Question

$\text{The output y(t) of a causal LTI system is related to the input x(t) as,}\frac{dy\left(t\right)}{dt}+2y\left(t\right)=x\left(t\right).$
$\text{If the frequency response of output is}\frac{e^{-j2\omega }}{2+j\omega }\text{, then the value of}x\left(t\right)att=2\text{is}$

• A. 0.5
• B. 4
• C. 1
• D. 2

Answer 1

The correct option is C 1

Given, Causal LTI system,
$\frac{dy\left(t\right)}{dt}+2y\left(t\right)=x\left(t\right)$

By using Fourier transform,
$j\omega Y\left(\omega \right)+2Y\left(\omega \right)=X\left(\omega \right)$

$Y\left(\omega \right)[2+j\omega ]=X\left(\omega \right)$

$Y\left(\omega \right)=\frac{X\left(\omega \right)}{2+j\omega }$

By comparing above equation with the given output frequency response.
$X\left(\omega \right)=e^{-j2\omega }$

We know that,
$\delta (t-t_0)\stackrel{F.T}{⟷}{e}^{-j\omega t_0}$

$\delta (t-2)\stackrel{F.T}{⟷}{e}^{-j2\omega }$

$\therefore x\left(t\right)=\delta (t-2)$

$\text{at}t=2;x\left(2\right)=\delta (t-2)=\delta \left(0\right)=1$