ECE2205: Circuits and Systems I 6 1 Series RC and RL Time Domain Solutions In the last chapter, we saw that capacitors and inductors had element relations that are differential equations: i c (t) = C d v c(t), and v L (t) = L d i L(t). We can analyze circuits containing capacitors and inductors using KVL, KCL, and node analysis, for example, but the equations that we will ultimately need to solve are differential equations. This chapter discusses how to solve differential equations in the time domain. We will derive a general method; This method gives insight, but is tedious for large problems; Next chapters will introduce alternate methods. Example Motivational Circuits In order to motivate discussion, consider the series RC circuit seen previously. KVL gives v in (t) = v r (t) + v c (t). Ohm s law: v r (t) = Ri(t). i(t) R Capacitor model: i(t) = Cdv c (t)/. Overall, v in (t) = RC dv c(t) + v c (t). v in (t) C v c (t)
ECE2205, Series RC and RL Time Domain Solutions 6 2 If we can solve this equation, we can find v c (t) and subsequently i(t). Alternately, consider the series RL circuit seen before. KVL: v in (t) = v r (t) + v L (t). Ohm s law: v r (t) = Ri(t). R i(t) Inductor model: v L (t) = Ldi(t)/. v in (t) L v L (t) Overall: v in (t) = Ri(t) + Ldi(t)/. Again, we have a differential equation to solve. In general, when circuits contain capacitors and/or inductors, we will need to solve differential equations to get the desired circuit quantities. One saving grace is that these equations are linear constantcoefficient ordinary differential equations (LCCODE), which are (relatively) easy to solve. LCCODEs A general LCCODE with input x(t) and output y(t) is d n y(t) dy(t) d m x(t) dx(t) a n + + a n 1 + a 0 y(t) = b m + + b m 1 + b 0 x(t). This is an nth-order LCCODE (n m). It is linear: all variables and derivatives to first power (no y 2 (t)... ) Constant coefficient: all parameters (e.g., a 1 ) are constants. Ordinary: no partial derivatives (all derivatives are with respect to single variable t).
ECE2205, Series RC and RL Time Domain Solutions 6 3 Fact of life: The solution to an LCCODE is the sum of two parts: the natural solution y n (t) and the particular solution y p (t). y(t) = y n (t) + y p (t). We are interested in methods to construct (not guess) y n (t) and y p (t). Natural solution to LCCODEs The natural solution is the solution to the homogeneous differential equation with input x(t) and all its derivatives zero. d n y(t) dy(t) a n + + a n 1 + a 0 y(t) = 0. Fact of life: A parametric function of the form ke st, k = 0, is a solution to y n (t) for certain choices of exponential frequency s. For example, consider the RC circuit RC dv c(t) + v c (t) = v in (t). Substitute v in (t) = 0 and v c (t) = ke st to find the natural solution RC d(kest ) + ke st = 0 (s RC + 1)ke st = 0, which is true in general for k = 0 if frequency s = 1 RC. Note that any value of k solves this equation (so far), but we will later chose it to match initial conditions in the circuit.
ECE2205, Series RC and RL Time Domain Solutions 6 4 The characteristic equation If we substitute ke st into the general homogeneous equation we get a useful result. a n d n ke st n + a n 1 d n 1 ke st n 1 + + a 1 dke st + a 0 ke st = 0 a n s n ke st + a n 1 s n 1 ke st + + a 1 s 1 ke st + a 0 ke st = 0 a n s n + a n 1 s n 1 + + a 1 s + a 0 = 0. This last equation is called the characteristic equation of the LCCODE. For example, the RC circuit had characteristic equation s RC + 1 = 0. Some observations: The natural solution solves the homogeneous differential equation. The natural solution does not depend on the input signal. Note the correspondence between the coefficients of the characteristic equation and those of the input-output LCCODE and the order of the derivatives in the input-output LCCODE and the powers of s in the characteristic equation. The characteristic equation specifies those values of s that are valid in ke st in the homogeneous equation. No other values work. The characteristic equation is algebraic and can be solved for its roots. In MATLAB, we solve s 3 + 3s 2 + 2s + 1 = 0 via roots([1 3 2 1]) The values of s that satisfy the characteristic equation are the natural frequencies.
ECE2205, Series RC and RL Time Domain Solutions 6 5 An nth-order LCCODE will have n roots (not necessarily distinct and possibly complex) (s s 1 )(s s 2 ) (s s n ) = 0. EXAMPLE: A second-order circuit with numeric values. LCCODE: d2 y(t) 2 + 5 dy(t) Homogeneous DE: d2 y(t) 2 + 6y(t) = 4x(t). + 5 dy(t) Characteristic equation: s 2 + 5s + 6 = 0. Factored form: (s + 2)(s + 3) = 0. Natural frequencies: s 1 = 2, s 2 = 3. + 6y(t) = 0. The overall natural solution (when natural frequencies are not repeated) comprises n terms y n (t) = k 1 e s 1t + k 2 e s 2t + + k n e s nt. In this particular example The particular solution y n (t) = k 1 e 2t + k 2 e 3t. The particular solution satisfies both the LHS and RHS of the LCCODE. For arbitrary x(t) it can be quite difficult to find. However, if the input is of the form Xe s pt and s p is not a root of the characteristic equation, it isn t too bad. This is not as limiting as it seems, as nearly any input comprises sums of frequencies...
ECE2205, Series RC and RL Time Domain Solutions 6 6 To see how this works, plug Xe s pt into the LCCODE. Let y p (t) = Y e s pt. d n Y e s pt a n + + a n 0 Y e spt d m Xe spt = b m + + b m 0 Xe s pt Y e [ s pt a n s n p + + a ] 1s p + a [ 0 = Xe s p t b m s m p + + b ] 1s p + b 0 Y = X ( bm s m p + + b ) 1s p + b 0 a n s n p + + a. 1s p + a 0 This gives a unique value for Y for input X. The rational polynomial Y X = b(s) = H(s) is called the transfer a(s) function of the circuit. The transfer function simplifies notation greatly: Y = X H(s p ) y p (t) = X H(s p )e s pt. The natural and particular solution sum to compute the overall output signal. In general, if x(t) = Xe spt then n y n (t) + y p (t) = k i e sit + X H(s p )e spt. For the LCCODE given above, H(s) = i=1 4 s 2 + 5s + 6. If x(t) = 2e t u(t) then ( ) y p (t) = 2H( 1)e t 4 u(t) = 2 e t u(t) = 4e t u(t). 1 5 + 6 y(t) = k 1 e 2t + k 2 e 3t + 4e t, t 0.
ECE2205, Series RC and RL Time Domain Solutions 6 7 Insights into the transfer function H(s) is a ratio of polynomials in the complex s-domain, where s = σ + jω. We look at real-valued s for now. The transfer function is evaluated at the source frequency s p, so it is essential to identify that frequency. The transfer function is a property of the circuit itself, not of the input signal. If the input signal is an exponential, the particular solution will be an exponential of the same frequency. Insights into the natural solution The natural solution depends on the roots of the characteristic polynomial, which is the denominator of the transfer function. The natural solution is also a system property, not an input-signal property. The parameters of the natural solution k 1 k n are required in the complete solution, and are determined by boundary conditions. These boundary conditions are usually specified in terms of initial element voltages or currents. In our example, we might be told that y(0 ) = 0 and dy(0 )/ = 1. Then y(0) = k 1 + k 2 + 4 = 0 dy(t) = 2k 1 e 2t 3k 2 e 3t 4e t t=0 t=0 = 2k 1 3k 2 4 = 1. We have two equations and two unknowns (we know all about solving those!)
ECE2205, Series RC and RL Time Domain Solutions 6 8 The solution gives k 1 = 7 and k 2 = 3. The unique solution to this problem is now known y(t) = 7e 2t + 3e 3t + 4e t, t 0. ZIR, ZSR, and ISR So far, this chapter has been more about solving differential equations than it has been about circuits. We now change directions slightly and spend time applying these concepts to RC and RL circuits. First, we define some engineering terms that are closely related to the mathematical concepts presented so far. RESPONSE: Engineers talk about the response of a circuit to an input (even the zero input) rather than the solution of a differential equation, even though the response may be computed by solving a DE. ZERO-INPUT RESPONSE (ZIR): The output of the circuit when the input signal and its derivatives are zero. This is computed via the DE homogeneous solution. ZERO-STATE RESPONSE (ZSR): The output of the circuit when all initial voltages/currents/etc (the system initial state ) are zero. This is computed via the particular solution to the DE. INITIAL-STATE RESPONSE (ISR): The output of the circuit when there is initial stored energy and a source is applied. It is the sum of the ZIR and the ZSR, and is computed as the full solution to the DE.
ECE2205, Series RC and RL Time Domain Solutions 6 9 Note that a DE is an abstract mathematical concept that can be solved with no reference to any physical system, such as a circuit. The DE is a model of a system; e.g., a circuit. The terms defined above relate directly to the physical system and can often be measured as well as computed. While subtle, this distinction is important a real circuit and its response will never be exactly the same as its model and model response. A real circuit is not built with ideal components, etc, but we can still talk about its ZIR, ZSR, and ISR. Solving the Series RC Circuit We will now focus the techniques that we have developed on a very important simple circuit the series RC circuit. We have already found the differential equation for this circuit: RC dv c(t) + v c (t) = v in (t). First, we look at the ZIR, then the ZSR, finally the ISR. Series RC circuit ZIR With the input turned off, the DE for capacitor voltage becomes RC dv c(t) + v c (t) = 0. The characteristic equation is (s RC + 1) = 0. The natural frequency is s = 1 RC. The natural solution is v c,n (t) = ke t RC, t 0. Because the capacitor voltage is continuous, v c (0 ) = v c (0 + ) = v o.
ECE2205, Series RC and RL Time Domain Solutions 6 10 Matching v c,n (0) = k with v o we get that k = v o. Therefore, the ZIR solution is v zir (t) = v o e t RC, t 0. Recall that we defined the time constant of a series RC circuit to be τ = RC. The larger the time constant (large R or C), the longer the capacitor takes to discharge. The smaller the time constant (small R or C), the shorter the capacitor takes to discharge. Now that we know ZIR voltage, we can find ZIR current. The simplest way is to use the element relationship i c (t) = C dv c(t) d t = Cv o RC u(t) e = Cv o [ 1 = v o R e t RC e RC, t 0. ] t RC u(t) + e t RC δ(t) Note: the δ(t) term explains how the initial charge came about on the capacitor and is an artifact of modeling the circuit for t 0 only. The δ(t) mathematically charges the capacitor from v c (0 ) = 0 to v c (0 + ) = v o. Since this initial charge really came about through other means, we drop the δ(t) from the i c (t) term. Series RC circuit ZSR We now consider the ZSR of the same RC circuit v c (0) = 0.
ECE2205, Series RC and RL Time Domain Solutions 6 11 Let us assume for now that v in (t) = x(t) = Xe s pt. We have already found the transfer function for the circuit, so we know that Series RC circuit ISR 1 H(s) = s RC + 1 y zsr (t) = X H(s p )e s pt = Xes pt s p RC + 1. The overall solution is the sum of the ZIR and ZSR. Consider RC = 5 and v in (t) = 20e 4t u(t) and v c (0) = 1. v isr (t) = [ K e t/5 + 20H( 4)e 4t] u(t) [ = K e t/5 + 20 ] 19 e 4t u(t). We find K by matching v isr (0) = v c (0) = 1. 1 = K 20 19 K = 39 19. So, the total response is v isr (t) = [ 39 19 e t/5 20 ] 19 e 4t u(t). STEP RESPONSE: Another important example is the step response of the system for initial voltages equal to zero. Note that a step input can be found as u(t) = lim sp 0 es pt u(t),
ECE2205, Series RC and RL Time Domain Solutions 6 12 so the particular solution has the form y zsr (t) = H(0)u(t). The ZIR solution is as before y zir (t) = ke t RC u(t), so overall we have v c (t) = ( ke t RC + 1 ) u(t) with initial condition v c (0) = 0, giving ( ) v c (t) = 1 e t RC u(t), which is what we claimed last chapter. The step response determines, for example, how quickly a digital logic signal can change from a 0 to a 1 and vice versa. The time constant is determined by circuit resistance and capacitance often stray quantities from the circuit-board layout. At very high frequency operation, traces must be made very short, and device capacitances very small. EXERCISE: Find the ISR of a circuit to a step input with nonzero initial capacitor voltage. Solving the Series RL Circuit As a second major example, we consider the series RL circuit. We have already found the DE to be di L (t) + i L (t) R L = v in(t) L.
ECE2205, Series RC and RL Time Domain Solutions 6 13 Series RL circuit ZIR The homogeneous response is found when v in (t) = 0 di L (t) + i L (t) R L = 0, which gives characteristic equation ( s + R ) = 0, L which has natural frequency s = R/L. Therefore, i zir (t) = ke R L t u(t). Notice that τ = L/R so either large L or small R implies slow response; either small L or large R implies fast response. Series RL circuit ZSR The ZSR is found via the transfer-function relationship 1 H(s) = sl + R. We can find the inductor voltage via element relations v L (t) = L di L(t) = k Re R L t u(t). Inductor voltage can change instantly, but inductor current cannot opposite behavior to capacitor. Therefore, for an input v in (t) = x(t) = Xe s pt, Series RL circuit ISR i zsr (t) = y zsr (t) = Xes pt s p L + R. The overall solution is the sum of the ZIR and ZSR.
ECE2205, Series RC and RL Time Domain Solutions 6 14 The unit-step response, for example, is i L (t) = i zir (t) + i zsr (t) [ = ke R L t + 1 ] u(t) R with i L (0) = 0, giving i L (t) = 1 R [ 1 e R L t] u(t). The inductor voltage is v L (t) = L di L(t) = L ( R R L e = e R L t u(t). R L t u(t) + Another way of seeing this is to consider v L (t) = v in (t) v r (t) Some more examples = u(t) i L R = e R L t u(t). [ 1 e R t] ) L δ(t) EXAMPLE: Consider the circuit shown with R = 1 k and C = 10 µf. First, we find the differential equation for the circuit in LCCODE form. R a C x(t) R This equation must be in terms of x(t) and y(t) only (with constant terms included). The equations must not have signals like v c (t), although such signals may be important intermediate results. y(t)
ECE2205, Series RC and RL Time Domain Solutions 6 15 First, write a KCL equation at a: e a x + C d(e a y) = 0. R Next, write a KCL equation at y: C d(e a y) = y R ė a = ẏ + 1 RC y. We would like to substitute this into the first KCL equation, but to do so we first need to differentiate that first equation: ė a ẋ + RCë a RC ÿ = 0 [ ẏ + 1 ] [ RC y ẋ + RC ÿ + 1 ] RC ẏ RC ÿ = 0 2ẏ + 1 RC y = ẋ 2ẏ + 100y = x. Now that we have the LCCODE, we find the natural solution to the circuit s response. Set 2ẏ + 100y = 0 and let y = ke st (2s + 100) = 0 s = 50 y n (t) = ke 50t, t 0. To find a particular solution, we will need to find a transfer function s H(s) = 2s + 100. We might be interested in the step response of the circuit, which is found for the particular input x(t) = lim sp 0 e s pt. y p (t) = H(0)e 0t = 0.
ECE2205, Series RC and RL Time Domain Solutions 6 16 The full solution for the step response is y(t) = y n (t) + y p (t) = ke 50t, t 0. But, how to find k? We need to know y(0), but how? Note that we assume initial rest to find step response, but how does that help us? Note that we can re-draw the circuit in a manner valid for t = 0 only, understanding that capacitor voltages and inductor currents do not change instantaneously for finite inputs. We recognize this as a voltage-divider circuit, so y(0) = 1/2 = k. Therefore, the step response of the circuit is y(t) = 1 2 e 50t, t 0. What about the response to another input, such as x(t) = e t u(t) with initial capacitor voltage v c (0) = 1 V? The natural solution retains the same form. The particular solution is y p (t) = H( 1)e t u(t). y(t) = 1 98 e t + ke 50t, t 0. Again, we must find k. Again, we re-draw the circuit for t = 0 only. By x(t) KVL we see that y(0) = 0 so k = 1/98. Overall, y(t) = 1 ( e 50t e t), t 0. 98 EXAMPLE: Consider the circuit shown with R 1 = 10 k, R 2 = 5 k and L = 0.1 H. First, we find the differential equation for the circuit in LCCODE form. R L a 1 V R x(t) R 1 R 2 y(t) y(t)
ECE2205, Series RC and RL Time Domain Solutions 6 17 Write a KCL equation at y: i L + x y = y R 1 R 2 di L + ẋ ( 1 = + 1 ) ẏ. R 1 R 1 R 2 Note: v L = x y = L di L x y + ẋ ( 1 = + 1 ) ẏ L R 1 R 1 R 2 R 1 R 2 x + R 2 Lẋ = L(R 1 + R 2 )ẏ + R 1 R 2 y 5 10 7 x + 500ẋ = 1500ẏ + 5 10 7 y. The natural solution to this system is 0 = 1500ẏ + 5 10 7 y 0 = 1500s + 5 10 7 s = 33333 y n (t) = ke 33333t, t 0. The transfer function for this system is 500s + 5 107 H(s) = 1500s + 5 10 = s + 100000 7 3s + 100000. To find the step response of this system, we note that H(0) = 1 so y p (t) = u(t) and y(t) = ke 33333t + 1. To find k, we note that initial inductor current is zero, so R 1 and R 2 R 2 form a voltage divider. y(0) = x(0) = 1 = 1 + k. Therefore, R 1 + R 2 3 k = 2/3. y(t) = 1 2 3 e 33333t, t 0.
ECE2205, Series RC and RL Time Domain Solutions 6 18 We can also find the response of the circuit to the input x(t) = e t u(t), for i L (0) = 1 A (in the indicated direction). Note that y p (t) = H( 1)e t u(t) = 99999 99997 e t u(t) ( y(t) = ke 33333t + 99999 ) 99997 e t u(t). 1 A x(t) R 1 R 2 y(t) To find k, we again redraw the circuit for t = 0, only. KCL at y: y 1 10000 1 + y 0 5000 = 0 3y = 10000 y(0) = 3333. We substitute 3333 = k + 99999 99997 k 3334 [ y(t) 3334e 33333t + 99999 ] 99997 e t EXAMPLE: (Challenging) Consider the circuit shown with R 1 = 1 k, R 2 = 10 k, C 1 = 0.5 F, and C 2 = 1 mf. First, we find the differential equation for the circuit in LCCODE form., t 0. R 1 C a 1 x(t) R 2 First, write a KCL equation at a: e a x d(e a y) + C 1 R 1 = 0. C 2 y(t)
ECE2205, Series RC and RL Time Domain Solutions 6 19 Next, write a KCL equation at y: d(e a y) C 1 = y dy + C 2 R 2 ė a = y C 1 R 2 + ( 1 + C 2 C 1 ) ẏ. Substitute this into the first KCL equation after first differentiating the first equation [ y + 1 C 1 R 1 R 2 R 1 [ ẏ C 1 + C 1 R 1 R 2 ( 1 + C 2 y C 1 R 2 + ė a ẋ + C 1 ë a C 1 ÿ = 0 R 1 R 1 ( 1 + C ) ] 2 ẏ x + R 1 C 1 y 5000 C 1 ( 1 + C 2 ) C 1 ) ] ÿ C 1 ÿ = 0 ẏ + R 1 R 2 ẏ + R 1 C 2 ÿ = ẋ + 1.002ẏ + 0.1ẏ + ÿ = ẋ 5000ÿ + 5510ẏ + y = 5000ẋ. Now, we proceed to find the natural solution. The characteristic equation is 5000s 2 + 5510s + 1 = 0 which has roots at s 1 = 0.0002 and s 2 = 1.1018. Therefore y n (t) = k 1 e 0.0002t + k 2 e 1.1018t, t 0. The transfer function for this circuit is 5000s H(s) = 5000s 2 + 5510s + 1.
ECE2205, Series RC and RL Time Domain Solutions 6 20 To find the step response of the circuit, we first note that y p (t) = H(0)u(t) = 0. Therefore, y(t) = k 1 e 0.0002t + k 2 e 1.1018t, t 0. We know that the capacitors are initially uncharged, so y(0) = v c2 (0) = 0 which gives k 1 + k 2 = 0 or k 1 = k 2 = K. We still need an equation to find this unknown. As we saw (much earlier), we differentiate y(t) to get this equation ẏ(0) = 0.002k 1 1.1018k 2, but, what is ẏ(0)? Note that ẏ(0) = v c2 (0) = i c2 (0)/C 2 by the capacitor v-i relation. To find i c2 (0) we redraw the circuit for t = 0 only. R 1 a 0 Note that x(0) y(0) i R1 = = 1 = 0.001 A R 1 R 1. x(t) R 2 0 This same current must all flow through C 2 (as no current flows through R 2 ). Therefore, ẏ(0) = 0.001 = 1. Finally, we get our second 0.001 equation 1 = 1.1016K K = 0.9078. y(t) And y(t) = 0.9078 [ e 0.0002t e 1.1018t], t 0. We conclude this example by considering the circuit response to x(t) = e t u(t) for v c1 (0) = 1 V and v c2 (0) = 2 V. y(t) = k 1 e 0.0002t + k 2 e 1.1018t + H( 1)e t, t 0
ECE2205, Series RC and RL Time Domain Solutions 6 21 where H( 1) = 9.8232. We find that y(0) = v c2 (0) = 2 = k 1 + k 2 + 9.8232 ẏ(0) = i c 2 (0) C 2. R 1 a 1 V Redraw again: x(t) R 2 2 V y(t) Analyzing like before: x(0) (y(0) + 1) i R1 = = R 1 1 (2 + 1) 1000 = 0.002 ẏ(0) = 2. We now have a second equation: ẏ(0) = 2 = 0.0002k 1 1.1018k 2 9.8232. Putting in a matrix form to solve: [ 1 1 0.0002 1.1018 ] [ k 1 k 2 ] = [ 7.8232 7.8232 ] giving k 1 = 0.7230 and k 2 = 7.1002. Our final result is y(t) = 0.7230e 0.0002t 7.1002e 1.1018t + 9.8232e t, t 0. Yeah, but what if...? The Next Step Here, we have focused on time-domain solutions to determine ZSR, ZIR, and ISR for simple exponential inputs. What if the system is more complex (e.g., many capacitors and/or inductors)? These methods will be tedious. What if the input is not exponential? One solution is to
ECE2205, Series RC and RL Time Domain Solutions 6 22 Find step response of system Differentiate step response to find impulse response Convolve impulse response with input to find output Also tedious. What if input frequencies duplicate natural frequencies, of if natural frequencies are duplicated? What if natural frequencies are complex? The next step is to look at the Laplace transform, a mathematical tool for analyzing systems in a complex frequency domain. In many ways it is similar to the z-transform we used in ECE2610 to analyze discrete-time systems. All of these analyses will be made easier.
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