\-. [)fe-~ LDO ~ (puct)

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-';.' 122 System Responses Chapter 4 The inverse Laplace transform is not derived here (see Problem 4.8); however, assuming for the moment that the poles of G(s) are complex, the result is I -Coo (.

5 -';.' 122 System Responses Chapter 4 The inverse Laplace transform is not derived here (see Problem 4.8); however, assuming for the moment that the poles of G(s) are complex, the result is I -Coo (. (A e c(t) = I- ~e - sm "'(Ont + ) (4-20) where p = Jl- S2 and e = tan-i (~/S). In this response, 1: = 1IS(On is the time constant of the exponentially damped sinusoid in seconds (we can usually ignore this term after approximately four time constants). Also, l3(on is the frequency of the damped sinusoid. We wish now to show typical step responses for a second-order system. The step response given by (4-20) is a function of both S and (On' Ifwe specify S, we still cannot plot c(t) without specifying (On- To simplify the plots, we give c(t) for a specified S as a function of (Ont. A family of such curves for various values of S is very useful and is given in Figure 4.4 for 0 ~ S ~ 2. Note that for 0 < S < 1, the response is a damped sinusoid. For S = 0, the sinusoid is undamped, or of sustained amplitude. For S 2: 1, the oscillations have ceased. It is apparent from (4-20) that for S < 0, the response grows without limit. We consider only the case that I;; 2: 0 in this chapter. A MATLAB program that calculates some of the step responses of Figure 4.4 is given by zeta = [ ), for k = 1:4 G =tf ([1),[1 2*zeta(k) I); step(g) hold on end hold off The two poles of the transfer function G(s) in (4-18) occur at s = -~(On±jro,.~ For I;; > 1, these poles are real and unequal, and the damped sinusoid portion of c(t) is replaced by the weighted sum of two exponential functions; that is, (4-21 ) where 1:1 = l/ (S (On + (OnJr/ -1 ), 1:2 = 11(I;;(On-(OnJl;;2 - I ) are the two system time constants. For S = I, the poles of G(s) are real and equal, so that cet) = l+kle-li1+k2te-llt,1:= I/<.On For 0 < 1;; < 1, the system is said to be underdamped, and for 1;; = 0 it is said to be undamped. For I;; = I, the system is said to be critically damped, and for I;; > 1, the system is overdamped. For a linear time-invariant system, C(s) = G(s)R(s) For the case that r(t) is a unit impulse function, R(s) = I and e(t) = -I[G(S)] = get) (4-22) (4-23) where g (t) is the unit impulse response, or weighting function, of a system with the transfer function G(s). Then, by the convolution integral (see Appendix B), for a general '1'- t ma,#t

'''-' Section 4.2 Time Response of Second-order Systems 123 c(t) 2.0-1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 I I, ~ 0 2 4 6 10 12 wnt Figure 4.4 Step response for second-order system (4-20).

6 '''-' Section 4.2 Time Response of Second-order Systems 123 c(t) I I, ~ wnt Figure 4.4 Step response for second-order system (4-20). input r(t), eel) = {g("t)r(t - or) dt (4-24) from (4-22). [In (4-24), or is the variable of integration and is not related to the time constant.] Hence, all response information for a general input is contained in the impulse response g(t). Recall also from Section 4.1 that an initial condition on a first-order system can be modeled as an impulse function input. While the initial condition excitation of higher-order systems cannot be modeled as simply as that of the first-order system, the impulse response of any system does give an indication of the nature of the initial-condition response, and thus the transient response, of the system. The unit-impulse response of the second-order system (4-18) is given in Figure 4.5. This figure is a plot of the function c(t) = -I( ffi~ ) = ffine-{w.lsin/3ffinl = get) (4-25) s2 + 2i;ffi n S + ffi~ f3 Compare Figure 4.4 with Figure 4.5 and note the similarity of the information. In fact, the unit impulse response of a system is the derivative of the unit step response (see Problem 4.9). The impulse response of the second-order system can also be considered to be the response to certain initial conditions, with r(t) = 0 (see Problem 4.9). of

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