el(t), the corresponding time signal to E1(s)is a scaled step. E2(s) has two poles and they remain in the left-half plane as long as K>0. In this case e2(t), the corresponding time signal to E2(s) will decay to 0 as t-00. Thus, e(oo)=k, i. e, the steady state error will not be On the other hand, if K <0. Then, E2(s)now has at least one of the poles in the right-half plane.Thus,e2(∞)→∞ast→∞; therefore e(o)→∞. (c)For this part, first we would like to get the transfer function from a (t) to y(t). Let's denote the output of Kf, af(t), then X(s)=KyX(s) 1 Y(s (s+10) r (s)+KsX(s-y(s) s(s+10) Y(s) (s+10 (K+Ks)X(s) (Kf+Ks) X 1 X(s) (s+10)+1 Since e(t)=r(t)-y(t), we can compute the Laplace transform of e(t), E(s)as follows E(s)=X(s-Y(s) E(s The input a(t) to the system is a ramp, so X(s=. Thus, E(s) is Kf+K 10s+1-(Kf+K) 82(s2+10s+1 +10s+1-(Kf+K) s(s2+10s+1) To be able to apply the final value theorem, we want to choose K,(s)and Ks(s)such that the poles of sE() are in the left-half plane. In addition, we need to choose K(s)and Ks() such lim SE(s)=0 o ensure that the stead state error to the ramp input is zero. Combining these two conditions we see that we need to make the numerator of sE(s)equal to s. From this conclusion alone, we only know that K(s)+Ks(s)=10s+ 1. Assuming that neither is 0, then we can choose� = . e1(t), the corresponding time signal to E1(s) is a scaled step. E2(s) has two poles and they remain in the left-half plane as long as K > 0. In this case e2(t), the corresponding time signal to E2(s) will decay to 0 as t ∩ ∗. Thus, e(∗) = 10 K , i.e., the steady state error will not be 0. On the other hand, if K → 0. Then, E2(s) now has at least one of the poles in the right-half plane. Thus, e2(∗) ∩ ∗ as t ∩ ∗; therefore e(∗) ∩ ∗. (c) For this part, first we would like to get the transfer function from x(t) to y(t). Let’s denote the output of Kf , xf (t), then Xf (s) = KfX(s). 1 Y (s) = s(s + 10)(Xf (s) + KsX(s) − Y (s)) � 1 � 1 1 + s(s + 10) Y (s) = s(s + 10)(Kf + Ks)X(s) 1 s(s+10) Y (Kf + Ks) (s) = X(s) 1 + 1 s(s+10) Y (s) Kf + Ks X(s) s(s + 10) + 1 Since e(t) = x(t) − y(t), we can compute the Laplace transform of e(t), E(s) as follows: E(s) = X(s) − Y (s) � Y (s) � E(s) = 1 − X(s) X(s). The input x(t) to the system is a ramp, so X(s) = s 1 2 . Thus, E(s) is: � Kf + Ks � 1 2 E(s) = 1 − s(s + 10) + 1 s s2 + 10s + 1 − (Kf + Ks) = s2(s2 + 10s + 1) ∩ sE(s) = s2 + 10s + 1 − (Kf + Ks) . s(s2 + 10s + 1) To be able to apply the final value theorem, we want to choose Kf (s) and Ks(s) such that the poles of sE(s) are in the left-half plane. In addition, we need to choose Kf (s) and Ks(s) such that lim sE(s) = 0 s�0 to ensure that the stead state error to the ramp input is zero. Combining these two conditions, we see that we need to make the numerator of sE(s) equal to s2. From this conclusion alone, we only know that Kf (s) + Ks(s) = 10s + 1. Assuming that neither is 0, then we can choose 8