A" 1L J CHAPTER CONTENTS 9.1 The Sinusoidal Source p. 308 9.2 The Sinusoidal Response p. 311 9.3 The Phasor p. 312 9.4 The Passive Circuit Elements in the Frequency Domain p. 317 9.5 Kirchhoff's Laws in the Frequency Domain p. 321 9.6 Series, Parallel, and Delta-to-Wye Simplifications p. 322 9.7 Source Transformations and Thevenin-Norton Equivalent Circuits p. 329 9.8 The Node-Voltage Method p. 332 9.9 The Mesh-Current Method p. 333 9.10 The Transformer p. 334 9.11 The Ideal Transformer p. 338 9.12 Phasor Diagrams p. 344 1 Understand phasor concepts and be able to perform a phasor transform and an inverse phasor transform. 2 Be able to transform a circuit with a sinusoidal source into the frequency domain using phasor concepts. 3 Know how to use the following circuit analysis techniques to solve a circuit in the frequency domain: • Kirchhoffs laws; • Series, parallel, and delta-to-wye simplifications; • Voltage and current division; • Thevenin and Norton equivalents; • Node-voltage method; and • Mesh-current method. 4 Be able to analyze circuits containing linear transformers using phasor methods. 5 Understand the ideal transformer constraints and be able to analyze circuits containing ideal transformers using phasor methods. 306 Sinusoidal Steady-State Analysis Thus far, we have focused on circuits with constant sources; in this chapter we are now ready to consider circuits energized by time-varying voltage or current sources. In particular, we are inter- ested in sources in which the value of the voltage or current varies sinusoidally. Sinusoidal sources and their effect on circuit behavior form an important area of study for several reasons. First, the gen- eration, transmission, distribution, and consumption of electric energy occur under essentially sinusoidal steady-state conditions. Second, an understanding of sinusoidal behavior makes it possible to predict the behavior of circuits with nonsinusoidal sources. Third, steady-state sinusoidal behavior often simplifies the design of electrical systems. Thus a designer can spell out specifications in terms of a desired steady-state sinusoidal response and design the circuit or system to meet those characteristics. If the device satis- fies the specifications, the designer knows that the circuit will respond satisfactorily to nonsinusoidal inputs. The subsequent chapters of this book are largely based on a thorough understanding of the techniques needed to analyze cir- cuits driven by sinusoidal sources. Fortunately, the circuit analysis and simplification techniques first introduced in Chapters 1-4 work for circuits with sinusoidal as well as dc sources, so some of the material in this chapter will be very familiar to you. The chal- lenges in first approaching sinusoidal analysis include developing the appropriate modeling equations and working in the mathe- matical realm of complex numbers. Practical Perspective A Household Distribution Circuit Power systems that generate, transmit, and distribute electri- cal power are designed to operate in the sinusoidal steady state. The standard household distribution circuit used in the United States is the three-wire, 240/120 V circuit shown in the accompanying figure. The transformer is used to reduce the utility distribution voltage from 13.2 kV to 240 V. The center tap on the second- ary winding provides the 120 V service. The operating fre- quency of power systems in the United States is 60 Hz. Both 50 and 60 Hz systems are found outside the United States. ^oy 103 The voltage ratings alluded to above are rms values. The rea- son for defining an rms value of a time-varying signal is explained in Chapter 10. 307 9.1 The Sinusoidal Source A sinusoidal voltage source (independent or dependent) produces a volt- age that varies sinusoidally with time. A sinusoidal current source (inde- pendent or dependent) produces a current that varies sinusoidally with time. In reviewing the sinusoidal function, we use a voltage source, but our observations also apply to current sources. We can express a sinusoidally varying function with either the sine function or the cosine function. Although either works equally well, we cannot use both functional forms simultaneously. We will use the cosine function throughout our discussion. Hence, we write a sinusoidally varying voltage as v = V m cos (art + 4>). (9.1) To aid discussion of the parameters in Eq. 9.1, we show the voltage versus time plot in Fig. 9.1. Note that the sinusoidal function repeats at regular intervals. Such a function is called periodic. One parameter of interest is the length of time required for the sinusoidal function to pass through all its possible values. This time is referred to as the period of the function and is denoted T. It is measured in seconds. The reciprocal of T gives the number of cycles per second, or the frequency, of the sine function and is denoted /, or f = f- (9.2) A cycle per second is referred to as a hertz, abbreviated Hz. (The term cycles per second rarely is used in contemporary technical literature.) The coefficient of t in Eq. 9.1 contains the numerical value of Torf. Omega (co) represents the angular frequency of the sinusoidal function, or co = 2-77-/ = 2-77-/7 1 (radians/second). (9.3) Equation 9.3 is based on the fact that the cosine (or sine) function passes through a complete set of values each time its argument, cot, passes through 2-7T rad (360°). From Eq. 9.3, note that, whenever t is an integral multiple of T, the argument cot increases by an integral multiple of 2TT rad. The coefficient V m gives the maximum amplitude of the sinusoidal voltage. Because ±1 bounds the cosine function, ±V m bounds the ampli- tude. Figure 9.1 shows these characteristics. The angle cp in Eq. 9.1 is known as the phase angle of the sinusoidal voltage. It determines the value of the sinusoidal function at t = 0; there- fore, it fixes the point on the periodic wave at which we start measuring time. Changing the phase angle cp shifts the sinusoidal function along the time axis but has no effect on either the amplitude (V m ) or the angular fre- quency (co). Note, for example, that reducing cp to zero shifts the sinusoidal function shown in Fig. 9.1 cp/co time units to the right, as shown in Fig. 9.2. Note also that if cp is positive, the sinusoidal function shifts to the left, whereas if cp is negative, the function shifts to the right. (See Problem 9.5.) A comment with regard to the phase angle is in order: cot and cp must carry the same units, because they are added together in the argument of the sinusoidal function. With cot expressed in radians, you would expect cp to be also. However, cp normally is given in degrees, and cot is converted from radians to degrees before the two quantities are added. We continue 9.1 The Sinusoidal Source 309 this bias toward degrees by expressing the phase angle in degrees. Recall from your studies of trigonometry that the conversion from radians to degrees is given by (number of degrees) 18CT TT (number of radians). (9.4) Another important characteristic of the sinusoidal voltage (or cur- rent) is its rms value. The rms value of a periodic function is defined as the square root of the mean value of the squared function. Hence, if v = V m cos (cot + 4>), the rms value of v is t»+ T V 2 m COS 2 (cot + ¢) dt. (9.5) Note from Eq. 9.5 that we obtain the mean value of the squared voltage by integrating v 2 over one period (that is, from t 0 to t Q + T) and then dividing by the range of integration, T. Note further that the starting point for the integration t (] is arbitrary. The quantity under the radical sign in Eq. 9.5 reduces to V 2 „/2. (See Problem 9.6.) Hence the rms value of v is V - y rms Vm vr (9.6) M rms value of a sinusoidal voltage source The rms value of the sinusoidal voltage depends only on the maximum amplitude of v, namely, V m . The rms value is not a function of either the frequency or the phase angle. We stress the importance of the rms value as it relates to power calculations in Chapter 10 (see Section 10.3). Thus, we can completely describe a specific sinusoidal signal if we know its frequency, phase angle, and amplitude (either the maximum or the rms value). Examples 9.1, 9.2, and 9.3 illustrate these basic properties of the sinusoidal function. In Example 9.4, we calculate the rms value of a periodic function, and in so doing we clarify the meaning of root mean square. Example 9.1 Finding the Characteristics of a Sinusoidal Current A sinusoidal current has a maximum amplitude of 20 A. The current passes through one complete cycle in 1 ms. The magnitude of the current at zero time is 10 A. a) What is the frequency of the current in hertz? b) What is the frequency in radians per second? c) Write the expression for i(t) using the cosine function. Express <£ in degrees. d) What is the rms value of the current? Solution a) From the statement of the problem, T = 1 ms; hence/ = 1/T = 1000 Hz. b) to « 277-/ = 2000TT rad/s. c) We have i(t) = I m cos (<ot + ¢) = 20 COS(2000TT/ + <f>), but /(0) = 10 A. Therefore 10 = 20 cos 4> and cl> = 60°. Thus the expression for i(t) becomes /(f) = 20cos(20007rf + 60°). d) From the derivation of Eq. 9.6, the rms value of a sinusoidal current is /„,/V2. Therefore the rms value is 20/V2, or 14.14 A. 310 Sinusoidal Steady-State Analysis Finding the Characteristics of a Sinusoidal Voltage A sinusoidal voltage is given by the expression v = 300 cos (12()77/ + 30°). a) What is the period of the voltage in milliseconds? b) What is the frequency in hertz? c) What is the magnitude of v at t = 2.778 ms? d) What is the rms value of V? Solution a) From the expression for v, to = 12077 rad/s. Because (0 = 2TT/7\ T = 2TT/<O = ^ s, or 16.667 ms. b) The frequency is 1/7 1 , or 60 Hz. c) From (a), co = 2 77-/ 16.667; thus, at t = 2.778 ms, at is nearly 1.047 rad, or 60°. Therefore, y(2.778ms) = 300 cos (60° + 30°) = 0 V. d)V ms = 300/ V2 = 212.13 V. Example 9.3 Translating a Sine Expression to a Cosine Expression We can translate the sine function to the cosine function by subtracting 90° (TT/2 rad) from the argu- ment of the sine function. a) Verify this translation by showing that sin (tot + 0) = cos (tot + 8 - 90°). b) Use the result in (a) to express sin (cot + 30°) as a cosine function. Solution a) Verification involves direct application of the trigonometric identity cos(a — /3) = cos a cos /3 + sin a sin /3. We let a = ait + 0 and /3 = 90°. As cos 90° = 0 and sin 90° = 1, we have cos(a -/3)= sin a = sin(atf + 0) = cos(a>/ + 0 - 90°). b) From (a) we have sin(wr + 30°) = cos(o>/ + 30° - 90°) = cos(atf - 60°). Example 9.4 Calculating the rms Value of a Triangular Waveform Calculate the rms value of the periodic triangular current shown in Fig. 9.3. Express your answer in terms of the peak current I p . -772\ - Figure 9.3 A Periodic triangular current. Solution From Eq. 9.5, the rms value of i is ^rms \l -T- Interpreting the integral under the radical sign as the area under the squared function for an interval of one period is helpful in finding the rms value. The squared function with the area between 0 and T shaded is shown in Fig. 9.4, which also indicates that for this particular function, the area under the 9.2 The Sinusoidal Response 311 squared current for an interval of one period is equal to four times the area under the squared cur- rent for the interval 0 to TfA seconds; that is, t«+T /.7/4 i 2 dt = 4 / i 2 dt. etc. -7/2-7/4 0 Figure 9.4 • r versus t. 7/4 7/2 37/4 7 The analytical expression for /' in the interval 0 to 774is 47 i = -jrt, 0 < / < 774. The area under the squared function for one period is / i 2 dt = 4 / Tlj r 2 3 The mean, or average, value of the function is simply the area for one period divided by the period. Thus 1 ^ - i,2 — in' 7 3 3 p ' The rms value of the current is the square root of this mean value. Hence rms V3' NOTE: Assess your understanding of this material by trying Chapter Problems 9.1, 9.4, 9.8. 9.2 The Sinusoidal Response Before focusing on the steady-state response to sinusoidal sources, let's consider the problem in broader terms, that is, in terms of the total response. Such an overview will help you keep the steady-state solution in perspective. The circuit shown in Fig. 9.5 describes the general nature of the problem. There, v s is a sinusoidal voltage, or v s = V m cos i^t + </>)• (9.7) For convenience, we assume the initial current in the circuit to be zero and measure time from the moment the switch is closed. The task is to derive the expression for /(0 when t > 0. It is similar to finding the step response of an RL circuit, as in Chapter 7. The only difference is that the voltage source is now a time-varying sinusoidal voltage rather than a constant, or dc, voltage. Direct application of Kirchhoffs voltage law to the circuit shown in Fig. 9.5 leads to the ordinary differential equation Figure 9.5 • An RL circuit excited by a sinusoidal voltage source. L— + Ri = V m cos {a)t + 4>\ (9.8) the formal solution of which is discussed in an introductory course in dif- ferential equations. We ask those of you who have not yet studied differ- ential equations to accept that the solution for / is -v., VR 2 + <o 2 L 2 cos (0 -6)e~W L)t + V„ VR 2 + <o 2 L 2 cos (cot + 4> - 0), (9.9) where 0 is defined as the angle whose tangent is coL/R. Thus we can easily determine 0 for a circuit driven by a sinusoidal source of known frequency. We can check the validity of Eq. 9.9 by determining that it satisfies Eq. 9.8 for all values of t > 0; this exercise is left for your exploration in Problem 9.10. The first term on the right-hand side of Eq. 9.9 is referred to as the transient component of the current because it becomes infinitesimal as time elapses. The second term on the right-hand side is known as the steady-state component of the solution. It exists as long as the switch remains closed and the source continues to supply the sinusoidal voltage. In this chapter, we develop a technique for calculating the steady-state response directly, thus avoiding the problem of solving the differential equation. However, in using this technique we forfeit obtaining either the transient component or the total response, which is the sum of the tran- sient and steady-state components. We now focus on the steady-state portion of Eq. 9.9. It is important to remember the following characteristics of the steady-state solution: 1. The steady-state solution is a sinusoidal function. 2. The frequency of the response signal is identical to the frequency of the source signal. This condition is always true in a linear circuit when the circuit parameters, R, L, and C, are constant. (If frequen- cies in the response signals are not present in the source signals, there is a nonlinear element in the circuit.) 3. The maximum amplitude of the steady-state response, in general, differs from the maximum amplitude of the source. For the circuit being discussed, the maximum amplitude of the response signal is VJ\/R 2 + arL 2 , and the maximum amplitude of the signal source is V m . 4. The phase angle of the response signal, in general, differs from the phase angle of the source. For the circuit being discussed, the phase angle of the current is 4> - 0 and that of the voltage source is <f>. These characteristics are worth remembering because they help you understand the motivation for the phasor method, which we introduce in Section 9.3. In particular, note that once the decision has been made to find only the steady-state response, the task is reduced to finding the max- imum amplitude and phase angle of the response signal. The waveform and frequency of the response are already known. NOTE: Assess your understanding of this material by trying Chapter Problem 9.9. 9.3 The Phasor The phasor is a complex number that carries the amplitude and phase angle information of a sinusoidal function. 1 The phasor concept is rooted in Euler's identity, which relates the exponential function to the trigono- metric function: e ±jd = cos6» ± /sin0. (9.10) Equation 9.10 is important here because it gives us another way of express- ing the cosine and sine functions. We can think of the cosine function as the If you feel a bit uneasy about complex numbers, peruse Appendix B. 9.3 The Phasor 313 real part of the exponential function and the sine function as the imaginary part of the exponential function; that is, cosfl = Sfc{tf**}, (9.11) and sin0 = 3{<?' 0 }, (9.12) where 5ft means "the real part of 1 and S means "the imaginary part of." Because we have already chosen to use the cosine function in analyz- ing the sinusoidal steady state (see Section 9.1), we can apply Eq. 9.11 directly. In particular, we write the sinusoidal voltage function given by Eq. 9.1 in the form suggested by Eq. 9.11: v = V m cos (cot 4- (f>) = V m $t{e**efi}. (9.13) We can move the coefficient V m inside the argument of the real part of the function without altering the result. We can also reverse the order of the two exponential functions inside the argument and write Eq. 9.13 as V = U{V m e jtj> ei 101 }. (9.14) In Eq. 9.14, note that the quantity V m e® is a complex number that carries the amplitude and phase angle of the given sinusoidal function. This complex number is by definition the phasor representation, or phasor transform, of the given sinusoidal function. Thus V = V m ef* = V{V m cos(cot + ¢)), (9,15) < Phasor transform where the notation V{V m cos (cot + <f>)} is read "the phasor transform of V m cos (cot 4- <£)."Thus the phasor transform transfers the sinusoidal func- tion from the time domain to the complex-number domain, which is also called the frequency domain, since the response depends, in general, on to. As in Eq. 9.15, throughout this book we represent a phasor quantity by using a boldface letter. Equation 9.15 is the polar form of a phasor, but we also can express a phasor in rectangular form. Thus we rewrite Eq. 9.15 as V = V m cos $ + jV m sin </>. (9.16) Both polar and rectangular forms are useful in circuit applications of the phasor concept. One additional comment regarding Eq. 9.15 is in order. The frequent occurrence of the exponential function e^ has led to an abbreviation that lends itself to text material. This abbreviation is the angle notation We use this notation extensively in the material that follows. Inverse Phasor Transform So far we have emphasized moving from the sinusoidal function to its pha- sor transform. However, we may also reverse the process. That is, for a phasor we may write the expression for the sinusoidal function. Thus for V = l()0/-26°, the expression for v is 100cos (a)t - 26°) because we have decided to use the cosine function for all sinusoids. Observe that we cannot deduce the value of co from the phasor. The phasor carries only amplitude and phase information. The step of going from the phasor transform to the time-domain expression is referred to as finding the inverse phasor transform and is formalized by the equation V~ x {V m e^} = $t{V m e i4 >e> m }, (9.17) where the notation V~ l {V m e^} is read as "the inverse phasor transform of V m e'^." Equation 9.17 indicates that to find the inverse phasor transform, we multiply the phasor by <? /W and then extract the real part of the product. The phasor transform is useful in circuit analysis because it reduces the task of finding the maximum amplitude and phase angle of the steady- state sinusoidal response to the algebra of complex numbers. The follow- ing observations verify this conclusion: 1. The transient component vanishes as time elapses, so the steady- state component of the solution must also satisfy the differential equation. (See Problem 9.10[b].) 2. In a linear circuit driven by sinusoidal sources, the steady-state response also is sinusoidal, and the frequency of the sinusoidal response is the same as the frequency of the sinusoidal source. 3. Using the notation introduced in Eq. 9.11, we can postulate that the steady-state solution is of the form lR{Ae' li e JM '}, where A is the maximum amplitude of the response and /3 is the phase angle of the response. 4. When we substitute the postulated steady-state solution into the differential equation, the exponential term e ,u>l cancels out, leaving the solution for A and ft in the domain of complex numbers. We illustrate these observations with the circuit shown in Fig. 9.5 (see p. 311). We know that the steady-state solution for the current i is of the form WO = K{//fl, (9.18) where the subscript u ss" emphasizes that we are dealing with the steady- state solution. When we substitute Eq. 9.18 into Eq. 9.8, we generate the expression ®.{}<oLI m e®ei«*} + n{RI m e iP e J(M } = ^{V m e^ wt \. (9.19) In deriving Eq. 9.19 we recognized that both differentiation and multiplica- tion by a constant can be taken inside the real part of an operation. We also rewrote the right-hand side of Eq. 9.8, using the notation of Eq. 9.11. From the algebra of complex numbers, we know that the sum of the real parts is the same as the real part of the sum. Therefore we may reduce the left-hand side of Eq. 9.19 to a single term: 9R{(/wL + K^'V*} = ftlV^V*"}. (9.20) Recall that our decision to use the cosine function in analyzing the response of a circuit in the sinusoidal steady state results in the use of the 5ft operator in deriving Eq. 9.20. If instead we had chosen to use the sine function in our sinusoidal steady-state analysis, we would have applied Eq. 9.12 directly, in place of Eq. 9.11, and the result would be Eq.9.21: Q{(j(ol + R)I m e ifi e^} = ^{V m e j ^ wt }. (9.21) Note that the complex quantities on either side of Eq. 9.21 are identical to those on either side of Eq. 9.20. When both the real and imaginary parts of two complex quantities are equal, then the complex quantities are them- selves equal. Therefore, from Eqs. 9.20 and 9.21, (J(oL + R)I m e® = V m e i<!> or <-* - OiZ (9 - 22 > Note that e Ja>t has been eliminated from the determination of the ampli- tude (/,„) and phase angle (/3) of the response. Thus, for this circuit, the task of finding /,„ and /3 involves the algebraic manipulation of the com- plex quantities V m e® and R + jwL. Note that we encountered both polar and rectangular forms. An important warning is in order: The phasor transform, along with the inverse phasor transform, allows you to go back and forth between the time domain and the frequency domain. Therefore, when you obtain a solution, you are either in the time domain or the frequency domain. You cannot be in both domains simultaneously. Any solution that con- tains a mixture of time domain and phasor domain nomenclature is nonsensical. The phasor transform is also useful in circuit analysis because it applies directly to the sum of sinusoidal functions. Circuit analysis involves sum- ming currents and voltages, so the importance of this observation is obvi- ous. We can formalize this property as follows: If v = Vi + Vj + ••• + v n (9.23) where all the voltages on the right-hand side are sinusoidal voltages of the same frequency, then v = v, -+-v 2 + ••• + v„. (9.24) . for several reasons. First, the gen- eration, transmission, distribution, and consumption of electric energy occur under essentially sinusoidal steady-state conditions. Second, an understanding. nonsinusoidal sources. Third, steady-state sinusoidal behavior often simplifies the design of electrical systems. Thus a designer can spell out specifications in terms of a desired steady-state