
-$$
--(50)I1 - (j30)I2 + (50 - j10)I3 = 0
-$$
-
-Loop 4
-
-$$
--(0.01)I_1 - (0.01)I_2 + (0.03)I_4 = 0
-$$
-
->>Z = [(54.01+10i),(-2-5i),-50,-0.01;(-2-5i), (4.01+40i),-30i,-0.01;-50,-30i,(50-10i), 0;-0.01,-0.01,0,0.03] Z = 54.0100 + 10.0000i -2.0000-5.0000i, -50.0000 -0.0100 -2.0000-5.0000i 4.0100-40.0000i 0-30.0000i 0.0100 -50.0000 0-30.0000i 50.0000-10.0000i 0 -0.0100 -0.0100 0 0.0300 >>V = [(204.8 + 36.12i);(-71.14-195.46i);0;0]
-
-```
-V =
-1.0e+002*
-2.0480+0.3612i
--0.7114-1.9546i
- 0
- 0
->>I = inv(Z)*V
-I =
-8.9309+2.6973i
-0.0093+4.5159i
-5.4623+3.7954i
-2.9801+2.4044i
- Iab = −I1 + I4 = −(8.931 + j2.697) + (2.98 + j2.404)
-= −5.951 − j0.293 = 5.958⧸−177.18° A. Answer checks.
-```
-
-6. **Satisfactory?** We have a satisfactory solution and an adequate check for the solution. We can now present the results as a solution to the problem.
-
-Practice Problem 12.12 For the unbalanced circuit in Fig. 12.32, use *PSpice* to find the generator current **I***ca*, the line current **I***cC*, and the phase current **I***AB*.
-
-**Answer:** 24.68⧸−90° A, 37.25⧸ 83.79° A, 15.55⧸−75.01° A.
-
-# **12.10** Applications
-
-Both wye and delta source connections have important practical applications. The wye source connection is used for long distance transmission of electric power, where resistive losses ( *I* 2 *R*) should be minimal. This
-
-is due to the fact that the wye connection gives a line voltage that is √ \_\_ 3 greater than the delta connection; hence, for the same power, the line current is √ \_\_ 3smaller. In addition, delta connected are also undesirable due to the potential of having disastrous circulating currents. Sometimes, using transformers, we create the equivalent of delta connect source. This conversion from three-phase to single-phase is required in residential wiring, because household lighting and appliances use single-phase power. Three-phase power is used in industrial wiring where a large power is required. In some applications, it is immaterial whether the load is wye- or delta-connected. For example, both connections are satisfac tory with induction motors. In fact, some manufacturers connect a motor in delta for 220 V and in wye for 440 V so that one line of motors can be readily adapted to two different voltages.
-
-Here we consider two practical applications of those concepts co vered in this chapter: power measurement in three-phase circuits and residential wiring.
-
-# **12.10.1** Three-Phase Power Measurement
-
-Section 11.9 presented the wattmeter as the instrument for measuring the average (or real) power in single-phase circuits. A single wattmeter can also measure the average power in a three-phase system that is bal anced, so that *P*1 = *P*2 = *P*3; the total power is three times the reading of that one wattmeter. However, two or three single-phase wattmeters are necessary to measure power if the system is unbalanced. The *threewattmeter method* of power measurement, shown in Fig. 12.33, will work regardless of whether the load is balanced or unbalanced, wye- or delta-connected. The three-wattmeter method is well suited for power measurement in a three-phase system where the power factor is con stantly changing. The total average power is the algebraic sum of the three wattmeter readings,
-
-$$
-P_T = P_1 + P_2 + P_3 \tag{12.61}
-$$
-
-where *P*1, *P*2, and *P*3 correspond to the readings of wattmeters *W*1, *W*2, and *W*3, respectively. Notice that the common or reference point *o* in Fig. 12.33 is selected arbitrarily. If the load is wye-connected, point *o* can be connected to the neutral point *n*. For a delta-connected load, point *o* can be connected to any point. If point *o* is connected to point *b*, for example, the voltage coil in wattmeter *W*2 reads zero and *P*2 = 0, indicating that wattmeter *W*2 is not necessary. Thus, two wattmeters are sufficient to measure the total power.
-
-The *two-wattmeter method* is the most commonly used method for three-phase power measurement. The two wattmeters must be properly connected to an y two phases, as sho wn typically in Fig. 12.34. Notice that the current coil of each w attmeter measures the line current, while the respective voltage coil is connected between the line and the third line and measures the line v oltage. Also notice that the ± terminal of the voltage coil is connected to the line to which the corresponding current coil is connected. Although the individual wattmeters no longer read the power taken by any particular phase, the algebraic sum of the two wattmeter readings equals the total a verage power absorbed by the load, regardless of whether it is wye- or delta-connected, balanced or
-
-unbalanced. The total real power is equal to the algebraic sum of the two wattmeter readings,
-
-$$
-P_T = P_1 + P_2 \tag{12.62}
-$$
-
-We will show here that the method works for a balanced three-phase system.
-
-Consider the balanced, wye-connected load in Fig. 12.35. Our objective is to apply the two-wattmeter method to find the average power absorbed by the load. Assume the source is in the *abc* sequence and the load impedance **Z***Y* = Z*Y*⧸*θ*. Due to the load impedance, each voltage coil leads its current coil by *θ*, so that the po wer factor is cos *θ*. We recall that each line voltage leads the corresponding phase voltage by 30°. Thus, the total phase dif ference between the phase current **I***a* and line v oltage **V***ab* is *θ* + 30°, and the a verage po wer read by wattmeter *W*1 is
-
-$$
-P_1 = \text{Re}[\mathbf{V}_{ab}\mathbf{I}_a^*] = V_{ab}I_a \cos(\theta + 30^\circ) = V_L I_L \cos(\theta + 30^\circ) \qquad (12.63)
-$$
-
-**Figure 12.35** Two-wattmeter method applied to a balanced wye load.
-
-Similarly, we can show that the average power read by wattmeter 2 is
-
-$$
-P_2 = \text{Re}[\mathbf{V}_{cb}\mathbf{I}_c^*] = V_{cb}I_c \cos(\theta - 30^\circ) = V_L I_L \cos(\theta - 30^\circ) \qquad (12.64)
-$$
-
-We now use the trigonometric identities
-
-$$
-\cos(A + B) = \cos A \cos B - \sin A \sin B
-$$
-
-\n
-$$
-\cos(A - B) = \cos A \cos B + \sin A \sin B
-$$
- (12.65)
-
-to find the sum and the difference of the two wattmeter readings in Eqs. (12.63) and (12.64):
-
-$$
-P_1 + P_2 = V_L I_L [\cos(\theta + 30^\circ) + \cos(\theta - 30^\circ)]
-$$
-
-= $V_L I_L (\cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ$
-+ $\cos \theta \cos 30^\circ + \sin \theta \sin 30^\circ)$
-= $V_L I_L 2 \cos 30^\circ \cos \theta = \sqrt{3} V_L I_L \cos \theta$ (12.66)
-
-since 2 cos 30° = √ \_\_ 3 . Comparing Eq. (12.66) with Eq. (12.50) shows that the sum of the wattmeter readings gives the total average power,
-
-$$
-P_T = P_1 + P_2 \t\t(12.67)
-$$
-
-Similarly,
-
-$$
-P_1 - P_2 = V_L I_L [\cos(\theta + 30^\circ) - \cos(\theta - 30^\circ)]
-$$
-
-= $V_L I_L (\cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ$
- $- \cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ)$ (12.68)
-= $-V_L I_L 2 \sin 30^\circ \sin \theta$
- $P_2 - P_1 = V_L I_L \sin \theta$
-
-since 2 sin 30° = 1. Comparing Eq. (12.68) with Eq. (12.51) shows that the difference of the wattmeter readings is proportional to the total reactive power, or
-
-$$
-Q_T = \sqrt{3}(P_2 - P_1)
-$$
- (12.69)
-
-From Eqs. (12.67) and (12.69), the total apparent power can be obtained as
-
-$$
-S_T = \sqrt{P_T^2 + Q_T^2}
-$$
- (12.70)
-
-Dividing Eq. (12.69) by Eq. (12.67) gives the tangent of the power fac tor angle as
-
-$$
-\tan \theta = \frac{Q_T}{P_T} = \sqrt{3} \frac{P_2 - P_1}{P_2 + P_1}
-$$
-\n(12.71)
-
-from which we can obtain the power factor as pf = cos *θ*. Thus, the two-wattmeter method not only provides the total real and reactive powers, it can also be used to compute the power factor. From Eqs. (12.67), (12.69), and (12.71), we conclude that:
-
-- 1. If *P*2 = *P*1, the load is resistive.
-- 2. If *P*2 > *P*1, the load is inductive.
-- 3. If *P*2 < *P*1, the load is capacitive.
-
-Although these results are derived from a balanced wye-connected load, they are equally valid for a balanced delta-connected load. However, the two-wattmeter method cannot be used for power measurement in a three-phase four-wire system unless the current through the neutral line is zero. We use the three-wattmeter method to measure the real power in a three-phase four-wire system.
-
-# Three wattmeters *W*1, *W*2, and *W*3 are connected, respectively, to phases *a, b,* and *c* to measure the total power absorbed by the unbalanced wye-connected load in Example 12.9 (see Fig. 12.23). (a) Predict the wattmeter readings. (b) Find the total power absorbed.
-
-# **Solution:**
-
-# Example 12.13
-
-Part of the problem is already solved in Example 12.9. Assume that the wattmeters are properly connected as in Fig. 12.36.
-
-**Figure 12.36** For Example 12.13.
-
-(a) From Example 12.9,
-
-$$
-\mathbf{V}_{AN} = 100 \underline{\text{/}0^{\circ}}, \qquad \mathbf{V}_{BN} = 100 \underline{\text{/}120^{\circ}}, \qquad \mathbf{V}_{CN} = 100 \underline{\text{/} -120^{\circ}} \text{ V}
-$$
-
-while
-
-$$
-\mathbf{I}_a = 6.67 \underline{\bigcirc}^{\circ}, \qquad \mathbf{I}_b = 8.94 \underline{\bigcirc} 3.44^{\circ}, \qquad \mathbf{I}_c = 10 \underline{\bigcirc} -66.87^{\circ} \text{ A}
-$$
-
-We calculate the wattmeter readings as follows:
-
-$$
-P_1 = \text{Re}(\mathbf{V}_{AN}\mathbf{I}_{a}^{*}) = V_{AN}I_a \cos(\theta_{\mathbf{V}_{AN}} - \theta_{\mathbf{I}_a})
-$$
-
-= 100 × 6.67 × cos(0° – 0°) = 667 W
-$$
-P_2 = \text{Re}(\mathbf{V}_{BN}\mathbf{I}_{b}^{*}) = V_{BN}I_b \cos(\theta_{\mathbf{V}_{BN}} - \theta_{\mathbf{I}_b})
-$$
-
-= 100 × 8.94 × cos(120° – 93.44°) = 800 W
-$$
-P_3 = \text{Re}(\mathbf{V}_{CN}\mathbf{I}_{c}^{*}) = V_{CN}I_c \cos(\theta_{\mathbf{V}_{CN}} - \theta_{\mathbf{I}_c})
-$$
-
-= 100 × 10 × cos(-120° + 66.87°) = 600 W
-
-(b) The total power absorbed is
-
-$$
-P_T = P_1 + P_2 + P_3 = 667 + 800 + 600 = 2067
-$$
- W
-
-We can find the power absorbed by the resistors in Fig. 12.36 and use that to check or confirm this result
-
-$$
-P_T = |I_a|^2(15) + |I_b|^2(10) + |I_c|^2(6)
-$$
-
-= 6.672(15) + 8.942(10) + 102(6)
-= 667 + 800 + 600 = 2067 W
-
-which is exactly the same thing.
-
-| Practice Problem 12.13 | Repeat Example 12.13 for the network in Fig. 12.24 (see Practice
Prob. 12.9). Hint: Connect the reference point o in Fig. 12.33 to point B.
Answer: (a) 13.175 kW, 0 W, 29.91 kW, (b) 43.08 kW. |
-|------------------------|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
-| Example 12.14 | The two-wattmeter method produces w attmeter readings P1 = 1560 W
and P2 = 2100 W when connected to a delta-connected load. If the line
voltage is 220 V, calculate: (a) the per-phase average power, (b) the per
phase reactive power, (c) the power factor, and (d) the phase impedance. |
-
-# **Solution:**
-
-We can apply the given results to the delta-connected load. (a) The total real or average power is
-
-$$
-P_T = P_1 + P_2 = 1560 + 2100 = 3660
-$$
- W
-
-The per-phase average power is then
-
-$$
-P_p = \frac{1}{3}P_T = 1220 \text{ W}
-$$
-
-(b) The total reactive power is
-
-$$
-Q_T = \sqrt{3}(P_2 - P_1) = \sqrt{3}(2100 - 1560) = 935.3 \text{ VAR}
-$$
-
-so that the per-phase reactive power is
-
-$$
-Q_p = \frac{1}{3}Q_T = 311.77 \text{ VAR}
-$$
-
-(c) The power angle is
-
-$$
-\theta = \tan^{-1} \frac{Q_T}{P_T} = \tan^{-1} \frac{935.3}{3660} = 14.33^{\circ}
-$$
-
-Hence, the power factor is
-
-$$
-\cos \theta = 0.9689(\text{lagging})
-$$
-
-It is a lagging pf because *QT* is positive or *P*2 > *P*1. (d) The phase impedance is **Z***p* = *Zp*⧸*θ*. We know that *θ* is the same as the pf angle; that is, *θ* = 14.33°.
-
-$$
-Z_p = \frac{V_p}{I_p}
-$$
-
-We recall that for a delta-connected load, *Vp* = *VL* = 220 V. From Eq. (12.46),
-
-2.46),
-\n
-$$
-P_p = V_p I_p \cos \theta \implies I_p = \frac{1220}{220 \times 0.9689} = 5.723 \text{ A}
-$$
-
-Hence,
-
-$$
-Z_p = \frac{V_p}{I_p} = \frac{220}{5.723} = 38.44 \ \Omega
-$$
-
-and
-
-$$
-Z_p = 38.44 / 14.33^{\circ} \,\Omega
-$$
-
-Let the line voltage *VL* = 208 V and the wattmeter readings of the balanced system in Fig. 12.35 be *P*1 = −560 W and *P*2 = 800 W. Determine:
-
-(a) the total average power
-
-- (b) the total reactive power
-- (c) the power factor
-
-(d) the phase impedance
-
-Is the impedance inductive or capacitive?
-
-**Answer:** (a) 240 W, (b) 2.356 kV AR, (c) 0.1014, (d) 18.25 ⧸ 84.18° Ω, inductive.
-
-Practice Problem 12.14
-
-Example 12.15
-
-The three-phase balanced load in Fig. 12.35 has impedance per phase of **Z***Y* = 8 + *j*6 Ω. If the load is connected to 208-V lines, predict the read ings of the wattmeters *W*1 and *W*2. Find *PT* and *QT*.
-
-# **Solution:**
-
-The impedance per phase is
-
-$$
-\mathbf{Z}_{Y} = 8 + j6 = 10/36.87^{\circ} \,\Omega
-$$
-
-so that the pf angle is 36.87°. Since the line voltage *VL* = 208 V, the line current is \_\_
-
-$$
-I_L = \frac{V_p}{|\mathbf{Z}_Y|} = \frac{208/\sqrt{3}}{10} = 12 \text{ A}
-$$
-
-Then
-
-$$
-P_1 = V_L I_L \cos(\theta + 30^\circ) = 208 \times 12 \times \cos(36.87^\circ + 30^\circ)
-$$
-
-= 980.48 W
-$$
-P_2 = V_L I_L \cos(\theta - 30^\circ) = 208 \times 12 \times \cos(36.87^\circ - 30^\circ)
-$$
-
-= 2478.1 W
-
-Thus, wattmeter 1 reads 980.48 W, while wattmeter 2 reads 2478.1 W. Since *P*2 > *P*1, the load is inductive. This is evident from the load **Z***Y* itself. Next,
-
-$$
-P_T = P_1 + P_2 = 3.459 \text{ kW}
-$$
-
-and
-
-$$
-Q_T = \sqrt{3}(P_2 - P_1) = \sqrt{3}(1497.6)
-$$
- VAR = 2.594 kVAR
-
-If the load in Fig. 12.35 is delta-connected with impedance per phase of **Z***p* = 30 + *j*40 Ω and *VL* = 220 V, predict the readings of the wattmeters *W*1 and *W*2. Calculate *PT* and *QT*. Practice Problem 12.15
-
-**Answer:** 200.6 W, 1.5418 kW, 1.7424 kW, 2.323 kVAR.
-
-# **12.10.2** Residential Wiring
-
-In the United States, most household lighting and appliances operate on 120-V, 60-Hz, single-phase alternating current. (The electricity may also be supplied at 110, 115, or 117 V, depending on the location.) The local power company supplies the house with a three-wire ac system. Typically, as in Fig. 12.37, the line voltage of, say, 12,000 V is stepped down to 120/240 V with a transformer (more details on transformers in the next chapter). The three wires coming from the transformer are typically colored red (hot), black (hot), and white (neutral). As shown in Fig. 12.38, the two 120-V voltages are opposite in phase and hence add up to zero.
-
-That is,
-$$
-\mathbf{V}_W = 0/\underline{0}^\circ
-$$
-, $\mathbf{V}_B = 120/\underline{0}^\circ$ , $\mathbf{V}_R = 120/\underline{180}^\circ = -\mathbf{V}_B$ .
-\n
-$$
-\mathbf{V}_{BR} = \mathbf{V}_B - \mathbf{V}_R = \mathbf{V}_B - (-\mathbf{V}_B) = 2\mathbf{V}_B = 240/\underline{0}^\circ \qquad (12.72)
-$$
-
-# **Figure 12.37**
-
-A 120/240 household power system. Source: A. Marcus and C. M. Thomson, *Electricity for Technicians,* 2nd edition, © 1975, p. 324. Pearson Education, Inc., Upper Saddle River, NJ.
-
-# **Figure 12.38**
-
-Single-phase three-wire residential wiring.
-
-Since most appliances are designed to operate with 120 V, the light ing and appliances are connected to the 120-V lines, as illustrated in Fig. 12.39 for a room. Notice in Fig. 12.37 that all appliances are connected in parallel. Hea vy appliances that consume lar ge currents, such as air conditioners, dishwashers, ovens, and laundry machines, are con nected to the 240-V power line.
-
-Because of the dangers of electricity , house wiring is carefully regulated by a code dra wn by local ordinances and by the National Electrical Code (NEC). To avoid trouble, insulation, grounding, fuses, and circuit breakers are used. Modern wiring codes require a third wire for a separate ground. The ground wire does not carry po wer like the neutral wire but enables appliances to have a separate ground connection. Figure 12.40 shows the connection of the receptacle to a 120-V rms line and to the ground. As shown in the figure, the neutral line is connected to the ground (the earth) at man y critical locations. Although the ground line seems redundant, grounding is important for many reasons. First, it is required by NEC. Second, grounding provides
-
-# **Figure 12.39**
-
-A typical wiring diagram of a room. Source: A. Marcus and C. M. Thomson, *Electricity for Technicians,* 2nd edition, © 1975, p. 325. Pearson Education, Inc., Upper Saddle River, NJ.
-
-a convenient path to ground for lightning that strik es the po wer line. Third, grounds minimize the risk of electric shock. What causes shock is the passage of current from one part of the body to another . The human body is lik e a big resistor *R*. If *V* is the potential dif ference between the body and the ground, the current through the body is determined by Ohm's law as
-
-$$
-I = \frac{V}{R} \tag{12.73}
-$$
-
-The value of *R* varies from person to person and depends on whether the body is wet or dry. How great or how deadly the shock is depends on the amount of current, the pathway of the current through the body, and the length of time the body is e xposed to the current. Currents less than 1 mA may not be harmful to the body , but currents greater than 10 mA can cause se vere shock. A modern safety de vice is the *ground-fault circuit interrupter* (GFCI), used in outdoor circuits and in bathrooms, where the risk of electric shock is greatest. It is essentially a circuit break er that opens when the sum of the currents *iR*, *iW*, and *iB* through the red, white, and the black lines is not equal to zero, or *iR* + *iW* + *iB* ≠ 0.
-
-The best w ay to a void electric shock is to follo w safety guide lines concerning electrical systems and appliances. Here are some of them:
-
-- Never assume that an electrical circuit is dead. Always check to be sure.
-- Use safety de vices when necessary , and wear suitable clothing (insulated shoes, gloves, etc.).
-- Never use tw o hands when testing high-v oltage circuits, since the current through one hand to the other hand has a direct path through your chest and heart.
-- Do not touch an electrical appliance when you are wet. Remember that water conducts electricity.
-- Be extremely careful when working with electronic appliances such as radio and TV because these appliances ha ve lar ge capacitors in them. The capacitors tak e time to dischar ge after the po wer is disconnected.
-- Always ha ve another person present when w orking on a wiring system, just in case of an accident.
-
-# **12.11** Summary
-
-- 1. The phase sequence is the order in which the phase v oltages of a three-phase generator occur with respect to time. In an *abc* sequence of balanced source v oltages, **V***an* leads **V***bn* by 120°, which in turn leads **V***cn* by 120°. In an *acb* sequence of balanced v oltages, **V***an* leads **V***cn* by 120°, which in turn leads **V***bn* by 120°.
-- 2. A balanced wye- or delta-connected load is one in which the threephase impedances are equal.
-- 3. The easiest way to analyze a balanced three-phase circuit is to transform both the source and the load to a Y-Y system and then analyze the single-phase equivalent circuit. Table 12.1 presents a summary of the formulas for phase currents and voltages and line currents and voltages for the four possible configurations.
-- 4. The line current *IL* is the current flowing from the generator to the load in each transmission line in a three-phase system. The line voltage *VL* is the v oltage between each pair of lines, e xcluding the neutral line if it e xists. The phase current *Ip* is the current flowing through each phase in a three-phase load. The phase voltage *Vp* is the voltage of each phase. For a wye-connected load,
-
-$$
-V_L = \sqrt{3} V_p \qquad \text{and} \qquad I_L = I_p
-$$
-
-For a delta-connected load,
-
-$$
-V_L = V_p \qquad \text{and} \qquad I_L = \sqrt{3}I_p
-$$
-
-- 5. The total instantaneous po wer in a balanced three-phase system is constant and equal to the average power.
-- 6. The total comple x po wer absorbed by a balanced three-phase Y-connected or ∆-connected load is
-
-$$
-\mathbf{S} = P + jQ = \sqrt{3} V_L I_L \underline{\theta}
-$$
-
-where *θ* is the angle of the load impedances.
-
-- 7. An unbalanced three-phase system can be analyzed using nodal or mesh analysis.
-- 8. *PSpice* is used to analyze three-phase circuits in the same w ay as it is used for analyzing single-phase circuits.
-- 9. The total real power is measured in three-phase systems using either the three-wattmeter method or the two-wattmeter method.
-- 10. Residential wiring uses a 120/240-V, single-phase, three-wire system.
-
-# Review Questions
-
-**12.1** What is the phase sequence of a three-phase motor for which **V***AN* = 220⧸−100° V and **V***BN* = 220⧸ 140° V?
-
-(a) *abc* (b) *acb*
-
-**12.2** If in an *acb* phase sequence, *Van* = 100⧸−20°, then **V***cn* is:
-
-(a)
-$$
-100 \div 140^{\circ}
-$$
- (b) $100 \div 100^{\circ}$
-
-(c)
-$$
-100\div 50^{\circ}
-$$
- (d) $100\div 10^{\circ}$
-
-**12.3** Which of these is not a required condition for a balanced system:
-
-$$
-(a) |\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}|
-$$
-
-(b)
-$$
-\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c = 0
-$$
-
-(c)
-$$
-V_{an} + V_{bn} + V_{cn} = 0
-$$
-
-- (d) Source voltages are 120° out of phase with each other.
-- (e) Load impedances for the three phases are equal.
-
-**12.4** In a Y-connected load, the line current and phase current are equal.
-
-(a) True (b) False
-
-**12.5** In a ∆-connected load, the line current and phase current are equal.
-
-(a) True (b) False
-
-**12.6** In a Y-Y system, a line voltage of 220 V produces a phase voltage of:
-
-(a) 381 V (b) 311 V (c) 220 V (d) 156 V (e) 127 V
-
-**12.7** In a ∆-∆ system, a phase voltage of 100 V produces a line voltage of:
-
-(a) 58 V (b) 71 V (c) 100 V (d) 173 V (e) 141 V
-
-Problems1
-
-# Section 12.2 Balanced Three-Phase Voltages
-
-**12.1** If **V***ab* = 400 V in a balanced Y-connected threephase generator, find the phase voltages, assuming the phase sequence is:
-
-(*a*) *abc* (*b*) *acb*
-
-- **12.2** What is the phase sequence of a balanced threephase circuit for which **V***an* = 120⧸30° V and **V***cn* = 120⧸−90° V? Find **V***bn*.
-- **12.3** Given a balanced Y-connected three-phase generator with a line-to-line voltage of **V***ab* = 100⧸45° V and **V***bc* = 100⧸165° V, determine the phase sequence and the value of **V***ca*.
-- **12.4** A three-phase system with *abc* sequence and *VL* = 440 V feeds a Y-connected load with *ZL* = 40⧸30° Ω. Find the line currents.
-- **12.5** For a Y-connected load, the time-domain expressions for three line-to-neutral voltages at the terminals are:
-
-*vAN* = 120 cos(*ωt* + 32°) V *vBN* = 120 cos(*ωt* – 88°) V *vCN* = 120 cos(*ωt* + 152°) V
-
- Write the time-domain expressions for the line-toline voltages *vAB*, *vBC*, and *vCA*.
-
-# Section 12.3 Balanced Wye-Wye Connection
-
-**12.6** Using Fig. 12.41, design a problem to help other students better understand balanced wye-wye connected circuits.
-
-**12.8** When a Y-connected load is supplied by voltages in *abc* phase sequence, the line voltages lag the corresponding phase voltages by 30°.
-
-(a) True (b) False
-
-**12.9** In a balanced three-phase circuit, the total instantaneous power is equal to the average power.
-
-(a) True (b) False
-
-**12.10** The total power supplied to a balanced ∆-load is found in the same way as for a balanced Y-load.
-
-(a) True (b) False
-
-*Answers: 12.1a, 12.2a, 12.3c, 12.4a, 12.5b, 12.6e, 12.7c, 12.8b, 12.9a, 12.10a.*
-
-# **Figure 12.41**
-
-For Prob. 12.6.
-
-- **12.7** Obtain the line currents in the three-phase circuit of Fig. 12.42 on the next page.
-- **12.8** In a balanced three-phase Y-Y system, the source is an *acb* sequence of voltages and **V***cn* = 120⧸ 35° V rms. The line impedance per phase is (1 + *j*2) Ω, while the per-phase impedance of the load is (11 + *j*14) Ω. Calculate the line currents and the load voltages.
-- **12.9** A balanced Y-Y four-wire system has phase voltages
-
-$$
-\mathbf{V}_{an} = 120 \underline{\text{O}^{\circ}}, \qquad \mathbf{V}_{bn} = 120 \underline{\text{O} - 120^{\circ}}
-$$
-$$
-\mathbf{V}_{cn} = 120 \underline{\text{O} \cdot 120^{\circ}} \text{ V}
-$$
-
- The load impedance per phase is 19 + *j*13 Ω, and the line impedance per phase is 1 + *j*2 Ω. Solve for the line currents and neutral current.
-
-1 Remember that unless stated otherwise, all given voltages and currents are rms values.
-
-Problems **543**
-
-**12.10** For the circuit in Fig. 12.43, determine the current in the neutral line.
-
-For Prob. 12.10.
-
-# Section 12.4 Balanced Wye-Delta Connection
-
-**12.11** In the Y-∆ system shown in Fig. 12.44, the source is a positive sequence with **V***an* = 440⧸ 0° V and phase impedance **Z***p* = 2 – *j*3 Ω. Calculate the line voltage **V***L* and the line current **I***L*.
-
-**Figure 12.44** For Prob. 12.11.
-
-**Figure 12.46** For Prob. 12.13.
-
-**12.14** Obtain the line currents in the three-phase circuit of Fig. 12.47 on the next page.
-
-For Prob. 12.14.
-
-**12.15** The circuit in Fig. 12.48 is excited by a balanced three-phase source with a line voltage of 210 V. If **Z***l* = 1 + *j*1 Ω, **Z**∆ = 24 − *j*30 Ω, and **Z***Y* = 12 + *j*5 Ω, determine the magnitude of the line current of the combined loads.
-
-For Prob. 12.15.
-
-- **12.16** A balanced delta-connected load has a phase current **I***AC* = 5⧸−30° A.
- - (a) Determine the three line currents assuming that the circuit operates in the positive phase sequence.
- - (b) Calculate the load impedance if the line voltage is **V***AB* = 440⧸ 0° V.
-- **12.17** A positive sequence wye-connected source where **V***an* = 120⧸ 90° V, is connected to a delta-connected load where **Z***L* = (60 + *j*45) Ω. Determine the line currents.
-- **12.18** If **V***an* = 220⧸ 60° V in the network of Fig. 12.49, find the load phase currents **I***AB*, **I***BC*, and **I***CA*.
-
-**Figure 12.49** For Prob. 12.18.
-
-# Section 12.5 Balanced Delta-Delta Connection
-
-**12.19** For the ∆-∆ circuit of Fig. 12.50, calculate the phase and line currents.
-
-# **Figure 12.50**
-
-For Prob. 12.19.
-
-**12.20** Using Fig. 12.51, design a problem to help other students better understand balanced delta-delta connected circuits.
-
-For Prob. 12.20.
-
-**12.21** Three 440-V generators form a delta-connected source that is connected to a balanced deltaconnected load of *ZL* = (8.66 + *j*5) Ω per phase as shown in Fig. 12.52. Determine the value of *IBC* and *IaA*. What is the pf of the load?
-
-b
-
-+
-
-+ ‒
-
-- **12.22** Find the line currents *IaA*, *IbB*, and *IcC* in the three-phase network of Fig. 12.53. Take **Z***L* = (114 + *j*87) Ω and **Z***l* = (2 + *j*) Ω.
-- **12.23** A balanced delta connected source is connected to a balanced delta connected load where *ZL* = (80 + *j*60) Ω and *Zl* = (2 + *j*) Ω. Given that the load voltages are **V***AB* = 100⧸ 0° V, **V***BC* = 100⧸ 120° V, and **V***CA* = 100⧸−120° V. Calculate the source voltages **V***ab*, **V***bc*, and **V***ca*.
-- **12.24** A balanced delta-connected source has phase voltage **V***ab* = 880⧸ 30° V and a positive phase sequence. If this is connected to a balanced delta-connected load, find the line and phase currents. Take the load impedance per phase as 60⧸ 30° Ω and line impedance per phase as 1 + *j*1 Ω.
-- **12.26** Using Fig. 12.55, design a problem to help other students better understand balanced delta connected sources delivering power to balanced wye connected loads.
-
-**Figure 12.55** For Prob. 12.26.
-
-10 ‒ j8 Ω
-
-10 ‒ j8 Ω
-
-# Section 12.6 Balanced Delta-Wye Connection
-
-12.25 In the circuit of Fig. 12.54, if
-$$
-\mathbf{V}_{ab} = 440/10^{\circ}
-$$
-,
-\n $V_{bc} = 440/-110^{\circ}$ , $V_{ca} = 440/130^{\circ}$ V, find the line currents.
-
-3 + j2 Ω 10 ‒ j8 W
-
-**I**b
-
-**I**a
-
-a 3+ j2 W
-
-Vab
-
-- **12.27** A ∆-connected source supplies power to a Yconnected load in a three-phase balanced system. Given that the line impedance is 2 + *j*1 Ω per phase while the load impedance is 6 + *j*4 Ω per phase, find the magnitude of the line voltage at the load. Assume the source phase voltage **V***ab* = 208⧸ 0° V rms.
-- **12.28** The line-to-line voltages in a Y-load have a magnitude of 880 V and are in the positive sequence at 60 Hz. If the loads are balanced with *Z*1 = *Z*2 = *Z*3 = 25⧸30°, find all line currents and phase voltages.
-
-# Section 12.7 Power in a Balanced System
-
-- **12.29** A balanced three-phase Y-∆ system has **V***an* = 240⧸ 0° V rms and **Z**∆ = 51 + *j*45 Ω. If the line impedance per phase is 0.4 + *j*1.2 Ω, find the total complex power delivered to the load.
-- **12.30** In Fig. 12.56, the rms value of the line voltage is 208 V. Find the average power delivered to the load.
-
-**Figure 12.56**
-
-For Prob. 12.30.
-
-- **12.31** A balanced delta-connected load is supplied by a 60-Hz three-phase source with a line voltage of 480 V. Each load phase draws 24 kW at a lagging power factor of 0.8. Find:
- - (a) the load impedance per phase
- - (b) the line current
- - (c) the value of capacitance needed to be connected in parallel with each load phase to minimize the current from the source
-
-- **12.32** Design a problem to help other students better understand power in a balanced three-phase system.
-- **12.33** A three-phase source delivers 4.8 kVA to a wyeconnected load with a phase voltage of 208 V and a power factor of 0.9 lagging. Calculate the source line current and the source line voltage.
-- **12.34** A balanced wye-connected load with a phase impedance of 10 – *j*16 Ω is connected to a balanced three-phase generator with a line voltage of 220 V. Determine the line current and the complex power absorbed by the load.
-- **12.35** Three equal impedances, 60 + *j*30 Ω each, are delta-connected to a 230-V rms, three-phase circuit. Another three equal impedances, 40 + *j*10 Ω each, are wye-connected across the same circuit at the same points. Determine:
- - (a) the line current
- - (b) the total complex power supplied to the two loads
- - (c) the power factor of the two loads combined
-- **12.36** A 4200-V, three-phase transmission line has an impedance of 4 + *j* Ω per phase. If it supplies a load of 1 MVA at 0.75 power factor (lagging), find:
- - (a) the complex power
- - (b) the power loss in the line
- - (c) the voltage at the sending end
-- **12.37** The total power measured in a three-phase system feeding a balanced wye-connected load is 12 kW at a power factor of 0.6 leading. If the line voltage is 440 V, calculate the line current *IL* and the load impedance **Z***Y*.
-- **12.38** Given the circuit in Fig. 12.57 below, find the total complex power absorbed by the load.
-
-**Figure 12.57** For Prob. 12.38.
-
-# For Prob. 12.39.
-
-**12.40** For the three-phase circuit in Fig. 12.59, find the average power absorbed by the delta-connected load with **Z**∆ = 21 + *j*24 Ω.
-
-# **Figure 12.59** For Prob. 12.40.
-
-- **12.41** A balanced delta-connected load draws 5 kW at a power factor of 0.8 lagging. If the three-phase system has an effective line voltage of 400 V, find the line current.
-- **12.42** A balanced three-phase generator delivers 7.2 kW to a wye-connected load with impedance 30 – *j*40 Ω per phase. Find the line current *IL* and the line voltage *VL*.
-- **12.43** Refer to Fig. 12.48. Obtain the complex power absorbed by the combined loads.
-- **12.44** A three-phase line has an impedance of 1 + *j*3 Ω per phase. The line feeds a balanced delta-connected load, which absorbs a total complex power of 12 + *j*5 kVA. If the line voltage at the load end has a magnitude of 240 V, calculate the magnitude of the line voltage at the source end and the source power factor.
-- **12.45** A balanced wye-connected load is connected to the generator by a balanced transmission line with an impedance of 0.5 + *j*2 Ω per phase. If the load is rated at 450 kW, 0.708 power factor lagging, 440-V line voltage, find the line voltage at the generator.
-- **12.46** A three-phase load consists of three 100-Ω resistors that can be wye- or delta-connected. Determine which connection will absorb the most average
-
-power from a three-phase source with a line voltage of 110 V. Assume zero line impedance.
-
-**12.47** The following three parallel-connected three-phase loads are fed by a balanced three-phase source:
-
-> Load 1: 250 kVA, 0.8 pf lagging Load 2: 300 kVA, 0.95 pf leading Load 3: 450 kVA, unity pf
-
- If the line voltage is 13.8 kV, calculate the line current and the power factor of the source. Assume that the line impedance is zero.
-
-- **12.48** A balanced, positive-sequence wye-connected source has **V***an* = 240⧸ 0° V rms and supplies an unbalanced delta-connected load via a transmission line with impedance 2 + *j*3 Ω per phase.
- - (a) Calculate the line currents if **Z***AB* = 40 + *j*15 Ω, **Z***BC* = 60 Ω, **Z***CA* = 18 – *j*12 Ω.
- - (b) Find the complex power supplied by the source.
-- **12.49** Each phase load consists of a 20-Ω resistor and a 10-Ω inductive reactance. With a line voltage of 480 V rms, calculate the average power taken by the load if:
-
-(a) the three-phase loads are delta-connected (b) the loads are wye-connected
-
-**12.50** A balanced three-phase source with **V***L* = 240 V rms is supplying 8 kVA at 0.6 power factor lagging to two wye-connected parallel loads. If one load draws 3 kW at unity power factor, calculate the impedance per phase of the second load.
-
-# Section 12.8 Unbalanced Three-Phase Systems
-
-**12.51** Consider the wye-delta system shown in Fig. 12.60. Let **Z**1 = 100 Ω, **Z**2 = *j*100 Ω, and **Z**3 = –*j*100 Ω. Determine the phase currents, **I***AB*, **I***BC*, and **I***CA*, and the line currents, **I***aA*, **I***bB* , and **I***cC*.
-
-**Figure 12.60** For Prob. 12.51.
-
-$$
-\mathbf{V}_{an} = 220/120^{\circ}, \qquad \mathbf{V}_{bn} = 220/0^{\circ}
-$$
-$$
-\mathbf{V}_{cn} = 220/-120^{\circ} \text{ V}
-$$
-
-If the impedances are
-
-$$
-\mathbf{Z}_{AN} = 20/60^\circ, \qquad \mathbf{Z}_{BN} = 30/0^\circ
-$$
-$$
-\mathbf{Z}_{cn} = 40/30^\circ \ \Omega
-$$
-
-find the current in the neutral line.
-
-**12.53** Using Fig. 12.61, design a problem that will help other students better understand unbalanced threephase systems.
-
-- For Prob. 12.53.
-- **12.54** A balanced three-phase Y-source with *VP* = 880 V rms drives a Y-connected three-phase load with phase impedance **Z***A* = 80 Ω, **Z***B* = 60 + *j*90 Ω, and **Z***C* = *j*80 Ω. Calculate the line currents and total complex power delivered to the load. Assume that the neutrals are connected.
-- **12.55** A three-phase supply, with the line-to-line voltage of 240 V rms, has the unbalanced load as shown in Fig. 12.62. Find the line currents and the total complex power delivered to the load.
-
-**Figure 12.62** For Prob. 12.55.
-
-**12.56** Using Fig. 12.63, design a problem to help other students to better understand unbalanced three-phase systems.
-
-# **Figure 12.63**
-
-For Prob. 12.56.
-
-**12.57** Determine the line currents for the three-phase circuit of Fig. 12.64. Let **V***a* = 220⧸ 0°, **V***b* = 220⧸−120°, **V***c* = 220⧸ 120° V.
-
-For Prob. 12.57.
-
-# Section 12.9 PSpice for Three-Phase Circuits
-
-- **12.58** Solve Prob. 12.10 using *PSpice or MultiSim*.
-- **12.59** The source in Fig. 12.65 is balanced and exhibits a positive phase sequence. If *f* = 60 Hz, use *PSpice or MultiSim* to find **V***AN*,**V***BN*, and **V***CN*.
-
-For Prob. 12.59.
-
-**12.60** Use *PSpice or MultiSim* to determine **I***o* in the single-phase, three-wire circuit of Fig. 12.66. Let **Z**1 = 15 – *j*10 Ω, **Z**2 = 30 + *j*20 Ω, and **Z**3 = 12 + *j*5 Ω.
-
-**12.61** Given the circuit in Fig. 12.67, use *PSpice or MultiSim* to determine currents **I***aA* and voltage **V***BN*.
-
-**Figure 12.67**
-
-For Prob. 12.61.
-
-**12.62** Using Fig. 12.68, design a problem to help other students better understand how to use *PSpice or MultiSim* to analyze three-phase circuits.
-
-**12.63** Use *PSpice or MultiSim* to find currents **I***aA* and **I***AC* in the unbal anced three-phase system shown in Fig. 12.69. Let
-
-$$
-Zl = 2 + j, \t Z1 = 40 + j20 Ω,\nZ2 = 50 - j30 Ω, \t Z3 = 25 Ω
-$$
-
-# **Figure 12.69**
-
-For Prob. 12.63.
-
-- **12.64** For the circuit in Fig. 12.58, use *PSpice or MultiSim* to find the line currents and the phase currents.
-- **12.65** A balanced three-phase circuit is shown in Fig. 12.70 on the next page. Use *PSpice or MultiSim* to find the line currents **I***aA*, **I***bB*, and **I***cC*.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/143_12.10 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/143_12.10 Applications.md
deleted file mode 100644
index d5152bbab82ecd70cf98ff6fc2ece2eb44152826..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/143_12.10 Applications.md
+++ /dev/null
@@ -1,53 +0,0 @@
-# Section 12.10 Applications
-
-- **12.66** A three-phase, four-wire system operating with a 480-V line voltage is shown in Fig. 12.71. The source voltages are balanced. The power absorbed by the resistive wye-connected load is measured by the three-wattmeter method. Calculate:
- - (a) the voltage to neutral
- - (b) the currents **I**1, **I**2, **I**3, and **I***n*
- - (c) the readings of the wattmeters
- - (d) the total power absorbed by the load
-- **12.67** As shown in Fig. 12.72, a three-phase four-wire line with a phase voltage of 120 V rms and positive phase sequence supplies a balanced motor load at 260 kVA at 0.85 pf lagging. The motor load is connected to the three main lines marked *a, b,* and *c*. In addition, incandescent lamps (unity pf) are connected as follows: 24 kW from line *c* to the neutral, 15 kW from line b to the neutral, and 9 kW from line *c* to the neutral.
- - (a) If three wattmeters are arranged to measure the power in each line, calculate the reading of each meter.
- - (b) Find the magnitude of the current in the neutral line.
-
-\* An asterisk indicates a challenging problem.
-
-**Figure 12.68** For Prob. 12.62.
-
-**Figure 12.70** For Prob. 12.65.
-
-- **12.68** Meter readings for a three-phase wye-connected alternator supplying power to a motor indicate that the line voltages are 330 V, the line currents are 8.4 A, and the total line power is 4.5 kW. Find:
- - (a) the load in VA
- - (b) the load pf
- - (c) the phase current
- - (d) the phase voltage
-- **12.69** A certain store contains three balanced three-phase loads. The three loads are:
-
-Load 1: 16 kVA at 0.85 pf lagging Load 2: 12 kVA at 0.6 pf lagging Load 3: 8 kW at unity pf
-
- The line voltage at the load is 208 V rms at 60 Hz, and the line impedance is 0.4 + *j*0.8 Ω. Determine the line current and the complex power delivered to the loads.
-
-# **Figure 12.72**
-
-For Prob. 12.67.
-
-- **12.70** The two-wattmeter method gives *P*1 = 1200 W and *P*2 = −400 W for a three-phase motor running on a 240-V line. Assume that the motor load is wye- connected and that it draws a line current of 6 A. Calculate the pf of the motor and its phase impedance.
-- **12.71** In Fig. 12.73, two wattmeters are properly connected to the unbalanced load supplied by a balanced source such that **V***ab* = 208⧸ 0° V with positive phase sequence.
- - (a) Determine the reading of each wattmeter.
- - (b) Calculate the total apparent power absorbed by the load.
-
-**Figure 12.73** For Prob. 12.71.
-
-- **12.72** If wattmeters *W*1 and *W*2 are properly connected respectively between lines *a* and *b* and lines *b* and *c* to measure the power absorbed by the deltaconnected load in Fig. 12.44, predict their readings.
-- **12.73** For the circuit displayed in Fig. 12.74, find the wattmeter readings.
-
-# **Figure 12.74**
-
-For Prob. 12.73.
-
-**12.74** Predict the wattmeter readings for the circuit in Fig. 12.75.
-
-# **Figure 12.75** For Prob. 12.74.
-
-- **12.75** A man has a body resistance of 600 Ω. How much current flows through his ungrounded body:
- - (a) when he touches the terminals of a 12-V autobattery?
- - (b) when he sticks his finger into a 120-V light socket?
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/144_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/144_Problems.md
deleted file mode 100644
index 46bccc2e8429c2c693003054c88c81103809d897..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/144_Problems.md
+++ /dev/null
@@ -1,39 +0,0 @@
-# Comprehensive Problems
-
-- **12.77** A three-phase generator supplied 10 kVA at a power factor of 0.85 lagging. If 7,500 W are delivered to the load and line losses are 160 W per phase, what are the losses in the generator?
-- **12.78** A three-phase 440-V, 51-kW, 60-kVA inductive load operates at 60 Hz and is wye-connected. It is desired to correct the power factor to 0.95 lagging. What value of capacitor should be placed in parallel with each load impedance?
-- **12.79** A balanced three-phase generator has an *abc* phase sequence with phase voltage **V***an* = 554.3⧸ 0° V. The generator feeds an induction motor which may be represented by a balanced Y-connected load with an impedance of 12 + *j*5 Ω per phase. Find the line currents and the load voltages. Assume a line impedance of 2 Ω per phase.
-- **12.80** A balanced three-phase source furnishes power to the following three loads:
-
-Load 1: 6 kVA at 0.83 pf lagging Load 2: unknown Load 3: 8 kW at 0.7071 pf leading
-
- If the line current is 84.6 A rms, the line voltage at the load is 208 V rms, and the combined load has a 0.8 pf lagging, determine the unknown load.
-
-**12.81** A professional center is supplied by a balanced three-phase source. The center has four balanced three-phase loads as follows:
-
-> Load 1: 150 kVA at 0.8 pf leading Load 2: 100 kW at unity pf Load 3: 200 kVA at 0.6 pf lagging Load 4: 80 kW and 95 kVAR (inductive)
-
- If the line impedance is 0.02 + *j*0.05 Ω per phase and the line voltage at the loads is 480 V, find the magnitude of the line voltage at the source.
-
-- **12.82** A balanced three-phase system has a distribution wire with impedance 2 + *j*6 Ω per phase. The system supplies two three-phase loads that are connected in parallel. The first is a balanced wye-connected load that absorbs 400 kVA at a power factor of 0.8 lagging. The second load is a balanced delta-connected load with impedance of 10 + *j*8 Ω per phase. If the magnitude of the line voltage at the loads is 2400 V rms, calculate the magnitude of the line voltage at the source and the total complex power supplied to the two loads.
-- **12.83** A commercially available three-phase inductive motor operates at a full load of 120 hp (1 hp = 746 W) at 95 percent efficiency at a lagging power
-
-factor of 0.707. The motor is connected in parallel to a 80-kW balanced three-phase heater at unity power factor. If the magnitude of the line voltage is 480 V rms, calculate the line current.
-
-**12.84** Figure 12.76 displays a three-phase delta-connected motor load which is connected to a line voltage of 440 V and draws 4 kVA at a power factor of 72 per cent lagging. In addition, a single 1.8 kVAR capacitor is connected between lines *a* and *b*, while a 800-W lighting load is connected between line *c* and neutral. Assuming the *abc* sequence and taking **V***an* = *Vp*⧸0°, find the magnitude and phase angle of currents **I***a*, **I***b*, **I***c*, and **I***n*.
-
-**Figure 12.76**
-
-For Prob. 12.84.
-
-**12.85** Design a three-phase heater with suitable symmetric loads using wye-connected pure resistance. Assume that the heater is supplied by a 240-V line voltage and is to give 27 kW of heat.
-
-**12.86** For the single-phase three-wire system in Fig. 12.77, find currents **I***aA*, **I***bB*, and **I***nN*.
-
-# **Figure 12.77**
-
-For Prob. 12.86.
-
-**12.87** Consider the single-phase three-wire system shown in Fig. 12.78. Find the current in the neutral wire and the complex power supplied by each source. Take **V***s* as a 220⧸ 0°-V, 60-Hz source.
-
-**Figure 12.78** For Prob. 12.87.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/145_Chapter 13 - Magnetically Coupled Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/145_Chapter 13 - Magnetically Coupled Circuits.md
deleted file mode 100644
index e897b9f95ef3da768e0564fdf9b456957d71d298..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/145_Chapter 13 - Magnetically Coupled Circuits.md
+++ /dev/null
@@ -1,37 +0,0 @@
-# Magnetically Coupled Circuits
-
-*If you would increase your happiness and prolong your life, forget your neighbor's faults. . . . Forget the peculiarities of your friends, and only remember the good points which make you fond of them. . . . Obliterate everything disagreeable from yesterday; write upon today's clean sheet those things lovely and lovable.*
-
-—Anonymous
-
-# **chapter**
-
-13
-
-# Enhancing Your Career
-
-# **Career in Electromagnetics**
-
-Electromagnetics (EM) is the branch of electrical engineering (or ph ysics) that deals with the analysis and application of electric and magnetic fields. In electromagnetics, electric circuit analysis is applied at low frequencies.
-
-The principles of EM are applied in various allied disciplines, such as electric machines, electromechanical ener gy conversion, radar meteorology, remote sensing, satellite communications, bioelectromagnetics, electromagnetic interference and compatibility, plasmas, and fiber optics. EM devices include electric motors and generators, transformers, electromagnets, magnetic levitation, antennas, radars, microwave ovens, microwave dishes, superconductors, and electrocardiograms. The design of these de vices requires a thorough knowledge of the laws and principles of EM.
-
-EM is regarded as one of the more dif ficult disciplines in electrical engineering. One reason is that EM phenomena are rather abstract. But if one enjoys working with mathematics and can visualize the invisible, one should consider being a specialist in EM, inasmuch as few electrical engineers specialize in this area. Electrical engineers who specialize in EM are needed in micro wave industries, radio/TV broadcasting stations, electromagnetic research laboratories, and se veral communications industries.
-
-Telemetry receiving station for space satellites. © DV169/Getty Images RF
-
-# Historical
-
-© Bettmann/Corbis
-
-**James Clerk Maxwell** (1831–1879), a graduate in mathematics from Cambridge University, in 1865 wrote a most remarkable paper in which he mathematically unified the laws of Faraday and Ampere. This relationship between the electric field and magnetic field served as the basis for what was later called electromagnetic fields and waves, a major field of study in electrical engineering. The Institute of Electrical and Electron ics Engineers (IEEE) uses a graphical representation of this principle in its logo, in which a straight arrow represents current and a curved arrow represents the electromagnetic field. This relationship is commonly known as *the right-hand rule* . Maxwell was a very active theoretician and scientist. He is best known for the "Maxwell equations." The max well, a unit of magnetic flux, was named after him.
-
-# Learning Objectives
-
-By using the information and exercises in this chapter you will be able to:
-
-- 1. Understand the physics behind mutually coupled circuits and how to analyze circuits containing mutually coupled inductors.
-- 2. Understand how energy is stored in mutually coupled circuits.
-- 3. Understand how linear transformers work and how to analyze circuits containing them.
-- 4. Understand how ideal transformers work and how to analyze circuits containing them.
-- 5. Understand how ideal auto transformers work and know how to analyze them when used in a variety of circuits.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/146_13.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/146_13.1 Introduction.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/146_13.1 Introduction.md
+++ /dev/null
@@ -1,67 +0,0 @@
-# **13.1** Introduction
-
-The circuits we have considered so far may be regarded as *conductively coupled*, because one loop affects the neighboring loop through current conduction. When tw o loops with or without contacts between them affect each other through the magnetic field generated by one of them, they are said to be *magnetically coupled*.
-
-The transformer is an electrical de vice designed on the basis of the concept of magnetic coupling. It uses magnetically coupled coils to transfer energy from one circuit to another. Transformers are key circuit elements. They are used in power systems for stepping up or stepping down ac voltages or currents. They are used in electronic circuits such as radio and television receivers for such purposes as impedance matching, isolating one part of a circuit from another, and again for stepping up or down ac voltages and currents.
-
-We will begin with the concept of mutual inductance and introduce the dot convention used for determining the v oltage polarities of inductively coupled components. Based on the notion of mutual inductance,
-
-we then introduce the circuit element known as the *transformer*. We will consider the linear transformer, the ideal transformer, the ideal autotransformer, and the three-phase transformer. Finally, among their important applications, we look at transformers as isolating and matching devices and their use in power distribution.
-
-# **13.2** Mutual Inductance
-
-When two inductors (or coils) are in a close proximity to each other, the magnetic flux caused by current in one coil links with the other coil, thereby inducing v oltage in the latter . This phenomenon is kno wn as *mutual inductance*.
-
-Let us first consider a single inductor, a coil with *N* turns. When current *i* flows through the coil, a magnetic flux *ϕ* is produced around it (Fig. 13.1). According to Faraday's law, the voltage *v* induced in the coil is proportional to the number of turns *N* and the time rate of change of the magnetic flux *ϕ*; that is,
-
-$$
-v = N \frac{d\phi}{dt} \tag{13.1}
-$$
-
-But the flux *ϕ* is produced by current *i* so that any change in *ϕ* is caused by a change in the current. Hence, Eq. (13.1) can be written as
-
-$$
-v = N \frac{d\phi}{di} \frac{di}{dt}
-$$
- (13.2)
-
-or
-
-$$
-v = L \frac{di}{dt}
-$$
- (13.3)
-
-which is the voltage-current relationship for the inductor. From Eqs. (13.2) and (13.3), the inductance *L* of the inductor is thus given by
-
-$$
-L = N \frac{d\phi}{di}
-$$
- (13.4)
-
-This inductance is commonly called *self-inductance,* because it relates the voltage induced in a coil by a time-varying current in the same coil.
-
-Now consider two coils with self-inductances *L*1 and *L*2 that are in close proximity with each other (Fig. 13.2). Coil 1 has *N*1 turns, while coil 2 has *N*2 turns. F or the sak e of simplicity, assume that the second inductor carries no current. The magnetic flux *ϕ*1 emanating from coil 1 has two components: One component *ϕ*11 links only coil 1, and another component *ϕ*12 links both coils. Hence,
-
-$$
-\phi_1 = \phi_{11} + \phi_{12} \tag{13.5}
-$$
-
-Although the tw o coils are ph ysically separated, the y are said to be *magnetically coupled*. Since the entire flux *ϕ*1 links coil 1, the v oltage induced in coil 1 is
-
-$$
-v_1 = N_1 \frac{d\phi_1}{dt} \tag{13.6}
-$$
-
-Only flux *ϕ*12 links coil 2, so the voltage induced in coil 2 is
-
-$$
-v_2 = N_2 \frac{d\phi_{12}}{dt}
-$$
- (13.7)
-
-**Figure 13.1** Magnetic flux produced by a single coil with *N* turns.
-
-Again, as the fluxes are caused by the current *i*1 flowing in coil 1, Eq. (13.6) can be written as
-
-$$
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/147_13.2 Mutual Inductance.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/147_13.2 Mutual Inductance.md
deleted file mode 100644
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+++ /dev/null
@@ -1,398 +0,0 @@
-v_1 = N_1 \frac{d\phi_1}{di_1} \frac{di_1}{dt} = L_1 \frac{di_1}{dt}
-$$
- (13.8)
-
-where *L*1 = *N*1 *dϕ*1∕*di*1 is the self-inductance of coil 1. Similarly, Eq. (13.7) can be written as
-
-$$
-v_2 = N_2 \frac{d\phi_{12}}{di_1} \frac{di_1}{dt} = M_{21} \frac{di_1}{dt}
-$$
- (13.9)
-
-where
-
-$$
-M_{21} = N_2 \frac{d\phi_{12}}{di_1}
-$$
- (13.10)
-
-*M*21 is known as the *mutual inductance* of coil 2 with respect to coil 1. Subscript 21 indicates that the inductance *M*21 relates the voltage induced in coil 2 to the current in coil 1. Thus, the open-circuit *mutual voltage* (or induced voltage) across coil 2 is
-
-$$
-v_2 = M_{21} \frac{di_1}{dt}
-$$
- (13.11)
-
-Suppose we now let current *i*2 flow in coil 2, while coil 1 carries no current (Fig. 13.3). The magnetic flux *ϕ*2 emanating from coil 2 comprises flux *ϕ*22 that links only coil 2 and flux *ϕ*21 that links both coils. Hence,
-
-$$
-\phi_2 = \phi_{21} + \phi_{22} \tag{13.12}
-$$
-
-The entire flux *ϕ*2 links coil 2, so the voltage induced in coil 2 is
-
-$$
-v_2 = N_2 \frac{d\phi_2}{dt} = N_2 \frac{d\phi_2}{di_2} \frac{di_2}{dt} = L_2 \frac{di_2}{dt}
-$$
- (13.13)
-
-where *L*2 = *N*2 *dϕ*2∕*di*2 is the self-inductance of coil 2. Since only flux *ϕ*21 links coil 1, the voltage induced in coil 1 is
-
-$$
-v_1 = N_1 \frac{d\phi_{21}}{dt} = N_1 \frac{d\phi_{21}}{di_2} \frac{di_2}{dt} = M_{12} \frac{di_2}{dt}
-$$
- (13.14)
-
-where
-
-$$
-M_{12} = N_1 \frac{d\phi_{21}}{di_2}
-$$
- (13.15)
-
-which is the *mutual inductance* of coil 1 with respect to coil 2. Thus, the open-circuit *mutual voltage* across coil 1 is
-
-$$
-v_1 = M_{12} \frac{di_2}{dt}
-$$
- (13.16)
-
-We will see in the next section that *M*12 and *M*21 are equal; that is,
-
-$$
-M_{12} = M_{21} = M \tag{13.17}
-$$
-
-and we refer to *M* as the mutual inductance between the tw o coils. Like self-inductance *L*, mutual inductance *M* is measured in henrys (H). Keep in mind that mutual coupling only exists when the inductors or coils are in close proximity, and the circuits are dri ven by time-varying sources. We recall that inductors act like short circuits to dc.
-
-From the two cases in Figs. 13.2 and 13.3, we conclude that mutual inductance results if a v oltage is induced by a time-v arying current in another circuit. It is the property of an inductor to produce a v oltage in reaction to a time-varying current in another inductor near it. Thus,
-
-Mutual inductance *M*12 of coil 1 with respect to coil 2.
-
-Mutual inductance is the ability of one inductor to induce a voltage across a neighboring inductor, measured in henrys (H).
-
-Although mutual inductance *M* is al ways a positi ve quantity, the mutual voltage *M di*∕*dt* may be negative or positive, just like the self-induced voltage *L di*∕*dt*. However, unlike the self-induced *L di*∕*dt*, whose polarity is determined by the reference direction of the current and the reference polarity of the v oltage (according to the passi ve sign convention), the polarity of mutual v oltage *M di*∕*dt* is not easy to determine, because four terminals are involved. The choice of the correct polarity for *M di*∕*dt* is made by examining the orientation or particular way in which both coils are physically wound and applying Lenz's law in conjunction with the right-hand rule. Since it is inconvenient to show the construction details of coils on a circuit schematic, we apply the *dot convention* in circuit analysis. By this convention, a dot is placed in the circuit at one end of each of the two magnetically coupled coils to indicate the direction of the magnetic flux if current enters that dotted terminal of the coil. This is illustrated in Fig. 13.4. Given a circuit, the dots are already placed beside the coils so that we need not bother about how to place them. The dots are used along with the dot con vention to determine the polarity of the mutual voltage. The dot convention is stated as follows:
-
-If a current enters the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is positive at the dotted terminal of the second coil.
-
-# Alternatively,
-
-If a current leaves the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is negative at the dotted terminal of the second coil.
-
-Thus, the reference polarity of the mutual v oltage depends on the ref erence direction of the inducing current and the dots on the coupled coils. Application of the dot con vention is illustrated in the four pairs of mutually coupled coils in Fig. 13.5. Fo r the coupled coils in Fig. 13.5(a), the sign of the mutual voltage *v*2 is determined by the reference polarity for *v*2 and the direction of *i*1. Since *i*1 enters the dotted terminal of coil 1 and *v*2 is positive at the dotted terminal of coil 2, the mutual voltage is +*M di*1∕*dt*. For the coils in Fig. 13.5(b), the current *i*1 enters
-
-Illustration of the dot convention.
-
-**Figure 13.5** Examples illustrating how to apply the dot convention.
-
-the dotted terminal of coil 1 and *v*2 is ne gative at the dotted terminal of coil 2. Hence, the mutual v oltage is −*M di*1∕*dt*. The same reasoning applies to the coils in Figs. 13.5(c) and 13.5(d).
-
-Figure 13.6 shows the dot convention for coupled coils in series. For the coils in Fig. 13.6(a), the total inductance is
-
-$$
-L = L_1 + L_2 + 2M
-$$
- (Series-aiding connection) (13.18)
-
-For the coils in Fig. 13.6(b),
-
-$$
-L = L_1 + L_2 - 2M
-$$
- (Series-opposing connection) (13.19)
-
-Now that we know how to determine the polarity of the mutual voltage, we are prepared to analyze circuits involving mutual inductance. As the first example, consider the circuit in Fig. 13.7(a). Applying KVL to coil 1 gives
-
-$$
-v_1 = i_1 R_1 + L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}
-$$
- (13.20a)
-
-For coil 2, KVL gives
-
-$$
-v_2 = i_2 R_2 + L_2 \frac{di_2}{dt} + M \frac{di_1}{dt}
-$$
- (13.20b)
-
-We can write Eq. (13.20) in the frequency domain as
-
-$$
-\mathbf{V}_1 = (R_1 + j\omega L_1)\mathbf{I}_1 + j\omega M \mathbf{I}_2 \tag{13.21a}
-$$
-
-$$
-\mathbf{V}_2 = j\omega M \mathbf{I}_1 + (R_2 + j\omega L_2) \mathbf{I}_2 \tag{13.21b}
-$$
-
-As a second e xample, consider the circuit in Fig. 13.7(b). We analyze this in the frequency domain. Applying KVL to coil 1, we get
-
-$$
-\mathbf{V} = (\mathbf{Z}_1 + j\omega L_1)\mathbf{I}_1 - j\omega M\mathbf{I}_2
-$$
- (13.22a)
-
-For coil 2, KVL yields
-
-$$
-0 = -j\omega M I_1 + (Z_L + j\omega L_2)I_2 \tag{13.22b}
-$$
-
-Equations (13.21) and (13.22) are solv ed in the usual manner to deter mine the currents.
-
-One of the most important things in making sure one solv es problems accurately is to be able to check each step during the solution pro cess and to mak e sure assumptions can be v erified. Too often, solving mutually coupled circuits requires the problem solv er to track tw o or more steps made at once re garding the sign and v alues of the mutually induced voltages.
-
-# **Figure 13.7**
-
-Time-domain analysis of a circuit containing coupled coils (a) and frequency-domain analysis of a circuit containing coupled coils (b).
-
-# **Figure 13.6**
-
-Dot convention for coils in series; the sign indicates the polarity of the mutual voltage: (a) seriesaiding connection, (b) seriesopposing connection.
-
-Model that makes analysis of mutually coupled easier to solve.
-
-Experience has sho wn that if we break the problem into steps of solving for the value and the sign into separate steps, the decisions made are easier to track. We suggest that model (Figure 13.8 (b)) be used when analyzing circuits containing a mutually c oupled circuit shown in Figure 13.8(a):
-
-Notice that we have not included the signs in the model. The reason for that is that we first determine the value of the induced voltages and then we determine the appropriate signs. Clearly, I1 induces a voltage within the second coil represented by the value *jω*I1 and I2 induces a voltage of *jω*I2 in the first coil. Once we have the values we next use both circuits to find the correct signs for the dependent sources as shown in Figure 13.8(c).
-
-Since I1 enters *L*1 at the dotted end, it induces a voltage in *L*2 that tries to force a current out of the dotted end of *L*2 which means that the source must have a plus on top and a minus on the bottom as sho wn in Figure 13.8(c). I2 leaves the dotted end of *L*2 which means that it induces a voltage in *L*1 which tries to force a current into the dotted end of *L*1 requiring a dependent source that has a plus on the bottom and a minus on top as shown in Figure 13.8(c). No w all we have to do is to analyze a circuit with two dependent sources. This process allows you to check each of your assumptions.
-
-At this introductory level we are not concerned with the determination of the mutual inductances of the coils and their dot placements. Lik e *R*, *L*, and *C*, calculation of *M* would involve applying the theory of elect romagnetics to the actual ph ysical properties of the coils. In this te xt, we assume that the mutual inductance and the placement of the dots are the "gi vens'' of the circuit problem, like the circuit components *R*, *L*, and *C*.
-
-For Example 13.1.
-
-$$
--12 + (-j4 + j5)\mathbf{I}_1 - j3\mathbf{I}_2 = 0
-$$
-
-$$
-jI_1 - j3I_2 = 12 \tag{13.1.1}
-$$
-
-For loop 2, KVL gives
-
-$$
--j3\mathbf{I}_1 + (12 + j6)\mathbf{I}_2 = 0
-$$
-
-or
-
-$$
-\mathbf{I}_1 = \frac{(12 + j6)\mathbf{I}_2}{j3} = (2 - j4)\mathbf{I}_2
-$$
- (13.1.2)
-
-Substituting this in Eq. (13.1.1), we get
-
-( *j*2 + 4 − *j*3)**I**2 = (4 − *j*)**I**2 = 12
-
-or
-
-$$
-\mathbf{I}_2 = \frac{12}{4 - j} = 2.91 \underline{ / 14.04^{\circ}} A \tag{13.1.3}
-$$
-
-From Eqs. (13.1.2) and (13.1.3),
-
-$$
-\mathbf{I}_1 = (2 - j4)\mathbf{I}_2 = (4.472 \angle -63.43^\circ)(2.91 \angle 14.04^\circ)
-$$
-
-= 13.01 \angle -49.39^\circ A
-
-Practice Problem 13.1 Determine the voltage **V***o* in the circuit of Fig. 13.10.
-
-**Figure 13.10** For Practice Prob. 13.1.
-
-**Answer:** 12⧸ −45° V.
-
-Example 13.2 Calculate the mesh currents in the circuit of Fig. 13.11.
-
-**Figure 13.11** For Example 13.2.
-
-# **Solution:**
-
-The key to analyzing a magnetically coupled circuit is knowing the polarity of the mutual voltage. We need to apply the dot rule. In Fig. 13.11, suppose coil 1 is the one whose reactance is 6 Ω, and coil 2 is the one whose reactance is 8 Ω. To figure out the polarity of the mutual voltage in coil 1 due to current **I**2, we observe that **I**2 leaves the dotted terminal of coil 2. Since we are applying KVL in the clockwise direction, it implies that the mutual voltage is negative, that is, −*j*2**I**2.
-
-Alternatively, it might be best to figure out the mutual voltage by redrawing the rele vant portion of the circuit, as sho wn in Fig. 13.12, where it becomes clear that the mutual voltage is **V**1 = −2*j* **I**2.
-
-Thus, for mesh 1 in Fig. 13.11, KVL gives
-
-$$
--100 + \mathbf{I}_1(4 - j3 + j6) - j6\mathbf{I}_2 - j2\mathbf{I}_2 = 0
-$$
-
-or
-
-$$
-100 = (4+j3)\mathbf{I}_1 - j8\mathbf{I}_2 \tag{13.2.1}
-$$
-
-Similarly, to figure out the mutual voltage in coil 2 due to current **I**1, consider the relevant portion of the circuit, as shown in Fig. 13.12. Applying the dot convention gives the mutual voltage as **V**2 = −2*j***I**1. Also, current **I**2 sees the two coupled coils in series in Fig. 13.11; since it leaves the dotted terminals in both coils, Eq. (13.18) applies. Therefore, for mesh 2 in Fig. 13.11, KVL gives
-
-$$
-0 = -2jI_1 - j6I_1 + (j6 + j8 + j2 \times 2 + 5)I_2
-$$
-
-or
-
-$$
-0 = -j8I_1 + (5+j18)I_2 \tag{13.2.2}
-$$
-
-Putting Eqs. (13.2.1) and (13.2.2) in matrix form, we get
-
-$$
-\begin{bmatrix} 100 \\ 0 \end{bmatrix} = \begin{bmatrix} 4+j3 & -j8 \\ -j8 & 5+j18 \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix}
-$$
-
-The determinants are
-
-$$
-\Delta = \begin{vmatrix} 4+j3 & -j8 \\ -j8 & 5+j18 \end{vmatrix} = 30 + j87
-$$
-
-\n
-$$
-\Delta_1 = \begin{vmatrix} 100 & -j8 \\ 0 & 5+j18 \end{vmatrix} = 100(5+j18)
-$$
-
-\n
-$$
-\Delta_2 = \begin{vmatrix} 4+j3 & 100 \\ -j8 & 0 \end{vmatrix} = j800
-$$
-
-Thus, we obtain the mesh currents as
-
-Thus, we obtain the mesh currents as
-\n
-$$
-\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{100(5 + j18)}{30 + j87} = \frac{1,868.2/74.5^{\circ}}{92.03/71^{\circ}} = 20.3/3.5^{\circ} \text{ A}
-$$
-\n
-$$
-\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{j800}{30 + j87} = \frac{800/90^{\circ}}{92.03/71^{\circ}} = 8.693/19^{\circ} \text{ A}
-$$
-
-Determine the phasor currents **I**1 and **I**2 in the circuit of Fig. 13.13. Practice Problem 13.2
-
-**Figure 13.13** For Practice Prob. 13.2.
-
-**Answer:** I1 = 17.889⧸86.57° A, I2 = 26.83⧸86.57° A.
-
-**Figure 13.12** Model for Example 13.2 showing the polarity of the induced voltages.
-
-# **13.3** Energy in a Coupled Circuit
-
-In Chapter 6, we saw that the energy stored in an inductor is given by
-
-$$
-w = \frac{1}{2} L i^2
-$$
- (13.23)
-
-We now want to determine the energy stored in magnetically coupled coils.
-
-Consider the circuit in Fig. 13.14. We assume that currents *i*1 and *i*2 are zero initially, so that the ener gy stored in the coils is zero. If we let *i*1 increase from zero to *I*1 while maintaining *i*2 = 0, the power in coil 1 is
-
-$$
-p_1(t) = v_1 i_1 = i_1 L_1 \frac{di_1}{dt}
-$$
- (13.24)
-
-and the energy stored in the circuit is
-
-$$
-w_1 = \int p_1 dt = L_1 \int_0^{I_1} i_1 dt_1 = \frac{1}{2} L_1 I_1^2
-$$
- (13.25)
-
-If we now maintain *i*1 = *I*1 and increase *i*2 from zero to *I*2, the mutual voltage induced in coil 1 is *M*12 *di*2∕*dt*, while the mutual v oltage induced in coil 2 is zero, since *i*1 does not change. The power in the coils is now
-
-$$
-p_2(t) = i_1 M_{12} \frac{di_2}{dt} + i_2 v_2 = I_1 M_{12} \frac{di_2}{dt} + i_2 L_2 \frac{di_2}{dt}
-$$
- (13.26)
-
-and the energy stored in the circuit is
-
-$$
-w_2 = \int p_2 dt = M_{12} I_1 \int_0^{I_2} di_2 + L_2 \int_0^{I_2} i_2 di_2
-$$
-$$
-= M_{12} I_1 I_2 + \frac{1}{2} L_2 I_2^2
-$$
-(13.27)
-
-The total ener gy stored in the coils when both *i*1 and *i*2 have reached constant values is
-
-$$
-w = w_1 + w_2 = \frac{1}{2} L_1 I_1^2 + \frac{1}{2} L_2 I_2^2 + M_{12} I_1 I_2
-$$
- (13.28)
-
-If we reverse the order by which the currents reach their final values, that is, if we first increase *i*2 from zero to *I*2 and later increase *i*1 from zero to *I*1, the total energy stored in the coils is
-
-$$
-w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 + M_{21}I_1I_2
-$$
-\n(13.29)
-
-Because the total energy stored should be the same regardless of how we reach the final conditions, comparing Eqs. (13.28) and (13.29) leads us to conclude that
-
-$$
-M_{12} = M_{21} = M \tag{13.30a}
-$$
-
-and
-
-$$
-w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 + MI_1I_2
-$$
- (13.30b)
-
-This equation was derived based on the assumption that the coil currents both entered the dotted terminals. If one current enters one dotted
-
-**Figure 13.14** The circuit for deriving energy stored in a coupled circuit.
-
-terminal while the other current leaves the other dotted terminal, the mutual voltage is negative, so that the mutual energy *MI*1*I*2 is also negative. In that case,
-
-$$
-w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 - MI_1I_2
-$$
- (13.31)
-
-Also, because *I*1 and *I*2 are arbitrary v alues, they may be replaced by *i*1 and *i*2, which gives the instantaneous ener gy stored in the circuit the general expression
-
-$$
-w = \frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 \pm Mi_1i_2
-$$
- (13.32)
-
-The positive sign is selected for the mutual term if both currents enter or leave the dotted terminals of the coils; the ne gative sign is selected otherwise.
-
-We will now establish an upper limit for the mutual inductance M. The energy stored in the circuit cannot be negative because the circuit is passive. This means that the quantity 1∕2*L*1 *i* 1 2 + 1∕2*L*2 *i* 2 2 − *Mi*1*i*2 must be greater than or equal to zero:
-
-$$
-\frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 - Mi_1i_2 \ge 0
-$$
-\n(13.33)
-
-To complete the square, we both add and subtract the term *i*1*i*2( √ *L*1*L*2 on the right-hand side of Eq. (13.33) and obtain
-
-$$
-\frac{1}{2}(i_1\sqrt{L_1} - i_2\sqrt{L_2})^2 + i_1i_2(\sqrt{L_1L_2} - M) \ge 0
-$$
- (13.34)
-
-The squared term is ne ver negative; at its least it is zero. Therefore, the sec ond term on the right-hand side of Eq. (13.34) must be greater than zero; that is,
-
-√
-
-$$
-\overline{L_1 L_2} - M \ge 0
-$$
-
-$$
-M \le \sqrt{L_1 L_2}
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/148_13.3 Energy in a Coupled Circuit.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/148_13.3 Energy in a Coupled Circuit.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/148_13.3 Energy in a Coupled Circuit.md
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@@ -1,131 +0,0 @@
-$$
- (13.35)
-
-Thus, the mutual inductance cannot be greater than the geometric mean of the self-inductances of the coils. The e xtent to which the mutual inductance *M* approaches the upper limit is specified by the *coefficient of coupling k*, given by
-
-$$
-k = \frac{M}{\sqrt{L_1 L_2}}\tag{13.36}
-$$
-
-$$
-M = k\sqrt{L_1 L_2} \tag{13.37}
-$$
-
-where 0 ≤ *k* ≤ 1 or equivalently 0 ≤ *M* ≤ √ \_\_\_\_ *L*1*L*2 . The coupling coefficient is the fraction of the total flux emanating from one coil that links the other coil. For example, in Fig. 13.2,
-
-$$
-k = \frac{\phi_{12}}{\phi_1} = \frac{\phi_{12}}{\phi_{11} + \phi_{12}}
-$$
- (13.38)
-
-or
-
-or
-
-and in Fig. 13.3,
-
-# **Figure 13.15**
-
-Windings: (a) loosely coupled, (b) tightly coupled; cutaway view demonstrates both windings.
-
-$$
-k = \frac{\phi_{21}}{\phi_2} = \frac{\phi_{21}}{\phi_{21} + \phi_{22}}
-$$
- (13.39)
-
-If the entire flux produced by one coil links another coil, then *k* = 1 and we have 100 percent coupling, or the coils are said to be *perfectly coupled*. For *k* < 0.5, coils are said to be *loosely coupled*; and for *k* > 0.5, they are said to be *tightly coupled*. Thus,
-
-The coupling coefficient k is a measure of the magnetic coupling between two coils; 0 ≤ k ≤ 1.
-
-We expect *k* to depend on the closeness of the two coils, their core, their orientation, and their windings. Figure 13.15 sho ws loosely coupled windings and tightly coupled windings. The air -core trans formers used in radio frequenc y circuits are loosely coupled, whereas iron-core transformers used in po wer systems are tightly coupled. The linear transformers discussed in Section 3.4 are mostly air -core; the ideal transformers discussed in Sections 13.5 and 13.6 are principally iron-core.
-
-**Figure 13.16** For Example 13.3.
-
-Consider the circuit in Fig. 13.16. Determine the coupling coef ficient. Calculate the ener gy stored in the coupled inductors at time *t* = 1 s if *v* = 60 cos(4*t* + 30°) V.
-
-# **Solution:**
-
-The coupling coefficient is
-
-$$
-k = \frac{M}{\sqrt{L_1 L_2}} = \frac{2.5}{\sqrt{20}} = 0.56
-$$
-
-indicating that the inductors are tightly coupled. To find the energy stored, we need to calculate the current. To find the current, we need to obtain the frequency-domain equivalent of the circuit.
-
-60 cos(4*t* + 30°)
-$$
-\Rightarrow
-$$
- 60/30°, $\omega = 4$ rad/s
-\n5 H $\Rightarrow$ $j\omega L_1 = j20 \Omega$
-\n2.5 H $\Rightarrow$ $j\omega M = j10 \Omega$
-\n4 H $\Rightarrow$ $j\omega L_2 = j16 \Omega$
-\n $\frac{1}{16}F \Rightarrow \frac{1}{j\omega C} = -j4 \Omega$
-
-The frequency-domain equivalent is shown in Fig. 13.17. We now apply mesh analysis. For mesh 1,
-
-$$
-(10 + j20)\mathbf{I}_1 + j10\mathbf{I}_2 = 60/30^{\circ}
-$$
- (13.3.1)
-
-For mesh 2,
-
-$$
-j10\mathbf{I}_1 + (j16 - j4)\mathbf{I}_2 = 0
-$$
-
-$$
-\overline{a}
-$$
-
-$$
-I_1 = -1.2I_2 \tag{13.3.2}
-$$
-
-Substituting this into Eq. (13.3.1) yields
-
-$$
-I_2(-12 - j14) = 60/30^{\circ} \qquad \Rightarrow \qquad I_2 = 3.254/160.6^{\circ} \text{ A}
-$$
-
-and
-
-$$
-I_1 = -1.2I_2 = 3.905 \div 19.4^{\circ}
-$$
- A
-
-In the time-domain,
-
-$$
-i_1 = 3.905 \cos(4t - 19.4^\circ),
-$$
- $i_2 = 3.254 \cos(4t + 160.6^\circ)$
-
-At time *t* = 1 s, 4*t* = 4 rad = 229.2°, and
-
-$$
-i_1 = 3.905 \cos(229.2^\circ - 19.4^\circ) = -3.389 \text{ A}
-$$
-
-$$
-i_2 = 3.254 \cos(229.2^\circ + 160.6^\circ) = 2.824 \text{ A}
-$$
-
-The total energy stored in the coupled inductors is
-
-$$
-w = \frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 + Mi_1i_2
-$$
-
-= $\frac{1}{2}(5)(-3.389)^2 + \frac{1}{2}(4)(2.824)^2 + 2.5(-3.389)(2.824) = 20.73 \text{ J}$
-
-**Figure 13.17**
-
-Frequency-domain equivalent of the circuit in Fig. 13.16.
-
-For the circuit in Fig. 13.18, determine the coupling coefficient and the energy stored in the coupled inductors at *t* = 1.5 s. Practice Problem 13.3
-
-**Answer:** 0.7071, 246.2 J.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/149_13.4 Linear Transformers.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/149_13.4 Linear Transformers.md
deleted file mode 100644
index 906609f4bbf2f8acad53fcefeb736526212a1201..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/149_13.4 Linear Transformers.md
+++ /dev/null
@@ -1,221 +0,0 @@
-# **13.4** Linear Transformers
-
-Here we introduce the transformer as a ne w circuit element. A transformer is a magnetic device that takes advantage of the phenomenon of mutual inductance.
-
-A transformer is generally a four-terminal device comprising two (or more) magnetically coupled coils.
-
-As shown in Fig. 13.19, the coil that is directly connected to the voltage source is called the *primary winding*. The coil connected to the load is called the *secondary winding*. The resistances *R*1 and *R*2 are included to account for the losses (po wer dissipation) in the coils. The transformer is said to be *linear* if the coils are w ound on a magnetically linear material—a material for which the magnetic permeability is constant. Such materials include air, plastic, Bakelite, and wood. In fact, most materials are magnetically linear. Linear transformers are sometimes called *air-core transformers*, although not all of them are necessarily air -core. They are used in radio and TV sets. Figure 13.20 portrays different types of transformers.
-
-# **Figure 13.20**
-
-Different types of transformers: (a) copper wound dry power transformer, (b) audio transformers. (a) © Electric Service Co., Cincinnati Ohio, (b) © Jensen Transformers, Inc., Chatsworth, CA
-
-A linear transformer may also be regarded as one whose flux is proportional to the currents in its windings.
-
-It should be noted that the result in Eq. (13.41) or (13.42) is not affected by the location of the dots on the transformer, because the same result is produced when *M* is replaced by −*M*.
-
-The little bit of experience gained in Sections 13.2 and 13.3 in analyzing magnetically coupled circuits is enough to con vince anyone that analyzing these circuits is not as easy as circuits in previous chapters. For this reason, it is sometimes convenient to replace a magnetically coupled circuit by an equi valent circuit with no magnetic coupling. We want to replace the linear transformer in Fig. 13.21 by an equi valent T or Π circuit, a circuit that would have no mutual inductance.
-
-The v oltage-current relationships for the primary and secondary coils give the matrix equation
-
-$$
-\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} j\omega L_1 & j\omega M \\ j\omega M & j\omega L_2 \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix}
-$$
- (13.43)
-
-By matrix inversion, this can be written as
-
-version, this can be written as
-\n
-$$
-\begin{bmatrix}\nI_1 \\
-I_2\n\end{bmatrix} = \begin{bmatrix}\n\frac{L_2}{j\omega(L_1L_2 - M^2)} & \frac{-M}{j\omega(L_1L_2 - M^2)} \\
-\frac{-M}{j\omega(L_1L_2 - M^2)} & \frac{L_1}{j\omega(L_1L_2 - M^2)}\n\end{bmatrix} \begin{bmatrix}\nV_1 \\
-V_2\n\end{bmatrix}
-$$
-\n(13.44)
-
-Our goal is to match Eqs. (13.43) and (13.44) with the corresponding equations for the T and Π networks.
-
-For the T (or Y) network of Fig. 13.22, mesh analysis pro vides the terminal equations as
-
-$$
-\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} j\omega(L_a + L_c) & j\omega L_c \\ j\omega L_c & j\omega(L_b + L_c) \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix}
-$$
- (13.45)
-
-# **Figure 13.21** Determining the equivalent circuit of a linear transformer.
-
-**Figure 13.22** An equivalent T circuit.
-
-If the circuits in Figs. 13.21 and 13.22 are equivalents, Eqs. (13.43) and (13.45) must be identical. Equating terms in the impedance matrices of Eqs. (13.43) and (13.45) leads to
-
-$$
-L_a = L_1 - M, \qquad L_b = L_2 - M, \qquad L_c = M \qquad (13.46)
-$$
-
-For the Π (or Δ) network in Fig. 13.23, nodal analysis gi ves the terminal equations as
-
-$$
-\begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \frac{1}{j\omega L_A} + \frac{1}{j\omega L_C} & -\frac{1}{j\omega L_C} \\ -\frac{1}{j\omega L_C} & \frac{1}{j\omega L_B} + \frac{1}{j\omega L_C} \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix}
-$$
-(13.47)
-
-Equating terms in admittance matrices of Eqs. (13.44) and (13.47), we obtain
-
-$$
-L_{A} = \frac{L_{1}L_{2} - M^{2}}{L_{2} - M}, \qquad L_{B} = \frac{L_{1}L_{2} - M^{2}}{L_{1} - M}
-$$
-$$
-L_{C} = \frac{L_{1}L_{2} - M^{2}}{M}
-$$
-(13.48)
-
-Note that in Figs. 13.22 and 13.23, the inductors are not magnetically coupled. Also note that changing the locations of the dots in Fig. 13.21 can cause *M* to become − *M*. As Example 13.6 illustrates, a ne gative value of *M* is ph ysically unrealizable b ut the equi valent model is still mathematically valid.
-
-In the circuit of Fig. 13.24, calculate the input impedance and current **I**1. Take **Z**1 = 60 − *j*100 Ω, **Z**2 = 30 + *j*40 Ω, and **Z***L* = 80 + *j*60 Ω.
-
-# **Figure 13.24**
-
-# **Solution:**
-
-From Eq. (13.41),
-
-$$
-Zin = Z1 + j20 + \frac{(5)2}{j40 + Z2 + ZL}
-$$
-
-= 60 - j100 + j20 + $\frac{25}{110 + j140}$
-= 60 - j80 + 0.14/ $-51.84^{\circ}$
-= 60.09 - j80.11 = 100.14/ $-53.1^{\circ}$ Ω
-
-**Figure 13.23** An equivalent Π circuit.
-
-Example 13.4
-
-For Example 13.4.
-
-Thus,
-
-$$
-\mathbf{I}_1 = \frac{\mathbf{V}}{\mathbf{Z}_{in}} = \frac{50/60^{\circ}}{100.14 / -53.1^{\circ}} = 0.5 / 113.1^{\circ} \text{ A}
-$$
-
-Find the input impedance of the circuit in Fig. 13.25 and the current from the voltage source. Practice Problem 13.4
-
-**Answer:** 8.58⧸ 58.05° Ω, 4.662⧸− 58.05° A.
-
-2 H
-
-Determine the T-equivalent circuit of the linear transformer in Fig. 13.26(a).
-
-> (a) 10 H 4 H a b c d (b) a b c d 2 H **I**1 **I**2 8 H 2 H
-
-# **Solution:**
-
-Given that *L*1 = 10, *L*2 = 4, and *M* = 2, the T-equivalent network has the following parameters:
-
-$$
-L_a = L_1 - M = 10 - 2 = 8
-$$
-H
-$$
-L_b = L_2 - M = 4 - 2 = 2
-$$
-H,
-$$
-L_c = M = 2
-$$
-H
-
-The T-equivalent circuit is shown in Fig. 13.26(b). We have assumed that reference directions for currents and voltage polarities in the primary and secondary windings conform to those in Fig. 13.21. Otherwise, we may need to replace *M* with −*M*, as Example 13.6 illustrates.
-
-Example 13.5
-
-Practice Problem 13.5
-
-For the linear transformer in Fig. 13.26(a), find the Π equivalent network.
-
-**Answer:** *LA* = 18 H, *LB* = 4.5 H, *LC* = 18 H.
-
-| Example 13.6 | | | |
-|--------------|--|--|--|
-|--------------|--|--|--|
-
-Solve for **I**1, **I**2, and **V***o* in Fig. 13.27 (the same circuit as for Practice Prob. 13.1) using the T-equivalent circuit for the linear transformer. Example 13.6
-
-# **Solution:**
-
-Notice that the circuit in Fig. 13.27 is the same as that in Fig. 13.10 except that the reference direction for current **I**2 has been reversed, just to make the reference directions for the currents for the magnetically coupled coils conform with those in Fig. 13.21.
-
-We need to replace the magnetically coupled coils with the T-equivalent circuit. The relevant portion of the circuit in Fig. 13.27 is shown in Fig. 13.28(a). Comparing Fig. 13.28(a) with Fig. 13.21 sho ws that there are tw o differences. First, due to the current reference direc tions and voltage polarities, we need to replace *M* by −*M* to make Fig. 13.28(a) conform with Fig. 13.21. Second, the circuit in Fig. 13.21 is in the time-domain, whereas the circuit in Fig. 13.28(a) is in the frequencydomain. The difference is the factor *jω*; that is, *L* in Fig. 13.21 has been replaced with *jωL* and *M* with *jωM*. Since *ω* is not specified, we can assume *ω* = 1 rad/s or any other value; it really does not matter. With these two differences in mind,
-
-$$
-L_a = L_1 - (-M) = 8 + 1 = 9 \text{ H}
-$$
-
-$$
-L_b = L_2 - (-M) = 5 + 1 = 6 \text{ H}, \qquad L_c = -M = -1 \text{ H}
-$$
-
-Thus, the T-equivalent circuit for the coupled coils is as shown in Fig. 13.28(b).
-
-Inserting the T-equivalent circuit in Fig. 13.28(b) to replace the two coils in Fig. 13.27 gives the equivalent circuit in Fig. 13.29, which can be solved using nodal or mesh analysis. Applying mesh analysis, we obtain
-
-$$
-j6 = I_1(4 + j9 - j1) + I_2(-j1)
-$$
- (13.6.1)
-
-and
-
-$$
-0 = I1(-j1) + I2(10 + j6 - j1)
-$$
- (13.6.2)
-
-From Eq. (13.6.2),
-
-$$
-\mathbf{I}_1 = \frac{(10 + j5)}{j} \mathbf{I}_2 = (5 - j10)\mathbf{I}_2
-$$
- (13.6.3)
-
-# **Figure 13.28**
-
-For Example 13.6: (a) circuit for coupled coils of Fig. 13.27, (b) T-equivalent circuit.
-
-For Example 13.6.
-
-Substituting Eq. (13.6.3) into Eq. (13.6.1) gives
-
-$$
-j6 = (4 + j8)(5 - j10)\mathbf{I}_2 - j\mathbf{I}_2 = (100 - j)\mathbf{I}_2 \simeq 100\mathbf{I}_2
-$$
-
-Since 100 is very large compared with 1, the imaginary part of (100 − *j*) can be ignored so that 100 − *j* ≃ 100. Hence,
-
-$$
-I_2 = \frac{j6}{100} = j0.06 = 0.06 / 90^{\circ} A
-$$
-
-From Eq. (13.6.3),
-
-$$
-I_1 = (5 - j10)j0.06 = 0.6 + j0.3
-$$
- A
-
-and
-
-$$
-\mathbf{V}_o = -10\mathbf{I}_2 = -j0.6 = 0.6 \div 90^\circ \text{ V}
-$$
-
-This agrees with the answer to Practice Prob. 13.1. Of course, the direction of **I**2 in Fig. 13.10 is opposite to that in Fig. 13.27. This will not affect **V***o*, but the value of **I**2 in this example is the negative of that of **I**2 in Practice Prob. 13.1. The advantage of using the T-equivalent model for the magnetically coupled coils is that in Fig. 13.29 we do not need to bother with the dot on the coupled coils.
-
-Solve the problem in Example 13.1 (see Fig. 13.9) using the T-equivalent model for the magnetically coupled coils.
-
-Practice Problem 13.6
-
-**Answer:** 13⧸ −49.4° A, 2.91⧸ 14.04° A.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/150_13.5 Ideal Transformers.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/150_13.5 Ideal Transformers.md
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@@ -1,397 +0,0 @@
-# **13.5** Ideal Transformers
-
-An ideal transformer is one with perfect coupling ( *k* =1). It consists of two (or more) coils with a lar ge number of turns w ound on a common core of high permeability. Because of this high permeability of the core, the flux links all the turns of both coils, thereby resulting in a perfect coupling.
-
-To see ho w an ideal transformer is the limiting case of tw o coupled inductors where the inductances approach infinity and the coupling is perfect, let us reexamine the circuit in Fig. 13.14. In the frequency domain,
-
-$$
-V_1 = j\omega L_1 I_1 + j\omega M I_2
-$$
-\n
-$$
-V_2 = j\omega M I_1 + j\omega L_2 I_2
-$$
-\n(13.49b)
-
-\_\_\_\_\_
-
-From Eq. (13.49a), **I**1 = (**V**1 − *jωM***I**2)∕*jωL*1 (we could have also used this equation to develop the current ratios instead of using the conservation of power which we will do shortly). Substituting this in Eq. (13.49b) gives
-
-$$
-\mathbf{V}_2 = j\omega L_2 \mathbf{I}_2 + \frac{M\mathbf{V}_1}{L_1} - \frac{j\omega M^2 \mathbf{I}_2}{L_1}
-$$
-
-But *M* = √ \_\_\_\_ *L*1*L*2 for perfect coupling (*k* = 1). Hence,
-
-$$
-\mathbf{V}_2 = j\omega L_2 \mathbf{I}_2 + \frac{\sqrt{L_1 L_2} \mathbf{V}_1}{L_1} - \frac{j\omega L_1 L_2 \mathbf{I}_2}{L_1} = \sqrt{\frac{L_2}{L_1}} \mathbf{V}_1 = n\mathbf{V}_1
-$$
-
-where *n* = √ *L*2∕*L*1 and is called the *turns ratio.* As *L*1, *L*2, *M* → ∞ such that *n* remains the same, the coupled coils become an ideal transformer. A transformer is said to be ideal if it has the following properties:
-
-- 1. Coils have very large reactances (*L*1, *L*2, *M* → ∞).
-- 2. Coupling coefficient is equal to unity (*k* = 1).
-- 3. Primary and secondary coils are lossless (*R*1 = 0 = *R*2).
-
-An ideal transformer is a unity-coupled, lossless transformer in which the primary and secondary coils have infinite self-inductances.
-
-Iron-core transformers are close approximations to ideal transformers. These are used in power systems and electronics.
-
-as shown in Fig. 13.31, the same magnetic flux *ϕ* goes through both windings. According to F araday's law, the v oltage across the primary winding is
-
-$$
-v_1 = N_1 \frac{d\phi}{dt} \tag{13.50a}
-$$
-
-while that across the secondary winding is
-
-$$
-v_2 = N_2 \frac{d\phi}{dt} \tag{13.50b}
-$$
-
-Dividing Eq. (13.50b) by Eq. (13.50a), we get
-
-$$
-\mathsf{w}\mathsf{c}\mathsf{g}\mathsf{c}\mathsf{t}
-$$
-
-$$
-\frac{v_2}{v_1} = \frac{N_2}{N_1} = n \tag{13.51}
-$$
-
-where *n* is, again, the *turns ratio* or *transformation ratio*. We can use the phasor voltages **V**1 and **V**2 rather than the instantaneous values *v*1 and *v*2. Thus, Eq. (13.51) may be written as
-
-$$
-\frac{\mathbf{V}_2}{\mathbf{V}_1} = \frac{N_2}{N_1} = \mathbf{0}
-$$
- (13.52)
-
-**Figure 13.30** (a) Ideal transformer, (b) circuit symbol for an ideal transformer.
-
-**Figure 13.31** Relating primary and secondary quantities in an ideal transformer.
-
-For the reason of po wer conservation, the ener gy supplied to the pri mary must equal the ener gy absorbed by the secondary , since there are no losses in an ideal transformer. This implies that
-
-$$
-\sum_{i=1}^{n} v_i = v_1 i_1 = v_2 i_2 \quad \text{(power conserved)} \quad (13.53)
-$$
-
-In phasor form, Eq. (13.53) in conjunction with Eq. (13.52) becomes
-
-$$
-\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{\mathbf{V}_2}{\mathbf{V}_1} = \mathbf{\hat{u}}
-$$
- (13.54)
-
-$$
-\frac{\mathbf{I}_2}{\mathbf{I}_1} = \frac{N_1}{N_2} = \frac{1}{n}
-$$
- (13.55)
-
-When *n* = 1, we generally call the transformer an *isolation transformer*. The reason will become ob vious in Section 13.9.1. If *n* > 1, we ha ve a *step-up transformer*, as the voltage is increased from primary to secondary (**V**2 > **V**1). On the other hand, if *n* < 1, the transformer is a *step-down transformer*, since the v oltage is decreased from primary to secondary
-
-A step-down transformer is one whose secondary voltage is less than its primary voltage.
-
-A step-up transformer is one whose secondary voltage is greater than its primary voltage.
-
-The ratings of transformers are usually specified as *V*1∕*V*2. A transformer with rating 2400∕120 V should have 2400 V on the primary and 120 in the secondary (i.e., a step-down transformer). Keep in mind that the voltage ratings are in rms.
-
-Power companies often generate at some convenient voltage and use a step-up transformer to increase the v oltage so that the po wer can be transmitted at very high voltage and low current over transmission lines, resulting in significant cost savings. Near residential consumer premises, step-down transformers are used to bring the voltage down to 120 V. Section 13.9.3 will elaborate on this.
-
-It is important that we know how to get the proper polarity of the voltages and the direction of the currents for the transformer in Fig. 13.31. If the polarity of **V**1 or **V**2 or the direction of **I**1 or **I**2 is changed, n in Eqs. (13.51) to (13.55) may need to be replaced by − *n*. The two simple rules to follow are:
-
-- 1. If **V**1 and **V**2 are *both* positive or both negative at the dotted termi nals, use *+n* in Eq. (13.52). Otherwise, use −*n*.
-- 2. If **I**1 and **I**2 *both* enter into or both lea ve the dotted terminals, use −*n* in Eq. (13.55). Otherwise, use *+n*.
-
-The rules are demonstrated with the four circuits in Fig. 13.32.
-
-# **Figure 13.32** Typical circuits illustrating proper voltage polarities and current directions in an ideal transformer.
-
-Using Eqs. (13.52) and (13.55), we can al ways express **V**1 in terms of **V**2 and **I**1 in terms of **I**2, or vice versa:
-
-$$
-V_1 = \frac{V_2}{n}
-$$
- or $V_2 = nV_1$ (13.56)
-
-$$
-\mathbf{I}_1 = n\mathbf{I}_2 \qquad \text{or} \qquad \mathbf{I}_2 = \frac{\mathbf{I}_1}{n} \tag{13.57}
-$$
-
-The complex power in the primary winding is
-
-$$
-\hat{\mathbf{S}}_1 = \mathbf{V}_1 \mathbf{I}_1^* = \frac{\mathbf{V}_2}{n} (n \mathbf{I}_2)^* = \mathbf{V}_2 \mathbf{I}_2^* = \mathbf{S}_2 \tag{13.58}
-$$
-
-The input impedance as seen by the source in Fig. 13.31 is found from
-
-$$
-Z_{\text{in}} = \frac{V_1}{I_1} = \frac{1}{n^2} \frac{V_2}{I_2}
-$$
- (13.59)
-
-$$
-V_2/I_2 = Z_L
-$$
-, so that
-
-$$
-\mathbf{Z}_{\text{in}} = \frac{\mathbf{Z}_L}{n^2}
-$$
- (13.60)
-
-The input impedance is also called the *reflected impedance,* inasmuch
-
-as it appears as if the load impedance is reflected to the primary side. This ability of the transformer to transform a gi ven impedance into another impedance provides us a means of *impedance matching* to ensure maximum power transfer. The idea of impedance matching is v ery useful in practice and will be discussed more in Section 13.9.2.
-
-In analyzing a circuit containing an ideal transformer , it is com mon practice to eliminate the transformer by reflecting impedances and sources from one side of the transformer to the other . In the circuit of Fig. 13.33, suppose we want to reflect the secondary side of the circuit to the primary side. We find the Thevenin equi valent of the circuit to the right of the terminals *a*-*b*. We obtain **V**Th as the open-circuit voltage at terminals *a*-*b*, as shown in Fig. 13.34(a).
-
-**Figure 13.33** Ideal transformer circuit whose equivalent circuits are to be found.
-
-Notice that an ideal transformer reflects an impedance as the square of the turns ratio.
-
-# **Figure 13.34**
-
-n2
-
-by n.
-
-(a) Obtaining **V**Th for the circuit in Fig. 13.33, (b) obtaining **Z**Th for the circuit in Fig. 13.33.
-
-Because terminals *a*-*b* are open, **I**1 = 0 = **I**2 so that **V**2 = **V***s*2. Hence,
-
-$$
-V_{\text{Th}} = V_1 = \frac{V_2}{n} = \frac{(V_{s2})}{n}
-$$
- (13.61)
-
-$$
-I_1 = nI_2
-$$
- and $V_1 = V_2/n$ , so that
-
-$$
-Z_{\text{Th}} = \frac{V_1}{I_1} = \frac{V_2/n}{nI_2} = \frac{Z_2}{n^2} \qquad V_2 = Z_2 I_2 \tag{13.62}
-$$
-
-**Figure 13.35**
-
-by reflecting the secondary circuit to the primary side.
-
-Equivalent circuit for Fig. 13.33 obtained
-
-The general rule for eliminating the transformer and reflecting the second-
-
-n2**Z**1 **Z**2 n**V**s1 **V**s2 c d **V**2 + ‒ ‒ + ‒
-
-by reflecting the primary circuit to the secondary side.
-
-Equivalent circuit for Fig. 13.33 obtained
-
-According to Eq. (13.58), the po wer remains the same, whether calcu lated on the primary or the secondary side. But realize that this reflection approach only applies if there are no e xternal connections between the primary and secondary windings. When we have external connections between the primary and secondary windings, we simply use regular mesh and nodal analysis. Examples of circuits where there are external connections between the primary and secondary windings are in Figs. 13.39 and 13.40. Also note that if the locations of the dots in Fig. 13.33 are changed, we might have to replace *n* by −*n* in order to obey the dot rule, illustrated in Fig. 13.32.
-
-, multiply the primary voltage by n, and divide the primary current
-
-Example 13.7
-
-An ideal transformer is rated at 2400 ∕120 V, 9.6 kVA, and has 50 turns on the secondary side. Calculate: (a) the turns ratio, (b) the number of turns on the primary side, and (c) the current ratings for the primary and secondary windings.
-
-# **Solution:**
-
-(a) This is a step-down transformer, since *V*1 = 2,400 V > *V*2 = 120 V.
-
-$$
-n = \frac{V_2}{V_1} = \frac{120}{2,400} = 0.05
-$$
-
-⇒ 0.05 = \_\_\_ 50
-
-*N*1
-
-(b)
-
-or
-
-$$
-N_1 = \frac{50}{0.05} = 1,000 \text{ turns}
-$$
-
-(c)
-$$
-S = V_1 I_1 = V_2 I_2 = 9.6
-$$
- kVA. Hence,
-
-*n* = \_\_\_ *N*2 *N*1
-
-$$
-I_1 = \frac{9,600}{V_1} = \frac{9,600}{2,400} = 4 \text{ A}
-$$
-$$
-I_2 = \frac{9,600}{V_2} = \frac{9,600}{120} = 80 \text{ A} \qquad \text{or} \qquad I_2 = \frac{I_1}{n} = \frac{4}{0.05} = 80 \text{ A}
-$$
-
-Practice Problem 13.7 The primary current to an ideal transformer rated at 2200∕110 V is 25 A. Calculate: (a) the turns ratio, (b) the kVA rating, (c) the secondary current.
-
-**Answer:** (a) 1∕20, (b) 55 kVA, (c) 500 A.
-
-For the ideal transformer circuit of Fig. 13.37, find: (a) the source current **I**1, (b) the output voltage **V***o*, and (c) the complex power supplied by the source. Example 13.8
-
-# **Solution:**
-
-(a) The 20-Ω impedance can be reflected to the primary side and we get
-
-$$
-Z_R = \frac{20}{n^2} = \frac{20}{4} = 5 \ \Omega
-$$
-
-Thus,
-
-$$
-\mathbf{Z}_{in} = 4 - j6 + \mathbf{Z}_R = 9 - j6 = 10.82 \underline{/ -33.69^\circ} \,\Omega
-$$
-\n
-$$
-\mathbf{I}_1 = \frac{120 \underline{/ 0^\circ}}{\mathbf{Z}_{in}} = \frac{120 \underline{/ 0^\circ}}{10.82 \underline{/ -33.69^\circ}} = 11.09 \underline{/ 33.69^\circ} \,\text{A}
-$$
-
-(b) Because both **I**1 and **I**2 leave the dotted terminals,
-
-$$
-\mathbf{I}_2 = -\frac{1}{n}\mathbf{I}_1 = -5.545/33.69^\circ \text{ A}
-$$
-
-$$
-\mathbf{V}_o = 20\mathbf{I}_2 = 110.9/213.69^\circ \text{ V}
-$$
-
-(c) The complex power supplied is
-
-$$
-S = V_s I_1^* = (120/0°)(11.09/-33.69°) = 1,330.8/-33.69° VA
-$$
-
-In the ideal transformer circuit of Fig. 13.38, find **V***o* and the complex power supplied by the source. Practice Problem 13.8
-
-**Answer:** 429.4⧸ 116.57° V, 17.174⧸ −26.57° kVA.
-
-Calculate the power supplied to the 10-Ω resistor in the ideal transformer circuit of Fig. 13.39.
-
-Example 13.9
-
-# **Solution:**
-
-Reflection to the secondary or primary side cannot be done with this circuit: There is direct connection between the primary and
-
-secondary sides due to the 30-Ω resistor. We apply mesh analysis. For mesh 1,
-
-$$
--120 + (20 + 30)I_1 - 30I_2 + V_1 = 0
-$$
-
-or
-
-$$
-50I_1 - 30I_2 + V_1 = 120 \tag{13.9.1}
-$$
-
-For mesh 2,
-
-$$
--\mathbf{V}_2 + (10 + 30)\mathbf{I}_2 - 30\mathbf{I}_1 = 0
-$$
-
-or
-
-$$
--30I_1 + 40I_2 - V_2 = 0 \tag{13.9.2}
-$$
-
-At the transformer terminals,
-
-$$
-V_2 = -\frac{1}{2} V_1 \tag{13.9.3}
-$$
-
-$$
-\mathbf{I}_2 = -2\mathbf{I}_1 \tag{13.9.4}
-$$
-
-(Note that *n* = 1∕2.) We now have four equations and four unknowns, but our goal is to get **I**2. So we substitute for **V**1 and **I**1 in terms of **V**2 and **I**2 in Eqs. (13.9.1) and (13.9.2). Equation (13.9.1) becomes
-
-$$
--55I_2 - 2V_2 = 120 \tag{13.9.5}
-$$
-
-and Eq. (13.9.2) becomes
-
-$$
-15I_2 + 40I_2 - V_2 = 0 \Rightarrow V_2 = 55I_2 \qquad (13.9.6)
-$$
-
-Substituting Eq. (13.9.6) in Eq. (13.9.5),
-
-$$
--165I_2 = 120 \qquad \Rightarrow \qquad I_2 = -\frac{120}{165} = -0.7272 \text{ A}
-$$
-
-The power absorbed by the 10-Ω resistor is
-
-$$
-P = (-0.7272)^{2}(10) = 5.3
-$$
- W
-
-# Find **V***o* Practice Problem 13.9 in the circuit of Fig. 13.40.
-
-For Practice Prob. 13.9.
-
-**Answer:** 96 V.
-
-# **13.6** Ideal Autotransformers
-
-Unlike the conventional two-winding transformer we have considered so far, an *autotransformer* has a single continuous winding with a connection point called a *tap* between the primary and secondary sides. The tap is often adjustable so as to provide the desired turns ratio for stepping up or stepping down the voltage. This way, a variable voltage is provided to the load connected to the autotransformer.
-
-An autotransformer is a transformer in which both the primary and the secondary are in a single winding.
-
-Figure 13.41 sho ws a typical autotransformer . As sho wn in Fig. 13.42, the autotransformer can operate in the step-down or step-up mode. The autotransformer is a type of po wer transformer. Its major advantage o ver the tw o-winding transformer is its ability to transfer larger apparent po wer. Example 13.10 will demonstrate this. Another advantage is that an autotransformer is smaller and lighter than an equivalent two-winding transformer. However, since both the primary and secondary windings are one winding, *electrical isolation* (no direct electrical connection) is lost. (We will see how the property of electri cal isolation in the conventional transformer is practically employed in Section 13.9.1.) The lack of electrical isolation between the primary and secondary windings is a major disadvantage of the autotransformer.
-
-Some of the formulas we deri ved for ideal transformers apply to ideal autotransformers as well. F or the step-do wn autotransformer cir cuit of Fig. 13.42(a), Eq. (13.52) gives
-
-$$
-\frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{N_1 + N_2}{N_2} = 1 + \frac{N_1}{N_2}
-$$
- (13.63)
-
-As an ideal autotransformer, there are no losses, so the comple x power remains the same in the primary and secondary windings:
-
-$$
-S_1 = V_1 I_1^* = S_2 = V_2 I_2^*
-$$
- (13.64)
-
-Equation (13.64) can also be expressed as
-
-$$
-V_1I_1=V_2I_2
-$$
-
-or
-
-$$
-\frac{V_2}{V_1} = \frac{I_1}{I_2}
-$$
- (13.65)
-
-Thus, the current relationship is
-
-$$
-\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{N_2}{N_1 + N_2}
-$$
- (13.66)
-
-For the step-up autotransformer circuit of Fig. 13.42(b),
-
-$$
-\frac{\mathbf{V}_1}{N_1} = \frac{\mathbf{V}_2}{N_1 + N_2}
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/151_13.6 Ideal Autotransformers.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/151_13.6 Ideal Autotransformers.md
deleted file mode 100644
index e3d8c9c58513d54b38fedf37c09677e48b7dd1b4..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/151_13.6 Ideal Autotransformers.md
+++ /dev/null
@@ -1,99 +0,0 @@
-$$
-
-**Figure 13.41** A typical autotransformer. © Todd Systems, Inc.
-
-# **Figure 13.42**
-
-(a) Step-down autotransformer, (b) step-up autotransformer.
-
-(b)
-
-$$
-\frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{N_1}{N_1 + N_2} \tag{13.67}
-$$
-
-$$
-\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{N_1 + N_2}{N_1} = 1 + \frac{N_2}{N_1}
-$$
-\n(13.68)
-
-A major dif ference between con ventional transformers and auto transformers is that the primary and secondary sides of the autotrans former are not only coupled magnetically b ut also coupled conductively. The autotransformer can be used in place of a con ventional transformer when electrical isolation is not required.
-
-Example 13.10
-
-Compare the po wer ratings of the tw o-winding transformer in Fig. 13.43(a) and the autotransformer in Fig. 13.43(b).
-
-# **Solution:**
-
-Although the primary and secondary windings of the autotransformer are together as a continuous winding, they are separated in Fig. 13.43(b) for clarity. We note that the current and voltage of each winding of the autotransformer in Fig. 13.43(b) are the same as those for the two-winding transformer in Fig. 13.43(a). This is the basis of comparing their power ratings.
-
-For the two-winding transformer, the power rating is
-
-*S*1 = 0.2(240) = 48 VA or *S*2 = 4(12) = 48 VA
-
-For the autotransformer, the power rating is
-
-*S*1 = 4.2(240) = 1,008 VA or *S*2 = 4(252) = 1,008 VA
-
-which is 21 times the power rating of the two-winding transformer.
-
-Refer to Fig. 13.43. If the two-winding transformer is a 60-VA, 120 V∕10 V transformer, what is the power rating of the autotransformer? Practice Problem 13.10
-
-**Answer:** 780 VA.
-
-Refer to the autotransformer circuit in Fig. 13.44. Calculate: (a) **I**1, **I**2, and **I***o* if **Z***L* = 8 + *j*6 Ω, and (b) the complex power supplied to the load. Example 13.11
-
-# **Solution:**
-
-(a) This is a step-up autotransformer with *N*1 = 80, *N*2 = 120, **V**1 = 120⧸ 30°, so Eq. (13.67) can be used to find **V**2 by
-
-$$
-\frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{N_1}{N_1 + N_2} = \frac{80}{200}
-$$
-
-or
-
-$$
-\mathbf{V}_2 = \frac{200}{80} \mathbf{V}_1 = \frac{200}{80} (120/30^\circ) = 300/30^\circ \text{ V}
-$$
-$$
-\mathbf{I}_2 = \frac{\mathbf{V}_2}{\mathbf{Z}_L} = \frac{300/30^\circ}{8 + j6} = \frac{300/30^\circ}{10/36.87^\circ} = 30/-6.87^\circ \text{ A}
-$$
-
-But
-
-$$
-\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{N_1 + N_2}{N_1} = \frac{200}{80}
-$$
-
-or
-
-$$
-\mathbf{I}_1 = \frac{200}{80} \mathbf{I}_2 = \frac{200}{80} (30 \angle -6.87^\circ) = 75 \angle -6.87^\circ \text{ A}
-$$
-
-At the tap, KCL gives
-
-$$
-\mathbf{I}_1 + \mathbf{I}_o = \mathbf{I}_2
-$$
-
-or
-
-$$
-I_o = I_2 - I_1 = 30 \underline{\textstyle{\frac{1}{6.87^\circ}} - 75 \underline{\textstyle{\frac{1}{6.87^\circ}}}} = 45 \underline{\textstyle{\frac{173.13^\circ}{173.13^\circ}}}
-$$
-
-(b) The complex power supplied to the load is
-
-$$
-\mathbf{S}_2 = \mathbf{V}_2 \mathbf{I}_2^* = |\mathbf{I}_2|^2 \mathbf{Z}_L = (30)^2 (10/36.87^\circ) = 9/36.87^\circ \text{ kVA}
-$$
-
-# Practice Problem 13.11
-
-For Practice Prob. 13.11.
-
-In the autotransformer circuit of Fig. 13.45, find currents **I**1, **I**2, and **I***o*. Take **V**1 = 8 kV, **V**2 = 2 kV.
-
-**Answer:** 2 A, 8 A, 6 A.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/152_13.7 Three-Phase Transformers.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/152_13.7 Three-Phase Transformers.md
deleted file mode 100644
index 6e79d49be4143adcb6e61e4d0414349319526d89..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/152_13.7 Three-Phase Transformers.md
+++ /dev/null
@@ -1,94 +0,0 @@
-# **13.7** Three-Phase Transformers
-
-To meet the demand for three-phase po wer transmission, transformer connections compatible with three-phase operations are needed. We can achieve the transformer connections in tw o ways: by connecting three single-phase transformers, thereby forming a so-called *transformer bank*, or by using a special three-phase transformer . For the same kVA rating, a three-phase transformer is always smaller and cheaper than three single-phase transformers. When single-phase transformers are used, one must ensure that the y have the same turns ratio *n* to achie ve a balanced three-phase system. There are four standard w ays of con necting three single-phase transformers or a three-phase transformer for three-phase operations: Y-Y, Δ-Δ, Y-Δ, and Δ-Y.
-
-For any of the four connections, the total apparent po wer *ST*, real power *PT*, and reactive power *QT* are obtained as
-
-$$
-S_T = \sqrt{3} V_L I_L \tag{13.69a}
-$$
-
-$$
-P_T = S_T \cos \theta = \sqrt{3} V_L I_L \cos \theta \tag{13.69b}
-$$
-
-$$
-Q_T = S_T \sin \theta = \sqrt{3} V_L I_L \sin \theta \qquad (13.69c)
-$$
-
-where *VL* and *IL* are, respecti vely, equal to the line v oltage *VLp* and the line current *ILp* for the primary side, or the line v oltage *VLs* and the line current *ILs* for the secondary side. Notice From Eq. (13.69) that for each of the four connections, *VLs ILs* = *VLp ILp*, since power must be conserved in an ideal transformer.
-
-For the Y-Y connection (Fig. 13.46), the line v oltage *VLp* at the primary side, the line v oltage *VLs* on the secondary side, the line current *ILp* on the primary side, and the line current *ILs* on the secondary side are related to the transformer per phase turns ratio *n* according to Eqs. (13.52) and (13.55) as
-
-$$
-V_{Ls} = nV_{Lp} \tag{13.70a}
-$$
-
-$$
-I_{Ls} = \frac{I_{Lp}}{n} \tag{13.70b}
-$$
-
-For the Δ-Δ connection (Fig. 13.47), Eq. (13.70) also applies for the line voltages and line currents. This connection is unique in the sense
-
-**Figure 13.46** Y-Y three-phase transformer connection.
-
-**Figure 13.47** Δ-Δ three-phase transformer connection.
-
-that if one of the transformers is removed for repair or maintenance, the other two form an *open delta,* which can provide three-phase voltages at a reduced level of the original three-phase transformer. \_\_
-
-For the Y-Δ connection (Fig. 13.48), there is a factor of √ 3 arising from the line-phase values in addition to the transformer per phase turns ratio *n*. Thus,
-
-\_\_
-
-*n* √ \_\_ 3
-
-$$
-V_{Ls} = \frac{nV_{Lp}}{\sqrt{3}}
-$$
-(13.71a)
-
-$$
-I_{Ls} = \frac{\sqrt{3}I_{Lp}}{n} \tag{13.71b}
-$$
-
-Similarly, for the Δ-Y connection (Fig. 13.49),
-
-$$
-V_{Ls} = n\sqrt{3} V_{Lp}
-$$
- (13.72a)
-$$
-I_{Ls} = \frac{I_{Lp}}{\sqrt{3}}
-$$
- (13.72b)
-
-Y-Δ three-phase transformer connection.
-
-**Figure 13.49** Δ-Y three-phase transformer connection.
-
-Example 13.12
-
-The 42-kVA balanced load depicted in Fig. 13.50 is supplied by a threephase transformer. (a) Determine the type of transformer connections. (b) Find the line voltage and current on the primary side. (c) Determine the kVA rating of each transformer used in the transformer bank. Assume that the transformers are ideal.
-
-# **Solution:**
-
-(a) A careful observation of Fig. 13.50 shows that the primary side is Y-connected, while the secondary side is Δ-connected. Thus, the threephase transformer is Y-Δ, similar to the one shown in Fig. 13.48. (b) Given a load with total apparent power *ST* = 42 kVA, the turns ratio *n* = 5, and the secondary line voltage *VLs* = 240 V, we can find the secondary line current using Eq. (13.69a), by
-
-$$
-I_{Ls} = \frac{S_T}{\sqrt{3} V_{Ls}} = \frac{42,000}{\sqrt{3}(240)} = 101 \text{ A}
-$$
-
-From Eq. (13.71),
-
-$$
-I_{Lp} = \frac{n}{\sqrt{3}} I_{Ls} = \frac{5 \times 101}{\sqrt{3}} = 292 \text{ A}
-$$
-$$
-V_{Lp} = \frac{\sqrt{3}}{n} V_{Ls} = \frac{\sqrt{3} \times 240}{5} = 83.14 \text{ V}
-$$
-
-(c) Because the load is balanced, each transformer equally shares the total load and since there are no losses (assuming ideal transformers), the kVA rating of each transformer is *S* = *ST*∕3 = 14 kVA. Alternatively, the transformer rating can be determined by the product of the phase current and phase voltage of the primary or secondary side. For the pri mary side, for example, we have a delta connection, so that the phase voltage is the same as the line voltage of 240 V, while the phase current is *ILp*∕√ \_\_ 3 = 58.34 A. Hence, *S* = 240 × 58.34 = 14 kVA.
-
-A three-phase Δ-Δ transformer is used to step down a line voltage of 625 kV, to supply a plant operating at a line voltage of 12.5 kV. The plant draws 40 MW with a lagging power factor of 85 percent. Find: (a) the current drawn by the plant, (b) the turns ratio, (c) the current on the primary side of the transformer, and (d) the load carried by each transformer. Practice Problem 13.12
-
-**Answer:** (a) 2.174 kA, (b) 0.02, (c) 43.47 A, (d) 15.69 MVA.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/153_13.8 PSpice Analysis of Magnetically Coupled Circuits.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/153_13.8 PSpice Analysis of Magnetically Coupled Circuits.md
deleted file mode 100644
index da1c17750162953f39c9081f973664383131accb..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/153_13.8 PSpice Analysis of Magnetically Coupled Circuits.md
+++ /dev/null
@@ -1,179 +0,0 @@
-# **13.8** PSpice Analysis of Magnetically Coupled Circuits
-
-*PSpice* analyzes magnetically coupled circuits just lik e inductor cir cuits except that the dot convention must be followed. In *PSpice* Schematic, the dot (not sho wn) is al ways next to pin 1, which is the lefthand terminal of the inductor when the inductor with part name L is placed (horizontally) without rotation on a schematic. Thus, the dot or pin 1 will be at the bottom after one 90 ° counterclockwise rotation, since rotation is al ways about pin 1. Once the magnetically coupled inductors are arranged with the dot convention in mind and their value attributes are set in henries, we use the coupling symbol K\_LINEAR to define the coupling. For each pair of coupled inductors, take the following steps:
-
-- 1. Select **Draw/Get New Part** and type K\_LINEAR.
-- 2. Hit or click **OK** and place the K\_LINEAR symbol on the schematic, as shown in Fig. 13.51. (Notice that K\_LINEAR is not a component and therefore has no pins.)
-- 3. **DCLICKL** on COUPLING and set the v alue of the coupling coefficient *k*.
-- 4. **DCLICKL** on the boxe d K (the coupling symbol) and enter the reference designator names for the coupled inductors as v alues of Li, *i* = 1, 2, …, 6. For example, if inductors L20 and L23 are coupled, we set L1 = L20 and L2 = L23. L1 and at least one other Li must be assigned values; other Li's may be left blank.
-
-In step 4, up to six coupled inductors with equal coupling can be specified.
-
-For the air -core transformer, the partname is XFRM\_LINEAR. It can be inserted in a circuit by selecting **Draw/Get P art Name** and then typing in the part name or by selecting the part name from the analog.slb library. As shown typically in Fig. 13.52(a), the main attributes of the linear transformer are the coupling coef ficient *k* and the inductance v alues L1 and L2 in henries. If the mutual induc tance *M* is specified, its value must be used along with L1 and L2 to calculate *k*. Keep in mind that the v alue of *k* should lie between 0 and 1.
-
-For the ideal transformer , the part name is XFRM\_NONLINEAR and is located in the breako ut.slb library. Select it by clicking **Draw/ Get Part Name** and then typing in the part name. Its attrib utes are the coupling coefficient and the numbers of turns associated with L1 and L2, as illustrated typically in Fig. 13.52(b). The value of the coef ficient of mutual coupling *k* = 1.
-
-*PSpice* has some additional transformer configurations that we will not discuss here.
-
-# **Figure 13.51**
-
-K\_Linear for defining coupling.
-
-COUPLING = 0.5 L1\_VALUE = 1 mH L2\_VALUE = 25 mH (a)
-
-**Figure 13.52** (a) Linear transformer XFRM\_LINEAR, (b) ideal transformer XFRM\_NONLINEAR.
-
-Use *PSpice* to find *i*1, *i*2, and *i*3 in the circuit displayed in Fig. 13.53.
-
-For Example 13.13.
-
-# **Solution:**
-
-The coupling coefficients of the three coupled inductors are determined as follows:
-
-$$
-k_{12} = \frac{M_{12}}{\sqrt{L_1 L_2}} = \frac{1}{\sqrt{3 \times 3}} = 0.3333
-$$
-$$
-k_{13} = \frac{M_{13}}{\sqrt{L_1 L_3}} = \frac{1.5}{\sqrt{3 \times 4}} = 0.433
-$$
-$$
-k_{23} = \frac{M_{23}}{\sqrt{L_2 L_3}} = \frac{2}{\sqrt{3 \times 4}} = 0.5774
-$$
-
-The operating frequency *f* is obtained from Fig. 13.53 as *ω*= 12 *π*= 2 *πf* → *f* = 6 Hz.
-
-The schematic of the circuit is portrayed in Fig. 13.54. Notice ho w the dot convention is adhered to. For L2, the dot (not shown) is on pin 1 (the left-hand terminal) and is therefore placed without rotation. For L1, in order for the dot to be on the right-hand side of the inductor, the inductor must be rotated through 180 °. For L3, the inductor must be rotated through 90° so that the dot will be at the bottom. Note that the 2-H in ductor (*L*4) is not coupled. To handle the three coupled inductors, we use three K\_LINEAR parts provided in the analog library and set the following attributes (by double-clicking on the symbol K in the box):
-
-**Figure 13.54** Schematic of the circuit of Fig. 13.53.
-
-designators of the inductors on the schematic.
-
-Three IPRINT pseudocomponents are inserted in the appropriate branches to obtain the required currents *i*1, *i*2, and *i*3. As an AC singlefrequency analysis, we select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 6, and *Final Freq* = 6. After saving the schematic, we select **Analysis/Simulate** to simulate it. The output file includes:
-
-| FREQ | IM(V_PRINT2) | IP(V_PRINT2) |
-|-----------|--------------|--------------|
-| 6.000E+00 | 2.114E-01 | -7.575E+01 |
-| FREQ | IM(V_PRINT1) | IP(V_PRINT1) |
-| 6.000E+00 | 4.654E-01 | -7.025E+01 |
-| FREQ | IM(V_PRINT3) | IP(V_PRINT3) |
-| 6.000E+00 | 1.095E-01 | 1.715E+01 |
-
-From this we obtain
-
-$$
-\mathbf{I}_1 = 0.4654 \underline{\smash{\big)}\xspace - 70.25^{\circ}}
-$$
-\n
-$$
-\mathbf{I}_2 = 0.2114 \underline{\smash{\big)}\xspace - 75.75^{\circ}}, \qquad \mathbf{I}_3 = 0.1095 \underline{\smash{\big)}\xspace 17.15^{\circ}}
-$$
-
-Thus,
-
-$$
-i_1 = 0.4654 \cos (12 \pi t - 70.25^\circ) \text{ A}
-$$
-
-$$
-i_2 = 0.2114 \cos(12 \pi t - 75.75^\circ) \text{ A}
-$$
-
-$$
-i_3 = 0.1095 \cos(12 \pi t + 17.15^\circ) \text{ A}
-$$
-
-Find *io* in the circuit of Fig. 13.55, using *PSpice*.
-
-For Practice Prob. 13.13.
-
-**Answer:** 2.012 cos(4*t* + 68.52°) A.
-
-Find **V**1 and **V**2 in the ideal transformer circuit of Fig. 13.56, using *PSpice*.
-
-# Practice Problem 13.13
-
-# Example 13.14
-
-# **Solution:**
-
-- 1. **Define.** The problem is clearly defined and we can proceed to the next step.
-- 2. **Present.** We have an ideal transformer and we are to find the input and the output voltages for that transformer. In addition, we are to use *PSpice* to solve for the voltages.
-- 3. **Alternative.** We are required to use *PSpice*. We can use mesh analysis to perform a check.
-- 4. **Attempt.** As usual, we assume *ω* = 1 and find the corresponding values of capacitance and inductance of the elements:
-
-$$
-j10 = j\omega L \qquad \Rightarrow \qquad L = 10 \text{ H}
-$$
-$$
--j40 = \frac{1}{j\omega C} \qquad \Rightarrow \qquad C = 25 \text{ mF}
-$$
-
-Figure 13.57 shows the schematic. For the ideal transformer, we set the coupling factor to 0.99999 and the numbers of turns to 400,000 and 100,000. The two VPRINT2 pseudocomponents are connected across the transformer terminals to obtain **V**1 and **V**2. As a single-frequency analysis, we select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 0.1592, and *Final Freq* = 0.1592. After saving the schematic, we select **Analysis/Simulate** to simulate it. The output file includes:
-
-| FREQ | | VM(\$N_0003,\$N_0006) VP(\$N_0003,\$N_0006) |
-|------|---------------------|---------------------------------------------|
-| | 1.592E-01 9.112E+01 | 3.792E+01 |
-| FREQ | | VM(\$N_0006,\$N_0005) VP(\$N_0006,\$N_0005) |
-| | 1.592E-01 2.278E+01 | -1.421E+02 |
-
-This can be written as
-
-$$
-V_1 = 91.12 \times 37.92^{\circ}
-$$
- V and $V_2 = 22.78 \times 142.1^{\circ}$ V
-
-5. **Evaluate.** We can check the answer by using mesh analysis as follows:
-
-Loop 1
-$$
--120\sqrt{30^{\circ}} + (80 - j40)I_1 + V_1 + 20(I_1 - I_2) = 0
-$$
-
-Loop 2
-$$
-20(-I_1 + I_2) - V_2 + (6 + j10)I_2 = 0
-$$
-
-**Figure 13.57** The schematic for the circuit in Fig. 13.56.
-
-Reminder: For an ideal transformer, the inductances of both the primary and secondary windings are infinitely large.
-
-But
-$$
-V_2 = -V_1/4
-$$
- and $I_2 = -4I_1$ . This leads to
-\n $-120/30^\circ + (80 - j40)I_1 + V_1 + 20(I_1 + 4I_1) = 0$
-\n $(180 - j40)I_1 + V_1 = 120/30^\circ$
-\n $20(-I_1 - 4I_1) + V_1/4 + (6 + j10)(-4I_1) = 0$
-\n $(-124 - j40)I_1 + 0.25V_1 = 0$ or $I_1 = V_1/(496 + j160)$
-
-Substituting this into the first equation yields
-
-$$
-(180 - j40)V_1/(496 + j160) + V_1 = 120/30^{\circ}
-$$
-$$
-(184.39/-12.53^{\circ}/521.2/17.88^{\circ})V_1 + V_1
-$$
-$$
-= (0.3538/-30.41^{\circ} + 1)V_1 = (0.3051 + 1 - j0.17909)V_1 = 120/30^{\circ}
-$$
-$$
-V_1 = 120/30^{\circ}/1.3173/-7.81^{\circ} = 91.1/37.81^{\circ} V \qquad \text{and}
-$$
-$$
-V_2 = 22.78/-142.19^{\circ} V
-$$
-
-Both answers check.
-
-6. **Satisfactory?** We have satisfactorily answered the problem and checked the solution. We can now present the entire solution to the problem.
-
-Obtain **V**1 and **V**2 in the circuit of Fig. 13.58 using *PSpice*.
-
-For Practice Prob. 13.14.
-
-**Answer:** V1 = 153⧸ 2.18° V, V2 = 230.2⧸ 2.09° V.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/154_13.9 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/154_13.9 Applications.md
deleted file mode 100644
index 5758b5aeb88678db5ae7250fb708734eb7b54ccb..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/154_13.9 Applications.md
+++ /dev/null
@@ -1,227 +0,0 @@
-# **13.9** Applications
-
-Transformers are the largest, the heaviest, and often the costliest of cir cuit components. Nevertheless, they are indispensable passive devices in electric circuits. They are among the most efficient machines, 95 percent efficiency being common and 99 percent being achie vable. They have numerous applications. For example, transformers are used:
-
-- To step up or step down voltage and current, making them useful for power transmission and distribution.
-- To isolate one portion of a circuit from another (i.e., to transfer power without any electrical connection).
-- As an impedance-matching device for maximum power transfer.
-- In frequenc y-selective circuits whose operation depends on the response of inductances.
-
-Practice Problem 13.14
-
-For more information on the many kinds of transformers, a good text is W. M. Flanagan, Handbook of Transformer Design and Applications, 2nd ed. (New York: McGraw-Hill, 1993).
-
-# **Figure 13.59**
-
-A transformer used to isolate an ac supply from a rectifier.
-
-Because of these di verse uses, there are man y special designs for transformers (only some of which are discussed in this chapter): v oltage transformers, current transformers, power transformers, distribution transformers, impedance-matching transformers, audio transformers, single-phase transformers, three-phase transformers, rectifier transformers, inverter transformers, and more. In this section, we consider three important applications: transformer as an isolation de vice, transformer as a matching device, and power distribution system.
-
-# **13.9.1** Transformer as an Isolation Device
-
-Electrical isolation is said to exist between two devices when there is no physical connection between them. In a transformer, energy is transferred by magnetic coupling, without electrical connection between the primary circuit and secondary circuit. We no w consider three simple practical examples of how we take advantage of this property.
-
-First, consider the circuit in Fig. 13.59. A rectifier is an electronic circuit that converts an ac supply to a dc supply. A transformer is often used to couple the ac supply to the rectifier. The transformer serves two purposes. First, it steps up or steps do wn the voltage. Second, it pro vides electrical isolation between the ac po wer supply and the rectifier, thereby reducing the risk of shock hazard in handling the electronic device.
-
-As a second e xample, a transformer is often used to couple tw o stages of an amplifier, to prevent any dc voltage in one stage from affecting the dc bias of the next stage. Biasing is the application of a dc voltage to a transistor amplifier or any other electronic device in order to produce a desired mode of operation. Each amplifier stage is biased separately to operate in a particular mode; the desired mode of operation will be compromised without a transformer providing dc isolation. As shown in Fig. 13.60, only the ac signal is coupled through the transformer from one stage to the next. We recall that magnetic coupling does not exist with a dc voltage source. Transformers are used in radio and TV receivers to couple stages of high-frequency amplifiers. When the sole purpose of a transformer is to pro vide isolation, its turns ratio *n* is made unity . Thus, an isolation transformer has *n* = 1.
-
-As a third example, consider measuring the voltage across 13.2-kV lines. It is obviously not safe to connect a voltmeter directly to such highvoltage lines. A transformer can be used both to electrically isolate the line power from the v oltmeter and to step do wn the v oltage to a safe level, as shown in Fig. 13.61. Once the voltmeter is used to measure the
-
-**Figure 13.60** A transformer providing dc isolation between two amplifier stages.
-
-# **Figure 13.61**
-
-A transformer providing isolation between the power lines and the voltmeter.
-
-secondary voltage, the turns ratio is used to determine the line v oltage on the primary side.
-
-Determine the voltage across the load in Fig. 13.62.
-
-# **Solution:**
-
-We can apply the superposition principle to find the load voltage. Let *vL* = *vL*1 + *vL*2, where *vL*1 is due to the dc source and *vL*2 is due to the ac source. We consider the dc and ac sources separately, as shown in Fig. 13.63. The load voltage due to the dc source is zero, because a timevarying voltage is necessary in the primary circuit to induce a voltage in the secondary circuit. Thus, *vL*1 = 0. For the ac source and a value of *Rs* so small it can be neglected,
-
-$$
-\frac{\mathbf{V}_2}{\mathbf{V}_1} = \frac{\mathbf{V}_2}{120} = \frac{1}{3} \quad \text{or} \quad \mathbf{V}_2 = \frac{120}{3} = 40 \text{ V}
-$$
-
-Hence, **V***L*2 = 40 V ac or *vL*2 = 40 cos *ωt*; that is, only the ac voltage is passed to the load by the transformer. This example shows how the transformer provides dc isolation.
-
-For Example 13.15: (a) dc source, (b) ac source.
-
-Refer to Fig. 13.61. Calculate the turns ratio required to step down the 14.4-kV line voltage to a safe level of 120 V.
-
-Practice Problem 13.15
-
-**Answer:** 120.
-
-# **13.9.2** Transformer as a Matching Device
-
-We recall that for maximum po wer transfer, the load resistance *RL* must be matched with the source resistance *Rs*. In most cases, the two resistances are not matched; both are fixed and cannot be altered. However, an iron-core transformer can be used to match the load resistance to the source resistance. This is called *impedance matching*. For example, to connect a loudspeaker to an audio power amplifier requires a transformer, because the speak er's resistance is only a few ohms while the internal resistance of the amplifier is several thousand ohms.
-
-Consider the circuit shown in Fig. 13.64. We recall from Eq. (13.60) that the ideal transformer reflects its load back to the primary with a
-
-# **Figure 13.64**
-
-Transformer used as a matching device.
-
-**Figure 13.62** For Example 13.15.
-
-12 V dc
-
-+ ‒
-
-scaling factor of *n*2 . To match this reflected load *RL*∕*n*2 with the source resistance *Rs*, we set them equal,
-
-$$
-R_s = \frac{R_L}{n^2} \tag{13.73}
-$$
-
-Equation (13.73) can be satisfied by proper selection of the turns ratio *n*. From Eq. (13.73), we notice that a step-down transformer (*n* < 1) is needed as the matching de vice when *Rs* > *RL*, and a step-up ( *n* > 1) is required when *Rs* < *RL*.
-
-The ideal transformer in Fig. 13.65 is used to match the amplifier circuit to the loudspeak er to achie ve maximum po wer transfer. The Thevenin (or output) impedance of the amplifier is 192 Ω, and the internal impedance of the speaker is 12 Ω. Determine the required turns ratio.
-
-# **Solution:**
-
-Speaker
-
-We replace the amplifier circuit with the Thevenin equivalent and reflect the impedance **Z***L* = 12 Ω of the speaker to the primary side of the ideal transformer. Figure 13.66 shows the result. For maximum power transfer,
-
-$$
-Z_{\text{Th}} = \frac{Z_L}{n^2}
-$$
- or $n^2 = \frac{Z_L}{Z_{\text{Th}}} = \frac{12}{192} = \frac{1}{16}$
-
-Thus, the turns ratio is *n* = 1∕4 = 0.25.
-
-Using *P* = *I* 2 *R*, we can sho w that indeed the po wer delivered to the speaker is much lar ger than without the ideal transformer. Without the ideal transformer, the amplifier is directly connected to the speaker. The power delivered to the speaker is
-
-$$
-P_L = \left(\frac{\mathbf{V}_{\text{Th}}}{\mathbf{Z}_{\text{Th}} + \mathbf{Z}_L}\right)^2 \mathbf{Z}_L = 288 \text{ V}_{\text{Th}}^2 \mu \text{W}
-$$
-
-With the transformer in place, the primary and secondary currents are
-
-$$
-I_p = \frac{\mathbf{V}_{\text{Th}}}{\mathbf{Z}_{\text{Th}} + \mathbf{Z}_L/n^2}, \qquad I_s = \frac{I_p}{n}
-$$
-
-Hence,
-
-$$
-P_L = I_s^2 \mathbf{Z}_L = \left(\frac{\mathbf{V}_{\text{Th}}/n}{\mathbf{Z}_{\text{Th}} + \mathbf{Z}_L/n^2}\right)^2 \mathbf{Z}_L
-$$
-$$
-= \left(\frac{n\mathbf{V}_{\text{Th}}}{n^2 \mathbf{Z}_{\text{Th}} + \mathbf{Z}_L}\right)^2 \mathbf{Z}_L = 1,302 \mathbf{V}_{\text{Th}}^2 \mu \mathbf{W}
-$$
-
-confirming what was said earlier.
-
-Calculate the turns ratio of an ideal transformer required to match a 8-Ω load to a source with internal impedance of 800 Ω. Find the load voltage when the source voltage is 300 V.
-
-**Answer:** 0.1, 15 V.
-
-**Figure 13.66** Equivalent circuit of the circuit in Fig. 13.65; for Example 13.16.
-
-Using an ideal transformer to match the speaker to the amplifier; for
-
-**Figure 13.65**
-
-Example 13.16.
-
-# Practice Problem 13.16
-
-# **13.9.3** Power Distribution
-
-A po wer system basically consists of three components: generation, transmission, and distrib ution. The local electric compan y operates a plant that generates se veral hundreds of me gavolt-amperes (MV A), typically at about 18 kV . As Fig. 13.67 illustrates, three-phase step-up transformers are used to feed the generated po wer to the transmission line. Why do we need the transformer? Suppose we need to transmit 100,000 VA over a distance of 50 km. Since *S* = *VI*, using a line voltage of 1,000 V implies that the transmission line must carry 100 A and this requires a transmission line of a large diameter. If, on the other hand, we use a line voltage of 10,000 V, the current is only 10 A. The smaller current reduces the required conductor size, producing considerable savings as well as minimizing transmission line *I* 2 *R* losses. To minimize losses requires a step-up transformer. Without the transformer, the majority of the power generated would be lost on the transmission line. The ability of the transformer to step up or step down voltage and distribute power economically is one of the major reasons for generating ac rather than dc. Thus, for a gi ven power, the lar ger the v oltage, the better . Today, 1 MV is the lar gest voltage in use; the le vel may increase as a result of research and experiments.
-
-# **Figure 13.67**
-
-A typical power distribution system.
-
-Source: A. Marcus and C. M. Thomson, *Electricity for Technicians,* 2nd edition, © 1975, p. 337. Pearson Education, Inc., Upper Saddle River, NJ.
-
-Beyond the generation plant, the power is transmitted for hundreds of miles through an electric netw ork called the *power grid*. The threephase power in the po wer grid is con veyed by transmission lines hung overhead from steel towers which come in a v ariety of sizes and shapes. The (aluminum-conductor, steel-reinforced) lines typically ha ve overall diameters up to about 40 mm and can carry current of up to 1,380 A.
-
-At the substations, distrib ution transformers are used to step do wn the voltage. The step-do wn process is usually carried out in stages. Power may be distributed throughout a locality by means of either overhead or under ground cables. The substations distrib ute the po wer to residential, commercial, and industrial customers. At the receiving end, a residential customer is e ventually supplied with 120 ∕240 V, while industrial or commercial customers are fed with higher voltages such as One may ask, How would increasing the voltage not increase the current, thereby increasing I 2 R losses? Keep in mind that I = Vℓ∕R, where Vℓ is the potential difference between the sending and receiving ends of the line. The voltage that is stepped up is the sending end voltage V, not Vℓ. If the receiving end is VR, then Vℓ = V − VR. Since V and VR are close to each other, Vℓ is small even when V is stepped up. 460∕208 V. Residential customers are usually supplied by distrib ution transformers often mounted on the poles of the electric utility company. When direct current is needed, the alternating current is converted to dc electronically.
-
-# Example 13.17
-
-A distribution transformer is used to supply a household as in Fig. 13.68. The load consists of eight 100-W bulbs, a 350-W TV, and a 15-kW kitchen range. If the secondary side of the transformer has 72 turns, calculate: (a) the number of turns of the primary winding, and (b) the current *Ip* in the primary winding.
-
-**Figure 13.68** For Example 13.17.
-
-# **Solution:**
-
-(a) The dot locations on the winding are not important, since we are only interested in the magnitudes of the variables involved. Since
-
-$$
-\frac{N_p}{N_s} = \frac{V_p}{V_s}
-$$
-
-we get
-
-$$
-N_p = N_s \frac{V_p}{V_s} = 72 \frac{2,400}{240} = 720 \text{ turns}
-$$
-
-(b) The total power absorbed by the load is
-
-$$
-S = 8 \times 100 + 350 + 15{,}000 = 16.15
-$$
- kW
-
-But *S* = *VpIp* = *VsIs*, so that
-
-$$
-I_p = \frac{S}{V_p} = \frac{16,150}{2,400} = 6.729 \text{ A}
-$$
-
-# Practice Problem 13.17
-
-In Example 13.17, if the eight 100-W bulbs are replaced by twelve 60-W bulbs and the kitchen range is replaced by a 4.5-kW air- conditioner, find: (a) the total power supplied, (b) the current *Ip* in the primary winding.
-
-**Answer:** (a) 5.57 kW, (b) 2.321 A.
-
-# **13.10** Summary
-
-1. Two coils are said to be mutually coupled if the magnetic flux *ϕ* emanating from one passes through the other . The mutual induc tance between the two coils is given by
-
-$$
-M = k\sqrt{L_1 L_2}
-$$
-
-where *k* is the coupling coefficient, 0 < *k* < 1.
-
-2. If *v*1 and *i*1 are the voltage and current in coil 1, while *v*2 and *i*2 are the voltage and current in coil 2, then
-
-$$
-v_1 = L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}
-$$
- and $v_2 = L_2 \frac{di_2}{dt} + M \frac{di_1}{dt}$
-
- Thus, the voltage induced in a coupled coil consists of self-induced voltage and mutual voltage.
-
-- 3. The polarity of the mutually-induced v oltage is e xpressed in the schematic by the dot convention.
-- 4. The energy stored in two coupled coils is
-
-$$
-\frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 \pm Mi_1i_2
-$$
-
-- 5. A transformer is a four -terminal de vice containing tw o or more magnetically coupled coils. It is used in changing the current, v oltage, or impedance level in a circuit.
-- 6. A linear (or loosely coupled) transformer has its coils wound on a magnetically linear material. It can be replaced by an equi valent T or Π network for the purposes of analysis.
-- 7. An ideal (or iron-core) transformer is a lossless (*R*1 = *R*2 = 0) transformer with unity coupling coef ficient (*k* = 1) and infinite inductances (*L*1, *L*2, *M* → ∞).
-- 8. For an ideal transformer,
-
-$$
-\mathbf{V}_2 = n\mathbf{V}_1, \qquad \mathbf{I}_2 = \frac{\mathbf{I}_1}{n}, \qquad \mathbf{S}_1 = \mathbf{S}_2, \qquad \mathbf{Z}_R = \frac{\mathbf{Z}_L}{n^2}
-$$
-
- where *n* = *N*2∕*N*1 is the turns ratio. *N*1 is the number of turns of the primary winding and *N*2 is the number of turns of the second ary winding. The transformer steps up the primary voltage when *n* > 1, steps it do wn when *n* < 1, or serv es as a matching de vice when *n* = 1.
-
-- 9. An autotransformer is a transformer with a single winding common to both the primary and the secondary circuits.
-- 10. *PSpice* is a useful tool for analyzing magnetically coupled circuits.
-- 11. Transformers are necessary in all stages of po wer distribution systems. Three-phase voltages may be stepped up or do wn by threephase transformers.
-- 12. Important uses of transformers in electronics applications are as electrical isolation devices and impedance-matching devices.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/155_13.10 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/155_13.10 Summary.md
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-# Review Questions
-
-**13.1** Refer to the two magnetically coupled coils of Fig. 13.69(a). The polarity of the mutual voltage is:
-
-(a) Positive (b) Negative
-
-# **Figure 13.69**
-
-For Review Questions 13.1 and 13.2.
-
-**13.2** For the two magnetically coupled coils of Fig. 13.69(b), the polarity of the mutual voltage is:
-
-(a) Positive (b) Negative
-
-**13.3** The coefficient of coupling for two coils having *L*1 = 2 H, *L*2 = 8 H, *M* = 3 H is:
-
-| (a) 0.1875 | (b) 0.75 |
-|------------|-----------|
-| (c) 1.333 | (d) 5.333 |
-
-- **13.4** A transformer is used in stepping down or stepping up:
- - (a) dc voltages (b) ac voltages (c) both dc and ac voltages
-- **13.5** The ideal transformer in Fig. 13.70(a) has *N*2∕*N*1 = 10. The ratio *V*2∕*V*1 is:
-
-(a) 10 (b) 0.1 (c) −0.1 (d) −10
-
-**Figure 13.70** For Review Questions 13.5 and 13.6.
-
-- **13.6** For the ideal transformer in Fig. 13.70(b), *N*2∕*N*1 = 10. The ratio *i*2∕*I*1 is:
- - (a) 10 (b) 0.1 (c) −0.1 (d) −10
-- **13.7** A three-winding transformer is connected as portrayed in Fig. 13.71(a). The value of the output voltage *Vo* is:
-
-(a) 10 (b) 6 (c) −6 (d) −10
-
-# **Figure 13.71**
-
-For Review Questions 13.7 and 13.8.
-
-- **13.8** If the three-winding transformer is connected as in Fig. 13.71(b), the value of the output voltage *Vo* is: (a) 10 (b) 6 (c) −6 (d) −10
-- **13.9** In order to match a source with internal impedance of 500 Ω to a 15-Ω load, what is needed is:
- - (a) step-up linear transformer
- - (b) step-down linear transformer
- - (c) step-up ideal transformer
- - (d) step-down ideal transformer
- - (e) autotransformer
-- **13.10** Which of these transformers can be used as an isolation device?
- - (a) linear transformer (b) ideal transformer (c) autotransformer (d) all of the above
-
-*Answers: 13.1b, 13.2a, 13.3b, 13.4b, 13.5d, 13.6b, 13.7c, 13.8a, 13.9d, 13.10b.*
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/156_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/156_Review Questions.md
deleted file mode 100644
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-# Problems1
-
-# Section 13.2 Mutual Inductance
-
-**13.1** For the three coupled coils in Fig. 13.72, calculate the total inductance.
-
-**Figure 13.72**
-
-For Prob. 13.1.
-
-**13.2** Using Fig. 13.73, design a problem to help other students better understand mutual inductance.
-
-For Prob. 13.2.
-
-- **13.3** Two coils connected in series-aiding fashion have a total inductance of 500 mH. When connected in a series-opposing configuration, the coils have a total inductance of 300 mH. If the inductance of one coil (*L*1) is three times the other, find *L*1, *L*2, and *M*. What is the coupling coefficient?
-- **13.4** (a) For the coupled coils in Fig. 13.74(a), show that
-
-$$
-L_{\text{eq}} = L_1 + L_2 + 2M
-$$
-
-(b) For the coupled coils in Fig. 13.74(b), show that
-
-eq: Cous in Fig. 13.74(
-$$
-L_{eq} = \frac{L_1 L_2 - M^2}{L_1 + L_2 - 2M}
-$$
-
-**13.5** Two coils are mutually coupled, with *L*1 = 50 mH, *L*2 = 120 mH, and *k* = 0.5. Calculate the maximum possible equivalent inductance if:
-
-> (a) the two coils are connected in series (b) the coils are connected in parallel
-
-**13.6** Given the circuit shown in Fig. 13.75, determine the value of V1 and I2.
-
-# **Figure 13.76**
-
-For Prob. 13.7.
-
-**Figure 13.77**
-
-For Prob. 13.8.
-
-**13.9** Find **V***x* in the network shown in Fig. 13.78.
-
-1Remember, unless otherwise specified, assume all values of currents and voltages are rms.
-
-For Prob. 13.10.
-
-**13.11** Use mesh analysis to find *ix* in Fig. 13.80, where *is* = 4 cos(600*t*) A and *vs* = 110 cos(600*t* + 30°)
-
-**Figure 13.80** For Prob. 13.11.
-
-**13.12** Determine the equivalent *L*eq in the circuit of Fig. 13.81.
-
-For Prob. 13.13.
-
-in Fig. 13.83 at terminals *a*-*b*.
-
-# **Figure 13.83**
-
-For Prob. 13.14.
-
-**13.15** Find the Norton equivalent for the circuit in Fig. 13.84 at terminals *a*-*b*.
-
-# **Figure 13.84**
-
-For Prob. 13.15.
-
-**13.16** Obtain the Norton equivalent at terminals *a*-*b* of the circuit in Fig. 13.85.
-
-# **Figure 13.85** For Prob. 13.16.
-
-**13.17** In the circuit of Fig. 13.86, Z*L* is a 15-mH inductor having an impedance of *j*40 Ω. Determine *Z*in when *k* = 0.6.
-
-**Figure 13.86** For Prob. 13.17.
-
-**13.18** Find the Thevenin equivalent to the left of the load Z in the circuit of Fig. 13.87.
-
-# **Figure 13.87**
-
-For Prob. 13.18.
-
-**13.19** Determine an equivalent T-section that can be used to replace the transformer in Fig. 13.88.
-
-**Figure 13.88**
-
-# For Prob. 13.19.
-
-# Section 13.3 Energy in a Coupled Circuit
-
-**13.20** Determine currents **I**1, **I**2, and **I**3 in the circuit of Fig. 13.89. Find the energy stored in the coupled coils at *t* = 2 ms. Take *ω* = 1,000 rad/s.
-
-# **Figure 13.89** For Prob. 13.20.
-
-**13.21** Using Fig. 13.90, design a problem to help other students better understand energy in a coupled circuit.
-
-**Figure 13.90** For Prob. 13.21.
-
-# **Figure 13.91** For Prob. 13.22.
-
-**13.23** Let *is* = 5 cos (100*t*) A. Calculate the voltage across the capacitor, *vc*. Also calculate the value of the energy stored in the coupled coils at *t* = 2.5*π* ms.
-
-# **Figure 13.92**
-
-For Prob. 13.23.
-
-**13.24** In the circuit of Fig. 13.93,
-
-- (a) find the coupling coefficient, (b) calculate *vo*,
-- (c) determine the energy stored in the coupled inductors at *t* = 2 s.
-
-\* An asterisk indicates a challenging problem.
-
-For Prob. 13.25.
-
-**13.26** Find **I***o* in the circuit of Fig. 13.95. Switch the dot on the winding on the right and calculate **I***o* again.
-
-For Prob. 13.26.
-
-# **Figure 13.96**
-
-For Prob. 13.27.
-
-For Prob. 13.28.
-
-# Section 13.4 Linear Transformers
-
-**13.29** In the circuit of Fig. 13.98, find the value of the coupling coefficient *k* that will make the 10-Ω resistor dissipate 1.28 kW. For this value of *k*, find the energy stored in the coupled coils at *t* = 1.5 s.
-
-# **Figure 13.98**
-
-For Prob. 13.29.
-
-- **13.30** (a) Find the input impedance of the circuit in Fig. 13.99 using the concept of reflected impedance.
- - (b) Obtain the input impedance by replacing the linear transformer by its T equivalent.
-
-# **Figure 13.100** For Prob. 13.31.
-
-**\*13.32** Two linear transformers are cascaded as shown in Fig. 13.101. Show that
-
-$$
-\omega^{2}R(L_{a}^{2} + L_{a}L_{b} - M_{a}^{2})
-$$
-$$
-Z_{in} = \frac{+j\omega^{3}(L_{a}^{2}L_{b} + L_{a}L_{b}^{2} - L_{a}M_{b}^{2} - L_{b}M_{a}^{2})}{\omega^{2}(L_{a}L_{b} + L_{b}^{2} - M_{b}^{2}) - j\omega R(L_{a} + L_{b})}
-$$
-
-For Prob. 13.32.
-
-**13.33** Determine the input impedance of the air-core transformer circuit of Fig. 13.102.
-
-**Figure 13.102** For Prob. 13.33.
-
-**13.34** Using Fig. 13.103, design a problem to help other students better understand how to find the input impedance of circuits with transformers.
-
-**Figure 13.103** For Prob. 13.34.
-
-# Section 13.5 Ideal Transformers
-
-**13.36** As done in Fig. 13.32, obtain the relationships between terminal voltages and currents for each of the ideal transformers in Fig. 13.105.
-
-For Prob. 13.36.
-
-- **13.37** A 240∕2,400-V rms step-up ideal transformer delivers 50 kW to a resistive load. Calculate:
- - (a) the turns ratio
- - (b) the primary current
- - (c) the secondary current
-
-**13.38** Design a problem to help other students better understand ideal transformers.
-
-**13.39** A 1,200∕240-V rms transformer has impedance 60⧸−30° Ω on the high-voltage side. If the transformer is connected to a 0.8⧸10°-Ω load on the low-voltage side, determine the primary and secondary currents when the transformer is
-
-**13.40** The primary of an ideal transformer with a turns ratio of 5 is connected to a voltage source with Thevenin parameters *v*Th = 10 cos 2000*t* V and *R*Th = 100 Ω. Determine the average power delivered to a 200-Ω load connected across the secondary winding.
-
-For Prob. 13.41.
-
-**13.42** For the circuit in Fig. 13.107, determine the power absorbed by the 2-Ω resistor. Assume the 120 V is an rms value.
-
-**Figure 13.107** For Prob. 13.42.
-
-For Prob. 13.43.
-
-**13.45** For the circuit shown in Fig. 13.110, find the value of the average power absorbed by the 8-Ω resistor.
-
-# **Figure 13.110**
-
-For Prob. 13.45.
-
-**13.46** (a) Find **I**1 and **I**2 in the circuit of Fig. 13.111 below. (b) Switch the dot on one of the windings. Find **I**1 and **I**2 again.
-
-For Prob. 13.47.
-
-For Prob. 13.46.
-
-# **Figure 13.113** For Prob. 13.48.
-
-**13.49** Find current *ix* in the ideal transformer circuit shown in Fig. 13.114.
-
-For Prob. 13.49.
-
-**13.50** Calculate the input impedance for the network in Fig. 13.115.
-
-**13.53** Refer to the network in Fig. 13.118.
-
-- (a) Find *n* for maximum power supplied to the 200-Ω load.
-- (b) Determine the power in the 200-Ω load if *n* = 10.
-
-**Figure 13.118** For Prob. 13.53.
-
-- **13.51** Use the concept of reflected impedance to find
-- the input impedance and current **I**1 in
-
-Fig. 13.116.
-
-**13.54** A transformer is used to match an amplifier with
-
-an 8-Ω load as shown in Fig. 13.119. The Thevenin equivalent of the amplifier is: *V*Th = 10 V, ZTh = 128 Ω.
-
-- (a) Find the required turns ratio for maximum energy power transfer.
-- (b) Determine the primary and secondary currents.
-- (c) Calculate the primary and secondary voltages.
-
-**Figure 13.119**
-
-For Prob. 13.54.
-
-**13.56** Find the power absorbed by the 100-Ω resistor in the ideal transformer circuit of Fig. 13.121.
-
-**Figure 13.121** For Prob. 13.56.
-
-**13.57** For the ideal transformer circuit of Fig. 13.122 below, find:
-
-- (b) **V**1, **V**2, and **V***o*,
-- (c) the complex power supplied by the source.
-
-**13.58** Determine the average power absorbed by each resistor in the circuit of Fig. 13.123.
-
-# **Figure 13.123**
-
-For Prob. 13.58.
-
-# **Figure 13.124**
-
-For Prob. 13.59.
-
-**13.60** Refer to the circuit in Fig. 13.125 on the following page.
-
-(a) Find currents **I**1, **I**2, and **I**3.
-
-**Figure 13.125** For Prob. 13.60.
-
-**\*13.61** For the circuit in Fig. 13.126, find **I**1, **I**2, and **V***o*.
-
-**13.62** For the network in Fig. 13.127, find: (a) the complex power supplied by the source, (b) the average power delivered to the 18-Ω resistor.
-
-# **Figure 13.128**
-
-For Prob. 13.63.
-
-**13.64** For the circuit in Fig. 13.129, find the turns ratio so that the maximum power is delivered to the 30-kΩ resistor.
-
-**\*13.65** Calculate the average power dissipated by the 20-Ω resistor in Fig. 13.130.
-
-# Section 13.6 Ideal Autotransformers
-
-**13.66** Design a problem to help other students better understand how the ideal autotransformer works.
-
-**13.67** An autotransformer with a 40 percent tap is supplied by an 880-V, 60-Hz source and is used for stepdown operation. A 5-kVA load operating at unity power factor is connected to the secondary terminals. Find:
-
-(a) the secondary voltage,
-
-- (b) the secondary current,
-- (c) the primary current.
-- **13.68** In the ideal autotransformer of Fig. 13.131, calculate **I**1, **I**2, and **I***o*. Find the average power delivered to the load.
-
-For Prob. 13.68.
-
-**\*13.69** In the circuit of Fig. 13.131, *N*1 = 190 turns and *N*2 = 10 turns. Determine the Thevenin equivalent circuit looking into terminals *a* and *b*. What would be the value of **Z***L* that would absorb maximum power from the circuit?
-
-For Prob. 13.69.
-
-**13.70** In the ideal transformer circuit shown in Fig. 13.133, determine the average power delivered to the load.
-
-# **Figure 13.133**
-
-For Prob. 13.70.
-
-**13.71** When individuals travel, their electrical appliances need to have converters to match the voltages required by their appliances to the local voltage available to power their appliances. Today these converters use power electronics to convert voltages. In the past these converters were autotransformers. The autotransformer shown in Fig. 13.134 is used to convert 115 to 220 V. What is the value of the turns? If the maximum current available from the 115 V source is 15 A, what will be the maximum current available for the 220-V appliance?
-
-# **Figure 13.134**
-
-For Prob. 13.71.
-
-# Section 13.7 Three-Phase Transformers
-
-**13.72** In order to meet an emergency, three single-phase transformers with 12,470∕7,200 V rms are connected in Δ-Y to form a three-phase transformer which is fed by a 12,470-V transmission line. If the transformer supplies 60 MVA to a load, find:
-
-(a) the turns ratio for each transformer,
-
-- (b) the currents in the primary and secondary windings of the transformer,
-- (c) the incoming and outgoing transmission line currents.
-
-**13.73** Figure 13.135 on the next page shows a three-phase transformer that supplies a Y-connected load.
-
-- (a) Identify the transformer connection.
-- (b) Calculate currents **I**2 and **I***c*.
-- (c) Find the average power absorbed by the load.
-
-**Figure 13.135** For Prob. 13.73.
-
-- **13.74** Consider the three-phase transformer shown in Fig. 13.136. The primary is fed by a three-phase source with line voltage of 2.4 kV rms, while the secondary supplies a three-phase 120-kW balanced load at pf of 0.8. Determine:
- - (a) the type of transformer connections,
-
-(b) the values of *ILS* and *IPS*,
-
-(c) the values of *ILP* and *IPP*,
-
-- (d) the kVA rating of each phase of the transformer.
-- **13.75** A balanced three-phase transformer bank with the Δ-Y connection depicted in Fig. 13.137 is used to step down line voltages from 4,500 V rms to 900 V rms. If the transformer feeds a 120-kVA load, find:
- - (a) the turns ratio for the transformer,
- - (b) the line currents at the primary and secondary sides.
-
-**Figure 13.137** For Prob. 13.75.
-
-**13.76** Using Fig. 13.138, design a problem to help other students better understand a Y-Δ, three-phase transformer and how they work.
-
-**Figure 13.138** For Prob. 13.76.
-
-- **13.77** The three-phase system of a town distributes power with a line voltage of 13.2 kV. A pole transformer connected to single wire and ground steps down the high-voltage wire to 120 V rms and serves a house as shown in Fig. 13.139.
- - (a) Calculate the turns ratio of the pole transformer to get 120 V.
- - (b) Determine how much current a 100-W lamp connected to the 120-V hot line draws from the high-voltage line.
-
-**Figure 13.139** For Prob. 13.77.
-
-# Section 13.8 PSpice Analysis of Magnetically Coupled Circuits
-
-**13.78** Use *PSpice* or *MultiSim* to determine the mesh currents in the circuit of Fig. 13.140. Take *ω* = 1 rad/s. Use *k* = 0.5 when solving this problem.
-
-For Prob. 13.78.
-
-**13.79** Use *PSpice* or *MultiSim* to find **I**1, **I**2, and **I**3 in the circuit of Fig. 13.141.
-
-# **Figure 13.141** For Prob. 13.79.
-
-- **13.80** Rework Prob. 13.22 using *PSpice* or *Multisim*.
-- **13.81** Use *PSpice* or *MultiSim* to find **I**1, **I**2, and **I**3 in the circuit of Fig. 13.142.
-
-# **Figure 13.142**
-
-For Prob. 13.81.
-
-**13.82** Use *PSpice* or *MultiSim* to find **V**1, **V**2, and **I***o* in the circuit of Fig. 13.143.
-
-# **Figure 13.143**
-
-For Prob. 13.82.
-
-**13.83** Find **I***x* and **V***x* in the circuit of Fig. 13.144 using *PSpice* or *MultiSim*.
-
-**13.84** Determine **I**1, **I**2, and **I**3 in the ideal transformer circuit of Fig. 13.145 using *PSpice* or *MultiSim*.
-
-# Section 13.9 Applications
-
-- **13.85** A stereo amplifier circuit with an output impedance of 7.2 kΩ is to be matched to a speaker with an input impedance of 8 Ω by a transformer whose primary side has 3,000 turns. Calculate the number of turns required on the secondary side.
-- **13.86** A transformer having 2,400 turns on the primary and 48 turns on the secondary is used as an impedancematching device. What is the reflected value of a 3-Ω load connected to the secondary?
-- **13.87** A radio receiver has an input resistance of 300 Ω. When it is connected directly to an antenna system with a characteristic impedance of 75 Ω, an
-
-impedance mismatch occurs. By inserting an impedance-matching transformer ahead of the receiver, maximum power can be realized. Calculate the required turns ratio.
-
-- **13.88** A step-down power transformer with a turns ratio of *n* = 0.1 supplies 12.6 V rms to a resistive load. If the primary current is 2.5 A rms, how much power is delivered to the load?
-- **13.89** A 240∕120-V rms power transformer is rated at 10 kVA. Determine the turns ratio, the primary current, and the secondary current.
-- **13.90** A 4-kVA, 2,400∕240-V rms transformer has 250 turns on the primary side. Calculate:
- - (a) the turns ratio,
- - (b) the number of turns on the secondary side,
- - (c) the primary and secondary currents.
-- **13.91** A 25,000∕240-V rms distribution transformer has a primary current rating of 75 A.
- - (a) Find the transformer kVA rating.
- - (b) Calculate the secondary current.
-- **13.92** A 4,800-V rms transmission line feeds a distribution transformer with 1,200 turns on the primary and 28 turns on the secondary. When a 10-Ω load is connected across the secondary, find:
- - (a) the secondary voltage,
- - (b) the primary and secondary currents,
- - (c) the power supplied to the load.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/157_Comprehensive Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/157_Comprehensive Problems.md
deleted file mode 100644
index 36ac9578af9c57d41bb10404e35b7e1a48b06233..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/157_Comprehensive Problems.md
+++ /dev/null
@@ -1,26 +0,0 @@
-# Comprehensive Problems
-
-- **13.93** A four-winding transformer (Fig. 13.146) is often used in equipment (e.g., PCs, VCRs) that may be operated from either 110 or 220 V. This makes the equipment suitable for both domestic and foreign use. Show which connections are necessary to provide:
- - (a) an output of 14 V with an input of 110 V, (b) an output of 50 V with an input of 220 V.
-
-# **Figure 13.146**
-
-For Prob. 13.93.
-
-**\*13.94** A 440∕110-V ideal transformer can be connected to become a 550∕440-V ideal autotransformer. There
-
-are four possible connections, two of which are wrong. Find the output voltage of:
-
-- (a) a wrong connection,
-- (b) the right connection.
-- **13.95** Ten bulbs in parallel are supplied by a 7,200∕120-V transformer as shown in Fig. 13.147, where the bulbs are modeled by the 144-Ω resistors. Find:
- - (a) the turns ratio *n*,
- - (b) the current through the primary winding.
-
-For Prob. 13.95.
-
-**\*13.96** Some modern power transmission systems now have major high-voltage DC transmission segments. There are a lot of good reasons for doing this but we will not go into them here. To go from the AC to DC, power electronics are used. We start with three-phase AC and then rectify it (using a full-wave rectifier). It was found that using a delta to wye and delta combination connected secondary would give us a much smaller ripple after the full-wave rectifier. How is this accomplished? Remember that these are real devices and are wound on common cores.
-
-*Hint:* Use Figs. 13.47 and 13.49, and the fact that each coil of the wye connected secondary and each coil of the delta connected secondary are wound around the same core of each coil of the delta connected primary so the voltage of each of the corresponding coils are in phase. When the output leads of both secondaries are connected through fullwave rectifiers with the same load, you will see that the ripple is now greatly reduced. Please consult the instructor for more help if necessary.
-
-# **chapter**
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/158_Chapter 14 - Frequency Response.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/158_Chapter 14 - Frequency Response.md
deleted file mode 100644
index 0d884ef84b37566cb5bab4a42fdda34bde001dbc..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/158_Chapter 14 - Frequency Response.md
+++ /dev/null
@@ -1,29 +0,0 @@
-# Frequency 14 Response
-
-*Dost thou love Life? Then do not squander Time; for that is the stuff Life is made.*
-
-—Benjamin Franklin
-
-# Enhancing Your Career
-
-# **Career in Control Systems**
-
-Control systems are another area of electrical engineering where circuit analysis is used. A control system is designed to re gulate the beha vior of one or more variables in some desired manner. Control systems play major roles in our everyday life. Household appliances such as heating and air -conditioning systems, switch-controlled thermostats, w ashers and dryers, cruise controllers in automobiles, elevators, traffic lights, manufacturing plants, navigation systems—all utilize control systems. In the aerospace field, precision guidance of space probes, the wide range of operational modes of the space shuttle, and the ability to maneuv er space vehicles remotely from earth all require knowledge of control systems. In the manuf acturing sector, repetitive production line operations are increasingly performed by robots, which are programmable control systems designed to operate for many hours without fatigue.
-
-Control engineering inte grates circuit theory and communication theory. It is not limited to an y specific engineering discipline but may involve en vironmental, chemical, aeronautical, mechanical, ci vil, and electrical engineering. For example, a typical task for a control system engineer might be to design a speed regulator for a disk drive head.
-
-A thorough understanding of control systems techniques is essen tial to the electrical engineer and is of great v alue for designing control systems to perform the desired task.
-
-A welding robot. © Vol. 1 PhotoDisc/Getty Images RF
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Understand what transfer functions are and how to determine them.
-- 2. Understand the decibel scale, why we use it, and how to use it.
-- 3. Understand Bode plots, and know why we use them and how to determine them.
-- 4. Understand series and parallel resonance, why they are important, and how to find them.
-- 5. Understand passive filters.
-- 6. Understand active filters.
-- 7. Discuss magnitude and frequency scaling and why they are important.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/159_14.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/159_14.1 Introduction.md
deleted file mode 100644
index 6a90df000c605a53ef8825bd1ca28a5b1a56aca8..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/159_14.1 Introduction.md
+++ /dev/null
@@ -1,9 +0,0 @@
-# **14.1** Introduction
-
-In our sinusoidal circuit analysis, we ha ve learned ho w to find voltages and currents in a circuit with a constant frequency source. If we let the amplitude of the sinusoidal source remain constant and v ary the frequency, we obtain the circuit' s *frequency r esponse.* The frequency response may be re garded as a complete description of the sinusoidal steady-state behavior of a circuit as a function of frequency.
-
-The frequency response of a circuit is the variation in its behavior with change in signal frequency.
-
-The sinusoidal steady-state frequenc y responses of circuits are of significance in many applications, especially in communications and control systems. A specific application is in electric filters that block out or eliminate signals with unw anted frequencies and pass signals of the desired frequencies. Filters are used in radio, TV, and telephone systems to separate one broadcast frequency from another.
-
-We begin this chapter by considering the frequency response of simple circuits using their transfer functions. We then consider Bode plots, which are the industry-standard w ay of presenting frequenc y response. We also consider series and parallel resonant circuits and encounter important concepts such as resonance, quality f actor, cutoff frequency, and bandwidth. We discuss different kinds of filters and network scaling. In the last section, we consider one practical application of resonant circuits and two applications of filters.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/160_14.2 Transfer Function.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/160_14.2 Transfer Function.md
deleted file mode 100644
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+++ /dev/null
@@ -1,133 +0,0 @@
-# **14.2** Transfer Function
-
-The transfer function **H**(*ω*) (also called the *network function*) is a useful analytical tool for finding the frequency response of a circuit. In fact, the frequency response of a circuit is the plot of the circuit' s transfer function **H**(*ω*) versus *ω*, with *ω* varying from *ω* = 0 to *ω* = ∞.
-
-A transfer function is the frequenc y-dependent ratio of a forced function to a forcing function (or of an output to an input). The idea of a transfer function was implicit when we used the concepts of impedance
-
-The frequency response of a circuit may also be considered as the variation of the gain and phase with frequency.
-
-and admittance to relate v oltage and current. In general, a linear net work can be represented by the block diagram shown in Fig. 14.1.
-
-The transfer function **H**(*ω*) of a circuit is the frequency-dependent ratio of a phasor output **Y**(*ω*) (an element voltage or current) to a phasor input **X**(*ω*) (source voltage or current).
-
-Thus,
-
-$$
-\mathbf{H}(\omega) = \frac{\mathbf{Y}(\omega)}{\mathbf{X}(\omega)}\tag{14.1}
-$$
-
-assuming zero initial conditions. Since the input and output can be ei ther voltage or current at any place in the circuit, there are four possible transfer functions:
-
-$$
-\mathbf{H}(\omega) = \text{Voltage gain} = \frac{\mathbf{V}_o(\omega)}{\mathbf{V}_i(\omega)}\tag{14.2a}
-$$
-
-$$
-\mathbf{H}(\omega) = \text{Current gain} = \frac{\mathbf{I}_o(\omega)}{\mathbf{I}_i(\omega)}\tag{14.2b}
-$$
-
-$$
-\mathbf{H}(\omega) = \text{Transfer impedance} = \frac{\mathbf{V}_o(\omega)}{\mathbf{I}_i(\omega)} \tag{14.2c}
-$$
-
-$$
-\mathbf{H}(\omega) = \text{Transfer admittance} = \frac{\mathbf{I}_o(\omega)}{\mathbf{V}_i(\omega)}\tag{14.2d}
-$$
-
-where subscripts *i* and *o* denote input and output values. Being a complex quantity, **H**(*ω*) has a magnitude *H*(*ω*) and a phase *ϕ*; that is, **H**(*ω*) = *H*(*ω*)⧸*ϕ*.
-
-To obtain the transfer function using Eq. (14.2), we first obtain the frequenc y-domain equi valent of the circuit by replacing resistors, inductors, and capacitors with their impedances *R*, *jωL*, and 1∕*jωC*. We then use an y circuit technique(s) to obtain the appropriate quantity in Eq. (14.2). We can obtain the frequency response of the circuit by plot ting the magnitude and phase of the transfer function as the frequenc y varies. A computer is a real time-saver for plotting the transfer function.
-
-The transfer function **H**(*ω*) can be expressed in terms of its numerator polynomial **N**(*ω*) and denominator polynomial **D**(*ω*) as
-
-$$
-\mathbf{H}(\omega) = \frac{\mathbf{N}(\omega)}{\mathbf{D}(\omega)}
-$$
- (14.3)
-
-where **N**(*ω*) and **D**(*ω*) are not necessarily the same e xpressions for the input and output functions, respecti vely. The representation of **H**(*ω*) in Eq. (14.3) assumes that common numerator and denominator f actors in **H**(*ω*) have canceled, reducing the ratio to lo west terms. The roots of **N**(*ω*) = 0 are called the *zeros* of **H**(*ω*) and are usually represented as *jω* = *z*1, *z*2, …. Similarly, the roots of **D**(*ω*) = 0 are the *poles* of **H**(*ω*) and are represented as *jω* = *p*1, *p*2,….
-
-A zero, as a root of the numerator polynomial, is a value that results in a zero value of the function. A pole, as a root of the denominator polynomial, is a value for which the function is infinite.
-
-To avoid complex algebra, it is expedient to replace *jω* temporarily with *s* when working with **H**(*ω*) and replace *s* with *jω* at the end.
-
-# **Figure 14.1**
-
-A block diagram representation of a linear network.
-
-In this context, **X**(*ω*) and **Y**(*ω*) denote the input and output phasors of a network; they should not be confused with the same symbolism used for reactance and admittance. The multiple usage of symbols is conventionally permissible due to lack of enough letters in the English language to express all circuit variables distinctly.
-
-Some authors use **H**( j*ω*) for transfer instead of **H**(*ω*), since *ω* and j are an inseparable pair.
-
-A zero may also be regarded as the value of s = j*ω* that makes H(s) zero, and a pole as the value of s = j*ω* that makes H(s) infinite.
-
-# **Solution:**
-
-The frequency-domain equivalent of the circuit is in Fig. 14.2(b). By voltage division, the transfer function is given by
-
-# **Figure 14.2** For Example 14.1: (a) time-domain *RC* circuit,
-
-Comparing this with Eq. (9.18e), we obtain the magnitude and phase of **H**(*ω*) as
-
-$$
-H = \frac{1}{\sqrt{1 + (\omega/\omega_0)^2}}, \qquad \phi = -\tan^{-1}\frac{\omega}{\omega_0}
-$$
-
-where *ω*0 = 1∕*RC*. To plot *H* and *ϕ* for 0 < *ω*< ∞, we obtain their values at some critical points and then sketch.
-
- At *ω* = 0, *H* = 1 and *ϕ* = 0. At *ω* = ∞, *H* = 0 and *ϕ* = −90°. Also, at *ω* = *ω*0, *H* = 1∕ √ \_\_ 2 and *ϕ* = −45°. With these and a few more points as shown in Table 14.1, we find that the frequency response is as shown in Fig. 14.3. Additional features of the frequency response in Fig. 14.3 will be explained in Section 14.6.1 on low-pass filters.
-
-# **TABLE 14.1** For Example 14.1. *ω*∕*ω***0** *H ϕ ω*∕*ω***0** *H ϕ* 0 1 0 10 0.1 −84° 1 0.71 −45° 20 0.05 −87° 2 0.45 −63° 100 0.01 −89°
-
-Practice Problem 14.1 Obtain the transfer function **V***o*∕**V***s* of the *RL* circuit in Fig. 14.4, assuming *vs* = *Vm* cos *ωt*. Sketch its frequency response.
-
-3 0.32 −72° ∞ 0 −90°
-
-**Answer:** *jωL*∕(*R* + *jωL*); see Fig. 14.5 for the response.
-
-H 1
-
-0.707
-
-Frequency response of the *RC* circuit: (a) amplitude response, (b) phase response.
-
-**Figure 14.4** *RL* circuit for Practice Prob. 14.1.
-
-(b) frequency-domain *RC* circuit.
-
-For the circuit in Fig. 14.6, calculate the g ain **I***o*(*ω*)∕**I***i*(*ω*) and its poles Example 14.2 and zeros.
-
-# **Solution:**
-
-By current division,
-
-$$
-\mathbf{I}_o(\omega) = \frac{4 + j2\omega}{4 + j2\omega + 1/j0.5\omega} \mathbf{I}_i(\omega)
-$$
-
-or
-
-$$
-\frac{\mathbf{I}_o(\omega)}{\mathbf{I}_i(\omega)} = \frac{j0.5\omega(4+j2\omega)}{1+j2\omega + (j\omega)^2} = \frac{s(s+2)}{s^2 + 2s + 1}, \qquad s = j\omega
-$$
-
-The zeros are at
-
-*s*(*s* + 2) =0 ⇒ *z*1 = 0, *z*2 = −2
-
-The poles are at
-
-$$
-s^2 + 2s + 1 = (s + 1)^2 = 0
-$$
-
-Thus, there is a repeated pole (or double pole) at *p* = −1.
-
-Find the transfer function **V***o*(*ω*)∕**I***i*(*ω*) for the circuit in Fig. 14.7. Obtain Practice Problem 14.2 its zeros and poles.
-
-its zeros and poles.
-\n**Answer:**
-$$
-\frac{10(s + 2)(s + 5)}{s^2 + 10s + 10}
-$$
-, $s = j\omega$ ; zeros: -2, -5; poles: -1.127, -8.873.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/161_14.3 The Decibel Scale.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/161_14.3 The Decibel Scale.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/161_14.3 The Decibel Scale.md
+++ /dev/null
@@ -1,85 +0,0 @@
-# **14.3** The Decibel Scale
-
-It is not always easy to get a quick plot of the magnitude and phase of the transfer function as we did above. A more systematic way of obtaining the frequency response is to use Bode plots. Before we begin to construct Bode plots, we should take care of two important issues: the use of logarithms and decibels in expressing gain.
-
-For Example 14.2.
-
-**Figure 14.7** For Practice Prob. 14.2.
-
-© Ingram Publishing RF
-
-# Historical
-
-**Alexander Graham Bell** (1847–1922) inventor of the telephone, was a Scottish-American scientist.
-
-Bell was born in Edinburgh, Scotland, a son of Alexander Melville Bell, a well-known speech teacher. Alexander the younger also became a speech teacher after graduating from the University of Edinburgh and the University of London. In 1866 he became interested in transmitting speech electrically. After his older brother died of tuberculosis, his father decided to move to Canada. Alexander was asked to come to Boston to work at the School for the Deaf. There he met Thomas A. Watson, who became his assistant in his electromagnetic transmitter experiment. On March 10, 1876, Alexander sent the famous first telephone message: "Watson, come here I want you." The bel, the logarithmic unit intro duced in this chapter, is named in his honor.
-
-Since Bode plots are based on log arithms, it is important that we keep the following properties of logarithms in mind:
-
-1. log *P*1*P*2 = log *P*1 + log *P*2 2. log *P*1∕*P*2 = log *P*1 − log *P*2 3. log *Pn* = *n* log *P* 4. log 1 = 0
-
-Historical note: The bel is named after Alexander Graham Bell, the inventor of the telephone.
-
-In communications systems, g ain is measured in *bels*. Historically, the bel is used to measure the ratio of two levels of power or power gain *G*; that is,
-
-$$
-G = \text{Number of bels} = \log_{10} \frac{P_2}{P_1} \tag{14.4}
-$$
-
-The *decibel* (dB) provides us with a unit of less magnitude. It is 1∕10th of a bel and is given by
-
-$$
-G_{\text{dB}} = 10 \log_{10} \frac{P_2}{P_1} \tag{14.5}
-$$
-
-When *P*1 = *P*2, there is no change in po wer and the g ain is 0 dB. If *P*2 = 2*P*1, the gain is
-
-$$
-G_{\rm dB} = 10 \log_{10} 2 \simeq 3 \text{ dB} \tag{14.6}
-$$
-
-and when *P*2 = 0.5*P*1, the gain is
-
-$$
-G_{\rm dB} = 10 \log_{10} 0.5 \simeq -3 \text{ dB} \tag{14.7}
-$$
-
-Equations (14.6) and (14.7) sho w another reason wh y log arithms are greatly used: The logarithm of the reciprocal of a quantity is simply negative the logarithm of that quantity.
-
-Alternatively, the g ain *G* can be e xpressed in terms of v oltage or current ratio. To do so, consider the network shown in Fig. 14.8. If *P*1 is the input power, *P*2 is the output (load) power, *R*1 is the input resistance,
-
-Voltage-current relationships for a fourterminal network.
-
-and *R*2 is the load resistance, then *P*1 = 0.5*V*2 1∕*R*1 and *P*2 = 0.5*V*2 2 ∕*R*2, and Eq. (14.5) becomes
-
-$$
-G_{\text{dB}} = 10 \log_{10} \frac{P_2}{P_1} = 10 \log_{10} \frac{V_2^2 / R_2}{V_1^2 / R_1}
-$$
-
-= 10 \log\_{10} \left(\frac{V\_2}{V\_1}\right)^2 + 10 \log\_{10} \frac{R\_1}{R\_2} (14.8)
-
-$$
-G_{\text{dB}} = 20 \log_{10} \frac{V_2}{V_1} - 10 \log_{10} \frac{R_2}{R_1}
-$$
- (14.9)
-
-For the case when *R*2 = *R*1, a condition that is often assumed when comparing voltage levels, Eq. (14.9) becomes
-
-$$
-G_{\text{dB}} = 20 \log_{10} \frac{V_2}{V_1}
-$$
- (14.10)
-
-Instead, if *P*1 = *I* 1 2 *R*1 and *P*2 = *I*2 2 *R*2, for *R*1 = *R*2, we obtain
-
-$$
-G_{\text{dB}} = 20 \log_{10} \frac{I_2}{I_1} \tag{14.11}
-$$
-
-Three things are important to note from Eqs. (14.5), (14.10), and (14.11):
-
-- 1. That 10 log 10 is used for po wer, while 20 log 10 is used for v oltage or current, because of the square relationship between them (*P* = *V*2 ∕*R* = *I* 2 *R*).
-- 2. That the dB value is a logarithmic measurement of the *ratio* of one variable to another *of the same type*. Therefore, it applies in expressing the transfer function *H* in Eqs. (14.2a) and (14.2b), which are dimensionless quantities, but not in expressing *H* in Eqs. (14.2c) and (14.2d).
-- 3. It is important to note that we only use voltage and current magnitudes in Eqs. (14.10) and (14.11). Negative signs and angles will be handled independently as we will see in Section 14.4.
-
-With this in mind, we now apply the concepts of logarithms and decibels to construct Bode plots.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/162_14.4 Bode Plots.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/162_14.4 Bode Plots.md
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-# **14.4** Bode Plots
-
-Obtaining the frequenc y response from the transfer function as we did in Section 14.2 is an uphill task. The frequency range required in fre quency response is often so wide that it is incon venient to use a linear scale for the frequenc y axis. Also, there is a more systematic w ay of locating the important features of the magnitude and phase plots of the transfer function. For these reasons, it has become standard practice to plot the transfer function on a pair of semilogarithmic plots: The magnitude in decibels is plotted against the logarithm of the frequency; on a separate plot, the phase in degrees is plotted against the logarithm of the frequency. Such semilogarithmic plots of the transfer function—kno wn as *Bode plots*—have become the industry standard.
-
-Bode plots are semilog plots of the magnitude (in decibels) and phase (in degrees) of a transfer function versus frequency.
-
-Historical note: Named after Hendrik W. Bode (1905–1982), an engineer with the Bell Telephone Laboratories, for his pioneering work in the 1930s and 1940s.
-
-Bode plots contain the same information as the nonlogarithmic plots discussed in the previous section, but they are much easier to construct, as we shall see shortly.
-
-The transfer function can be written as
-
-$$
-\mathbf{H} = H/\phi = He^{j\phi} \tag{14.12}
-$$
-
-Taking the natural logarithm of both sides,
-
-$$
-\ln H = \ln H + \ln e^{j\phi} = \ln H + j\phi \tag{14.13}
-$$
-
-Thus, the real part of ln**H** is a function of the magnitude while the imaginary part is the phase. In a Bode magnitude plot, the gain
-
-$$
-H_{\rm dB} = 20 \log_{10} H \tag{14.14}
-$$
-
-is plotted in decibels (dB) v ersus frequency. Table 14.2 provides a few values of *H* with the corresponding v alues in decibels. In a Bode phase plot, *ϕ* is plotted in degrees versus frequency. Both magnitude and phase plots are made on semilog graph paper.
-
-A transfer function in the form of Eq. (14.3) may be written in terms of factors that have real and imaginary parts. One such representation might be
-
-A transfer function in the form of Eq. (14.3) may be written in terms of
-\nors that have real and imaginary parts. One such representation might be
-\n
-$$
-\mathbf{H}(\omega) = \frac{K(j\omega)^{\pm 1}(1 + j\omega/z_1)[1 + j2\zeta_1\omega/\omega_k + (j\omega/\omega_k)^2] \cdots}{(1 + j\omega/p_1)[1 + j2\zeta_2\omega/\omega_n + (j\omega/\omega_n)^2] \cdots}
-$$
-\n(14.15)
-
-which is obtained by di viding out the poles and zeros in **H**(*ω*). The representation of **H**(*ω*) as in Eq. (14.15) is called the *standard form.* **H**(*ω*) may include up to seven types of different factors that can appear in various combinations in a transfer function. These are:
-
-- 1. A gain *K*
-- 2. A pole (*jω*) −1 or zero (*jω*) at the origin
-- 3. A simple pole 1∕(1 + *jω*∕*p*1) or zero (1 + *jω*∕*z*1)
-- 4. A quadratic pole 1 ∕[1 + *j*2*ζ*2*ω*∕*ωn* + ( *jω*∕*ωn*) 2 ] or zero [1 + *j*2*ζ*1*ω*∕*ωk* + ( *jω*∕*ωk*) 2 ]
-
-In constructing a Bode plot, we plot each factor separately and then add them graphically. The factors can be considered one at a time and then combined additively because of the logarithms involved. It is this mathematical convenience of the logarithm that makes Bode plots a powerful engineering tool. 0 *ϕ*
-
-We will now make straight-line plots of the factors listed above. We shall find that these straight-line plots known as Bode plots approximate the actual plots to a reasonable degree of accuracy. 0.1 1 10 100 *ω*
-
-**Constant term:** For the g ain *K*, the magnitude is 20 log 10 *K* and the phase is 0°; both are constant with frequenc y. Thus, the magnitude and phase plots of the gain are shown in Fig. 14.9. If *K* is negative, the magnitude remains 20 log10 ∣*K*∣ but the phase is ±180°.
-
-**Pole/zero at the origin:** For the zero (*jω*) at the origin, the magnitude is 20 log10 *ω* and the phase is 90°. These are plotted in Fig. 14.10, where we notice that the slope of the magnitude plot is 20 dB/decade, while the phase is constant with frequency.
-
-The Bode plots for the pole (*jω*) −1 are similar except that the slope of the magnitude plot is −20 dB/decade while the phase is −90°. In general,
-
-# **TABLE 14.2**
-
-Specific gain and their decibel values.\*
-
-| Magnitude H | 20 log10
H (dB) |
-|--------------|--------------------|
-| 0.001 | −60 |
-| 0.01 | −40 |
-| 0.1 | −20 |
-| 0.5
__ | −6 |
-| 1∕ √
2 | −3 |
-| 1 | 0 |
-| __
√
2 | 3 |
-| 2 | 6 |
-| 10 | 20 |
-| 20 | 26 |
-| 100 | 40 |
-| 1000 | 60 |
-| | |
-
-\* Some of these values are approximate.
-
-The origin is where *ω* = 1 or log *ω* = 0 and the gain is zero.
-
-**Figure 14.9**
-
-Bode plots for gain *K*: (a) magnitude plot, (b) phase plot.
-
-for (*jω*) *N*, where *N* is an integer, the magnitude plot will have a slope of 20*N* dB/decade, while the phase is 90*N* degrees.
-
-**Simple pole/zero:** For the simple zero (1 + *jω*∕*z*1), the magnitude is 20 log10 ∣1 + *jω*∕*z*1∣ and the phase is tan−1 *ω*∕*z*1. We notice that
-
-$$
-H_{\text{dB}} = 20 \log_{10} \left| 1 + \frac{j\omega}{z_1} \right| \qquad \Rightarrow \qquad 20 \log_{10} 1 = 0 \qquad \textbf{(14.16)}
-$$
-\n
-$$
-\text{as} \quad \omega \to 0
-$$
-\n
-$$
-H_{\text{dB}} = 20 \log_{10} \left| 1 + \frac{j\omega}{z_1} \right| \qquad \Rightarrow \qquad 20 \log_{10} \frac{\omega}{z_1} \qquad \textbf{(14.17)}
-$$
-\n
-$$
-\text{as} \quad \omega \to \infty
-$$
-
-showing that we can approximate the magnitude as zero (a straight line with zero slope) for small v alues of *ω* and by a straight line with slope 20 dB/decade for large values of *ω*. The frequency *ω* = *z*1 where the two asymptotic lines meet is called the *corner frequency* or *break frequency*. Thus, the approximate magnitude plot is sho wn in Fig. 14.11(a), where the actual plot is also shown. Notice that the approximate plot is close to the actual plot except at the break frequency, where *ω* = *z*1 and the deviation is 20 log10 ∣(1 + *j*1)∣ = 20 log10 √ \_\_ 2 ≃ 3 dB.
-
-The phase tan−1 (*ω*∕*z*1) can be expressed as
-
-$$
-\phi = \tan^{-1}\left(\frac{\omega}{z_1}\right) = \begin{cases} 0, & \omega = 0 \\ 45^\circ, & \omega = z_1 \\ 90^\circ, & \omega \to \infty \end{cases}
-$$
- (14.18)
-
-As a straight-line approximation, we let *ϕ*≃ 0 for *ω*≤ *z*1∕10, *ϕ* ≃ 45° for *ω* = *z*1, and *ϕ*≃ 90° for *ω* ≥ 10*z*1. As shown in Fig. 14.11(b) along with the actual plot, the straight-line plot has a slope of 45° per decade.
-
-The Bode plots for the pole 1 ∕(1 + *jω*∕*p*1) are similar to those in Fig. 14.11 except that the corner frequency is at *ω* = *p*1, the magnitude has a slope of −20 dB/decade, and the phase has a slope of −45° per decade.
-
-**Quadratic pole/zer o:** The magnitude of the quadratic pole 1 ∕[1 + *j*2*ζ*2*ω*∕*ωn* + (*jω*∕*ωn*) 2 ] is −20 log10∣1 + *j*2*ζ*2*ω*∕*ωn* + ( *jω*∕*ωn*) 2 ∣ and the phase is −tan−1 (2*ζ*2*ω*∕*ωn*)∕(1 − *ω*2 ∕*ωn* 2 ). But
-
-A decade is an interval between two frequencies with a ratio of 10; e.g., between *ω*0 and 10*ω*0, or between 10 and 100 Hz. Thus, 20 dB/decade means that the magnitude changes 20 dB whenever the frequency changes tenfold or one decade.
-
-The special case of dc (*ω* = 0) does not appear on Bode plots because log 0 = −∞, implying that zero frequency is infinitely far to the left of the origin of Bode plots.
-
-# **Figure 14.10**
-
-Bode plot for a zero ( *jω*) at the origin: (a) magnitude plot, (b) phase plot.
-
-**(14.19)**
-
-Bode plots of zero (1 + *jω*∕*z*1): (a) magnitude plot, (b) phase plot.
-
-and
-
-$$
-H_{\text{dB}} = -20 \log_{10} \left| 1 + \frac{j2\zeta_2 \omega}{\omega_n} + \left( \frac{j\omega}{\omega_n} \right)^2 \right| \qquad \Rightarrow \qquad -40 \log_{10} \frac{\omega}{\omega_n}
-$$
-as $\omega \to \infty$ (14.20)
-
-Thus, the amplitude plot consists of tw o straight asymptotic lines: one with zero slope for *ω*< *ωn* and the other with slope −40 dB/decade for *ω* > *ωn*, with *ωn* as the corner frequenc y. Figure 14.12(a) sho ws the approximate and actual amplitude plots. Note that the actual plot depends on the damping factor *ζ*2 as well as the corner frequency *ωn*. The significant peaking in the neighborhood of the corner frequency should be added to the straight-line approximation if a high le vel of accurac y is desired. However, we will use the straight-line approximation for the sake of simplicity.
-
-**Figure 14.12** Bode plots of quadratic pole [1 + *j*2*ζω*∕*ωn* − *ω*2 ∕ *ωn* 2 ] −1 : (a) magnitude plot, (b) phase plot.
-
-There is another procedure for obtaining Bode plots that is faster and perhaps more efficient than the one we have just discussed. It consists in realizing that zeros cause an increase in slope, while poles cause a decrease. By starting with the low-frequency asymptote of the Bode plot, moving along the frequency axis, and increasing or decreasing the slope at each corner frequency, one can sketch the Bode plot immediately from the transfer function without the effort of making individual plots and adding them. This procedure can be used once you become proficient in the one discussed here.
-
- Digital computers have rendered the procedure discussed here almost obsolete. Several software packages such as PSpice, MATLAB, Mathcad, and Micro-Cap can be used to generate frequency response plots. We will discuss PSpice later in the chapter.
-
-# The phase can be expressed as
-
-$$
-\phi = -\tan^{-1} \frac{2\zeta_2 \omega / \omega_n}{1 - \omega^2 / \omega_n^2} = \begin{cases} 0, & \omega = 0 \\ -90^\circ, & \omega = \omega_n \\ -180^\circ, & \omega \to \infty \end{cases}
-$$
- (14.21)
-
-The phase plot is a straight line with a slope of −90° per decade starting at *ωn*∕10 and ending at 10*ωn*, as shown in Fig. 14.12(b). We see again that the difference between the actual plot and the straight-line plot is due to the damping f actor. Notice that the straight-line approximations for both magnitude and phase plots for the quadratic pole are the same as those for a double pole, that is, (1 + *jω*∕*ωn*) −2 . We should e xpect this because the double pole (1 + *jω*∕*ωn*) −2 equals the quadratic pole 1∕[1 + *j*2*ζ*2*ω*∕*ωn* + (*jω*∕*ωn*) 2 ] when *ζ*2 = 1. Thus, the quadratic pole can be treated as a double pole as far as straight-line approximation is concerned.
-
-For the quadratic zero [1 + *j*2*ζ*1*ω*∕*ωk* + (*jω*∕*ωk*) 2 ], the plots in Fig. 14.12 are in verted because the magnitude plot has a slope of 40 dB/decade while the phase plot has a slope of 90° per decade.
-
-Table 14.3 presents a summary of Bode plots for the se ven factors. Of course, not every transfer function has all seven factors. To sketch the Bode plots for a function **H**(*ω*) in the form of Eq. (14.15), for e xample, we first record the corner frequencies on the semilog graph paper, sketch the factors one at a time as discussed above, and then combine additively
-
-the graphs of the f actors. The combined graph is often dra wn from left to right, changing slopes appropriately each time a corner frequenc y is encountered. The following examples illustrate this procedure.
-
-Example 14.3 Construct the Bode plots for the transfer function
-
-ots for the transfer function
-\n
-$$
-\mathbf{H}(\omega) = \frac{200j\omega}{(j\omega + 2)(j\omega + 10)}
-$$
-
-# **Solution:**
-
-We first put **H**(*ω*) in the standard form by dividing out the poles and zeros. Thus,
-
-We first put
-$$
-\mathbf{H}(\omega)
-$$
- in the standard form by dividing out the poles and
-zeros. Thus,
-$$
-\mathbf{H}(\omega) = \frac{10j\omega}{(1 + j\omega/2)(1 + j\omega/10)}
-$$
-$$
-= \frac{10 |j\omega|}{|1 + j\omega/2||1 + j\omega/10|} \frac{(90^\circ - \tan^{-1} \omega/2 - \tan^{-1} \omega/10)}{1 - j\omega/10}
-$$
-
-Hence, the magnitude and phase are
-
-$$
-H_{\text{dB}} = 20 \log_{10} 10 + 20 \log_{10} |j\omega| - 20 \log_{10} \left| 1 + \frac{j\omega}{2} \right|
-$$
-$$
-- 20 \log_{10} \left| 1 + \frac{j\omega}{10} \right|
-$$
-$$
-\phi = 90^{\circ} - \tan^{-1} \frac{\omega}{2} - \tan^{-1} \frac{\omega}{10}
-$$
-
-We notice that there are two corner frequencies at *ω* = 2,10. For both the magnitude and phase plots, we sketch each term as shown by the dotted lines in Fig. 14.13. We add them up graphically to obtain the overall plots shown by the solid curves.
-
-**Figure 14.13** For Example 14.3: (a) magnitude plot, (b) phase plot.
-
-$$
-\mathbf{H}(\omega) = \frac{5(j\omega + 2)}{j\omega(j\omega + 10)}
-$$
-
-**Answer:** See Fig. 14.14.
-
-$$
-\mathbf{H}(\omega) = \frac{j\omega + 10}{j\omega(j\omega + 5)^2}
-$$
-
-# **Solution:**
-
-Putting **H**(*ω*) in the standard form, we get
-
-$$
-\text{H}(\omega) = \frac{0.4(1 + j\omega/10)}{j\omega(1 + j\omega/5)^2}
-$$
-
-From this, we obtain the magnitude and phase as
-
-$$
-H_{\text{dB}} = 20 \log_{10} 0.4 + 20 \log_{10} \left| 1 + \frac{j\omega}{10} \right| - 20 \log_{10} |j\omega|
-$$
-$$
-- 40 \log_{10} \left| 1 + \frac{j\omega}{5} \right|
-$$
-$$
-\phi = 0^{\circ} + \tan^{-1} \frac{\omega}{10} - 90^{\circ} - 2 \tan^{-1} \frac{\omega}{5}
-$$
-
-There are two corner frequencies at *ω* = 5, 10 rad/s. For the pole with corner frequency at *ω* = 5, the slope of the magnitude plot is −40 dB/decade and that of the phase plot is −90° per decade due to the power of 2. The
-
-Obtain the Bode plots for Example 14.4
-
-magnitude and the phase plots for the individual terms (in dotted lines) and the entire **H**( *jω*) (in solid lines) are in Fig. 14.15.
-
-# **Figure 14.15**
-
-Bode plots for Example 14.4: (a) magnitude plot, (b) phase plot.
-
-# **Figure 14.16**
-
-For Practice Prob. 14.4: (a) magnitude plot, (b) phase plot.
-
-Example 14.5 Draw the Bode plots for
-
-$$
-H(s) = \frac{s+1}{s^2 + 12s + 100}
-$$
-
-# **Solution:**
-
-- 1. **Define.** The problem is clearly stated and we follo w the technique outlined in the chapter.
-- 2. **Present.** We are to develop the approximate Bode plot for the given function, **H**(*s*).
-- 3. **Alternative.** The two most effective choices would be the approximation technique outlined in the chapter , which we will use here, and *MATLAB*, which can actually give us the exact Bode plots.
-
-# 4. **Attempt.** We express **H**(*s*) as
-
-express **H**(*s*) as
-$$
-\mathbf{H}(\omega) = \frac{1/100(1 + j\omega)}{1 + j\omega 1.2/10 + (j\omega/10)^2}
-$$
-
- For the quadratic pole, *ωn* = 10 rad/s, which serves as the corner frequency. The magnitude and phase are
-
-$$
-H_{\text{dB}} = -20 \log_{10} 100 + 20 \log_{10} |1 + j\omega|
-$$
-$$
-- 20 \log_{10} \left| 1 + \frac{j\omega 1.2}{10} - \frac{\omega^2}{100} \right|
-$$
-$$
-\phi = 0^\circ + \tan^{-1} \omega - \tan^{-1} \left[ \frac{\omega 1.2/10}{1 - \omega^2/100} \right]
-$$
-
- Figure 14.17 shows the Bode plots. Notice that the quadratic pole is treated as a repeated pole at *ωk*, that is, (1 + *jω*∕*ωk*) 2 , which is an approximation.
-
-**Figure 14.17** Bode plots for Example 14.5: (a) magnitude plot, (b) phase plot.
-
-5. **Evaluate.** Although we could use *MATLAB* to validate the solution, we will use a more straightforward approach. First, we must realize that the denominator assumes that *ζ* = 0 for the approximation, so we will use the following equation to check our answer:
-
-$$
-\mathbf{H}(s) \simeq \frac{s+1}{s^2 + 10^2}
-$$
-
- We also note that we need to actually solve for *H*dB and the corresponding phase angle *ϕ*. First, let *ω* = 0.
-
-$$
-H_{\text{dB}} = 20 \log_{10}(1/100) = -40
-$$
- and $\phi = 0^{\circ}$
-
-Now try *ω* = 1.
-
-$$
-H_{\rm dB} = 20 \log_{10}(1.4142/99) = -36.9 \text{ dB}
-$$
-
-which is the expected 3 dB up from the corner frequency.
-
-$$
-\phi = 45^{\circ}
-$$
- from $\mathbf{H}(j) = \frac{j+1}{-1+100}$
-
-Now try *ω* = 100.
-
-*H*dB = 20 log10 (100) − 20 log10 (9900) = 39.91 dB
-
-*ϕ* is 90° from the numerator minus 180°, which gives −90°. We now have checked three different points and got close agreement, and, because this is an approximation, we can feel confident that we have worked the problem successfully.
-
- You can reasonably ask why did we not check at *ω* = 10? If we just use the approximate value we used above, we end up with an infinite value, which is to be expected from *ζ* = 0 (see Fig. 14.12a). If we used the actual value of **H**( *j*10) we will still end up being far from the approximate values, since *ζ* = 0.6 and Fig. 14.12a shows a significant deviation from the approximation. We could have reworked the problem with *ζ* = 0.707, which would have gotten us closer to the approximation. However, we really have enough points without doing this.
-
-6. **Satisfactory?** We are satisfied the problem has been worked successfully and we can present the results as a solution to the problem.
-
-Practice Problem 14.5 Construct the Bode plots for
-
-**Answer:** See Fig. 14.18.
-
-$$
-H(s) = \frac{10}{s(s^2 + 80s + 400)}
-$$
-
-For Practice Prob. 14.5: (a) magnitude plot, (b) phase plot.
-
-Example 14.6 Given the Bode plot in Fig. 14.19, obtain the transfer function **H**(*ω*).
-
-# **Solution:**
-
-To obtain **H**(*ω*) from the Bode plot, we keep in mind that a zero al ways causes an upward turn at a corner frequency, while a pole causes
-
-40 dB
-
-H
-
-0
-
-**Figure 14.19** For Example 14.6.
-
-a downward turn. We notice from Fig. 14.19 that there is a zero *jω* at the origin, which should have intersected the frequency axis at *ω* = 1. This is indicated by the straight line with slope +20 dB/decade. The fact that this straight line is shifted by 40 dB indicates that there is a 40-dB gain; that is,
-
-$$
-40 = 20 \log_{10} K \qquad \Rightarrow \qquad \log_{10} K = 2
-$$
-
-or
-
-$$
-K = 10^2 = 100
-$$
-
-In addition to the zero *jω* at the origin, we notice that there are three factors with corner frequencies at *ω* = 1, 5, and 20 rad/s. Thus, we have:
-
-- 1. A pole at *p* = 1 with slope −20 dB/decade to cause a downward turn and counteract the zero at the origin. The pole at *p* = 1 is determined as 1∕(1 + *jω*∕1).
-- 2. Another pole at *p* = 5 with slope −20 dB/decade causing a do wnward turn. The pole is 1∕(1 + *jω*∕5).
-- 3. A third pole at *p* = 20 with slope −20 dB/decade causing a further downward turn. The pole is 1∕(1 + *jω*∕20).
-
-Putting all these together gives the corresponding transfer function as
-
-H turn. The pole is
-$$
-1/(1 + j\omega/20)
-$$
-.
-\nthese together gives the corresponding transfer in
-\n
-$$
-\mathbf{H}(\omega) = \frac{100 j\omega}{(1 + j\omega/1)(1 + j\omega/5)(1 + j\omega/20)}
-$$
-\n
-$$
-= \frac{j\omega 10^4}{(j\omega + 1)(j\omega + 5)(j\omega + 20)}
-$$
-
-or
-
-$$
-(j\omega + 1)(j\omega + 3)(j\omega + 20)
-$$
-
-$$
-\mathbf{H}(s) = \frac{10^4 s}{(s+1)(s+5)(s+20)}, \qquad s = j\omega
-$$
-
-Obtain the transfer function H( *ω*) corresponding to the Bode plot in Practice Problem 14.6 Fig. 14.20.
-
-**Answer: H**(*ω*) = 2,000,000(*s* + 5) \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*s* + 10)(*s* + 100)2 .
-
-To see how to use MATLAB to produce Bode plots, refer to Section 14.11.
-
-0.1 1 5 10 20 100
-
-‒40 dB/decade
-
-+20 dB/decade
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/163_14.5 Series Resonance.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/163_14.5 Series Resonance.md
deleted file mode 100644
index 7a6839a484e51825358442b0b76b77b1e360d9c4..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/163_14.5 Series Resonance.md
+++ /dev/null
@@ -1,337 +0,0 @@
-# **14.5** Series Resonance
-
-The most prominent feature of the frequenc y response of a circuit may be the sharp peak (or *resonant peak* ) exhibited in its amplitude char acteristic. The concept of resonance applies in se veral areas of science and engineering. Resonance occurs in an y system that has a comple x conjugate pair of poles; it is the cause of oscillations of stored ener gy from one form to another . It is the phenomenon that allo ws frequency
-
-*ω*
-
-‒20 dB/decade
-
-discrimination in communications netw orks. Resonance occurs in an y circuit that has at least one inductor and one capacitor.
-
-Resonance is a condition in an RLC circuit in which the capacitive and inductive reactances are equal in magnitude, thereby resulting in a purely resistive impedance.
-
-Resonant circuits (series or parallel) are useful for constructing filters, as their transfer functions can be highly frequency selective. They are used in many applications such as selecting the desired stations in radio and TV receivers.
-
-Consider the series *RLC* circuit shown in Fig. 14.21 in the frequency domain. The input impedance is
-
-$$
-\mathbf{Z} = \mathbf{H}(\omega) = \frac{\mathbf{V}_s}{\mathbf{I}} = R + j\omega L + \frac{1}{j\omega C}
-$$
- (14.22)
-
-or
-
-$$
-\mathbf{Z} = R + j \left( \omega L - \frac{1}{\omega C} \right) \tag{14.23}
-$$
-
-Resonance results when the imaginary part of the transfer function is zero, or
-
-$$
-\operatorname{Im}(\mathbf{Z}) = \omega L - \frac{1}{\omega C} = 0 \tag{14.24}
-$$
-
-The value of *ω* that satisfies this condition is called the *resonant frequency ω*0. Thus, the resonance condition is
-
-$$
-\omega_0 L = \frac{1}{\omega_0 C} \tag{14.25}
-$$
-
-or
-
-$$
-\omega_0 = \frac{1}{\sqrt{LC}} \text{rad/s} \tag{14.26}
-$$
-
-Since *ω*0 = 2 *π f*0,
-
-$$
-f_0 = \frac{1}{2\pi\sqrt{LC}} \text{Hz}
-$$
- (14.27)
-
-Note that at resonance:
-
-- 1. The impedance is purely resistive, thus, **Z** = *R*. In other words, the *LC* series combination acts like a short circuit, and the entire voltage is across *R*.
-- 2. The voltage **V***s* and the current **I** are in phase, so that the po wer factor is unity.
-- 3. The magnitude of the transfer function **H**(*ω*) = **Z**(*ω*) is minimum.
-- 4. The inductor v oltage and capacitor v oltage can be much more than the source voltage.
-
-The frequency response of the circuit's current magnitude
-
-response of the circuit's current magnitude
-\n
-$$
-I = |\mathbf{I}| = \frac{V_m}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}}
-$$
-\n(14.28)
-
-$$
-|\mathbf{V}_L| = \frac{V_m}{R} \omega_0 L = Q V_m
-$$
-$$
-|\mathbf{V}_C| = \frac{V_m}{R} \frac{1}{\omega_0 C} = Q V_m
-$$
-
-where Q is the quality factor, defined in Eq. (14.38).
-
-The series resonant circuit.
-
-is shown in Fig. 14.22; the plot only sho ws the symmetry illustrated in this graph when the frequenc y axis is a log arithm. The average power dissipated by the *RLC* circuit is
-
-$$
-P(\omega) = \frac{1}{2} \hat{I}^2 R \tag{14.29}
-$$
-
-The highest po wer dissipated occurs at resonance, when *I* = *Vm*∕*R*, so that
-
-$$
-P(\omega_0) = \frac{1}{2} \frac{V_m^2}{R}
-$$
- (14.30)
-
-At certain frequencies *ω* = *ω*1, *ω*2, the dissipated power is half the maximum value; that is,
-
-$$
-P(\omega_1) = P(\omega_2) = \frac{(V_m/\sqrt{2})^2}{2R} = \frac{V_m^2}{4R}
-$$
- (14.31)
-
-Hence, *ω*1 and *ω*2 are called the *half-power frequencies.*
-
-The half-power frequencies are obtained by setting *Z* equal to √ \_\_ 2 *R*, and writing
-
-equences are obtained by setting Z equal to V2R,
-\n
-$$
-\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} = \sqrt{2}R
-$$
-\n(14.32)
-
-Solving for *ω*, we obtain
-
-$$
-\omega_1 = -\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}}
-$$
-\n
-$$
-\omega_2 = \frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}}
-$$
-\n(14.33)
-
-We can relate the half-po wer frequencies with the resonant frequenc y. From Eqs. (14.26) and (14.33),
-
-$$
-\omega_0 = \sqrt{\omega_1 \omega_2} \tag{14.34}
-$$
-
-showing that the resonant frequenc y is the geometric mean of the halfpower frequencies. Notice that *ω*1 and *ω*2 are in general not symmetrical around the resonant frequency *ω*0, because the frequency response is not generally symmetrical. However, as will be explained shortly, symmetry of the half-power frequencies around the resonant frequenc y is often a reasonable approximation.
-
-Although the height of the curv e in Fig. 14.22 is determined by *R*, the width of the curv e depends on other f actors. The width of the re sponse curve depends on the *bandwidth B*, which is defined as the difference between the two half-power frequencies,
-
-$$
-B = \omega_2 - \omega_1 \tag{14.35}
-$$
-
-This definition of bandwidth is just one of several that are commonly used. Strictly speaking, *B* in Eq. (14.35) is a half-po wer bandwidth, because it is the width of the frequenc y band between the half-po wer frequencies.
-
-The "sharpness" of the resonance in a resonant circuit is measured quantitatively by the *quality factor Q*. At resonance, the reactive energy
-
-# **Figure 14.22**
-
-The current amplitude versus frequency for the series resonant circuit of Fig. 14.21.
-
-Although the same symbol Q is used for the reactive power, the two are not equal and should not be confused. Q here is dimensionless, whereas reactive power Q is in VAR. This may help distinguish between the two.
-
-in the circuit oscillates between the inductor and the capacitor. The quality factor relates the maximum or peak ener gy stored to the energy dissipated in the circuit per cycle of oscillation:
-
-is the maximum or peak energy stored to the energy dis-
-rcuit per cycle of oscillation:
-
-\n
-$$
-Q = 2\pi \frac{\text{Peak energy stored in the circuit}}{\text{Energy dissipated by the circuit}} \qquad (14.36)
-$$
-\nin one period at resonance
-
-It is also regarded as a measure of the energy storage property of a circuit in relation to its energy dissipation property. In the series *RLC* circuit, the peak energy stored is \_\_1 2 *LI*2 , while the energy dissipated in one period is \_\_1 2 (*I* 2 *R*)(1∕*f*0). Hence,
-
-$$
-Q = 2\pi \frac{\frac{1}{2}LI^2}{\frac{1}{2}I^2R(1/f_0)} = \frac{2\pi f_0L}{R}
-$$
- (14.37)
-
-B3 Q3 (greatest selectivity) Q2 (medium selectivity) Q1 (least selectivity) B2 B1 *ω*
-
-# Amplitude
-
-The higher the circuit *Q*, the smaller the bandwidth.
-
-The quality factor is a measure of the selectivity (or "sharpness" of resonance) of the circuit.
-
-or
-
-$$
-Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR}
-$$
- (14.38)
-
-Notice that the quality factor is dimensionless. The relationship between the bandwidth *B* and the quality f actor *Q* is obtained by substituting Eq. (14.33) into Eq. (14.35) and utilizing Eq. (14.38).
-
-$$
-B = \frac{R}{L} = \frac{\omega_0}{Q} \tag{14.39}
-$$
-
-or *B* = *ω*0 2 *CR*. Thus,
-
-> The quality factor of a resonant circuit is the ratio of its resonant frequency to its bandwidth.
-
-Keep in mind that Eqs. (14.33), (14.38), and (14.39) only apply to a series *RLC* circuit.
-
-As illustrated in Fig. 14.23, the higher the v alue of *Q*, the more selective the circuit is but the smaller the bandwidth. The *selectivity* of an *RLC* circuit is the ability of the circuit to respond to a certain frequenc y and discriminate against all other frequencies. If the band of frequencies to be selected or rejected is narrow, the quality f actor of the resonant circuit must be high. If the band of frequencies is wide, the quality factor must be low.
-
-A resonant circuit is designed to operate at or near its resonant fre quency. It is said to be a *high-Q circuit* when its quality factor is equal to or greater than 10. F or high *-Q* circuits (*Q* ≥ 10), the half- power frequencies are, for all practical purposes, symmetrical around the resonant frequency and can be approximated as
-
-$$
-\omega_1 \simeq \omega_0 - \frac{B}{2}, \qquad \omega_2 \simeq \omega_0 + \frac{B}{2} \qquad (14.40)
-$$
-
-High-*Q* circuits are used often in communications networks.
-
-We see that a resonant circuit is characterized by five related parameters: the two half-power frequencies *ω*1 and *ω*2, the resonant frequency *ω*0, the bandwidth *B*, and the quality factor *Q*.
-
-In the circuit of Fig. 14.24, *R* = 2 Ω, *L* = 1 mH, and *C* = 0.4 *μ*F. (a) Find Example 14.7 the resonant frequency and the half-power frequencies. (b) Calculate the quality factor and bandwidth. (c) Determine the amplitude of the current at *ω*0, *ω*1, and *ω*2.
-
-# **Solution:**
-
-(a) The resonant frequency is
-
-ant frequency is
-\n
-$$
-\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-3} \times 0.4 \times 10^{-6}}} = 50 \text{ krad/s}
-$$
-
-■ **METHOD 1** The lower half-power frequency is
-
-$$
-\omega_1 = -\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}}
-$$
-
-= $-\frac{2}{2 \times 10^{-3}} + \sqrt{(10^3)^2 + (50 \times 10^3)^2}$
-= $-1 + \sqrt{1 + 2500}$ krad/s = 49 krad/s
-
-Similarly, the upper half-power frequency is
-
-$$
-\omega_2 = 1 + \sqrt{1 + 2500} \text{ krad/s} = 51 \text{ krad/s}
-$$
-
-(b) The bandwidth is
-
-$$
-B = \omega_2 - \omega_1 = 2 \text{ krad/s}
-$$
-
-or
-
-$$
-B = \frac{R}{L} = \frac{2}{10^{-3}} = 2 \text{ krad/s}
-$$
-
-The quality factor is
-
-$$
-Q = \frac{\omega_0}{B} = \frac{50}{2} = 25
-$$
-
-■ **METHOD 2** Alternatively, we could find
-
-Alternatively, we could find
-\n
-$$
-Q = \frac{\omega_0 L}{R} = \frac{50 \times 10^3 \times 10^{-3}}{2} = 25
-$$
-
-From *Q*, we find
-
-$$
-B = \frac{\omega_0}{Q} = \frac{50 \times 10^3}{25} = 2 \text{ krad/s}
-$$
-
-Since *Q* > 10, this is a high-*Q* circuit and we can obtain the half-power frequencies as
-
-$$
-\omega_1 = \omega_0 - \frac{B}{2} = 50 - 1 = 49 \text{ krad/s}
-$$
-$$
-\omega_2 = \omega_0 + \frac{B}{2} = 50 + 1 = 51 \text{ krad/s}
-$$
-
-as obtained earlier.
-
-(c) At *ω* = *ω*0,
-
-$$
-I = \frac{V_m}{R} = \frac{20}{2} = 10 \text{ A}
-$$
-
-At *ω* = *ω*1, *ω*2,
-
-$$
-I = \frac{V_m}{\sqrt{2} R} = \frac{10}{\sqrt{2}} = 7.071 \text{ A}
-$$
-
-Practice Problem 14.7 A series-connected circuit has *R* = 4 Ω and *L* = 25 mH. (a) Calculate the value of *C* that will produce a quality factor of 50. (b) Find *ω*1, *ω*2, and *B*. (c) Determine the average power dissipated at *ω* = *ω*0, *ω*1, *ω*2. Take *Vm* = 100 V.
-
-> **Answer:** (a) 0.625 *μ*F, (b) 7920 rad/s, 8080 rad/s, 160 rad/s, (c) 1.25 kW , 0.625 kW, 0.625 kW.
-
-# **14.6** Parallel Resonance
-
-The parallel *RLC* circuit in Fig. 14.25 is the dual of the series *RLC* circuit. So we will avoid needless repetition. The admittance is
-
-$$
-Y = H(\omega) = \frac{I}{V} = \frac{1}{R} + j\omega C + \frac{1}{j\omega L}
-$$
- (14.41)
-
-or
-
-or
-
-$$
-\mathbf{Y} = \frac{1}{R} + j \left( \omega C - \frac{1}{\omega L} \right) \tag{14.42}
-$$
-
-Resonance occurs when the imaginary part of **Y** is zero,
-
-$$
-\omega C - \frac{1}{\omega L} = 0 \tag{14.43}
-$$
-
-$$
-\omega_0 = \frac{1}{\sqrt{LC}} \text{ rad/s}
-$$
- (14.44)
-
-which is the same as Eq. (14.26) for the series resonant circuit. The voltage ∣**V**∣ is sketched in Fig. 14.26 as a function of frequenc y. Notice that at resonance, the parallel *LC* combination acts like an open circuit, so that the entire current flows through *R*. Also, the inductor and capacitor current can be much more than the source current at resonance.
-
-We exploit the duality between Figs. 14.21 and 14.25 by comparing Eq. (14.42) with Eq. (14.23). By replacing *R*, *L*, and *C* in the expressions
-
-**Figure 14.25**
-
-The parallel resonant circuit.
-
-The current amplitude versus frequency for the series resonant circuit of Fig. 14.25.
-
-We can see this from the fact that
-
-$$
-|\mathbf{I}_L| = \frac{I_m R}{\omega_0 L} = Q I_m
-$$
-
-$$
-|\mathbf{I}_C| = \omega_0 C I_m R = Q I_m
-$$
-
-where Q is the quality factor, defined in Eq. (14.47).
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/164_14.6 Parallel Resonance.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/164_14.6 Parallel Resonance.md
deleted file mode 100644
index 92771020349242b359286ccde3ddf1c7bffeab29..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/164_14.6 Parallel Resonance.md
+++ /dev/null
@@ -1,123 +0,0 @@
-
-for the series circuit with 1 ∕*R*, *C*, and *L* respectively, we obtain for the parallel circuit
-
-$$
-\omega_1 = -\frac{1}{2RC} + \sqrt{\left(\frac{1}{2RC}\right)^2 + \frac{1}{LC}}
-$$
-\n
-$$
-\omega_2 = \frac{1}{2RC} + \sqrt{\left(\frac{1}{2RC}\right)^2 + \frac{1}{LC}}
-$$
-\n(14.45)
-
-$$
-B = \omega_2 - \omega_1 = \frac{1}{RC}
-$$
- (14.46)
-
-$$
-Q = \frac{\omega_0}{B} = \omega_0 RC = \frac{R}{\omega_0 L}
-$$
- (14.47)
-
-It should be noted that Eqs. (14.45) to (14.47) apply only to a parallel *RLC* circuit. Using Eqs. (14.45) and (14.47), we can e xpress the halfpower frequencies in terms of the quality factor. The result is
-
-$$
-\omega_1 = \omega_0 \sqrt{1 + \left(\frac{1}{2Q}\right)^2} - \frac{\omega_0}{2Q}, \qquad \omega_2 = \omega_0 \sqrt{1 + \left(\frac{1}{2Q}\right)^2} + \frac{\omega_0}{2Q}
-$$
-\n(14.48)
-
-Again, for high-*Q* circuits (*Q* ≥ 10)
-
-$$
-\omega_1 \simeq \omega_0 - \frac{B}{2}, \qquad \omega_2 \simeq \omega_0 + \frac{B}{2}
-$$
- (14.49)
-
-Table 14.4 presents a summary of the characteristics of the series and parallel resonant circuits. Besides the series and parallel *RLC* considered here, other resonant circuits exist. Example 14.9 treats a typical example.
-
-## **TABLE 14.4**
-
-Summary of the characteristics of resonant RLC circuits.
-
-| Characteristic | Series circuit | Parallel circuit |
-|--------------------------------|--------------------------------------------------------|--------------------------------------------------------|
-| Resonant frequency, ω0 | ____ 1
___
√
LC | ____ 1
___
√
LC |
-| Quality factor, Q | ω0L ____
or _____ 1
ω0 RC
R | ____ R
or ω0RC
ω0 L |
-| Bandwidth, B | ω0
___
Q
__________ | ω0
___
Q
__________ |
-| Half-power frequencies, ω1, ω2 | ω0
___1
2
± ___
1 + (
ω0 √
2Q)
2Q | ω0
___1
2
± ___
1 + (
ω0 √
2Q)
2Q |
-| For Q ≥ 10, ω1, ω2 | B
± __
ω0
2 | B
± __
ω0
2 |
-
-For Example 14.8.
-
-Example 14.8 In the parallel *RLC* circuit of Fig. 14.27, let *R* = 8 kΩ, *L* = 0.2 mH, and *C* = 8 *μ*F. (a) Calculate *ω*0, *Q*, and *B*. (b) Find *ω*1 and *ω*2. (c) Determine the power dissipated at *ω*0, *ω*1, and *ω*2.
-
-# **Solution:**
-
-(a)
-
-$$
-\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.2 \times 10^{-3} \times 8 \times 10^{-6}}} = \frac{10^5}{4} = 25 \text{ krad/s}
-$$
-$$
-Q = \frac{R}{\omega_0 L} = \frac{8 \times 10^3}{25 \times 10^3 \times 0.2 \times 10^{-3}} = 1,600
-$$
-$$
-B = \frac{\omega_0}{Q} = 15.625 \text{ rad/s}
-$$
-
-(b) Due to the high value of *Q*, we can regard this as a high- *Q* circuit, Hence,
-
-$$
-\omega_1 = \omega_0 - \frac{B}{2} = 25,000 - 7.812 = 24,992 \text{ rad/s}
-$$
-$$
-\omega_2 = \omega_0 + \frac{B}{2} = 25,000 + 7.812 = 25,008 \text{ rad/s}
-$$
-
-(c) At *ω* = *ω*0, **Y** = 1∕*R* or **Z** = *R* = 8 kΩ. Then
-
-$$
-I_o = \frac{V}{Z} = \frac{10/-90^{\circ}}{8,000} = 1.25/-90^{\circ} \text{ mA}
-$$
-
-Since the entire current flows through *R* at resonance, the average power dissipated at *ω* = *ω*0 is
-
-$$
-P = \frac{1}{2} |\mathbf{I}_o|^2 R = \frac{1}{2} (1.25 \times 10^{-3})^2 (8 \times 10^3) = 6.25 \text{ mW}
-$$
-
-or
-
-$$
-P = \frac{V_m^2}{2R} = \frac{100}{2 \times 8 \times 10^3} = 6.25
-$$
- mW
-
-At *ω* = *ω*1, *ω*2,
-
-$$
-P = \frac{V_m^2}{4R} = 3.125 \text{ mW}
-$$
-
-Practice Problem 14.8 A parallel resonant circuit has *R* = 100 kΩ, *L* = 50 mH, and *C* = 2 nF. Calculate *ω*0, *ω*1, *ω*2, *Q*, and *B*.
-
-**Answer:** 100 krad/s, 97.5 krad/s,102.5 krad/s, 20, 5 krad/s.
-
-# **Solution:**
-
-The input admittance is
-
-$$
-\mathbf{Y} = j\omega 0.1 + \frac{1}{10} + \frac{1}{2 + j\omega 2} = 0.1 + j\omega 0.1 + \frac{2 - j\omega 2}{4 + 4\omega^2}
-$$
-
-At resonance, Im(**Y**) = 0 and
-
-$$
-\omega_0 0.1 - \frac{2\omega_0}{4 + 4\omega_0^2} = 0 \qquad \Rightarrow \qquad \omega_0 = 2 \text{ rad/s}
-$$
-
-Calculate the resonant frequency of the circuit in Fig. 14.29. Practice Problem 14.9
-
-**Answer:** 173.21 rad/s.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/165_14.7 Passive Filters.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/165_14.7 Passive Filters.md
deleted file mode 100644
index 04afcedd2da840b6088ff415b4cca706869106e7..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/165_14.7 Passive Filters.md
+++ /dev/null
@@ -1,288 +0,0 @@
-# **Figure 14.29 14.7** Passive Filters For Practice Prob. 14.9
-
-The concept of filters has been an integral part of the evolution of electrical engineering from the beginning. Several technological achievements would not have been possible without electrical filters. Because of this prominent role of filters, much effort has been expended on the theory, design, and construction of filters and many articles and books have been written on them. Our discussion in this chapter should be considered introductory.
-
-A filter is a circuit that is designed to pass signals with desired frequencies and reject or attenuate others.
-
-As a frequency-selective device, a filter can be used to limit the frequency spectrum of a signal to some specified band of frequencies. Filters are the circuits used in radio and TV receivers to allow us to select one desired signal out of a multitude of broadcast signals in the environment.
-
-A filter is a *passive filter* if it consists of only passive elements *R*, *L*, and *C*. It is said to be an *active filter* if it consists of acti ve elements (such as transistors and op amps) in addition to passi ve elements *R*, *L*, and *C*. We consider passive filters in this section and active filters in the next section. *LC* filters have been used in practical applications for more than eight decades. *LC* filter technology feeds related areas such as equalizers, impedance-matching networks, transformers, shaping networks, power dividers, attenuators, and directional couplers, and is continuously providing practicing engineers with opportunities to inno vate and experiment. Besides the *LC* filters we study in these sections, there are other kinds of filters—such as digital filters, electromechanical filters, and microwave filters—which are beyond the level of this text.
-
-0 (b) *ω*c *ω* 1 0 (a) *ω*c *ω* 1 0 (c) *ω*1 *ω*2 *ω* 1 0 Passband Passband Passband Stopband Stopband Stopband Passband Passband Stopband Stopband *ω*1 *ω*2 *ω* 1 │H(*ω*)│ │H(*ω*)│ │H(*ω*)│ │H(*ω*)│
-
-# **Figure 14.30**
-
-Ideal frequency response of four types of filters: (a) low-pass filter, (b) high-pass filter, (c) band-pass filter, (d) band-stop filter.
-
-(d)
-
-**Figure 14.31** A low-pass filter.
-
-**Figure 14.32** Ideal and actual frequency response of a low-pass filter.
-
-As shown in Fig. 14.30, there are four types of filters whether passive or active:
-
-- 1. A *low-pass filter* passes low frequencies and stops high frequencies, as shown ideally in Fig. 14.30(a).
-- 2. A *high-pass filter* passes high frequencies and rejects low frequencies, as shown ideally in Fig. 14.30(b).
-- 3. A *band-pass filter* passes frequencies within a frequenc y band and blocks or attenuates frequencies outside the band, as sho wn ideally in Fig. 14.30(c).
-- 4. A *band-stop filter* passes frequencies outside a frequenc y band and blocks or attenuates frequencies within the band, as sho wn ideally in Fig. 14.30(d).
-
-Table 14.5 presents a summary of the characteristics of these filters. Be aware that the characteristics in Table 14.5 are only valid for first- or second-order filters—but one should not have the impression that only these kinds of filter exist. We now consider typical circuits for realizing the filters shown in Table 14.5.
-
-# **TABLE 14.5**
-
-Summary of the characteristics of ideal filters.
-
-| Type of Filter | H(0) | H(∞) | H(ωc) or H(ω0) |
-|----------------|------|------|-----------------------|
-| Low-pass | 1 | 0 | __
1∕ √
2
__ |
-| High-pass | 0 | 1 | 1∕ √
2 |
-| Band-pass | 0 | 0 | 1 |
-| Band-stop | 1 | 1 | 0 |
-
-*ωc* is the cutoff frequency for low-pass and high-pass filters; *ω*0 is the center frequency for band-pass and band-stop filters.
-
-# **14.7.1** Low-Pass Filter
-
-A typical low-pass filter is formed when the output of an *RC* circuit is taken off the capacitor as shown in Fig. 14.31. The transfer function (see also Example 14.1) is
-
-$$
-\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \frac{1/j\omega C}{R + 1/j\omega C}
-$$
-$$
-\mathbf{H}(\omega) = \frac{1}{1 + j\omega RC}
-$$
-(14.50)
-
-Note that **H**(0) = 1, **H**(∞) = 0. Figure 14.32 shows the plot of ∣*H*(*ω*)∣, along with the ideal characteristic. The half-power frequency, which is equivalent to the corner frequency on the Bode plots but in the context of filters is usually known as the *cutoff frequency ωc*, is obtained by setting the magnitude of **H**(*ω*) equal to 1∕ √ \_\_ 2 , thus,
-
-$$
-H(\omega_c) = \frac{1}{\sqrt{1 + \omega_c^2 R^2 C^2}} = \frac{1}{\sqrt{2}}
-$$
-
-or
-
-$$
-\omega_c = \frac{1}{RC} \tag{14.51}
-$$
-
-The cutoff frequency is also called the *rolloff frequency*.
-
-A low-pass filter is designed to pass only frequencies from dc up to the cutoff frequency *ω*c.
-
-A low-pass filter can also be formed when the output of an *RL* circuit is taken off the resistor. Of course, there are many other circuits for low-pass filters.
-
-# **14.7.2** High-Pass Filter
-
-A high-pass filter is formed when the output of an *RC* circuit is taken off the resistor as shown in Fig. 14.33. The transfer function is
-
-$$
-\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \frac{R}{R + 1/j\omega C}
-$$
-$$
-\mathbf{H}(\omega) = \frac{j\omega RC}{1 + j\omega RC}
-$$
-(14.52)
-
-Note that **H**(0) = 0, **H**(∞) = 1. Figure 14.34 shows the plot of ∣*H*(*ω*)∣. Again, the corner or cutoff frequency is
-
-$$
-\omega_c = \frac{1}{RC} \tag{14.53}
-$$
-
-A high-pass filter is designed to pass all frequencies above its cutoff frequency *ω*c.
-
-A high-pass filter can also be formed when the output of an *RL* circuit is taken off the inductor.
-
-# **14.7.3** Band-Pass Filter
-
-The *RLC* series resonant circuit provides a band-pass filter when the output is taken off the resistor as shown in Fig. 14.35. The transfer function is
-
-$$
-\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \frac{R}{R + j(\omega L - 1/\omega C)}
-$$
-(14.54)
-
-We observe that **H**(0) = 0, **H**(∞) = 0. Figure 14.36 shows the plot of ∣*H*(*ω*)∣. The band-pass filter passes a band of frequencies (*ω*1 < *ω*< *ω*2) centered on *ω*0, the center frequency, which is given by
-
-$$
-\omega_0 = \frac{1}{\sqrt{LC}}\tag{14.55}
-$$
-
-A band-pass filter is designed to pass all frequencies within a band of frequencies, *ω*1 < *ω*< *ω*2.
-
-Because the band-pass filter in Fig. 14.35 is a series resonant circuit, the half-power frequencies, the bandwidth, and the quality factor are determined as in Section 14.5. A band-pass filter can also be formed by cascading the low-pass filter (where *ω*2 = *ωc*) in Fig. 14.31 with the The cutoff frequency is the frequency at which the transfer function **H** drops in magnitude to 70.71% of its maximum value. It is also regarded as the frequency at which the power dissipated in a circuit is half of its maximum value.
-
-# **Figure 14.34**
-
-Ideal and actual frequency response of a high-pass filter.
-
-**Figure 14.35**
-
-# **Figure 14.36** Ideal and actual frequency response of a band-pass filter.
-
-high-pass filter (where *ω*1 = *ωc*) in Fig. 14.33. However, the result would not be the same as just adding the output of the low-pass filter to the input of the high-pass filter, because one circuit loads the other and alters the desired transfer function.
-
-# **14.7.4** Band-Stop Filter
-
-A filter that prevents a band of frequencies between two designated values (*ω*1 and *ω*2) from passing is variably known as a *band-stop, bandreject*, or *notch* filter. A band-stop filter is formed when the output *RLC* series resonant circuit is taken off the *LC* series combination as shown in Fig. 14.37. The transfer function is
-
-nster function is
-\n
-$$
-\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \frac{j(\omega L - 1/\omega C)}{R + j(\omega L - 1/\omega C)}
-$$
-\n(14.56)
-
-Notice that **H**(0) = 1, **H**(∞) = 1. Figure 14.38 shows the plot of ∣*H*(*ω*)∣. Again, the center frequency is given by
-
-$$
-\omega_0 = \frac{1}{\sqrt{LC}}\tag{14.57}
-$$
-
-while the half-power frequencies, the bandwidth, and the quality factor are calculated using the formulas in Section 14.5 for a series reso nant circuit. Here, *ω*0 is called the *frequency of rejection*, while the corresponding bandwidth (*B* = *ω*2 − *ω*1) is known as the *bandwidth of rejection*. Thus,
-
-A band-stop filter is designed to stop or eliminate all frequencies within a band of frequencies, *ω*1 < *ω*< *ω*2.
-
-Notice that adding the transfer functions of the band-pass and the band-stop gives unity at any frequency for the same values of *R, L,* and *C*. Of course, this is not true in general b ut true for the circuits treated here. This is due to the fact that the characteristic of one is the inverse of the other.
-
-In concluding this section, we should note that:
-
-- 1. From Eqs. (14.50), (14.52), (14.54), and (14.56), the maximum gain of a passive filter is unity. To generate a gain greater than unity, one should use an active filter as the next section shows.
-- 2. There are other ways to get the types of filters treated in this section.
-- 3. The filters treated here are the simple types. Many other filters have sharper and complex frequency responses.
-
-Example 14.10 Determine what type of filter is shown in Fig. 14.39. Calculate the corner or cutoff frequency. Take *R* = 2 kΩ, *L* = 2 H, and *C* = 2 *μ*F.
-
-# **Solution:**
-
-The transfer function is
-
-ction is
-\n
-$$
-\mathbf{H}(s) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \frac{R||1/sC}{sL + R||1/sC}, \qquad s = j\omega
-$$
-\n(14.10.1)
-
-**Figure 14.37** A band-stop filter.
-
-band-stop filter.
-
-Ideal and actual frequency response of a
-
-But
-
-$$
-R\left\|\frac{1}{sC} = \frac{R/sC}{R+1/sC} = \frac{R}{1+sRC}
-$$
-
-Substituting this into Eq. (14.10.1) gives
-
-uting this into Eq. (14.10.1) gives
-\n
-$$
-\mathbf{H}(s) = \frac{R/(1+sRC)}{sL + R/(1+sRC)} = \frac{R}{s^2RLC + sL + R}, \qquad s = j\omega
-$$
-
-or
-
-$$
-R/(1 + sRC) \t s2RLC + sL + R'
-$$
-
-$$
-H(\omega) = \frac{R}{-\omega^2 RLC + j\omega L + R}
-$$
- (14.10.2)
-
-Because **H**(0) = 1 and **H**(∞) = 0, we conclude from Table 14.5 that the circuit in Fig. 14.39 is a second-order low-pass filter. The magnitude of **H** is *H* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ *R √*
-
-a second-order low-pass filter. The magnitude of
-$$
-H = \frac{R}{\sqrt{(R - \omega^2 R LC)^2 + \omega^2 L^2}}
-$$
-(14.10.3)
-
-The corner frequency is the same as the half-power frequency, that is, where **H** is reduced by a factor of 1 ∕ *√ \_\_* 2 . Because the dc value of *H*(*ω*) is 1, at the corner frequency, Eq. (14.10.3) becomes after squaring
-
-where **H** is reduced by a factor of 1 /
-$$
-\sqrt{2}
-$$
-. Because the d
-is 1, at the corner frequency, Eq. (14.10.3) becomes after
-$$
-H^2 = \frac{1}{2} = \frac{R^2}{(R - \omega_c^2 R LC)^2 + \omega_c^2 L^2}
-$$
-or
-
-$$
-2 = (1 - \omega_c^2 LC)^2 + \left(\frac{\omega_c L}{R}\right)^2
-$$
-
-Substituting the values of *R*, *L*, and *C*, we obtain
-
-$$
-2 = (1 - \omega_c^2 4 \times 10^{-6})^2 + (\omega_c 10^{-3})^2
-$$
-
-Assuming that *ωc* is in krad/s,
-
-$$
-2 = (1 - 4\omega_c^2)^2 + \omega_c^2 \qquad \text{or} \qquad 16\omega_c^4 - 7\omega_c^2 - 1 = 0
-$$
-
-Solving the quadratic equation in *ωc* 2 , we get *ωc* 2 = 0.5509 and −0.1134. Because *ωc* is real,
-
-$$
-\omega_c = 0.742 \text{ krad/s} = 742 \text{ rad/s}
-$$
-
-For the circuit in Fig. 14.40, obtain the transfer function **V***o*(*ω*)∕**V***i*(*ω*). Practice Problem 14.10 Identify the type of filter the circuit represents and determine the corner frequency. Take *R*1 = 100 Ω = *R*2, *L* = 2 mH.
-
-Answer:
-$$
-\frac{R_2}{R_1 + R_2} \left( \frac{j\omega}{j\omega + \omega_c} \right)
-$$
-, high-pass filter
-$$
-\omega_c = \frac{R_1 R_2}{(R_1 + R_2)L} = 25 \text{ krad/s}.
-$$
-
-For Practice Prob. 14.10.
-
-Example 14.11 If the band-stop filter in Fig. 14.37 is to reject a 200-Hz sinusoid while passing other frequencies, calculate the v alues of *L* and *C*. Take *R* = 150 Ω and the bandwidth as 100 Hz.
-
-# **Solution:**
-
-We use the formulas for a series resonant circuit in Section 14.5.
-
-$$
-B = 2\pi(100) = 200\pi \text{ rad/s}
-$$
-
-But
-
-*B* = \_\_ *R L* ⇒ *L* = \_\_ *R B* = \_\_\_\_\_ 150 200*π* = 0.2387 H
-
-Rejection of the 200-Hz sinusoid means that *f*0 is 200 Hz, so that *ω*0 in Fig. 14.38 is
-
-$$
-\omega_0 = 2\pi f_0 = 2\pi (200) = 400\pi
-$$
-
-Given that *ω*0 = 1∕ √ \_\_\_ *LC* ,
-
-$$
-= 1/\sqrt{LC},
-$$
-
-\n
-$$
-C = \frac{1}{\omega_0^2 L} = \frac{1}{(400\pi)^2 (0.2387)} = 2.653 \pi F
-$$
-
-Practice Problem 14.11 Design a band-pass filter of the form in Fig. 14.35 with a lower cutoff frequency of 20.1 kHz and an upper cutoff frequency of 20.3 kHz. Take *R* = 30 kΩ. Calculate *L*, *C*, and *Q*.
-
-**Answer:** 23.87 H, 2.6 pF, 101.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/166_14.8 Active Filters.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/166_14.8 Active Filters.md
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@@ -1,339 +0,0 @@
-# **14.8** Active Filters
-
-There are three major limitations to the passive filters considered in the previous section. First, they cannot generate gain greater than 1; passive elements cannot add energy to the network. Second, they may require bulky and expensive inductors. Third, they perform poorly at frequencies below the audio frequency range (300 Hz < *f* < 3,000Hz). Nevertheless, passive filters are useful at high frequencies.
-
-Active filters consist of combinations of resistors, capacitors, and op amps. They offer some adv antages over passive *RLC* filters. First, they are often smaller and less expensive, because they do not require inductors. This makes feasible the integrated circuit realizations of filters. Second, they can provide amplifier gain in addition to pro viding the same frequenc y response as *RLC* filters. Third, active filters can be combined with buffer amplifiers (voltage followers) to isolate each stage of the filter from source and load impedance effects. This isolation allows designing the stages independently and then cascading them to realize the desired transfer function. (Bode plots, being log arithmic, may be added when transfer functions are cascaded.) Ho wever, active filters are less reliable and less stable. The practical limit of most active filters is about 100 kHz—most active filters operate well below that frequency.
-
-Filters are often classified according to their order (or number of poles) or their specific design type.
-
-# **14.8.1** First-Order Low-Pass Filter
-
-One type of first-order filter is shown in Fig. 14.41. The components selected for *Zi* and *Zf* determine whether the filter is low-pass or high-pass, but one of the components must be reactive.
-
-Figure 14.42 shows a typical active low-pass filter. For this filter, the transfer function is
-
-$$
-\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = -\frac{\mathbf{Z}_f}{\mathbf{Z}_i}
-$$
- (14.58)
-
-where **Z***i* = *Ri* and
-
-$$
-\mathbf{Z}_f = R_f \left\| \frac{1}{j\omega C_f} = \frac{R_f/j\omega C_f}{R_f + 1/j\omega C_f} = \frac{R_f}{1 + j\omega C_f R_f} \right. \tag{14.59}
-$$
-
-Therefore,
-
-$$
-\mathbf{H}(\omega) = -\frac{R_f}{R_i} \frac{1}{1 + j\omega C_f R_f}
-$$
-(14.60)
-
-We notice that Eq. (14.60) is similar to Eq. (14.50), except that there is a low frequency ( *ω* → 0) gain or dc gain of −*Rf*∕*Ri*. Also, the corner frequency is
-
-$$
-\omega_c = \frac{1}{R_f C_f} \tag{14.61}
-$$
-
-which does not depend on *Ri*. This means that several inputs with dif ferent *Ri* could be summed if required, and the corner frequency would remain the same for each input.
-
-# **14.8.2** First-Order High-Pass Filter
-
-Figure 14.43 shows a typical high-pass filter. As before,
-
-$$
-\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = -\frac{\mathbf{Z}_f}{\mathbf{Z}_i}
-$$
- (14.62)
-
-where **Z***i* = *Ri* + 1∕*jωCi* and **Z***f* = *Rf* so that
-
-$$
-\mathbf{H}(\omega) = -\frac{R_f}{R_i + 1/j\omega C_i} = -\frac{j\omega C_i R_f}{1 + j\omega C_i R_i}
-$$
-(14.63)
-
-This is similar to Eq. (14.52), except that at very high frequencies (*ω* → ∞), the gain tends to −*Rf*∕*Ri*. The corner frequency is
-
-$$
-\omega_c = \frac{1}{R_i C_i} \tag{14.64}
-$$
-
-# **14.8.3** Band-Pass Filter
-
-The circuit in Fig. 14.42 may be combined with that in Fig. 14.43 to form a band-pass filter that will have a gain *K* over the required range of frequencies. By cascading a unity-gain low-pass filter, a unity-gain
-
-Active first-order high-pass filter.
-
-This way of creating a band-pass filter, not necessarily the best, is perhaps the easiest to understand.
-
-**Figure 14.41** A general first-order active filter.
-
-**Figure 14.42** Active first-order low-pass filter.
-
-high-pass filter, and an inverter with gain −*Rf*∕*Ri*, as shown in the block diagram of Fig. 14.44(a), we can construct a band-pass filter whose frequency response is that in Fig. 14.44(b). The actual construction of the band-pass filter is shown in Fig. 14.45.
-
-# **Figure 14.44**
-
-Active band-pass filter: (a) block diagram, (b) frequency response.
-
-The analysis of the band-pass filter is relatively simple. Its transfer function is obtained by multiplying Eqs. (14.60) and (14.63) with the gain of the inverter; that is,
-
-$$
-\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \left(-\frac{1}{1 + j\omega C_1 R}\right) \left(-\frac{j\omega C_2 R}{1 + j\omega C_2 R}\right) \left(-\frac{R_f}{R_i}\right)
-$$
-$$
-= -\frac{R_f}{R_i} \frac{1}{1 + j\omega C_1 R} \frac{j\omega C_2 R}{1 + j\omega C_2 R}
-$$
-(14.65)
-
-The low-pass section sets the upper corner frequency as
-
-$$
-\omega_2 = \frac{1}{RC_1} \tag{14.66}
-$$
-
-while the high-pass section sets the lower corner frequency as
-
-$$
-\omega_1 = \frac{1}{RC_2} \tag{14.67}
-$$
-
-With these values of *ω*1 and *ω*2, the center frequency, bandwidth, and quality factor are found as follows:
-
-$$
-\omega_0 = \sqrt{\omega_1 \omega_2} \tag{14.68}
-$$
-
-$$
-B = \omega_2 - \omega_1 \tag{14.69}
-$$
-
-$$
-Q = \frac{\omega_0}{B} \tag{14.70}
-$$
-
-To find the passband gain *K*, we write Eq. (14.65) in the standard form of Eq. (14.15),
-
-To find the passband gain *K*, we write Eq. (14.65) in the standard
-\nn of Eq. (14.15),
-\n
-$$
-\mathbf{H}(\omega) = -\frac{R_f}{R_i} \frac{j\omega/\omega_1}{(1 + j\omega/\omega_1)(1 + j\omega/\omega_2)} = -\frac{Rf}{R_i} \frac{j\omega\omega_2}{(\omega_1 + j\omega)(\omega_2 + j\omega)}
-$$
-\n(14.71)
-
-At the center frequency *ω*0 = √ \_\_\_\_\_ *ω*1*ω*2 , the magnitude of the transfer function is
-
-center frequency
-$$
-\omega_0 = \sqrt{\omega_1 \omega_2}
-$$
-, the magnitude of the transfer
-\nn is
-\n
-$$
-|\mathbf{H}(\omega_0)| = \left| \frac{R_f}{R_i} \frac{j \omega_0 \omega_2}{(\omega_1 + j \omega_0)(\omega_2 + j \omega_0)} \right| = \frac{R_f}{R_i} \frac{\omega_2}{\omega_1 + \omega_2}
-$$
-(14.72)
-
-Thus, the passband gain is
-
-$$
-K = \frac{R_f}{R_i} \frac{\omega_2}{\omega_1 + \omega_2} \tag{14.73}
-$$
-
-# **14.8.4** Band-Reject (or Notch) Filter
-
-A band-reject filter may be constructed by parallel combination of a lowpass filter and a high-pass filter and a summing amplifier, as shown in the block diagram of Fig. 14.46(a). The circuit is designed such that the lower cutoff frequency *ω*1 is set by the low-pass filter while the upper cutoff frequency *ω*2 is set by the high-pass filter. The gap between *ω*1 and *ω*2 is the bandwidth of the filter. As shown in Fig. 14.46(b), the filter passes frequencies below *ω*1 and above *ω*2. The block diagram in Fig. 14.46(a) is actually constructed as shown in Fig. 14.47. The transfer function is
-
-$$
-\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = -\frac{R_f}{R_i} \left( -\frac{1}{1 + j\omega C_1 R} - \frac{j\omega C_2 R}{1 + j\omega C_2 R} \right) \tag{14.74}
-$$
-
-**Figure 14.46** Active band-reject filter: (a) block diagram, (b) frequency response.
-
-Active band-reject filter.
-
-The formulas for calculating the values of *ω*1, *ω*2, the center frequency, bandwidth, and quality factor are the same as in Eqs. (14.66) to (14.70).
-
-To determine the passband gain *K* of the filter, we can write Eq. (14.74) in terms of the upper and lower corner frequencies as
-
-$$
-\mathbf{H}(\omega) = \frac{R_f}{R_i} \left( \frac{1}{1 + j\omega/\omega_2} + \frac{j\omega/\omega_1}{1 + j\omega/\omega_1} \right)
-$$
-
-= $\frac{R_f}{R_i} \frac{(1 + j2\omega/\omega_1 + (j\omega)^2/\omega_1\omega_1)}{(1 + j\omega/\omega_2)(1 + j\omega/\omega_1)}$ (14.75)
-
-Comparing this with the standard form in Eq. (14.15) indicates that in the two passbands (*ω* → 0 and *ω* → ∞) the gain is
-
-$$
-K = \frac{R_f}{R_i} \tag{14.76}
-$$
-
-We can also find the gain at the center frequency by finding the magnitude of the transfer function at *ω*0 = √ \_\_\_\_\_ *ω*1*ω*2 , writing
-
-and the gain at the center frequency by finding the magni-
-\nnsfer function at
-$$
-\omega_0 = \sqrt{\omega_1 \omega_2}
-$$
-, writing
-\n
-$$
-H(\omega_0) = \left| \frac{R_f (1 + j2\omega_0/\omega_1 + (j\omega_0)^2/\omega_1 \omega_1)}{R_i (1 + j\omega_0/\omega_2)(1 + j\omega_0/\omega_1)} \right|
-$$
-\n
-$$
-= \frac{R_f}{R_i} \frac{2\omega_1}{\omega_1 + \omega_2}
-$$
-\n(14.77)
-
-Again, the filters treated in this section are only typical. There are many other active filters that are more complex.
-
-Example 14.12 Design a low-pass active filter with a dc gain of 4 and a corner frequency of 500 Hz.
-
-# **Solution:**
-
-From Eq. (14.61), we find
-
-$$
-\omega_c = 2\pi f_c = 2\pi (500) = \frac{1}{R_f C_f} \tag{14.12.1}
-$$
-
-The dc gain is
-
-$$
-H(0) = -\frac{R_f}{R_i} = -4\tag{14.12.2}
-$$
-
-We have two equations and three unknowns. If we select *Cf* = 0.2*μ*F, then
-
-tions and three unknowns. If we se
-$$
-R_f = \frac{1}{2\pi (500) 0.2 \times 10^{-6}} = 1.59 \text{ k}\Omega
-$$
-
-and
-
-$$
-R_i = \frac{R_f}{4} = 397.5 \ \Omega
-$$
-
-We use a 1.6-kΩ resistor for *Rf* and a 400-Ω resistor for *Ri* . Figure 14.42 shows the filter.
-
-Design a high-pass filter with a high-frequency gain of 5 and a corner Practice Problem 14.12 frequency of 2 kHz. Use a 50-nF capacitor in your design.
-
-**Answer:** *Ri* = 1,600 Ω and *Rf* = 8 kΩ.
-
-Design a band-pass filter in the form of Fig. 14.45 to pass frequencies Example 14.13 between 250 and 3,000 Hz and with *K* = 10. Select *R* = 20 kΩ.
-
-# **Solution:**
-
-- 1. **Define.** The problem is clearly stated and the circuit to be used in the design is specified.
-- 2. **Present.** We are ask ed to use the op amp circuit specified in Fig. 14.45 to design a band-pass filter. We are given the value of *R* to use (20 kΩ). In addition, the frequency range of the signals to be passed is 250 Hz to 3 kHz.
-- 3. **Alternative.** We will use the equations developed in Section 14.8.3 to obtain a solution. We will then use the resulting transfer function to validate the answer.
-- 4. **Attempt.** Because *ω*1 = 1∕*RC*2, we obtain
-
-te the answer.
-\n**apt.** Because
-$$
-\omega_1 = 1/RC_2
-$$
-, we obtain
-\n
-$$
-C_2 = \frac{1}{R\omega_1} = \frac{1}{2\pi f_1 R} = \frac{1}{2\pi \times 250 \times 20 \times 10^3} = 31.83 \text{ nF}
-$$
-
-Similarly, since *ω*2 = 1∕*RC*1,
-
-$$
-K\omega_1 = 2\pi f_1 K - 2\pi \times 250 \times 20 \times 10^8
-$$
-
-arly, since $\omega_2 = 1/RC_1$ ,
-$$
-C_1 = \frac{1}{R\omega_2} = \frac{1}{2\pi f_2 R} = \frac{1}{2\pi \times 3,000 \times 20 \times 10^3} = 2.65 \text{ nF}
-$$
-
-From Eq. (14.73),
-
-$$
-\frac{R_f}{R_i} = K \frac{\omega_1 + \omega_2}{\omega_2} = K \frac{f_1 + f_2}{f_2} = \frac{10(3,250)}{3,000} = 10.83
-$$
-
-If we select *Ri* = **10 k**Ω, then *Rf* = 10.83*Ri* ≃ **108.3 k**Ω. 5. **Evaluate.** The output of the first op amp is given by
-
-ct
-$$
-R_i = 10 \text{ k}\Omega
-$$
-, then $R_f = 10.83R_i \approx 108.3 \text{ k}\Omega$ .
-The output of the first op amp is given by
-$$
-\frac{V_i - 0}{20 \text{ k}\Omega} + \frac{V_1 - 0}{20 \text{ k}\Omega} + \frac{s2.65 \times 10^{-9} (V_1 - 0)}{1}
-$$
-$$
-= 0 \rightarrow V_1 = -\frac{V_i}{1 + 5.3 \times 10^{-5} s}
-$$
-
-The output of the second op amp is given by
-
-$$
-t \text{ of the second op amp is given by}
-$$
-\n
-$$
-\frac{V_1 - 0}{20 \text{ k}\Omega + \frac{1}{s31.83 \text{ nF}}} + \frac{V_2 - 0}{20 \text{ k}\Omega} = 0 \rightarrow
-$$
-\n
-$$
-V_2 = -\frac{6.366 \times 10^{-4} s V_1}{1 + 6.366 \times 10^{-4} s}
-$$
-\n
-$$
-= \frac{6.366 \times 10^{-4} s V_i}{(1 + 6.366 \times 10^{-4} s)(1 + 5.3 \times 10^{-5} s)}
-$$
-
-The output of the third op amp is given by
-
-$$
-\frac{V_2 - 0}{10 \text{ k}\Omega} + \frac{V_o - 0}{108.3 \text{ k}\Omega} = 0 \to V_o = 10.83 V_2 \to j2\pi \times 25^\circ
-$$
-$$
-V_o = -\frac{6.894 \times 10^{-3} sV_i}{(1 + 6.366 \times 10^{-4} s)(1 + 5.3 \times 10^{-5} s)}
-$$
-
-Let *j*2*π* × 25° and solve for the magnitude of *Vo*∕*Vi*.
-
-$$
-\frac{V_o}{V_i} = \frac{-j10.829}{(1+j1)(1)}
-$$
-
- ∣*Vo*∕*Vi* ∣ = **(0.7071)10.829**, which is the lower corner frequency point. Let *s* = *j*2*π* × 3000 = *j*18.849 kΩ. We then get
-
-Let
-$$
-s = j2\pi \times 3000 = j18.849 \text{ k}\Omega
-$$
-. We then get
-\n
-$$
-\frac{V_o}{V_i} = \frac{-j129.94}{(1+j12)(1+j1)}
-$$
-\n
-$$
-= \frac{129.94/-90^{\circ}}{(12.042/85.24^{\circ})(1.4142/45^{\circ})} = (0.7071)10.791/-18.61^{\circ}
-$$
-
-Clearly this is the upper corner frequency and the answer checks.
-
-6. **Satisfactory?** We have satisfactorily designed the circuit and can present the results as a solution to the problem.
-
-| Practice Problem 14.13 | Design a notch filter based on Fig. 14.47 for ω0 = 20 krad/s, K = 5, and |
-|------------------------|--------------------------------------------------------------------------|
-| | Q = 10. Use R = Ri = 10 kΩ. |
-
-**Answer:** *C*1 = 4.762 nF, *C*2 = 5.263 nF, and *Rf* = 50 kΩ.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/167_14.9 Scaling.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/167_14.9 Scaling.md
deleted file mode 100644
index 0e34f74e36d288fa3bf781a2565549526c27d286..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/167_14.9 Scaling.md
+++ /dev/null
@@ -1,153 +0,0 @@
-# **14.9** Scaling
-
-In designing and analyzing filters and resonant circuits or in circuit analysis in general, it is sometimes convenient to work with element values of 1 Ω, 1 H, or 1 F, and then transform the values to realistic values by
-
-*scaling*. We have taken advantage of this idea by not using realistic element values in most of our examples and problems; mastering circuit analysis is made easy by using convenient component values. We have thus eased calculations, knowing that we could use scaling to then make the values realistic.
-
-There are two ways of scaling a circuit: *magnitude* or *impedance scaling*, and *frequency scaling* . Both are useful in scaling responses and circuit elements to values within the practical ranges. While magnitude scaling leaves the frequency response of a circuit unaltered, frequency scaling shifts the frequency response up or down the frequency spectrum.
-
-# **14.9.1** Magnitude Scaling
-
-Magnitude scaling is the process of increasing all impedances in a network by a factor, the frequency response remaining unchanged.
-
-Recall that impedances of indi vidual elements *R*, *L*, and *C* are given by
-
-$$
-\mathbf{Z}_R = R, \qquad \mathbf{Z}_L = j\omega L, \qquad \mathbf{Z}_C = \frac{1}{j\omega C} \tag{14.78}
-$$
-
-In magnitude scaling, we multiply the impedance of each circuit element by a factor *Km* and let the frequency remain constant. This gives the new impedances as
-
-$$
-\mathbf{Z}'_R = K_m \mathbf{Z}_R = K_m R, \qquad \mathbf{Z}'_L = K_m \mathbf{Z}_L = j\omega K_m L
-$$
-$$
-\mathbf{Z}'_C = K_m \mathbf{Z}_C = \frac{1}{j\omega C/K_m}
-$$
-(14.79)
-
-Comparing Eq. (14.79) with Eq. (14.78), we notice the following changes in the element values: *R* → *KmR*, *L* → *Km L*, and *C* → *C*∕*Km*. Thus, in magnitude scaling, the new values of the elements and frequency are
-
-$$
-R' = K_m R, \qquad L' = K_m L
-$$
-
-$$
-C' = \frac{C}{K_m}, \qquad \omega' = \omega
-$$
- (14.80)
-
-The primed variables are the new values and the unprimed variables are the old values. Consider the series or parallel *RLC* circuit. We now have
-
-$$
-\omega_0' = \frac{1}{\sqrt{LC'}} = \frac{1}{\sqrt{K_m LC/K_m}} = \frac{1}{\sqrt{LC}} = \omega_0
-$$
-(14.81)
-
-showing that the resonant frequency, as expected, has not changed. Similarly, the quality factor and the bandwidth are not affected by magnitude scaling. Also, magnitude scaling does not affect transfer functions in the forms of Eqs. (14.2a) and (14.2b), which are dimen sionless quantities.
-
-# **14.9.2** Frequency Scaling
-
-Frequency scaling is equivalent to relabeling the frequency axis of a frequency response plot. It is needed when translating frequencies such as a resonant frequency, a corner frequency, a bandwidth, etc., to a realistic level. It can be used to bring capacitance and inductance values into a range that is convenient to work with.
-
-Frequency scaling is the process of shifting the frequency response of a network up or down the frequency axis while leaving the impedance the same.
-
-We achieve frequency scaling by multiplying the frequency by a factor *Kf* while keeping the impedance the same.
-
-From Eq. (14.78), we see that the impedances of *L* and *C* are frequency-dependent. If we apply frequenc y scaling to **Z***L*(*ω*) and **Z***C*(*ω*) in Eq. (14.78), we obtain
-
-$$
-\mathbf{Z}_L = j(\omega K_f)L' = j\omega L \qquad \Rightarrow \qquad L' = \frac{L}{K_f} \tag{14.82a}
-$$
-
-$$
-Z_C = \frac{1}{j(\omega K_f)C'} = \frac{1}{j\omega C} \qquad \Rightarrow \qquad C' = \frac{C}{K_f} \tag{14.82b}
-$$
-
-since the impedance of the inductor and capacitor must remain the same after frequency scaling. We notice the following changes in the element values: *L* → *L*∕*Kf* and *C* → *C*∕*Kf*. The value of *R* is not affected, since its impedance does not depend on frequency. Thus, in frequency scaling, the new values of the elements and frequency are
-
-$$
-R' = R, \qquad L' = \frac{L}{K_f}
-$$
-
-$$
-C' = \frac{C}{K_f}, \qquad \omega' = K_f \omega
-$$
- (14.83)
-
-Again, if we consider the series or parallel *RLC* circuit, for the resonant frequency
-
-$$
-\omega'_{0} = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(L/K_{f})(C/K_{f})}} = \frac{K_{f}}{\sqrt{LC}} = K_{f}\omega_{0}
-$$
-(14.84)
-
-and for the bandwidth
-
-$$
-B' = K_f B \tag{14.85}
-$$
-
-but the quality factor remains the same (*Q*′ = *Q*).
-
-# **14.9.3** Magnitude and Frequency Scaling
-
-If a circuit is scaled in magnitude and frequency at the same time, then
-
-$$
-R' = K_m R, \qquad L' = \frac{K_m}{K_f} L
-$$
-
-$$
-C' = \frac{1}{K_m K_f} C, \qquad \omega' = K_f \omega
-$$
- (14.86)
-
-These are more general formulas than those in Eqs. (14.80) and (14.83). We set *Km* = 1 in Eq. (14.86) when there is no magnitude scaling or *Kf* = 1 when there is no frequency scaling.
-
-A fourth-order Butterworth low-pass filter is shown in Fig. 14.48(a). The Example 14.14 filter is designed such that the cutoff frequency *ωc* = 1 rad/s. Scale the circuit for a cutoff frequency of 50 kHz using 10-kΩ resistors.
-
-# **Figure 14.48**
-
-For Example 14.14: (a) Normalized Butterworth low-pass filter, (b) scaled version of the same low-pass filter.
-
-# **Solution:**
-
-If the cutoff frequency is to shift from *ωc* = 1 rad/s to *ω*′ *c* = 2*π*(50) krad/s, then the frequency scale factor is
-
-$$
-K_f = \frac{\omega_c'}{\omega_c} = \frac{100\pi \times 10^3}{1} = \pi \times 10^5
-$$
-
-Also, if each 1-Ω resistor is to be replaced by a 10-k Ω resistor, then the magnitude scale factor must be
-
-$$
-K_m = \frac{R'}{R} = \frac{10 \times 10^3}{1} = 10^4
-$$
-
-Using Eq. (14.86),
-
-$$
-L'_1 = \frac{K_m}{K_f} L_1 = \frac{10^4}{\pi \times 10^5} (1.848) = 58.82 \text{ mH}
-$$
-
-\n
-$$
-L'_2 = \frac{K_m}{K_f} L_2 = \frac{10^4}{\pi \times 10^5} (0.765) = 24.35 \text{ mH}
-$$
-
-\n
-$$
-C'_1 = \frac{C_1}{K_m K_f} = \frac{0.765}{\pi \times 10^9} = 243.5 \text{ pF}
-$$
-
-\n
-$$
-C'_2 = \frac{C_2}{K_m K_f} = \frac{1.848}{\pi \times 10^9} = 588.2 \text{ pF}
-$$
-
-The scaled circuit is shown in Fig. 14.48(b). This circuit uses practical values and will provide the same transfer function as the prototype in Fig. 14.48(a), but shifted in frequency.
-
-A third-order Butterworth filter normalized to *ωc* = 1 rad/s is shown Practice Problem 14.14 in Fig. 14.49. Scale the circuit to a cutoff frequency of 10 kHz. Use 15-nF capacitors.
-
-**Answer:** *R*′ 1 = *R*′ 2 = 1.061 kΩ, *C*′ 1 = *C*′ 2 = 15 nF, *L*′ = 33.77 mH.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/168_14.10 Frequency Response Using PSpice.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/168_14.10 Frequency Response Using PSpice.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/168_14.10 Frequency Response Using PSpice.md
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@@ -1,71 +0,0 @@
-# **14.10** Frequency Response Using PSpice
-
-*PSpice* is a useful tool in the hands of the modern circuit designer for obtaining the frequency response of circuits. The frequency response is obtained using the AC Sweep as discussed in Section D.5 (Appendix D). This requires that we specify in the AC Sweep dialog box *Total Pts, Start Freq, End Freq*, and the sweep type. *Total Pts* is the number of points in the frequency sweep, and *Start Freq* and *End Freq* are, respectively, the starting and final frequencies, in hertz. In order to know what frequencies to select for *Start Freq* and *End Freq*, one must have an idea of the frequency range of interest by making a rough sketch of the frequency response. In a complex circuit where this may not be possible, one may use a trial-and-error approach.
-
-There are three types of sweeps:
-
-- *Linear:* The frequency is varied linearly from *Start Freq* to *End Freq* with *Total Pts* equally spaced points (or responses).
-- *Octave:* The frequency is swept logarithmically by octaves from *Start Freq* to *End Freq* with *Total Pts* per octave. An octave is a factor of 2 (e.g., 2 to 4, 4 to 8, 8 to 16).
-- *Decade:* The frequency is varied logarithmically by decades from *Start Freq* to *End Freq* with *Total Pts* per decade. A decade is a factor of 10 (e.g., from 2 to 20 Hz, 20 to 200 Hz, 200 Hz to 2 kHz).
-
-It is best to use a linear sweep when displaying a narrow frequency range of interest, as a linear sweep displays the frequency range well in a nar row range. Conversely, it is best to use a logarithmic (octave or decade) sweep for displaying a wide frequency range of interest—if a linear sweep is used for a wide range, all the data will be crowded at the highor low-frequency end and insufficient data at the other end.
-
-With the abo ve specifications, *PSpice* performs a steady-state sinusoidal analysis of the circuit as the frequency of all the independent sources is varied (or swept) from *Start Freq* to *End Freq*.
-
-The *PSpice* A/D program produces a graphical output. The output data type may be specified in the *Trace Command Box* by adding one of the following suffixes to V or I:
-
-- M Amplitude of the sinusoid.
-- P Phase of the sinusoid.
-- dB Amplitude of the sinusoid in decibels, that is, 20 log 10 (amplitude).
-
-Example 14.15 Determine the frequency response of the circuit shown in Fig. 14.50.
-
-# **Solution:**
-
-We let the input voltage *vs* be a sinusoid of amplitude 1 V and phase 0°. Figure 14.51 is the schematic for the circuit. The capacitor is rotated 270° counterclockwise to ensure that pin 1 (the positive terminal) is on top. The voltage marker is inserted to the output voltage across the capacitor. To perform a linear sweep for 1 < *f* < 1,000 Hz with 50 points, we select **Analysis/Setup/AC Sweep, DCLICK** *Linear*, type 50 in the *Total Pts* box, type 1 in the *Start Freq* box, and type 1000 in the *End Freq* box. After saving the file, we select **Analysis/Simulate** to simulate the circuit. If there are no errors, the *PSpice* A/D window will
-
-**Figure 14.50** For Example 14.15.
-
-The schematic for the circuit in Fig. 14.50.
-
-display the plot of V(C1:1), which is the same as *Vo* or *H*(*ω*) = *Vo*∕1, as shown in Fig. 14.52(a). This is the magnitude plot, since V(C1:1) is the same as VM(C1:1). To obtain the phase plot, select **Trace/Add** in the *PSpice* A/D menu and type VP(C1:1) in the **Trace Command** box. Figure 14.52(b) shows the result. By hand, the transfer function is
-
-$$
-H(\omega) = \frac{V_o}{V_s} = \frac{1,000}{9,000 + j\omega 8}
-$$
-
-or
-
-$$
-H(\omega) = \frac{1}{9 + j16\pi \times 10^{-3}}
-$$
-
-showing that the circuit is a low-pass filter as demonstrated in Fig. 14.52. Notice that the plots in Fig. 14.52 are similar to those in Fig. 14.3 (note that the horizontal axis in Fig. 14.52 is logrithic while the horizontal axis in Fig. 14.3 is linear.)
-
-**Figure 14.53** For Practice Prob. 14.15.
-
-For Practice Problem 14.15: (a) magnitude plot, (b) phase plot of the frequency response.
-
-For Example 14.16.
-
-Example 14.16 Use *PSpice* to generate the gain and phase Bode plots of *V* in the circuit of Fig. 14.55.
-
-# **Solution:**
-
-The circuit treated in Example 14.15 is first-order while the one in this example is second-order. Since we are interested in Bode plots, we use decade frequency sweep for 300 < *f* < 3,000 Hz with 50 points per de cade. We select this range because we know that the resonant frequency of the circuit is within the range. Recall that
-
-$$
-\omega_0 = \frac{1}{\sqrt{LC}} = 5 \text{ krad/s} \qquad \text{or} \qquad f_0 = \frac{\omega}{2\pi} = 795.8 \text{ Hz}
-$$
-
-After drawing the circuit as in Fig. 14.55, we select **Analysis/Setup/AC Sweep, DCLICK** *Decade*, enter 50 in the *Total Pts* box, 300 as the *Start Freq*, and 3,000 in the *End Freq* box. Upon saving the file, we simulate it by selecting **Analysis/Simulate**. This will automatically bring up the *PSpice* A/D window and display V(C1:1) if there are no errors. Since we are interested in the Bode plot, we select **Trace/Add** in the *PSpice* A/D menu and type dB(V(C1:1)) in the **Trace Command** box. The result is the Bode magnitude plot in Fig. 14.56(a). For the phase plot, we select
-
-For Example 14.16: (a) Bode plot, (b) phase plot of the response.
-
-**Trace/Add** in the *PSpice* A/D menu and type VP(C1:1) in the **Trace Command** box. The result is the Bode phase plot of Fig. 14.56(b). Notice that the plots confirm the resonant frequency of 795.8 Hz.
-
-Consider the network in Fig. 14.57. Use *PSpice* to obtain the Bode plots Practice Problem 14.16 for *Vo* over a frequency from 1 to 100 kHz using 20 points per decade.
-
-For Practice Prob. 14.16.
-
-**Answer:** See Fig. 14.58.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/169_14.11 Computation Using MATLAB.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/169_14.11 Computation Using MATLAB.md
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-# **14.11** Computation Using MATLAB
-
-*MATLAB* is a software package that is widely used for engineering computation and simulation. A review of *MATLAB* is provided in Appen dix E for the beginner. This section shows how to use the software to numerically perform most of the operations presented in this chapter and Chapter 15. The key to describing a system in *MATLAB* is to specify the numerator (num) and denominator (den) of the transfer function of the system. Once this is done, we can use several *MATLAB* commands to obtain the system's Bode plots (frequency response) and the system's response to a given input.
-
-The command **bode** produces the Bode plots (both magnitude and phase) of a given transfer function *H*(*s*). The format of the com mand is **bode** (num, den), where num is the numerator of *H*(*s*) and den is its denominator. The frequency range and number of points are automatically selected. For example, consider the transfer function in Example 14.3. It is better to first write the numerator and denominator in polynomial forms.
-
-Thus,
-
-$$
-H(s) = \frac{200 j\omega}{(j\omega + 2)(j\omega + 10)} = \frac{200s}{s^2 + 12s + 20}, \qquad s = j\omega
-$$
-
-Using the following commands, the Bode plots are generated as shown in Fig. 14.59. If necessary, the command **logspace** can be included to generate a logarithmically spaced frequency and the command **semilogx** can be used to produce a semilog scale.
-
->> num = [200 0]; % specify the numerator of H(s) >> den = [1 12 20]; % specify the denominator of H(s) >> bode(num, den); % determine and draw Bode plots
-
-The step response *y*(*t*) of a system is the output when the input *x*(*t*) is the unit step function. The command **step** plots the step response of a system given the numerator and denominator of the transfer function of that sys tem. The time range and number of points are automatically selected. F or example, consider a second-order system with the transfer function
-
-$$
-H(s) = \frac{12}{s^2 + 3s + 12}
-$$
-
-We obtain the step response of the system shown in Fig. 14.60 by using the following commands.
-
->> n = 12; >> d = [1 3 12]; >> step(n,d);
-
-We can verify the plot in Fig. 14.60 by obtaining *y*(*t*) = *x*(*t*) \* *u*(*t*) or *Y*(*s*) = *X*(*s*)*H*(*s*).
-
-The command **lsim** is a more general command than **step**. It calculates the time response of a system to any arbitrary input signal. The format of the command is *y* = **lsim** (num, den, *x*, *t*), where *x*(*t*) is the input signal, *t* is the time vector, and *y*(*t*) is the output generated. For example, assume a system is described by the transfer function
-
-for, and
-$$
-y(t)
-$$
- is the output gen-
-cribed by the transfer function
-$$
-H(s) = \frac{s+4}{s^3 + 2s^2 + 5s + 10}
-$$
-
-To find the response *y*(*t*) of the system to input *x*(*t*) = 10*e*−*t u*(*t*), we use the following *MATLAB* commands. Both the response *y*(*t*) and the input *x*(*t*) are plotted in Fig. 14.61.
-
-```
->> t = 0:0.02:5; % time vector 0 < t < 5 with increment
- 0.02
->> x = 10*exp(-t);
->> num = [1 4];
->> den = [1 2 5 10];
->> y = lsim(num,den,x,t);
->> plot(t,x,t,y)
-```
-
-# **Figure 14.61**
-
-The response of the system described by *H*(*s*) = (*s* + 4)∕(*s* 2 + 2*s* 2 + 5*s* + 10) to an exponential input.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/170_14.12 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/170_14.12 Applications.md
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/170_14.12 Applications.md
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-# **14.12** Applications
-
-Resonant circuits and filters are widely used, particularly in electronics, power systems, and communications systems. For example, a Notch filter with a cutoff frequency at 60 Hz may be used to eliminate the 60-Hz power line noise in various communications electronics. Filtering of signals in communications systems is necessary in order to select the desired signal from a host of others in the same range (as in the case of radio receivers discussed next) and also to minimize the effects of noise and interference on the desired signal. In this section, we consider one practical application of resonant circuits and two applications of filters. The focus of each application is not to understand the details of how each device works but to see how the circuits considered in this chapter are applied in the practical devices.
-
-# **14.12.1** Radio Receiver
-
-Series and parallel resonant circuits are commonly used in radio and TV receivers to tune in stations and to separate the audio signal from the radio-frequency carrier wave. As an example, consider the block diagram of an AM radio receiver shown in Fig. 14.62. Incoming amplitudemodulated radio waves (thousands of them at different frequencies from different broadcasting stations) are received by the antenna. A resonant circuit (or a band-pass filter) is needed to select just one of the incoming waves. The selected signal is very weak and is amplified in stages in order to generate an audible audio-frequency wave. Thus, we have the radiofrequency (RF) amplifier to amplify the selected broadcast signal, the intermediate frequency (IF) amplifier to amplify an internally generated signal based on the RF signal, and the audio amplifier to amplify the audio signal just before it reaches the loudspeaker. It is much easier to amplify the signal at three stages than to build an amplifier to provide the same amplification for the entire band.
-
-# **Figure 14.62**
-
-A simplified block diagram of a superheterodyne AM radio receiver.
-
-The type of AM receiver shown in Fig. 14.62 is known as the *superheterodyne receiver*. In the early de velopment of radio, each amplification stage had to be tuned to the frequenc y of the incoming signal. This way, each stage must have several tuned circuits to cover the entire AM band (540 to 1600 kHz). To avoid the problem of having several resonant circuits, modern recei vers use a *frequency mixer* or *heterodyne* circuit, which always produces the same IF signal (445 kHz) but retains the audio frequencies carried on the incoming signal. To produce the constant IF frequency, the rotors of two separate variable capacitors are mechanically coupled with one another so that they can be rotated simultaneously with a single control; this is called *ganged tuning*. A *local oscillator* ganged with the RF amplifier produces an RF signal that is combined with the incoming wave by the frequenc y mixer to produce a n output signal that contains the sum and the difference frequencies of the two signals. For example, if the resonant circuit is tuned to recei ve an 800-kHz incoming signal, the local oscillator must produce a 1,255-kHz signal, so that the sum (1,255 + 800 = 2,055 kHz) and the difference (1,255 – 800 = 455 kHz) of frequencies are a vailable at the output of the mix er. However, only
-
-the difference, 455 kHz, is used in practice. This is the only frequency to which all the IF amplifier stages are tuned, regardless of the station dialed. The original audio signal (containing the "intelligence") is e xtracted in the detector stage. The detector basically removes the IF signal, leaving the audio signal. The audio signal is amplified to drive the loudspeaker, which acts as a transducer converting the electrical signal to sound.
-
-Our major concern here is the tuning circuit for the AM radio re ceiver. The operation of the FM radio recei ver is different from that of the AM receiver discussed here, and in a much dif ferent range of fre quencies, but the tuning is similar.
-
-The resonant or tuner circuit of an AM radio is portrayed in Fig. 14.63. Example 14.17 Given that *L* = 1*μ*H, what must be the range of *C* to have the resonant frequency adjustable from one end of the AM band to another?
-
-# **Solution:**
-
-The frequency range for AM broadcasting is 540 to 1,600 kHz. We consider the low and high ends of the band. Since the resonant circuit in Fig. 14.63 is a parallel type, we apply the ideas in Section 14.6. From Eq. (14.44),
-
-$$
-\omega_0 = 2\pi f_0 = \frac{1}{\sqrt{LC}}
-$$
-
-or
-
-$$
-C = \frac{1}{4\pi^2 f_0^2 L}
-$$
-
-For the high end of the AM band, *f*0 = 1,600 kHz, and the corresponding *C* is
-
-$$
-4\pi J_0 L
-$$
-
-d of the AM band, $f_0 = 1,600$ kHz, and the
-$$
-C_1 = \frac{1}{4\pi^2 \times 1,600^2 \times 10^6 \times 10^{-6}} = 9.9
-$$
- nF
-
-For the low end of the AM band, *f*0 = 540 kHz, and the corresponding *C* is
-
-of the AM band,
-$$
-f_0 = 540
-$$
- kHz, and the co-
-$$
-C_2 = \frac{1}{4\pi^2 \times 540^2 \times 10^6 \times 10^{-6}} = 86.9
-$$
- nF
-
-Thus, *C* must be an adjustable (gang) capacitor varying from 9.9 to 86.9 nF.
-
-For an FM radio receiver, the incoming wave is in the frequency range Practice Problem 14.17 from 88 to 108 MHz. The tuner circuit is a parallel *RLC* circuit with a 4-*μ*H coil. Calculate the range of the variable capacitor necessary to cover the entire band.
-
-**Answer:** From 0.543 to 0.818 pF.
-
-**Figure 14.63** The tuner circuit for Example 14.17.
-
-# **14.12.2** Touch-Tone Telephone
-
-A typical application of filtering is the touch-tone telephone set shown in Fig. 14.64. The keypad has 12 buttons arranged in four rows and three columns. The arrangement provides 12 distinct signals by using seven tones divided into two groups: the low-frequency group (697 to 941 Hz) and the high-frequency group (1,209 to 1,477 Hz). Pressing a button generates a sum of two sinusoids corresponding to its unique pair of frequencies. For example, pressing the number 6 button generates sinusoidal tones with frequencies 770 and 1,477 Hz.
-
-**Figure 14.64** Frequency assignments for touch-tone dialing.
-
-When a caller dials a telephone number, a set of signals is transmitted to the telephone of fice, where the touch-tone signals are de coded by detecting the frequencies they contain. Figure 14.65 shows the block diagram for the detection scheme. The signals are f irst amplified and separated into their respective groups by the low-pass (LP) and high-pass (HP) f ilters. The limiters (L) are used to con vert the separated tones into square w aves. The individual tones are identified using seven band-pass (BP) filters, each filter passing one tone and rejecting other tones. Each f ilter is follow ed by a detec tor (D), which is energized when its input voltage exceeds a certain level. The outputs of the detectors pro vide the required dc signals needed by the switching system to connect the caller to the party being called.
-
-Block diagram of detection scheme.
-
-Using the standard 600-Ω resistor used in telephone circuits and a series Example 14.18 *RLC* circuit, design the band-pass filter *BP*2 in Fig. 14.65.
-
-# **Solution:**
-
-The band-pass filter is the series *RLC* circuit in Fig. 14.35. Inasmuch as *BP*2 passes frequencies 697 to 852 Hz and is centered at *f*0 = 770 Hz, its bandwidth is
-
-$$
-B = 2\pi(f_2 - f_1) = 2\pi(852 - 697) = 973.89 \text{ rad/s}
-$$
-
-From Eq. (14.39),
-
-$$
-L = \frac{R}{B} = \frac{600}{973.89} = 0.616 \text{ H}
-$$
-
-From Eq. (14.27) or (14.55),
-
-(14.27) or (14.55),
-\n
-$$
-C = \frac{1}{\omega_0^2 L} = \frac{1}{4\pi^2 f_0^2 L} = \frac{1}{4\pi^2 \times 770^2 \times 0.616} = 69.36 \text{ nF}
-$$
-
-Repeat Example 14.18 for band-pass filter *BP*6. Practice Problem 14.18
-
-**Answer:** 356 mH, 39.83 nF.
-
-# **14.12.3** Crossover Network
-
-Another typical application of filters is the *crossover network* that couples an audio amplifier to woofer and tweeter speakers, as shown in Fig. 14.66(a). The network basically consists of one high-pass *RC* filter
-
-**Figure 14.66**
-
-(a) A crossover network for two loudspeakers, (b) equivalent model.
-
-**Figure 14.67** Frequency responses of the crossover network in Fig. 14.66.
-
-and one low-pass *RL* filter. It routes frequencies higher than a prescribed crossover frequency *fc* to the tweeter (high-frequency l oudspeaker) and frequencies below *fc* into the woofer (low-frequency loudspeaker). These loudspeakers have been designed to accommodate certain frequency responses. A woofer is a low- frequency loudspeaker designed to repro duce the lower part of the frequency range, up to about 3 kHz. A tweeter can reproduce audio frequencies from about 3 kHz to about 20 kHz. The two speaker types can be combined to reproduce the entire audio range of interest and provide the optimum in frequency response.
-
-By replacing the amplifier with a voltage source, the approximate equivalent circuit of the crosso ver network is sho wn in Fig. 14.66(b), where the loudspeak ers are modeled by resistors. As a high-pass filter, the transfer function *V*1∕*Vs* is given by
-
-$$
-H_1(\omega) = \frac{V_1}{V_s} = \frac{j\omega R_1 C}{1 + j\omega R_1 C}
-$$
- (14.87)
-
-Similarly, the transfer function of the low-pass filter is given by
-
-$$
-H_2(\omega) = \frac{V_2}{V_s} = \frac{R_2}{R_2 + j\omega L}
-$$
- (14.88)
-
-The values of *R*1, *R*2, *L*, and *C* may be selected such that the two filters have the same cutoff frequency, known as the *crossover frequency*, as shown in Fig. 14.67.
-
-The principle behind the crossover network is also used in the resonant circuit for a TV receiver, where it is necessary to separate the video and audio bands of RF carrier frequencies. The lower-frequency band (picture information in the range from about 30 Hz to about 4 MHz) is channeled into the receiver's video amplifier, while the high-frequency band (sound information around 4.5 MHz) is channeled to the receiver's sound amplifier.
-
-Example 14.19 In the crossover network of Fig. 14.66, suppose each speak er acts as a 6-Ω resistance. Find *C* and *L* if the crossover frequency is 2.5 kHz.
-
-# **Solution:**
-
-For the high-pass filter,
-
-$$
-\omega_c = 2\pi f_c = \frac{1}{R_1 C}
-$$
-
-or
-
-or
-
-$$
-C = \frac{1}{2\pi f_c R_1} = \frac{1}{2\pi \times 2.5 \times 10^3 \times 6} = 10.61 \,\mu\text{F}
-$$
-
-For the low-pass filter,
-
-$$
-\omega_c = 2\pi f_c = \frac{R_2}{L}
-$$
-
-$$
-L = \frac{R_2}{2\pi f_c} = \frac{6}{2\pi \times 2.5 \times 10^3} = 382 \,\mu\text{H}
-$$
-
-If each speaker in Fig. 14.66 has an 8-Ω resistance and *C* = 10*μ*F, find *L* Practice Problem 14.19 and the crossover frequency.
-
-**Answer:** 0.64 mH, 1.989 kHz.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/171_14.13 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/171_14.13 Summary.md
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-# **14.13** Summary
-
-- 1. The transfer function **H**(*ω*) is the ratio of the output response **Y**(*ω*) to the input excitation **X**(*ω*); that is, **H**(*ω*) = **Y**(*ω*)∕**X**(*ω*).
-- 2. The frequency response is the variation of the transfer function with frequency.
-- 3. Zeros of a transfer function **H**(*s*) are the values of *s* = *jω* that make *H*(*s*) = 0, while poles are the values of *s* that make *H*(*s*) → ∞.
-- 4. The decibel is the unit of log arithmic gain. For a voltage or current gain *G*, its decibel equivalent is *G*dB = 20 log10 *G*.
-- 5. Bode plots are semilog plots of the magnitude and phase of the transfer function as it v aries with frequenc y. The straight-line approximations of *H* (in dB) and *ϕ* (in degrees) are constructed using the corner frequencies defined by the poles and zeros of **H**(*ω*).
-- 6. The resonant frequency is that frequency at which the imaginary part of a transfer function vanishes. For series and parallel *RLC* circuits.
-
-$$
-\omega_0=\frac{1}{\sqrt{LC}}
-$$
-
-7. The half-power frequencies (*ω*1, *ω*2) are those frequencies at which the power dissipated is one-half of that dissipated at the resonant frequency. The geometric mean between the half-power frequencies is the resonant frequency, or
-
-$$
-\omega_0 = \sqrt{\omega_1 \omega_2}
-$$
-
-8. The bandwidth is the frequenc y band between half-po wer frequencies:
-
-$$
-B=\omega_2-\omega_1
-$$
-
-9. The quality f actor is a measure of the sharpness of the resonance peak. It is the ratio of the resonant (angular) frequency to the bandwidth,
-
-$$
-Q = \frac{\omega_0}{B}
-$$
-
-- 10. A filter is a circuit designed to pass a band of frequencies and reject others. Passive filters are constructed with resistors, capacitors, and inductors. Active filters are constructed with resistors, capacitors, and an active device, usually an op amp.
-- 11. Four common types of filters are low-pass, high-pass, band-pass, and band-stop. A low-pass filter passes only signals whose frequencies are below the cutoff frequency *ωc*. A high-pass filter passes only signals whose frequencies are abo ve the cutof f frequency *ωc*. A band-pass filter passes only signals whose frequencies are within a prescribed
-
-range (*ω*1 < *ω*< *ω*2). A band-stop filter passes only signals whose frequencies are outside a prescribed range (*ω*1 > *ω*> *ω*2).
-
-12. Scaling is the process whereby unrealistic element v alues are magnitude-scaled by a factor *Km* and/or frequency-scaled by a factor *Kf* to produce realistic values.
-
-$$
-R' = K_m R, \qquad L' = \frac{K_m}{K_f} L, \qquad C' = \frac{1}{K_m K_f} C
-$$
-
-- 13. *PSpice* can be used to obtain the frequency response of a circuit if a frequency range for the response and the desired number of points within the range are specified in the AC Sweep.
-- 14. The radio receiver—one practical application of resonant circuits employs a band-pass resonant circuit to tune in one frequenc y among all the broadcast signals picked up by the antenna.
-- 15. The touch-tone telephone and the crosso ver network are tw o typical applications of filters. The touch-tone telephone system employs filters to separate tones of different frequencies to activate electronic switches. The crossover network separates signals in dif ferent frequency ranges so that they can be delivered to different devices such as tweeters and woofers in a loudspeaker system.
-
-# Review Questions
-
-**14.1** A zero of the transfer function
-
-$$
-H(s) = \frac{10(s + 1)}{(s + 2)(s + 3)}
-$$
-
-is at
-
-(a) 10 (b) −1 (c) −2 (d) −3
-
-- **14.2** On the Bode magnitude plot, the slope of 1∕(5 + *jω*) 2 for large values of *ω* is
-- (a) 20 dB/decade (b) 40 dB/decade
-
-(c) −40 dB/decade (d) −20 dB/decade
-
- **14.3** On the Bode phase plot for 0.5 < *ω*< 50, the slope of [1 + *j*10*ω*− *ω*2 ∕25]2 is
-
-| (a) 45°/decade | (b) 90°/decade |
-|----------------|----------------|
-| | |
-
-- (c) 135°/decade (d) 180°/decade
- - **14.4** How much inductance is needed to resonate at 5 kHz with a capacitance of 12 nF?
-
-| (a) 2,652 H | (b) 11.844 H |
-|-------------|--------------|
-| (c) 3.333 H | (d) 84.43 mH |
-
- **14.5** The difference between the half-power frequencies is called the:
-
-| (a) quality factor | (b) resonant frequency |
-|--------------------|------------------------|
-|--------------------|------------------------|
-
-- (c) bandwidth (d) cutoff frequency
- - **14.6** In a series *RLC* circuit, which of these quality factors has the steepest magnitude response curve near resonance?
-
-| (a) Q = 20 | (b) Q = 12 |
-|------------|------------|
-| (c) Q = 8 | (d) Q = 4 |
-
- **14.7** In a parallel *RLC* circuit, the bandwidth *B* is directly proportional to *R*.
-
-(a) True (b) False
-
- **14.8** When the elements of an *RLC* circuit are both magnitude-scaled and frequency-scaled, which quality is unaffected?
-
-| (a) resistor | (b) resonant frequency |
-|---------------|------------------------|
-| (c) bandwidth | (d) quality factor |
-
- **14.9** What kind of filter can be used to select a signal of one particular radio station?
-
-(a) low-pass (b) high-pass
-
-- (c) band-pass (d) band-stop
-- **14.10** A voltage source supplies a signal of constant amplitude, from 0 to 40 kHz, to an RC low-pass filter. A load resistor, connected in parallel across the capacitor, experiences the maximum voltage at:
-
-| (a) dc | (b) 10 kHz |
-|------------|------------|
-| (c) 20 kHz | (d) 40 kHz |
-
-*Answers: 14.1b, 14.2c, 14.3d, 14.4d, 14.5c, 14.6a, 14.7b, 14.8d, 14.9c, 14.10a.*
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/172_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/172_Review Questions.md
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-# Problems
-
-# Section 14.2 Transfer Function
-
-**14.1** Find the transfer function *Io*∕*Ii* of the *RL* circuit in Fig. 14.68. Express it using *ω*0 = *R*∕*L*.
-
-**Figure 14.68** For Prob. 14.1.
-
-**14.2** Using Fig. 14.69, design a problem to help other students better understand how to determine transfer functions.
-
-# **Figure 14.69**
-
-- For Prob. 14.2.
-- **14.3** For the circuit shown in Fig. 14.70, find **H**(*s*) = **V***o*(*s*)/**I***i* (*s*).
-
-# **Figure 14.70**
-
-For Prob. 14.3.
-
-**14.4** Find the transfer function **H**(*s*) = **V***o*∕**V***i* of the circuit shown in Fig. 14.71.
-
-**14.5** For the circuit shown in Fig. 14.72, find **H**(*s*) = **V***o*∕**I***s*.
-
-# **Figure 14.72** For Prob. 14.5.
-
-**14.6** For the circuit shown in Fig. 14.73, find **H**(*s*) = **V***o*(*s*)∕**V***s*(*s*).
-
-**Figure 14.73** For Prob. 14.6.
-
-# Section 14.3 The Decibel Scale
-
-**14.7** Calculate ∣H(*ω*)∣ if *H*dB equals
-
-(a) 0.1 dB (b) −5 dB (c) 215 dB
-
-**14.8** Design a problem to help other students calculate the magnitude in dB and phase in degrees of a variety of transfer functions at a single value of *ω*.
-
-# Section 14.4 Bode Plots
-
-**14.9** A ladder network has a voltage gain of
-
-etwork has a voltage gain of
-
-\n
-$$
-\mathbf{H}(\omega) = \frac{10}{(1 + j\omega)(10 + j\omega)}
-$$
-
-Sketch the Bode plots for the gain.
-
-- **14.10** Design a problem to help other students better understand how to determine the Bode magnitude and phase plots of a given transfer function in terms of *jω*.
-- **14.11** Sketch the Bode plots for
-
-$$
-\mathbf{H}(\omega) = \frac{0.2(10 + j\omega)}{j\omega(2 + j\omega)}
-$$
-
- **14.12** A transfer function is given by
-
-$$
-T(s) = \frac{100(s+10)}{s(s+10)}
-$$
-
-Sketch the magnitude and phase Bode plots.
-
- **14.13** Construct the Bode plots for
-
-$$
-G(s) = \frac{0.1(s+1)}{s^2(s+10)}, \qquad s = j\omega
-$$
-
- **14.14** Draw the Bode plots for
-
-the Bode plots for
-\n
-$$
-\mathbf{H}(\omega) = \frac{250(j\omega + 1)}{j\omega(-\omega^2 + 10j\omega + 25)}
-$$
-
- **14.15** Construct the Bode magnitude and phase plots for
-
-$$
-H(s) = \frac{2(s+1)}{(s+2)(s+10)}, \qquad s = j\omega
-$$
-
- **14.16** Sketch Bode magnitude and phase plots for
-
-h Bode magnitude and phase pl
-$$
-H(s) = \frac{1.6}{s(s^2 + s + 16)}, \quad s = j\omega
-$$
-
- **14.17** Sketch the Bode plots for
-
-Let the Bode plots for
-
-\n
-$$
-G(s) = \frac{s}{(s+2)^2(s+1)}, \qquad s = j\omega
-$$
-
-**14.18** A linear network has this transfer function
-
-ar network has this transfer function
-\n
-$$
-H(s) = \frac{7s^2 + s + 4}{s^3 + 8s^2 + 14s + 5}, \qquad s = j\omega
-$$
-
- Use *MATLAB* or equivalent to plot the magnitude and phase (in degrees) of the transfer function. Take 0.1 < *ω*< 10 rad/s.
-
-**14.19** Sketch the asymptotic Bode plots of the magnitude and phase for \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*s* + 10)(*s* + 20)(*s* + 40) , *s* = *jω*
-
-$$
-H(s) = \frac{80s}{(s+10)(s+20)(s+40)}, \qquad s =
-$$
-
-**14.20** Design a more complex problem than given in Prob. 14.10, to help other students better understand how to determine the Bode magnitude and phase plots of a given transfer function in terms of *jω*. Include at least a second order repeated root.
-
-**14.21** Sketch the magnitude Bode plot for
-
-ch the magnitude Bode plot for
-\n
-$$
-H(s) = \frac{10s(s + 20)}{(s + 1)(s^2 + 60s + 400)}, \qquad s = j\omega
-$$
-
-**14.22** Find the transfer function **H**(*ω*) with the Bode magnitude plot shown in Fig. 14.74.
-
-**Figure 14.74** For Prob. 14.22.
-
-**14.23** The Bode magnitude plot of **H**(*ω*) is shown in Fig. 14.75. Find **H**(*ω*).
-
-# **Figure 14.75**
-
-- For Prob. 14.23.
- - **14.24** The magnitude plot in Fig. 14.76 represents the transfer function of a preamplifier. Find *H*(*s*).
-
-For Prob. 14.24.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/173_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/173_Problems.md
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-## Problems **665**
-
-# Section 14.5 Series Resonance
-
-- **14.25** A series *RLC* network has *R* = 2 kΩ, *L* = 40 mH, and *C* = 1*μ*F. Calculate the impedance at resonance and at one-fourth, one-half, twice, and four times the resonant frequency.
-
-**14.26** Design a problem to help other students better understand *ω*0, *Q*, and *B* at resonance in series *RLC* circuits.
-
-- **14.27** Design a series *RLC* resonant circuit with *ω*0 = 40 rad/s and *B* = 10 rad/s.
-- **14.28** Design a series *RLC* circuit with *B* = 20 rad/s and *ω*0 = 1,000 rad/s. Find the circuit's *Q*. Let *R* = 10 Ω.
-- **14.29** Let *vs* = 20 cos(*at*) V in the circuit of Fig. 14.77. Find *ω*0, *Q*, and *B*, as seen by the capacitor.
-
-**Figure 14.77** For Prob. 14.29.
-
-- **14.30** A circuit consisting of a coil with inductance 10 mH and resistance 20 Ω is connected in series with a capacitor and a generator with an rms voltage of 120 V. Find:
- - (a) the value of the capacitance that will cause the circuit to be in resonance at 15 kHz
- - (b) the current through the coil at resonance
- - (c) the *Q* of the circuit
-
-# Section 14.6 Parallel Resonance
-
-- **14.31** Design a parallel resonant *RLC* circuit with *ω*0 = 100 krad/s and a bandwidth of 10 krad/s. Additionally what is the value of *Q*?
-- **14.32** Design a problem to help other students better understand the quality factor, the resonant frequency, and bandwidth of a parallel *RLC* circuit.
-- **14.33** A parallel resonant circuit with a bandwidth of 40 krad/s and the half-power frequencies are *ω*1 = 4.98 Mrad/s and *ω*2 = 5.02 Mrad/s, calculate the quality factor and resonant frequency.
-- **14.34** A parallel *RLC* circuit has *R* = 100 kΩ, *L* = 100 mH, and a *C* = 10 *μ*F. Determine the value of *Q*, the resonant frequency, and the bandwidth. If
-
-*R* = 200 kΩ, how does that affect the values of *Q*, resonant frequency, and the bandwidth?
-
-- **14.35** A parallel *RLC* circuit has *R* = 10 kΩ, *L* = 100 mH, and a resonant frequency of 200 krad/s. Calculate the value of *C*, the value of the quality factor, and the bandwidth.
-- **14.36** It is expected that a parallel *RLC* resonant circuit has a midband admittance of 25 × 10−3 S, quality factor of 120, and a resonant frequency of 200 krad/s. Calculate the values of *R*, *L*, and *C*. Find the bandwidth and the half-power frequencies.
-- **14.37** Rework Prob. 14.25 if the elements are connected in parallel.
-- **14.38** Find the resonant frequency of the circuit in Fig. 14.78.
-
-**Figure 14.78** For Prob. 14.38.
-
-**14.39** For the "tank" circuit in Fig. 14.79, find the resonant frequency.
-
-# **Figure 14.79**
-
-- **14.40** A parallel resonance circuit has a resistance of 2 kΩ and half-power frequencies of 86 kHz and 90 kHz. Determine:
- - (a) the capacitance
- - (b) the inductance
- - (c) the resonant frequency
- - (d) the bandwidth
- - (e) the quality factor
-- **14.41** Using Fig. 14.80, design a problem to help other students better understand the quality factor, the resonant frequency, and bandwidth of *RLC* circuits.
-
-**Figure 14.80**
-
-For Prob. 14.41.
-
-**14.42** For the circuits in Fig. 14.81, find the resonant frequency *ω*0, the quality factor *Q*, and the bandwidth *B*.
-
-For Prob. 14.42.
-
-**14.43** Calculate the resonant frequency of each of the circuits in Fig. 14.82.
-
-**Figure 14.82** For Prob. 14.43.
-
-**\*14.44** For the circuit in Fig. 14.83, find:
-
-(a) the resonant frequency *ω*0
-
-(b) **Z**in(*ω*0)
-
-**Figure 14.83** For Prob. 14.44.
-
-> **14.45** For the circuit shown in Fig. 14.84, find *ω*0, *B*, and *Q*, as seen by the voltage across the inductor.
-
-**Figure 14.84** For Prob. 14.45.
-
-- **14.46** For the network illustrated in Fig. 14.85, find
- - (a) the transfer function **H**(*ω*) = **V***o*(*ω*)∕**I**(*ω*),
-
-(b) the magnitude of **H** at *ω*0 = 1 rad/s.
-
-**Figure 14.85**
-
-For Probs. 14.46, 14.78, and 14.92.
-
-# Section 14.7 Passive Filters
-
-- **14.47** Show that a series *LR* circuit is a low-pass filter if the output is taken across the resistor. Calculate the corner frequency *fc* if *L* = 2 mH and *R* = 10 kΩ.
-- **14.48** Find the transfer function **V***o*∕**V***s* of the circuit in Fig. 14.86. Show that the circuit is a low-pass filter.
-
-For Prob. 14.48.
-
-**14.49** Design a problem to help other students better understand low-pass filters described by transfer functions.
-
-**14.50** Determine what type of filter is in Fig. 14.87. Calculate the corner frequency *fc*.
-
-**Figure 14.87** For Prob. 14.50.
-
-\* An asterisk indicates a challenging problem.
-
-Problems **667**
-
-- **14.55** Determine the range of frequencies that will be passed by a series *RLC* band-pass filter with *R* = 10 Ω, *L* = 25 mH, and *C* = 0.4*μ*F. Find the quality factor.
-- **14.56** (a) Show that for a band-pass filter,
-
-$$
-\mathbf{H}(s) = \frac{sB}{s^2 + sB + \omega_0^2}, \qquad s = j\omega
-$$
-
- where *B* = bandwidth of the filter and *ω*0 is the center frequency.
-
-(b) Similarly, show that for a band-stop filter,
-
-$$
-\mathbf{H}(s) = \frac{s^2 + \omega_0^2}{s^2 + s + \omega_0^2}, \qquad s = j\omega
-$$
-
-**14.57** Determine the center frequency and bandwidth of the band-pass filters in Fig. 14.88.
-
-# **Figure 14.88**
-
-For Prob. 14.57.
-
-- **14.58** The circuit parameters for a series *RLC* bandstop filter are *R* = 250 Ω, *L* = 1 mH, *C* = 40 pF. Calculate:
- - (a) the center frequency
- - (b) the half-power frequencies
- - (c) the quality factor
-- **14.59** Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89.
-
-# **Figure 14.89** For Prob. 14.59.
-
-# Section 14.8 Active Filters
-
-- **14.60** Obtain the transfer function of a high-pass filter with a passband gain of 100 and a cutoff frequency of 40 rad/s.
-- **14.61** Find the transfer function for each of the active filters in Fig. 14.90.
-
-# **Figure 14.90**
-
-**14.62** The filter in Fig. 14.90(b) has a 3-dB cutoff frequency at 1 kHz. If its input is connected to a 120-mV variable frequency signal, find the output voltage at:
-
-(a) 200 Hz (b) 2 kHz (c) 10 kHz
-
-**14.63** Design an active first-order high-pass filter with
-
-$$
-\mathbf{H}(s) = -\frac{100s}{s+10}, \qquad s = j\omega
-$$
-
-Use a 1-*μ*F capacitor.
-
-**14.64** Obtain the transfer function of the active filter in Fig. 14.91 on the next page. What kind of filter is it?
-
-For Prob. 14.64.
-
-**14.65** A high-pass filter is shown in Fig. 14.92. Show that the transfer function is
-
-**Figure 14.92** For Prob. 14.65.
-
-- **14.66** A "general" first-order filter is shown in Fig. 14.93.
- - (a) Show that the transfer function is
-
-(a) Show that the transfer function is
-\n
-$$
-\mathbf{H}(s) = \frac{R_4}{R_3 + R_4} \times \frac{s + (1/R_1C)[R_1/R_2 - R_3/R_4]}{s + 1/R_2C},
-$$
-\n
-$$
-s = j\omega
-$$
-
-- (b) What condition must be satisfied for the circuit to operate as a high-pass filter?
-- (c) What condition must be satisfied for the circuit to operate as a low-pass filter?
-
-**14.67** Design an active low-pass filter with dc gain of 0.25 and a corner frequency of 500 Hz.
-
- **14.68** Design a problem to help other students better understand the design of active high-pass filters when specifying a high-frequency gain and a corner frequency.
-
-**14.69** Design the filter in Fig. 14.94 to meet the following requirements:
-
-- (a) It must attenuate a signal at 2 kHz by 3 dB compared with its value at 10 MHz.
-- (b) It must provide a steady-state output of *vo*(*t*) = 10 sin(2*π* × 108 *t* + 180°) V for an input *vs*(*t*) = 4 sin(2*π* × 108 *t*) V.
-
-# **Figure 14.94**
-
-For Prob. 14.69.
-
-- **\*14.70** A second-order active filter known as a Butterworth filter is shown in Fig. 14.95.
- - (a) Find the transfer function **V***o*∕**V***i*.
- - (b) Show that it is a low-pass filter.
-
-# **Figure 14.95**
-
-For Prob. 14.70.
-
-# Section 14.9 Scaling
-
-**14.71** Use magnitude and frequency scaling on the circuit of Fig. 14.79 to obtain an equivalent circuit in which the inductor and capacitor have magnitude 1 H and 1 F respectively.
-
-**14.72** Design a problem to help other students better understand magnitude and frequency scaling.
-
-**14.73** Calculate the values of *R*, *L*, and *C* that will result in *R* = 12 kΩ, *L* = 40*μ*H, and *C* = 300 nF respectively when magnitude-scaled by 800 and frequency-scaled by 1000.
-
-Problems **669**
-
-- **14.74** A circuit has *R*1 = 3 Ω, *R*2 = 10 Ω, *L* = 2H, and *C* = 1∕10 F. After the circuit is magnitude-scaled by 100 and frequency-scaled by 106 , find the new values of the circuit elements.
-- **14.75** In an *RLC* circuit, *R* = 20 Ω, *L* = 4 H, and *C* = 1 F. The circuit is magnitude-scaled by 10 and frequency-scaled by 105 . Calculate the new values of the elements.
-- **14.76** Given a parallel *RLC* circuit with *R* = 5 kΩ, *L* = 10 mH, and *C* = 20*μ*F, if the circuit is magnitude-scaled by *Km* = 500 and frequencyscaled by *Kf* = 105 , find the resulting values of *R*, *L*, and *C*.
-- **14.77** A series *RLC* circuit has *R* = 10 Ω, *ω*0 = 40 rad/s, and *B* = 5 rad/s. Find *L* and *C* when the circuit is scaled:
- - (a) in magnitude by a factor of 600,
- - (b) in frequency by a factor of 1,000,
- - (c) in magnitude by a factor of 400 and in frequency by a factor of 105 .
-- **14.78** Redesign the circuit in Fig. 14.85 so that all resistive elements are scaled by a factor of 1,000 and all frequency-sensitive elements are frequency-scaled by a factor of 104 .
-- **\*14.79** Refer to the network in Fig. 14.96.
- - (a) Find **Z**in(*s*).
- - (b) Scale the elements by *Km* = 10 and *Kf* = 100. Find **Z**in(*s*) and *ω*0.
-
-# **Figure 14.96**
-
-For Prob. 14.79.
-
-- **14.80** (a) For the circuit in Fig. 14.97, draw the new circuit after it has been scaled by *Km* = 200 and *Kf* = 104 .
- - (b) Obtain the Thevenin equivalent impedance at terminals *a-b* of the scaled circuit at *ω* = 104 rad/s.
-
-**Figure 14.97** For Prob. 14.80.
-
- **14.81** The circuit shown in Fig. 14.98 has the impedance
-
-The circuit shown in Fig. 14.98 has the impedance
-$$
-Z(s) = \frac{1,000(s + 1)}{(s + 1 + j50)(s + 1 - j50)}, \qquad s = j\omega
-$$
-
-Find:
-
-- (a) the values of *R*, *L*, *C*, and *G*
-- (b) the element values that will raise the resonant frequency by a factor of 103 by frequency scaling
-
-# **Figure 14.98**
-
-For Prob. 14.81.
-
-**14.82** Scale the low-pass active filter in Fig. 14.99 so that its corner frequency increases from 1 rad/s to 200 rad/s. Use a 1-*μ*F capacitor.
-
-# **Figure 14.99**
-
-For Prob. 14.82.
-
-**14.83** The op amp circuit in Fig. 14.100 is to be magnitude-scaled by 100 and frequency-scaled by 105 . Find the resulting element values.
-
-# **Figure 14.100**
-
-For Prob. 14.83.
-
-# Section 14.10 Frequency Response Using PSpice
-
-**14.84** Using *PSpice or MultiSim,* obtain the frequency response of the circuit in Fig. 14.101 on the next page.
-
-# **Figure 14.101**
-
-**14.85** Use *PSpice or MultiSim* to obtain the magnitude and phase plots of **V***o*∕**I***s* of the circuit in Fig. 14.102.
-
-# **Figure 14.102**
-
-For Prob. 14.85.
-
-**14.86** Using Fig. 14.103, design a problem to help other students better understand how to use *PSpice* to obtain the frequency response (magnitude and phase of I) in electrical circuits.
-
-**Figure 14.103**
-
-- For Prob. 14.86.
- - **14.87** In the interval 0.1 < *f* < 100 Hz, plot the response of the network in Fig. 14.104. Classify this filter and obtain *ω*0.
-
-**Figure 14.104** For Prob. 14.87.
-
-**14.88** Use *PSpice or MultiSim* to generate the magnitude and phase Bode plots of **V***o* in the circuit of Fig. 14.105.
-
-For Prob. 14.88.
-
- **14.89** Obtain the magnitude plot of the response **V***o* in the network of Fig. 14.106 for the frequency interval 100 < *f* < 1,000 Hz.
-
-# **Figure 14.106**
-
-For Prob. 14.89.
-
-- **14.90** Obtain the frequency response of the circuit in Fig. 14.40 (see Practice Problem 14.10). Take *R*1 = *R*2 = 100 Ω, *L* = 2 mH. Use 1 < *f* < 100,000 Hz.
-- **14.91** For the "tank" circuit of Fig. 14.79, obtain the frequency response (voltage across the capacitor) using *PSpice or MultiSim*. Determine the resonant frequency of the circuit.
-- **14.92** Using *PSpice or MultiSim*, plot the magnitude of the frequency response of the circuit in Fig. 14.85.
-
-# Section 14.12 Applications
-
-**14.93** For the phase shifter circuit shown in Fig. 14.107, find H = *Vo*∕*Vs*.
-
-# **Figure 14.107**
-
-For Prob. 14.93.
-
-**14.94** For an emergency situation, an engineer needs to make an *RC* high-pass filter. He has one 10-pF capacitor, one 30-pF capacitor, one 1.8-kΩ resistor, and one 3.3-kΩ resistor available. Find the greatest cutoff frequency possible using these elements.
-
-**14.95** A series-tuned antenna circuit consists of a variable capacitor (40 pF to 360 pF) and a 240-*μ*H antenna coil that has a dc resistance of 12 Ω.
-
-- (a) Find the frequency range of radio signals to which the radio is tunable.
-- (b) Determine the value of *Q* at each end of the frequency range.
-
-**14.96** The crossover circuit in Fig. 14.108 is a low-pass filter that is connected to a woofer. Find the transfer function **H**(*ω*) = **V***o*(*ω*)∕**V***i*(*ω*).
-
-For Prob. 14.96.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/174_Comprehensive Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/174_Comprehensive Problems.md
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-# Comprehensive Problems
-
-- **14.98** A certain electronic test circuit produced a resonant curve with half-power points at 432 Hz and 454 Hz. If *Q* = 20, what is the resonant frequency of the circuit?
-- **14.99** In an electronic device, a series circuit is employed that has a resistance of 100 Ω, a capacitive reactance of 5 kΩ, and an inductive reactance of 300 Ω when used at 2 MHz. Find the resonant frequency and bandwidth of the circuit.
-- **14.100** In a certain application, a simple *RC* low-pass filter is designed to reduce high frequency noise. If the desired corner frequency is 20 kHz and *C* = 0.5*μ*F, find the value of *R*.
-- **14.101** In an amplifier circuit, a simple *RC* high-pass filter is needed to block the dc component while passing the time-varying component. If the desired rolloff frequency is 15 Hz and *C* = 10*μ*F, find the value of *R*.
-- **14.102** Practical *RC* filter design should allow for source and load resistances as shown in Fig. 14.110. Let *R* = 4 kΩ and *C* = 40-nF. Obtain the cutoff frequency when:
-
-**Figure 14.110** For Prob. 14.102.
-
-**14.103** The *RC* circuit in Fig. 14.111 is used for a lead compensator in a system design. Obtain the transfer function of the circuit.
-
-# **Figure 14.111** For Prob. 14.103.
-
- **14.104** A low-quality-factor, double-tuned band-pass filter is shown in Fig. 14.112. Use *PSpice or MultiSim* to generate the magnitude plot of **V***o*(*ω*).
-
-**Figure 14.112** For Prob. 14.104.
-
-# **PART THREE**
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/175_PART 3 - Advanced Circuit Analysis.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/175_PART 3 - Advanced Circuit Analysis.md
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-# Advanced Circuit Analysis
-
-# OUTLINE
-
-- 15 Introduction to the Laplace Transform
-- 16 Applications of the Laplace Transform
-- 17 The Fourier Series
-- 18 Fourier Transform
-- 19 Two-Port Networks
-
-# **chapter**
-
-# 15
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/176_Chapter 15 - Introduction to the Laplace Transform.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/176_Chapter 15 - Introduction to the Laplace Transform.md
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@@ -1,42 +0,0 @@
-# Introduction to the Laplace Transform
-
-*The important thing about a problem is not its solution, but the strength we gain in finding the solution.*
-
-—Anonymous
-
-# Enhancing Your Skills and Your Career
-
-# **ABET EC 2000 criteria (3.h), "the broad education necessary to understand the impact of engineering solutions in a global and societal context."**
-
-As a student, you must mak e sure you acquire "the broad education necessary to understand the impact of engineering solutions in a global and societal context." To some extent, if you are already enrolled in an ABET-accredited engineering program, then some of the courses you are required to take must meet this criteria. My recommendation is that even if you are in such a program, you look at all the elective courses you take to make sure that you e xpand your awareness of global issues and societal concerns. The engineers of the future must fully understand that they and their activities affect all of us in one way or another.
-
-# **ABET EC 2000 criteria (3.i), "need for, and an ability to engage in life-long learning."**
-
-You must be fully aware of and recognize the "need for, and an ability to engage in life-long learning. " It almost seems absurd that this need and ability must be stated. Yet, you w ould be surprised at ho w m any engineers do not really understand this concept. The only way to be really able to keep up with the explosion in technology we are facing now and will be f acing in the future is through constant learning. This learning must include nontechnical issues as well as the latest technology in your field.
-
-The best w ay to k eep up with the state of the art in your field is through your colleagues and association with indi viduals you meet through your technical organization or organizations (especially IEEE). Reading state-of-the-art technical articles is the next best way to stay current.
-
-Photo by Charles Alexander
-
-**Pierre Simon Laplace** (1749–1827), a French astronomer and mathematician, first presented the transform that bears his name and its applications to differential equations in 1779.
-
-Born of humble origins in Beaumont-en-Auge, Normandy, France, Laplace became a professor of mathematics at the age of 20. His math ematical abilities inspired the f amous mathematician Simeon Poisson, who called Laplace the Isaac Newton of France. He made important contributions in potential theory, probability theory, astronomy, and celestial mechanics. He w as widely kno wn for his w ork, *Traite de Mecanique Celeste (Celestial Mechanics)*, which supplemented the work of Newton on astronomy . The Laplace transform, the subject of this chapter , is named after him.
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Understand the Laplace transform, its importance in circuit analysis, and how to determine the Laplace transform of functions common to circuit analysis.
-- 2. Understand the properties of the Laplace transform.
-- 3. Understand the inverse Laplace transform and how to determine its given functions in the s-domain.
-- 4. Understand the convolution integral and how to use it in the time domain and its equivalence in the s-domain.
-
-# **15.1** Introduction
-
-Our goal in this and the follo wing chapters is to develop techniques for analyzing circuits with a wide variety of inputs and responses. Such circuits are modeled by *differential equations* whose solutions describe the total response behavior of the circuits. Mathematical methods have been devised to systematically determine the solutions of dif ferential equa tions. We now introduce the po werful method of *Laplace transformation*, which involves turning differential equations into *algebraic equations*, thus greatly facilitating the solution process.
-
-The idea of transformation should be f amiliar by now. When using phasors for the analysis of circuits, we transform the circuit from the time domain to the frequency or phasor domain. Once we obtain the phasor result, we transform it back to the time domain. The Laplace transform method follows the same process: We use the Laplace transformation to transform the circuit from the time domain to the frequenc y domain, obtain the solution, and apply the inverse Laplace transform to the result to transform it back to the time domain.
-
-The Laplace transform is significant for a number of reasons. First, it can be applied to a wider variety of inputs than phasor analysis. Second, it provides an easy way to solve circuit problems involving initial conditions, because it allo ws us to w ork with algebraic equations instead of differential equations. Third, the Laplace transform is capable of providing us, in one single operation, the total response of the circuit comprising both the natural and forced responses.
-
-We begin with the definition of the Laplace transform which gives rise to its most essential properties. By e xamining these properties, we shall see ho w and wh y the method w orks. This also helps us to better appreciate the idea of mathematical transformations. We also consider some properties of the Laplace transform that are very helpful in circuit analysis. We then consider the inverse Laplace transform, transfer functions, and convolution. In this chapter , we will focus on the mechanics of the Laplace transformation. In Chapter 16 we will e xamine how the Laplace transform is applied in circuit analysis, netw ork stability, and network synthesis.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/177_15.2 Definition of the Laplace Transform.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/177_15.2 Definition of the Laplace Transform.md
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@@ -1,129 +0,0 @@
-# **15.2** Definition of the Laplace Transform
-
-Given a function *f*(*t*), its Laplace transform, denoted by *F*(*s*) or [ *f*(*t*)], is defined by
-
-$$
-\mathcal{L}[f(t)] = F(s) = \int_{0^{-}}^{\infty} f(t)e^{-st} dt
-$$
- (15.1)
-
-where *s* is a complex variable given by
-
-$$
-s = \sigma + j\omega \tag{15.2}
-$$
-
-Because the argument *st* of the exponent *e* in Eq. (15.1) must be dimensionless, it follows that *s* has the dimensions of frequenc y and units of inverse seconds (s −1 ) or "frequenc y." In Eq. (15.1), the lo wer limit is specified as 0− to indicate a time just before *t* = 0. We use 0− as the lower limit to include the origin and capture an y discontinuity of *f*(*t*) at *t* = 0; this will accommodate functions—such as singularity functions—that may be discontinuous at *t* = 0.
-
-It should be noted that the integral in Eq. (15.1) is a definite integral with respect to time. Hence, the result of inte gration is indepen dent of time and only involves the variable "*s*."
-
-Equation (15.1) illustrates the general concept of transformation. The function *f*(*t*) is transformed into the function *F*(*s*). Whereas the former function involves *t* as its argument, the latter involves *s*. We say the transformation is from *t*-domain to *s*-domain. Gi ven the interpre tation of *s* as frequenc y, we arri ve at the follo wing description of the Laplace transform:
-
-The Laplace transform is an integral transformation of a function f(t) from the time domain into the complex frequency domain, giving F(s).
-
-When the Laplace transform is applied to circuit analysis, the differential equations represent the circuit in the time domain. The terms in the differential equations take the place of *f*(*t*). Their Laplace transform, which corresponds to *F*(*s*), constitutes algebraic equations representing the circuit in the frequency domain.
-
-For an ordinary function f(t), the lower limit can be replaced by 0.
-
-We assume in Eq. (15.1) that *f* (*t*) is ignored for *t* < 0. To ensure that this is the case, a function is often multiplied by the unit step. Thus, *f* (*t*) is written as *f* (*t*)*u*(*t*) or *f* (*t*), *t* ≥ 0.
-
-The Laplace transform in Eq. (15.1) is kno wn as the *one-sided* (or *unilateral*) Laplace transform. The *two-sided* (or *bilateral*) Laplace transform is given by
-
-$$
-F(s) = \int_{-\infty}^{\infty} f(t)e^{-st} dt
-$$
- (15.3)
-
-The one-sided Laplace transform in Eq. (15.1), being adequate for our purposes, is the only type of Laplace transform that we will treat in this book.
-
-A function *f* (*t*) may not ha ve a Laplace transform. F or *f* (*t*) to have a Laplace transform, the inte gral in Eq. (15.1) must con verge to a finite value. Because |*ejωt* | = 1 for any value of *t*, the integral converges when
-
-$$
-\int_0^\infty e^{-\sigma t} |f(t)| \, dt < \infty \tag{15.4}
-$$
-
-for some real v alue *σ* = *σc*. Thus, the re gion of con vergence for the Laplace transform is Re(*s*) = *σ* > *σc*, as shown in Fig. 15.1. In thisregion, |*F*(*s*)| < ∞ and *F*(*s*) exists. *F*(*s*) is undefined outside the region of convergence. Fortunately, all functions of interest in circuit analysis satisfy the convergence criterion in Eq. (15.4) and have Laplace transforms. Therefore, it is not necessary to specify *σc* in what follows.
-
-A companion to the direct Laplace transform in Eq. (15.1) is the *inverse* Laplace transform given by
-
-$$
-\mathcal{L}^{-1}[F(s)] = f(t) = \frac{1}{2\pi j} \int_{\sigma_1 - j\infty}^{\sigma_1 + j\infty} F(s)e^{st} ds
-$$
- (15.5)
-
-where the integration is performed along a straight line (*σ*1 + *jω*, −∞ < *ω* < ∞) in the region of convergence, *σ*1 > *σc*. See Fig. 15.1. The direct application of Eq. (15.5) in volves some kno wledge about comple x analysis beyond the scope of this book. F or this reason, we will not use Eq. (15.5) to find the inverse Laplace transform. We will rather use a look-up table, to be de veloped in Section 15.3. The functions *f*(*t*) and *F*(*s*) are regarded as a Laplace transform pair where
-
-$$
-f(t) \qquad \Leftrightarrow \qquad F(s) \tag{15.6}
-$$
-
-meaning that there is one-to-one correspondence between *f*(*t*) and *F*(*s*). The following examples derive the Laplace transforms of some important functions.
-
-Example 15.1
-
-Determine the Laplace transform of each of the following functions: (a) *u*(*t*), (b) *e*−*at u*(*t*), *a* ≥ 0, and (c) *δ*(*t*).
-
-# **Solution:**
-
-(a) For the unit step function *u*(*t*), shown in Fig. 15.2(a), the Laplace transform is
-
-$$
-\mathcal{L}[u(t)] = \int_{0^{-}}^{\infty} 1e^{-st} dt = -\frac{1}{s} e^{-st} \Big|_{0}^{\infty}
-$$
-
-= $-\frac{1}{s}(0) + \frac{1}{s}(1) = \frac{1}{s}$ (15.1.1)
-
-*ω*t + sin*2*
-
-*ω*t = 1
-
-**Figure 15.1** Region of convergence for the Laplace transform.
-
-|ej*ω*t
-
-| = cos*2*
-
-(b) F or the e xponential function, sho wn in Fig. 15.2(b), the Laplace transform is
-
-$$
-\mathcal{L}[e^{-at} u(t)] = \int_{0^{-}}^{\infty} e^{-at} e^{-st} dt
-$$
-
-= $-\frac{1}{s+a} e^{-(s+a)t} \Big|_{0}^{\infty} = \frac{1}{s+a}$ (15.1.2)
-
-(c) For the unit impulse function, shown in Fig. 15.2(c),
-
-$$
-\mathcal{L}[\delta(t)] = \int_{0^{-}}^{\infty} \delta(t)e^{-st} dt = e^{-0} = 1
-$$
- (15.1.3)
-
-since the impulse function *δ*(*t*) is zero e verywhere except at *t* = 0. The sifting property in Eq. (7.33) has been applied in Eq. (15.1.3).
-
-# **Figure 15.2**
-
-For Example 15.1: (a) unit step function, (b) exponential function, (c) unit impulse function.
-
-Find the Laplace transforms of these functions: *r*(*t*) = *tu*(*t*), that is, the Practice Problem 15.1 ramp function; *Ae*−*at u*(*t*); and *Be*−*jωt u*(*t*).
-
-**Answer:** 1∕*s 2* , *A*∕(*s* + *a*), *B*∕(*s* + *jω*).
-
-Determine the Laplace transform of *f*(*t*) = sin *ωt u*(*t*). Example 15.2
-
-# **Solution:**
-
-Using Eq. (B.27) in addition to Eq. (15.1), we obtain the Laplace trans form of the sine function as
-
-$$
-F(s) = \mathcal{L} \left[ \sin \omega t \right] = \int_0^\infty (\sin \omega t) e^{-st} dt = \int_0^\infty \left( \frac{e^{j\omega t} - e^{-j\omega t}}{2j} \right) e^{-st} dt
-$$
-$$
-= \frac{1}{2j} \int_0^\infty (e^{-(s-j\omega)t} - e^{-(s+j\omega)t}) dt
-$$
-$$
-= \frac{1}{2j} \left( \frac{1}{s - j\omega} - \frac{1}{s + j\omega} \right) = \frac{\omega}{s^2 + \omega^2}
-$$
-
-Find the Laplace transform of *f*(*t*) = 15 cos(3 *t*) using the e xponential Practice Problem 15.2 representation for the cosine function.
-
-**Answer:** 15*s*∕(*s* 2 + 9).
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/178_15.3 Properties of the Laplace Transform.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/178_15.3 Properties of the Laplace Transform.md
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-# **15.3** Properties of the Laplace Transform
-
-The properties of the Laplace transform help us to obtain transform pairs without directly using Eq. (15.1) as we did in Examples 15.1 and 15.2. As we derive each of these properties, we should keep in mind the definition of the Laplace transform in Eq. (15.1).
-
-# **Linearity**
-
-If *F*1(*s*) and *F*2(*s*) are, respectively, the Laplace transforms of *f*1(*t*) and *f*2(*t*), then
-
-$$
-\mathcal{L}[a_1 f_1(t) + a_2 f_2(t)] = a_1 F_1(s) + a_2 F_2(s)
-$$
- (15.7)
-
-where *a*1 and *a*2 are constants. Equation 15.7 e xpresses the linearity property of the Laplace transform. The proof of Eq. (15.7) follows readily from the definition of the Laplace transform in Eq. (15.1).
-
-For example, by the linearity property in Eq. (15.7), we may write
-
-$$
-\mathcal{L}[\cos \omega t \, u(t)] = \mathcal{L} \left[ \frac{1}{2} (e^{j\omega t} + e^{-j\omega t}) \right] = \frac{1}{2} \mathcal{L} [e^{j\omega t}] + \frac{1}{2} \mathcal{L} [e^{-j\omega t}] \tag{15.8}
-$$
-
-But from Example 15.1(b), [*e*−*at*] = 1∕(*s* + *a*). Hence,
-
-$$
-\mathcal{L}[\cos \omega t \, u(t)] = \frac{1}{2} \left( \frac{1}{s - j\omega} + \frac{1}{s + j\omega} \right) = \frac{s}{s^2 + \omega^2}
-$$
- (15.9)
-
-# **Scaling**
-
-If *F*(*s*) is the Laplace transform of *f*(*t*), then
-
-$$
-\mathcal{L}[f(at)] = \int_{0-}^{\infty} f(at)e^{-st} dt
-$$
- (15.10)
-
-where *a* is a constant and *a* > 0. If we let *x* = *at*, *dx* = *a dt*, then
-
-$$
-\mathcal{L}[f(at)] = \int_{0-}^{\infty} f(x) e^{-x(s/a)} \frac{dx}{a} = \frac{1}{a} \int_{0-}^{\infty} f(x) e^{-x(s/a)} dx \tag{15.11}
-$$
-
-Comparing this inte gral with the definition of the Laplace transform in Eq. (15.1) shows that *s* in Eq. (15.1) must be replaced by *s*∕*a* while the dummy v ariable *t* is replaced by *x*. Hence, we obtain the scaling property as
-
-$$
-\mathcal{L}[f(at)] = \frac{1}{a} F(\frac{s}{a})
-$$
-\n(15.12)
-
-For example, we know from Example 15.2 that
-
-$$
-\mathcal{L}[\sin \omega t \, u(t)] = \frac{\omega}{s^2 + \omega^2} \tag{15.13}
-$$
-
-Using the scaling property in Eq. (15.12),
-
-$$
-\mathcal{L}[\sin 2\omega t \, u(t)] = \frac{1}{2} \frac{\omega}{\left(s/2\right)^2 + \omega^2} = \frac{2\omega}{s^2 + 4\omega^2} \tag{15.14}
-$$
-
-which may also be obtained from Eq. (15.13) by replacing *ω* with 2*ω*.
-
-# **Time Shift**
-
-If *F*(*s*) is the Laplace transform of *f*(*t*), then
-
-$$
-\mathcal{L}[f(t-a)u(t-a)] = \int_{0-}^{\infty} f(t-a)u(t-a)e^{-st} dt
-$$
- (15.15)
-
-But *u*(*t* − *a*) = 0 for *t* < *a* and *u*(*t* − *a*) = 1 for *t* > *a*. Hence,
-
-$$
-\mathcal{L}[f(t-a)u(t-a)] = \int_{a}^{\infty} f(t-a)e^{-st} dt
-$$
- (15.16)
-
-If we let *x* = *t* − *a*, then *dx* = *dt* and *t* = *x* + *a*. As *t* → *a*, *x* → 0 and as *t* → ∞, *x* → ∞. Thus,
-
-$$
-\mathcal{L}[f(t-a)u(t-a)] = \int_0^\infty f(x)e^{-s(x+a)} dx
-$$
-$$
-= e^{-as} \int_0^\infty f(x)e^{-sx} dx = e^{-as} F(s)
-$$
-
-or
-
-$$
-\mathcal{L}[f(t-a)u(t-a)] = e^{-as} F(s)
-$$
- (15.17)
-
-In other w ords, if a function is delayed in time by *a*, the result in the *s*-domain is found by multiplying the Laplace transform of the function (without the delay) by *e*−*as*. This is called the *time-delay* or *time-shift property* of the Laplace transform.
-
-As an example, we know from Eq. (15.9) that
-
-$$
-\mathcal{L}[\cos \omega t \, u(t)] = \frac{s}{s^2 + \omega^2}
-$$
-
-Using the time-shift property in Eq. (15.17),
-
-$$
-\mathcal{L}[\cos \omega(t-a)u(t-a)] = e^{-as} \frac{s}{s^2 + \omega^2}
-$$
- (15.18)
-
-# **Frequency Shift**
-
-If *F*(*s*) is the Laplace transform of *f*(*t*), then
-
-$$
-\mathcal{L}[e^{-at}f(t)u(t)] = \int_0^\infty e^{-at}f(t)e^{-st} dt
-$$
-$$
-= \int_0^\infty f(t)e^{-(s+a)t} dt = F(s+a)
-$$
-
-or
-
-$$
-\mathcal{L}[e^{-at}f(t)u(t)] = F(s+a)
-$$
- (15.19)
-
-That is, the Laplace transform of *e*−*at f*(*t*) can be obtained from the Laplace transform of *f*(*t*) by replacing every *s* with *s* + *a*. This is known as *frequency shift* or *frequency translation.*
-
-As an example, we know that
-
-cos *ωt u*(*t*) ⇔ \_\_\_\_\_\_\_ *s s* 2 + *ω*2
-
-and
-
-**(15.20)**
-
-$$
-\sin \omega t \, u(t) \qquad \Leftrightarrow \qquad \frac{\omega}{s^2 + \omega^2}
-$$
-
-Using the shift property in Eq. (15.19), we obtain the Laplace transform of the damped sine and damped cosine functions as
-
-and damped cosine functions as
-\n
-$$
-\mathcal{L}[e^{-at} \cos \omega t \, u(t)] = \frac{s+a}{(s+a)^2 + \omega^2}
-$$
-\n(15.21a)
-
-$$
-(s + a)^{2} + \omega^{2}
-$$
-
-\n
-$$
-\mathcal{L}[e^{-at} \sin \omega t \, u(t)] = \frac{\omega}{(s + a)^{2} + \omega^{2}}
-$$
-\n(15.21b)
-
-# **Time Differentiation**
-
-Given that *F*(*s*) is the Laplace transform of *f*(*t*), the Laplace transform of its derivative is
-
-$$
-\mathcal{L}\left[\frac{df}{dt}u(t)\right] = \int_{0^{-}}^{\infty} \frac{df}{dt} e^{-st} dt
-$$
- (15.22)
-
-To inte grate this by parts, we let *u* = *e*−*st*, *du* = −*se*−*st dt*, and *dv* = (*df*∕*dt*) *dt* = *df*(*t*), *v* = *f*(*t*). Then
-
-$$
-\mathcal{L}\left[\frac{df}{dt}u(t)\right] = f(t)e^{-st}\Big|_{0^{-}}^{\infty} - \int_{0^{-}}^{\infty}f(t)[-se^{-st}] dt
-$$
-
-= 0 - f(0-) + s $\int_{0^{-}}^{\infty}f(t)e^{-st} dt = sF(s) - f(0^{-})$
-
-or
-
-$$
-\mathcal{L}[f'(t)] = sF(s) - f(0^{-})
-$$
-\n(15.23)
-
-The Laplace transform of the second deri vative of *f*(*t*) is a repeated application of Eq. (15.23) as
-
-$$
-\mathcal{L}\left[\frac{d^2f}{dt^2}\right] = s\mathcal{L}[f'(t)] - f'(0^-) = s[sF(s) - f(0^-)] - f'(0^-)
-$$
-
-= $s^2F(s) - sf(0^-) - f'(0^-)$
-
-or
-
-$$
-\mathcal{L}[f''(t)] = s^2 F(s) - sf(0^-) - f'(0^-)
-$$
- (15.24)
-
-Continuing in this manner , we can obtain the Laplace transform of the *n*th derivative of *f*(*t*) as
-
-$$
-\mathcal{L}\left[\frac{d^n f}{dt^n}\right] = s^n F(s) - s^{n-1} f(0^-) - s^{n-2} f'(0^-) - \dots - s^0 f^{(n-1)}(0^-)
-$$
-\n(15.25)
-
-As an example, we can use Eq. (15.23) to obtain the Laplace transform of the sine from that of the cosine. If we let *f*(*t*) = cos *ωt u*(*t*), then *f*(0) = 1 and *f* ′(*t*) = −*ω* sin *ωt u*(*t*). Using Eq. (15.23) and the scaling property,
-
-$$
-\mathcal{L}[\sin \omega t \, u(t)] = -\frac{1}{\omega} \mathcal{L}[f'(t)] = -\frac{1}{\omega} [sF(s) - f(0^{-})]
-$$
-$$
-= -\frac{1}{\omega} \left(s \frac{s}{s^{2} + \omega^{2}} - 1\right) = \frac{\omega}{s^{2} + \omega^{2}}
-$$
-(15.26)
-
-as expected.
-
-# **Time Integration**
-
-If *F*(*s*) is the Laplace transform of *f*(*t*), the Laplace transform of its integral is
-
-$$
-\mathcal{L}\bigg[\int_0^t f(x)dx\bigg] = \int_0^\infty \bigg[\int_0^t f(x)dx\bigg] e^{-st} dt \qquad (15.27)
-$$
-
-To integrate this by parts, we let
-
-$$
-u = \int_0^t f(x)dx, \qquad du = f(t)dt
-$$
-
-and
-
-$$
-dv = e^{-st} dt, \qquad v = -\frac{1}{s}e^{-st}
-$$
-
-Then
-
-$$
-C\left[\int_0^t f(x)dx\right] = \left[\int_0^t f(x)dx\right] \left(-\frac{1}{s}e^{-st}\right)\Big|_0^{\infty}
-$$
-$$
--\int_0^{\infty} \left(-\frac{1}{s}\right)e^{-st}f(t)dt
-$$
-
-For the first term on the right-hand side of the equation, evaluating the term at *t* = ∞ yields zero due to *e*−*s*∞ and evaluating it at *t* = 0 gives \_\_1 *s* ∫ 0 0 *f*(*x*) *dx* = 0. Thus, the first term is zero, and *t* ∞
-
-[∫ 0 *f*(*x*)*dx* ] = \_\_1 *s* ∫ 0− *f*(*t*)*e*−*st dt* = \_\_1 *s F*(*s*)
-
-or simply,
-
-$$
-\mathcal{L}\left[\int_0^t f(x)dx\right] = \frac{1}{s}F(s)
-$$
- (15.28)
-
-As an example, if we let *f*(*t*) = *u*(*t*), from Example 15.1(a), *F*(*s*) = 1∕*s*. Using Eq. (15.28),
-
-$$
-\mathcal{L}\bigg[\int_0^t f(x)dx\bigg] = \mathcal{L}[t] = \frac{1}{s}\bigg(\frac{1}{s}\bigg)
-$$
-
-Thus, the Laplace transform of the ramp function is
-
-$$
-\mathcal{L}[t] = \frac{1}{s^2} \tag{15.29}
-$$
-
-Applying Eq. (15.28), this gives
-
-$$
-\left[\int_0^t x dx\right] = \mathcal{L}\left[\frac{t^2}{2}\right] = \frac{1}{s} \frac{1}{s^2}
-$$
-
-$$
-\mathcal{L}[t^2] = \frac{2}{s^3} \tag{15.30}
-$$
-
-or
-
-Repeated applications of Eq. (15.28) lead to
-
-$$
-\mathcal{L}[t^n] = \frac{n!}{s^{n+1}} \tag{15.31}
-$$
-
-Similarly, using integration by parts, we can show that
-
-$$
-\mathcal{L}\bigg[\int_{-\infty}^{t} f(x)dx\bigg] = \frac{1}{s}F(s) + \frac{1}{s}f^{-1}(0^{-})
-$$
-\n(15.32)
-
-where
-
-$$
-f^{-1}(0^{-}) = \int_{-\infty}^{0^{-}} f(t)dt
-$$
-
-# **Frequency Differentiation**
-
-If *F*(*s*) is the Laplace transform of *f*(*t*), then
-
-$$
-F(s) = \int_{0^-}^{\infty} f(t)e^{-st} dt
-$$
-
-Taking the derivative with respect to *s*,
-
-$$
-\frac{dF(s)}{ds} = \int_{0^{-}}^{\infty} f(t)(-te^{-st}) dt = \int_{0^{-}}^{\infty} (-tf(t))e^{-st} dt = \mathcal{L}[-tf(t)]
-$$
-
-and the frequency differentiation property becomes
-
-$$
-\mathcal{L}[tf(t)] = -\frac{dF(s)}{ds}
-$$
- (15.33)
-
-Repeated applications of this equation lead to
-
-$$
-\mathcal{L}[t^n f(t)] = (-1)^n \frac{d^n F(s)}{ds^n}
-$$
- (15.34)
-
-For example, we know from Example 15.1(b) that [*e* −*at*] = 1∕(*s* + *a*). Using the property in Eq. (15.33),
-
-$$
-\mathcal{L}[te^{-at}u(t)] = -\frac{d}{ds}\left(\frac{1}{s+a}\right) = \frac{1}{(s+a)^2}
-$$
-(15.35)
-
-Note that if *a* = 0, we obtain [*t*] = 1∕*s* 2 as in Eq. (15.29), and repeated applications of Eq. (15.33) will yield Eq. (15.31).
-
-# **Time Periodicity**
-
-If function *f*(*t*) is a periodic function such as sho wn in Fig. 15.3, it can be represented as the sum of time-shifted functions sho wn in Fig. 15.4. Thus,
-
-$$
-f(t) = f_1(t) + f_2(t) + f_3(t) + \cdots
-$$
-
-= $f_1(t) + f_1(t - T)u(t - T)$
-+ $f_1(t - 2T)u(t - 2T) + \cdots$ (15.36)
-
-where *f*1(*t*) is the same as the function *f*(*t*) g ated o ver the interv al 0 < *t* < *T*, that is,
-
-$$
-f_1(t) = f(t)[u(t) - u(t - T)]
-$$
- (15.37a)
-
-A periodic function.
-
-$$
-f_1(t) = \begin{cases} f(t), & 0 < t < T \\ 0, & \text{otherwise} \end{cases} \tag{15.37b}
-$$
-
-We no w transform each term in Eq. (15.36) and apply the time-shift property in Eq. (15.17). We obtain
-
-$$
-F(s) = F_1(s) + F_1(s)e^{-Ts} + F_1(s)e^{-2Ts} + F_1(s)e^{-3Ts} + \cdots
-$$
-
-= $F_1(s)[1 + e^{-Ts} + e^{-2Ts} + e^{-3Ts} + \cdots]$ (15.38)
-
-But
-
-$$
-1 + x + x2 + x3 + \dots = \frac{1}{1 - x}
-$$
- (15.39)
-
-if |*x*| < 1. Hence,
-
-$$
-F(s) = \frac{F_1(s)}{1 - e^{-Ts}} \tag{15.40}
-$$
-
-where *F*1(*s*) is the Laplace transform of *f*1(*t*); in other words, *F*1(*s*) is the transform *f*(*t*) defined over its first period only. Equation (15.40) shows that the Laplace transform of a periodic function is the transform of the first period of the function divided by 1 − *e*−*Ts*.
-
-# **Initial and Final Values**
-
-The initial-value and final-value properties allo w us to find the initial value *f*(0) and the final value *f*(∞) of *f*(*t*) directly from its Laplace transform *F*(*s*). To obtain these properties, we be gin with the dif ferentiation property in Eq. (15.23), namely,
-
-$$
-sF(s) - f(0) = \mathcal{L}\left[\frac{df}{dt}\right] = \int_{0^-}^{\infty} \frac{df}{dt} e^{-st} dt
-$$
- (15.41)
-
-If we let *s* → ∞, the integrand in Eq. (15.41) vanishes due to the damping exponential factor, and Eq. (15.41) becomes
-
-$$
-\lim_{s \to \infty} [sF(s) - f(0)] = 0
-$$
-
-Because *f* (0) is independent of *s*, we can write
-
-$$
-f(0) = \lim_{s \to \infty} sF(s) \tag{15.42}
-$$
-
-This is known as the *initial-value theorem*. For example, we know from Eq. (15.21a) that
-
-5.21a) that
-\n
-$$
-f(t) = e^{-2t} \cos 10t \, u(t) \qquad \Leftrightarrow \qquad F(s) = \frac{s+2}{(s+2)^2 + 10^2} \quad \textbf{(15.43)}
-$$
-
-Using the initial-value theorem,
-
-$$
-f(0) = \lim_{s \to \infty} sF(s) = \lim_{s \to \infty} \frac{s^2 + 2s}{s^2 + 4s + 104}
-$$
-$$
-= \lim_{s \to \infty} \frac{1 + 2/s}{1 + 4/s + 104/s^2} = 1
-$$
-
-which confirms what we would expect from the given *f*(*t*).
-
-In Eq. (15.41), we let *s* → 0; then
-
-$$
-\lim_{s \to 0} [sF(s) - f(0^{-})] = \int_{0^{-}}^{\infty} \frac{df}{dt} e^{0t} dt = \int_{0^{-}}^{\infty} df = f(\infty) - f(0^{-})
-$$
-
-or
-
-$$
-f(\infty) = \lim_{s \to 0} sF(s) \tag{15.44}
-$$
-
-This is referred to as the *final-value theorem*. In order for the final-value theorem to hold, all poles of *F*(*s*) must be located in the left half of the *s* plane (see Fig. 15.1 or 15.9); that is, the poles must have negative real parts. The only e xception to this requirement is the case in which *F*(*s*) has a simple pole at *s* = 0, because the effect of 1∕*s* will be nullified by *sF*(*s*) in Eq. (15.44). For example, from Eq. (15.21b),
-
-$$
-f(t) = e^{-2t} \sin 5t \, u(t) \qquad \Leftrightarrow \qquad F(s) = \frac{5}{(s+2)^2 + 5^2} \quad \textbf{(15.45)}
-$$
-
-Applying the final-value theorem,
-
-$$
-f(\infty) = \lim_{s \to 0} s F(s) = \lim_{s \to 0} \frac{5s}{s^2 + 4s + 29} = 0
-$$
-
-as expected from the given *f*(*t*). As another example,
-
-$$
-f(t) = \sin t \, u(t) \qquad \Leftrightarrow \qquad f(s) = \frac{1}{s^2 + 1}
-$$
- (15.46)
-
-so that
-
-$$
-f(\infty) = \lim_{s \to 0} sF(s) = \lim_{s \to 0} \frac{s}{s^2 + 1} = 0
-$$
-
-This is incorrect, because *f*(*t*) = sin *t* oscillates between +1 and −1 and does not ha ve a limit as *t* → ∞. Thus, the final-value theorem cannot be used to find the final value of *f*(*t*) = sin *t*, because *F*(*s*) has poles at *s* = ±*j*, which are not in the left half of the *s* plane. In gen eral, the final-value theorem does not apply in finding the final values of sinusoidal functions—these functions oscillate forever and do not have final values.
-
-The initial-value and final-value theorems depict the relationship between the origin and infinity in the time domain and the *s*-domain. They serve as useful checks on Laplace transforms.
-
-Table 15.1 provides a list of the properties of the Laplace trans form. The last property (on con volution) will be pro ved in Sec tion 15.5. There are other properties, b ut these are enough for pres ent purposes. Table 15.2 summarizes the Laplace transforms of some common functions. We have omitted the f actor *u*(*t*) except where it is necessary.
-
-We should mention that many software packages, such as Mathcad, *MATLAB*, Maple, and Mathematica, offer symbolic math. For example, Mathcad has symbolic math for the Laplace, F ourier, and Z transforms as well as the inverse function.
-
-| TABLE 15.1 | | | TABLE 15.2 | |
-|--------------------------------------|------------------------|-------------------------------------------------------------------------------|------------------------------------------|-------------------------------------------------------|
-| Properties of the Laplace transform. | | Laplace transform pairs.* | | |
-| Property | f(t) | F(s) | f(t) | F(s) |
-| Linearity | a1f1(t) + a2f2(t) | a1F1(s) + a2F2(s) | δ(t) | 1 |
-| Scaling | f(at) | __1
__s
a F(
a ) | u(t) | __1
s |
-| Time shift | f(t − a)u(t − a) | e−as F(s) | e−at | _____ 1
s + a |
-| Frequency shift | e−at f(t) | F(s + a) | t | __1 |
-| Time differentiation | df
__
dt | sF(s) − f(0−) | t n | 2
s
____ n!
n+1 |
-| | d2
f ___
dt2 | 2
F(s) − sf(0−) −
f ′(0−)
s | te−at | s
_______ 1
2
(s + a) |
-| | d3
f ___
dt3 | 3
2
f(0−) −
sf ′(0−) −
f ″(0−)
s
F(s) − s | n
e−at
t | ________ n!
n+1
(s + a) |
-| | n
d
f ___
dtn | n
n−1 f(0−) −
n−2 f ′(0−)
s
F(s) − s
s
(n−1)(0−)
− ⋯ − f | sin ωt | _______ ω
2
+ ω2
s |
-| Time integration | t
f(x)dx
∫
0 | __1
F(s)
s | cos ωt | _______ s
2
+ ω2
s |
-| Frequency
differentiation | tf(t) | − __d
ds F(s) | sin(ωt + θ) | s sin θ + ω cos θ
______________
2
+ ω2
s |
-| Frequency
integration | f(t) ___
t | ∞
F(s)ds
∫
s | cos(ωt + θ) | s cos θ − ω sin θ
______________
2
+ ω2
s |
-| Time periodicity | f(t) = f(t + nT) | F1(s) _______
1 − e−sT | e−at sin ωt | ___________ ω
2
+ ω2
(s + a) |
-| Initial value | f(0) | s→∞ sF(s)
lim | e−at cos ωt | ___________ s + a
2
+ ω2
(s + a) |
-| Final value | f(∞) | lim
sF(s)
s→0 | | |
-| Convolution | f1(t) * f2(t) | F1(s)F2(s) | *Defined for t ≥ 0; f(t) = 0, for t < 0. | |
-
-Obtain the Laplace transform of *f*(*t*) = *δ*(*t*) + 2*u*(*t*) − 3*e* Example 15.3 −2*t u*(*t*).
-
-# **Solution:**
-
-By the linearity property,
-
-$$
-F(s) = \mathcal{L}[\delta(t)] + 2\mathcal{L}[u(t)] - 3\mathcal{L}[e^{-2t} u(t)]
-$$
-$$
-= 1 + 2\frac{1}{s} - 3\frac{1}{s+2} = \frac{s^2 + s + 4}{s(s+2)}
-$$
-
-Find the Laplace transform of *f*(*t*) = (cos (2*t*) + *e* Practice Problem 15.3 −4*t* )*u*(*t*).
-
-Find the Laplace transform
-\n**Answer:**
-$$
-\frac{2s^2 + 4s + 4}{(s + 4)(s^2 + 4)}
-$$
-
-Example 15.4 Determine the Laplace transform of *f*(*t*) = *t* 2 sin 2*t u*(*t*).
-
-# **Solution:**
-
-We know that
-
-$$
-[\sin 2t] = \frac{2}{s^2 + 2^2}
-$$
-
-Using frequency differentiation in Eq. (15.34),
-
-$$
-F(s) = \mathcal{L}[t^2 \sin 2t] = (-1)^2 \frac{d^2}{ds^2} \left(\frac{2}{s^2 + 4}\right)
-$$
-$$
-= \frac{d}{ds} \left(\frac{-4s}{(s^2 + 4)^2}\right) = \frac{12s^2 - 16}{(s^2 + 4)^3}
-$$
-
-Practice Problem 15.4 Find the Laplace transform of *f*(*t*) = *t* 2 cos 3*t u*(*t*).
-
-**Answer:**
-$$
-\frac{2s(s^2 - 27)}{(s^2 + 9)^3}
-$$
-
-**Answer:**
-
-\_\_\_ 10 *s*
-
-(2 − *e*−4*s* − *e*−8*s*
-
-We can express the gate function in Fig. 15.5 as
-
-$$
-g(t) = 10[u(t-2) - u(t-3)]
-$$
-
-Given that we kno w the Laplace transform of *u*(*t*), we apply the timeshift property and obtain
-
-$$
-G(s) = 10\left(\frac{e^{-2s}}{s} - \frac{e^{-3s}}{s}\right) = \frac{10}{s}(e^{-2s} - e^{-3s})
-$$
-
-The gate function; for Example 15.5.
-
-Practice Problem 15.5 Find the Laplace transform of the function *h*(*t*) in Fig. 15.6.
-
-).
-
-| $h(t)$ | | | |
-|--------|---|---|---|
-| 20 | | | |
-| 10 | | | |
-| 0 | 4 | 8 | t |
-
-\n**Figure 15.6**
-
-For Practice Prob. 15.5.
-
-$$
-\mathbf{36}^{\dagger}
-$$
-
-Calculate the Laplace transform of the periodic function in Fig. 15.7. Example 15.6
-
-# **Solution:**
-
-The period of the function is *T* = 2. To apply Eq. (15.40), we first obtain the transform of the first period of the function.
-
-$$
-f_1(t) = 2t[u(t) - u(t-1)] = 2tu(t) - 2tu(t-1)
-$$
-
-= 2tu(t) - 2(t - 1 + 1)u(t - 1)
-= 2tu(t) - 2(t - 1)u(t-1) - 2u(t-1)
-
-Using the time-shift property,
-
-$$
-F_1(s) = \frac{2}{s^2} - 2\frac{e^{-s}}{s^2} - \frac{2}{s}e^{-s} = \frac{2}{s^2}(1 - e^{-s} - se^{-s})
-$$
-
-Thus, the transform of the periodic function in Fig. 15.7 is
-
-$$
-F(s) = \frac{F_1(s)}{1 - e^{-Ts}} = \frac{2}{s^2(1 - e^{-2s})}(1 - e^{-s} - se^{-s})
-$$
-
-Determine the Laplace transform of the periodic function in Fig. 15.8. Practice Problem 15.6
-
-For Practice Prob. 15.6.
-
-Find the initial and final values of the function whose Laplace trans- Example 15.7 form is
-
-inal values of the function who
-$$
-H(s) = \frac{20}{(s+3)(s^2+8s+25)}
-$$
-
-# **Solution:**
-
-Applying the initial-value theorem,
-
-**lution:**
-\nplying the initial-value theorem,
-\n
-$$
-h(0) = \lim_{s \to \infty} sH(s) = \lim_{s \to \infty} \frac{20s}{(s+3)(s^2+8s+25)}
-$$
-\n
-$$
-= \lim_{s \to \infty} \frac{20/s^2}{(1+3/s)(1+8/s+25/s^2)} = \frac{0}{(1+0)(1+0+0)} = 0
-$$
-
-To be sure that the final-value theorem is applicable, we check where the poles of *H*(*s*) are located. The poles of *H*(*s*) are *s* = −3, −4 ± *j*3, which all have negative real parts: They are all located on the left half of the *s* plane (Fig. 15.9). Hence, the final-value theorem applies and \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ 20*s*
-
-$$
-h(\infty) = \lim_{s \to 0} sH(s) = \lim_{s \to 0} \frac{20s}{(s+3)(s^2+8s+25)}
-$$
-$$
-= \frac{0}{(0+3)(0+0+25)} = 0
-$$
-
-**Figure 15.9** For Example 15.7: Poles of *H*(*s*).
-
-**Figure 15.7** For Example 15.6. Both the initial and final values could be determined from *h*(*t*) if we knew it. See Example 15.11, where *h*(*t*) is given.
-
-Practice Problem 15.7 Obtain the initial and the final values of
-
-he final values of
-\n
-$$
-G(s) = \frac{3s^3 + 2s + 6}{s(s+1)^2(s+1.5)}
-$$
-
-**Answer:** 3, 4.
-
-# **15.4** The Inverse Laplace Transform
-
-Given *F*(*s*), how do we transform it back to the time domain and obtain the corresponding *f*(*t*)? By matching entries in Table 15.2, we a void using Eq. (15.5) to find *f*(*t*).
-
-Suppose *F*(*s*) has the general form of
-
-$$
-F(s) = \frac{N(s)}{D(s)}
-$$
-(15.47)
-
-where *N*(*s*) is the numerator polynomial and *D*(*s*) is the denominator polynomial. The roots of *N*(*s*) = 0 are called the *zeros* of *F*(*s*), while the roots of *D*(*s*) = 0 are the *poles* of *F*(*s*). Although Eq. (15.47) is similar in form to Eq. (14.3), here *F*(*s*) is the Laplace transform of a function, which is not necessarily a transfer function. We use *partial fraction expansion* to break *F*(*s*) down into simple terms whose inverse transform we obtain from Table 15.2. Thus, finding the inverse Laplace transform of *F*(*s*) involves two steps.
-
-Steps to Find the Inverse Laplace Transform:
-
-- 1. Decompose *F*(*s*) into simple terms using partial fraction expansion.
-- 2. Find the inverse of each term by matching entries in Table 15.2.
-
-Let us consider the three possible forms *F*(*s*) may take and how to apply the two steps to each form.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/179_15.4 The Inverse Laplace Transform.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/179_15.4 The Inverse Laplace Transform.md
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-# **15.4.1** Simple Poles
-
-Recall from Chapter 14 that a simple pole is a first-order pole. If *F*(*s*) has only simple poles, then *D*(*s*) becomes a product of factors, so that
-
-then
-$$
-D(s)
-$$
- becomes a product of factors, so that
-\n
-$$
-F(s) = \frac{N(s)}{(s+p_1)(s+p_2)\cdots(s+p_n)}
-$$
-\n(15.48)
-
-where *s* = −*p*1, −*p*2, … , −*pn* are the simple poles, and *pi* ≠ *pj* for all *i* ≠ *j* (i.e., the poles are distinct). Assuming that the degree of *N*(*s*) is
-
-Software packages such as MATLAB, Mathcad, and Maple are capable of finding partial fraction expansions quite easily.
-
-Otherwise, we must first apply long division so that F(s) = N(s)∕D(s) = Q(s) + R(s)∕D(s), where the degree of R(s), the remainder of the long division, is less than the degree of D(s).
-
-less than the degree of *D*(*s*), we use partial fraction expansion to decompose *F*(*s*) in Eq. (15.48) as
-
-$$
-F(s) = \frac{k_1}{s + p_1} + \frac{k_2}{s + p_2} + \dots + \frac{k_n}{s + p_n}
-$$
- (15.49)
-
-The expansion coefficients *k*1, *k*2, … , *kn* are kno wn as the *residues* of *F*(*s*). There are man y ways of finding the expansion coefficients. One way is using the *residue method*. If we multiply both sides of Eq. (15.49) by (*s* + *p*1), we obtain
-
-$$
-(s+p_1)F(s) = k_1 + \frac{(s+p_1)k_2}{s+p_2} + \dots + \frac{(s+p_1)k_n}{s+p_n}
-$$
- (15.50)
-
-Because *pi* ≠ *pj*, setting *s* = −*p*1 in Eq. (15.50) leaves only *k*1 on the righthand side of Eq. (15.50). Hence,
-
-$$
-(s+p_1)F(s)\big|_{s=-p_1} = k_1\tag{15.51}
-$$
-
-Thus, in general,
-
-$$
-k_i = (s + p_i)F(s) \Big|_{s = -p_i}
-$$
- (15.52)
-
-This is known as *Heaviside's theorem*. Once the values of *ki* are known, we proceed to find the inverse of *F*(*s*) using Eq. (15.49). Since the inverse transform of each term in Eq. (15.49) is −1 [*k*∕(*s* + *a*)] = *ke*−*at u*(*t*), then, from Table 15.2,
-
-$$
-f(t) = (k_1 e^{-p_1 t} + k_2 e^{-p_2 t} + \dots + k_n e^{-p_n t}) u(t)
-$$
- (15.53)
-
-**15.4.2** Repeated Poles
-
-Suppose *F*(*s*) has *n* repeated poles at *s* = −*p*. Then we may represent *F*(*s*) as
-
-$$
-F(s) = \frac{k_n}{(s+p)^n} + \frac{k_{n-1}}{(s+p)^{n-1}} + \dots + \frac{k_2}{(s+p)^2} + \frac{k_1}{s+p} + F_1(s)
-$$
-\n(15.54)
-
-where *F*1(*s*) is the remaining part of *F*(*s*) that does not ha ve a pole at *s* = −*p*. We determine the expansion coefficient *kn* as
-
-$$
-k_n = (s+p)^n F(s) \Big|_{s=-p} \tag{15.55}
-$$
-
-as we did above. To determine *kn*−1, we multiply each term in Eq. (15.54) by (*s* + *p*) *n* and differentiate to get rid of *kn*, then evaluate the result at *s* = −*p* to get rid of the other coefficients except *kn*−1. Thus, we obtain
-
-$$
-k_{n-1} = \frac{d}{ds} [(s+p)^n F(s)] \Big|_{s=-p}
-$$
- (15.56)
-
-Repeating this gives
-
-$$
-k_{n-2} = \frac{1}{2!} \frac{d^2}{ds^2} [(s+p)^n F(s)] \Big|_{s=-p}
-$$
- (15.57)
-
-Historical note: Named after Oliver Heaviside (1850–1925), an English engineer, the pioneer of operational calculus.
-
-The *m*th term becomes
-
-$$
-k_{n-m} = \frac{1}{m!} \frac{d^m}{ds^m} [(s+p)^n F(s)] \Big|_{s=-p}
-$$
- (15.58)
-
-where *m* = 1, 2, … , *n* − 1. One can expect the differentiation to be difficult to handle as *m* increases. Once we obtain the values of *k*1, *k*2, … , *kn* by partial fraction expansion, we apply the inverse transform
-
-$$
-\mathcal{L}^{-1}\left[\frac{1}{(s+a)^n}\right] = \frac{t^{n-1}e^{-at}}{(n-1)!}u(t)
-$$
-\n(15.59)
-
-to each term on the right-hand side of Eq. (15.54) and obtain
-
-$$
-f(t) = \left(k_1 e^{-pt} + k_2 t e^{-pt} + \frac{k_3}{2!} t^2 e^{-pt} + \dots + \frac{k_n}{(n-1)!} t^{n-1} e^{-pt} \right) u(t) + f_1(t)
-$$
-\n(15.60)
-
-# **15.4.3** Complex Poles
-
-A pair of comple x poles is simple if it is not repeated; it is a double or multiple pole if repeated. Simple comple x poles may be handled the same way as simple real poles, but because complex algebra is involved the result is always cumbersome. An easier approach is a method known as *completing the square*. The idea is to express each complex pole pair (or quadratic term) in *D*(*s*) as a complete square such as (*s* + *α*) 2 + *β*2 and then use Table 15.2 to find the inverse of the term.
-
-Because *N*(*s*) and *D*(*s*) always have real coefficients and we know that the complex roots of polynomials with real coef ficients must occur in conjugate pairs, *F*(*s*) may have the general form
-
-$$
-F(s) = \frac{A_1 s + A_2}{s^2 + as + b} + F_1(s)
-$$
-\n(15.61)
-
-where *F*1(*s*) is the remaining part of *F*(*s*) that does not ha ve this pair of complex poles. If we complete the square by letting
-
-$$
-s^{2} + as + b = s^{2} + 2as + a^{2} + \beta^{2} = (s + a)^{2} + \beta^{2}
-$$
- (15.62)
-
-and we also let
-
-$$
-A_1s + A_2 = A_1(s + \alpha) + B_1\beta \tag{15.63}
-$$
-
-then Eq. (15.61) becomes
-
-$$
-F(s) = \frac{A_1(s+\alpha)}{(s+\alpha)^2 + \beta^2} + \frac{B_1\beta}{(s+\alpha)^2 + \beta^2} + F_1(s)
-$$
-(15.64)
-
-From Table 15.2, the inverse transform is
-
-$$
-f(t) = (A_1 e^{-\alpha t} \cos \beta t + B_1 e^{-\alpha t} \sin \beta t) u(t) + f_1(t)
-$$
- (15.65)
-
-The sine and cosine terms can be combined using Eq. (9.11).
-
-Whether the pole is simple, repeated, or comple x, a general approach that can al ways be used in finding the expansion coefficients is the *method of algebra*, illustrated in Examples 15.9 to 15.11. To apply the method, we first set *F*(*s*) = *N*(*s*)∕*D*(*s*) equal to an expansion containing unknown constants. We multiply the result through by a common denominator. Then we determine the unkno wn constants by equating coefficients (i.e., by algebraically solving a set of simultaneous equations for these coefficients at like powers of *s*).
-
-Another general approach is to substitute specific, convenient values of *s* to obtain as man y simultaneous equations as the number of unknown coefficients, and then solve for the unkno wn coefficients. We must make sure that each selected v alue of *s* is not one of the poles of *F*(*s*). Example 15.11 illustrates this idea.
-
-Find the inverse Laplace transform of Example 15.8
-
-$$
-F(s) = \frac{3}{s} - \frac{5}{s+1} + \frac{6}{s^2+4}
-$$
-
-# **Solution:**
-
-The inverse transform is given by
-
-$$
-f(t) = \mathcal{L}^{-1}[F(s)] = \mathcal{L}^{-1}\left(\frac{3}{s}\right) - \mathcal{L}^{-1}\left(\frac{5}{s+1}\right) + \mathcal{L}^{-1}\left(\frac{6}{s^2+4}\right)
-$$
-$$
-= (3 - 5e^{-t} + 3\sin 2t)u(t), \qquad t \ge 0
-$$
-
-where Table 15.2 has been consulted for the inverse of each term.
-
-$$
-F(s) = 5 + \frac{6}{s+4} - \frac{7s}{s^2 + 25}
-$$
-
-**Answer:** 5*δ*(*t*) + (6*e*−4*t* − 7 cos(5*t*))*u*(*t*).
-
-$$
-F(s) = \frac{s^2 + 12}{s(s+2)(s+3)}
-$$
-
-# **Solution:**
-
-Unlike in the pre vious example where the partial fractions ha ve been provided, we first need to determine the partial fractions. Given that there are three poles, we let
-
-les, we let
-\n
-$$
-\frac{s^2 + 12}{s(s+2)(s+3)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+3}
-$$
-\n(15.9.1)
-
-where *A*, *B*, and *C* are the constants to be determined. We can find the constants using two approaches.
-
-Determine the inverse Laplace transform of Practice Problem 15.8
-
-Find *f*(*t*) given that Example 15.9
-
-■ **METHOD 1 Residue method:**
-
-**PROOF Residue method:**
-\n
-$$
-A = sF(s) \Big|_{s=0} = \frac{s^2 + 12}{(s+2)(s+3)} \Big|_{s=0} = \frac{12}{(2)(3)} = 2
-$$
-\n
-$$
-B = (s+2)F(s) \Big|_{s=-2} = \frac{s^2 + 12}{s(s+3)} \Big|_{s=-2} = \frac{4+12}{(-2)(1)} = -8
-$$
-\n
-$$
-C = (s+3)F(s) \Big|_{s=-3} = \frac{s^2 + 12}{s(s+2)} \Big|_{s=-3} = \frac{9+12}{(-3)(-1)} = 7
-$$
-
-■ **METHOD 2 Algebraic method:** Multiplying both sides of Eq. (15.9.1) by *s*(*s* + 2)(*s* + 3) gives
-
-$$
-s^2 + 12 = A(s+2)(s+3) + Bs(s+3) + Cs(s+2)
-$$
-
-or
-
-$$
-s^2 + 12 = A(s^2 + 5s + 6) + B(s^2 + 3s) + C(s^2 + 2s)
-$$
-
-Equating the coefficients of like powers of *s* gives
-
-Constant:
-$$
-12 = 6A \Rightarrow A = 2
-$$
-
-\ns: $0 = 5A + 3B + 2C \Rightarrow 3B + 2C = -10$
-
-\ns: $1 = A + B + C \Rightarrow B + C = -1$
-
-Thus, *A* = 2, *B* = −8, *C* = 7, and Eq. (15.9.1) becomes
-
-$$
-F(s) = \frac{2}{s} - \frac{8}{s+2} + \frac{7}{s+3}
-$$
-
-By finding the inverse transform of each term, we obtain
-
-$$
-f(t) = (2 - 8e^{-2t} + 7e^{-3t})u(t)
-$$
-
-Practice Problem 15.9 Find *f*(*t*) if
-
-$$
-F(s) = \frac{48(s+2)}{(s+1)(s+3)(s+4)}
-$$
-
-**Answer:**
-$$
-f(t) = (8e^{-t} + 24e^{-3t} - 32e^{-4t})u(t)
-$$
-.
-
-Example 15.10 Calculate *v*(*t*) given that
-
-$$
-V(s) = \frac{10s^2 + 4}{s(s+1)(s+2)^2}
-$$
-
-# **Solution:**
-
-While the pre vious e xample is on simple roots, this e xample is on repeated roots. Let
-
-$$
-V(s) = \frac{10s^2 + 4}{s(s+1)(s+2)^2}
-$$
-
-= $\frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+2)^2} + \frac{D}{s+2}$ (15.10.1)
-
-# ■ **METHOD 1 Residue method:**
-
-**THEOREM 1** Residue method:
-\n
-$$
-A = sV(s) \Big|_{s=0} = \frac{10s^2 + 4}{(s+1)(s+2)^2} \Big|_{s=0} = \frac{4}{(1)(2)^2} = 1
-$$
-\n
-$$
-B = (s+1)V(s) \Big|_{s=-1} = \frac{10s^2 + 4}{s(s+2)^2} \Big|_{s=-1} = \frac{14}{(-1)(1)^2} = -14
-$$
-\n
-$$
-C = (s+2)^2 V(s) \Big|_{s=-2} = \frac{10s^2 + 4}{s(s+1)} \Big|_{s=-2} = \frac{44}{(-2)(-1)} = 22
-$$
-\n
-$$
-D = \frac{d}{ds} [(s+2)^2 V(s)] \Big|_{s=-2} = \frac{d}{ds} \left( \frac{10s^2 + 4}{s^2 + s} \right) \Big|_{s=-2} = \frac{(s^2 + s)(20s) - (10s^2 + 4)(2s + 1)}{(s^2 + s)^2} \Big|_{s=-2} = \frac{52}{4} = 13
-$$
-
-■ **METHOD 2 Algebraic method:** Multiplying Eq. (15.10.1) by *s*(*s* + 1)(*s* + 2)2 , we obtain
-
-$$
-10s2 + 4 = A(s + 1)(s + 2)2 + Bs(s + 2)2 + Cs(s + 1) + Ds(s + 1)(s + 2)
-$$
-
-or
-
-$$
-10s2 + 4 = A(s3 + 5s2 + 8s + 4) + B(s3 + 4s2 + 4s)
-$$
-
-+ C(s2 + s) + D(s3 + 3s2 + 2s)
-
-Equating coefficients,
-
-Constant:
-$$
-4 = 4A
-$$
- $\Rightarrow$ $A = 1$
-
-\ns: $0 = 8A + 4B + C + 2D$ $\Rightarrow$ $4B + C + 2D = -8$
-
-\ns2: $10 = 5A + 4B + C + 3D$ $\Rightarrow$ $4B + C + 3D = 5$
-
-\ns3: $0 = A + B + D$ $\Rightarrow$ $B + D = -1$
-
-Solving these simultaneous equations gi ves *A* = 1, *B* = −14, *C* = 22, *D* = 13, so that
-
-$$
-V(s) = \frac{1}{s} - \frac{14}{s+1} + \frac{13}{s+2} + \frac{22}{(s+2)^2}
-$$
-
-Taking the inverse transform of each term, we get
-
-$$
-v(t) = (1 - 14e^{-t} + 13e^{-2t} + 22te^{-2t})u(t)
-$$
-
-Obtain *g*(*t*) if Practice Problem 15.10
-
-$$
-G(s) = \frac{s^3 + 2s + 6}{s(s+1)^2(s+3)}
-$$
-
-**Answer:** (2 − 3.25*e*−*t* − 1.5*te*−*t* + 2.25*e*−3*t* )*u*(*t*).
-
-Find the in verse transform of the frequenc y-domain function in Example 15.11 Example 15.7: *H*(*s*) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ 20 (*s* + 3)(*s*
-
-$$
-H(s) = \frac{20}{(s+3)(s^2+8s+25)}
-$$
-
-# **Solution:**
-
-In this example, *H*(*s*) has a pair of complex poles at *s* 2 + 8*s* + 25 = 0 or *s* = −4 ± *j*3. We let
-
-is example,
-$$
-H(s)
-$$
- has a pair of complex poles at $s^2 + 8s + 25 = 0$ or
--4 ± j3. We let
-$$
-H(s) = \frac{20}{(s+3)(s^2+8s+25)} = \frac{A}{s+3} + \frac{Bs+C}{(s^2+8s+25)}
-$$
-(15.11.1)
-
-We now determine the expansion coefficients in two ways.
-
-■ **METHOD 1 Combination of methods:** We can obtain *A* using the method of residue,
-
-$$
-A = (s + 3)H(s)\Big|_{s=-3} = \frac{20}{s^2 + 8s + 25}\Big|_{s=-3} = \frac{20}{10} = 2
-$$
-
-Although *B* and *C* can be obtained using the method of residue, we will not do so, to a void comple x algebra. Rather , we can substitute two specific values of *s* [say *s* = 0, 1, which are not poles of *F*(*s*)] into Eq. (15.11.1). This will give us two simultaneous equations from which to find *B* and *C*. If we let *s* = 0 in Eq. (15.11.1), we obtain
-
-$$
-\frac{20}{75} = \frac{A}{3} + \frac{C}{25}
-$$
-
-or
-
-$$
-20 = 25A + 3C \tag{15.11.2}
-$$
-
-Because *A* = 2, Eq. (15.11.2) gi ves *C* = −10. Substituting *s* = 1 into Eq. (15.11.1) gives
-
-$$
-\frac{20}{(4)(34)} = \frac{A}{4} + \frac{B+C}{34}
-$$
-
-or
-
-$$
-20 = 34A + 4B + 4C \tag{15.11.3}
-$$
-
-But *A* = 2, *C* = −10, so that Eq. (15.11.3) gives *B* = −2.
-
-■ **METHOD 2 Algebraic method:** Multiplying both sides of Eq. (15.11.1) by (*s* + 3)(*s* 2 + 8*s* + 25) yields
-
-$$
-20 = A(s2 + 8s + 25) + (Bs + C)(s + 3)
-$$
-
-= A(s2 + 8s + 25) + B(s2 + 3s) + C(s + 3) (15.11.4)
-
-Equating coefficients gives
-
-*s* 2 : 0 = *A* + *B* ⇒ *A* = −*B s*: 0 = 8*A* + 3*B* + *C* = 5*A* + *C* ⇒ *C* = −5*A* Constant: 20 = 25*A* + 3*C* = 25*A* − 15*A* ⇒ *A* = 2
-
-That is, *B* = −2, *C* = −10. Thus,
-
-$$
-B = -2, C = -10. \text{ Thus,}
-$$
-
-\n
-$$
-H(s) = \frac{2}{s+3} - \frac{2s+10}{(s^2+8s+25)} = \frac{2}{s+3} - \frac{2(s+4)+2}{(s+4)^2+9}
-$$
-
-\n
-$$
-= \frac{2}{s+3} - \frac{2(s+4)}{(s+4)^2+9} - \frac{2}{3} \frac{3}{(s+4)^2+9}
-$$
-
-Taking the inverse of each term, we obtain
-
-$$
-h(t) = \left(2e^{-3t} - 2e^{-4t}\cos 3t - \frac{2}{3}e^{-4t}\sin 3t\right)u(t) \tag{15.11.5}
-$$
-
-It is alright to lea ve the result this w ay. However, we can combine the cosine and sine terms as
-
-$$
-h(t) = (2e^{-3t} - Re^{-4t}\cos(3t - \theta))u(t)
-$$
- (15.11.6)
-
-To obtain Eq. (15.11.6) from Eq. (15.11.5), we apply Eq. (9.11). Next, we determine the coefficient *R* and the phase angle *θ*:
-
-$$
-R = \sqrt{2^2 + \left(\frac{2}{3}\right)^2} = 2.108, \qquad \theta = \tan^{-1} \frac{\frac{2}{3}}{2} = 18.43^{\circ}
-$$
-
-Thus,
-
-$$
-h(t) = (2e^{-3t} - 2.108e^{-4t}\cos(3t - 18.43^\circ))u(t)
-$$
-
-$$
-G(s) = \frac{20}{(s+1)(s^2+4s+13)}
-$$
-
-**Answer:** 2*e*−*t* − 2*e*−2*t* cos 3*t* − 0.6667*e*−2*t* sin 3*t*, *t* ≥ 0.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/180_15.5 The Convolution Integral.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/180_15.5 The Convolution Integral.md
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-# **15.5** The Convolution Integral
-
-The term *convolution* means "folding." Convolution is an invaluable tool to the engineer because it provides a means of viewing and characterizing physical systems. For example, it is used in finding the response *y*(*t*) of a system to an e xcitation *x*(*t*), knowing the system impulse response *h*(*t*). This is achieved through the *convolution integral*, defined as
-
-$$
-y(t) = \int_{-\infty}^{\infty} x(\lambda)h(t - \lambda) d\lambda
-$$
- (15.66)
-
-or simply
-
-$$
-y(t) = x(t) * h(t)
-$$
- (15.67)
-
-where *λ* is a dummy v ariable and the asterisk denotes con volution. Equation (15.66) or (15.67) states that the output is equal to the input convolved with the unit impulse response. The convolution process is commutative:
-
-$$
-y(t) = x(t) * h(t) = h(t) * x(t)
-$$
-\n(15.68a)
-
-or
-
-$$
-y(t) = \int_{-\infty}^{\infty} x(\lambda)h(t-\lambda) d\lambda = \int_{-\infty}^{\infty} h(\lambda)x(t-\lambda) d\lambda
-$$
- (15.68b)
-
-This implies that the order in which the tw o functions are convolved is immaterial. We will see shortly how to take advantage of this commutative property when performing graphical computation of the convolution integral.
-
-Find *g*(*t*) given that Practice Problem 15.11
-
-The convolution of two signals consists of time-reversing one of the signals, shifting it, and multiplying it point by point with the second signal, and integrating the product.
-
-The convolution integral in Eq. (15.66) is the general one; it applies to any linear system. However, the convolution integral can be simplified if we assume that a system has two properties. First, if *x*(*t*) = 0 for *t* < 0, then
-
-$$
-y(t) = \int_{-\infty}^{\infty} x(\lambda)h(t-\lambda) \, d\lambda = \int_{0}^{\infty} x(\lambda)h(t-\lambda) \, d\lambda \tag{15.69}
-$$
-
-Second, if the system's impulse response is *causal* (i.e., *h*(*t*) = 0 for *t* < 0), then *h*(*t* − *λ*) = 0 for *t* − *λ* < 0 or *λ* > *t*, so that Eq. (15.69) becomes
-
-$$
-y(t) = h(t) * x(t) = \int_0^t x(\lambda)h(t - \lambda)d\lambda
-$$
- (15.70)
-
-Here are some properties of the convolution integral.
-
-1. *x*(*t*) \* *h*(*t*) = *h*(*t*) \* *x*(*t*) (Commutative) 2. *f* (*t*) \* [*x*(*t*) + *y*(*t*)] = *f* (*t*) \* *x*(*t*) + *f* (*t*) \* *y*(*t*) (Distributive) 3. *f* (*t*) \* [*x*(*t*) \* *y*(*t*)] = [ *f*(*t*) \* *x*(*t*)] \* *y*(*t*) (Associative) 4. *f* (*t*) \* *δ*(*t*) = ∫ −∞ ∞ *f* (*λ*)*δ*(*t* − *λ*) *dλ* = *f* (*t*) 5. *f* (*t*) \* *δ*(*t* − *to*) = *f* (*t* − *to*) 6. *f* (*t*) \* *δ*′(*t*) = ∫ −∞ ∞ *f* (*λ*)*δ*′(*t* − *λ*) *dλ* = *f* ′(*t*) 7. *f* (*t*) \* *u*(*t*) = ∫ −∞ ∞ *f* (*λ*)*u*(*t* − *λ*) *dλ* = ∫ −∞ *t f* (*λ*) *dλ*
-
-Before learning ho w to e valuate the con volution inte gral in Eq. (15.70), let us establish the link between the Laplace transform and the convolution integral. Given two functions *f*1(*t*) and *f*2(*t*) with Laplace transforms *F*1(*s*) and *F*2(*s*), respectively, their convolution is
-
-$$
-f(t) = f_1(t) * f_2(t) = \int_0^t f_1(\lambda) f_2(t - \lambda) d\lambda
-$$
- (15.71)
-
-Taking the Laplace transform gives
-
-$$
-F(s) = \mathcal{L}[f_1(t) * f_2(t)] = F_1(s)F_2(s)
-$$
-\n(15.72)
-
-To prove that Eq. (15.72) is true, we begin with the fact that *F*1(*s*) is defined as
-
-$$
-F_1(s) = \int_{0^-}^{\infty} f_1(\lambda) e^{-s\lambda} d\lambda \qquad (15.73)
-$$
-
-Multiplying this with *F*2(*s*) gives
-
-$$
-F_1(s)F_2(s) = \int_{0^-}^{\infty} f_1(\lambda)[F_2(s)e^{-s\lambda}] d\lambda
-$$
- (15.74)
-
-We recall from the time shift property in Eq. (15.17) that the term in brackets can be written as
-
-$$
-F_2(s)e^{-s\lambda} = \mathcal{L}[f_2(t-\lambda)u(t-\lambda)]
-$$
-
-=
-$$
-\int_0^\infty f_2(t-\lambda)u(t-\lambda)e^{-st} dt
-$$
- (15.75)
-
-Substituting Eq. (15.75) into Eq. (15.74) gives
-
-$$
-F_1(s)F_2(s) = \int_0^\infty f_1(\lambda) \left[ \int_0^\infty f_2(t-\lambda)u(t-\lambda)e^{-st} dt \right] d\lambda \qquad (15.76)
-$$
-
-Interchanging the order of integration results in
-
-$$
-F_1(s)F_2(s) = \int_0^\infty \left[ \int_0^t f_1(\lambda)f_2(t-\lambda) d\lambda \right] e^{-st} dt \qquad (15.77)
-$$
-
-The integral in brackets extends only from 0 to *t* because the delayed unit step *u*(*t* − *λ*) = 1 for *λ* < *t* and *u*(*t* − *λ*) = 0 for *λ* > *t*. We notice that the integral is the convolution of *f*1(*t*) and *f*2(*t*) as in Eq. (15.71). Hence,
-
-$$
-F_1(s)F_2(s) = \mathcal{L}[f_1(t) * f_2(t)]
-$$
- (15.78)
-
-as desired. This indicates that convolution in the time domain is equivalent to multiplication in the *s*-domain. For example, if *x*(*t*) = 4*e*−*t* and *h*(*t*) = 5*e*−2*t* , applying the property in Eq. (15.78), we get
-
-$$
-h(t) * x(t) = \mathcal{L}^{-1}[H(s)X(s)] = \mathcal{L}^{-1}\left[\left(\frac{5}{s+2}\right)\left(\frac{4}{s+1}\right)\right]
-$$
-$$
-= \mathcal{L}^{-1}\left[\frac{20}{s+1} + \frac{-20}{s+2}\right]
-$$
-(15.79)
-$$
-= 20(e^{-t} - e^{-2t}), \qquad t \ge 0
-$$
-
-Although we can find the convolution of tw o signals using Eq. (15.78), as we have just done, if the product *F*1(*s*)*F*2(*s*) is very complicated, finding the inverse may be tough. Also, there are situations in which *f*1(*t*) and *f*2(*t*) are a vailable in the form of e xperimental data and there are no explicit Laplace transforms. In these cases, one must do the convolution in the time domain.
-
-The process of convolving two signals in the time domain is better appreciated from a graphical point of view. The graphical procedure for evaluating the convolution integral in Eq. (15.70) usually in volves four steps.
-
-Steps to Evaluate the Convolution Integral:
-
-- 1. Folding: Take the mirror image of *h*(*λ*) about the ordinate axis to obtain *h*(−*λ*).
-- 2. Displacement: Shift or delay *h*(−*λ*) by *t* to obtain *h*(*t* − *λ*).
-- 3. Multiplication: Find the product of *h*(*t* − *λ*) and *x*(*λ*).
-- 4. Integration: F or a gi ven time *t*, calculate the area under the product *h*(*t* − *λ*)*x*(*λ*) for 0 < *λ* < *t* to get *y*(*t*) at *t*.
-
-The folding operation in step 1 is the reason for the term *convolution*. The function *h*(*t* − *λ*) scans or slides over *x*(*λ*). In view of this superposition procedure, the convolution integral is also known as the *superposition integral*.
-
-To apply the four steps, it is necessary to be able to sk etch *x*(*λ*) and *h*(*t* − *λ*). To get *x*(*λ*) from the original function *x*(*t*) involves merely replacing every *t* with *λ*. Sketching *h*(*t* − *λ*) is the k ey to the con volution process. It involves reflecting *h*(*λ*) about the vertical axis and shifting it by *t*. Analytically, we obtain *h*(*t* − *λ*) by replacing every *t* in *h*(*t*) by *t* − *λ*. Given that con volution is commutati ve, it may be more con venient to apply steps 1 and 2 to *x*(*t*) instead of *h*(*t*). The best way to illustrate the procedure is with some examples.
-
-Example 15.12 Find the convolution of the two signals in Fig. 15.10.
-
-# **Solution:**
-
-We follow the four steps to get *y*(*t*) = *x*1(*t*) \* *x*2(*t*). First, we fold *x*1(*t*) as shown in Fig. 15.11(a) and shift it by *t* as sho wn in Fig. 15.11(b). F or different values of *t*, we now multiply the two functions and integrate to determine the area of the overlapping region.
-
-For 0 < *t* < 1, there is no overlap of the two functions, as shown in Fig. 15.12(a). Hence,
-
-$$
-y(t) = x_1(t) * x_2(t) = 0, \qquad 0 < t < 1 \tag{15.12.1}
-$$
-
-For 1 < *t* < 2, the tw o signals o verlap between 1 and *t*, as sho wn in Fig. 15.12(b).
-
-$$
-y(t) = \int_1^t (2)(1) \, d\lambda = 2\lambda \bigg|_1^t = 2(t - 1), \qquad 1 < t < 2 \tag{15.12.2}
-$$
-
-For 2 < *t* < 3, the two signals completely overlap between (*t* − 1) and *t*, as shown in Fig. 15.12(c). It is easy to see that the area under the curve is 2. Or
-
-$$
-y(t) = \int_{t-1}^{t} (2)(1) \, d\lambda = 2\lambda \Big|_{t-1}^{t} = 2, \qquad 2 < t < 3 \tag{15.12.3}
-$$
-
-For 3 < *t* < 4, the two signals overlap between (*t* − 1) and 3, as sho wn in Fig. 15.12(d).
-
-$$
-y(t) = \int_{t-1}^{3} (2)(1) \, d\lambda = 2\lambda \Big|_{t-1}^{3}
-$$
-\n
-$$
-= 2(3 - t + 1) = 8 - 2t, \qquad 3 < t < 4 \tag{15.12.4}
-$$
-
-For *t* > 4, the two signals do not overlap [Fig. 15.12(e)], and
-
-$$
-y(t) = 0, \qquad t > 4 \tag{15.12.5}
-$$
-
-Combining Eqs. (15.12.1) to (15.12.5), we obtain
-
-$$
-y(t) = \begin{cases} 0, & 0 \le t \le 1 \\ 2t - 2, & 1 \le t \le 2 \\ 2, & 2 \le t \le 3 \\ 8 - 2t, & 3 \le t \le 4 \\ 0, & t \ge 4 \end{cases}
-$$
-(15.12.6)
-
-0 1 t t
-
-0 1 2 3
-
-1
-
-# **Figure 15.11**
-
-x1(t) x2(t)
-
-**Figure 15.10** For Example 15.12.
-
-2
-
-which is sketched in Fig. 15.13. Notice that *y*(*t*) in this equation is continuous. This fact can be used to check the results as we move from one range of *t* to another. The result in Eq. (15.12.6) can be obtained without using the graphical procedure—by directly using Eq. (15.70) and the properties of step functions. This will be illustrated in Example 15.14.
-
-Convolution of signals *x*1(*t*) and *x*2(*t*) in Fig. 15.10.
-
-Graphically con volve the two functions in Fig. 15.14. To sho w ho w Practice Problem 15.12 powerful working in the *s*-domain is, verify your answer by performing the equivalent operation in the *s*-domain.
-
-**Answer:** The result of the convolution *y*(*t*) is shown in Fig. 15.15, where
-
-For Example 15.13.
-
-Example 15.13 Graphically convolve *g*(*t*) and *u*(*t*) shown in Fig. 15.16.
-
-# **Solution:**
-
-Let *y*(*t*) = *g*(*t*) \* *u*(*t*). We can find *y*(*t*) in two ways.
-
-■ **METHOD 1** Suppose we fold *g*(*t*), as in Fig. 15.17(a), and shift it by *t*, as in Fig. 15.17(b). Since *g*(*t*) = *t*, 0 < *t* < 1 originally, we expect that *g*(*t* − *λ*) = *t* − *λ*, 0 < *t* − *λ* < 1 or *t* − 1 < *λ* < *t*. There is no overlap of the two functions when *t* < 0 so that *y*(0) = 0 for this case.
-
-**Figure 15.17** Convolution of *g*(*t*) and *u*(*t*) in Fig. 15.16 with *g*(*t*) folded.
-
-For 0 < *t* < 1, *g*(*t* − *λ*) and *u*(*λ*) o verlap from 0 to *t*, as e vident in Fig. 15.17(b). Therefore,
-
-$$
-y(t) = \int_0^t (1)(t - \lambda) d\lambda = \left(t\lambda - \frac{1}{2}\lambda^2\right)\Big|_0^t
-$$
-
-= $t^2 - \frac{t^2}{2} = \frac{t^2}{2}$ , $0 \le t \le 1$ (15.13.1)
-
-For *t* > 1, the two functions overlap completely between (*t* − 1) and *t* [see Fig. 15.17(c)]. Hence,
-
-$$
-y(t) = \int_{t-1}^{t} (1)(t - \lambda) d\lambda
-$$
-
-= $\left(t\lambda - \frac{1}{2}\lambda^2\right)\Big|_{t-1}^{t} = \frac{1}{2}, \qquad t \ge 1$ (15.13.2)
-
-Thus, from Eqs. (15.13.1) and (15.13.2),
-
-$$
-y(t) = \begin{cases} \frac{1}{2}t^2, & 0 \le t \le 1\\ \frac{1}{2}, & t \ge 1 \end{cases}
-$$
-
-■ **METHOD 2** Instead of folding *g,* suppose we fold the unit step function *u*(*t*), as in Fig. 15.18(a), and then shift it by *t*, as in Fig. 15.18(b). Because *u*(*t*) = 1 for *t* > 0, *u*(*t* − *λ*) = 1 for *t* − *λ* > 0 or *λ* < *t*, the two functions overlap from 0 to *t*, so that
-
-$$
-y(t) = \int_0^t (1)\lambda \, d\lambda = \frac{1}{2}\lambda^2 \bigg|_0^t = \frac{t^2}{2}, \qquad 0 \le t \le 1 \quad (15.13.3)
-$$
-
-# **Figure 15.18**
-
-Convolution of *g*(*t*) and *u*(*t*) in Fig. 15.16 with *u*(*t*) folded.
-
-For *t* > 1, the tw o functions o verlap between 0 and 1, as sho wn in Fig. 15.18(c). Hence,
-
-$$
-y(t) = \int_0^1 (1)\lambda \, d\lambda = \frac{1}{2}\lambda^2 \bigg|_0^1 = \frac{1}{2}, \qquad t \ge 1 \tag{15.13.4}
-$$
-
-And, from Eqs. (15.13.3) and (15.13.4),
-
-$$
-y(t) = \begin{cases} \frac{1}{2}t^2, & 0 \le t \le 1\\ \frac{1}{2}, & t \ge 1 \end{cases}
-$$
-
-Although the two methods give the same result, as expected, notice that it is more convenient to fold the unit step function *u*(*t*) than fold *g*(*t*) in this example. Figure 15.19 shows *y*(*t*).
-
-Result of Example 15.13.
-
-Given *g*(*t*) and *f*(*t*) in Fig. 15.20, graphically find *y*(*t*) = *g*(*t*) \* *f*(*t*). Practice Problem 15.13
-
-Answer:
-$$
-y(t) = \begin{cases} 3(1 - e^{-t}), & 0 \le t \le 1 \\ 3(e - 1)e^{-t}, & t \ge 1 \\ 0, & \text{elsewhere.} \end{cases}
-$$
-
-For the *RL* circuit in Fig. 15.21(a), use the convolution integral to find the Example 15.14 response *io*(*t*) due to the excitation shown in Fig. 15.21(b).
-
-# **Solution:**
-
-1. **Define.** The problem is clearly stated and the method of solution is also specified.
-
-**Figure 15.21** For Example 15.14.
-
-For the circuit in Fig. 15.21(a): (a) its *s*-domain equivalent, (b) its impulse response.
-
-- 2. **Present.** We are to use the con volution inte gral to solv e for the response *io*(*t*) due to *is*(*t*) shown in Fig. 15.21(b).
-- 3. **Alternative.** We have learned to do con volution by using the con volution integral and how to do it graphically. In addition, we could always work in the *s*-domain to solve for the current. We will solve for the current using the convolution integral and then check it using the graphical approach.
-- 4. **Attempt.** As we stated, this problem can be solv ed in tw o ways: directly using the con volution inte gral or using the graphical technique. To use either approach, we first need the unit impulse response *h*(*t*) of the circuit. In the *s*-domain, applying the current division principle to the circuit in Fig. 15.22(a) gives
-
-$$
-I_o = \frac{1}{s+1} I_s
-$$
-
-Hence,
-
-$$
-H(s) = \frac{I_o}{I_s} = \frac{1}{s+1}
-$$
- (15.14.1)
-
-and the inverse Laplace transform of this gives
-
-$$
-h(t) = e^{-t} u(t)
-$$
- (15.14.2)
-
-Figure 15.22(b) shows the impulse response *h*(*t*) of the circuit.
-
-To use the convolution integral directly, recall that the response is given in the *s*-domain as
-
-$$
-I_o(s) = H(s)I_s(s)
-$$
-
-With the given *is*(*t*) in Fig. 15.21(b),
-
-$$
-i_s(t) = u(t) - u(t-2)
-$$
-
-so that
-
-$$
-i_o(t) = h(t) * i_s(t) = \int_0^t i_s(\lambda)h(t - \lambda) d\lambda
-$$
-
-=
-$$
-\int_0^t [u(\lambda) - u(\lambda - 2)]e^{-(t - \lambda)} d\lambda
-$$
- (15.14.3)
-
-Since *u*(*λ* − 2) = 0 for 0 < *λ* < 2, the inte grand involving *u*(*λ*) is nonzero for all *λ* > 0, whereas the inte grand involving *u*(*λ* − 2) is nonzero only for λ > 2. The best way to handle the integral is to do the two parts separately. For 0 < *t* < 2,
-
-$$
-i'_{o}(t) = \int_{0}^{t} (1)e^{-(t-\lambda)} d\lambda = e^{-t} \int_{0}^{t} (1)e^{\lambda} d\lambda
-$$
-
-= $e^{-t}(e^{t} - 1) = 1 - e^{-t}, \qquad 0 < t < 2$ (15.14.4)
-
-For *t* > 2,
-
-$$
-i''_o(t) = \int_2^t (1)e^{-(t-\lambda)} d\lambda = e^{-t} \int_2^t e^{\lambda} d\lambda
-$$
-
-= $e^{-t}(e^t - e^2) = 1 - e^2 e^{-t}, \qquad t > 2$ (15.14.5)
-
-Substituting Eqs. (15.14.4) and (15.14.5) into Eq. (15.14.3) gives
-
-$$
-i_o(t) = i'_o(t) - i''_o(t)
-$$
-
-= $(1 - e^{-t})[u(t - 2) - u(t)] - (1 - e^2 e^{-t})u(t - 2)$
-= $\begin{cases} 1 - e^{-t}A, & 0 < t < 2 \\ (e^2 - 1)e^{-t}A, & t > 2 \end{cases}$ (15.14.6)
-
-5. **Evaluate.** To use the graphical technique, we may fold *is*(*t*) in Fig. 15.21(b) and shift by *t*, as shown in Fig. 15.23(a). For 0 < *t* < 2, the overlap between *is*(*t* − *λ*) and *h*(*λ*) is from 0 to *t*, so that
-
-$$
-i_o(t) = \int_0^t (1)e^{-\lambda} d\lambda = -e^{-\lambda} \Big|_0^t = (1 - e^{-t}) \mathbf{A}, \qquad 0 \le t \le 2 \quad (15.14.7)
-$$
-
-For *t* > 2, the tw o functions o verlap between ( *t* − 2) and *t*, as in Fig. 15.23(b). Hence,
-
-$$
-i_o(t) = \int_{t-2}^{t} (1)e^{-\lambda} d\lambda = -e^{-\lambda} \Big|_{t-2}^{t} = -e^{-t} + e^{-(t-2)}
-$$
-
-= $(e^2 - 1)e^{-t} A$ , $t \ge 0$ (15.14.8)
-
-From Eqs. (15.14.7) and (15.14.8), the response is
-
-15.14.7) and (15.14.8), the response is
-\n
-$$
-i_o(t) = \begin{cases} 1 - e^{-t} A, & 0 \le t \le 2 \\ (e^2 - 1)e^{-t} A, & t \ge 2 \end{cases}
-$$
-\n(15.14.9)
-
-which is the same as in Eq. (15.14.6). Thus, the response *io*(*t*) along the excitation *is*(*t*) is as shown in Fig. 15.24.
-
-6. **Satisfactory?** We have satisf actorily solv ed the problem and can present the results as a solution to the problem.
-
-Use convolution to find *vo*(*t*) in the circuit of Fig. 15.25(a) when the exci- Practice Problem 15.14 tation is the signal shown in Fig. 15.25(b). To show how powerful working in the *s*-domain is, verify your answer by performing the equivalent operation in the *s*-domain.
-
-**Answer:**
-$$
-20(e^{-t} - e^{-2t})u(t)
-$$
- V.
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-# **15.6** Application to Integrodifferential Equations
-
-The Laplace transform is useful in solving linear inte grodifferential equations. Using the differentiation and integration properties of Laplace transforms, each term in the integrodifferential equation is transformed.
-
-**Figure 15.23** For Example 15.14.
-
-# **Figure 15.24**
-
-For Example 15.14; excitation and response.
-
-Initial conditions are automatically taken into account. We solve the resulting algebraic equation in the *s*-domain. We then con vert the solution back to the time domain by using the in verse transform. The following examples illustrate the process.
-
-Example 15.15 Use the Laplace transform to solve the differential equation
-
-$$
-\frac{d^2v(t)}{dt^2} + 6\frac{dv(t)}{dt} + 8v(t) = 2u(t)
-$$
-
-subject to *v*(0) = 1, *v*′(0) = −2.
-
-# **Solution:**
-
-We tak e the Laplace transform of each term in the gi ven dif ferential equation and obtain
-
-$$
-[s^2V(s) - sv(0) - v'(0)] + 6[sV(s) - v(0)] + 8V(s) = \frac{2}{s}
-$$
-
-Substituting *v*(0) = 1, *v*′(0) = −2,
-
-$$
-s^2V(s) - s + 2 + 6sV(s) - 6 + 8V(s) = \frac{2}{s}
-$$
-
-or
-
-$$
-(s2 + 6s + 8)V(s) = s + 4 + \frac{2}{s} = \frac{s2 + 4s + 2}{s}
-$$
-
-Hence,
-
-$$
-V(s) = \frac{s^2 + 4s + 2}{s(s+2)(s+4)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+4}
-$$
-
-where
-
-$$
-A = sV(s) \Big|_{s=0} = \frac{s^2 + 4s + 2}{(s+2)(s+4)} \Big|_{s=0} = \frac{2}{(2)(4)} = \frac{1}{4}
-$$
-$$
-B = (s+2)V(s) \Big|_{s=-2} = \frac{s^2 + 4s + 2}{s(s+4)} \Big|_{s=-2} = \frac{-2}{(-2)(2)} = \frac{1}{2}
-$$
-$$
-C = (s+4)V(s) \Big|_{s=-4} = \frac{s^2 + 4s + 2}{s(s+2)} \Big|_{s=-4} = \frac{2}{(-4)(-2)} = \frac{1}{4}
-$$
-
-Hence,
-
-$$
-V(s) = \frac{\frac{1}{4}}{s} + \frac{\frac{1}{2}}{s+2} + \frac{\frac{1}{4}}{s+4}
-$$
-
-By the inverse Laplace transform,
-
-$$
-v(t) = \frac{1}{4}(1 + 2e^{-2t} + e^{-4t})u(t)
-$$
-
-Solve the follo wing differential equation using the Laplace transform Practice Problem 15.15 method.
-
-$$
-\frac{d^2v(t)}{dt^2} + 4\frac{dv(t)}{dt} + 4v(t) = 7e^{-t}
-$$
-
-if *v*(0) = *v*′(0) = 2.
-
-**Answer:**
-$$
-(7e^{-t} - 5e^{-2t} - te^{-2t})u(t)
-$$
-.
-
-Solve for the response *y*(*t*) in the following integrodifferential equation. Example 15.16
-
-$$
-\frac{dy}{dt} + 5y(t) + 6 \int_0^t y(\tau)d\tau = u(t), \qquad y(0) = 2
-$$
-
-# **Solution:**
-
-Taking the Laplace transform of each term, we get
-
-$$
-[sY(s) - y(0)] + 5Y(s) + \frac{6}{s}Y(s) = \frac{1}{s}
-$$
-
-Substituting *y*(0) = 2 and multiplying through by *s*,
-
-$$
-Y(s)(s^2 + 5s + 6) = 1 + 2s
-$$
-
-or
-
-$$
-Y(s) = \frac{2s+1}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3}
-$$
-
-where
-
-$$
-A = (s + 2)Y(s) \Big|_{s=-2} = \frac{2s+1}{s+3} \Big|_{s=-2} = \frac{-3}{1} = -3
-$$
-$$
-B = (s+3)Y(s) \Big|_{s=-3} = \frac{2s+1}{s+2} \Big|_{s=-3} = \frac{-5}{-1} = 5
-$$
-
-*s* + 2
-
-−1
-
-Thus,
-
-$$
-Y(s) = \frac{-3}{s+2} + \frac{5}{s+3}
-$$
-
-Its inverse transform is
-
-$$
-y(t) = (-3e^{-2t} + 5e^{-3t})u(t)
-$$
-
-Use the Laplace transform to solve the integrodifferential equation Practice Problem 15.16
-
-$$
-\frac{dy}{dt} + 3y(t) + 2\int_0^t y(\tau)d\tau = 2e^{-3t}, \qquad y(0) = 0
-$$
-
-**Answer:** (−*e*−*t* + 4*e*−2*t* − 3*e*−3*t* )*u*(*t*).
-
-# **15.7** Summary
-
-1. The Laplace transform allows a signal represented by a function in the time domain to be analyzed in the *s*-domain (or comple x frequency domain). It is defined as
-
-$$
-\mathcal{L}[f(t)] = F(s) = \int_0^\infty f(t)e^{-st} dt
-$$
-
-- 2. Properties of the Laplace transform are listed in Table 15.1, while the Laplace transforms of basic common functions are listed in Table 15.2.
-- 3. The inverse Laplace transform can be found using partial fraction expansions and using the Laplace transform pairs in Table 15.2 as a look-up table. Real poles lead to exponential functions and complex poles to damped sinusoids.
-- 4. The convolution of tw o signals consists of time-re versing one of the signals, shifting it, multiplying it point by point with the second signal, and integrating the product. The convolution integral relates the convolution of two signals in the time domain to the in verse of the product of their Laplace transforms:
-
-$$
-\mathcal{L}^{-1}[F_1(s)F_2(s)] = f_1(t) * f_2(t) = \int_0^t f_1(\lambda)f_2(t - \lambda) d\lambda
-$$
-
-5. In the time domain, the output *y*(*t*) of the network is the convolution of the impulse response with the input *x*(*t*),
-
-$$
-y(t) = h(t) * x(t)
-$$
-
-Convolution may be re garded as the flip-shift-multiply-time-area method.
-
-6. The Laplace transform can be used to solve a linear integrodifferential equation.
-
-# Review Questions
-
-**15.1** Every function *f*(*t*) has a Laplace transform.
-
-(a) True (b) False
-
-**15.2** The variable *s* in the Laplace transform *H*(*s*) is called
-
-| (a) complex frequency | (b) transfer function |
-|-----------------------|-----------------------|
-| (c) zero | (d) pole |
-
-**15.3** The Laplace transform of *u*(*t* − 2) is:
-
-(a)
-$$
-\frac{1}{s+2}
-$$
- (b) $\frac{1}{s-2}$
-(c) $\frac{e^{2s}}{s}$ (d) $\frac{e^{-2s}}{s}$
-
-**15.4** The zero of the function
-
-of the function
-$$
-F(s) = \frac{s+1}{(s+2)(s+3)(s+4)}
-$$
-
-| is at | |
-|--------|--------|
-| (a) −4 | (b) −3 |
-| (c) −2 | (d) −1 |
-
-**15.5** The poles of the function
-
-$$
-F(s) = \frac{s+1}{(s+2)(s+3)(s+4)}
-$$
-
-are at
-
-| (a) −4 | (b) −3 |
-|--------|--------|
-| (c) −2 | (d) −1 |
-
-**15.6** If *F*(*s*) = 1∕(*s* + 2), then *f*(*t*) is
-
-(a) *e*2*t u*(*t*) (b) *e*−2*t u*(*t*) (c) *u*(*t* − 2) (d) *u*(*t* + 2)
-
-Problems **707**
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/182_15.7 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/182_15.7 Summary.md
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-**15.7** Given that *F*(*s*) = *e*−2*s* ∕(*s* + 1), then *f*(*t*) is (a) *e*−2(*t*−1)*u*(*t* − 1) (b) *e*−(*t*−2)*u*(*t* − 2) (c) *e*−*t u*(*t* − 2) (d) *e*−*t u*(*t* + 1) (e) *e*−(*t*−2)*u*(*t*)
-
-**15.8** The initial value of *f*(*t*) with transform
-
-$$
-F(s) = \frac{s+1}{(s+2)(s+3)}
-$$
-
-is:
-
-| (a) nonexistent | (b) ∞ | (c) 0 |
-|-----------------|--------------|-------|
-| (d) 1 | (e) __1
6 | |
-
-**15.9** The inverse Laplace transform of
-
-$$
-\frac{s+2}{\left(s+2\right)^2+1}
-$$
-
-# Problems
-
-# Sections 15.2 and 15.3 Definition and Properties of the Laplace Transform
-
-- **15.1** Find the Laplace transform of 5 sin(*at*) cos(*bt*). (*Hint*: Using the exponential representation for both functions may make this problem easier.)
-- **15.2** Determine the Laplace transform of 3.5 cos (5*t* − 45°).
-- **15.3** Obtain the Laplace transform of each of the following functions:
-
-| (a) e−2t | (b) e−2t |
-|------------------------|-------------|
-| cos 3tu(t) | sin 4 tu(t) |
-| (c) e−3t | (d) e−4t |
-| cosh 2tu(t) | sinh tu(t) |
-| (e) te−t
sin 2tu(t) | |
-
-- **15.4** Design a problem to help other students better understand how to find the Laplace transform of different time varying functions.
-- **15.5** Find the Laplace transform of each of the following functions:
- - (a) *t* 2 cos(2*t* + 30°)*u*(*t*)
-
-(b)
-$$
-3t^4e^{-2t}u(t)
-$$
-
-(c)
-$$
-2tu(t) - 4\frac{d}{dt}\delta(t)
-$$
-
-(d)
-$$
-2e^{-(t-1)}u(t)
-$$
-
-$$
-(e) 5u(t/2)
-$$
-
-(f)
-$$
-6e^{-t/3} u(t)
-$$
-
-(g) *dn* \_\_\_ *dtn δ*(*t*)
-
-is:
-\n(a)
-$$
-e^{-t} \cos 2t
-$$
-
-\n(b) $e^{-t} \sin 2t$
-\n(c) $e^{-2t} \cos t$
-\n(d) $e^{-2t} \sin 2t$
-\n(e) none of the above
-
-**15.10** The result of *u*(*t*) \* *u*(*t*) is:
-
-| (a) u2
(t) | (b) tu(t) |
-|--------------------|-----------|
-| 2
(c) t
u(t) | (d) δ(t) |
-
-*Answers: 15.1b, 15.2a, 15.3d, 15.4d, 15.5a,b,c, 15.6b, 15.7b, 15.8c, 15.9c, 15.10b.*
-
-- **15.6** Find *G*(*s*) given that *g*(*t*) = 2*r*(*t*) 2*r*(*t* − 2).
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/183_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/183_Problems.md
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-- **15.7** Find the Laplace transform of the following signals:
-
-$$
-(a) f(t) = (2t + 4)u(t)
-$$
-
-(b)
-$$
-g(t) = (4 + 3e^{-2t})u(t)
-$$
-
-- (c) *h*(*t*) = (6 sin(3*t*) + 8 cos(3*t*))*u*(*t*)
-- (d) *x*(*t*) = (*e*−2*t* cosh(4*t*))*u*(*t*)
-- **15.8** Find the Laplace transform *F*(*s*), given that *f*(*t*) is:
- - (a) 2*tu*(*t* − 4) (b) 5 cos(*t*) *δ*(*t* − 2) (c) *e*−*t u*(*t* − *t*) (d) sin(2*t*)*u*(*t* − *τ*)
-- **15.9** Determine the Laplace transforms of these functions:
-
-(a)
-$$
-f(t) = (t - 4)u(t - 2)
-$$
-
-\n(b) $g(t) = 2e^{-4t}u(t - 1)$
-\n(c) $h(t) = 5 \cos(2t - 1)u(t)$
-
-- (d) *p*(*t*) = 6[*u*(*t* − 2) − *u*(*t* − 4)]
-- **15.10** In two different ways, find the Laplace transform of
-
-$$
-g(t) = \frac{d}{dt}(te^{-t}\cos t)
-$$
-
-**15.11** Find *F*(*s*) if:
-
-(a)
-$$
-f(t) = 6e^{-t} \cosh 2t
-$$
- (b) $f(t) = 3te^{-2t} \sinh 4t$
-(c) $f(t) = 8e^{-3t} \cosh tu(t - 2)$
-
-- **15.12** If *g*(*t*) = 4*e*−2*t* cos 4*t*, find *G*(*s*).
-- **15.13** Find the Laplace transform of the following functions: (a) *t* cos *tu*(*t*) (b) *e*−*t t* sin *tu*(*t*)
-
-(a)
-$$
-t \cos u(t)
-$$
-
-(c) $\frac{\sin \beta t}{t} u(t)$
-
-*t*
-
-**15.14** Find the Laplace transform of the signal in Fig. 15.26.
-
-# **Figure 15.26**
-
-For Prob. 15.14.
-
-**15.15** Determine the Laplace transform of the function in Fig. 15.27.
-
-For Prob. 15.15.
-
-**15.16** Obtain the Laplace transform of *f*(*t*) in Fig. 15.28.
-
-**Figure 15.29**
-
-For Prob. 15.17.
-
-**15.18** Obtain the Laplace transforms of the functions in Fig. 15.30.
-
-**Figure 15.30**
-
-For Prob. 15.18.
-
-**15.19** Calculate the Laplace transform of the infinite train of unit impulses in Fig. 15.31.
-
-**Figure 15.31** For Prob. 15.19.
-
-**15.17** Using Fig. 15.29, design a problem to help other students better understand the Laplace transform of a simple, non-periodic waveshape.
-
-**Figure 15.32** For Prob. 15.20.
-
-**15.21** Obtain the Laplace transform of the periodic waveform in Fig. 15.33.
-
-# **Figure 15.33**
-
-For Prob. 15.21.
-
-**15.22** Find the Laplace transforms of the functions in Fig. 15.34.
-
-**15.23** Determine the Laplace transforms of the periodic functions in Fig. 15.35.
-
-# **Figure 15.35**
-
-For Prob. 15.23.
-
-**15.24** Design a problem to help other students better understand how to find the initial and final value of a transfer function.
-
-**15.25** Let
-
-$$
-F(s) = \frac{18(s+1)}{(s+2)(s+3)}
-$$
-
-- (a) Use the initial and final value theorems to find *f*(0) and *f* (∞).
-- (b) Verify your answer in part (a) by finding *f*(*t*), using partial fractions.
-
-**15.26** Determine the initial and final values of *f*(*t*), if they exist, given that:
-
-(a)
-$$
-F(s) = \frac{5s^2 + 3}{s^3 + 4s^2 + 6}
-$$
-
-\n(b) $F(s) = \frac{s^2 - 2s + 1}{4(s - 2)(s^2 + 2s + 4)}$
-
-# Section 15.4 The Inverse Laplace Transform
-
-**15.27** Determine the inverse Laplace transform of each of the following functions:
-
-Problems 709
-\nDetermine the initial and final values of
-$$
-f(t)
-$$
-, if they exist, given that:
-\n(a) $F(s) = \frac{s^2 + 3}{s^3 + 4s^2 + 6}$
-\n(b) $F(s) = \frac{s^2 - 2s + 1}{4(s - 2)(s^2 + 2s + 4)}$
-\nOn 15.4 The Inverse Laplace Transform
-\n7 Determine the inverse Laplace transform of each of the following functions:
-\n(a) $F(s) = \frac{1}{s} + \frac{2}{s + 1}$
-\n(b) $G(s) = \frac{3s + 1}{s + 4}$
-\n(c) $H(s) = \frac{12}{(s + 1)(s + 3)}$
-\n(d) $J(s) = \frac{12}{(s + 2)^2(s + 4)}$
-\nDesign a problem to help other students better understand how to find the inverse Laplace transform.
-\nFind the inverse Laplace transform of:
-\n $F(s) = \frac{s^2 + 2}{s^3 + 2s^2 + 2s}$
-\nFind the inverse Laplace transform of:
-\n(a) $F_1(s) = \frac{6s^2 + 8s + 3}{s(s^2 + 2s + 5)}$
-\n(b) $F_2(s) = \frac{s^2 + 5s + 6}{(s + 1)^2(s + 4)}$
-\n(c) $F_3(s) = \frac{10}{(s + 1)(s^2 + 4s + 8)}$
-\nFind $f(t)$ for each $F(s)$ :
-\n(a) $\frac{10s}{(s + 1)(s + 2)^3}$
-\n(b) $\frac{2s^2 + 4s + 1}{(s + 1)(s + 2)^3}$
-\n(c) $\frac{s + 1}{(s + 2)(s^2 + 2s + 5)}$
-\nDetermine the inverse Laplace transform of each of the following functions:
-\n(a) $\frac{8(s + 1)(s + 2)}{s(s + 2)(s + 4)}$
-\n(b) $\frac{s^2 - 2s + 4}{s(s + 1)(s + 2)^2}$
-\n(c) $\frac{s^2 + 1}{(s + 1)(s + 2)^2}$
-
-- **15.28** Design a problem to help other students better understand how to find the inverse Laplace transform.
-- **15.29** Find the inverse Laplace transform of:
-
-$$
-F(s) = \frac{s^2 + 2}{s^3 + 2s^2 + 2s}
-$$
-
-**15.30** Find the inverse Laplace transform of:
-
-Find the inverse Laplace transform
-\n(a)
-$$
-F_1(s) = \frac{6s^2 + 8s + 3}{s(s^2 + 2s + 5)}
-$$
-
-\n(b) $F_2(s) = \frac{s^2 + 5s + 6}{(s+1)^2(s+4)}$
-\n(c) $F_3(s) = \frac{10}{(s+1)(s^2 + 4s + 8)}$
-
-**15.31** Find *f*(*t*) for each *F*(*s*):
-
-Find
-$$
-f(t)
-$$
- for each $F(s)$ :
-\n(a) $\frac{10s}{(s + 1)(s + 2)(s + 3)}$
-\n(b) $\frac{2s^2 + 4s + 1}{(s + 1)(s + 2)^3}$
-\n(c) $\frac{s + 1}{(s + 2)(s^2 + 2s + 5)}$
-
-**15.32** Determine the inverse Laplace transform of each of the following functions:
-
-the following functions:
-\n(a)
-$$
-\frac{8(s + 1)(s + 3)}{s(s + 2)(s + 4)}
-$$
-\n(b)
-$$
-\frac{s^2 - 2s + 4}{(s + 1)(s + 2)^2}
-$$
-\n(c)
-$$
-\frac{s^2 + 1}{(s + 3)(s^2 + 4s + 5)}
-$$
-
-**15.33** Calculate the inverse Laplace transform of:
-
-(a)
-$$
-\frac{6(s-1)}{s^4 - 1}
-$$
- (b) $\frac{se^{-\pi s}}{s^2 + 1}$
-(c) $\frac{8}{s(s+1)^3}$
-
-**15.34** Find the time functions that have the following Laplace transforms:
-
-(a)
-$$
-F(s) = 10 + \frac{s^2 + 1}{s^2 + 4}
-$$
-
-\n(b) $G(s) = \frac{e^{-s} + 4e^{-2s}}{s^2 + 6s + 8}$
-\n(c) $H(s) = \frac{(s + 1)e^{-2s}}{s(s + 3)(s + 4)}$
-
-**15.35** Obtain *f*(*t*) for the following transforms:
-
-Obtain
-$$
-f(t)
-$$
- for the following
-\n(a) $F(s) = \frac{(s+3)e^{-6s}}{(s+1)(s+2)}$
-\n(b) $F(s) = \frac{4 - e^{-2s}}{s^2 + 5s + 4}$
-\n(c) $F(s) = \frac{se^{-s}}{(s+3)(s^2 + 4)}$
-
-**15.36** Obtain the inverse Laplace transforms of the following functions:
-
-following functions:
-\n(a)
-$$
-X(s) = \frac{3}{s^2(s+2)(s+3)}
-$$
-
-\n(b) $Y(s) = \frac{2}{s(s+1)^2}$
-\n(c) $Z(s) = \frac{5}{s(s+1)(s^2+6s+10)}$
-
-Laplace transform of:
-\n**15.37** Find the inverse Laplace transform of:
-\n**(b)**
-$$
-\frac{se^{-xs}}{s^2 + 1}
-$$
-
-\n**(c)** $F(s) = \frac{s^2 + 4s + 5}{(s + 2)(s^2 + 2s + 2)}$
-\n**(d)** $D(s) = \frac{10s}{(s^2 + 1)(s^2 + 4)}$
-\n**15.38** Find $f(t)$ given that:
-\n**(a)** $F(s) = \frac{10s}{(s^2 + 1)(s^2 + 4)}$
-\n**15.38** Find $f(t)$ given that:
-\n**(a)** $F(s) = \frac{s^2 + 4s}{s^2 + 10s + 26}$
-\n**(b)** $F(s) = \frac{5s^2 + 7s + 29}{s(s^2 + 4s + 29)}$
-\nIlowing transforms:
-\n**(a)** $F(s) = \frac{5s^2 + 7s + 29}{(s^2 + 2s + 17)(s^2 + 4s + 20)}$
-\n**(b)** $F(s) = \frac{2s^3 + 4s^2 + 1}{(s^2 + 2s + 17)(s^2 + 6s + 3)}$
-\n**(c)** $F(s) = \frac{2s^3 + 4s^2 + 1}{(s^2 + 2s + 17)(s^2 + 6s + 3)}$
-\n**(d)** $F(s) = \frac{s^2 + 4}{(s^2 + 2s + 17)(s^2 + 4s + 20)}$
-\n**(e)** $F(s) = \frac{s^2 + 4}{(s^2 + 2s + 17)(s^2 + 6s + 3)}$
-\n**15.40** Show that
-\n**(f)** $F(s) = \frac{s^2 + 4}{(s^2 + 2s + 17)(s^2 + 6s + 3)}$
-\n**(g)** Show that
-\n**(h)** $F(s) = \frac{s^2 + 4}{(s^2 + 2s + 17)(s^2 + 6s + 3)}$
-\n**(i)** $F(s) = \frac{s^2 + 4}{(s^2 + 2s + 17)(s^2 +$
-
-**15.38** Find *f*(*t*) given that:
-
-Find
-$$
-f(t)
-$$
- given that:
-\n(a) $F(s) = \frac{s^2 + 4s}{s^2 + 10s + 26}$
-\n(b) $F(s) = \frac{5s^2 + 7s + 29}{s(s^2 + 4s + 29)}$
-
-**15.39** Determine *f*(*t*) if: \*
-
-$$
-s(s^{2} + 4s + 29)
-$$
-
-Determine $f(t)$ if:
-(a) $F(s) = \frac{2s^{3} + 4s^{2} + 1}{(s^{2} + 2s + 17)(s^{2} + 4s + 20)}$
-(b) $F(s) = \frac{s^{2} + 4}{(s^{2} + 9)(s^{2} + 6s + 3)}$
-
-**15.40** Show that
-
-at
-\n
-$$
-\mathcal{L}^{-1} \left[ \frac{4s^2 + 7s + 13}{(s+2)(s^2 + 2s + 5)} \right] =
-$$
-\n
-$$
-\left[ \sqrt{2}e^{-t} \cos(2t + 45^\circ) + 3e^{-2t} \right] u(t)
-$$
-
-# Section 15.5 The Convolution Integral
-
-**15.41** Let *x*(*t*) and *y*(*t*) be as shown in Fig. 15.36. Find *z*(*t*) = *x*(*t*) \* *y*(*t*). \*
-
-**15.42** Design a problem to help other students better understand how to convolve two functions together.
-
-For Prob. 15.41.
-
-\* An asterisk indicates a challenging problem.
-
-**Figure 15.37**
-
-- For Prob. 15.43.
-- **15.44** Obtain the convolution of the pairs of signals in Fig. 15.38.
-
-**15.45** Given *h*(*t*) = 4*e*−2*t u*(*t*) and *x*(*t*) = *δ*(*t*) − 2*e*−2*t u*(*t*), find *y*(*t*) = *x*(*t*) \* *h*(*t*).
-
-**15.46** Given the following functions
-
-*x*(*t*) = 2*δ*(*t*), *y*(*t*) = 4*u*(*t*), *z*(*t*) = *e*−2*t u*(*t*), evaluate the following convolution operations.
-
-\n- (a)
- $$
- x(t) \ast y(t)
- $$
-\n- (b) $x(t) \ast z(t)$
-\n- (c) $y(t) \ast z(t)$
-\n- (d) $y(t) \ast [y(t) + z(t)]$
-\n
-
-**15.47** A system has the transfer function
-
-as the transfer function
-$$
-H(s) = \frac{6s}{(s+1)(s+2)}
-$$
-
-- (a) Find the impulse response of the system.
-- (b) Determine the output *y*(*t*), given that the input is *x*(*t*) = *u*(*t*).
-- **15.48** Find *f*(*t*) using convolution given that:
-
-Find
-$$
-f(t)
-$$
- using convolution
-\n(a) $F(s) = \frac{4}{(s^2 + 2s + 5)^2}$
-\n(b) $F(s) = \frac{2s}{(s + 1)(s^2 + 4)}$
-
-**15.49** Use the convolution integral to find: \*
-
-(a)
-$$
-t * e^{at}u(t)
-$$
-
-(b) $\cos(t) * \cos(t)u(t)$
-
-# Section 15.6 Application to Integrodifferential Equations
-
-**15.50** Use the Laplace transform to solve the differential equation
-
-$$
-\frac{d^2v(t)}{dt^2} + 2\frac{dv(t)}{dt} + 10v(t) = 3\cos 2t
-$$
-
-subject to *v*(0) = 1, *dv*(0)∕*dt* = −2.
-
-**15.51** Given that *v*(0) = 5 and *dv*(0)∕*dt* = 10, solve
-
-$$
-\frac{d^2v}{dt^2} + 5\frac{dv}{dt} + 6v = 25e^{-t}u(t)
-$$
-
-**15.52** Use the Laplace transform to find *i*(*t*) for *t* > 0 if
-
-$$
-\frac{d^2i}{dt^2} + 3\frac{di}{dt} + 2i + \delta(t) = 0,
-$$
-
-$$
-i(0) = 0, \qquad i'(0) = 3
-$$
-
-**15.53** Use Laplace transforms to solve for *x*(*t*) in \*
-
-$$
-x(t) = \cos t + \int_0^t e^{\lambda - t} x(\lambda) d\lambda
-$$
-
-**15.54** Design a problem to help other students better understand solving second order differential equations with a time varying input.
-
-**15.55** Solve for *y*(*t*) in the following differential equation if the initial conditions are zero.
-
-$$
-\frac{d^3y}{dt^3} + 6\frac{d^2y}{dt^2} + 8\frac{dy}{dt} = e^{-t}\cos 2t
-$$
-
-**15.56** Solve for *v*(*t*) in the integrodifferential equation
-
-$$
-12\frac{dv}{dt} + 36 \int_0^t v \, d\tau = 0
-$$
-
-given that *v*(0) = 2.
-
-**15.57** Design a problem to help other students better understand solving integrodifferential equations with a periodic input, using Laplace transforms.
-
-**15.58** Given that
-
-$$
-\frac{dv}{dt} + 2v + 5 \int_0^t v(\lambda) d\lambda = 4u(t)
-$$
-
-with *v*(0) = −1, determine *v*(*t*) for *t* > 0.
-
-**15.59** Solve the integrodifferential equation
-
-$$
-\frac{dy}{dt} + 4y + 3 \int_0^t y \, d\tau = 18e^{-2t} u(t), \qquad y(0) = -3
-$$
-
-**15.60** Solve the following integrodifferential equation
-
-$$
-2\frac{dx}{dt} + 5x + 3\int_0^t x\,dt + 4 = \sin 4t, \qquad x(0) = 1
-$$
-
-- **15.61** Solve the following differential equations subject to the specified initial conditions.
- - (a) *d*2 *v*/*dt*2 + 4*v* = 12, *v*(0) = 0, *dv*(0)/*dt* = 2 (b) *d*2 *i*/*dt*2 + 5*di*/*dt* + 4*i* = 8, *i*(0) = −1, *di*(0)/*dt* = 0 (c) *d*2 *v*/*dt*2 + 2*dv*/*dt* + *v* = 3, *v*(0) = 5, *dv*(0)/*dt* = 1 (d) *d*2 *i*/*dt*2 + 2*di*/*dt* + 5*i* = 10, *i*(0) = 4, *di*(0)/*dt* = −2
-
-# **chapter**
-
-# 16
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/184_Chapter 16 - Applications of the Laplace Transform.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/184_Chapter 16 - Applications of the Laplace Transform.md
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-# Applications of the Laplace Transform
-
-*Communication skills are the most important skills any engineer can have. A very critical element in this tool set is the ability to ask a ques tion and understand the answer, a very simple thing and yet it may make the difference between success and failure!*
-
-—James A. Watson
-
-# Enhancing Your Skills and Your Career
-
-# **Asking Questions**
-
-In more than 30 years of teaching, I ha ve struggled with determining ho w best to help students learn. Regardless of how much time students spend in studying for a course, the most helpful activity for students is learning how to ask questions in class and then asking those questions. The student, by asking questions, becomes actively involved in the learning process and no longer is merely a passive receptor of information. I think this acti ve involvement contributes so much to the learning process that it is probably the single most important aspect to the de velopment of a modern engineer . In fact, asking questions is the basis of science. As Charles P. Steinmetz rightly said, "No man really becomes a fool until he stops asking questions."
-
-It seems very straightforward and quite easy to ask questions. Have we not been doing that all our li ves? The truth is to ask questions in an appropriate manner and to maximize the learning process tak es some thought and preparation.
-
-I am sure that there are se veral models one could ef fectively use. Let me share what has w orked for me. The most important thing to k eep in mind is that you do not ha ve to form a perfect question. Because the question-and-answer format allows the question to be developed in an iterative manner, the original question can easily be refined as you go. I frequently tell students that they are most welcome to read their questions in class.
-
-Here are three things you should keep in mind when asking questions. First, prepare your question. If you are lik e many students who are either shy or ha ve not learned to ask questions in class, you may wish to start with a question you ha ve written down outside of class. Second, w ait for an appropriate time to ask the question. Simply use your judgment on that. Third, be prepared to clarify your question by paraphrasing it or saying it in a different way in case you are asked to repeat the question.
-
-Photo by Charles Alexander
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Understand and use effectively circuit element models in the *s*-domain.
-- 2. Understand how to perform circuit analysis in the *s*-domain and how to transform the results back into the time domain.
-- 3. Understand what a transfer function is and how it is used.
-- 4. Understand state variables and how to apply and use them in circuit analysis.
-
-One last comment: Not all professors like students to ask questions in class even though they may say the y do. You need to find out which professors like classroom questions. Good luck in enhancing one of your most important skills as an engineer.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/185_16.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/185_16.1 Introduction.md
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-# **16.1** Introduction
-
-Now that we have introduced the Laplace transform, let us see what we can do with it. Keep in mind that with the Laplace transform we actually have one of the most powerful mathematical tools for analysis, synthesis, and design. Being able to look at circuits and systems in the *s*-domain can help us to understand ho w our circuits and systems really function. In this chapter we will tak e an in-depth look at ho w easy it is to w ork with circuits in the *s*-domain. In addition, we will briefly look at physical systems. We are sure you have studied some mechanical systems and may have used the same differential equations to describe them as we use to describe our electric circuits. Actually that is a wonderful thing about the physical universe in which we li ve; the same dif ferential equations can be used to describe any linear circuit, system, or process. The key is the term *linear*.
-
-A system is a mathematical model of a physical process relating the input to the output.
-
-It is entirely appropriate to consider circuits as systems. Historically, circuits have been discussed as a separate topic from systems, so we will actually talk about circuits and systems in this chapter realizing that circuits are nothing more than a class of electrical systems.
-
-The most important thing to remember is that everything we discussed in the last chapter and in this chapter applies to any linear system. In the last chapter, we saw how we can use Laplace transforms to solv e linear differential equations and integral equations. In this chapter, we introduce the concept of modeling circuits in the *s*-domain. We can use that principle to help us solv e just about an y kind of linear circuit. W e will take a quick look at how state variables can be used to analyze systems with multiple inputs and multiple outputs. Finally, we examine how the Laplace transform is used in netw ork stability analysis and in netw ork synthesis.
-
-# **16.2** Circuit Element Models
-
-Having mastered how to obtain the Laplace transform and its inverse, we are now prepared to employ the Laplace transform to analyze circuits. This usually involves three steps.
-
-# Steps in Applying the Laplace Transform:
-
-- 1. Transform the circuit from the time domain to the *s*-domain.
-- 2. Solve the circuit using nodal analysis, mesh analysis, source transformation, superposition, or any circuit analysis technique with which we are familiar.
-- 3. Take the inverse transform of the solution and thus obtain the solution in the time domain.
-
-Only the first step is new and will be discussed here. As we did in phasor analysis, we transform a circuit in the time domain to the frequenc y or *s*-domain by Laplace transforming each term in the circuit.
-
-For a resistor, the voltage-current relationship in the time domain is
-
-$$
-v(t) = Ri(t) \tag{16.1}
-$$
-
-Taking the Laplace transform, we get
-
-$$
-V(s) = RI(s) \tag{16.2}
-$$
-
-For an inductor,
-
-$$
-v(t) = L \frac{di(t)}{dt}
-$$
- (16.3)
-
-Taking the Laplace transform of both sides gives
-
-$$
-V(s) = L[sI(s) - i(0^{-})] = sLI(s) - Li(0^{-})
-$$
-\n(16.4)
-
-or
-
-$$
-I(s) = \frac{1}{sL} V(s) + \frac{i(0^{-})}{s}
-$$
- (16.5)
-
-The *s*-domain equivalents are shown in Fig. 16.1, where the initial condition is modeled as a voltage or current source.
-
-For a capacitor,
-
-$$
-i(t) = C \frac{dv(t)}{dt}
-$$
- (16.6)
-
-which transforms into the *s*-domain as
-
-$$
-I(s) = C[sV(s) - v(0^{-})] = sCV(s) - Cv(0^{-})
-$$
-\n(16.7)
-
-$$
-\sum_{i=1}^{n} x_i
-$$
-
-$$
-V(s) = \frac{1}{sC} I(s) + \frac{v(0^{-})}{s}
-$$
- (16.8)
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/186_16.2 Circuit Element Models.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/186_16.2 Circuit Element Models.md
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@@ -1,355 +0,0 @@
-i(t) + ‒ *v*(t) i(0) L (a) I(s) + ‒ V(s) sL (b) Li(0‒) (c) V(s) I(s) + ‒ sL i(0‒) s + ‒ **Figure 16.1**
-
-Representation of an inductor: (a) timedomain, (b,c) *s*-domain equivalents.
-
-As one can infer from step 2, all the circuit analysis techniques applied for dc circuits are applicable to the s-domain.
-
-The elegance of using the Laplace transform in circuit analysis lies in the automatic inclusion of the initial conditions in the transformation process, thus providing a complete (transient and steady-state) solution.
-
-# **Figure 16.3**
-
-Time-domain and *s*-domain representations of passive elements under zero initial conditions.
-
-The *s*-domain equi valents are sho wn in Fig. 16.2. With the *s*-domain equivalents, the Laplace transform can be used readily to solv e first- and second-order circuits such as those we considered in Chapters 7 and 8. We should observe from Eqs. (16.3) to (16.8) that the initial conditions are part of the transformation. This is one advantage of using the Laplace transform in circuit analysis. Another advantage is that a complete re sponse—transient and steady state—of a netw ork is obtained. We will illustrate this with Examples 16.2 and 16.3. Also, observe the duality of Eqs. (16.5) and (16.8), confirming what we already know from Chapter 8 (see Table 8.1), namely, that *L* and *C, I*(*s*) and *V*(*s*), and *v*(0) and *i*(0) are dual pairs.
-
-If we assume zero initial conditions for the inductor and the capacitor, the above equations reduce to:
-
-Resistor:
-$$
-V(s) = RI(s)
-$$
-
-Inductor: $V(s) = sLI(s)$ (16.9)
-Capacitor: $V(s) = \frac{1}{sC}I(s)$
-
-The *s*-domain equivalents are shown in Fig. 16.3.
-
-We define the impedance in the *s*-domain as the ratio of the voltage transform to the current transform under zero initial conditions; that is,
-
-$$
-Z(s) = \frac{V(s)}{I(s)}
-$$
-(16.10)
-
-Thus, the impedances of the three circuit elements are
-
-Resistor:
-$$
-Z(s) = R
-$$
-
-Inductor: $Z(s) = sL$ (16.11)
-Capacitor: $Z(s) = \frac{1}{sC}$
-
-# Table 16.1 summarizes these. The admittance in the *s*-domain is the reciprocal of the impedance, or
-
-$$
-Y(s) = \frac{1}{Z(s)} = \frac{I(s)}{V(s)}
-$$
-(16.12)
-
-The use of the Laplace transform in circuit analysis f acilitates the use of various signal sources such as impulse, step, ramp, e xponential, and sinusoidal.
-
-The models for dependent sources and op amps are easy to develop drawing from the simple fact that if the Laplace transform of *f*(*t*) is *F*(*s*),
-
-# **TABLE 16.1**
-
-Impedance of an element in the s-domain.\*
-
-| Element | Z(s) = V(s)∕I(s) |
-|-----------|------------------|
-| Resistor | R |
-| Inductor | sL |
-| Capacitor | 1∕sC |
-| | |
-
-\* Assuming zero initial conditions
-
-then the Laplace transform of *af*(*t*) is *aF*(*s*)—the linearity property. The dependent source model is a little easier in that we deal with a single value. The dependent source can have only two controlling values, a constant times either a voltage or a current. Thus,
-
-$$
-\mathcal{L}[av(t)] = aV(s) \tag{16.13}
-$$
-
-$$
-\mathcal{L}[ai(t)] = aI(s) \tag{16.14}
-$$
-
-The ideal op amp can be treated just lik e a resistor. Nothing within an op amp, either real or ideal, does an ything more than multiply a voltage by a constant. Thus, we only need to write the equations as we always do using the constraint that the input voltage to the op amp has to be zero and the input current has to be zero.
-
-Find *vo*(*t*) in the circuit of Fig. 16.4, assuming zero initial conditions. Example 16.1
-
-# **Solution:**
-
-We first transform the circuit from the time domain to the *s*-domain.
-
-*u*(*t*) ⇒ \_\_1 *s* 1 H ⇒ *sL* = *s* \_\_1 3 F⇒ \_\_\_1 *sC* = \_\_3 *s*
-
-The resulting *s*-domain circuit is in Fig. 16.5. We now apply mesh analysis. For mesh 1,
-
-$$
-\frac{1}{s} = \left(1 + \frac{3}{s}\right)I_1 - \frac{3}{s}I_2\tag{16.1.1}
-$$
-
-For mesh 2,
-
-$$
-0 = -\frac{3}{s}I_1 + \left(s + 5 + \frac{3}{s}\right)I_2
-$$
-
-or
-
-$$
-I_1 = \frac{1}{3}(s^2 + 5s + 3)I_2
-$$
- (16.1.2)
-
-Substituting this into Eq. (16.1.1),
-
-$$
-\frac{1}{s} = \left(1 + \frac{3}{s}\right)\frac{1}{3}(s^2 + 5s + 3)I_2 - \frac{3}{s}I_2
-$$
-
-Multiplying through by 3*s* gives
-
-ng through by 3*s* gives
-\n
-$$
-3 = (s^3 + 8s^2 + 18s)I_2 \implies I_2 = \frac{3}{s^3 + 8s^2 + 18s}
-$$
-\n
-$$
-V_o(s) = sI_2 = \frac{3}{s^2 + 8s + 18} = \frac{3}{\sqrt{2}} \frac{\sqrt{2}}{(s + 4)^2 + (\sqrt{2})^2}
-$$
-
-Taking the inverse transform yields
-
-$$
-v_o(t) = \frac{3}{\sqrt{2}} e^{-4t} \sin \sqrt{2t} \text{ V}, \qquad t \ge 0
-$$
-
-**Figure 16.4** For Example 16.1.
-
-# **Figure 16.5**
-
-Mesh analysis of the frequency-domain equivalent of the same circuit.
-
-+ ‒ 4 Ω *v*o(t) 1 H F 2.5u(t) V 1 4
-
-+
-
-u(t) V (t) 2*δ*(t) A +
-
-‒
-
-0.1 F
-
-**Answer:** 10(1 − *e*−2*t* − 2*te*−2*t* )*u*(*t*) V.
-
-**Figure 16.6**
-
-For Practice Prob. 16.1.
-
-Example 16.2 Find *vo*(*t*) in the circuit of Fig. 16.7. Assume *vo*(0) = 5 V.
-
-10 Ω
-
-10 Ω *v*o 10e‒t
-
-‒
-
-**Figure 16.7** For Example 16.2.
-
-# **Solution:**
-
-We transform the circuit to the *s*-domain as shown in Fig. 16.8. The initial condition is included in the form of the current source *Cvo*(0) = 0.1(5) = 0.5 A. [See Fig. 16.2(c).] We apply nodal analysis. At the top node,
-
-16.2(c).] We apply nodal analysis. At the to
-\n
-$$
-\frac{10/(s+1) - V_o}{10} + 2 + 0.5 = \frac{V_o}{10} + \frac{V_o}{10/s}
-$$
-
-or
-
-$$
-\frac{1}{s+1} + 2.5 = \frac{2V_o}{10} + \frac{sV_o}{10} = \frac{1}{10}V_o(s+2)
-$$
-
-Multiplying through by 10,
-
-$$
-\frac{10}{s+1} + 25 = V_o(s+2)
-$$
-
-or
-
-$$
-V_o = \frac{25s + 35}{(s + 1)(s + 2)} = \frac{A}{s + 1} + \frac{B}{s + 2}
-$$
-
-where
-
-$$
-A = (s+1)V_o(s) \Big|_{s=-1} = \frac{25s+35}{(s+2)} \Big|_{s=-1} = \frac{10}{1} = 10
-$$
-
-$$
-B = (s+2)V_o(s) \Big|_{s=-2} = \frac{25s+35}{(s+1)} \Big|_{s=-2} = \frac{-15}{-1} = 15
-$$
-
-Thus,
-
-$$
-V_o(s) = \frac{10}{s+1} + \frac{15}{s+2}
-$$
-
-Taking the inverse Laplace transform, we obtain
-
-$$
-v_o(t) = (10e^{-t} + 15e^{-2t})u(t) \text{ V}
-$$
-
-Find *vo*(*t*) in the circuit shown in Fig. 16.9. Note that, since the voltage input is multiplied by *u*(*t*), the voltage source is a short for all *t* < 0 and *iL*(0) = 0.
-
-**Answer:**
-$$
-(12e^{-2t} - 2e^{-t/3})u(t)
-$$
- V.
-
-**Figure 16.9** For Practice Prob. 16.2.
-
-**Figure 16.10** For Example 16.3.
-
-In the circuit of Fig. 16.10(a), the switch moves from position *a* to posi- Example 16.3 tion *b* at *t* = 0. Find *i*(*t*) for *t* > 0.
-
-# **Solution:**
-
-The initial current through the inductor is *i*(0) = *Io*. For *t* > 0, Fig. 16.10(b) shows the circuit transformed to the *s*-domain. The initial condition is incorporated in the form of a voltage source as *Li*(0) = *LIo*. Using mesh analysis,
-
-$$
-I(s)(R + sL) - L I_o - \frac{V_o}{s} = 0
-$$
-\n(16.3.1)
-
-or
-
-$$
-I(s) = \frac{L I_o}{R + sL} + \frac{V_o}{s(R + sL)} = \frac{I_o}{s + R/L} + \frac{V_o/L}{s(s + R/L)}
-$$
-(16.3.2)
-
-Applying partial fraction expansion on the second term on the right-hand side of Eq. (16.3.2) yields
-
-$$
-I(s) = \frac{I_o}{s + R/L} + \frac{V_o/R}{s} - \frac{V_o/R}{(s + R/L)}
-$$
-(16.3.3)
-
-The inverse Laplace transform of this gives
-
-$$
-i(t) = \left(I_o - \frac{V_o}{R}\right)e^{-t/\tau} + \frac{V_o}{R}, \qquad t \ge 0
-$$
- (16.3.4)
-
-where *τ* = *R*∕*L*. The term in parentheses is the transient response, while the second term is the steady-state response. In other words, the final value is *i*(∞) = *Vo*∕*R*, which we could have predicted by applying the final-value theorem on Eq. (16.3.2) or (16.3.3); that is,
-
-$$
-\lim_{s \to 0} sI(s) = \lim_{s \to 0} \left( \frac{sI_o}{s + R/L} + \frac{V_o/L}{s + R/L} \right) = \frac{V_o}{R}
-$$
- (16.3.5)
-
-Equation (16.3.4) may also be written as
-
-$$
-i(t) = I_0 e^{-t/\tau} + \frac{V_o}{R} (1 - e^{-t/\tau}), \qquad t \ge 0
-$$
- (16.3.6)
-
-The first term is the natural response, while the second term is the forced response. If the initial condition *Io* = 0, Eq. (16.3.6) becomes
-
-$$
-i(t) = \frac{V_o}{R}(1 - e^{-t/\tau}), \qquad t \ge 0
-$$
- (16.3.7)
-
-which is the step response, since it is due to the step input *Vo* with no initial energy.
-
-**Figure 16.11** For Practice Prob. 16.3.
-
-Practice Problem 16.3 The switch in Fig. 16.11 has been in position *b* for a long time. It is moved to position *a* at *t* = 0. Determine *v*(*t*) for *t* > 0.
-
-> **Answer:** *v*(*t*) = (*Vo* − *IoR*)*e*−*t*∕*τ* + *IoR*, *t* > 0, where *τ* = *RC*.
-
-# **16.3** Circuit Analysis
-
-Circuit analysis is again relatively easy to do when we are in the *s*-domain. We merely need to transform a complicated set of mathematical relationships in the time domain into the *s*-domain where we convert operators (derivatives and integrals) into simple multipliers of *s* and 1∕*s*. This now allows us to use algebra to set up and solv e our circuit equations. The exciting thing about this is that *all* of the circuit theorems and relation ships we developed for dc circuits are perfectly valid in the *s*-domain.
-
-Remember, equivalent circuits, with capacitors and inductors, only exist in the s-domain; they cannot be transformed back into the time domain.
-
-Example 16.4 Consider the circuit in Fig. 16.12(a). Find the value of the voltage across the capacitor assuming that the value of *vs*(*t*) = 10*u*(*t*) V and assume that at *t* = 0, −1 A flows through the inductor and +5 V is across the capacitor.
-
-# **Solution:**
-
-Figure 16.12(b) represents the entire circuit in the *s*-domain with the initial conditions incorporated. We now have a straightforward nodal analysis problem. Because the value of *V*1 is also the value of the capacitor voltage in the time domain and is the only unknown node voltage, we only need to write one equation.
-
-$$
-\frac{V_1 - 10/s}{10/3} + \frac{V_1 - 0}{5s} + \frac{i(0)}{s} + \frac{V_1 - [v(0)/s]}{1/(0.1s)} = 0
-$$
- (16.4.1)
-
-or
-
-$$
-0.1\left(s+3+\frac{2}{s}\right)V_1 = \frac{3}{s} + \frac{1}{s} + 0.5\tag{16.4.2}
-$$
-
-where *v*(0) = 5 V and *i*(0) = −1 A. Simplifying we get
-
-$$
-(s^2 + 3s + 2) V_1 = 40 + 5s
-$$
-
-or
-
-$$
-V_1 = \frac{40 + 5s}{(s+1)(s+2)} = \frac{35}{s+1} - \frac{30}{s+2}
-$$
- (16.4.3)
-
-Taking the inverse Laplace transform yields
-
-$$
-v_1(t) = (35e^{-t} - 30e^{-2t})u(t) \text{ V}
-$$
- (16.4.4)
-
-For the circuit shown in Fig. 16.12 with the same initial conditions, find Practice Problem 16.4 the current through the inductor for all time *t* > 0.
-
-**Answer:**
-$$
-i(t) = (3 - 7e^{-t} + 3e^{-2t})u(t)
-$$
- A.
-
-For the circuit sho wn in Fig. 16.12, and the initial conditions used Example 16.5 in Example 16.4, use superposition to find the value of the capacitor voltage.
-
-# **Solution:**
-
-Inasmuch as the circuit in the *s*-domain actually has three independent sources, we can look at the solution one source at a time. Figure 16.13 presents the circuits in the *s*-domain considering one source at a time. We now have three nodal analysis problems. First, let us solve for the capacitor voltage in the circuit shown in Fig. 16.13(a).
-
-$$
-\frac{V_1 - 10/s}{10/3} + \frac{V_1 - 0}{5s} + 0 + \frac{V_1 - 0}{1/(0.1s)} = 0
-$$
-
-or
-
-$$
-0.1\left(s+3+\frac{2}{s}\right)V_1 = \frac{3}{s}
-$$
-
-Simplifying we get
-
-2
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/187_16.3 Circuit Analysis.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/187_16.3 Circuit Analysis.md
deleted file mode 100644
index 259cebcbbbcf822626a4a98920af1d4609c4e312..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/187_16.3 Circuit Analysis.md
+++ /dev/null
@@ -1,279 +0,0 @@
-(*s*
-
- + 3*s* + 2)*V*1 = 30 *V*1 = \_\_\_\_\_\_\_\_\_\_\_\_ 30 (*s* + 1)(*s* + 2) = \_\_\_\_\_ 30 *s* + 1 − \_\_\_\_\_ 30 *s* + 2
-
-or
-
-$$
-v_1(t) = (30e^{-t} - 30e^{-2t})u(t) \text{ V} \tag{16.5.1}
-$$
-
-For Fig. 16.13(b) we get,
-
-$$
-\frac{V_2 - 0}{10/3} + \frac{V_2 - 0}{5s} - \frac{1}{s} + \frac{V_2 - 0}{1/(0.1s)} = 0
-$$
-
-or
-
-$$
-0.1\left(s+3+\frac{2}{s}\right)V_2 = \frac{1}{s}
-$$
-
-This leads to
-
-$$
-V_2 = \frac{10}{(s+1)(s+2)} = \frac{10}{s+1} - \frac{10}{s+2}
-$$
-
-Taking the inverse Laplace transform, we get
-
-$$
-v_2(t) = (10e^{-t} - 10e^{-2t})u(t) \text{ V}
-$$
- (16.5.2)
-
-For Fig. 16.13(c),
-
-$$
-\frac{V_3 - 0}{10/3} + \frac{V_3 - 0}{5s} - 0 + \frac{V_3 - 5/s}{1/(0.1s)} = 0
-$$
-
-$$
-0.1\left(s+3+\frac{2}{s}\right)V_3 = 0.5
-$$
-$$
-V_3 = \frac{5s}{(s+1)(s+2)} = \frac{-5}{s+1} + \frac{10}{s+2}
-$$
-
-This leads to
-
-$$
-v_3(t) = (-5e^{-t} + 10e^{-2t})u(t) \text{ V}
-$$
- (16.5.3)
-
-Now all we need to do is to add Eqs. (16.5.1), (16.5.2), and (16.5.3):
-
-$$
-v(t) = v_1(t) + v_2(t) + v_3(t)
-$$
-
-= { (30 + 10 - 5)e-t + (-30 + 10 - 10)e-2t}u(t) V
-
-or
-
-$$
-v(t) = (35e^{-t} - 30e^{-2t})u(t) \text{ V}
-$$
-
-which agrees with our answer in Example 16.4.
-
-Practice Problem 16.5 For the circuit shown in Fig. 16.12, and the same initial conditions in Example 16.4, find the current through the inductor for all time *t* > 0 using superposition.
-
-**Answer:**
-$$
-i(t) = (3 - 7e^{-t} + 3e^{-2t})u(t)
-$$
- A.
-
-**Figure 16.14** For Example 16.6.
-
-Example 16.6 Assume that there is no initial ener gy stored in the circuit of Fig. 16.14 at *t* = 0 and that *is* = 10*u*(*t*) A. (a) Find *Vo*(*s*) using Thevenin's theorem. (b) Apply the initial- and final-value theorems to find *vo*(0+) and *vo*(∞). (c) Determine *vo*(*t*).
-
-# **Solution:**
-
-Because there is no initial energy stored in the circuit, we assume that the initial inductor current and initial capacitor voltage are zero at *t* = 0.
-
-(a) To find the Thevenin equivalent circuit, we remove the 5-Ω resistor and then find *V*oc (*V*Th) and *I*sc. To find *V*Th, we use the Laplacetransformed circuit in Fig. 16.15(a). Since *Ix* = 0, the dependent voltage source contributes nothing, so
-
-$$
-V_{\text{oc}} = V_{\text{Th}} = 5\left(\frac{10}{s}\right) = \frac{50}{s}
-$$
-
-To find *Z*Th, we consider the circuit in Fig. 16.15(b), where we first find *I*sc. We can use nodal analysis to solve for *V*1 which then leads to *I*sc(*I*sc = *Ix* = *V*1∕2*s*).
-
-s).
-$$
--\frac{10}{s} + \frac{(V_1 - 2I_x) - 0}{5} + \frac{V_1 - 0}{2s} = 0
-$$
-
-along with
-
-$$
-I_x = \frac{V_1}{2s}
-$$
-
-leads to
-
-$$
-V_1 = \frac{100}{2s + 3}
-$$
-
-Hence,
-
-$$
-I_{\rm sc} = \frac{V_1}{2s} = \frac{100/(2s+3)}{2s} = \frac{50}{s(2s+3)}
-$$
-
-and
-
-$$
-Z_{\text{Th}} = \frac{V_{\text{oc}}}{I_{\text{sc}}} = \frac{50/s}{50/[s(2s+3)]} = 2s + 3
-$$
-
-The given circuit is replaced by its Thevenin equivalent at terminals *a*-*b* as shown in Fig. 16.16. From Fig. 16.16,
-
-$$
-V_o = \frac{5}{5 + Z_{\text{Th}}} V_{\text{Th}} = \frac{5}{5 + 2s + 3} \left(\frac{50}{s}\right) = \frac{250}{s(2s + 8)} = \frac{125}{s(s + 4)}
-$$
-
-(b) Using the initial-value theorem we find
-
-$$
-v_o(0) = \lim_{s \to \infty} sV_o(s) = \lim_{s \to \infty} \frac{125}{s+4} = \lim_{s \to \infty} \frac{125/s}{1+4/s} = \frac{0}{1} = 0
-$$
-
-Using the final-value theorem we find
-
-$$
-v_o(\infty) = \lim_{s \to 0} sV_o(s) = \lim_{s \to 0} \frac{125}{s+4} = \frac{125}{4} = 31.25 \text{ V}
-$$
-
-(c) By partial fraction,
-
-$$
-V_o = \frac{125}{s(s+4)} = \frac{A}{s} + \frac{B}{s+4}
-$$
-
-\n
-$$
-A = sV_o(s) \Big|_{s=0} = \frac{125}{s+4} \Big|_{s=0} = 31.25
-$$
-
-\n
-$$
-B = (s+4)V_o(s) \Big|_{s=-4} = \frac{125}{s} \Big|_{s=-4} = -31.25
-$$
-
-\n
-$$
-V_o = \frac{31.25}{s} - \frac{31.25}{s+4}
-$$
-
-Taking the inverse Laplace transform gives
-
-$$
-v_o(t) = 31.25(1 - e^{-4t})u(t)
-$$
- V
-
-Notice that the values of *vo*(0) and *vo*(∞) obtained in part (b) are confirmed.
-
-The initial energy in the circuit of Fig. 16.17 is zero at *t* = 0. Assume that Practice Problem 16.6 *vs* = 360*u*(*t*) V. (a) Find *Vo*(*s*) using the Thevenin theorem. (b) Apply the initial- and final-value theorems to find *vo*(0) and *vo*(∞). (c) Obtain *vo*(*t*).
-
-**Answer:** (a)
-$$
-V_o(s) = \frac{288(s+0.25)}{s(s+0.3)}
-$$
-, (b) 288 V, 240 V,
-(c) $(240 + 48e^{-0.3t})u(t)$ V.
-
-(b)
-
-# **Figure 16.15** For Example 16.6: (a) finding *V*Th, (b) determining *Z*Th.
-
-**Figure 16.16** The Thevenin equivalent of the circuit in Fig. 16.14 in the *s*-domain.
-
-**Figure 16.17** For Practice Prob. 16.6.
-
-# **16.4** Transfer Functions
-
-The *transfer function* is a key concept in signal processing because it indicates how a signal is processed as it passes through a network. It is a fitting tool for finding the network response, determining (or designing for) network stability, and network synthesis. The transfer function of a network describes how the output behaves with respect to the input. It specifies the transfer from the input to the output in the *s*-domain, assuming no initial energy.
-
-The transfer function H(s) is the ratio of the output response Y(s) to the input excitation X(s), assuming all initial conditions are zero.
-
-Thus,
-
-$$
-H(s) = \frac{Y(s)}{X(s)}
-$$
-\n(16.15)
-
-The transfer function depends on what we define as input and output. Because the input and output can be either current or voltage at any place in the circuit, there are four possible transfer functions:
-
-$$
-H(s) = \text{Voltage gain} = \frac{V_o(s)}{V_i(s)}\tag{16.16a}
-$$
-
-$$
-H(s) = \text{Current gain} = \frac{I_o(s)}{I_i(s)}\tag{16.16b}
-$$
-
-$$
-H(s) = \text{Impedance} = \frac{V(s)}{I(s)}\tag{16.16c}
-$$
-
-$$
-H(s) = \text{Admittance} = \frac{I(s)}{V(s)}\tag{16.16d}
-$$
-
-Thus, a circuit can ha ve man y transfer functions. Note that *H*(*s*) is dimensionless in Eqs. (16.16a) and (16.16b).
-
-Each of the transfer functions in Eq. (16.16) can be found in two ways. One w ay is to assume any convenient input *X*(*s*), use any circuit analysis technique (such as current or v oltage division, nodal or mesh analysis) to find the output *Y*(*s*), and then obtain the ratio of the two. The other approach is to apply the *ladder method*, which involves walking our way through the circuit. By this approach, we assume that the output is 1 V or 1 A as appropriate and use the basic laws of Ohm and Kirchhoff (KCL only) to obtain the input. The transfer function becomes unity divided by the input. This approach may be more convenient to use when the circuit has many loops or nodes so that applying nodal or mesh analysis becomes cumbersome. In the first method, we assume an input and find the output; in the second method, we assume the output and find the input. In both methods, we calculate *H*(*s*) as the ratio of output to input transforms. The two methods rely on the linearity property, since we only deal with linear circuits in this book. Example 16.8 illustrates these methods.
-
-Some authors would not consider Eqs. (16.16c) and (16.16d) transfer functions.
-
-For electrical networks, the transfer function is also known as the network
-
-function.
-
-Equation (16.15) assumes that both *X*(*s*) and *Y*(*s*) are known. Sometimes, we know the input *X*(*s*) and the transfer function *H*(*s*). We find the output *Y*(*s*) as
-
-$$
-Y(s) = H(s)X(s) \tag{16.17}
-$$
-
-and take the inverse transform to get *y*(*t*). A special case is when the input is the unit impulse function, *x*(*t*) = *δ*(*t*), so that *X*(*s*) = 1. For this case,
-
-$$
-Y(s) = H(s)
-$$
- or $y(t) = h(t)$ (16.18)
-
-where
-
-$$
-h(t) = \mathcal{L}^{-1}[H(s)]
-$$
- (16.19)
-
-The term *h*(*t*) represents the *unit impulse response*—it is the time-domain response of the netw ork to a unit impulse. Thus, Eq. (16.19) pro vides a new interpretation for the transfer function: *H*(*s*) is the Laplace transform of the unit impulse response of the netw ork. Once we kno w the impulse response *h*(*t*) of a network, we can obtain the response of the netw ork to *any* input signal using Eq. (16.17) in the *s*-domain or using the convolution integral (section 15.5) in the time domain.
-
-The output of a linear system is *y*(*t*) = 10*e* Example 16.7 −*t* cos 4*t u*(*t*) when the input is *x*(*t*) = *e*−*t u*(*t*). Find the transfer function of the system and its impulse response.
-
-# **Solution:**
-
-If *x*(*t*) = *e*−*t u*(*t*) and *y*(*t*) = 10*e*−*t* cos 4*t u*(*t*), then
-
-$$
-X(s) = \frac{1}{s+1}
-$$
- and $Y(s) = \frac{10(s+1)}{(s+1)^2 + 4^2}$
-
-Hence,
-
-$$
-H(s) = \frac{Y(s)}{X(s)} = \frac{10(s+1)^2}{(s+1)^2 + 16} = \frac{10(s^2 + 2s + 1)}{s^2 + 2s + 17}
-$$
-
-To find *h*(*t*), we write *H*(*s*) as
-
-$$
-H(s) = 10 - 40 \frac{4}{(s+1)^2 + 4^2}
-$$
-
-From Table 15.2, we obtain
-
-$$
-h(t) = 10\delta(t) - 40e^{-t}\sin 4t \,u(t)
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/188_16.4 Transfer Functions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/188_16.4 Transfer Functions.md
deleted file mode 100644
index b264ac8fb8ba9d0f4510ed909b892bceee9788c4..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/188_16.4 Transfer Functions.md
+++ /dev/null
@@ -1,224 +0,0 @@
-$$
-
-The transfer function of a linear system is Practice Problem 16.7
-
-$$
-H(s) = \frac{2s}{s+6}
-$$
-
-Find the output *y*(*t*) due to the input 45*e* −3*t u*(*t*) and its impulse response.
-
-**Answer:** −90*e*−3*t* + 180*e*−6*t* , *t* ≥ 0, 2*δ*(*t*) − 12*e*−6*t u*(*t*). The unit impulse response is the output response of a circuit when the input is a unit impulse.
-
-I 1
-
-Example 16.8 Determine the transfer function *H*(*s*) = *Vo*(*s*)∕*Io*(*s*) of the circuit in Fig. 16.18.
-
-# **Solution:**
-
-■ **METHOD 1** By current division,
-
-$$
-I_2 = \frac{(s+4)I_o}{s+4+2+1/2s}
-$$
-
-But
-
-+ ‒ Vo
-
-2 Ω
-
-4 Ω
-
-s
-
-$$
-V_o = 2I_2 = \frac{2(s+4)I_o}{s+6+1/2s}
-$$
-
-Hence,
-
-$$
-H(s) = \frac{V_o(s)}{I_o(s)} = \frac{4s(s+4)}{2s^2 + 12s + 1}
-$$
-
-■ **METHOD 2** We can apply the ladder method. We let *Vo* = 1 V. By Ohm's law, *I*2 = *Vo*∕2 = 1∕2 A. The voltage across the (2 + 1∕2*s*) impedance is
-
-$$
-V_1 = I_2 \left( 2 + \frac{1}{2s} \right) = 1 + \frac{1}{4s} = \frac{4s + 1}{4s}
-$$
-
-This is the same as the voltage across the (*s* + 4) impedance. Hence,
-
-$$
-I_1 = \frac{V_1}{s+4} = \frac{4s+1}{4s(s+4)}
-$$
-
-Applying KCL at the top node yields
-
-$$
-I_o = I_1 + I_2 = \frac{4s + 1}{4s(s + 4)} + \frac{1}{2} = \frac{2s^2 + 12s + 1}{4s(s + 4)}
-$$
-
-Hence,
-
-$$
-H(s) = \frac{V_o}{I_o} = \frac{1}{I_o} = \frac{4s(s+4)}{2s^2 + 12s + 1}
-$$
-
-as before.
-
-Practice Problem 16.8 Find the transfer function *H*(*s*) = *I*1(*s*)∕*Io*(*s*) in the circuit of Fig. 16.18.
-
-Answer:
-$$
-\frac{4s+1}{2s^2+12s+1}
-$$
-.
-
-V(s)
-
-+ ‒
-
-**Figure 16.18** For Example 16.8.
-
-# **Solution:**
-
-(a) Using voltage division,
-
-$$
-V_o = \frac{1}{s+1} V_{ab}
-$$
- (16.9.1)
-
-But
-
-$$
-V_{ab} = \frac{1|| (s+1)}{1+1|| (s+1)} V_i = \frac{(s+1)/(s+2)}{1+(s+1)/(s+2)} V_i
-$$
-
-or
-
-$$
-V_{ab} = \frac{s+1}{2s+3} V_i
-$$
- (16.9.2)
-
-Substituting Eq. (16.9.2) into Eq. (16.9.1) results in
-
-$$
-V_o = \frac{V_i}{2s + 3}
-$$
-
-Thus, the transfer function is
-
-$$
-H(s) = \frac{V_o}{V_i} = \frac{1}{2s + 3}
-$$
-
-(b) We may write *H*(*s*) as
-
-$$
-H(s) = \frac{1}{2} \frac{1}{s + \frac{3}{2}}
-$$
-
-Its inverse Laplace transform is the required impulse response:
-
-$$
-h(t) = \frac{1}{2}e^{-3t/2}u(t)
-$$
-
-(c) When *vi*(*t*) = *u*(*t*), *Vi*(*s*) = 1∕*s*, and
-
-$$
-V_o(s) = H(s)V_i(s) = \frac{1}{2s(s + \frac{3}{2})} = \frac{A}{s} + \frac{B}{s + \frac{3}{2}}
-$$
-
-where
-
-$$
-A = sV_o(s)|_{s=0} = \frac{1}{2(s + \frac{3}{2})}|_{s=0} = \frac{1}{3}
-$$
-$$
-B = \left(s + \frac{3}{2}\right) V_o(s)|_{s=-3/2} = \frac{1}{2s}|_{s=-3/2} = -\frac{1}{3}
-$$
-
-Hence, for *vi*(*t*) = *u*(*t*),
-
-$$
-V_o(s) = \frac{1}{3} \left( \frac{1}{s} - \frac{1}{s + \frac{3}{2}} \right)
-$$
-
-and its inverse Laplace transform is
-
-$$
-v_o(t) = \frac{1}{3}(1 - e^{-3t/2})u(t) \text{ V}
-$$
-
-For Example 16.9.
-
-(d) When
-$$
-v_i(t) = 8 \cos 2t
-$$
-, then $V_i(s) = \frac{8s}{s^2 + 4}$ , and
-\n
-$$
-V_o(s) = H(s)V_i(s) = \frac{4s}{(s + \frac{3}{2})(s^2 + 4)}
-$$
-\n
-$$
-= \frac{A}{s + \frac{3}{2}} + \frac{Bs + C}{s^2 + 4}
-$$
-\n(16.9.3)
-
-where
-
-$$
-A = \left(s + \frac{3}{2}\right) V_o(s) \Big|_{s = -3/2} = \frac{4s}{s^2 + 4} \Big|_{s = -3/2} = -\frac{24}{25}
-$$
-
-To get *B* and *C*, we multiply Eq. (16.9.3) by (*s* + 3∕2)(*s* 2 + 4). We get
-
-$$
-4s = A(s^{2} + 4) + B(s^{2} + \frac{3}{2}s) + C(s + \frac{3}{2})
-$$
-
-Equating coefficients,
-
-Constant:
-$$
-0 = 4A + \frac{3}{2}C
-$$
- $\Rightarrow$ $C = -\frac{8}{3}A$
-
-\ns: $4 = \frac{3}{2}B + C$
-
-\ns2: $0 = A + B$ $\Rightarrow$ $B = -A$
-
-Solving these gives *A* = −24∕25, *B* = 24∕25, *C* = 64∕25. Hence, for *vi*(*t*) = 8 cos 2*t* V,
-
-$$
-V_o(s) = \frac{-\frac{24}{25}}{s + \frac{3}{2}} + \frac{24}{25} \frac{s}{s^2 + 4} + \frac{32}{25} \frac{2}{s^2 + 4}
-$$
-
-and its inverse is
-
-$$
-v_o(t) = \frac{24}{25} \left( -e^{-3t/2} + \cos 2t + \frac{4}{3} \sin 2t \right) u(t) \text{ V}
-$$
-
-**Figure 16.20**
-
-**Figure 16.21**
-
-A linear system with *m* inputs and *p* outputs.
-
-Practice Problem 16.9 Rework Example 16.9 for the circuit shown in Fig. 16.20.
-
-**Answer:** (a)
-$$
-2/(s + 4)
-$$
-, (b) $2e^{-4t}u(t)$ , (c) $\frac{1}{2}(1 - e^{-4t})u(t)$ V,
-(d) $3.2(-e^{-4t} + \cos 2t + \frac{1}{2} \sin 2t)u(t)$ V.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/189_16.5 State Variables.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/189_16.5 State Variables.md
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-# For Practice Prob. 16.9. **16.5** State Variables
-
-Thus far in this book we have considered techniques for analyzing systems with only one input and only one output. Man y engineering systems have many inputs and many outputs, as shown in Fig. 16.21. The state variable method is a v ery important tool in analyzing systems and understanding such highly complex systems. Thus, the state variable model is more gen eral than the single-input, single-output model, such as a transfer function. Although the topic cannot be adequately co vered in one chapter, let alone one section of a chapter, we will cover it briefly at this point.
-
-In the state variable model, we specify a collection of variables that describe the internal behavior of the system. These variables are known as the *state variables* of the system. They are the variables that determine the future behavior of a system when the present state of the system and the input signals are kno wn. In other w ords, the y are those v ariables which, if known, allow all other system parameters to be determined by using only algebraic equations.
-
-A state variable is a physical property that characterizes the state of a system, regardless of how the system got to that state.
-
-Common examples of state variables are the pressure, volume, and temperature. In an electric circuit, the state v ariables are the inductor current and capacitor voltage since they collectively describe the energy state of the system.
-
-The standard way to represent the state equations is to arrange them as a set of first-order differential equations:
-
-$$
-\dot{x} = Ax + Bz \tag{16.20}
-$$
-
-where
-
-$$
-\dot{\mathbf{x}}(t) = \begin{bmatrix} x_1(t) \\ x_2(t) \\ \vdots \\ x_n(t) \end{bmatrix} = \text{state vector representing } n \text{ state vectors}
-$$
-
-and the dot represents the first derivative with respect to time, i.e.,
-
-$$
-\dot{\mathbf{x}}(t) = \begin{bmatrix} \dot{x}_1(t) \\ \dot{x}_2(t) \\ \vdots \\ \dot{x}_n(t) \end{bmatrix}
-$$
-
-and
-
-$$
-\mathbf{z}(t) = \begin{bmatrix} z_1(t) \\ z_2(t) \\ \vdots \\ z_m(t) \end{bmatrix} = \text{input vector representing } m \text{ inputs}
-$$
-
-**A** and **B** are respectively *n* × *n* and *n* × *m* matrices. In addition to the state equation in Eq. (16.20), we need the output equation. The complete state model or state space is
-
-$$
-\dot{x} = Ax + Bz
-$$
- (16.21a)
-y = Cx + Dz (16.21b)
-
-where
-
-$$
-\mathbf{y}(t) = \begin{bmatrix} y_1(t) \\ y_2(t) \\ \vdots \\ y_p(t) \end{bmatrix} = \text{the output vector representing } p \text{ outputs}
-$$
-
-and **C** and **D** are, respectively, *p* × *n* and *p* × *m* matrices. For the special case of single-input single-output, *n* = *m* = *p* = 1.
-
-Assuming zero initial conditions, the transfer function of the system is found by taking the Laplace transform of Eq. (16.21a); we obtain
-
-$$
-s\mathbf{X}(s) = \mathbf{A}\mathbf{X}(s) + \mathbf{B}\mathbf{Z}(s) \qquad \rightarrow \qquad (s\mathbf{I} - \mathbf{A})\mathbf{X}(s) = \mathbf{B}\mathbf{Z}(s)
-$$
-
-or
-
-$$
-\mathbf{X}(s) = (s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} \mathbf{Z}(s)
-$$
- (16.22)
-
-where **I** is the identity matrix. Taking the Laplace transform of Eq. (16.21b) yields
-
-$$
-\mathbf{Y}(s) = \mathbf{C}\mathbf{X}(s) + \mathbf{D}\mathbf{Z}(s) \tag{16.23}
-$$
-
-Substituting Eq. (16.22) into Eq. (16.23) and di viding by **Z**(*s*) gives the transfer function as
-
-$$
-H(s) = \frac{Y(s)}{Z(s)} = C(sI - A)^{-1}B + D
-$$
- (16.24)
-
-where
-
-**A** = system matrix **B** = input coupling matrix **C** = output matrix **D** = feedforward matrix
-
-In most cases, **D** = **0**, so the degree of the numerator of *H*(*s*) in Eq. (16.24) is less than that of the denominator. Thus,
-
-$$
-H(s) = C(sI - A)^{-1}B
-$$
- (16.25)
-
-Because of the matrix computation in volved, *MATLAB* can be used to find the transfer function.
-
-To apply state variable analysis to a circuit, we follow the following three steps.
-
-Steps to Apply the State Variable Method to Circuit Analysis:
-
-- 1. Select the inductor current *i* and capacitor voltage *v* as the state variables, making sure they are consistent with the passive sign convention.
-- 2. Apply KCL and KVL to the circuit and obtain circuit variables (voltages and currents) in terms of the state v ariables. This should lead to a set of first-order differential equations necessary and sufficient to determine all state variables.
-- 3. Obtain the output equation and put the final result in state-space representation.
-
-Steps 1 and 3 are usually straightforward; the major task is in step 2. We will illustrate this with examples.
-
-# **Solution:**
-
-We select the inductor current *i* and capacitor voltage *v* as the state variables.
-
-$$
-v_L = L \frac{di}{dt} \tag{16.10.1}
-$$
-
-$$
-i_C = C \frac{dv}{dt} \tag{16.10.2}
-$$
-
-Applying KCL at node 1 gives
-
-$$
-i = i_x + i_C
-$$
- $\rightarrow$ $C\frac{dv}{dt} = i - \frac{v}{R}$
-
-or
-
-$$
-\dot{v} = -\frac{v}{RC} + \frac{i}{C}
-$$
- (16.10.3)
-
-since the same voltage *v* is across both *R* and *C*. Applying KVL around the outer loop yields
-
-$$
-v_s = v_L + v \rightarrow L\frac{di}{dt} = -v + v_s
-$$
-
-$$
-\dot{i} = -\frac{v}{L} + \frac{v_s}{L}
-$$
- (16.10.4)
-
-Equations (16.10.3) and (16.10.4) constitute the state equations. If we regard *ix* as the output,
-
-$$
-i_x = \frac{v}{R} \tag{16.10.5}
-$$
-
-Putting Eqs. (16.10.3), (16.10.4), and (16.10.5) in the standard form leads to
-
-$$
-\begin{bmatrix} \dot{v} \\ \dot{i} \end{bmatrix} = \begin{bmatrix} \frac{-1}{RC} & \frac{1}{C} \\ \frac{-1}{L} & 0 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix} + \begin{bmatrix} 0 \\ \frac{1}{L} \end{bmatrix} v_s
-$$
- (16.10.6a)
-$$
-i_x = \begin{bmatrix} \frac{1}{R} & 0 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix}
-$$
- (16.10.6b)
-
-If *R* = 1, *C* = \_1 4 , and *L* = \_1 2 , we obtain from Eq. (16.10.6) matrices
-
-$$
-\mathbf{A} = \begin{bmatrix} \frac{-1}{RC} & \frac{1}{C} \\ \frac{-1}{L} & 0 \end{bmatrix} = \begin{bmatrix} -4 & 4 \\ -2 & 0 \end{bmatrix}, \qquad \mathbf{B} = \begin{bmatrix} 0 \\ \frac{1}{L} \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \end{bmatrix},
-$$
-$$
-\mathbf{C} = \begin{bmatrix} \frac{1}{R} & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \end{bmatrix}
-$$
-$$
-s\mathbf{I} - \mathbf{A} = \begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix} - \begin{bmatrix} -4 & 4 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} s+4 & -4 \\ 2 & s \end{bmatrix}
-$$
-
-Taking the inverse of this gives
-
-inverse of this gives
-\n
-$$
-(s\mathbf{I} - \mathbf{A})^{-1} = \frac{\text{adjoint of } \mathbf{A}}{\text{determinant of } \mathbf{A}} = \frac{\begin{bmatrix} s & 4\\ -2 & s+4 \end{bmatrix}}{s^2 + 4s + 8}
-$$
-
-Thus, the transfer function is given by
-
-Thus, the transfer function is given by
-
-\n
-$$
-\mathbf{H}(s) = \mathbf{C}(s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} = \frac{\begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} s & 4 \\ -2 & s+4 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \end{bmatrix}}{s^2 + 4s + 8} = \frac{8}{s^2 + 4s + 8}
-$$
-\n
-$$
-= \frac{8}{s^2 + 4s + 8}
-$$
-
-which is the same thing we would get by directly Laplace transforming the circuit and obtaining **H**(*s*) = *Ix*(*s*)∕*Vs*(*s*). The real advantage of the state variable approach comes with multiple inputs and multiple outputs. In this case, we have one input *vs* and one output *ix*. In the next example, we will have two inputs and two outputs.
-
-Example 16.11 Consider the circuit in Fig. 16.24, which may be regarded as a two- input, two-output system. Determine the state v ariable model and find the transfer function of the system.
-
-**Figure 16.24** For Example 16.11.
-
-# **Solution:**
-
-In this case, we have two inputs *vs* and *vi* and two outputs *vo* and *io*. Again, we select the inductor current *i* and capacitor voltage *v* as the state variables. Applying KVL around the left-hand loop gives
-
-$$
--v_s + i_1 + \frac{1}{6}i = 0 \quad \to \quad i = 6v_s - 6i_1 \quad (16.11.1)
-$$
-
-We need to eliminate *i*1. Applying KVL around the loop containing *vs*,1-Ω resistor, 2-Ω resistor, and \_1 3 -F capacitor yields
-
-$$
-v_s = i_1 + v_o + v \tag{16.11.2}
-$$
-
-But at node 1, KCL gives
-
-$$
-i_1 = i + \frac{v_o}{2} \rightarrow v_o = 2(i_1 - i)
-$$
- (16.11.3)
-
-For Practice Prob. 16.10.
-
-Substituting this in Eq. (16.11.2),
-
-$$
-v_s = 3i_1 + v - 2i \rightarrow i_1 = \frac{2i - v + v_s}{3}
-$$
- (16.11.4)
-
-Substituting this in Eq. (16.11.1) gives
-
-$$
-\dot{i} = 2v - 4i + 4v_s \tag{16.11.5}
-$$
-
-which is one state equation. To obtain the second one, we apply KCL at node 2.
-
-$$
-\frac{v_o}{2} = \frac{1}{3} \dot{v} + i_o \rightarrow \dot{v} = \frac{3}{2} v_o - 3i_o \tag{16.11.6}
-$$
-
-We need to eliminate *vo* and *io*. From the right-hand loop, it is evident that
-
-$$
-i_o = \frac{v - v_i}{3} \tag{16.11.7}
-$$
-
-Substituting Eq. (16.11.4) into Eq. (16.11.3) gives
-
-*i*
-
-$$
-v_o = 2\left(\frac{2i - v + v_s}{3} - i\right) = -\frac{2}{3}(v + i - v_s) \tag{16.11.8}
-$$
-
-Substituting Eqs. (16.11.7) and (16.11.8) into Eq. (16.11.6) yields the second state equation as
-
-$$
-\dot{v} = -2v - i + v_s + v_i \tag{16.11.9}
-$$
-
-The two output equations are already obtained in Eqs. (16.11.7) and (16.11.8). Putting Eqs. (16.11.5) and (16.11.7) to (16.11.9) together in the standard form leads to the state model for the circuit, namely,
-
-$$
-\begin{bmatrix} \dot{v} \\ \dot{i} \end{bmatrix} = \begin{bmatrix} -2 & -1 \\ 2 & -4 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix} + \begin{bmatrix} 1 & 1 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} v_s \\ v_i \end{bmatrix}
-$$
-(16.11.10a)
-$$
-\begin{bmatrix} v_o \\ i_o \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} & -\frac{2}{3} \\ \frac{1}{3} & 0 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix} + \begin{bmatrix} \frac{2}{3} & 0 \\ 0 & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} v_s \\ v_i \end{bmatrix}
-$$
-(16.11.10b)
-
-For the electric circuit in Fig. 16.25, determine the state model. Take *vo* Practice Problem 16.11 and *io* as the output variables.
-
-**Answer:**
-
-**Figure 16.25** For Practice Prob. 16.11.
-
-Example 16.12 Assume we have a system where the output is *y*(*t*) and the input is *z*(*t*). Let the following differential equation describe the relationship between the input and the output.
-
-$$
-\frac{d^2y(t)}{dt^2} + 3\frac{dy(t)}{dt} + 2y(t) = 5z(t)
-$$
- (16.12.1)
-
-Obtain the state model and the transfer function of the system.
-
-*x .*
-
-# **Solution:**
-
-First, we select the state variables. Let *x*1 = *y*(*t*), therefore
-
-$$
-x_1 = \dot{y}(t) \tag{16.12.2}
-$$
-
-Now let
-
-$$
-x_2 = \dot{x}_1 = \dot{y}(t) \tag{16.12.3}
-$$
-
-Note that at this time we are looking at a second-order system that would normally have two first-order terms in the solution.
-
-Now we have *x .* 2 = *y ..*(*t*), where we can find the value *x .* 2 from Eq. (16.12.1), i.e.,
-
-$$
-\dot{x}_2 = \ddot{y}(t) = -2y(t) - 3\dot{y}(t) + 5z(t) = -2x_1 - 3x_2 + 5z(t)
-$$
- (16.12.4)
-
-From Eqs. (16.12.2) to (16.12.4), we can now write the following matrix equations:
-
-$$
-\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 0 \\ 5 \end{bmatrix} z(t)
-$$
- (16.12.5)
-
-$$
-\mathbf{y}(t) = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \tag{16.12.6}
-$$
-
-We now obtain the transfer function.
-
-$$
-s\mathbf{I} - \mathbf{A} = s \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} = \begin{bmatrix} s & -1 \\ 2 & s + 3 \end{bmatrix}
-$$
-
-The inverse is
-
-$$
-(s\mathbf{I} - \mathbf{A})^{-1} = \frac{\begin{bmatrix} s+3 & 1\\ -2 & s \end{bmatrix}}{s(s+3)+2}
-$$
-
-The transfer function is
-
-ne transfer function is
-\n
-$$
-\mathbf{H}(s) = \mathbf{C}(s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} = \frac{(1 \quad 0) \begin{bmatrix} s+3 & 1 \\ -2 & s \end{bmatrix} \begin{pmatrix} 0 \\ 5 \end{pmatrix}}{s(s+3)+2} = \frac{(1 \quad 0) \begin{pmatrix} 5 \\ 5s \end{pmatrix}}{(s+1)(s+2)}
-$$
-
-To check this, we directly apply the Laplace transfer to each term in Eq. (16.12.1). Given that initial conditions are zero, we get
-
-$$
-[s2 + 3s + 2]Y(s) = 5Z(s) \rightarrow H(s) = \frac{Y(s)}{Z(s)} = \frac{5}{s2 + 3s + 2}
-$$
-
-which is in agreement with what we got previously.
-
-$$
-\frac{d^3y}{dt^3} + 18\frac{d^2y}{dt^2} + 20\frac{dy}{dt} + 5y = z(t)
-$$
-
-**Answer:**
-
-$$
-\mathbf{A} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -20 & -18 \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \quad \mathbf{C} = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}.
-$$
-
-# **16.6** Applications
-
-So far we have considered three applications of Laplace's transform: circuit analysis in general, obtaining transfer functions, and solving linear integrodifferential equations. The Laplace transform also finds application in other areas in circuit analysis, signal processing, and control systems. Here we will consider tw o more important applications: netw ork stability and network synthesis.
-
-# **16.6.1** Network Stability
-
-A circuit is *stable* if its impulse response *h*(*t*) is bounded (i.e., *h*(*t*) converges to a finite value) as *t* → ∞; it is *unstable* if *h*(*t*) grows without bound as *t* → ∞. In mathematical terms, a circuit is stable when
-
-$$
-\lim_{t \to \infty} |h(t)| = \text{finite} \tag{16.26}
-$$
-
-Because the transfer function *H*(*s*) is the Laplace transform of the impulse response *h*(*t*), *H*(*s*) must meet certain requirements for Eq. (16.26) to hold. Recall that *H*(*s*) may be written as
-
-$$
-H(s) = \frac{N(s)}{D(s)}
-$$
-(16.27)
-
-where the roots of *N*(*s*) = 0 are called the *zeros* of *H*(*s*) because the y make *H*(*s*) = 0, while the roots of *D*(*s*) = 0 are called the *poles* of *H*(*s*) since they cause *H*(*s*) → ∞. The zeros and poles of *H*(*s*) are often located in the *s* plane as sho wn in Fig. 16.26(a). Recall from Eqs. (15.47) and (15.48) that *H*(*s*) may also be written in terms of its poles as \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*s* + *p*1)(*s* + *p*2) ⋯ (*s* + *pn*)
-
-$$
-H(s) = \frac{N(s)}{D(s)} = \frac{N(s)}{(s+p_1)(s+p_2)\cdots(s+p_n)}
-$$
-(16.28)
-
-*H*(*s*) must meet tw o requirements for the circuit to be stable. First, the degree of *N*(*s*) must be less than the degree of *D*(*s*); otherwise, long division would produce
-
-$$
-H(s) = k_n s^n + k_{n-1} s^{n-1} + \dots + k_1 s + k_0 + \frac{R(s)}{D(s)}
-$$
-(16.29)
-
-where the de gree of *R*(*s*), the remainder of the long di vision, is less than the degree of *D*(*s*). The inverse of *H*(*s*) in Eq. (16.29) does not meet the condition in Eq. (16.26). Second, all the poles of *H*(*s*) in
-
-# **Figure 16.26** The complex *s* plane: (a) poles and zeros plotted, (b) left-half plane.
-
-Eq. (16.27) (i.e., all the roots of *D*(*s*) = 0) must have negative real parts; in other words, all the poles must lie in the left half of the *s* plane, as shown typically in Fig. 16.26(b). The reason for this will be apparent if we take the inverse Laplace transform of *H*(*s*) in Eq. (16.27). Because Eq. (16.27) is similar to Eq. (15.48), its partial fraction expansion is similar to the one in Eq. (15.49) so that the inverse of *H*(*s*) is similar to that in Eq. (15.53). Hence,
-
-$$
-h(t) = (k_1 e^{-p_1 t} + k_2 e^{-p_2 t} + \dots + k_n e^{-p_n t}) u(t)
-$$
-\n(16.30)
-
-We see from this equation that each pole *pi* must be positive (i.e., pole *s* = −*pi* in the left-half plane) for *e*−*pi t* to decrease with increasing *t*. Thus,
-
-A circuit is stable when all the poles of its transfer function H(s) lie in the left half of the s plane.
-
-An unstable circuit never reaches steady state because the transient response does not decay to zero. Consequently , steady-state analysis is only applicable to stable circuits.
-
-A circuit made up exclusively of passive elements (*R*, *L*, and *C*) and independent sources cannot be unstable, because that w ould imply that some branch currents or v oltages would grow indefinitely with sources set to zero. Passive elements cannot generate such indefinite growth. Passive circuits either are stable or have poles with zero real parts. To show that this is the case, consider the series *RLC* circuit in Fig. 16.27. The transfer function is given by
-
-$$
-H(s) = \frac{V_o}{V_s} = \frac{1/sC}{R + sL + 1/sC}
-$$
-
-$$
-H(s) = \frac{1/L}{s^2 + sR/L + 1/LC}
-$$
- (16.31)
-
-Notice that *D*(*s*) = *s* 2 + *sR*∕*L* + 1∕*LC* = 0 is the same as the characteristic equation obtained for the series *RLC* circuit in Eq. (8.8). The circuit has poles at \_\_\_\_\_\_\_
-
-$$
-p_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}
-$$
- (16.32)
-
-where
-
-or
-
-$$
-\alpha = \frac{R}{2L}, \qquad \omega_0 = \frac{1}{LC}
-$$
-
-For *R*, *L*, *C* > 0, the tw o poles always lie in the left half of the *s* plane, implying that the circuit is al ways stable. However, when *R* = 0, *α* = 0 and the circuit becomes unstable. Although ideally this is possible, it does not really happen, because *R* is never zero.
-
-On the other hand, active circuits or passive circuits with controlled sources can supply energy, and they can be unstable. In fact, an oscillator is a typical example of a circuit designed to be unstable. An oscillator is designed such that its transfer function is of the form
-
-imple of a circuit designed to be unstable. An oscula-
-ch that its transfer function is of the form
-$$
-H(s) = \frac{N(s)}{s^2 + \omega_0^2} = \frac{N(s)}{(s + j\omega_0)(s - j\omega_0)}
-$$
-(16.33)
-
-so that its output is sinusoidal.
-
-**Figure 16.27** A typical *RLC* circuit.
-
-Determine the values of *k* for which the circuit in Fig. 16.28 is stable. Example 16.13
-
-# **Solution:**
-
-Applying mesh analysis to the first-order circuit in Fig. 16.28 gives
-
-$$
-V_i = \left(R + \frac{1}{sC}\right)I_1 - \frac{I_2}{sC}
-$$
- (16.13.1)
-
-and
-
-$$
-0 = -kI_1 + \left(R + \frac{1}{sC}\right)I_2 - \frac{I_1}{sC}
-$$
-
-or
-
-$$
-0 = -\left(k + \frac{1}{sC}\right)I_1 + \left(R + \frac{1}{sC}\right)I_2\tag{16.13.2}
-$$
-
-We can write Eqs. (16.13.1) and (16.13.2) in matrix form as
-
-$$
-\begin{bmatrix} V_i \\ 0 \end{bmatrix} = \begin{bmatrix} \left(R + \frac{1}{sC}\right) & -\frac{1}{sC} \\ -\left(k + \frac{1}{sC}\right) & \left(R + \frac{1}{sC}\right) \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix}
-$$
-
-The determinant is
-
-inant is
-\n
-$$
-\Delta = \left(R + \frac{1}{sC}\right)^2 - \frac{k}{sC} - \frac{1}{s^2C^2} = \frac{sR^2C + 2R - k}{sC}
-$$
-\n(16.13.3)
-
-The characteristic equation (∆ = 0) gives the single pole as
-
-$$
-p = \frac{k - 2R}{R^2 C}
-$$
-
-which is negative when *k* < 2*R*. Thus, we conclude the circuit is stable when *k* < 2*R* and unstable for *k* > 2*R*.
-
-**Answer:** *β* > −1∕*R*.
-
-For Practice Prob. 16.13.
-
-An active filter has the transfer function Example 16.14
-
-$$
-H(s) = \frac{k}{s^2 + s(4 - k) + 1}
-$$
-
-For what values of *k* is the filter stable?
-
-+
-
-1
-
-I1 I2 kI1
-
-R R
-
-‒ +
-
-# **Solution:**
-
-As a second-order circuit, *H*(*s*) may be written as
-
-$$
-H(s) = \frac{N(s)}{s^2 + bs + c}
-$$
-
-where *b* = 4 − *k*, *c* = 1, and *N*(*s*) = *k*. This has poles at *p*2 + *bp* + *c* = 0; that is,
-
-$$
-p_{1,2} = \frac{-b \pm \sqrt{b^2 - 4c}}{2}
-$$
-
-For the circuit to be stable, the poles must be located in the left half of the *s* plane. This implies that *b* > 0.
-
-Applying this to the given *H*(*s*) means that for the circuit to be stable, 4 − *k* > 0 or *k* < 4.
-
-Practice Problem 16.14 A second-order active circuit has the transfer function
-
-circuit has the transfer function
-$$
-H(s) = \frac{1}{s^2 + s(25 + \alpha) + 25}
-$$
-
-Find the range of the values of *α* for which the circuit is stable. What is the value of *α* that will cause oscillation?
-
-**Answer:** *α* > −25, *α* = −25.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/190_16.6 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/190_16.6 Applications.md
deleted file mode 100644
index 6c944db0452fdf4296a7ae2472ac9bc4a0d62986..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/190_16.6 Applications.md
+++ /dev/null
@@ -1,307 +0,0 @@
-# **16.6.2** Network Synthesis
-
-Network synthesis may be regarded as the process of obtaining an appropriate network to represent a given transfer function. Network synthesis is easier in the *s*-domain than in the time domain.
-
-In network analysis, we find the transfer function of a given network. In netw ork synthesis, we re verse the approach: Gi ven a transfer function, we are required to find a suitable network.
-
-Network synthesis is finding a network that represents a given transfer function.
-
-Keep in mind that in synthesis, there may be man y dif ferent answers—or possibly no answers—because there are many circuits that can be used to represent the same transfer function; in network analysis, there is only one answer.
-
-Network synthesis is an exciting field of prime engineering importance. Being able to look at a transfer function and come up with the type of circuit it represents is a great asset to a circuit de signer. Although network synthesis constitutes a whole course by itself and requires some e xperience, the follo wing e xamples are meant to stimulate your appetite.
-
-Given the transfer function Example 16.15
-
-$$
-H(s) = \frac{V_o(s)}{V_i(s)} = \frac{10}{s^2 + 3s + 10}
-$$
-
-realize the function using the circuit in Fig. 16.30(a). (a) Select *R* = 5 Ω, and find *L* and *C*. (b) Select *R* = 1 Ω, and find *L* and *C*.
-
-# **Solution:**
-
-1. **Define.** The problem is clearly and completely defined. This problem is what we call a synthesis problem: Given a transfer function, synthesize a circuit that produces the given transfer function. However, to keep the problem more manageable, we give a circuit that produces the desired transfer function.
-
-Had one of the variables, *R* in this case, not been given a value, then the problem would have had an infinite number of answers. An open-ended problem of this kind would require some additional assumptions that would have narrowed the set of solutions.
-
-- 2. **Present.** A transfer function of the voltage out versus the voltage in is equal to 10∕(*s* 2 + 3*s* + 10). A circuit, Fig. 16.30, is also given that should be able to produce the required transfer function. Two different values of *R*, 5 and 1 Ω, are to be used to calculate the values of *L* and *C* that produce the given transfer function.
-- 3. **Alternative.** All solution paths involve determining the transfer function of Fig. 16.30 and then matching the various terms of the transfer function. Two approaches would be to use mesh analysis or nodal analysis. Because we are looking for a ratio of voltages, nodal analysis makes the most sense.
-- 4. **Attempt.** Using nodal analysis leads to
-
-$$
-\frac{V_o(s) - V_i(s)}{sL} + \frac{V_o(s) - 0}{1/(sC)} + \frac{V_o(s) - 0}{R} = 0
-$$
-
-Now multiply through by *sLR*:
-
-$$
-RV_o(s) - RV_i(s) + s^2 R LCV_o(s) + sLV_o(s) = 0
-$$
-
-Collecting terms we get
-
-$$
-(s^2RLC + sL + R)V_o(s) = RV_i(s)
-$$
-
-or
-
-$$
-(s2RLC + sL + R)Vo(s) = RVi(s)
-$$
-$$
-\frac{Vo(s)}{Vi(s)} = \frac{1/(LC)}{s2 + [1/(RC)]s + 1/(LC)}
-$$
-
- Matching the two transfer functions produces two equations with three unknowns.
-
-$$
-LC = 0.1 \qquad \text{or} \qquad L = \frac{0.1}{C}
-$$
-
-and
-
-$$
-RC = \frac{1}{3} \qquad \text{or} \qquad C = \frac{1}{3R}
-$$
-
-We have a constraint equation, *R* = 5 Ω for (a) and = 1 Ω for (b).
-
-(a) *C* = 1∕(3 × 5) = **66.67 mF** and *L* = **1.5 H** (b) *C* = 1∕(3 × 1) = **333.3 mF** and *L* = **300 mH**
-
-**Figure 16.30** For Example 16.15.
-
-5. **Evaluate.** There are different ways of checking the answer. Solving for the transfer function by using mesh analysis seems the most straightforward and the approach we could use here. However, it should be pointed out that this is mathematically more complex and will take longer than the original nodal analysis approach. Other approaches also exist. We can assume an input for *vi*(*t*), *vi*(*t*) = *u*(*t*) V and, using either nodal analysis or mesh analysis, see if we get the same answer we would get with just using the transfer function. That is the approach we will try using mesh analysis.
-
-Let *vi*(*t*) = *u*(*t*) *V* or *Vi*(*s*) = 1∕*s*. This will produce
-
-$$
-V_o(s) = 10/(s^3 + 3s^2 + 10s)
-$$
-
-Based on Fig. 16.30, mesh analysis leads to (a) For loop 1,
-
-$$
--(1/s) + 1.5sI_1 + [1/(0.06667s)] (I_1 - I_2) = 0
-$$
-
-or
-
-$$
-(1.5s^2 + 15)I_1 - 15I_2 = 1
-$$
-
-For loop 2,
-
-$$
-(15/s)(I_2 - I_1) + 5I_2 = 0
-$$
-
-or
-
-$$
--15I_1 + (5s + 15)I_2 = 0 \qquad \text{or} \qquad I_1 = (0.3333s + 1)I_2
-$$
-
-Substituting into the first equation we get
-
-$$
-(0.5s3 + 1.5s2 + 5s + 15)I2 - 15I2 = 1
-$$
-
-or
-
-$$
-I_2 = 2/(s^3 + 3s^2 + 10s)
-$$
-
-but
-
-$$
-V_o(s) = 5I_2 = 10/(s^3 + 3s^2 + 10s)
-$$
-
-and the answer checks.
-
-(b) For loop 1,
-
-$$
--(1/s) + 0.3sI_1 + [1/(0.3333s)] (I_1 - I_2) = 0
-$$
-
-or
-
-$$
-(0.3s^2 + 3)I_1 - 3I_2 = 1
-$$
-
-For loop 2,
-
-$$
-(3/s)(I_2 - I_1) + I_2 = 0
-$$
-
-or
-
-$$
--3I_1 + (s+3)I_2 = 0
-$$
- or $I_1 = (0.3333s + 1)I_2$
-
-Substituting into the first equation we get
-
-$$
-(0.09999s3 + 0.3s2 + s + 3)I2 - 3I2 = 1
-$$
-
-or
-
-$$
-I_2 = 10/(s^3 + 3s^2 + 10s)
-$$
-
-but *Vo*(*s*) = 1 × *I*2 = 10∕(*s* 3 + 3*s* 2 + 10*s*) and the answer checks.
-
-6. **Satisfactory?** We have clearly identified values of *L* and *C* for each of the conditions. In addition, we have carefully checked the answers to see if they are correct. The problem has been adequately solved. The results can now be presented as a solution to the problem.
-
-Realize the function Practice Problem 16.15
-
-$$
-G(s) = \frac{V_o(s)}{V_i(s)} = \frac{4s}{s^2 + 4s + 20}
-$$
-
-using the circuit in Fig. 16.31. Select *R* = 2 Ω, and determine *L* and *C*.
-
-**Answer:** 500 mH, 100 mF.
-
-Synthesize the function Example 16.16
-
-ion
-$$
-T(s) = \frac{V_o(s)}{V_s(s)} = \frac{10^6}{s^2 + 100s + 10^6}
-$$
-
-using the topology in Fig. 16.32.
-
-For Example 16.16.
-
-# **Solution:**
-
-We apply nodal analysis to nodes 1 and 2. At node 1,
-
-$$
-(V_s - V_1)Y_1 = (V_1 - V_o)Y_2 + (V_1 - V_2)Y_3 \tag{16.16.1}
-$$
-
-At node 2,
-
-$$
-(V_1 - V_2)Y_3 = (V_2 - 0)Y_4
-$$
- (16.16.2)
-
-But *V*2 = *Vo*, so Eq. (16.16.1) becomes
-
-$$
-Y_1 V_s = (Y_1 + Y_2 + Y_3) V_1 - (Y_2 + Y_3) V_o \tag{16.16.3}
-$$
-
-and Eq. (16.16.2) becomes
-
-$$
-V_1 Y_3 = (Y_3 + Y_4) V_o
-$$
-
-or
-
-$$
-V_1 = \frac{1}{Y_3} (Y_3 + Y_4) V_o
-$$
- (16.16.4)
-
-Substituting Eq. (16.16.4) into Eq. (16.16.3) gives
-
-$$
-Y_1 V_s = (Y_1 + Y_2 + Y_3) \frac{1}{Y_3} (Y_3 + Y_4) V_o - (Y_2 + Y_3) V_o
-$$
-
-or
-
-$$
-Y_1Y_3V_s = [Y_1Y_3 + Y_4(Y_1 + Y_2 + Y_3)]V_o
-$$
-
-Thus,
-
-$$
-V_s = [Y_1Y_3 + Y_4(Y_1 + Y_2 + Y_3)]V_o
-$$
-
-$$
-\frac{V_o}{V_s} = \frac{Y_1Y_3}{Y_1Y_3 + Y_4(Y_1 + Y_2 + Y_3)}
-$$
-(16.16.5)
-
-To synthesize the given transfer function *T*(*s*), compare it with the one in Eq. (16.16.5). Notice two things: (1) *Y*1*Y*3 must not involve *s* because the numerator of *T*(*s*) is constant; (2) the given transfer function is second-order, which implies that we must have two capacitors. Therefore, we must make *Y*1 and *Y*3 resistive, while *Y*2 and *Y*4 are capacitive. So we select
-
-$$
-Y_1 = \frac{1}{R_1}
-$$
-, $Y_2 = sC_1$ , $Y_3 = \frac{1}{R_2}$ , $Y_4 = sC_2$ (16.16.6)
-
-Substituting Eq. (16.16.6) into Eq. (16.16.5) gives
-
-q. (16.16.6) into Eq. (16.16.5) gives
-\n
-$$
-\frac{V_o}{V_s} = \frac{1/(R_1R_2)}{1/(R_1R_2) + sC_2(1/R_1 + 1/R_2 + sC_1)}
-$$
-\n
-$$
-= \frac{1/(R_1R_2C_1C_2)}{s^2 + s(R_1 + R_2)/(R_1R_2C_1) + 1/(R_1R_2C_1C_2)}
-$$
-
-Comparing this with the given transfer function *T*(*s*), we notice that
-
-$$
-\frac{1}{R_1 R_2 C_1 C_2} = 10^6, \qquad \frac{R_1 + R_2}{R_1 R_2 C_1} = 100
-$$
-
-If we select *R*1 = *R*2 = 10 kΩ, then
-
-$$
-R_1 = R_2 = 10 \text{ k}\Omega, \text{ then}
-$$
-
-\n
-$$
-C_1 = \frac{R_1 + R_2}{100R_1R_2} = \frac{20 \times 10^3}{100 \times 100 \times 10^6} = 2 \text{ }\mu\text{F}
-$$
-
-\n
-$$
-C_2 = \frac{10^{-6}}{R_1R_2C_1} = \frac{10^{-6}}{100 \times 10^6 \times 2 \times 10^{-6}} = 5 \text{ nF}
-$$
-
-Thus, the given transfer function is realized using the circuit shown in Fig. 16.33.
-
-For Example 16.16.
-
-Synthesize the function Practice Problem 16.16
-
-$$
-\frac{V_o(s)}{V_{\text{in}}} = \frac{-2s}{s^2 + 6s + 10}
-$$
-
-using the op amp circuit shown in Fig. 16.34. Select
-
-$$
-Y_1 = \frac{1}{R_1}
-$$
-, $Y_2 = sC_1$ , $Y_3 = sC_2$ , $Y_4 = \frac{1}{R_2}$
-
-Let *R*1 = 1 kΩ, and determine *C*1, *C*2, and *R*2.
-
-For Practice Prob. 16.16.
-
-**Answer:** 100 *µ*F, 500 *µ*F, 2 kΩ.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/191_16.7 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/191_16.7 Summary.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/191_16.7 Summary.md
+++ /dev/null
@@ -1,32 +0,0 @@
-# **16.7** Summary
-
-- 1. The Laplace transform can be used to analyze a circuit. We convert each element from the time domain to the *s*-domain, solve the problem using an y circuit analysis technique, and con vert the result to the time domain using the inverse transform.
-- 2. In the *s*-domain, the circuit elements are replaced with the initial condition at *t* = 0 as follo ws. (Please note, v oltage models are
-
-given below, but the corresponding current models w ork equally well.):
-
-Resistor:
-$$
-v_R = Ri \rightarrow V_R = RI
-$$
-
-\nInductor: $v_L = L\frac{di}{dt} \rightarrow V_L = sLI - Li(0^-)$
-\nCapacitor: $v_C = \int i \, dt \rightarrow V_C = \frac{1}{sC} - \frac{v(0^-)}{s}$
-
-- 3. Using the Laplace transform to analyze a circuit results in a com plete (both transient and steady state) response because the initial conditions are incorporated in the transformation process.
-- 4. The transfer function *H*(*s*) of a network is the Laplace transform of the impulse response *h*(*t*).
-- 5. In the *s*-domain, the transfer function *H*(*s*) relates the output response *Y*(*s*) and an input excitation *X*(*s*); that is, *H*(*s*) = *Y*(*s*)∕*X*(*s*).
-- 6. The state v ariable model is a useful tool for analyzing comple x systems with several inputs and outputs. State variable analysis is a powerful technique that is most popularly used in circuit theory and control. The state of a system is the smallest set of quanti ties (known as state v ariables) that we must kno w to determine its future response to an y given input. The state equation in state variable form is
-
-$$
-\dot{\mathbf{x}} = \mathbf{A}x + \mathbf{B}z
-$$
-
-while the output equation is
-
-$$
-\mathbf{y} = \mathbf{C}x + \mathbf{D}z
-$$
-
-- 7. For an electric circuit, we first select capacitor voltages and inductor current as state v ariables. We then apply KCL and KVL to obtain the state equations.
-- 8. Two other areas of applications of the Laplace transform co vered in this chapter are circuit stability and synthesis. A circuit is stable when all the poles of its transfer function lie in the left half of the *s* plane. Network synthesis is the process of obtaining an appropriate network to represent a given transfer function for which analysis in the *s*-domain is well suited.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/192_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/192_Review Questions.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/192_Review Questions.md
+++ /dev/null
@@ -1,65 +0,0 @@
-# Review Questions
-
-**16.1** The voltage through a resistor with current *i*(*t*) in the *s*-domain is *sRI*(*s*).
-
-(a) True (b) False
-
-**16.2** The current through an *RL* series circuit with input voltage *v*(*t*) is given in the *s*-domain as:
-
-(a)
-$$
-V(s) \left[ R + \frac{1}{sL} \right]
-$$
- (b) $V(s)(R + sL)$
-(c) $\frac{V(s)}{R + 1/sL}$ (d) $\frac{V(s)}{R + sL}$
-
-**16.3** The impedance of a 10-F capacitor is:
-
-(a) 10∕*s* (b) *s*∕10 (c) 1∕10*s* (d) 10*s*
-
-**16.4** We can usually obtain the Thevenin equivalent in the time domain.
-
-(a) True (b) False
-
-**16.5** A transfer function is defined only when all initial conditions are zero.
-
-(a) True (b) False
-
-Problems **745**
-
-**16.6** If the input to a linear system is *δ*(*t*) and the output is *e*−2*t u*(*t*), the transfer function of the system is:
-
-(a)
-$$
-\frac{1}{s+2}
-$$
- (b) $\frac{1}{s-2}$ (c) $\frac{s}{s+2}$ (d) $\frac{s}{s-2}$
-
-(e) None of the above
-
-**16.7** If the transfer function of a system is
-
-sfer function of a system is
-\n
-$$
-H(s) = \frac{s^2 + s + 2}{s^3 + 4s^2 + 5s + 1}
-$$
-
-it follows that the input is *X*(*s*) = *s* 3 + 4*s* 2 + 5*s* + 1, while the output is *Y*(*s*) = *s* 2 + *s* + 2.
-
-$$
-(a) True \t\t (b) False
-$$
-
-**16.8** A network has its transfer function as
-
-has its transfer function a
-$$
-H(s) = \frac{s+1}{(s-2)(s+3)}
-$$
-
-The network is stable.
-
-$$
-(a) True \t\t (b) False
-$$
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/193_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/193_Problems.md
deleted file mode 100644
index e242afacc87c1a2f53ed5fe7f70704ea3d5af56b..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/193_Problems.md
+++ /dev/null
@@ -1,605 +0,0 @@
-# Problems
-
-- Sections 16.2 and 16.3 Circuit Element Models and Circuit Analysis
-- **16.1** The current in an *RLC* circuit is described by
-
-$$
-\frac{d^2i}{dt^2} + 10\frac{di}{dt} + 25i = 0
-$$
-
-If
-$$
-i(0) = 7
-$$
- A and $di(0)/dt = 0$ , find $i(t)$ for $t > 0$ .
-
-**16.2** The differential equation that describes the voltage in an *RLC* network is
-
-$$
-3\frac{d^2i}{dt^2} + 15\frac{di}{dt} + 12i = 0
-$$
-
-Given that *v*(0) = 0, *dv*(0)/*dt* = 6 mA∕s, obtain *i*(*t*).
-
-**16.3** The natural response of an *RLC* circuit is described by the differential equation
-
-$$
-\frac{d^2v}{dt^2} + 2\frac{dv}{dt} + v = 0
-$$
-
- for which the initial conditions are *v*(0) = 350 V and *dv*(0)/*dt* = 0. Solve for *v*(*t*).
-
-- **16.4** If *R* = 20 Ω, *L* = 0.6 H, what value of *C* will make an *RLC* series circuit:
- - (a) overdamped?
- - (b) critically damped?
- - (c) underdamped?
-
-**16.9** Which of the following equations is called the state equation?
-
-(a)
-$$
-\dot{\mathbf{x}} = \mathbf{A}\mathbf{x} + \mathbf{B}\mathbf{z}
-$$
-
-\n(b) $\mathbf{y} = \mathbf{C}\mathbf{x} + \mathbf{D}\mathbf{z}$
-\n(c) $\mathbf{H}(s) = \mathbf{Y}(s)/\mathbf{Z}(s)$
-\n(d) $\mathbf{H}(s) = \mathbf{C}(s\mathbf{I} - \mathbf{A})^{-1}$
-
-**16.10** A single-input, single-output system is described by the state model as:
-
-**B**
-
-$$
-\dot{x}_1 = 2x_1 - x_2 + 3z
-$$
-
-\n
-$$
-\dot{x}_2 = -4x_2 - z
-$$
-
-\n
-$$
-y = 3x_1 - 2x_2 + z
-$$
-
-Which of the following matrices is incorrect?
-
-(a)
-$$
-A = \begin{bmatrix} 2 & -1 \\ 0 & -4 \end{bmatrix}
-$$
- (b) $B = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$
-(c) $C = \begin{bmatrix} 3 & -2 \end{bmatrix}$ (d) $D = 0$
-
-*Answers: 16.1b, 16.2d, 16.3c, 16.4b, 16.5b, 16.6a, 16.7b, 16.8b, 16.9a, 16.10d.*
-
-**16.5** The responses of a series *RLC* circuit are
-
-$$
-v_c(t) = [30 - 10e^{-20t} + 30e^{-10t}]u(t)
-$$
-
-$$
-i_L(t) = [40e^{-20t} - 60e^{-10t}]u(t)
-$$
-mA
-
- where *vC*(*t*) and *iL*(*t*) are the capacitor voltage and inductor current, respectively. Determine the values of *R*, *L*, and *C*.
-
-**16.6** Design a parallel *RLC* circuit that has the characteristic equation
-
-$$
-s^2 + 100s + 10^6 = 0.
-$$
-
-**16.7** The step response of an *RLC* circuit is given by
-
-$$
-\frac{d^2i}{dt^2} + 2\frac{di}{dt} + 5i = 30
-$$
-
- Given that *i*(0) = 18 A and *di*(0)/*dt* = 36 A/s, solve for *i*(*t*).
-
-**16.8** A branch voltage in an *RLC* circuit is described by
-
-$$
-\frac{d^2v}{dt^2} + 4\frac{dv}{dt} + 8v = 120
-$$
-
- If the initial conditions are *v*(0) = 0 = *dv*(0)/*dt*, find *v*(*t*).
-
-**16.9** A series *RLC* circuit is described by
-
-$$
-L\frac{d^2i(t)}{dt} + R\frac{di(t)}{dt} + \frac{i(t)}{C} = 15
-$$
-
- Find the response when *L* = 0.5*H*, *R* = 4Ω, and *C* = 0.2 *F*. Let *i*(0−) = 7.5 A and [*di*(0−)/*dt*] = 0. **16.10** The step responses of a series *RLC* circuit are
-
-$$
-V_c = 40 - 10e^{-2000t} - 10e^{-4000t} \text{ V}, t > 0
-$$
-$$
-i_L(t) = 3e^{-2000t} + 6e^{-4000t} mA, t > 0
-$$
-
-- (a) Find *C*.
- - (b) Determine what type of damping is exhibited by the circuit.
-- **16.11** The step response of a parallel RLC circuit is
-
-$$
-v = 10 + 20e^{-300t} (\cos 400t - 2 \sin 400t) \text{V}, t \ge 0
-$$
-
-when the inductor is 50 mH. Find *R* and *C*.
-
-**16.12** Determine *i*(*t*) in the circuit of Fig. 16.35 by means of the Laplace transform.
-
-# **Figure 16.35**
-
-For Prob. 16.12.
-
-**16.13** Using Fig. 16.36, design a problem to help other students better understand circuit analysis using Laplace transforms.
-
-- **16.14** Find *i*(*t*) for *t* > 0 for the circuit in Fig. 16.37. Assume *is*(*t*) = [6(*t*) + 3*δ*(*t*)]mA.
-
-**16.15** For the circuit in Fig. 16.38, calculate the value of *R* needed to have a critically damped response.
-
-# **Figure 16.38**
-
-**16.16** The capacitor in the circuit of Fig. 16.39 is initially uncharged. Find *v*0(*t*) for *t* > 0.
-
-# **Figure 16.39**
-
-**16.17** If *is*(*t*) = 7.5*e* −2*t u*(*t*) A in the circuit shown in Fig. 16.40, find the value of *io*(*t*).
-
-**Figure 16.40**
-
-For Prob. 16.17.
-
-**16.18** Find *v*(*t*), *t* > 0 in the circuit of Fig. 16.41. Let *vs* = 12 V.
-
-# **Figure 16.41**
-
-For Prob. 16.18.
-
-**16.19** The switch in Fig. 16.42 moves from position A to position B at *t* = 0 (please note that the switch must connect to point B before it breaks the connection at A, a make before break switch). Find *v*(*t*) for *t* > 0.
-
-For Prob. 16.19.
-
-**16.20** Find *i*(*t*) for *t* > 0 in the circuit of Fig. 16.43.
-
-**16.21** In the circuit of Fig. 16.44, the switch moves (make before break switch) from position *A* to *B* at *t* = 0. Find *v*(*t*) for all *t* ≥ 0.
-
-**16.22** Find the voltage across the capacitor as a function of time for *t* > 0 for the circuit in Fig. 16.45. Assume steady-state conditions exist at *t* = 0−.
-
-**Figure 16.45** For Prob. 16.22.
-
-**16.23** Obtain *v*(*t*) for *t* > 0 in the circuit of Fig. 16.46.
-
-**Figure 16.46** For Prob. 16.23.
-
-**16.24** The switch in the circuit of Fig. 16.47 has been closed for a long time but is opened at *t* = 0. Determine *i*(*t*) for *t* > 0.
-
-**Figure 16.47** For Prob. 16.24.
-
-**16.25** Calculate *v*(*t*) for *t* > 0 in the circuit of Fig. 16.48.
-
-**Figure 16.48** For Prob. 16.25.
-
-**16.26** The switch in Fig. 16.49 moves from position A to position B at *t* = 0 (please note that the switch must connect to point B before it breaks the connection at A, a make before break switch). Determine *i*(*t*) for *t* > 0. Also assume that the initial voltage on the capacitor is zero.
-
-**Figure 16.49** For Prob. 16.26.
-
-**16.27** Find *v*(*t*) for *t* > 0 in the circuit in Fig. 16.50.
-
-**16.28** For the circuit in Fig. 16.51, find *v*(*t*) for *t* > 0.
-
-**Figure 16.55** For Prob. 16.32.
-
-**16.33** Using Fig. 16.56, design a problem to help other students understand how to use Thevenin's theorem (in the *s*-domain) to aid in circuit analysis.
-
-**Figure 16.56** For Prob. 16.33.
-
-**16.30** Find *vo*(*t*), for all *t* > 0, in the circuit of Fig. 16.53.
-
-**16.31** Obtain *v*(*t*) and *i*(*t*) for *t* > 0 in the circuit in Fig. 16.54.
-
-**Figure 16.54** For Prob. 16.31.
-
-**16.34** Solve for the mesh currents in the circuit of Fig. 16.57. You may leave your results in the *s*-domain.
-
-**Figure 16.57** For Prob. 16.34.
-
-**16.35** Find *vo*(*t*) in the circuit of Fig. 16.58.
-
-**Figure 16.58** For Prob. 16.35.
-
-**16.36** Refer to the circuit in Fig. 16.59. Calculate *i*(*t*) for *t* > 0.
-
-# **Figure 16.59**
-
-For Prob. 16.36.
-
-**16.37** Determine *v* for *t* > 0 in the circuit in Fig. 16.60.
-
-For Prob. 16.37.
-
-**16.38** The switch in the circuit of Fig. 16.61 is moved from position *a* to *b* (a make before break switch) at *t* = 0. Determine *i*(*t*) for *t* > 0.
-
-# **Figure 16.61**
-
-For Prob. 16.38.
-
-**16.39** For the network in Fig. 16.62, find *i*(*t*) for *t* > 0.
-
-**Figure 16.62** For Prob. 16.39.
-
-**16.40** In the circuit of Fig. 16.63, find *v*(*t*) and *i*(*t*) for *t* > 0. Assume *v*(0) = 0 V and *i*(0) = 1.25 A.
-
-# **Figure 16.63** For Prob. 16.40.
-
-**16.41** Find the output voltage *vo*(*t*) in the circuit of Fig. 16.64.
-
-For Prob. 16.41.
-
-**16.42** Given the circuit in Fig. 16.65, find *i*(*t*) and *v*(*t*) for *t* > 0.
-
-# **Figure 16.65** For Prob. 16.42.
-
-**16.43** Determine *i*(*t*) for *t* > 0 in the circuit of Fig. 16.66.
-
-**Figure 16.67**
-
-**16.45** Find *v*(*t*) for *t* > 0 in the circuit in Fig. 16.68.
-
-For Prob. 16.45.
-
-**16.46** Determine *io*(*t*) in the circuit in Fig. 16.69.
-
-# **Figure 16.69**
-
-For Prob. 16.46.
-
-**16.47** Determine *io*(*t*) in the network shown in Fig. 16.70.
-
-# **Figure 16.70**
-
-For Prob. 16.47.
-
-**Figure 16.71** For Prob. 16.48.
-
-**16.49** Find *i*0(*t*) for *t* > 0 in the circuit in Fig. 16.72.
-
-# **Figure 16.72**
-
-For Prob. 16.49.
-
-**16.50** For the circuit in Fig. 16.73, find *v*(*t*) for *t* > 0. Assume that *i*(0) = 2 A.
-
-# **Figure 16.73**
-
-For Prob. 16.50.
-
-# **Figure 16.74**
-
-For Prob. 16.51.
-
-**16.52** Given the circuit shown in Fig. 16.75, determine the values for *i*(*t*) and *v*(*t*) for all *t* > 0.
-
-# **Figure 16.75**
-
-For Prob. 16.52.
-
-**16.53** In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to position 2 at *t* = 0. Find:
-
-> (a) *v*(0+), *dv*(0+)/*dt* (b) *v*(*t*) for *t* ≥ 0.
-
-For Prob. 16.44.
-
-Problems **751**
-
-**Figure 16.76**
-
-For Prob. 16.53.
-
-**16.54** The switch in Fig. 16.77 has been in position 1 for *t* < 0. At *t* = 0, it is moved from position 1 to the top of the capacitor at *t* = 0. Please note that the switch is a make before break switch; it stays in contact with position 1 until it makes contact with the top of the capacitor and then breaks the contact at position 1. Determine *v*(*t*).
-
-**16.55** Obtain *i*1 and *i*2 for *t* > 0 in the circuit of Fig. 16.78.
-
-**Figure 16.78** For Prob. 16.55.
-
-**16.56** Calculate *io*(*t*) for *t* > 0 in the network of Fig. 16.79.
-
-**16.57** (a) Find the Laplace transform of the voltage shown in Fig. 16.80(a). (b) Using that value of *vs*(*t*) in the circuit shown in Fig. 16.80(b), find the value of *vo*(*t*).
-
-**Figure 16.80** For Prob. 16.57.
-
-**16.58** Using Fig. 16.81, design a problem to help other students better understand circuit analysis in the *s*-domain with circuits that have dependent sources.
-
-**Figure 16.81** For Prob. 16.58.
-
-**16.59** Find *vo*(*t*) in the circuit of Fig. 16.82 if *vx*(0) = 10 V and *i*(0) = 5 A.
-
-# **Figure 16.82**
-
-For Prob. 16.59.
-
-For Prob. 16.60.
-
-**16.60** Find the response *v*(*t*) for *t* > 0 in the circuit in Fig. 16.83. Let *R* = 8 Ω, *L* = 2 H, and *C* = 125 mF.
-
-**16.61** Find the voltage *vo*(*t*) in the circuit of Fig. 16.84 by means of the Laplace transform. \*
-
-# **Figure 16.84**
-
-For Prob. 16.61.
-
-**16.62** Using Fig. 16.85, design a problem to help other
-
-students better understand solving for node voltages by working in the *s*-domain.
-
-# **Figure 16.85**
-
-For Prob. 16.62.
-
-**16.63** Consider the parallel *RLC* circuit of Fig. 16.86. Find *v*(*t*) and *i*(*t*) given that *v*(0) = 7.5 V and *i*(0) = −3 A.
-
-# **Figure 16.86**
-
-For Prob. 16.63.
-
-**16.64** The switch in Fig. 16.87 moves from position 1 to position 2 at *t* = 0. Find *v*(*t*), for all *t* > 0.
-
-# **Figure 16.87**
-
-For Prob. 16.64.
-
-**16.65** For the *RLC* circuit shown in Fig. 16.88, find the complete response if *v*(0) = 100 V when the switch is closed.
-
-# **Figure 16.88**
-
-For Prob. 16.65.
-
-\* An asterisk indicates a challenging problem. For Prob. 16.69.
-
-**16.66** For the op amp circuit in Fig. 16.89, find *v*0(*t*) for *t* > 0. Take *vs* = 12 *e*−5*t u*(*t*) V.
-
-# **Figure 16.89**
-
-For Prob. 16.66.
-
-**16.67** Given the op amp circuit in Fig. 16.90, if *v*1(0+) = 2 V and *v*2(0+) = 0 V, find *v*0 for *t* > 0. Let *R* = 100 kΩ and *C* = 1 *μ*F.
-
-# **Figure 16.90**
-
-For Prob. 16.67.
-
-**16.68** Obtain *V*0/*Vs* in the op amp circuit in Fig. 16.91.
-
-# **Figure 16.91**
-
-For Prob. 16.68.
-
-**16.69** Find *I*1(*s*) and *I*2(*s*) in the circuit of Fig. 16.92.
-
-# **Figure 16.93**
-
-For Prob. 16.70.
-
-**16.71** For the ideal transformer circuit in Fig. 16.94, determine *io*(*t*).
-
-# **Figure 16.94**
-
-For Prob. 16.71.
-
-# Section 16.4 Transfer Functions
-
-**16.72** The transfer function of a system is
-
-$$
-H(s) = \frac{s^2}{3s+1}
-$$
-
- Find the output when the system has an input of 14*e*−*t*∕3 *u*(*t*).
-
-- **16.73** When the input to a system is a unit step function, the response is 120 cos 2*tu*(*t*). Obtain the transfer function of the system.
-- **16.74** Design a problem to help other students better
-- understand how to find outputs when given a transfer function and an input.
-- **16.75** When a unit step is applied to a system at *t* = 0, its response is
-
-*y*(*t*) = [6 + 0.75 *e*−3*t* − *e*−2*t* (3 cos 4*t* + 4.5 sin 4*t*)]*u*(*t*)
-
-What is the transfer function of the system?
-
-**16.76** For the circuit in Fig. 16.95, find *H*(*s*) = *Vo*(*s*)∕*Vs*(*s*). Assume zero initial conditions.
-
-**Figure 16.95** For Prob. 16.76.
-
-**16.77** Obtain the transfer function *H*(*s*) = *Vo*∕*Vs* for the circuit of Fig. 16.96.
-
-# **Figure 16.96**
-
-For Prob. 16.77.
-
-**16.78** The transfer function of a certain circuit is
-
-$$
-H(s) = \frac{10}{s+1} - \frac{6}{s+2} + \frac{12}{s+4}
-$$
-
-Find the impulse response of the circuit.
-
-**16.79** For the circuit in Fig. 16.97, find:
-
-# **Figure 16.97**
-
-For Prob. 16.79.
-
-**16.80** Refer to the network in Fig. 16.98. Find the following transfer functions:
-
-(a)
-$$
-H_1(s) = V_o(s)/V_s(s)
-$$
-
-\n(b) $H_2(s) = V_o(s)/I_s(s)$
-\n(c) $H_3(s) = I_o(s)/I_s(s)$
-\n(d) $H_4(s) = I_o(s)/V_s(s)$
-
-$$
-v_{s} \xrightarrow{I_{s}} 1 \Omega \xrightarrow{1 H} 1 \Omega \xrightarrow{I_{0}}
-$$
-\n
-$$
-1 \Gamma \xrightarrow{1 \Gamma} 1 \Gamma \xrightarrow{1 \Omega} \xleftarrow{1} v_{0}
-$$
-
-# **Figure 16.98**
-
-For Prob. 16.80.
-
-**16.81** For the op-amp circuit in Fig. 16.99, find the transfer function, *T*(*s*) = *I*(*s*)/*Vs*(*s*). Assume all initial conditions are zero.
-
-**Figure 16.99** For Prob. 16.81.
-
-**16.82** Calculate the gain *H*(*s*) = *Vo*∕*Vs* in the op amp circuit of Fig. 16.100.
-
-- **16.83** Refer to the *RL* circuit in Fig. 16.101. Find:
- - (a) the impulse response *h*(*t*) of the circuit.
- - (b) the unit step response of the circuit.
-
-**Figure 16.101**
-
-- For Prob. 16.83.
-- **16.84** A parallel *RL* circuit has *R* = 4 Ω and *L* = 1 H. The input to the circuit is *is*(*t*) = 1.4*e*−*t u*(*t*) A. Find the inductor current *iL*(*t*) for all *t* > 0 and assume that *iL*(0) = −1.4 A.
-- **16.85** A circuit has a transfer function
-
-s a transfer function
-$$
-H(s) = \frac{3(s + 4)}{(s + 1)(s + 2)^2}
-$$
-
-Find the impulse response.
-
-# Section 16.5 State Variables
-
-- **16.86** Develop the state equations for Prob. 16.12.
-- **16.87** Develop the state equations for the problem you designed in Prob. 16.13.
-- **16.88** Develop the state equations for the circuit shown in Fig. 16.102.
-
-**Figure 16.102** For Prob. 16.88.
-
-**16.89** Develop the state equations for the circuit shown in Fig. 16.103.
-
-**16.90** Develop the state equations for the circuit shown in Fig. 16.104.
-
-For Prob. 16.90.
-
-**16.91** Develop the state equations for the following differential equation.
-
-$$
-\frac{d^2y(t)}{dt^2} + \frac{6\ dy(t)}{dt} + 7y(t) = z(t)
-$$
-
-**16.92** Develop the state equations for the following differential equation. \*
-
-$$
-\frac{d^2y(t)}{dt^2} + \frac{7\,dy(t)}{dt} + 9y(t) = \frac{dz(t)}{dt} + z(t)
-$$
-
-**16.93** Develop the state equations for the following differential equation. \*
-
-$$
-\frac{d^3y(t)}{dt^3} + \frac{6 \ d^2y(t)}{dt^2} + \frac{11 \ dy(t)}{dt} + 6y(t) = z(t)
-$$
-
-**16.94** Given the following state equation, solve for *y*(*t*): \*
-
-$$
-\dot{\mathbf{x}} = \begin{bmatrix} -4 & 4 \\ -2 & 0 \end{bmatrix} x + \begin{bmatrix} 0 \\ 2 \end{bmatrix} u(t)
-$$
-$$
-\mathbf{y}(t) = \begin{bmatrix} 1 & 0 \end{bmatrix} x
-$$
-
-**16.95** Given the following state equation, solve for *y*1(*t*) and *y*2(*t*). \*
-
-$$
-\dot{\mathbf{x}} = \begin{bmatrix} -2 & -1 \\ 2 & -4 \end{bmatrix} x + \begin{bmatrix} 1 & 1 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} u(t) \\ 2u(t) \end{bmatrix}
-$$
-$$
-\mathbf{y} = \begin{bmatrix} -2 & -2 \\ 1 & 0 \end{bmatrix} x + \begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} u(t) \\ 2u(t) \end{bmatrix}
-$$
-
-# Section 16.6 Applications
-
-**16.96** Show that the parallel *RLC* circuit shown in Fig. 16.105 is stable.
-
-# **Figure 16.105**
-
-For Prob. 16.96.
-
-**16.97** A system is formed by cascading two systems as shown in Fig. 16.106. Given that the impulse responses of the systems are
-
-$$
-h_1(t) = 21e^{-t}u(t), \qquad h_2(t) = e^{-4t}u(t)
-$$
-
-- (a) Obtain the impulse response of the overall system.
-- (b) Check if the overall system is stable.
-
-**Figure 16.106**
-
-For Prob. 16.97.
-
-**16.98** Determine whether the op amp circuit in Fig. 16.107 is stable.
-
-**Figure 16.107**
-
-For Prob. 16.98.
-
-**16.99** It is desired to realize the transfer function
-
-$$
-\frac{V_2(s)}{V_1(s)} = \frac{2s}{s^2 + 2s + 6}
-$$
-
- using the circuit in Fig. 16.108. Choose *R* = 1 kΩ and find *L* and *C*.
-
-**Figure 16.108** For Prob. 16.99.
-
-**16.100** Design an op amp circuit, using Fig. 16.109, that will realize the following transfer function:
-
-$$
-\frac{V_o(s)}{V_i(s)} = -\frac{s + 1000}{2(s + 4000)}
-$$
-
-Choose *C*1 = 10 *μ*F; determine *R*1, *R*2, and *C*2.
-
-# **Figure 16.109**
-
-For Prob. 16.100.
-
-**16.101** Realize the transfer function
-
-$$
-\frac{V_o(s)}{V_s(s)} = -\frac{s}{s+10}
-$$
-
- using the circuit in Fig. 16.110. Let *Y*1 = *sC*1, *Y*2 = 1∕*R*1, *Y*3 = *sC*2. Choose *R*1 = 1 kΩ and determine *C*1 and *C*2.
-
-# **Figure 16.110**
-
-For Prob. 16.101.
-
-**16.102** Synthesize the transfer function
-
-ize the transfer function
-\n
-$$
-\frac{V_o(s)}{V_{in}(s)} = \frac{10^6}{s^2 + 100s + 10^6}
-$$
-
- using the topology of Fig. 16.111. Let *Y*1 = 1∕*R*1, *Y*2 = 1∕*R*2, *Y*3 = *sC*1, *Y*4 = *sC*2. Choose *R*1 = 1 kΩ and determine *C*1, *C*2, and *R*2.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/194_Comprehensive Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/194_Comprehensive Problems.md
deleted file mode 100644
index a6c24513fe1705bbf8a68324feaa3210ee7868b0..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/194_Comprehensive Problems.md
+++ /dev/null
@@ -1,22 +0,0 @@
-# Comprehensive Problems
-
-**16.103** Obtain the transfer function of the op amp circuit in Fig. 16.112 in the form of
-
-$$
-\frac{V_o(s)}{V_i(s)} = \frac{as}{s^2 + bs + c}
-$$
-
- where *a*, *b*, and *c* are constants. Determine the constants.
-
-**16.104** A certain network has an input admittance *Y*(*s*). The admittance has a pole at *s* = −3, a zero at *s* = −1, and *Y*(∞) = 0.25 S.
-
-- (a) Find *Y*(*s*).
-- (b) An 8-V battery is connected to the network via a switch. If the switch is closed at *t* = 0, find the current *i*(*t*) through *Y*(*s*) using the Laplace transform.
-
-**16.105** A gyrator is a device for simulating an inductor in a network. A basic gyrator circuit is shown in
-
-**Figure 16.113** For Prob. 16.105.
-
-# **chapter**
-
-17
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/195_Chapter 17 - The Fourier Series.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/195_Chapter 17 - The Fourier Series.md
deleted file mode 100644
index c598e01929c6fb56ff2715d966537a6359b24197..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/195_Chapter 17 - The Fourier Series.md
+++ /dev/null
@@ -1,35 +0,0 @@
-# The Fourier Series
-
-*Research is to see what everybody else has seen, and think what nobody has thought.*
-
-—Albert Szent Györgyi
-
-# Enhancing Your Skills and Your Career
-
-# **ABET EC 2000 criteria (3.j), "a knowledge of contemporary issues."**
-
-Engineers must have knowledge of contemporary issues. To have a truly meaningful career in the twenty-first century, you must have knowledge of contemporary issues, especially those that may directly af fect your job and/or work. One of the easiest ways to achieve this is to read a lot newspapers, magazines, and contemporary books. As a student enrolled in an ABET-accredited program, some of the courses you take will be directed toward meeting this criteria.
-
-# **ABET EC 2000 criteria (3.k), "an ability to use the techniques, skills, and modern engineering tools necessary for engineering practice."**
-
-The successful engineer must ha ve the "ability to use the techniques, skills, and modern engineering tools necessary for engineering practice." Clearly, a major focus of this te xtbook is to do just that. Learning to use skillfully the tools that facilitate your working in a modern "knowledge capturing integrated design environment" (KCIDE) is fundamental to your performance as an engineer . The ability to w ork in a modern KCIDE environment requires a thorough understanding of the tools associated with that environment.
-
-The successful engineer , therefore, must k eep abreast of the ne w design, analysis, and simulation tools. That engineer must also use those tools until he or she is comfortable with using them. The engineer also must make sure software results are consistent with real-world actualities. It is probably in this area that most engineers have the greatest difficulty. Thus, successful use of these tools requires constant learning and relearning the fundamentals of the area in which the engineer is working.
-
-Photo by Charles Alexander
-
-# Historical
-
-**Jean Baptiste Joseph Fourier** (1768–1830), a French mathematician, first presented the series and transform that bear his name. Fourier's results were not en thusiastically received by the scientific world. He could not even get his work published as a paper.
-
-Born in Auxerre, France, Fourier was orphaned at age 8. He attended a local military college run by Benedictine monks, where hedemonstrated great proficiency in mathematics. Like most of his contemporaries, Fourier was swept into the politics of the French Revolution. He played an important role in Napoleon's expeditions to Egypt in the later 1790s. Due to his political involvement, he narrowly escaped death twice.
-
-# Learning Objectives
-
-By using the information and exercises in this chapter you will be able to:
-
-- 1. Understand the trigonometric Fourier series and know how to determine the Fourier series with a variety of periodic functions.
-- 2. Effectively use the Fourier series to analyze the response of circuits to a variety of periodic sources.
-- 3. Know how the symmetrical characteristics of some wave shapes can make determining the Fourier series of classes of periodic functions easier to determine.
-- 4. Understand how to determine average power and rms values associated with periodic functions.
-- 5. Understand the use of the discrete Fourier transform and the fast Fourier transform.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/196_17.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/196_17.1 Introduction.md
deleted file mode 100644
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--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/196_17.1 Introduction.md
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-# **17.1** Introduction
-
-We have spent a considerable amount of time on the analysis of circuits with sinusoidal sources. This chapter is concerned with a means of ana lyzing circuits with periodic, nonsinusoidal e xcitations. The notion of periodic functions w as introduced in Chapter 9; it w as mentioned there that the sinusoid is the most simple and useful periodic function. This chapter introduces the F ourier series, a technique for e xpressing a periodic function in terms of sinusoids. Once the source function is expressed in terms of sinusoids, we can apply the phasor method to analyze circuits.
-
-The F ourier series is named after Jean Baptiste Joseph F ourier (1768–1830). In 1822, F ourier's genius came up with the insight that any practical periodic function can be represented as a sum of sinusoids. Such a representation, along with the superposition theorem, allo ws us to find the response of circuits to arbitrary periodic inputs using phasor techniques.
-
-We begin with the trigonometric F ourier series. Later we consider the exponential Fourier series. We then apply F ourier series in circuit analysis. Finally, practical applications of Fourier series in spectrum analyzers and filters are demonstrated.
-
-# **17.2** Trigonometric Fourier Series
-
-While studying heat flow, Fourier discovered that a nonsinusoidal periodic function can be expressed as an infinite sum of sinusoidal functions. Recall that a periodic function is one that repeats every *T* seconds. In other words, a periodic function *f* (*t*) satisfies
-
-$$
-f(t) = f(t + nT)
-$$
- (17.1)
-
-where *n* is an integer and *T* is the period of the function.
-
-According to the *Fourier theorem*, any practical periodic function of angular frequenc y *ω*0 can be e xpressed as an infinite sum of sine or cosine functions that are inte gral multiples of *ω*0. Thus, *f*(*t*) can be expressed as
-
-$$
-f(t) = a_0 + a_1 \cos \omega_0 t + b_1 \sin \omega_0 t + a_2 \cos 2\omega_0 t + b_2 \sin 2\omega_0 t + a_3 \cos 3\omega_0 t + b_3 \sin 3\omega_0 t + \cdots
-$$
- (17.2)
-
-or
-
-$$
-f(t) = a_0 + \sum_{n=1}^{\infty} (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t)
-$$
- (17.3)
-
-where *ω*0 = 2*π*∕*T* is called the *fundamental angular frequency* in radians per second. The sinusoid sin *nω*0*t* or cos *nω*0*t* is called the *n*th harmonic of *f*(*t*); it is an odd harmonic if *n* is odd and an even harmonic if *n* is even. Equation 17.3 is called the *trigonometric Fourier series* of *f*(*t*). The constants *an* and *bn* are the *Fourier coefficients*. The coefficient *a*0 is the dc component or the average value of *f*(*t*). (Recall that sinusoids have zero average values.) The coefficients *an* and *bn* (for *n* ≠ 0) are the amplitudes of the sinusoids in the ac component. Thus,
-
-The Fourier series of a periodic function f (t) is a representation that resolves f (t) into a dc component and an ac component comprising an infinite series of harmonic sinusoids.
-
-A function that can be represented by a Fourier series as in Eq. (17.3) must meet certain requirements, because the infinite series in Eq. (17.3) may or may not converge. These conditions on *f*(*t*) to yield a convergent Fourier series are as follows:
-
-1. *f*(*t*) is single-valued everywhere.
-
-- 2. *f*(*t*) has a finite number of finite discontinuities in any one period.
-- 3. *f*(*t*) has a finite number of maxima and minima in any one period.
-- 4. The integral ∫ *t*0 *t*0+*T* ∣ *f*(*t*)∣ *dt* < ∞ for any *t*0.
-
-The harmonic angular frequency *ω*n is an integer multiple of the fundamental angular frequency *ω*0, i.e., *ω*n = n*ω*0.
-
-Historical note: Although Fourier published his theorem in 1822, it was P. G. L. Dirichlet (1805–1859) who later supplied an acceptable proof of the theorem.
-
-A software package like Mathcad or Maple can be used to evaluate the Fourier coefficients.
-
-These conditions are called *Dirichlet conditions*. Although they are not necessary conditions, the y are sufficient conditions for a Fourier series to exist.
-
-A major task in Fourier series is the determination of the Fourier coefficients *a*0, *an*, and *bn*. The process of determining the coefficients is called *Fourier analysis*. The following trigonometric inte grals are very helpful in Fourier analysis. For any integers *m* and *n*,
-
-$$
-\int_0^T \sin n\omega_0 t \, dt = 0 \tag{17.4a}
-$$
-
-$$
-\int_0^T \cos n\omega_0 t \, dt = 0 \tag{17.4b}
\ No newline at end of file
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/197_17.2 Trigonometric Fourier Series.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/197_17.2 Trigonometric Fourier Series.md
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@@ -1,372 +0,0 @@
-$$
-
-$$
-\int_0^T \sin n\omega_0 t \cos m\omega_0 t \, dt = 0 \tag{17.4c}
-$$
-
-$$
-\int_0^T \sin n\omega_0 t \sin m\omega_0 t \, dt = 0, \qquad (m \neq n) \tag{17.4d}
-$$
-
-$$
-\int_0^T \cos n\omega_0 t \cos m\omega_0 t \, dt = 0, \qquad (m \neq n) \tag{17.4e}
-$$
-
-$$
-\int_0^T \sin^2 n\omega_0 t \, dt = \frac{T}{2} \tag{17.4f}
-$$
-
-$$
-\int_0^T \cos^2 n\omega_0 t \, dt = \frac{T}{2} \tag{17.4g}
-$$
-
-Let us use these identities to evaluate the Fourier coefficients.
-
-We begin by finding *a*0. We integrate both sides of Eq. (17.3) o ver one period and obtain
-
-$$
-\int_0^T f(t) dt = \int_0^T \left[ a_0 + \sum_{n=1}^\infty (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t) \right] dt
-$$
-
-=
-$$
-\int_0^T a_0 dt + \sum_{n=1}^\infty \left[ \int_0^T a_n \cos n\omega_0 t dt + \int_0^T b_n \sin n\omega_0 t dt \right] dt
-$$
- (17.5)
-
-Invoking the identities of Eqs. (17.4a) and (17.4b), the tw o inte grals involving the ac terms vanish. Hence,
-
-$$
-\int_0^T f(t) \, dt = \int_0^T a_0 \, dt = a_0 \, T
-$$
-
-or
-
-$$
-a_0 = \frac{1}{T} \int_0^T f(t) \, dt \tag{17.6}
-$$
-
-showing that *a*0 is the average value of *f*(*t*).
-
-To evaluate *an*, we multiply both sides of Eq. (17.3) by cos *mω*0*t* and integrate over one period:
-
-$$
-\int_0^T f(t) \cos m\omega_0 t \, dt
-$$
-\n
-$$
-= \int_0^T \left[ a_0 + \sum_{n=1}^\infty (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t) \right] \cos m\omega_0 t \, dt
-$$
-\n
-$$
-= \int_0^T a_0 \cos m\omega_0 t \, dt + \sum_{n=1}^\infty \left[ \int_0^T a_n \cos n\omega_0 t \cos m\omega_0 t \, dt \right]
-$$
-\n
-$$
-+ \int_0^T b_n \sin n\omega_0 t \cos m\omega_0 t \, dt \right] dt \tag{17.7}
-$$
-
-The inte gral containing *a*0 is zero in vie w of Eq. (17.4b), while the integral containing *bn* vanishes according to Eq. (17.4c). The inte gral containing *an* will be zero e xcept when *m* = *n*, in which case it is *T*∕2, according to Eqs. (17.4e) and (17.4g). Thus,
-
-$$
-\int_0^T f(t) \cos m\omega_0 t \, dt = a_n \frac{T}{2}, \qquad \text{for } m = n
-$$
-
-or
-
-$$
-a_n = \frac{2}{T} \int_0^T f(t) \cos n\omega_0 t \, dt \qquad (17.8)
-$$
-
-In a similar vein, we obtain *bn* by multiplying both sides of Eq. (17.3) by sin *mω*0*t* and integrating over the period. The result is
-
-$$
-b_n = \frac{2}{T} \int_0^T f(t) \sin n\omega_0 t \, dt \qquad (17.9)
-$$
-
-Be aware that because *f*(*t*) is periodic, it may be more convenient to carry the integrations above from −*T*∕2 to *T*∕2 or generally from *t*0 to *t*0 + *T* instead of 0 to *T*. The result will be the same.
-
-An alternative form of Eq. (17.3) is the *amplitude-phase* form
-
-$$
-f(t) = a_0 + \sum_{n=1}^{\infty} A_n \cos(n\omega_0 t + \phi_n)
-$$
- (17.10)
-
-We can use Eqs. (9.11) and (9.12) to relate Eq. (17.3) to Eq. (17.10), or we can apply the trigonometric identity
-
-$$
-\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \tag{17.11}
-$$
-
-to the ac terms in Eq. (17.10) so that
-
-$$
-a_0 + \sum_{n=1}^{\infty} A_n \cos(n\omega_0 t + \phi_n) = a_0 + \sum_{n=1}^{\infty} (A_n \cos \phi_n) \cos n\omega_0 t
-$$
-
-
-$$
--(A_n \sin \phi_n) \sin n\omega_0 t
-$$
- (17.12)
-
-Equating the coef ficients of the series expansions in Eqs. (17.3) and (17.12) shows that
-
-$$
-a_n = A_n \cos \phi_n, \qquad b_n = -A_n \sin \phi_n \tag{17.13a}
-$$
-
-or
-
-$$
-A_n = \sqrt{a_n^2 + b_n^2}, \qquad \phi_n = -\tan^{-1}\frac{b_n}{a_n}
-$$
- (17.13b)
-
-To avoid any confusion in determining *n*, it may be better to relate the terms in complex form as
-
-$$
-A_n / \underline{\phi_n} = a_n - jb_n \tag{17.14}
-$$
-
-The convenience of this relationship will become evident in Section 17.6. The plot of the amplitude *An* of the harmonics v ersus *nω*0 is called the *amplitude spectrum* of *f*(*t*); the plot of the phase *n* v ersus *nω*0 is the *phase spectrum* of *f*(*t*). Both the amplitude and phase spectra form the *frequency spectrum* of *f*(*t*).
-
-The frequency spectrum of a signal consists of the plots of the amplitudes and phases of the harmonics versus frequency.
-
-Thus, the F ourier analysis is also a mathematical tool for finding the spectrum of a periodic signal. Section 17.6 will elaborate more on the spectrum of a signal.
-
-To evaluate the Fourier coefficients *a*0, *an*, and *bn*, we often need to apply the following integrals:
-
-$$
-\int \cos at \, dt = \frac{1}{a} \sin at \tag{17.15a}
-$$
-
-$$
-\int \sin at \, dt = -\frac{1}{a} \cos at \tag{17.15b}
-$$
-
-$$
-\int t \cos at \, dt = \frac{1}{a^2} \cos at + \frac{1}{a} t \sin at \tag{17.15c}
-$$
-
-$$
-\int t \sin at \, dt = \frac{1}{a^2} \sin at - \frac{1}{a} \, t \cos at \tag{17.15d}
-$$
-
-It is also useful to kno w the v alues of the cosine, sine, and e xponential functions for inte gral multiples of *π*. These are given in Table 17.1, where *n* is an integer.
-
-For Example 17.1; a square wave.
-
-Determine the Fourier series of the waveform shown in Fig. 17.1. Obtain the amplitude and phase spectra.
-
-# **Solution:**
-
-The Fourier series is given by Eq. (17.3), namely,
-
-$$
-f(t) = a_0 + \sum_{n=1}^{\infty} (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t)
-$$
- (17.1.1)
-
-The frequency spectrum is also known as the line spectrum in view of the discrete frequency components.
-
-# **TABLE 17.1**
-
-Values of cosine, sine, and exponential functions for integral multiples of *π*.
-
-| Function | Value |
-|-----------------|---------------------------------------------------------------|
-| cos 2nπ | 1 |
-| sin 2nπ | 0 |
-| cos nπ | (−1)n |
-| sin nπ | 0 |
-| cos ___ nπ
2 | (−1)n∕2
,
n = even
0,
n = odd
{ |
-| sin ___ nπ
2 | (−1)(n−1)∕2
,
n = odd
0,
{
n = even |
-| e j2nπ | 1 |
-| e jnπ | (−1)n |
-| e jnπ∕2 | (−1)n∕2
,
n = even
n = odd
{
j(−1)(n−1)∕2
, |
-
-Our goal is to obtain the Fourier coefficients *a*0, *an*, and *bn* using Eqs. (17.6), (17.8), and (17.9). First, we describe the waveform as
-
-$$
-f(t) = \begin{cases} 1, & 0 < t < 1 \\ 0, & 1 < t < 2 \end{cases} \tag{17.1.2}
-$$
-
-and *f*(*t*) = *f*(*t* + *T*). Because *T* = 2, *ω*0 = 2*π*∕*T* = *π*. Thus,
-
-$$
-a_0 = \frac{1}{T} \int_0^T f(t) dt = \frac{1}{2} \left[ \int_0^1 1 dt + \int_1^2 0 dt \right] = \frac{1}{2} t \Big|_0^1 = \frac{1}{2}
-$$
- (17.1.3)
-
-Using Eq. (17.8) along with Eq. (17.15a),
-
-$$
-a_n = \frac{2}{T} \int_0^T f(t) \cos n\omega_0 t \, dt
-$$
-
-= $\frac{2}{2} \left[ \int_0^1 1 \cos n\pi t \, dt + \int_1^2 0 \cos n\pi t \, dt \right]$
-= $\frac{1}{n\pi} \sin n\pi t \Big|_0^1 = \frac{1}{n\pi} [\sin n\pi - \sin(0)] = 0$ (17.1.4)
-
-From Eq. (17.9) with the aid of Eq. (17.15b),
-
-$$
-b_n = \frac{2}{T} \int_0^T f(t) \sin n\omega_0 t \, dt
-$$
-
-= $\frac{2}{2} \left[ \int_0^1 1 \sin n\pi t \, dt + \int_1^2 0 \sin n\pi t \, dt \right]$
-= $-\frac{1}{n\pi} \cos n\pi t \Big|_0^1$ (17.1.5)
-= $-\frac{1}{n\pi} (\cos n\pi - 1)$ , $\cos n\pi = (-1)^n$
-= $\frac{1}{n\pi} [1 - (-1)^n] = \begin{cases} \frac{2}{n\pi}, & n = \text{odd} \\ 0, & n = \text{even} \end{cases}$
-
-Substituting the Fourier coefficients in Eqs. (17.1.3) to (17.1.5) into Eq. (17.1.1) gives the Fourier series as
-
-$$
-f(t) = \frac{1}{2} + \frac{2}{\pi} \sin \pi t + \frac{2}{3\pi} \sin 3 \pi t + \frac{2}{5\pi} \sin 5 \pi t + \dots
-$$
- (17.1.6)
-
-Given that *f*(*t*) contains only the dc component and the sine terms with the fundamental component and odd harmonics, it may be written as
-
-$$
-f(t) = \frac{1}{2} + \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{1}{n} \sin n\pi t, \qquad n = 2k - 1
-$$
- (17.1.7)
-
-By summing the terms one by one as demonstrated in Fig. 17.2, we notice how superposition of the terms can evolve into the original square. As more and more F ourier components are added, the sum gets closer and closer to the square wave. However, it is not possible in practice to sum the series in Eq. (17.1.6) or (17.1.7) to infinity. Only a partial sum (*n* = 1, 2, 3, … , *N*, where *N* is finite) is possible. If we plot the partial sum (or truncated series) o ver one period for a lar ge *N* as in
-
-Sum of first three ac components
-
-# **Figure 17.2**
-
-Evolution of a square wave from its Fourier components.
-
- Summing the Fourier terms by hand calculation may be tedious. A computer is helpful to compute the terms and plot the sum like those shown in Fig. 17.2.
-
-# **Figure 17.4**
-
-For Example 17.1: (a) amplitude and (b) phase spectrum of the function shown in Fig. 17.1.
-
-Truncating the Fourier series at *N* = 11; Gibbs phenomenon.
-
-Fig. 17.3, we notice that the partial sum oscillates abo ve and below the actual value of *f*(*t*). At the neighborhood of the points of discontinuity (*x* = 0, 1, 2, …), there is o vershoot and damped oscillation. In f act, an overshoot of about 9 percent of the peak value is always present, regardless of the number of terms used to approximate *f*(*t*). This is called the *Gibbs phenomenon*.
-
-Finally, let us obtain the amplitude and phase spectra for the signal in Fig. 17.1. Since *an* = 0,
-
-$$
-A_n = \sqrt{a_n^2 + b_n^2} = |b_n| = \begin{cases} \frac{2}{n\pi}, & n = \text{odd} \\ 0, & n = \text{even} \end{cases}
-$$
- (17.1.8)
-
-and
-
-$$
-\phi_n = -\tan^{-1} \frac{b_n}{a_n} = \begin{cases} -90^\circ, & n = \text{odd} \\ 0, & n = \text{even} \end{cases}
-$$
- (17.1.9)
-
-The plots of *An* and *n* for different values of *nω*0 = *nπ* provide the amplitude and phase spectra in Fig. 17.4. Notice that the amplitudes of the harmonics decay very fast with frequency.
-
-**Figure 17.5** For Practice Prob. 17.1.
-
-Find the Fourier series of the square wave in Fig. 17.5. Plot the ampli tude and phase spectra.
-
-# **Figure 17.6**
-
-the amplitude and phase spectra.
-
-# **Solution:**
-
-The function is described as
-
-$$
-f(t) = \begin{cases} t, & 0 < t < 1 \\ 0, & 1 < t < 2 \end{cases}
-$$
-
-Because *T* = 2, *ω*0 = 2*π*∕*T* = *π*. Then
-
-$$
-a_0 = \frac{1}{T} \int_0^T f(t) \, dt = \frac{1}{2} \left[ \int_0^1 t \, dt + \int_1^2 0 \, dt \right] = \frac{1}{2} \frac{t^2}{2} \Big|_0^1 = \frac{1}{4} \quad (17.2.1)
-$$
-
-To evaluate *an* and *bn*, we need the integrals in Eq. (17.15):
-
-$$
-a_n = \frac{2}{T} \int_0^T f(t) \cos n\omega_0 t \, dt
-$$
-
-= $\frac{2}{2} \left[ \int_0^1 t \cos n\pi t \, dt + \int_1^2 0 \cos n\pi t \, dt \right]$
-= $\left[ \frac{1}{n^2 \pi^2} \cos n\pi t + \frac{t}{n\pi} \sin n\pi t \right]_0^1$
-= $\frac{1}{n^2 \pi^2} (\cos n\pi - 1) + 0 = \frac{(-1)^n - 1}{n^2 \pi^2}$ (17.2.2)
-
-since cos *nπ* = (−1)*n* ; and
-
-$$
-b_n = \frac{2}{T} \int_0^T f(t) \sin n\omega_0 t \, dt
-$$
-
-= $\frac{2}{2} \left[ \int_0^1 t \sin n\pi t \, dt + \int_1^2 0 \sin n\pi t \, dt \right]$
-= $\left[ \frac{1}{n^2 \pi^2} \sin n\pi t - \frac{t}{n\pi} \cos n\pi t \right]_0^1$
-= $0 - \frac{\cos n\pi}{n\pi} = \frac{(-1)^{n+1}}{n\pi}$ (17.2.3)
-
-Substituting the Fourier coefficients just found into Eq. (17.3) yields
-
-$$
-f(t) = \frac{1}{4} + \sum_{n=1}^{\infty} \left[ \frac{[(-1)^n - 1]}{(n\pi)^2} \cos n\pi t + \frac{(-1)^{n+1}}{n\pi} \sin n\pi t \right]
-$$
-
-To obtain the amplitude and phase spectra, we notice that, for e ven harmonics, *an* = 0, *bn* = −1∕*nπ*, so that
-
-$$
-A_n / \underline{\phi_n} = a_n - jb_n = 0 + j\frac{1}{n\pi} \tag{17.2.4}
-$$
-
-Hence,
-
-$$
-A_n = |b_n| = \frac{1}{n\pi}, \qquad n = 2, 4, ...
-$$
-
-\n
-$$
-\phi_n = 90^\circ, \qquad n = 2, 4, ...
-$$
-\n(17.2.5)
-
-For odd harmonics, *an* = −2∕(*n*2 *π*2 ),*bn* = 1∕(*nπ*) so that
-
-$$
-A_n / \underline{\phi_n} = a_n - jb_n = -\frac{2}{n^2 \pi^2} - j \frac{1}{n \pi}
-$$
- (17.2.6)
-
-That is,
-
-$$
-A_n = \sqrt{a_n^2 + b_n^2} = \sqrt{\frac{4}{n^4 \pi^4} + \frac{1}{n^2 \pi^2}}
-$$
-
-= $\frac{1}{n^2 \pi^2} \sqrt{4 + n^2 \pi^2}$ , $n = 1, 3, ...$ (17.2.7)
-
-From Eq. (17.2.6), we observe that lies in the third quadrant, so that
-
-$$
-\phi_n = 180^\circ + \tan^{-1} \frac{n\pi}{2}, \qquad n = 1, 3, ... \tag{17.2.8}
-$$
-
-From Eqs. (17.2.5), (17.2.7), and (17.2.8), we plot *An* and *n* for different values of *nω*0 = *nπ* to obtain the amplitude spectrum and phase spectrum as shown in Fig. 17.8.
-
-**Figure 17.8** For Example 17.2: (a) amplitude spectrum, (b) phase spectrum.
-
-# Practice Problem 17.2
-
-**Figure 17.9** For Practice Prob. 17.2.
-
-Determine the Fourier series of the sawtooth waveform in Fig. 17.9.
-
-**Answer:**
-$$
-f(t) = 4.5 - \frac{9}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \sin 2 \pi nt
-$$
-.
diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/198_17.3 Symmetry Considerations.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/198_17.3 Symmetry Considerations.md
deleted file mode 100644
index f6865fc0b24938876c586b4c9237940a07703e7d..0000000000000000000000000000000000000000
--- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/198_17.3 Symmetry Considerations.md
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-# **17.3** Symmetry Considerations
-
-We noticed that the Fourier series of Example 17.1 consisted only of the sine terms. One may w onder if a method e xists whereby one can kno w in advance that some F ourier coefficients would be zero and a void the unnecessary work involved in the tedious process of calculating them. Such a method does e xist; it is based on recognizing the e xistence of symmetry. Here we discuss three types of symmetry: (1) even symmetry, (2) odd symmetry, (3) half-wave symmetry.
-
-# **17.3.1** Even Symmetry
-
-A function *f*(*t*) is *even* if its plot is symmetrical about the vertical axis; that is,
-
-$$
-f(t) = f(-t) \tag{17.16}
-$$
-
-**(17.18)**
-
-Examples of e ven functions are *t* 2 , *t* 4 , and cos *t*. Figure 17.10 sho ws more e xamples of periodic e ven functions. Note that each of these examples satisfies Eq. (17.16). A main property of an e ven function *fe*(*t*) is that:
-
-$$
-\int_{-T/2}^{T/2} f_e(t) \, dt = 2 \int_0^{T/2} f_e(t) \, dt \tag{17.17}
-$$
-
-because integrating from −*T*∕2 to 0 is the same as inte grating from 0 to *T*∕2. Utilizing this property, the Fourier coefficients for an even function become
-
-$$
-a_0 = \frac{2}{T} \int_0^{T/2} f(t)dt
-$$
-
-\n
-$$
-a_n = \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t dt
-$$
-
-\n
-$$
-b_n = 0
-$$
-
-Because *bn* = 0, Eq. (17.3) becomes a *Fourier cosine series.* This makes sense because the cosine function is itself e ven. It also mak es intuitive sense that an e ven function contains no sine terms gi ven that the sine function is odd.
-
-To confirm Eq. (17.18) quantitatively, we apply the property of an even function in Eq. (17.17) in evaluating the Fourier coefficients in Eqs. (17.6), (17.8), and (17.9). It is convenient in each case to integrate over the interval −*T*∕2 < *t* < *T*∕2, which is symmetrical about the origin. Thus,
-
-$$
-a_0 = \frac{1}{T} \int_{-T/2}^{T/2} f(t) dt = \frac{1}{T} \left[ \int_{-T/2}^{0} f(t) dt + \int_{0}^{T/2} f(t) dt \right]
-$$
- (17.19)
-
-We change variables for the inte gral over the interval −*T*∕2 < *t* < 0 by letting *t* = −*x*, so that *dt* = −*dx*, *f*(*t*) = *f*(−*t*) = *f*(*x*), since *f*(*t*) is an even function, and when *t* = −*T*∕2, *x* = *T*∕2. Then,
-
-$$
-a_0 = \frac{1}{T} \left[ \int_{T/2}^0 f(x)(-dx) + \int_0^{T/2} f(t) dt \right]
-$$
-
-=
-$$
-\frac{1}{T} \left[ \int_0^{T/2} f(x) dx + \int_0^{T/2} f(t) dt \right]
-$$
- (17.20)
-
-showing that the two integrals are identical. Hence,
-
-$$
-a_0 = \frac{2}{T} \int_0^{T/2} f(t) dt
-$$
- (17.21)
-
-as expected. Similarly, from Eq. (17.8),
-
-$$
-a_n = \frac{2}{T} \left[ \int_{-T/2}^{0} f(t) \cos n\omega_0 t \, dt + \int_{0}^{T/2} f(t) \cos n\omega_0 t \, dt \right]
-$$
- (17.22)
-
-Typical examples of even periodic functions.
-
-We make the same change of v ariables that led to Eq. (17.20) and note that both *f*(*t*) and cos *nω*0*t* are even functions, implying that *f*(−*t*) = *f*(*t*) and cos(−*nω*0*t*) = cos *nω*0*t*. Equation (17.22) becomes
-
-$$
-a_n = \frac{2}{T} \left[ \int_{T/2}^0 f(-x) \cos(-n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right]
-$$
-
-=
-$$
-\frac{2}{T} \left[ \int_{T/2}^0 f(x) \cos(n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right]
-$$
-
-=
-$$
-\frac{2}{T} \left[ \int_0^{T/2} f(x) \cos(n\omega_0 x) dx + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right]
-$$
-(17.23a)
-
-or
-
-t
-
-$$
-a_n = \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \tag{17.23b}
-$$
-
-as expected. For *bn*, we apply Eq. (17.9),
-
-$$
-b_n = \frac{2}{T} \left[ \int_{-T/2}^{0} f(t) \sin n\omega_0 t \, dt + \int_{0}^{T/2} f(t) \sin n\omega_0 t \, dt \right] \tag{17.24}
-$$
-
-We make the same change of variables but keep in mind that *f*(−*t*) = *f*(*t*) but sin(−*nω*0*t*) = −sin *nω*0*t*. Equation (17.24) yields
-
-$$
-b_n = \frac{2}{T} \left[ \int_{T/2}^0 f(-x) \sin(-n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]
-$$
-
-= $\frac{2}{T} \left[ \int_{T/2}^0 f(x) \sin n\omega_0 x dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]$
-= $\frac{2}{T} \left[ - \int_0^{T/2} f(x) \sin (n\omega_0 x) dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]$
-= 0 (17.25)
-
-confirming Eq. (17.18).
-
-# **17.3.2** Odd Symmetry
-
-A function *f*(*t*) is said to be *odd* if its plot is antisymmetrical about the vertical axis:
-
-$$
-f(-t) = -f(t) \tag{17.26}
-$$
-
-Examples of odd functions are *t*, *t* 3 , and sin *t*. Figure 17.11 sho ws more examples of periodic odd functions. All these e xamples satisfy Eq. (17.26). An odd function *fo*(*t*) has this major characteristic:
-
-∫
-
-$$
-f_{-T/2}^{T/2} f_o(t) dt = 0
-$$
-\n(17.27)
-
-f(t)
-
-(c)
-
-**Figure 17.11** Typical examples of odd periodic functions.
-
-because integration from −*T*∕2 to 0 is the negative of that from 0 to *T*∕2. With this property, the Fourier coefficients for an odd function become
-
-$$
-a_0 = 0, \t a_n = 0
-$$
-
-$$
-b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt
-$$
- (17.28)
-
-which give us a *Fourier sine series*. Again, this makes sense because the sine function is itself an odd function. Also, note that there is no dc term for the Fourier series expansion of an odd function.
-
-The quantitati ve proof of Eq. (17.28) follo ws the same proce dure tak en to pro ve Eq. (17.18) e xcept that *f*(*t*) is no w odd, so that *f*(*t*) = −*f*(*t*). With this fundamental b ut simple difference, it is easy to see that *a*0 = 0 in Eq. (17.20), *an* = 0 in Eq. (17.23a), and *bn* in Eq. (17.24) becomes
-
-$$
-b_n = \frac{2}{T} \left[ \int_{T/2}^0 f(-x) \sin(-n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]
-$$
-
-$$
-= \frac{2}{T} \left[ -\int_{T/2}^0 f(x) \sin n\omega_0 x \, dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]
-$$
-
-$$
-= \frac{2}{T} \left[ \int_0^{T/2} f(x) \sin(n\omega_0 x) \, dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]
-$$
-
-$$
-b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \qquad (17.29)
-$$
-
-as expected.
-
-It is interesting to note that an y periodic function *f*(*t*) with neither even nor odd symmetry may be decomposed into e ven and odd parts. Using the properties of even and odd functions from Eqs. (17.16) and (17.26), we can write
-
-$$
-f(t) = \underbrace{\frac{1}{2} [f(t) + f(-t)]}_{\text{even}} + \underbrace{\frac{1}{2} [f(t) - f(-t)]}_{\text{odd}} = f_e(t) + f_o(t)
-$$
-(17.30)
-
-Notice that *fe*(*t*) = \_\_1 2 [ *f*(*t*) + *f*(−*t*)] satisfies the property of an even function in Eq. (17.16), while *fo*(*t*) = \_\_1 2 [ *f*(*t*) − *f*(−*t*)] satisfies the property of an odd function in Eq. (17.26). The fact that *fe*(*t*) contains only the dc term and the cosine terms, while *fo*(*t*) has only the sine terms, can be e xploited in grouping the F ourier series expansion of *f*(*t*) as
-
-$$
-f(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + \sum_{n=1}^{\infty} b_n \sin n\omega_0 t = f_e(t) + f_o(t)
-$$
- (17.31)
-even
-odd
-
-It follows readily from Eq. (17.31) that when *f*(*t*) is e ven, *bn* = 0, and when *f*(*t*) is odd, *a*0 = 0 = *an*.
-
-Also, note the following properties of odd and even functions:
-
-- 1. The product of two even functions is also an even function.
-- 2. The product of two odd functions is an even function.
-- 3. The product of an e ven function and an odd function is an odd function.
-- 4. The sum (or dif ference) of two even functions is also an even function.
-- 5. The sum (or difference) of two odd functions is an odd function.
-- 6. The sum (or difference) of an even function and an odd function is neither even nor odd.
-
-Each of these properties can be proved using Eqs. (17.16) and (17.26).
-
-# **17.3.3** Half-Wave Symmetry
-
-A function is half-wave (odd) symmetric if
-
-$$
-f\left(t - \frac{T}{2}\right) = -f(t) \tag{17.32}
-$$
-
-which means that each half-c ycle is the mirror image of the ne xt halfcycle. Notice that functions cos *nω*0*t* and sin *nω*0*t* satisfy Eq. (17.32) for odd values of *n* and therefore possess half-wave symmetry when *n* is odd. Figure 17.12 sho ws other examples of half-wave symmetric func tions. The functions in Figs. 17.11(a) and 17.11(b) are also half-w ave symmetric. Notice that for each function, one half-c ycle is the inverted version of the adjacent half-cycle. The Fourier coefficients become
-
-**Figure 17.12** Typical examples of half-wave odd symmetric functions.
-
-showing that the Fourier series of a half-wave symmetric function con tains only odd harmonics.
-
-To deri ve Eq. (17.33), we apply the property of half-w ave sym metric functions in Eq. (17.32) in e valuating the Fourier coefficients in Eqs. (17.6), (17.8), and (17.9). Thus,
-
-$$
-a_0 = \frac{1}{T} \int_{-T/2}^{T/2} f(t) dt = \frac{1}{T} \left[ \int_{-T/2}^{0} f(t) dt + \int_{0}^{T/2} f(t) dt \right]
-$$
- (17.34)
-
-We change v ariables for the inte gral o ver the interv al −*T*∕2 < *t* < 0 by letting *x* = *t* + *T*∕2, so that *dx* = *dt*; when *t* = −*T*∕2, *x* = 0; and when *t* = 0, *x* = *T*∕2. Also, we k eep Eq. (17.32) in mind; that is, *f*(*x* − *T*∕2) = −*f* (*x*). Then,
-
-$$
-a_0 = \frac{1}{T} \left[ \int_0^{T/2} f\left(x - \frac{T}{2}\right) dx + \int_0^{T/2} f(t) dt \right]
-$$
-
-= $\frac{1}{T} \left[ - \int_0^{T/2} f(x) dx + \int_0^{T/2} f(t) dt \right] = 0$ (17.35)
-
-confirming the expression for *a*0 in Eq. (17.33). Similarly,
-
-$$
-a_n = \frac{2}{T} \left[ \int_{-T/2}^{0} f(t) \cos n\omega_0 t \, dt + \int_{0}^{T/2} f(t) \cos n\omega_0 t \, dt \right] \quad (17.36)
-$$
-
-We make the same change of variables that led to Eq. (17.35) so that Eq. (17.36) becomes
-
-$$
-a_n = \frac{2}{T} \left[ \int_0^{T/2} f\left(x - \frac{T}{2}\right) \cos n\omega_0 \left(x - \frac{T}{2}\right) dx + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right]
-$$
- (17.37)
-
-Because *f*(*x* − *T*∕2) = −*f*(*x*) and
-
-$$
-\cos n\omega_0 \left( x - \frac{T}{2} \right) = \cos(n\omega_0 t - n\pi)
-$$
-
-= $\cos n\omega_0 t \cos n\pi + \sin n\omega_0 t \sin n\pi$ (17.38)
-= $(-1)^n \cos n\omega_0 t$
-
-substituting these in Eq. (17.37) leads to
-
-$$
-a_n = \frac{2}{T} \left[ 1 - (-1)^n \right] \int_0^{T/2} f(t) \cos n\omega_0 t \, dt
-$$
-
-=
-$$
-\begin{cases} \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t \, dt, & \text{for } n \text{ odd} \\ 0, & \text{for } n \text{ even} \end{cases}
-$$
-(17.39)
-
-confirming Eq. (17.33). By following a similar procedure, we can derive *bn* as in Eq. (17.33).
-
-Table 17.2 summarizes the ef fects of these symmetries on the Fourier coef ficients. Table 17.3 pro vides the F ourier series of some common periodic functions.
-
-# **TABLE 17.2**
-
-# Effects of symmetry on Fourier coefficients.
-
-| Symmetry | a0 | an | bn | Remarks |
-|-----------|--------|-----------|-----------|---------------------------------------------------------------|
-| Even | a0 ≠ 0 | an ≠ 0 | bn = 0 | Integrate over T∕2 and multiply by 2 to get the coefficients. |
-| Odd | a0 = 0 | an = 0 | bn ≠ 0 | Integrate over T∕2 and multiply by 2 to get the coefficients. |
-| Half-wave | a0 = 0 | a2n = 0 | b2n = 0 | Integrate over T∕2 and multiply by 2 to get the coefficients. |
-| | | a2n+1 ≠ 0 | b2n+1 ≠ 0 | |
-
-# **TABLE 17.3**
-
-Find the Fourier series expansion of *f*(*t*) given in Fig. 17.13.
-
-**Figure 17.13** For Example 17.3.
-
-# **Solution:**
-
-The function *f*(*t*) is an odd function. Hence *a*0 = 0 = *an*. The period is *T* = 4, and *ω*0 = 2*π*∕*T* = *π*∕2, so that
-
-$$
-b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt
-$$
-
-= $\frac{4}{4} \left[ \int_0^1 1 \sin \frac{n\pi}{2} t \, dt + \int_1^2 0 \sin \frac{n\pi}{2} t \, dt \right]$
-= $-\frac{2}{n\pi} \cos \frac{n\pi t}{2} \Big|_0^1 = \frac{2}{n\pi} \left( 1 - \cos \frac{n\pi}{2} \right)$
-
-Hence,
-
-$$
-f(t) = \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \left( 1 - \cos \frac{n\pi}{2} \right) \sin \frac{n\pi}{2} t
-$$
-
-which is a Fourier sine series.
-
-Find the Fourier series of the function *f*(*t*) in Fig. 17.14.
-
-t f(t) ‒2*𝜋* ‒*𝜋 𝜋* 0 2*𝜋* 3*𝜋* 20 ‒20
-
-**Figure 17.14** For Practice Prob. 17.3.
-
-**Answer:**
-$$
-f(t) = -\frac{80}{\pi} \sum_{k=1}^{\infty} \frac{1}{n} \sin nt
-$$
-, $n = 2k - 1$ .
-
-Practice Problem 17.3
-
-Example 17.4
-
-Determine the Fourier series for the half-w ave rectified cosine function shown in Fig. 17.15.
-
-A half-wave rectified cosine function; for Example 17.4.
-
-# **Solution:**
-
-This is an even function so that *bn* = 0. Also, *T* = 4, *ω*0 = 2*π*∕*T* = *π*∕2. Over a period,
-
-$$
-f(t) = \begin{cases} 0, & -2 < t < -1 \\ \cos \frac{\pi}{2} t, & -1 < t < 1 \end{cases}
-$$
-
-$$
-a_0 = \frac{2}{T} \int_0^{T/2} f(t) dt = \frac{2}{4} \left[ \int_0^1 \cos \frac{\pi}{2} t dt + \int_1^2 0 dt \right]
-$$
-
-$$
-= \frac{1}{2} \frac{2}{\pi} \sin \frac{\pi}{2} t \Big|_0^1 = \frac{1}{\pi}
-$$
-
-$$
-a_n = \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t dt = \frac{4}{4} \left[ \int_0^1 \cos \frac{\pi}{2} t \cos \frac{n\pi t}{2} dt + 0 \right]
-$$
-
-Put $\cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)]$ . Then
-
-But cos *A* cos *B* = \_\_1 2 [cos(*A* + *B*) + cos(*A* − *B*)]. Then *an* = \_\_1 2 ∫ 0 1 [ cos \_\_ *π* 2 (*n* + 1)*t* + cos \_\_ *π* 2 (*n* − 1)*t* ] *dt*
-
-For *n* = 1,
-
-$$
-a_1 = \frac{1}{2} \int_0^1 \left[ \cos \pi t + 1 \right] dt = \frac{1}{2} \left[ \frac{\sin \pi t}{\pi} + t \right] \Big|_0^1 = \frac{1}{2}
-$$
-
-For *n* > 1,
-
-$$
-a_n = \frac{1}{\pi(n+1)} \sin \frac{\pi}{2} (n+1) + \frac{1}{\pi(n-1)} \sin \frac{\pi}{2} (n-1)
-$$
-
-For *n* = odd (*n* = 1, 3, 5, …), (*n* + 1) and (*n* − 1) are both even, so
-
-$$
-\sin \frac{\pi}{2}(n+1) = 0 = \sin \frac{\pi}{2}(n-1), \quad n = \text{odd}
-$$
-
-For *n* = even (*n* = 2, 4, 6, …), (*n* + 1) and (*n* − 1) are both odd. Also,
-
-$$
-\sin\frac{\pi}{2}(n+1) = -\sin\frac{\pi}{2}(n-1) = \cos\frac{n\pi}{2} = (-1)^{n/2}, \qquad n = \text{even}
-$$
-
-Hence,
-
-$$
-a_n = \frac{(-1)^{n/2}}{\pi(n+1)} + \frac{-(-1)^{n/2}}{\pi(n-1)} = \frac{-2(-1)^{n/2}}{\pi(n^2 - 1)}, \qquad n = \text{even}
-$$
-
-Thus,
-
-$$
-f(t) = \frac{1}{\pi} + \frac{1}{2}\cos\frac{\pi}{2}t - \frac{2}{\pi}\sum_{n=\text{even}}^{\infty}\frac{(-1)^{n/2}}{(n^2 - 1)}\cos\frac{n\pi}{2}t
-$$
-
-To avoid using *n* = 2, 4, 6, … and also to ease computation, we can replace *n* by 2*k*, where *k* = 1, 2, 3, … and obtain
-
-$$
-f(t) = \frac{1}{\pi} + \frac{1}{2}\cos\frac{\pi}{2}t - \frac{2}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^k}{(4k^2 - 1)}\cos k\pi t
-$$
-
-which is a Fourier cosine series.
-
-Find the Fourier series expansion of the function in Fig. 17.16.
-
-**Answer:**
-$$
-f(t) = 16 - \frac{128}{\pi^2} \sum_{k=1}^{\infty} \frac{1}{n^2} \cos nt, n = 2k - 1.
-$$
-
-**Figure 17.16** For Practice Prob. 17.4.
-
-Calculate the Fourier series for the function in Fig. 17.17.
-
-# **Solution:**
-
-The function in Fig. 17.17 is half-wave odd symmetric, so that *a*0 = 0 = *an*. It is described over half the period as
-
-$$
-f(t) = t, \qquad -1 < t < 1
-$$
-
-*T* = 4, *ω*0 = 2*π*∕*T* = *π*∕2. Hence,
-
-$$
-b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt
-$$
-
-Instead of integrating *f*(*t*) from 0 to 2, it is more convenient to integrate from −1 to 1. Applying Eq. (17.15d),
-
-$$
-b_n = \frac{4}{4} \int_{-1}^{1} t \sin \frac{n\pi t}{2} dt = \left[ \frac{\sin n\pi t/2}{n^2 \pi^2/4} - \frac{t \cos n\pi t/2}{n\pi/2} \right] \Big|_{-1}^{1}
-$$
-
-= $\frac{4}{n^2 \pi^2} \left[ \sin \frac{n\pi}{2} - \sin \left( -\frac{n\pi}{2} \right) \right] - \frac{2}{n\pi} \left[ \cos \frac{n\pi}{2} - \cos \left( -\frac{n\pi}{2} \right) \right]$
-= $\frac{8}{n^2 \pi^2} \sin \frac{n\pi}{2}$
-
-since sin(−*x*) = −sin *x* is an odd function, while cos( −*x*) = cos *x* is an even function. Using the identities for sin *nπ*∕2 in Table 17.1,
-
-$$
-b_n = \frac{8}{n^2 \pi^2} (-1)^{(n-1)/2}, \quad n = \text{odd} = 1, 3, 5, ...
-$$
-
-**Figure 17.17** For Example 17.5.
-
-Thus,
-
-$$
-f(t) = \sum_{n=1,3,5}^{\infty} b_n \sin \frac{n\pi}{2} t.
-$$
-
-# Practice Problem 17.5
-
-Determine the Fourier series of the function in Fig. 17.12(a). Take *A* = 8 and *T* = 2*π*.
-
-**Answer:**
-$$
-f(t) = \frac{16}{\pi} \sum_{k=1}^{\infty} \left( \frac{-2}{n^2 \pi} \cos nt + \frac{1}{n} \sin nt \right), n = 2k - 1.
-$$
-
-# **17.4** Circuit Applications
-
-We find that in practice, many circuits are driven by nonsinusoidal periodic functions. To find the steady-state response of a circuit to a nonsinusoidal periodic excitation requires the application of a Fourier series, ac phasor analysis, and the superposition principle. The procedure usually involves four steps.
-
-# Steps for Applying Fourier Series:
-
-- 1. Express the excitation as a Fourier series.
-- 2. Transform the circuit from the time domain to the frequenc y domain.
-- 3. Find the response of the dc and ac components in the Fourier series.
-- 4. Add the individual dc and ac responses using the superposition principle.
-
-The first step is to determine the Fourier series e xpansion of the excitation. For the periodic v oltage source sho wn in Fig. 17.18(a), for example, the Fourier series is expressed as
-
-$$
-v(t) = V_0 + \sum_{n=1}^{\infty} V_n \cos(n\omega_0 t + \theta_n)
-$$
- (17.40)
-
-(The same could be done for a periodic current source.) Equation (17.40) shows that *v*(*t*) consists of tw o parts: the dc component *V*0 and the ac component **V***n* = *Vn*⧸*θn* with several harmonics. This Fourier series representation may be re garded as a set of series-connected sinusoidal sources, with each source ha ving its own amplitude and frequency, as shown in Fig. 17.18(b).
-
-The third step is finding the response to each term in the Fourier series. The response to the dc component can be determined in the
-
-(a) Linear network excited by a periodic voltage source, (b) Fourier series representation (time-domain).
-
-frequency domain by setting *n* = 0 or *ω* = 0 as in Fig. 17.19(a), or in the time domain by replacing all inductors with short circuits and all capacitors with open circuits. The response to the ac component is obtained by applying the phasor techniques co vered in Chapter 9, as shown in Fig. 17.19(b). The network is represented by its impedance **Z**(*nω*0) or admittance **Y**(*nω*0). **Z**(*nω*0) is the input impedance at the source when *ω* is everywhere replaced by *nω*0, and **Y**(*nω*0) is the reciprocal of **Z**(*nω*0).
-
-Finally, following the principle of superposition, we add all the individual responses. For the case shown in Fig. 17.19,
-
-$$
-i(t) = i_0(t) + i_1(t) + i_2(t) + \cdots
-$$
-
-= $\mathbf{I}_0 + \sum_{n=1}^{\infty} |\mathbf{I}_n| \cos(n\omega_0 t + \psi_n)$ (17.41)
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-where each component **I***n* with frequenc y *nω*0 has been transformed to the time domain to get *in*(*t*), and *ψn* is the argument of **I***n*.
-
-# **Figure 17.19**
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-Steady-state responses: (a) dc component, (b) ac component (frequency domain).
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-Let the function *f*(*t*) in Example 17.1 be the v oltage source *vs*(*t*) in the circuit of Fig. 17.20. Find the response *vo*(*t*) of the circuit.
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-# **Solution:**