Question

A continuous time LTI system is described by $\frac{d^{2}y\left(t\right)}{d{t}^2}+4\frac{dy\left(t\right)}{dt}+3y\left(t\right)=2\frac{dx\left(t\right)}{dt}+4x\left(t\right)$
Assuming zero initial conditions, the response y(t) of the above system for the input $x\left(t\right)=e^{-2t}u\left(t\right)$ is given by

• A. $(e^{-t}-e^{-3t})u\left(t\right)$
• B. $(e^t-e^{3t})u\left(t\right)$
• C. $(e^{-t}+e^{-3t})u\left(t\right)$
• D. $(e^t+e^{3t})u\left(t\right)$

Answer 1

The correct option is A $(e^{-t}-e^{-3t})u\left(t\right)$

$\frac{d^{2}y\left(t\right)}{d{t}^2}+\frac{4dy\left(t\right)}{dt}+3y\left(t\right)=\frac{2dx\left(t\right)}{dt}+4x\left(t\right)$

Taking Laplace transform on both sides (assuming zero initial conditions),
$s^{2}Y\left(s\right)+4sY\left(s\right)+3Y\left(s\right)=2sX\left(s\right)+4X\left(s\right)$

or $\frac{Y\left(s\right)}{X\left(s\right)}=\frac{2s+4}{s^2+4s+3}$

$=\frac{2(s+2)}{(s+1)(s+3)}$

Given that, $x\left(t\right)=e^{-2t}u\left(t\right)$

$X\left(s\right)=\frac{1}{s+2}$

$Y\left(s\right)=\frac{2(s+2)}{(s+1)(s+3)(s+2)}$

$=\frac{2}{(s+1)(s+3)}$

$=\frac{1}{s+1}-\frac{1}{s+3}$

Taking inverse Laplace transform on both sides,
$y\left(t\right)=(e^{-t}-e^{-3t})u\left(t\right)$