-
The closed-loop transfer function of a control system is given by
C(s) = 2(s - 1) R(s) (s + 2)(s + 1)
For a unit step input the output is—
-
- – 3e–2t + 4e– 1 – 1
- – 3e–2t – 4e–t + 1
- zero
- infinity
- – 3e–2t + 4e– 1 – 1
Correct Option: A
Given r (t) = 1 then R (s) = 1/s
= | ||
R(s) | (s + 2)(s + 1) |
C(s) = R(s). | |
s(s + 2)(s + 1) |
= | |
s(s + 2)(s + 1) |
C(s) = | + | + | |||
s | s + 1 | s + 2 |
A = | |s=0 = -1 | |
(s + 1)(s + 2) |
B = | |s=1 = 4 | |
s(s + 2) |
C = | |s=2 = -3 | |
s(s + 1) |
on putting the value of A, B, C we get
C(s) = | + | - | |||
s | s + 1 | s + 2 |
Taking inverse Laplace transform, we get
C (t) = – 1 + 4e– t – 3e– 2t