Question

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

$\begin{array}{cc}\text{Input}& \text{Output}\\ \text{x(t)}& \text{y(t)}\\ \frac{d}{dt}x\left(t\right)& -3y\left(t\right)+e^{-2t}u\left(t\right)\end{array}$

Then the value of $h\left(t\right){|}_{t=1}$ is ____(Answer upto three decimal places)

• A. 0.184

Answer 1

The correct option is A 0.184
$x\left(t\right)=2e^{-5t}.u(t-1)$





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

$=-3.2e^{-3t}.u(t-1)+2.e^{-3t}.8(t-1)$

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

For the input $-3x\left(t\right)+2e^{-3}\delta (t-1)$ corresponding output of the system is $-3y\left(t\right)+e^{-2t}u\left(t\right)$. Here we can say $-3y\left(t\right)$ is output for the input $-3x\left(t\right)$ and $e^{-2t}u\left(t\right)$ is output for the input $2e^{-3.\delta (t-1)}$.

$x_1\left(t\right)=2e^{-3}.\delta (t-1)$ corresponding
output $y_1\left(t\right)=e^{-2t}u\left(t\right)$

$\therefore$ $\text{Impulse response,}H\left(s\right)=\frac{Y_1\left(s\right)}{X_1\left(s\right)}$

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

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

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

$\therefore h\left(1\right)=\frac{e^3}{2}{e}^{-2\left(2\right)}u\left(2\right)=\frac{e^3.e^{-4}}{2}=\frac{e^{-1}}{2}=0.184$