Question

The response of an initially relaxed linear constant parameter network to a unit impulse applied at $t=0$ is $4e^{-2t}u\left(t\right).$
The response of this network to unit step function will be

• A.
  $2[1-e^{-2t}]u\left(t\right)$
• B.
  $(1-4e^{-4t})u\left(t\right)$
• C.
  $4[e^{-t}-e^{-2t}]u\left(t\right)$
• D.
  Sin 2t

Answer 1

The correct option is A

$2[1-e^{-2t}]u\left(t\right)$
Given that, $h\left(t\right)=4e^{-2t}$
So, $h\left(s\right)=4\frac{1}{s+2}$

Given that
$x\left(t\right)=u\left(t\right)$
So, $X\left(s\right)=\frac{1}{s}$

We know that,
$H\left(s\right)=\frac{Y\left(s\right)}{X\left(s\right)}$

$Y\left(s\right)=H\left(s\right)X\left(s\right)$

$Y\left(s\right)=\frac{4}{s+2}\times \frac{1}{s}=2[\frac{1}{s}-\frac{1}{s+2}]$

Taking inverse Laplace transform on both sides.
$y\left(t\right)=2[1-e^{-2t}]u\left(t\right)$