Electric Circuits, 9th Edition P32 doc

286 Natural and Step Responses of RLC Circuits which we can rearrange as Characteristic equation—series RLC circuit • d 2 i R di i —r + —— + = 0. dr L dt LC (8.54) Comparing Eq. 8.54 with Eq. 8.3 reveals that they have the same form. Therefore, to find the solution of Eq. 8.54, we follow the same process that led us to the solution of Eq. 8.3. From Eq. 8.54, the characteristic equation for the series RLC circuit is 2 R 1^ (8.55) The roots of the characteristic equation are Neper frequency—series RLC circuit • Resonant radian frequency—series RLC circuit • *l,2 *1 R 2L i — ± -a U ±vV r - Oil- 1 LC or The neper frequency (a) for the series RLC circuit is R A, a = — rad/s, and the expression for the resonant radian frequency is 1 (8.56) (8.57) (8.58) w 0 = rad/s. (8.59) Current natural response forms in series RLC circuits F • Note that the equation for neper frequency of the series RLC circuit differs from that of the parallel RLC circuit, but the equations for resonant and damped radian frequencies are the same. The current response will be overdamped, underdamped, or critically damped according to whether col < a 2 , WQ > a 2 , or a>o = a 2 , respectively. Thus the three possible solutions for the current are as follows: /(f) = A t e Sit + A 2 e S2t (overdamped), (8.60) i(t) = B x e~ ai cos o) d t + B 2 e~ at sin oj d t (underdamped), (8.61) i(t) = Drfe'" 1 + L\e~ al (critically damped). (8.62) r = 0 -WW R + v c: Figure 8.15 • A circuit used to illustrate the step response of a series RLC circuit. When you have obtained the natural current response, you can find the natural voltage response across any circuit element. To verify that the procedure for finding the step response of a series RLC circuit is the same as that for a parallel RLC circuit, we show that the differential equation that describes the capacitor voltage in Fig. 8.15 has the same form as the differential equation that describes the inductor cur- rent in Fig. 8.11. For convenience, we assume that zero energy is stored in the circuit at the instant the switch is closed. Applying Kirchhoff s voltage law to the circuit shown in Fig. 8.15 gives V = Ri + L^- + v c . dt (8.63) 8.4 The Natural and Step Response of a Series RLC Circuit 287 The current (i) is related to the capacitor voltage (%•) by the expression , dv c i = C dt (8.64) from which d[ = d 2 v c dt dt 1 ' (8.65) Substitute Eqs. 8.64 and 8.65 into Eq. 8.63 and write the resulting expression as d 2 v c R dv c v c V dr L dt LC LC (8.66) Equation 8.66 has the same form as Eq. 8.41; therefore the procedure for finding v c parallels that for finding i L . The three possible solutions for v c are as follows: v c = V f + A\e Slt + A' 2 e S2 ' (overdamped), (8.67) v c = V f + B\e~ at cos to d t + B' 2 e~ at sin (o d t (underdamped), (8.68) 4 Capadtor voltage step response forms in v c = V f + D[te~ at + D' 2 e~ a1 (critically damped), (8.69) series RLC drcuits where Vf is the final value of v c . Hence, from the circuit shown in Fig. 8.15, the final value of v c is the dc source voltage V. Example 8.11 and 8.12 illustrate the mechanics of finding the natural and step responses of a series RLC circuit. Example 8.11 Finding the Underdamped Natural Response of a Series RLC Circuit The 0.1 /xF capacitor in the circuit shown in Fig. 8.16 is charged to 100 V. At t = 0 the capacitor is discharged through a series combination of a 100 mH inductor and a 560 fl resistor. a) Find i(t) for t > 0. b) Find v c (t) for t > 0. / = 0 100 mH Figure 8.16 A The circuit for Example 8.11. Solution a) The first step to finding /'(?) is to calculate the roots of the characteristic equation. For the given element values, 2 a = 1 LC (10 3 )(1Q 6 ) (100)(0.1) A 2L 560 10 8 , 2(100) = 2800 rad/s X 10 3 288 Natural and Step Responses of RLC Circuits Next, we compare a^ to a 2 and note that o)l > a 2 , because a 2 - 7.84 X 10 6 = 0.0784 X 10 8 . At this point, we know that the response is under- damped and that the solution for /(f) is of the form /(f) = B^'** 1 cos wj + B 2 e~ at sincojt, where a = 2800 rad/s and a) d = 9600 rad/s. The numerical values of B { and B 2 come from the initial conditions. The inductor current is zero before the switch has been closed, and hence it is zero immedi- ately after. Therefore /(0) = 0 = B { . To find B 2 , we evaluate di(() + )/dt. From the circuit, we note that, because /(0) = 0 immediately after the switch has been closed, there will be no voltage drop across the resistor. Thus the initial voltage on the capacitor appears across the terminals of the inductor, which leads to the expression, L dt % ' or di(0 + ) V 0 100 dt " L " 100 X 10 = 1000 A/s. Because B } = 0, di — = 4003 2 <r 28()0 '(24 cos 9600f - 7sin9600f). Thus di(Q + ) dt = 96005 2 , The solution for /(f) is /(f) = 0.1042<T 2800 ' sin 9600f A, f>0. b) To find Vc(t), we can use either of the following rela- tionships: 1 f l v c = — / idr + 100 or wo r di v c = iR + L—. dt Whichever expression is used (the second is recom- mended), the result is v c (t) = (100cos9600f + 29.17sin9600f)e- 280Ur V, t > 0. Example 8.12 Finding the Underdamped Step Response of a Series RLC Circuit No energy is stored in the 100 mH inductor or the 0.4 ^iF capacitor when the switch in the circuit shown in Fig. 8.17 is closed. Find v c (t) for t > 0. 48 V -\ 0.1 H f = 0 -*M, 28011 0.4 M F; + Figure 8.17 • The circuit for Example 8.12. Solution The roots of the characteristic equation are 280 280 0.2 10 f (0.1)(0.4) = (-1400 +/4800) rad/s, s 2 = (-1400 - ;4800) rad/s. The roots are complex, so the voltage response is underdamped. Thus v c (t) = 48 + Bie" 14()0 'cos4800f + £^T 1400 'sin4800f, f > 0. No energy is stored in the circuit initially, so both i; c (0) and dv c (0 + )/dt are zero. Then, v c (0) dv c (0 + ) dt = 0 = 48 + S' b = 0 = 4800B' 2 - 14005;. Solving for B\ and B' 2 yields B\ = -48 V, B 2 = -14 V. Therefore, the solution for v c (t) is v c {t) = (48 - 48<T 14()0 'cos4800f - 14e" 1400 'sin 48000 V, ' * 0. 8.5 A Circuit with Two Integrating Amplifiers 289 ^/ASSESSMENT PROBLEMS Objective 2—Be able to determine the natural response and the step response of series RLC circuits 8.7 The switch in the circuit shown has been in position a for a long time. At t = 0, it moves to position b. Find (a) /(0 + ); (b) v c (0 + ); (c) di(Q + )/dt; (d) 5,, s 2 ; and (e) i(t) for t > 0. Answer: (a) 0; (b) 50 V; (c) 10,000 A/s; (d) (-8000 + /6000) rad/s, (-8000 - /6000) rad/s; (e) (1.67<r 8000f sin 6000/) A for/ > 0. NOTE: Also try Chapter Problems 8.50-8.52. 9kfl 5mH 8.8 Find v c {t) for t > 0 for the circuit in Assessment Problem 8.7. Answer: [ 100 - <T 8000f (50 cos 6000? + 66.67 sin 6()00/)] V for t > 0. 8.5 A Circuit with Two Integrating Amplifiers A circuit containing two integrating amplifiers connected in cascade 1 is also a second-order circuit; that is, the output voltage of the second inte- grator is related to the input voltage of the first by a second-order differ- ential equation. We begin our analysis of a circuit containing two cascaded amplifiers with the circuit shown in Fig. 8.18. Figure 8.18 • Two integrating amplifiers connected in cascade. We assume that the op amps are ideal. The task is to derive the differ- ential equation that establishes the relationship between v ( , and v g . We begin the derivation by summing the currents at the inverting input termi- nal of the first integrator. Because the op amp is ideal. From Eq. 8.70, 0 - v„ d — + C,-(0 - » ol) = 0. dv o] dt v r . *.C, l (8.70) (8.71) ' In a cascade connection, the output signal of the first amplifier (v <a in Fig. 8.18) is the input signal for the second amplifier. 290 Natural and Step Responses of RLC Circuits Now we sum the currents away from the inverting input terminal of the second integrating amplifier: 0 — u„i d -jf + c 2 - ( o - %) - (., or dv„ dt Differentiating Eq. 8.73 gives d 2 v„ R-yCt h>V 1 dv oi dt 2 R 2 C 2 dt (8.72) (8.73) (8.74) We find the differential equation that governs the relationship between v a and v g by substituting Eq. 8.71 into Eq. 8.74: d 2 v,> 1 1 -?V (8.75) dt 2 R [ C l R 2 C 2 *' Example 8.13 illustrates the step response of a circuit containing two cas- caded integrating amplifiers. Example 8.13 Analyzing Two Cascaded Integrating Amplifiers No energy is stored in the circuit shown in Fig. 8.19 when the input voltage v„ jumps instantaneously from 0 to 25 mV. a) Derive the expression for v 0 (t) for 0 < t < / sat . b) How long is it before the circuit saturates? Solution a) Figure 8.19 indicates that the amplifier scaling factors are 0,1 /*F 1 1000 R [ C l (250)(0.1) 1 1000 = 40, R 2 C 2 (500)(1) = 2. Now, because v g = 25 mV for t > 0, Eq. 8.75 becomes d 2 v (> dt 2 To solve for v () , we let = (40)(2)(25 X 10" 3 ) = 2. then, dg(t) dt ,,v dv () g(0 = -^ 2, and dg(t) = 2dt. Figure 8.19 • The circuit for Example 8.13. Hence from which However, dy = 2 I dx, g(0) ./() g(t) ~ g(0) = 2t. dvJQ) m ^i-o. because the energy stored in the circuit ini- tially is zero, and the op amps are ideal. (See Problem 8.57.) Then, ^ = 2t, and v a = t 2 + v o (0). But v o (0) = 0, so the experssion for v a becomes v„ = t 2 , 0 < t < ; Sill . 8.5 A Circuit with Two Integrating Amplifiers 291 b) The second integrating amplifier saturates when v a reaches 9 V or t = 3 s. But it is possible that the first integrating amplifier saturates before t = 3 s. To explore this possibility, use Eq. 8.71 to find dv„i/dt: dVoi dt -40(25) X 10" 3 = -1. Solving for v nl yields %] = -t. Thus, at t = 3 s, v ()] = —3 V, and, because the power supply voltage on the first integrating amplifier is ±5 V, the circuit reaches saturation when the second amplifier saturates. When one of the op amps satu- rates, we no longer can use the linear model to predict the behavior of the circuit. NOTE: Assess your understanding of this material by trying Chapter Problem 8.63. Two Integrating Amplifiers with Feedback Resistors Figure 8.20 depicts a variation of the circuit shown in Fig. 8.18. Recall from Section 7.7 that the reason the op amp in the integrating amplifier satu- rates is the feedback capacitor's accumulation of charge. Here, a resistor is placed in parallel with each feedback capacitor (C { and C 2 ) to overcome this problem. We rederive the equation for the output voltage, v tr and determine the impact of these feedback resistors on the integrating ampli- fiers from Example 8.13. We begin the derivation of the second-order differentia] equation that relates v a] to v g by summing the currents at the inverting input node of the first integrator: () - Vg () - v „i d , (8.76) We simplify Eq. 8.76 to read dV a i , 1 _ -¾ dt RiC^" 1 K a C, (8.77) For convenience, we let T\ = R\C\ and write Eq. 8.77 as dt T] •Vo R n C ] (8.78) The next step is to sum the currents at the inverting input terminal of the second integrator: 0 - v,,i 0 - v n d (8.79) Figure 8.20 • Cascaded integrating amplifiers with feedback resistors. We rewrite Eq. 8.79 as dv p v a -v oi ^n + v 2 = i^: (8 - 80) where r 2 = R 2 C 2 . Differentiating Eq. 8.80 yields d 2 v a J_dv (> _ 1 dv ol It? + V2~d7 ~ ~W 2 ~^' (8 - 81) From Eq. 8.78, dv ol -v ol v g dt TJ R A C X (8.82) and from Eq. 8.80, dv.y R b Ci V 0 i = - R hC 2 -7r ~ -T^V 0 . (8.83) dt T 2 We use Eqs. 8.82 and 8.83 to eliminate dv a Jdt from Eq. 8.81 and obtain the desired relationship: d\ ( 1 1 W / 1 \ v R From Eq. 8.84, the characteristic equation is s s • — + —)s + = 0. (8.85) Ji T 2 J T\T 2 The roots of the characteristic equation are real, namely, Si = —, (8.86) r i -1 ft = . (8.87) Example 8.14 illustrates the analysis of the step response of two cascaded integrating amplifiers when the feedback capacitors are shunted with feedback resistors. 8.5 A Circuit with Two Integrating Amplifiers 293 Example 8.14 Analyzing Two Cascaded Integrating Amplifiers with Feedback Resistors The parameters for the circuit shown in Fig. 8.20 are R a = 100 kfi, R Y = 500 kO, C x = 0.1 ^F, R h = 25 kH, R 2 = 100 kO, and C 2 = 1 /xF. The power supply voltage for each op amp is ±6 V. The signal voltage (v g ) for the cascaded integrating amplifiers jumps from 0 to 250 mV at t = 0. No energy is stored in the feedback capacitors at the instant the signal is applied. a) Find the numerical expression of the differential equation for v Q . b) Find v () (t) for t > 0. c) Find the numerical expression of the differential equation for v a] . d) Find v (A (t) for/ > 0. The solution for v» thus takes the form: Solution a) From the numerical values of the circuit parame- ters, we have TJ = R\C] = 0.05 s; r 2 = R 2 C 2 = 0.10 s, and v g /R. A CiR h C 2 = 1000 V/s 2 . Substi- tuting these values into Eq. 8.84 gives 10f v a = 5 + A[e~ m + A' 2 e -2i.tr With v o (0) = 0 and dv o (0)/dt = 0, the numeri- cal values of A\ and A' 2 are A\ — —10 V and A 2 = 5 V. Therefore, the solution for v () is 2QC\ v 0 (t) = (5 - 10e~ lu ' + 5e" iU 0 V, f > 0. The solution assumes that neither op amp saturates. We have already noted that the final value of v a is 5 V, which is less than 6 V; hence the second op amp does not saturate. The final value of v ol is (250 X 10" 3 )(-500/100), or -1.25 V. Therefore, the first op amp does not saturate, and our assumption and solution are correct. c) Substituting the numerical values of the parame- ters into Eq. 8.78 generates the desired differen- tial equation: d v () dv a —£ + 30—^ + 200v o = 1000. dt 2 dt dv 0\ dt + 20v ol = -25. b) The roots of the characteristic equation are S] = -20rad/s and s 2 = -lOrad/s. The final value of v 0 is the input voltage times the gain of each stage, because the capacitors behave as open circuits as t —» oo. Thus, Vo(°°) (250 X 10 -3 )- -500) (-100) 100 25 5 V. d) We have already noted the initial and final val- ues of v 0 \, along with the time constant T\. Thus we write the solution in accordance with the technique developed in Section 7.4: >o\ -1.25 + [0 - (-1.25)]e -20/ m = -1.25 + 1.25<T /l "V, t > 0 NOTE: Assess your understanding of this material by trying Chapter Problem 8.64. 294 Natural and Step Responses of RLC Circuits Practical Perspective Figure 8.21 A The circuit diagram of the conven- tional automobile ignition system. An Ignition Circuit Now let us return to the conventional ignition system introduced at the beginning of the chapter. A circuit diagram of the system is shown in Fig. 8.21. Consider the circuit characteristics that provide the energy to ignite the fuel-air mixture in the cylinder. First, the maximum voltage avail- able at the spark plug, v sp , must be high enough to ignite the fuel. Second, the voltage across the capacitor must be limited to prevent arcing across the switch or distributor points. Third, the current in the primary winding of the autotransformer must cause sufficient energy to be stored in the system to ignite the fuel-air mixture in the cylinder. Remember that the energy stored in the circuit at the instant of switching is proportional to the primary current squared, that is, a> 0 = |L/ 2 (0). EXAMPLE a) Find the maximum voltage at the spark plug, assuming the following val- ues in the circuit of Fig. 8.21: V dc = 12 V, R = 4 ft, L = 3 mH, C - OAfiF, and a = 100. b) What distance must separate the switch contacts to prevent arcing at the time the voltage at the spark plug is maximum? Solution a) We analyze the circuit in Fig. 8.21 to find an expression for the spark plug voltage v sp . We limit our analysis to a study of the voltages in the circuit prior to the firing of the spark plug. We assume that the current in the primary winding at the time of switching has its maximum possi- ble value V&JR, where R is the total resistance in the primary circuit. We also assume that the ratio of the secondary voltage (v 2 ) to the pri- mary voltage (v{) is the same as the turns ratio N 2 /Ni. We can justify this assumption as follows. With the secondary circuit open, the voltage induced in the secondary winding is di dt and the voltage induced in the primary winding is (8.88) v l — L di It It follows from Eqs. 8.88 and 8.89 that «i M_ L" (8.89) (8.90) It is reasonable to assume that the permeance is the same for the fluxes 4>u and <f> 2 \ in the iron-core autotransformer; hence Eq. 8.90 reduces to V2 «1 N t N 2 ^ N 2 Nty Ni — a. (8.91) We are now ready to analyze the voltages in the ignition circuit. The values of R, L, and C are such that when the switch is opened, the primary coil current response is underdamped. Using the techniques developed in Section 8.4 and assuming t = 0 at the instant the switch is opened, the expression for the primary coil current is found to be I = Y*£ e -at R cosw d t + I — ] sm o) d t (8.92) where R_ 2 V a = "*"^LC - ° (See Problem 8.66(a).) The voltage induced in the primary winding of the autotransformer is V\ = L— = di -V Ac _ at e ut sin (o d t. dt co d RC (See Problem 8.66(b).) It follows from Eq. 8.91 that (8.93) v 2 = -«Kic _, io d RC e smw/ (8.94) The voltage across the capacitor can be derived either by using the relationship Vc = ~cl idx + Vc ^ (8.95) or by summing the voltages around the mesh containing the primary winding: v c = V dc - iR - L-} (8.96) at In either case, we find *V = Kfc[l _ e~ a ' cos co d t + Ke~ at sin <a d t], (8.97) where co d \RC J (See Problem 8.66(c).) As can be seen from Fig. 8.21, the voltage across the spark plug is T/ ttV ^ -at • = Vdc - ~^~^ e Sm Udt = Vr dc 1 - io d RC e sin a) d t (8.98)