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The transfer function of a linear time invariant system is given as
G(s) = 1 s2 + 3s + 2
The steady state value of the output of this system for a unit impulse input applied at time instant t = 1 will be
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- 0
- 0.5
- 1
- 2
Correct Option: B
| ∴ G(s) = | = | ||
| U(s) | s2 + 3s + 2 |
| ∴ y(s) = | U(s) | |
| s2 + 3s + 2 |
But U(t) = δ (t – 1)
and U(s) = e– s
| ∴ y(s) = | ||
| s2 + 3s + 2 |
For steady state,
| = | = | = 0.5 | |||
| s → 0 | s2 + 3s + 2 | 2 |
