ISRO EC 2025 Q.7: solution and answer

Official answer: CSource: ISRO's final answer key of 15 May 2026

1. Use the standard transform pair for a sinc pulse: $\frac{\sin (Wt)}{\pi t}⟷X(j\omega )=\{\begin{array}{ll}1,& |\omega |<W\\ 0,& |\omega |>W\end{array}$. This follows because the inverse transform of a unit rectangle on $|\omega |<W$ is $\frac{1}{2\pi }{\int }_{-W}^W{e}^{j\omega t}d\omega =\frac{\sin (Wt)}{\pi t}$.

2. Here $W=100\pi \text{rad/s}$, so $X(j\omega )=1$ for $|\omega |<100\pi$ and $X(j\omega )=0$ for $|\omega |>100\pi$.

3. The spectrum is exactly zero beyond $\omega =100\pi$ (that is, $f_{\max }=50\text{Hz}$), so the signal is band limited.

4. In the time domain, $\frac{\sin (100\pi t)}{\pi t}$ is non-zero except at the isolated points $t=k/100$ ($k\ne 0$), and its envelope only decays as $1/|t|$. It never becomes identically zero beyond any finite time, so it is not time limited.

5. A non-zero signal cannot be both time limited and band limited, so option (a) is impossible. Option (b) is wrong because the spectrum is bounded. Option (d) is wrong because the signal is band limited.

6. Hence the signal is not time limited but band limited: option (c).