sum is absolutely integrable. Because the ROC of this signal, d ht+5-n(t+ 6h(t)is the entire s-plane, there can be no poles except at oo. However, we know that H(s)has at least two poles. In order for the signal to have no poles, we must make sure that the poles of H(s)are cancelled by eros of the signal. Taking the Laplace transform of the signal gives us s2H(s)+5sH(s)+6(s)=(s+2)(s+3)H(s) Thus, H(s) can only have two poles, at s=-2 and s=-3 Combining all the information about the poles and zeros gives us H(s=K (s+2)(s+3) e{s}>-2. Finally, we choose K such that H(O)=3. That is, K=-2 and s-1 s) (s+2)(s+3) 界e{s}>-2. Problem 5 Consider the basic feedback system of Figure 11.3 (a)on p 819 of O&w. Determine the closed-loop system impulse response when 2 H(s)= G(s) We can use Black's Formula to compute the system function for the entire system, Q(s. This (s+5)(+2) (s+2) s2+7s+12 Next, we can use partial fraction expansion to write Q(s)as a sum of lst order terms Q(s) +78+12 (s+4)(s+3) Since this is a feedback system, we know it is causal. Thus, we find the inverse Laplace transform for the system function using the causal part to obtain the inpulse response as follows q(t)=2e u(t)-e-sutd2h(t) dh(t) sum is absolutely integrable. Because the ROC of this signal, + 5 + 6h(t) is the entire dt2 dt s-plane, there can be no poles except at ∼. However, we know that H(s) has at least two poles. In order for the signal to have no poles, we must make sure that the poles of H(s) are cancelled by zeros of the signal. Taking the Laplace transform of the signal gives us s2H(s) + 5sH(s) + 6H(s) = (s + 2)(s + 3)H(s). Thus, H(s) can only have two poles, at s = −2 and s = −3. Combining all the information about the poles and zeros gives us, H(s) = K s − 1 (s + 2)(s + 3), √e{s} > −2. Finally, we choose K such that H(0) = 1 3 . That is, K = −2 and H(s) = −2 s − 1 (s + 2)(s + 3), √e{s} > −2. Problem 5 Consider the basic feedback system of Figure 11.3 (a) on p.819 of O&W. Determine the closed-loop system impulse response when 1 2 H(s) = , G(s) = . s + 5 s + 2 We can use Black’s Formula to compute the system function for the entire system, Q(s). This is given by: 1 Q(s) = s+5 1 + 2 (s+5)(s+2) = (s + 2) s2 + 7s + 12 Next, we can use partial fraction expansion to write Q(s) as a sum of 1st order terms. s + 2 Q(s) = s2 + 7s + 12 s + 2 = (s + 4)(s + 3) 2 1 = s + 4 − s + 3 Since this is a feedback system, we know it is causal. Thus, we find the inverse Laplace transform for the system function using the causal part to obtain the inpulse response as follows: −3t q(t) = 2e−4t u(t) − e u(t) 10