Question

Consider the following table for an LTI system having an impulse response h(t).

Input	Output
x(t)	y(t)
$\frac{d}{dt}x\left(t\right)$	$-3y\left(t\right)+e^{-2t}u\left(t\right)$

where, $x\left(t\right)=2e^{-3t}u(t-1)$

Then the impulse response h(t) of the system is

• A. $\frac{e^{-3}}{2}{e}^{-2(t+1)}u(t+1)$
• B. $\frac{e^3}{2}{e}^{2(t+1)}u(t+1)$
• C. $\frac{e^{-3}}{2}{e}^{-2(t-1)}u(t-1)$
• D. $\frac{e^3}{2}{e}^{-2(t+1)}u(t+1)$

Answer 1

The correct option is D $\frac{e^3}{2}{e}^{-2(t+1)}u(t+1)$

System output is y(t) for the input x(t)

$\frac{d}{dt}\left[x\right(t\left)\right]=\frac{d}{dt}\left[2e^{-3t}u\right(t-1\left)\right]$

$=-3\left[2e^{-3t}u\right(t-1\left)\right]+2e^{-3t}\delta (t-1)$

$=-3x\left(t\right)+2e^{-3}\delta (t-1)$

$Fortheinput-3x\left(t\right)+2e^{-3}\delta (t-1)$

$Correspondingoutputofthesystemis-3y\left(t\right)+e^{-2t}u\left(t\right)$

$Herewecansay-3y\left(t\right)isoutputfortheinput-3x\left(t\right)and{e}^{-2t}u\left(t\right)isoutputforinput2e^{-3}\delta (t-1)$

$x_1\left(t\right)=2e^{-3}\delta (t-1)correspondingoutputis{y}_1\left(t\right)=e^{-2t}u\left(t\right)$

$\therefore H\left(s\right)=\frac{Y_1\left(s\right)}{X_1\left(s\right)}=\frac{\frac{1}{s+2}}{2e^{-3}{e}^{-s}}$

$H\left(s\right)=\frac{e^3}{2}\frac{e^s}{s+2}$

$h\left(t\right)=\frac{e^3}{2}{e}^{-2(t+1)}u(t+1)$