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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 – t – 1
- – 3e– 2t – 4e– t + 1
- zero
- infinity
Correct Option: A
| = | ; R(s) = | |||
| R(s) | (s + 2)(s + 1) | s |
| ∴ C(s) = | ||
| (s + 1)(s + 2) |
Expanding in partial fractions
| = | + | + | |||
| s | s + 1 | s + 3 |
| ∴ k1 = | = - 1 | |
| 2 |
| k2 = | |s=-1 = | = 4 | ||
| s(s + 2) | - 1(1) |
| k3 = | |s=-2 = | = - 3 | ||
| s(s + 2) | - 2( - 2) |
| Hence C(s) = | + | - | |||
| s | s + 1 | s + 2 |
and output, c(t) = [– 1 + 4e– t – 3e– 2t] u(t).