Electric Circuits, 9th Edition P30 pot

266 Natural and Step Responses of RLC Circuits c; Figure 8 response introducing these three forms, we show that the same forms apply to the step response of a parallel RLC circuit as well as to the natu- ral and step responses of series RLC circuits. ,1 A A circuit used to illustrate the natural of a parallel RLC circuit. Figure 8. response 2 A A circuit used to illustrate the step of a parallel RLC circuit. Figure 8.3 A A circuit used to illustrate the natural response of a series RLC circuit. ^V / = 0 c Figure 8.4 A A circuit used to illustrate the step response of a series RLC circuit. 8,1 Introduction to the Natural Response of a Parallel RLC Circuit The first step in finding the natural response of the circuit shown in Fig. 8.1 is to derive the differential equation that the voltage v must satisfy. We choose to find the voltage first, because it is the same for each component. After that, a branch current can be found by using the current-voltage relationship for the branch component. We easily obtain the differential equation for the voltage by summing the currents away from the top node, where each current is expressed as a function of the unknown voltage v: v If, r ^dv (8.1) We eliminate the integral in Eq. 8.1 by differentiating once with respect to t, and, because 7 0 is a constant, we get 1 dv v d v R dt L dt 2 (8.2) We now divide through Eq. 8.2 by the capacitance C and arrange the derivatives in descending order: d v 1 dv v ~di I + lRClt + Tc~ ' (8.3) Comparing Eq. 8.3 with the differential equations derived in Chapter 7 reveals that they differ by the presence of the term involving the second derivative. Equation 8.3 is an ordinary, second-order differential equation with constant coefficients. Circuits in this chapter contain both inductors and capacitors, so the differential equation describing these circuits is of the sec- ond order. Therefore, we sometimes call such circuits second-order circuits. The General Solution of the Second-Order Differential Equation We can't solve Eq. 8.3 by separating the variables and integrating as we were able to do with the first-order equations in Chapter 7. The classical approach to solving Eq. 8.3 is to assume that the solution is of exponential form, that is, to assume that the voltage is of the form v = Ae s (8.4) where A and s are unknown constants. Before showing how this assumption leads to the solution of Eq. 8.3, we need to show that it is rational. The strongest argument we can make in favor of Eq. 8.4 is to note from Eq. 8.3 that the second derivative of the 8.1 Introduction to the Natural Response of a Parallel RLC Circuit 267 solution, plus a constant times the first derivative, plus a constant times the solution itself, must sum to zero for all values of t. This can occur only if higher order derivatives of the solution have the same form as the solu- tion. The exponential function satisfies this criterion. A second argument in favor of Eq. 8.4 is that the solutions of all the first-order equations we derived in Chapter 7 were exponential. It seems reasonable to assume that the solution of the second-order equation also involves the exponential function. If Eq. 8.4 is a solution of Eq. 8.3, it must satisfy Eq. 8.3 for all values of t. Substituting Eq. 8.4 into Eq. 8.3 generates the expression , -, _, As v , Ae st As 2 e st + e" + = 0, RC LC or ! + ^ + 7Z' which can be satisfied for all values of t only if A is zero or the parentheti- cal term is zero, because e xt ¥• 0 for any finite values of st. We cannot use A = 0 as a general solution because to do so implies that the voltage is zero for all time —a physical impossibility if energy is stored in either the inductor or capacitor. Therefore, in order for Eq. 8.4 to be a solution of Eq. 8.3, the parenthetical term in Eq. 8.5 must be zero, or 2 s 1 s + —— + —— = 0. (8.6) A Characteristic equation, parallel KC LC RLC circuit Equation 8.6 is called the characteristic equation of the differential equa- tion because the roots of this quadratic equation determine the mathe- matical character of v(t). The two roots of Eq. 8.6 are 2RC V\2RC / LC ; T^ (8-7) * = -55c-\/l^] ^- (8-8) If either root is substituted into Eq. 8.4, the assumed solution satisfies the given differential equation, that is, Eq. 8.3. Note from Eq. 8.5 that this result holds regardless of the value of A Therefore, both v = A^* 1 'and v = Aif* 268 Natural and Step Responses of RLC Circuits satisfy Eq. 8.3. Denoting these two solutions v } and v 2 , respectively, we can show that their sum also is a solution. Specifically, if we let v = Vi + v 2 = A^ 1 ' + A 2 e Sl , (8.9) then dv — = Aw**' + A 2 s 2 e s > 1 , at (8.10) -T = Arfe* + A 2 s 2 2 e^. at (8.11) Substituting Eqs. 8.9-8.11 into Eq. 8.3 gives 1 ; l + ^ + Ic ) + A ^' H + i^ + ^l ' r But each parenthetical term is zero because by definition s t and s 2 are roots of the characteristic equation. Hence the natural response of the parallel RLC circuit shown in Fig. 8.1 is of the form v = A x e S]t + A 2 e $2t (8.13) Equation 8.13 is a repeat of the assumption made in Eq. 8.9. We have shown that v } is a solution, v 2 is a solution, and v x + v 2 is a solution. Therefore, the general solution of Eq. 8.3 has the form given in Eq. 8.13. The roots of the characteristic equation (^ and s 2 ) are determined by the circuit parameters /?, L, and C.The initial conditions determine the values of the constants A] and A 2 . Note that the form of Eq. 8.13 must be modi- fied if the two roots s\ and s 2 are equal. We discuss this modification when we turn to the critically damped voltage response in Section 8.2. The behavior of v(t) depends on the values of s-i and s 2 . Therefore the first step in finding the natural response is to determine the roots of the characteristic equation. We return to Eqs. 8.7 and 8.8 and rewrite them using a notation widely used in the literature: -a + Va 2 — a>j), .92 = —a — va 2 wf), (8.14) (8.15) where Neper frequency, parallel RLC circuit • a — 2RC 7 (8.16) Resonant radian frequency, parallel RLC circuit • w 0 = (8.17) These results are summarized in Table 8.1. 8.1 Introduction to the Natural Response of a Parallel RLC Circuit 269 TABLE 8.1 Natural Response Parameters of the Parallel RLC Circuit Parameter «u Terminology Value In Natural Response Characteristic roots Neper frequency Resonant radian frequency S X = -a + Va 2 s 2 = -a - Vcr a = 2RC 1 "°~ Vie - col - (4 The exponent of e must be dimensionless, so both ,vj and s 2 (and hence a and a> () ) must have the dimension of the reciprocal of time, or fre- quency. To distinguish among the frequencies s h s 2 , a, and w () , we use the following terminology: ,5¾ and s 2 are referred to as complex frequencies, a is called the neper frequency, and IOQ is the resonant radian frequency. The full significance of this terminology unfolds as we move through the remaining chapters of this book. All these frequencies have the dimen- sion of angular frequency per time. For complex frequencies, the neper frequency, and the resonant radian frequency, we specify values using the unit radians per second (rad/s). The nature of the roots s { and s 2 depends on the values of a and o) () . There are three possible outcomes. First, if <of) < a 2 , both roots will be real and distinct. For reasons to be discussed later, the voltage response is said to be overdamped in this case. Second, if col > a 2 , both s-\ and s 2 will be complex and, in addition, will be conju- gates of each other. In this situation, the voltage response is said to be underdamped. The third possible outcome is that co 2 ) = a 2 . In this case, $i and 52 will be real and equal. Here the voltage response is said to be critically damped. As we shall see, damping affects the way the voltage response reaches its final (or steady-state) value. We discuss each case separately in Section 8.2. Example 8.1 illustrates how the numerical values of S[ and s 2 are determined by the values of R, L, and C. Example 8.1 Finding the Roots of the Characteristic Equation of a Parallel RLC Circuit a) Find the roots of the characteristic equation that governs the transient behavior of the voltage shown in Fig. 8.5 if R - 200 O, L = 50 mH, and C = 0.2 ixF. b) Will the response be overdamped, underdamped, or critically damped? c) Repeat (a) and (b) for R = 312.5 H. d) What value of R causes the response to be criti- cally damped? Solution a) For the given values of R, L, and C, 1 10 f a = 2RC (400)(0.2) = 1.25 X 10 4 rad/s, 2 l (io 3 )(io 6 ) in8 2 , From Eqs. 8.14 and 8.15, s t = -1.25 X 10 4 + Vl.5625 X 10 8 - 10 8 = -12,500 + 7500 = -5000 rad/s, Figure 8.5 A A circuit used to illustrate the natural response of a parallel RLC circuit. fc = -1.25 x 10 4 - Vl.5625 X 10 8 - 10 8 270 Natural and Step Responses of RLC Circuits b) The voltage response is overdamped because o)() < a . c) Fori? = 312.511, 10 6 a = = 8000 rad/s. (625)(0.2) a 2 = 64 X 10 6 = 0.64 X 10 8 rad 2 /s 2 . As col remains at 10 8 rad 2 /s 2 , Si = -8000 + /6000 rad/s, s 2 = -8000 - /6000 rad/s. (In electrical engineering, the imaginary number V—T is represented by the letter /", because the letter /' represents current.) In this case, the voltage response is under- damped since ag > a 2 . d) For critical damping, a 2 = co 2 h so or = 10\ and R i v 2RC) 1 2RC 10 6 1 LC ' = 10 4 (2 X 10 4 )(0.2) = 250 Q. I/ASSESSMENT PROBLEM Objective 1—Be able to determine the natural response and the step response of parallel RLC circuits 8.1 The resistance and inductance of the circuit in Fig. 8.5 are 100 O and 20 mH, respectively. a) Find the value of C that makes the voltage response critically damped. b) If C is adjusted to give a neper frequency of 5 krad/s, find the value of C and the roots of the characteristic equation. c) If C is adjusted to give a resonant frequency of 20 krad/s, find the value of C and the roots of the characteristic equation. NOTE: Also try Chapter Problem 8.1. Answer: (a) 500 nF; (b) C = 1 juF, Si = -5000 + /5000 rad/s, s 2 = -5000 - y'5000 rad/s; (c) C = 125 nF, s?! = -5359 rad/s, s 2 = -74,641 rad/s. 8.2 The Forms of the Natural Response of a Parallel RLC Circuit So far we have seen that the behavior of a second-order RLC circuit depends on the values of .Vj and .v 2 , which in turn depend on the circuit parameters R, L, and C. Therefore, the first step in finding the natural response is to calcu- late these values and, relatedly, determine whether the response is over-, under-, or critically damped. Completing the description of the natural response requires finding two unknown coefficients, such as A { and A 2 in Eq. 8.13.The method used to do this is based on matching the solution for the natural response to the initial conditions imposed by the circuit, which are the initial value of the current (or voltage) and the initial value of the first derivative of the current (or voltage). Note that these same initial conditions, plus the final value of the variable, will also be needed when finding the step response of a second-order circuit. In this section, we analyze the natural response form for each of the three types of damping, beginning with the overdamped response. As we will see, the response equations, as well as the equations for evaluating the unknown coefficients, are slightly different for each of the three damping configurations. This is why we want to determine at the outset of the problem whether the response is over-, under-, or critically damped. 8.2 The Forms of the Natural Response of a Parallel RLC Circuit 271 The Overdamped Voltage Response When the roots of the characteristic equation are real and distinct, the volt- age response of a parallel RLC circuit is said to be overdamped. The solu- tion for the voltage is of the form A^ + A 2 e Sl \ (8.18) < Voltage natural response—overdamped parallel RLC circuit where S] and s 2 are the roots of the characteristic equation. The constants A] and A 2 are determined by the initial conditions, specifically from the values of v(Q + ) and dv(Q + )/dt, which in turn are determined from the ini- tial voltage on the capacitor, V (h and the initial current in the inductor, / () . Next, we show how to use the initial voltage on the capacitor and the initial current in the inductor to find A x and A 2 . First we note from Eq. 8.18 that A\ and A 2 . First we note from Eq. 8.18 that v(0 + ) = A, + A 2 , (8.19) dv(Q + ) di = s x Ax + s 2 A 2 . (8.20) With S\ and s 2 known, the task of finding A { and A 2 reduces to finding v(0 + ) and dv(0 + )/dt. The value of v(0 + ) is the initial voltage on the capac- itor V{\. We get the initial value of dv/dt by first finding the current in the capacitor branch at t = 0 + . Then, dv(0 + ) i c (0 + ) We use Kirchhoff s current law to find the initial current in the capac- itor branch. We know that the sum of the three branch currents at t - 0 + must be zero. The current in the resistive branch at t = () + is the initial voltage Vo divided by the resistance, and the current in the inductive branch is / () . Using the reference system depicted in Fig. 8.5, we obtain /c( ° +) = ~R~ h ' (8 - 22) After finding the numerical value of J*c(0 + )i we use Eq. 8.21 to find the ini- tial value of dv/dt. We can summarize the process for finding the overdamped response, v(t), as follows: 1. Find the roots of the characteristic equation, s } and s 2 , using the val- ues of /?, L, and C. 2. Find v(0 + ) and dv(0 + )/dt using circuit analysis. 3. Find the values of Ai and A 2 by solving Eqs. 8.23 and 8.24 simultaneously: v(0 + ) = A t + A 2 , (8.23) ——— = ——- = s x A x + s 2 A 2 . (8.24) 4. Substitute the values for s u s 2 , Aj, and A 2 into Eq. 8.18 to deter- mine the expression for v(t) for t > 0. Examples 8.2 and 8.3 illustrate how to find the overdamped response of a parallel RLC circuit. 272 Natural and Step Responses of RLC Circuits Example 8.2 Finding the Overdamped Natural Response of a Parallel RLC Circuit For the circuit in Fig. 8.6, v(0 + ) = 12 V, and '"L(0 + ) = 30 mA. a) Find the initial current in each branch of the circuit. b) Find the initial value of dv/dt, c) Find the expression for v(t). d) Sketch v(t) in the interval 0 < r < 250 ms. Solution a) The inductor prevents an instantaneous change in its current, so the initial value of the inductor current is 30 mA: fe(0") = fe(0) =/L(0 + ) = 30mA. The capacitor holds the initial voltage across the parallel elements to 12 V. Thus the initial current in the resistive branch, //?(0 + ), is 12/200, or 60 mA. Kirchhoffs current law requires the sum of the currents leaving the top node to equal zero at every instant. Hence / c (o + ) = -/ L (o + ) - ;*(o + ) = -90 mA. Note that if we assumed the inductor current and capacitor voltage had reached their dc values at the instant that energy begins to be released, *c(0 _ ) = 0. In other words, there is an instanta- neous change in the capacitor current alt = 0. b) Because ic — C(dv/dt), dv(() + ) -90 X 10" dt 0.2 X 10 -6 -450 kV/s. c) The roots of the characteristic equation come from the values of R, L, and C. For the values specified and from Eqs. 8.14 and 8.15 along with 8.16 and 8.17, -1.25 X 10 4 + Vl.5625 X 10 8 - 10 8 -12,500 + 7500 = -5000rad/s, s 2 = -1.25 X 10 4 - 2 1.5625 X 10 8 - 10 8 -12,500 - 7500 = -20,000 rad/s. Figure 8.6 A The circuit for Example 8.2. Because the roots are real and distinct, we know that the response is overdamped and hence has the form of Eq. 8.18. We find the co-efficients A x and A 2 from Eqs. 8.23 and 8.24. We've already determined s\, s 2 , v(0 + ), and dv(Q + )/dt, so 12 = A l + A 2 , -450 X 10 3 = -50()0^ - 20,000A 2 . We solve two equations for A r and A 2 to obtain Ai = -14 V and A 2 = 26 V. Substituting these values into Eq. 8.18 yields the overdamped volt- age response: v(t) = (-Ue' 5i)m + 26e- 20000 ') V, t > 0. As a check on these calculations, we note that the solution yields v(0) = 12 V and dv(0 + )/dt = -450,000 V/s. d) Figure 8.7 shows a plot of v(t) versus t over the interval 0 < t < 250 ms. t(/XS) Figure 8.7 • The voltage response for Example 8.2. 8.2 The Forms of the Natural Response of a Parallel RLC Circuit 273 Calculating Branch Currents in the Natural Response of a Parallel RLC Circuit Derive the expressions that describe the three branch currents i R , / L , and ic in Example 8.2 (Fig. 8.6) during the time the stored energy is being released. Solution We know the voltage across the three branches from the solution in Example 8.2, namely. ,-500()/ -20,000/ v(t) = (-14*?-™"" + 26<r^ uuu ') V, t > 0 The current in the resistive branch is then i R (t) =^- (-70e- 5m)l + 130*- 20 ' 000 ') mA, t > 0. There are two ways to find the current in the induc- tive branch. One way is to use the integral relation- ship that exists between the current and the voltage at the terminals of an inductor: hiO =7/ v L (x)dx + /(). A second approach is to find the current in the capacitive branch first and then use the fact that ia + ijL + 'c = 0. Let's use this approach. The cur- rent in the capacitive branch is i c (0 = c dv dt 0.2 X l(r 6 (70,000e" -5000/ 20,000/ 520,000<r z,WJUW ) (14c -5000/ 104^-20.000/ )mA ^ , > () + Note that *c(0 + ) = -90 mA, which agrees with the result in Example 8.2. Now we obtain the inductive branch current from the relationship k(0 = -<K(0 - W0 (56c -5000/ 2^-20.000/) mA? /s0 We leave it to you, in Assessment Problem 8.2, to show that the integral relation alluded to leads to the same result. Note that the expression for i L agrees with the initial inductor current, as it must. ^ASSESSMENT PROBLEMS Objective 1—Be able to determine the natural response and the step response of parallel RLC circuits 8.2 Use the integral relationship between i L and v to find the expression for i L in Fig. 8.6. ,-5000/ 20,000/N Answer: i L (t) = (56^^ - 2de~ mwx ) mA, t > 0. 8.3 The element values in the circuit shown are R = 2 kH, L = 250 mH, and C = 10 nF. The initial current / 0 in the inductor is —4 A, and the initial voltage on the capacitor is 0 V. The output signal is the voltage v. Find (a) //?(0 + ); (b) / c (0 + ); (c) dv(0 + )/dt; (d) A X ; (e) A 2 ; and (f) v{t) when t > 0. Answer: (a) 0; (b)4A; (c) 4 X 10 8 V/s; (d) 13,333 V; (e) -13,333 V; (f) 13,333(e- iao00 ' — e -40.000/ ) V. NOTE: Also try Chapter Problems 8.8, 8.11, and 8.18. Natural and Step Responses of RLC Circuits The Underdamped Voltage Response When IOQ > a 2 , the roots of the characteristic equation are complex, and the response is underdamped. For convenience, we express the roots jj and Si as (8.25) (8.26) where Damped radian frequency • co d = Vwo - a 2 . (8.27) The term (o d is called the damped radian frequency. We explain later the reason for this terminology. The underdamped voltage response of a parallel RLC circuit is Voltage natural response—underdamped parallel RLC circuits • v(t) = B\e~ at cos <o d t + B 2 e~ at sin w d t, (8.28) which follows from Eq. 8.18. In making the transition from Eq. 8.18 to Eq. 8.28, we use the Euler identity: e ±jo = cos 0 ± j sin 0 ^ 29 ) Thus, v{t) = A^- n+l "' l)l + A 2 e~ {a+Mt = Aie-^ei"* + A 2 e~ at e~^ = e~"'(Ai cos co d t + /Ai sin o>/ + A 2 coso) (t t - jA 2 sin (o d t) = e~°"[(Ai + A 2 ) cosco d t + j{A { - A 2 )smcD d t]. At this point in the transition from Eq. 8.18 to 8.28, replace the arbitrary constants A\ + A 2 and }(A\ - A 2 ) with new arbitrary constants denoted B x and B 2 to get v = e~ m (B] cosco d t + B 2 sma) d t) = B x e~°" cos co d t + B 2 e~ at sin a) d t. The constants B x and B 2 are real, not complex, because the voltage is a real function. Don't be misled by the fact that B 2 = j(Ai - A 2 ). In this underdamped case, A j and A 2 are complex conjugates, and thus B { and B 2 are real. (See Problems 8.12 and 8.13.) The reason for defining the under- damped response in terms of the coefficients B x and B 2 is that it yields a sim- 8.2 The Forms of the Natural Response of a Parallel RLC Circuit pier expression for the voltage, v. We determine B { and B 2 by the initial energy stored in the circuit, in the same way that we found A x and A 2 for the overdamped response: by evaluating v at t = 0 + and its derivative at t = 0 + . As with S\ and .y 2 , a and oj d are fixed by the circuit parameters R, L, and C. For the underdamped response, the two simultaneous equations that determine B { and B 2 are v(0 + ) = V Q = 0, (8.30) dv(0 + ) i c (0 + ) dl C = —aB\ + d>d^2- (8.31) Let's look at the general nature of the underdamped response. First, the trigonometric functions indicate that this response is oscillatory; that is, the voltage alternates between positive and negative values. The rate at which the voltage oscillates is fixed by a) d . Second, the amplitude of the oscillation decreases exponentially. The rate at which the amplitude falls off is determined by a. Because a determines how quickly the oscillations subside, it is also referred to as the damping factor or damping coefficient. That explains why a> d is called the damped radian frequency. If there is no damping, a = 0 and the frequency of oscillation is a) {) . Whenever there is a dissipative element, R, in the circuit, a is not zero and the frequency of oscillation, a> d , is less than o> 0 . Thus when a is not zero, the frequency of oscillation is said to be damped. The oscillatory behavior is possible because of the two types of energy- storage elements in the circuit: the inductor and the capacitor. (A mechan- ical analogy of this electric circuit is that of a mass suspended on a spring, where oscillation is possible because energy can be stored in both the spring and the moving mass.) We say more about the characteristics of the underdamped response following Example 8.4, which examines a circuit whose response is underdamped. In summary, note that the overall process for finding the underdamped response is the same as that for the overdamped response, although the response equations and the simulta- neous equations used to find the constants are slightly different. Example 8.4 Finding the Underdamped Natural Response of a Parallel RLC Circuit In the circuit shown in Fig. 8.8, V {) = 0, and / () = -12.25 mA. a) Calculate the roots of the characteristic equation. b) Calculate v and dv/dt at t = 0 + . c) Calculate the voltage response for t S: 0. d) Plot v{t) versus t for the time interval ()</<!! ms. 0.125 /xF Figure 8.8 • The circuit for Example 8.4. Solution a) Because 1 1(T 2RC 2(20) 10 3 (0.125) 0)() = we have 10 f (8)(0.125) 200 rad/s, 10 3 rad/s. a>5 > a~ . col remains at 10 8 rad 2 /s 2 , Si = -8000 + /6000 rad/s, s 2 = -8000 - /6000 rad/s. (In electrical engineering, the imaginary number V—T is represented by the letter /", because. energy- storage elements in the circuit: the inductor and the capacitor. (A mechan- ical analogy of this electric circuit is that of a mass suspended on a spring, where oscillation is possible because