GATE 2024 EE Q.37: solution and answer

Official answer: BSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Linearity (superposition). $x(\tau )$ appears only to the first power inside the integral, and integration is a linear operation. For the input $a{x}_1(t)+b{x}_2(t)$, $$e^{-t}{\int }_{-\infty }^t{e}^{\tau }[a{x}_1(\tau )+b{x}_2(\tau )]d\tau =a{y}_1(t)+b{y}_2(t).$$ The system is linear, which rules out (A).

2. Move $e^{-t}$ inside the integral. $t$ is a constant during the integration over $\tau$, so $$y(t)={\int }_{-\infty }^t{e}^{-(t-\tau )}x(\tau )d\tau .$$ The kernel depends on $t$ and $\tau$ only through the difference $t-\tau$.

3. Recognise a convolution. Take $h(t)=e^{-t}u(t)$. The factor $u(t-\tau )$ is zero for $\tau >t$, so it cuts the integral off at $\tau =t$: $$y(t)={\int }_{-\infty }^{\infty }x(\tau )e^{-(t-\tau )}u(t-\tau )d\tau =x(t)\ast h(t).$$ Any system that can be written as $y=x\ast h$ is linear and time-invariant.

4. Direct check of time invariance. Apply the delayed input $x(t-t_0)$: $$y_1(t)=e^{-t}{\int }_{-\infty }^t{e}^{\tau }x(\tau -t_0)d\tau .$$ Substitute $\sigma =\tau -t_0$, so $e^{\tau }=e^{t_0}{e}^{\sigma }$ and the upper limit becomes $t-t_0$: $$y_1(t)=e^{-t}{e}^{t_0}{\int }_{-\infty }^{t-t_0}{e}^{\sigma }x(\sigma )d\sigma =e^{-(t-t_0)}{\int }_{-\infty }^{t-t_0}{e}^{\sigma }x(\sigma )d\sigma =y(t-t_0).$$ Delaying the input only delays the output, so the system is time-invariant. This rules out (C).

5. Causality. The upper limit is $t$, so $y(t)$ uses only $x(\tau )$ for $\tau \le t$. Equivalently, $h(t)=e^{-t}u(t)=0$ for $t<0$. The system is causal, which rules out (D).

6. The system is linear and time-invariant. It is also causal, and BIBO stable because ${\int }_0^{\infty }{e}^{-t}dt=1$ is finite. Answer: (B).