GATE 2024 EC Q.21: solution and answer

Official answer: ASource: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Write the convolution integral for the original system: $$y(t)={\int }_{-\infty }^{\infty }h(\tau )x(t-\tau )d\tau$$

2. Write the output $z(t)$ of the new system. Its input is $x(0.5t)$ and its impulse response is $h(0.5t)$, so the shifted input is $x(0.5(t-\tau ))$: $$z(t)={\int }_{-\infty }^{\infty }h(0.5\tau )x(0.5(t-\tau ))d\tau ={\int }_{-\infty }^{\infty }h(0.5\tau )x(0.5t-0.5\tau )d\tau$$

3. Substitute $u=0.5\tau$, so $d\tau =2du$. The limits stay $-\infty$ to $\infty$ because the scale factor is positive: $$z(t)=2{\int }_{-\infty }^{\infty }h(u)x(0.5t-u)du$$

4. Compare with step 1. The integral is the original convolution with $t$ replaced by $0.5t$, so it equals $y(0.5t)$. Therefore $z(t)=2y(0.5t)$.

5. Frequency-domain check, using the time-scaling property $f(at)\leftrightarrow \frac{1}{|a|}F(\omega /a)$: $x(0.5t)\leftrightarrow 2X(2\omega )$ and $h(0.5t)\leftrightarrow 2H(2\omega )$, so $Z(\omega )=4X(2\omega )H(2\omega )=4Y(2\omega )$. Since $y(0.5t)\leftrightarrow 2Y(2\omega )$, this gives $Z(\omega )=2\mathcal{F}\{y(0.5t)\}$, which confirms $z(t)=2y(0.5t)$.

6. This matches option (A).