APPENDIX A The Phasor Solution Method This appendix is intended as a review of the important phasor methods of solving electric circuits and other linear systems in which the excitation is a singlefrequency sinusoidal waveform and the system is in steady state. This method of solution will be used extensively throughout this text. See [1] or any other standard electric circuit analysis text for a more thorough review of this method. A.1 SOLVING DIFFERENTIAL EQUATIONS FOR THEIR SINUSOIDAL, STEADY-STATE SOLUTION This important general problem is depicted in Fig. A.1a as a single-input, singleoutput linear system. It is important to note that this important method applies to a large class of physical systems (electrical, mechanical, chemical, etc.). The input x(t) is a sinusoid, arbitrarily shown as being a cosine, which is described as A cos(vt þ u). The output or response to this input excitation y(t) is also a sinusoid of the same frequency as the input: B cos(vt þ f), but having a different amplitude B and phase angle f. The radian frequency of the waveform v (radians per second) and the cyclic frequency f (Hertz or cycles per second) are related by v ¼ 2p f. The input x(t) and the output y(t) are related by a differential equation. For a lumped system such as an electric circuit, the differential equation is an ordinary differential equation, whereas for a distributed-parameter system such as an electromagnetic field problem, heat flow, or fluid flow, the differential equation is a partial differential equation. In order to illustrate the method, we will use the following example: d 2 y(t) dt 2 þ a dy(t) þ by(t) ¼ x(t) dt ¼ A cos(vt þ u) (A:1) Introduction to Electromagnetic Compatibility, Second Edition, by Clayton R. Paul Copyright # 2006 John Wiley & Sons, Inc. 859
860 THE PHASOR SOLUTION METHOD The general solution to this differential equation is the sum of a homogeneous or transient solution with x(t) ¼ 0, and a particular or steady-state solution with x(t) = 0. We are interested only in the steady-state solution. Hence the task is to determine a solution yðtþ that satisfies (A.1). The key to this phasor solution is Euler s identity e ju ¼ cos(u) þ j sin(u) (A:2) p where j ¼ ffiffiffiffiffiffi 1. In order to obtain the solution to (A.1), we solve a different but much easier problem d 2 ^Y(t) dt 2 þ a d ^Y(t) þ b ^Y(t) ¼ A u e jvt dt {z} ^A ¼ A cos(vt þ u) þ ja sin(vt þ u) (A:3) and we denote a complex-valued phasor quantity having magnitude and phase with a caret over the quantity ( ^A ¼ A u). We have used the important complex algebra equivalence of e ju ; 1 u (A:4) to write the right-hand side in an equivalent form. The reason we solve this different problem is that (1) the solution of (A.3) is considerably simpler than the solution of (A.1) and (2) the solution of (A.1) can be easily determined from the solution of (A.3). To show this, we assume a form of the solution to (A.3) of the same form as the right-hand side: ^Y(t) ¼ B f e jvt (A:5) Substituting (A.5) into (A.3) gives ( jv) 2 þ a( jv) þ b B f e jvt ¼ A u e jvt (A:6) Canceling e jvt on both sides and noting that j 2 ¼ 1, j 3 ¼ j, etc. gives B/f ¼ A u v 2 þ jav þ b (A:7) Since we presumably know the frequency of the input v as well as its magnitude A and phase u, we can determine the magnitude B and phase f of the solution to (A.3) using complex algebra. But how does the solution to (A.3) relate to the solution of the original problem in (A.1)? The answer is simple. The original input x(t) ¼ A cos(vt þ u) is the real part
A.1 SOLVING DIFFERENTIAL EQUATIONS 861 of the new input: 8 >< x(t) ¼ Re ^X(t) ¼ A u >: {z} ^A e jvt 9 >= ¼ Re{A cos(vt þ u) þ ja sin(vt þ u)} ¼ A cos(vt þ u) >; (A:8) Hence the solution to the original problem is y(t) ¼ Re ^Y(t) ¼ Re B f e jvt ¼ Re B cos(vt þ f) þ jb sin(vt þ f) ¼ B cos(vt þ f) (A:9) Hence, if the original input was a cosine, then the output will be a cosine, and if the original input had been a sine, then the output would be a sine, both having the same magnitude and phase: x(t) ¼ A cos(vt þ u) ) y(t) ¼ B cos(vt þ f) x(t) ¼ A sin(vt þ u) ) y(t) ¼ B sin(vt þ f) (A:10) But solving (A.3) using complex algebra is considerably simpler than solving (A.1). This is referred to as the phasor method and is the heart of the solution of electric circuits as well as all other linear systems. The method was originally developed in the 1930s by a pioneer in electrical engineering, Charles Steinmetz, while working at General Electric. Note that this solution method applies only to inputs that are sinusoidal and for solutions that are in the steady state, i.e., after all transients have died out. The next and final question is Why do we invest so much interest and time in analyzing systems whose inputs are single-frequency sinusoids? A sinusoidal waveform is only one type of possible waveform that could excite this linear system in the real world, so why devote so much interest to determining the response to it? The answer is very simple: superposition and the Fourier series. Any periodic waveform can be alternatively represented as an infinite sum of sinusoidal components with the Fourier series (see Chapter 3) as x(t) ¼ A 0 þ X1 n¼1 A n cos(nv 0 t þ u n ) (A:11) where the x(t) is periodic (but not sinusoidal) with period T and fundamental frequency v 0 ¼ 2p f 0 where f 0 ¼ 1=T. A function x(t) is periodic with period T if x(t + mt) ¼ x(t) for m ¼ 1, 2, 3,... Hence, because the system is stipulated to be linear (the phasor method does not work for nonlinear systems), we can use the important principle of superposition to pass each single-frequency sinusoidal
862 THE PHASOR SOLUTION METHOD FIGURE A.1 (a) The phasor analysis of linear circuits involves determination of the response to a single-frequency sinusoidal input (source); (b) any other input waveform can be decomposed with the Fourier series into the sum of single-frequency components, and superposition can be used, along with phasor analysis, to indirectly determine the output as the sum of the responses to the individual sinusoidal frequency components. component in (A.11) through the system to determine the response to this general periodic input waveform as the sum of the responses to those sinusoidal components y(t) ¼ B 0 þ X1 n¼1 B n cos(nv 0 t þ u n ) (A:12) as illustrated in Fig. A.1b. But each of these individual responses can be easily determined using the phasor method. Even if the input waveform is not periodic, a similar process can be used because any nonperiodic waveform can be represented as a smooth continuum of sinusoidal components with the Fourier transform. This is the essential reason why we invest so much time and effort in learning to analyze the response of a linear system to a single-frequency sinusoidal input. In addition, as we saw in Chapter 3, by breaking a general, periodic time-domain waveform into its sinusoidal components via the Fourier series, we can more readily understand how these individual sinusoidal components go together to shape the output waveform. This is the essence of the design of electric circuit filters. Example A.1 Determine the solution to the following differential equation: d 3 y(t) dt 3 þ d2 y(t) dt 2 þ 3 dy(t) þ 2y(t) ¼ 10 sin(2t þ 308) dt
A.2 SOLVING ELECTRIC CIRCUITS 863 Solution: Replacing the right-hand side with e j2t where v ¼ 2 and assuming a solution of ^Y(t) ¼ B f e j2t gives, after canceling the e j2t that is common to both sides, the following equation: Hence we identify the solution as B f ¼ ( j2) 3 þ ( j2) 2 þ 3( j2) þ 2 ¼ j8 4 þ j6 þ 2 ¼ 2 j2 ¼ p 2 ffiffiffi 2 1358 ¼ 3:54 1658 y(t) ¼ 3:54 sin(2t þ 1658) Review Exercise A.1 Determine the sinusoidal steady-state solution for the following differential equation: d 2 y(t) dt 2 þ 2 dy(t) þ 4y(t) ¼ 10 sin(4t 458) dt Answer: y(t) ¼ 0:69 sin(4t 191:318) ¼ 0:69 sin(4t þ 168:698). A.2 SOLVING ELECTRIC CIRCUITS FOR THEIR SINUSOIDAL, STEADY-STATE RESPONSE This phasor method can be used to directly solve for the response of an electric circuit to a sinusoidal source. The reason we investigate the solution for the response to a single-frequency sinusoidal source is, again, because the response to any other source waveform that is periodic can be found using the Fourier series and superposition. If the source waveform is not periodic, we use the Fourier transform to decompose the waveform into a continuum of sinusoids in a similar manner. We will acquire great deal of our understanding of EMC through analysis of the sinusoidal response of circuits, and the analysis method will be the phasor method. Hence the reader should become thoroughly familiar with this technique as applied to linear electric circuits. Although we may use this method for the analysis of electric circuits, so far it requires that we first derive the differential equation that relates the output
864 THE PHASOR SOLUTION METHOD (a circuit current or voltage) to the input source. We need not derive that differential equation and can apply the phasor method directly to the circuit in the following manner. Figure A.2 shows how to transform the circuit elements from the time domain to the frequency (phasor) domain. First consider transforming the independent voltage and current sources. These become complex-valued quantities as V S cos(vt þ u V ) () ^V S ¼ V S u V (A:13a) I S cos(vt þ u I ) () ^I S ¼ I S u I (A:13b) Next consider transforming the circuit elements of the resistor, the inductor, and the capacitor. The time-domain terminal relation for the resistor is Ohm s law: v(t) ¼ Ri(t). Substituting the phasor representation of the voltage and current as ^V(t) ¼ V f V e jvt and ^I(t) ¼ I f I e jvt and canceling the common e jvt factor yields the phasor relation: ^V ¼ R^I (A:14) FIGURE A.2 The transformation between the time-domain circuit and the frequencydomain (phasor) circuit.
A.2 SOLVING ELECTRIC CIRCUITS 865 and the resistor remains unchanged in the phasor circuit. Next consider the inductor whose terminal relation is v(t) ¼ L½di(t)=dtŠ. Substituting the phasor representation of the voltage and current as ^V(t) ¼ V f V e jvt and ^I(t) ¼ I f I e jvt and canceling the common e jvt factor yields the phasor relation: ^V ¼ jv L^I (A:15) Hence we may replace the inductor with a complex-valued resistor having an impedance of ^Z L ¼ jv L (A:16) Next consider the capacitor whose terminal relation is i(t) ¼ C½dv(t)=dtŠ. Substituting the phasor representation of the voltage and current as ^V(t) ¼ V f V e jvt and ^I(t) ¼ I f I e jvt and canceling the common e jvt factor yields the phasor relation ^I ¼ jv C ^V (A:17) or ^V ¼ 1 jv C ^I 1 ¼ j v C ^I (A:18) and we have used the complex algebra result of (1=j) ¼ (1=j)( j=j) ¼ j. Hence we may replace the capacitor with a complex-valued resistor having an impedance of 1 ^Z C ¼ j v C (A:19) Hence the phasor circuit is essentially a resistive circuit having complex-valued resistors of impedance ^Z R ¼ R, ^Z L ¼ jv L, and ^Z C ¼ jð1=v CÞ. Thus we only need to use resistive circuit solution techniques and complex algebra to solve this phasor circuit, which is much easier than solving the differential equation directly. Example A.2 Determine the solution for the current iðtþ in the circuit of Fig. A.3a. Solution: The transformed phasor circuit is shown in Fig. A.3b, where the source has radian frequency v ¼ 3. The inductor transforms as ^Z L ¼ jv L ¼ j(3)(3) ¼ j9 V
866 THE PHASOR SOLUTION METHOD FIGURE A.3 Example A.2. The capacitor transforms as 1 ^Z C ¼ j v C 1 ¼ j (3) 1 21 ¼ j7 V Treating this as a resistive circuit, the phasor current ^I is the source phasor voltage divided by the sum of the impedances in the loop: ^I ¼ 2 þ j9 j7 ¼ 2 þ j2 ¼ p 2 ffiffi 2 458 ¼ 3:54 158 A Therefore, since the source was a sine, the solution is iðtþ ¼3:54 sinð3t 158Þ A Review Exercise A.2 Determine the sinusoidal steady-state solution for the voltage, v(t), in the circuit of Fig. EA.2. Answer: v(t) ¼ 17:89 cos(2t 71:578)V.
PROBLEMS 867 FIGURE EA.2 Review Exercise A.2. PROBLEMS Solve the following problems by hand and confirm your result using PSPICE (See Appendix D). A.1 Determine the current i(t) in the circuit of Fig. PA.1. [3:54 sin(2t 758) A] FIGURE PA.1. A.2 Determine the voltage v(t) in the circuit of Fig. PA.2. [20 cos(t þ 458) V] FIGURE PA.2. A.3 Determine the current i(t) in the circuit of Fig. PA.3. [2 cos(2t þ 66:878) A] FIGURE PA.3. A.4 Determine the voltage v(t) in the circuit of Fig. PA.4. [4:81 sin(2t 116:318)V] FIGURE PA.4.
868 THE PHASOR SOLUTION METHOD A.5 Determine the voltage v(t) in the circuit of Fig. PA.5. [5 cos(2t 908) V] FIGURE PA.5. A.6 Determine the voltage v(t) in the circuit of Fig. PA.6. [7:95 sin(2t þ 6:348) V] FIGURE PA.6. A.7 Determine the current i(t) in the circuit of Fig. PA.7. [0:57 cos(3t 35:738)A] FIGURE PA.7. A.8 Determine the current i(t) in the circuit of Fig. PA.8. [5 sin(2t 308) A] FIGURE PA.8. A.9 Determine the voltage i(t) in the circuit of Fig. PA.9. [2:71 sin(2p 10 8 t 57:178) ma] FIGURE PA.9.
REFERENCE 869 A.10 Determine the voltage v(t) in the circuit of Fig. PA.10. [4:368 10 5 sin(4p 10 6 t þ 29:138) V] FIGURE PA.10. REFERENCE 1. C. R. Paul, Fundamentals of Electric Circuit Analysis, Wiley, New York, 2001.