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Signals and systems electrical engineering miscellaneous

  1. 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
    1. 2[1 – e– 2t]u(t)
    2. 4[e– t – e– 2t]u(t)
    3. sin 2t
    4. (1 – 4e– 4t)u(t)
Correct Option: A

hs(t) = 4e– 2t u(t); H(s) =
4
s + 2

For unit step, Y(s) = H(s) X(s)
Y(s) =
4
s(s + 2)

=
2
=
2
......since [X(s) = 1/s]
5s + 2

y(t) = 2[1 – e– 2t] u(t)



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