Unit 3 Response of SecondOrder Systems In this unit, we consider the natural and step responses of simple series and parallel circuits containing inductors, capacitors and resistors. The equations which result from this analysis will, in general, be secondorder ordinary differential equations (O.D.E. s) with constant coefficients. Before introducing a particular circuit example, we will consider how to determine the solution of such an equation. The solution of the general secondorder O.D.E. with constant coefficients may then be applied to any circuit whose equation fits that form. 3.1 Natural Response of a Parallel RLC Circuit 3.1.1 Review of SecondOrder O.D.E. with Constant Coefficients Consider the following secondoder O.D.E. in which x is a function of time and α and ω 0 are constants. The middle term is written with a factor of 2 for reasons that will become obvious shortly.... (1) When we get to the actual circuits, x will be current i or voltage v and the constants will depend on the the elements R, L and C and their particular configuration in the circuit. The typical approach in solving (1) is to assume that a solution of the form x = Ae st... (2) exists. Here, A and s are unknown constants. To determine A and s we begin by substituting (2) into (1) to get 1
which implies Ae [ st s 2 (2α)s ω0] 2 = 0... (3) From (3) we see that either A = 0 which is a trivial but useless case since our voltages and currents won t be zero for all time or s 2 (2α)s ω0 2 = 0... (4) which implies... (5) Equation (4) is the socalled characteristic equation because its roots (solutions) determine the character of x. These roots may be written explicitly as s 1 =... (6a) s 2 =... (6b) Since, from (3) we know that both (6a) and (6b) are solutions to the charactersitic equation regardless of the value of A, equation (2) indicates that both x 1 = and x 2 = are solutions of (1). Since x 1 and x 2 are linearly independent solutions of (1), their combination given by x = x 1 x 2 =... (7) is also a solution. As we shall see, s 1 and s 2 will depend on the circuit parameters R, L and C, while A 1 and A 2 will be determined from the initial conditions. Equation (6) suggests three possible kinds of solution: (1) α 2 > ω0 2 both characteristic roots are real [overdamped system] (2) α 2 < ω0 2 roots are complex conjugates[underdamped system] (3) α 2 = ω0 2 both roots are real and equal [critically damped system] The meanings of these system responses will be considered shortly, as will the significance of α (the socalled neper frequency of the system) and ω 0 (the socalled resonant radian frequency of the system). 2
3.1.2 The Parallel RLC Circuit Natural Response Consider the following parallel combination of R, L and C. Since the voltage is the same across each component we will seek that first. Applying KCL to the top node of the circuit, we have... (8) where we have allowed for an initial (constant) current I 0 Differentiating this equation gives through the inductor.... (9) and dividing by C gives d 2 v dt 1 dv 2 RC dt 1 LC v = 0... (10) Comparing (10) with the general form of the secondorder O.D.E. in (1), we immediately identify α = 1 2RC and ω 0 = 1 LC... (11) and therefore s 1 and s 2 can be immediately found from (6a) and (6b). From (7) we see that we must still find A 1 and A 2. In fact, while we proceed slightly differently for the three classes of systems mentioned above, the general procedure for finding v does not change fundamentally: (1) Find s 1 and s 2 using R, L, and C. (2) Determine the initial voltage across the elements (which is the initial voltage capacitor) i.e. v(0 ) = V 0. (3) Find the initial rate of change of the voltage dv(0 ) = i C(0 ) dt C. (4) Use (2) and (3) to find the remaining constants required in the solution for v. This last step will vary in application depending on the type of system as we will develop below. 3
(5) Once the voltage is established, we can easily determine the branch currents. Overdamped Voltage Response (α 2 > ω0) 2 For this case, s 1 and s 2 are real and may be found using (6a), (6b) and (11) (we won t carry out this step until we use a particular example). Next, from (7), v(0 ) = A 1 A 2... (12) while dv(0 ) dt = i C(0 ) C = s 1 A 1 s 2 A 2... (13) From (12) and (13), A 1 and A 2 may be readily found so that equation (7) is completely specified. Of course, in general, i C (0 ) will have to be determined by circuit analysis (or be otherwise specified). See example 8.2 of the text, pages 338339. Underdamped Voltage Response (α 2 < ω0) 2 In this case, s 1 and s 2 are complex conjugates and may be written as s 1 = = s 1 =... (14) where ω d = ω 2 0 α 2... (15) is referred to as the damped radian frequency to be explained shortly. Also, s 2 = α jω d... (16) Substituting these s values into (7), we get (recalling that here x = v(t)), 4
It is possible to show (see Problem Set 4 later) that A 1 and A 2 are complex conjugates so that B 1 and B 2 are real, as is required to obtain a real voltage. We note the following about equation (17): (1) The sine and cosine functions indicate an oscillatory response. The rate of the oscillation is determined by ω d, the damped radian frequency. (2) If there is a resistance, there will be an α and therefore a decay of the voltage over time by virtue of the e αt. For this reason, α is often referred to as the damping coefficient. Thus, for an underdamped RLC parallel circuit, the voltage response will be as shown below. Illustration: The oscillatory behavior occurs because of the presence of two energy storage elements (the inductor and capacitor) which alternately store and release energy howbeit in a decaying fashion if there is a resistance present. Notice that the oscillation occurs about a final value this phenomenon is referred to as ringing and it will be a very important consideration in later courses (eg., control systems and filters). Critically Damped Voltage Response (α 2 = ω0) 2 Finally, we consider what happens to the voltage response when s 1 = s 2 = α (i.e., the characteristic roots are equal). There is now a fundamental difference because the solution to the general equation changes from that given in (7) to x = D 1 te αt D 2 e αt... (18) where D 1 and D 2 are constants this is easily shown to be a solution of (1). In determining x (or voltage v for the case being discussed) we must find D 1 and D 2 from the initial conditions as before. That is, we use v(0 ) = V 0 = D 2 (from (18)) and dv(0 ) dt = i c(0 ) C = D 1 αd 2. 5
The critically damped response indicates a smooth approach to the final value as depicted. Illustration: Before considering the step response of the system for which we have just completed the natural response, we address the following three examples. Overdamped Response Drill exercise 8.3, page 340 of the text. Consider the following circuit in which L = 250 mh, C = 10 nf and R = 2 kω. The initial current I 0 in the inductor is 4 A while the initial voltage on the capacitor is 0 V. The output signal is the voltage v. Determine (a) i R (0 ); (b) i C (0 ); (c) dv(0 )/dt; (d) A 1 ; (e) A 2 ; and (f) v(t) for t 0. 6
Underdamped Response Drill exercise 8.4, page 344 of the text. Consider the following circuit in which L = 10 mh and C = 1 µf. R is adjusted so that the roots of the characteristic equation are 8000 ± j6000 rad/s. The initial current I 0 in the inductor is 80 ma while the initial voltage on the capacitor is 10 V. Determine (a) R; (b) dv(0 )/dt; (c) B 1 and B 2 in the solution for v; and (d) i L (t). 7
Critically Damped Response Drill exercise 8.5, page 346 of the text. Suppose that the resistance in the previous example is adjusted for critical damping. The inductance L = 0.4 H and the capacitance C = 10 µf. The initial energy stored in the circuit is 25 mj and is distributed equally between the inductor and capacitor. Determine (a) R; (b) V 0 ; (c) I 0 ; (d) D 1 and D 2 in the solution for v; and (e) i R for t 0. 8
3.2 Step Response of a Parallel RLC Circuit Consider the following circuit in which a step input is initiated by opening the switch at t = 0. For now we consider that there is no energy initially stored in the system. I When the switch is opened, the circuit is forced by the current I (the forcing function) from the ideal current source. We will concentrate first on determining the expression for i L, the current in the inductor. Using KCL, i L i R i C = I... (1) which, recalling the relationships between currents and voltages in the three components, we can write in terms of voltage and i L as... (2) Since, here, the voltage across each element of the parallel section is the same (for example, v = v L = L(di L /dt)) (2) may be divided by LC and written as d 2 i L dt 1 di L 2 RC dt i L LC = I LC... (3) This is clearly a secondorder inhomogeneous D.E. see Term 2 Math Course whose solution is the sum of the the solution to the homogeneous part (i.e. the the complementary function) plus any one solution (i.e. the particular solution). The homogeneous part of (3) looks just like equation (10) on page 3 of this unit with v there replaced by i L here. Furthermore, i L = I is obviously a solution (a particular solution) to (3). Thus, considering equations (7), (17) and (18) of Section 3.1, with the time functions replaced by i L here, the solution to (3) may be written as i L = I A 1 e s 1t A 2 e s 2t [overdamping]... (4) i L = I e αt [B 1 cos ω d t B 2 sin ω d t] [underdamping]... (5) i L = I D 1 te αt D 2 e αt [critical damping]... (6) 9
Of course, the A s, B s and D s here are arbitrary constants whose values won t be the same as those in the voltage equations we developed earlier. They may be found, in each case here from simultaneously solving the system which results from i L (0) and di L (0)/dt. If there is an initial stored energy that is, if there is in the above analysis a nonzero initial value for i L this will affect the solutions for the constants. See Example 8.10, page 354, of the text. By observation, we see that equations (4), (5) and (6) take the general form { } function of the same form g(t) = G f... (7) as the natural response where g(t) may be current or voltage and G f is the final value of the current or voltage response. Step Response Example:(Drill exercise 8.6, page 355 of text.) In the circuit shown, R = 500 Ω, L = 0.64 H, C = 1 µf, I 0 = 0.5 A (this is the initial current through the inductor), V 0 = 40 V (this is the initial voltage), and I = 1 A. We find various circuit parameters and responses below. I 10
3.3 Natural and Step Response of a Series RLC Circuit Natural Response In this section there are no new mathematical forms the equations look similar to those already seen. Consider the following series RLC circuit. At the instant the analysis begins, the initial current through the inductor is I 0 and the initial voltage across the capacitor is V 0. In general, the current in the circuit is i. Applying KVL around the path we have Ri L di dt 1 C t Differentiating again and dividing by L gives 0 idt V 0 = 0... (1)... (2) This looks the same as equation (10) page 3 of this unit with v replaced by i and 1/RC replaced by R/L. The solution to equation (2) is therefore of the form and while i(t) = Ae st α =... (3) ω 0 =... (4) As before, s 1 = α α 2 ω0 2 s 1 = α α 2 ω0 2... (5a)... (5b) 11
Note that the natural frequency of the series RLC circuit is the same as that for the parallel RLC circuit.the three possible solutions again take the form i = A 1 e s 1t A 2 e s 2t [overdamping]... (6) i = e αt [B 1 cos ω d t B 2 sin ω d t] [underdamping]... (7) i = D 1 te αt D 2 e αt [critical damping]... (8) Once i is known it is straightforward to calculate the voltage v across any element since v R = ir ; v L = L di dt ; i = C dv c dt... (9a,b,c). We ll consider an example following the step response analysis below. Step Response Consider the following circuit in which the switch has been open for a long time and then is abruptly closed at t = 0. The source voltage is V. Applying KVL around the loop gives V =... (10) Since the current is related to the capacitor voltage via (9c), equation (10) becomes and dividing by LC gives d 2 v c dt 2 R L dv c dt v c LC = V LC... (11) Equation (11) has exactly the same form as equation (3) of Section 3.2 with v c replacing i L, R/L replacing 1/RC, and V replacing I. It therefore follows that the three possible solutions are: v c = V A 1 e s 1t A 2 e s 2t [overdamping]... (4) v c = V e αt [B 1 cos ω d t B 2 sin ω d t] [underdamping]... (5) v c = V D 1 te αt D 2 e αt [critical damping]... (6) 12
Note that in all cases, the solutions take the general form given in equation (7) on page 10 of this unit. V is simply the final voltage across the capacitor here. Be sure to consider all text examples in Chapter 8 up to page 360. Example for Natural Response of a Series RLC Circuit: Problem 8.35, page 374 of the text. 13