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The response of an initially relaxed linear constant parameter network to a unit impulse applied at t = 0 is 4e– 2tu(t). The response of this network to a unit step function will be
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- 2[1 – e– 2t]u(t)
- 4[e– t – e– 2t]u(t)
- sin 2t
- (1 – 4e– 4t)u(t)
Correct Option: A
hs(t) = 4e– 2t u(t); H(s) = | ||
s + 2 |
For unit step, Y(s) = H(s) X(s)
Y(s) = | ||
s(s + 2) |
= | = | ......since [X(s) = 1/s] | ||
5 | s + 2 |
y(t) = 2[1 – e– 2t] u(t)