GATE 2025 EE Q.42: solution and answer

Official answer: BSource: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Impulse response. The response of an LTI system to $\delta (t)$ is, by definition, its impulse response, so $h(t)=u(t)$. Convolution then gives $$y(t)={\int }_{-\infty }^{\infty }x(\tau )u(t-\tau )d\tau ={\int }_{-\infty }^{t}x(\tau )d\tau$$ so $\mathbf{S}$ is an ideal integrator.

2. BIBO stability test (CT LTI). A continuous-time LTI system is BIBO stable if and only if $h(t)$ is absolutely integrable. Here $${\int }_{-\infty }^{\infty }|h(t)|dt={\int }_0^{\infty }1dt=\infty$$ The integral diverges, so $\mathbf{S}$ is not BIBO stable. This rules out (A).

3. A bounded input that gives an unbounded output (supports B). Take $x(t)=u(t)$, which is bounded ($|x(t)|\le 1$). Then $$y(t)=u(t)\ast u(t)={\int }_0^{t}1d\tau =tu(t)$$ This ramp grows without limit as $t\to \infty$, so (B) is true. The same example rules out (C), because here the output is not bounded.

4. Rule out (D). The input $x(t)=\delta (t)$ gives $y(t)=u(t)$, which is bounded. The zero input also gives zero output. So the output is not always unbounded.

5. Only (B) is true.