GATE 2022 IN Q.18: solution and answer

Official answer: CSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

1. Write the operation as a compression followed by a delay: $y(t)=x\left(2\left(t-\frac{5}{2}\right)\right)$. Let $g(t)=x(2t)$, so $y(t)=g\left(t-\frac{5}{2}\right)$.

2. Time-scaling property: if $x(t)\leftrightarrow X(f)$, then $x(at)\leftrightarrow \frac{1}{|a|}X\left(\frac{f}{a}\right)$. With $a=2$: $G(f)=\frac{1}{2}X\left(\frac{f}{2}\right)$.

3. Time-shift property: a delay only multiplies the spectrum by a phase factor, $Y(f)=G(f)e^{-j2\pi f(5/2)}=\frac{1}{2}X\left(\frac{f}{2}\right)e^{-j5\pi f}$. Since $\left|e^{-j5\pi f}\right|=1$, the delay does not change where the spectrum is zero or non-zero.

4. Band-limited between 100 Hz and 200 Hz means $X(f)=0$ unless $100\le |f|\le 200\text{Hz}$. So $Y(f)$ can be non-zero only where $100\le \left|\frac{f}{2}\right|\le 200$, i.e. $200\le |f|\le 400\text{Hz}$.

5. Compressing in time by a factor of 2 stretches the spectrum by a factor of 2, so both band edges double and $y(t)$ is band-limited between 200 Hz and 400 Hz. This is option (C). (B) would hold only for a pure delay, (A) is what an expansion $x(t/2)$ would give, and (D) is wrong because scaling and shifting never create frequencies outside the stretched band.