Sn Problem 2 (a) First let output of the plant G(s) s+10 be y(t. Then, by Black's formula, the clos transfer function from the input a(t)to y(t) can be found as Y G(sK(s 1+G(s)K(s) X(s)8(s+10)+K The error signal e(t)is the difference between r(t) and y(t). Thus from the linearity of Laplace rIms e(t)= (t-y(t) E(s (s)-Y(s) So, with Eqns(2) and 3), we can obtain the transfer function from the input (t) to the error× ≤m 2 −2 −1 ↓e Problem 2 1 (a) First let output of the plant G(s) = s(s+10) be y(t). Then, by Black’s formula, the closed transfer function from the input x(t) to y(t) can be found as: Y (s) = G(s)K(s) X(s) 1 + G(s)K(s) K = s(s+10) 1 + K s(s+10) � Y (s) X(s) = K s(s + 10) + K . (2) The error signal e(t) is the difference between x(t) and y(t). Thus from the linearity of Laplace transforms: e(t) = x(t) − y(t) E(s) = X(s) − Y (s). (3) So, with Eqns (2) and (3), we can obtain the transfer function from the input x(t) to the error e(t): 6