276 Natural and Step Responses of RLC Circuits Therefore, the response is underdamped. Now, w (i = V(4 ~ a 2 = VlO 6 - 4 X 10 4 = 100V96 = 979.80 rad/s, Sl = -a + joo d = -200 + /979.80 rad/s, s 2 = - a - j(o ti = -200 - /979.80 rad/s. For the underdamped case, we do not ordinarily solve for S] and s 2 because we do not use them explicitly. However, this example emphasizes why S] and s 2 are known as complex frequencies. b) Because v is the voltage across the terminals of a capacitor, we have v(0) = v(0 + ) = V 0 = 0. Because v(0 + ) = 0, the current in the resistive branch is zero at t = 0 + . Hence the current in the capacitor at t = 0 + is the negative of the inductor current: *c(0 + ) = -(-12.25) = 12.25 mA. Therefore the initial value of the derivative is dv(0 + ) (12.25)(10"" 3 ) , J = f- = 98,000 V/s. dt (0.125)(10 -6 ) ' c) From Eqs. 8.30 and 8.31,5, = 0 and 98,000 B 7 100 V. co d Substituting the numerical values of a, co d , B\ s and B 2 into the expression for v(t) gives v(t) = 100<r ZUUf sin 979.80? V, t > 0. d) Figure 8.9 shows the plot of v(t) versus t for the first 11 ms after the stored energy is released. It clearly indicates the damped oscillatory nature of the underdamped response. The voltage v(t) approaches its final value, alternating between values that are greater than and less than the final value. Furthermore, these swings about the final value decrease exponentially with time. v(V) t(ms) Figure 8.9 A The voltage response for Example 8.4. Characteristics of the Underdamped Response The underdamped response has several important characteristics. First, as the dissipative losses in the circuit decrease, the persistence of the oscilla- tions increases, and the frequency of the oscillations approaches w 0 . In other words, as R —> oo, the dissipation in the circuit in Fig. 8.8 approaches zero because /; = v 2 /R. As R —> oo, a —* 0, which tells us that m d —* w () . When a - 0, the maximum amplitude of the voltage remains constant; thus the oscillation at a> () is sustained. In Example 8.4, if R were increased to infinity, the solution for v(t) would become v(t) = 98sinl000rV, t > 0. Thus, in this case the oscillation is sustained, the maximum amplitude of the voltage is 98 V, and the frequency of oscillation is 1000 rad/s. We may now describe qualitatively the difference between an under- damped and an overdamped response. In an underdamped system, the response oscillates, or "bounces," about its final value. This oscillation is also referred to as ringing. In an overdamped system, the response approaches its final value without ringing or in what is sometimes described as a "sluggish" manner. When specifying the desired response of a second order system, you may want to reach the final value in the short- est time possible, and you may not be concerned with small oscillations about that final value. If so, you would design the system components to achieve an underdamped response. On the other hand, you may be con- cerned that the response not exceed its final value, perhaps to ensure that components are not damaged. In such a case, you would design the system components to achieve an overdamped response, and you would have to accept a relatively slow rise to the final value. 8.2 The Forms of the Natural Response of a Parallel RLC Circuit 277 ^ASSESSMENT PROBLEM Objective 1—Be able to determine the natural and the step response of parallel RLC circuits 8.4 A 10 mH inductor, a 1 /xF capacitor, and a vari- able resistor are connected in parallel in the circuit shown. The resistor is adjusted so that the roots of the characteristic equation are -8000 ± /6000 rad/s. The initial voltage on the capacitor is 10 V, and the initial current in the inductor is 80 mA. Find a) R; b) dv(0 + )/dt; c) B x and B 2 in the solution for v; and d) idt). NOTE: Also try Chapter Problems 8.7 and 8.19. Answer: (a) 62.5 ft; (b) -240,000 V/s; (c) B x = 10 V, B 2 = -80/3 V; (d)i L (t) = 10e- 80()0f [8 cos 6000? + (82/3) sin 6000f] mA when t The Critically Damped Voltage Response The second-order circuit in Fig. 8.8 is critically damped when coj, = a 2 , or tu () = a. When a circuit is critically damped, the response is on the verge of oscillating. In addition, the two roots of the characteristic equation are real and equal; that is. 1 Si = $ 2 —a = — 2RC (8.32) When this occurs, the solution for the voltage no longer takes the form of Eq. 8.18. This equation breaks down because if s% = .s^ = -a, it pre- dicts that {A x + A 2 )e~ at = A Q e (8.33) where A {) is an arbitrary constant. Equation 8.33 cannot satisfy two inde- pendent initial conditions (VQ, / 0 ) with only one arbitrary constant, A {) . Recall that the circuit parameters R and C fix a. We can trace this dilemma back to the assumption that the solution takes the form of Eq. 8.18. When the roots of the characteristic equation are equal, the solution for the differential equation takes a different form, namely v(t) = D { te~ a ' + D 2 e (8.34) Thus in the case of a repeated root, the solution involves a simple expo- nential term plus the product of a linear and an exponential term. The jus- tification of Eq. 8.34 is left for an introductory course in differential equations. Finding the solution involves obtaining D x and D 2 by following Ihe same pattern set in the overdamped and underdamped cases: We use the initial values of the voltage and the derivative of the voltage with respect to time to write two equations containing D\ and/or D 2 . < Voltage natural response—critically damped parallel RLC circuit 278 Natural and Step Responses of RLC Circuits From Eq. 8.34, the two simultaneous equations needed to determine D x and D 2 are ,,(0+) = V () = D 2 , dv(Q + ) / c (0 + ) dt c = Di - aDj. (8.35) (8.36) Example 8.5 As we can see, in the case of a critically damped response, both the equation for v(t) and the simultaneous equations for the constants D\ and D 2 differ from those for over- and underdamped responses, but the general approach is the same. You will rarely encounter critically damped systems in practice, largely because co 0 must equal a exactly. Both of these quanti- ties depend on circuit parameters, and in a real circuit it is very difficult to choose component values that satisfy an exact equality relationship. Example 8.5 illustrates the approach for finding the critically damped response of a parallel RLC circuit. Finding the Critically Damped Natural Response of a Parallel RLC Circuit a) For the circuit in Example 8.4 (Fig. 8.8), find the value of R that results in a critically damped volt- age response. b) Calculate v(t) for t > 0. c) Plot v(t) versus t for 0 < t =s 7 ms. Solution a) From Example 8.4, we know that o>o = 10 6 . Therefore for critical damping. or R a = 1()- 1 = 10 6 1 2i?C = 4000 a. (2000)(0.125) b) From the solution of Example 8.4, we know that v(0 + ) = 0 and dv{0 + )/dt = 98,000 V/s. From Eqs. 8.35 and 8.36, D 2 = 0 and D x = 98,000 V/s. Substituting these values for a, D l% and D 2 into Eq. 8.34 gives v(t) = 98,000*e _1000f V, t s> 0. c) Figure 8.10 shows a plot of v(t) versus t in the interval 0 < r ^ 7 ms. 12 3 4 5 6 7 Figure 8.10 • The voltage response for Example 8.5. L. t (ms) ^ASSESSMENT PROBLEM Objective 1—Be able to determine the natural and the step response of parallel RLC circuits 8.5 The resistor in the circuit in Assessment Problem 8.4 is adjusted for critical damping. The inductance and capacitance values are 0.4 H and 10 fiF, respectively. The initial energy stored in the circuit is 25 mJ and is distributed equally between the inductor and capacitor. Find (a) R\ (b) V 0 ; (c) / () ; (d) D x and D 2 in the solution for v; and (e) i R , t S 0 + . Answer: (a) 100 O; (b)50V; (c) 250 mA; (d) -50,000 V/s, 50 V; (e) i R (t) = (-500te~ 5m + 0.50e -500 ') A, t £= 0 + . NOTE: Also try Chapter Problems 8.9 and 8.20. 8.2 The Forms of the Natural Response of a Parallel RLC Circuit 279 A Summary of the Results We conclude our discussion of the parallel RLC circuit's natural response with a brief summary of the results. The first step in finding the natural response is to calculate the roots of the characteristic equation. You then know immediately whether the response is overdamped, underdamped, or critically damped. If the roots are real and distinct (c_ 2 , < a 2 ), the response is over- damped and the voltage is v(t) = A { e Sit + A 2 e s *, where S\ = —a + voT s 2 = —a — Va 2 1 2RC 2_ 1 "° " LC- - w { 2 ,, - <4h The values of A { and A 2 are determined by solving the following simulta- neous equations: v(0 + ) = Ai + A 2 , dv{0 + ) / C (0 + ) If the roots are complex OJ 2 } > a" the response is underdamped and the voltage is v(t) = B { e~ n ' cos o>/ + B 2 e~ a! sin ctjt, where to (l = vo){) — or. The values of B\ and B 2 are found by solving the following simultaneous equations: v(Q + ) - % = B h dv(0 + ) ic(0 + ) _ , _ _____ = _______ _ aB] + ^ If the roots of the characteristic equation are real and equal (OJQ = a"), the voltage response is v(t) = D^e-o" + D 2 e~ a \ where a is as in the other solution forms. To determine values for the con- stants D) and D 2 , solve the following simultaneous equations: v(0 + ) = V 0 = D 2 , dv(0 + ) i c (0 + ) _ -_-_ — __ A - aD, : 8,3 The Step Response of a Parallel RLC Circuit • Finding the step response of a parallel RLC circuit involves finding the voltage across the parallel branches or the current in the individual branches as a result of the sudden application of a dc current source. There may or may not be energy stored in the circuit when the current source is applied. The task is represented by the circuit shown in Fig. 8.11. To develop a general approach to finding the step response of a second- order circuit, we focus on finding the current in the inductive branch (//). This current is of particular interest because it does not approach zero as t increases. Rather, after the switch has been open for a long time, the inductor current equals the dc source current I. Because we want to focus on the technique for finding the step response, we assume that the initial energy stored in the circuit is zero. This assumption simplifies the calcula- tions and doesn't alter the basic process involved. In Example 8.10 we will see how the presence of initially stored energy enters into the general procedure. To find the inductor current i L , we must solve a second-order differ- ential equation equated to the forcing function /, which we derive as fol- lows. From Kirchhoffs current law, we have k + hi + k = U or Because v _ dv lL + - + C- = I. (8.37) dir v = L—-, (8.38) we get dt ' dv T d 2 i L Substituting Eqs. 8.38 and 8.39 into Eq. 8.37 gives L dir afit R dt dt For convenience, we divide through by LC and rearrange terms: d 2 i, 1 di L i L I , —£ + + —^ = . (8.41) dt 1 RC dt LC LC Comparing Eq. 8.41 with Eq. 8.3 reveals that the presence of a nonzero term on the right-hand side of the equation alters the task. Before show- ing how to solve Eq. 8.41 directly, we obtain the solution indirectly. When we know the solution of Eq. 8.41, explaining the direct approach will be easier. The Indirect Approach We can solve for i L indirectly by first finding the voltage v. We do this with the techniques introduced in Section 8.2, because the differential equation that v must satisfy is identical to Eq. 8.3. To see this, we simply return to Eq. 8.37 and express i L as a function of v; thus If, v „dv li vdT + J + c T t = '• (842) Differentiating Eq. 8.42 once with respect to t reduces the right-hand side to zero because I is a constant. Thus v 1 dv d 2 v — + — — + C—r = 0, L R dt dt 2 or d 2 v 1 dv v ^ + 7^ + ^ = 0 - (8 ' 43) As discussed in Section 8.2, the solution for v depends on the roots of the characteristic equation. Thus the three possible solutions are v = A x e Slt + A 2 e &2 \ (8.44) v = B x e~ a ' cos o) d t + B 2 e~ al sin a)/, (8.45) v = D x te~ a ' + D 2 e~ at . (8.46) A word of caution: Because there is a source in the circuit for t > 0, you must take into account the value of the source current at t = 0 + when you evaluate the coefficients in Eqs. 8.44-8.46. To find the three possible solutions for / L , we substitute Eqs. 8.44-8.46 into Eq. 8.37. You should be able to verify, when this has been done, that the three solutions for i L will be i L = I + A\e Sli + A' 2 e s - 1 , (8.47) //, = I + B\e~ at cos o) d t + B' 2 e~ at sin wj, (8.48) i L = I + D[te- at + D' 2 e~ at , (8.49) where A\, A 2 , B[, B 2 , D\, and D' 2 , are arbitrary constants. In each case, the primed constants can be found indirectly in terms of the arbitrary constants associated with the voltage solution. However, this approach is cumbersome. The Direct Approach It is much easier to find the primed constants directly in terms of the ini- tial values of the response function. For the circuit being discussed, we would find the primed constants from / L (0) and dii(0)/dt. The solution for a second-order differential equation with a constant forcing function equals the forced response plus a response function 282 Natural and Step Responses of RLC Circuits identical in form to the natural response. Thus we can always write the solution for the step response in the form ._ [function of the same form! ' \ as the natural response J' or v = V f + { function of the same form 1 as the natural response J' (8.50) (8.51) where If and Vf represent the final value of the response function. The final value may be zero, as was, for example, the case with the voltage v in the circuit in Fig. 8.8. Examples 8.6-8.10 illustrate the technique of finding the step response of a parallel RLC circuit using the direct approach. Example 8.6 Finding the Overdamped Step Response of a Parallel RLC Circuit The initial energy stored in the circuit in Fig. 8.12 is zero. At t = 0, a dc current source of 24 mA is applied to the circuit. The value of the resistor is 400 ft. a) What is the initial value of i L l b) What is the initial value of dijdtl c) What are the roots of the characteristic equation? d) What is the numerical expression for //,(/ 1 ) when t > 0? '(p'X Figure 8.12 • The circuit for Example 8.6. Solution a) No energy is stored in the circuit prior to the application of the dc current source, so the initial current in the inductor is zero. The inductor pro- hibits an instantaneous change in inductor cur- rent; therefore i L (0) = 0 immediately after the switch has been opened. b) The initial voltage on the capacitor is zero before the switch has been opened; therefore it will be zero immediately after. Now, because v = Ldifjdt, di dt -(0 + ) = 0. c) From the circuit elements, we obtain , 1 10 12 (-0() = a = LC (25)(25) 1 10 y 2RC (2)(400)(25) 16 x 10*, = 5 x 10 4 rad/s, or a 2 = 25 X 10 8 . Because a»o < a 2 , the roots of the characteristic equation are real and distinct. Thus Sl = -5 X 10 4 + 3 X 10 4 = -20,000 rad/s, s 2 = -5 X 10 4 - 3 X 10 4 = -80,000 rad/s. d) Because the roots of the characteristic equation are real and distinct, the inductor current response will be overdamped. Thus i L (t) takes the form of Eq. 8.47, namely, i L = I f + A[e Sit + A! 2 e s -'. • Inductor current in overdamped parallel RLC circuit step response Hence, from this solution, the two simultaneous equations that determine A\ and A^are * t (0) = If + A\ + A' 2 = 0, di L di (0) = s x A\ + s 2 A , 2 = 0. Solving for A\ and A' 2 gives A[ = -32 mA and A' 2 = 8 mA. The numerical solution for i L (t) is i L {t) = (24 - 32e- 20 - 000 ' + 8e-" )mk ) mA, t > 0. 8.3 The Step Response of a Parallel RLC Circuit 283 Finding the Underdamped Step Response of a Parallel RLC Circuit The resistor in the circuit in Example 8.6 (Fig. 8.12) is increased to 625 ft. Find i L (t) for t 2s 0. Solution Because L and C remain fixed, col has the same value as in Example 8.6; that is, col = 16 X 10 8 . Increasing R to 625 ft decreases a to 3.2 X 10 4 rad/s. With col > oc\ the roots of the characteristic equation are complex. Hence s x = -3.2 X 10 4 + /2.4 X 10 4 rad/s, s 2 = -3.2 X 10 4 - /2.4 X 10 4 rad/s. The current response is now underdamped and given by Eq. 8.48: kit) = I/ + B[e~°" cos co d t + B' 2 e~ at sin to d t. • Inductor current in underdamped parallel RLC circuit step response Here, a is 32,000 rad/s, co d is 24,000 rad/s, and If is 24 mA. As in Example 8.6, B[ and B' 2 are determined from the initial conditions. Thus the two simultane- ous equations are I'ZXO) = I f + B[ = 0, ^(0)=co (l B 2 -aB[ = Q. Then, and B\ = -24 mA 54 = -32 mA. The numerical solution for i L (t) is i L (t) = (24 - 24e- 3Z000 'cos24,000r - 32e" 32 ' 000 ' sin 24,000?) mA, t > 0. Example 8.8 Finding the Critically Damped Step Response of a Parallel RLC Circuit The resistor in the circuit in Example 8.6 (Fig. 8.12) is set at 500 ft. Find i L for t > 0. Solution We know that col remains at 16 X 10 8 . With R set at 500 ft, a becomes 4 X 10 4 s~\ which corresponds to critical damping. Therefore the solution for i L (t) takes the form of Eq. 8.49: kit) = I f + D\te~ at + D' 2 e~°". • Inductor current in critically damped parallel RLC circuit step response Again, D\ and D 2 are computed from initial conditions, or i L (0) = If + D' 2 = 0, ~(0) = D\ - aD'o = 0. at Thus D\ = -960,000 mA/s and D' 2 = -24 mA. The numerical expression for i L {i) is i L (t) = (24 - 960,000re- 4(X()0{) ' - 24e _40 ' 000 ') mA, t > 0. 284 Natural and Step Responses of RLC Circuits Example 8.9 Comparing the Three-Step Response Forms a) Plot on a single graph, over a range from 0 to 220 jus, the overdamped, underdamped, and critically damped responses derived in Examples 8.6-8.8. b) Use the plots of (a) to find the time required for i L to reach 90% of its final value. c) On the basis of the results obtained in (b), which response would you specify in a design that puts a premium on reaching 90% of the final value of the output in the shortest time? d) Which response would you specify in a design that must ensure that the final value of the cur- rent is never exceeded? i L (mA) Underdamped {R = 625 il) jOverdamped (R = 400 SI) I Critically damped {R = 500 Q) tQa) 0 20 60 100 140 180 Figure 8.13 A The current plots for Example 8.9. Solution a) See Fig. 8.13. b) The final value of i L is 24 mA, so we can read the times off the plots corresponding to i L = 21.6 mA. Thus t od = 130 ^is, t cd = 97 /xs, and t ud = 74 fis. c) The underdamped response reaches 90% of the final value in the fastest time, so it is the desired response type when speed is the most important design specification. d) From the plot, you can see that the under- damped response overshoots the final value of current, whereas neither the critically damped nor the overdamped response produces currents in excess of 24 mA. Although specifying either of the latter two responses would meet the design specification, it is best to use the overdamped response. It would be impractical to require a design to achieve the exact component values that ensure a critically damped response. Example 8.10 Finding Step Response of a Parallel RLC Circuit with Initial Stored Energy Energy is stored in the circuit in Example 8.8 (Fig. 8.12, with R = 500 n) at the instant the dc cur- rent source is applied. The initial current in the inductor is 29 mA, and the initial voltage across the capacitor is 50 V. Find (a) / L (0); (b) di L {0)/dt; (c) i L {t) for t > 0; (d) v(t) for t > 0. Solution a) There cannot be an instantaneous change of cur- rent in an inductor, so the initial value of i L in the first instant after the dc current source has been applied must be 29 mA. b) The capacitor holds the initial voltage across the inductor to 50 V. Therefore c) From the solution of Example 8.8, we know that the current response is critically damped. Thus idO = if + D\te~ at + D 2 e~ at , where 2RC = 40,000 rad/s and I f = 24 mA. Notice that the effect of the nonzero initial stored energy is on the calculations for the con- stants D[ and D' 2 , which we obtain from the ini- tial conditions. First we use the initial value of the inductor current: Lf«n = 50, ^(0+) = § X 10 3 = 2000 A/s. at 25 i L (0) = I f + D' 2 = 29 mA, from which we get D' 2 = 29 - 24 = 5 mA. 8.4 The Natural and Step Response of a Series RLC Circuit 285 The solution for D[ is ~(0 + ) = D{ - aD' 2 = 2000, or D\ = 2000 + aD' 2 = 2000 + (40,000)(5 X 10~ 3 ) = 2200 A/s = 2.2 X 10 6 m A/s. Thus the numerical expression for i L (t) is i L (t) = (24 + 2.2 X lO'Ve" 40 - 000 ' + 5e~ 4{UMH ) mA, t > 0. d) We can get the expression for v(t), t > 0 by using the relationship between the voltage and current in an inductor: dij v(t) = L^ = (25 X 10~ 3 )[(2.2 X 10 6 )(-40,000)te- 4ao,1() ' + 2.2 X ioV 40 - 000 ' + (5)(-40,000)<r 4(M)00 '] X 10~ 3 = -2.2 x lO 6 ^ 40 - 000 ' + 50e^ wu * V, / s> 0. To check this result, let's verify that the initial voltage across the inductor is 50 V: v(0) = -2.2 X 10 6 (0)(1) + 50(1) = 50 V. ^ASSESSMENT PROBLEM Objective 1—Be able to determine the natural response and the step response of parallel RLC circuits 8.6 In the circuit shown, R = 500 O, L = 0.64 H, C = 1 fxF, and I = —1 A. The initial voltage drop across the capacitor is 40 V and the initial inductor current is 0.5 A. Find (a) //?(0 + ); (b) i c (0 + ); (c) di L (0 + )/dt; (d) s t , s 2 ; (e) i L (t) for t > 0; and (f) v(t) for t > 0 + . Answer: (a) 80 mA; (b)-1.58 A; (c) 62.5 A/s; (d) (-1000 + ;750) rad/s, (-1000 - /750) rad/s; (e)[-l + <r 1000 '[1.5cos750f + 2.0833 sin 750f] A, for t > 0; (f) £r 1000 '(40cos750; - 2053.33 sin 750r) V, for t > 0 + . NOTE: Also try Chapter Problems 8.29-8.31. 8.4 The Natural and Step Response of a Series RLC Circuit The procedures for finding the natural or step responses of a series RLC circuit are the same as those used to find the natural or step responses of a parallel RLC circuit, because both circuits are described by differential equations that have the same form. We begin by summing the voltages around the closed path in the circuit shown in Fig. 8.14. Thus /* 1 p Ri + L~ + - idr + V 0 = 0. dt CJ () We now differentiate Eq. 8.52 once with respect to t to get n di , dri i R— + L— + — = 0, dt dt 2 C (8.52) (8.53) R •"•"N, iy L X ) d + ^V a Figure 8.14 A A circuit used to illustrate the natural response of a series RLC circuit.