Question

d(t) Figure 1: Figure for Question 3 (b) (5 pts) Suppose H is an integrator (ie, ,nd C is a first order system with transfer function 2 Is the closed-loop system stable? Obtain the asymptotic value of the error e when z and d are steps, respectively z au and d-Au, with α and β positive constants. Justify your steps.

Answer 1

The closed-loop transfer function is given by Y(s)/X(s) = HG/(1+HG) = 1/(s*(s+2)) / (1+1/(s*(s+2)) ) = 1/(s^2+2*s+1)

The poles of the closed loop trasnsfer function are -1, -1 which are on the LHS of the jw axis. Therefore the system is stable.

based on the above block diagram the output in the Laplace domain is given by,

Y(s) = X(s) * 1/(s^2+2*s+1) + D(s) * (s^2+2*s)/(s^2+2*s+1)

for x(t) = au(t) , X(s) = a/s

for d(t) = bu(t) , D(s) = b/s

Y(s) = a/s * 1/(s^2+2*s+1) + b/s * (s^2+2*s)/(s^2+2*s+1)

asymptotic value of y(t) = Lim s->0 sY(s) = a + 0

steady state error = a - a = 0.