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

u(t)” option2=”4[e-t – e-2t]u(t)” option3=”sin2t” option4=”(1 – 4e-4t)u(t)” correct=”option1″]

The correct answer is A. 2[1 – e-2t]u(t).

The response of an initially relaxed linear constant parameter network to a unit impulse applied at t = 0 is 4e-2tu(t). This means that the network’s output is 4e-2t for all values of t after t = 0.

The response of this network to a unit step function is the integral of the response to a unit impulse. The integral of 4e-2t is 2(1 – e-2t). Therefore, the response of this network to a unit step function is 2(1 – e-2t)u(t).

Option B is incorrect because it does not include the factor of 2. Option C is incorrect because it is not a step function. Option D is incorrect because it is not a decaying exponential.