e(s X(s E s(s+1)+K (s+10)+K s(s+10) X(s)82+10s+K Thus, using Eqn(4), E(s)for the unit step input s(t=u(t) is E s2+10s+K s(s+10)I +10s+K s2+10s+K so if K <0, we can see from Routh-Hurwitz criteria that E(s) has at least one of the poles in the right half pla the real part of the itive. hence. in th state tracking error diverges to oo Now, consider the case when K>0. Then, both poles of E(s)are in the left half plane and the order of the numerator is strictly less than that of the denominator (the latter condition is usually referred to as being strictly proper ). Thus we can apply the final value theorem to compute the steady state tracking error: lim e(t) s+10 +10s+K Thus, regardless of the value of K, as far as it is positive, e(oo)=0 (b) To compute the error signal to the ramp input, we can use Eqn(4)with now X(s= E s2+10s+Ks2 (s2+10s+K 2#s+1-E(s) X(s) = 1 − Y (s) X(s) � Eqn (3) K = 1 − s(s + 1) � Eqn (2) + K s(s + 10) = s(s + 10) + K E(s) s(s + 10) � = . (4) X(s) s2 + 10s + K Thus, using Eqn (4), E(s) for the unit step input x(t) = u(t) is: s(s + 10) E(s) = X(s) s2 + 10s + K s(s + 10) 1 = s2 + 10s + K s s + 10 = s2 + 10s + K , so if K → 0, we can see from Routh-Hurwitz criteria that E(s) has at least one of the poles in the right half plane, i.e, the real part of the pole is positive. Hence, in this case, the steady state tracking error diverges to ∗. Now, consider the case when K > 0. Then, both poles of E(s) are in the left half plane and the order of the numerator is strictly less than that of the denominator (the latter condition is usually referred to as being strictly proper) . Thus we can apply the final value theorem to compute the steady state tracking error: e(∗) = lim e(t) t�� = lim sE(s) s�0 s + 10 = lim s s�0 s2 + 10s + K = 0. Thus, regardless of the value of K, as far as it is positive, e(∗) = 0. (b) To compute the error signal to the ramp input, we can use Eqn(4) with now X(s) = s 1 2 : s(s + 10) 1 E(s) = 2 s2 + 10s + K s s + 10 = s(s2 + 10s + K) 10 10 s + 1 − 100 K + = − K K . s s ���� 2 + 10s + K � �� � E1(s) E2(s) 7