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Consider an LTI system with impulse response h(t) = e–5t u(t). If the output of the system is y(t) = e–3t u(t) then the input, x(t), is given by
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- e–3t u(t)
- 2–3t u(t)
- e–5t u(t)
- 2–5t u(t)
Correct Option: B
h(t) = e–5t u(t)
Laplace transform gives,
| H(s) = | ||
| s + 5 |
y(t) = e–3t u(t) – e–5t u(t)
Laplace transform gives,
| Y(s) = | - | = | |||
| s + 3 | s + 5 | (s + 3)(s + 5) |
| Then, X(s) = | = | = | |||
| H(s) | (s + 3)(s + 5) | s + 3 |
Inverse Laplace transform gives,
x(t) = L–1 X (s) = 2e–3t u(t)
