Kf(s)=10s and Ks =1. Then, SE(s +10s+1-(10s+1) s(s2+10s+1) (s2+10s+1) 2+10s+ Now, we can safely apply the final value theorem to obtain the steady state error and we indeed obtain o as lim SE(s= lim Problem 3(O&W 11.27) Given H()=2+25+4Cs)=K First, let's identify the poles and zeros of H(s). Since the system H(s) is a second order system there are two poles and two zeros. It is clear that one of the zeros is of finite at -2 and the other is at oo. The poles are the solutions of s+2s+4=0, s0-1++V3. Thus, the pole-zero diagram is as shown below Sm Since H(s)is causal, the ROC for H(s)is Res)>-1 The closed loop poles are the solution to the following equation 1+KH(s)=1+KKf (s) = 10s and Ks = 1. Then, s2 + 10s + 1 − (10s + 1) sE(s) = s(s2 + 10s + 1) 2s = s(s2 + 10s + 1) s � sE(s) = . s2 + 10s + 1 Now, we can safely apply the final value theorem to obtain the steady state error and we indeed obtain 0 as expected: s lim sE(s) = lim = 0. s�0 s�0 s2 + 10s + 1 Problem 3 (O&W 11.27) Given: s + 2 H(s) = s2 + 2s + 4 , G(s) = K. First, let’s identify the poles and zeros of H(s). Since the system H(s) is a second order system, is at ∗. The poles are the solutions of s is as shown below: − 2 s − ±⇒ there are two poles and two zeros. It is clear that one of the zeros is of finite at 2 and the other + 2 + 4 = 0, so 1 + 3. Thus, the pole-zero diagram ≤m ↓e × × ⇒3 − ⇒ 3 −2 −1 Since H(s) is causal, the ROC for H(s) is ↓e{s} > −1. The closed loop poles are the solution to the following equation: s + 2 1 + KH(s) = 1 + K = 0, s2 + 2s + 4 9