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The impulse response of a continuous time system is given by h(t) = δ(t – 1) + δ(t – 3). The value of the step response at t = 2 is
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- 0
- 1
- 2
- 3
Correct Option: B
Apply Laplace transform
h(s) = e –s + e –3s
for input step voltage →
| y(s) = h(s) | ||
| s |
| = [e–s + e–3s] | ||
| s |
y(t) = u(t – 1) + u(t – 3)
u(t) = 1 for t 30 = 0 prt < 0
∴ O/P in y(2) u(t – 1) + u(t – 3) = (4 – 1)
= 4(z – 1) + 4 (2 – 3)
= 4(1) + (4 – 1) = 0 + 1 = 1
