GATE 2022 EC Q.33: solution and answer

Official answer: 0.0Source: Official answer key, GATE 2022 EC (IIT Kharagpur)

1. $x_2(t)=u(t)-u(t-2)$ is a rectangular pulse of height 1 on $0\le t<2$. Keep the pulse fixed and slide the exponential: $y(t)={\int }_{-\infty }^{\infty }{x}_2(\tau )x_1(t-\tau )d\tau ={\int }_0^2{e}^{-(t-\tau )}u(t-\tau )d\tau$.

2. For $t>2$, every $\tau$ in $[0,2]$ satisfies $\tau <t$, so $u(t-\tau )=1$ over the whole pulse and $$y(t)=e^{-t}{\int }_0^2{e}^{\tau }d\tau =e^{-t}\left(e^2-1\right),t>2.$$

3. As $t\to \infty$, $e^{-t}(e^2-1)\to 0$, so ${\lim }_{t\to \infty }y(t)=0$.

4. Check with the final value theorem: $Y(s)=X_1(s)X_2(s)=\frac{1}{s+1}\cdot \frac{1-e^{-2s}}{s}$.

5. $sY(s)=\frac{1-e^{-2s}}{s+1}$ has its only pole at $s=-1$, so it has no poles in the closed right half-plane and the theorem applies: ${\lim }_{s\to 0}sY(s)=\frac{1-1}{1}=0$.

6. Physically, a stable system with impulse response $e^{-t}u(t)$ is driven by an input that switches off at $t=2$, so its output decays to zero. Answer: $0.0$.