The response of an initially relaxed linear constant parameter

The response of an initially relaxed linear constant parameter network to a unit impulse applied at t = 0 is 4e^{– 2t}u(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)

h_s(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]
	5		s + 2

y(t) = 2[1 – e^{– 2t}] u(t)