GATE 2023 EC Q.50: solution and answer

Official answer: DSource: Official answer key, GATE 2023 EC (IIT Kanpur)

1. From the figure, $y(t)=x_1(t)\cos (4\pi \times 10^{3}t)+x_2(t)\cos (12\pi \times 10^{3}t)$. Each $x_i(t)$ is low-pass with bandwidth $B$, so $X_i(\omega )=0$ for $|\omega |>4\pi \times 10^3$ rad/s.

2. Modulation property: $x(t)\cos ({\omega }_{c}t)\leftrightarrow \frac{1}{2}\left[X(\omega -{\omega }_c)+X(\omega +{\omega }_c)\right]$. Multiplying by the cosine moves a copy of the spectrum to $\pm {\omega }_c$, and each copy still extends $B$ either side of its centre.

3. Top branch, ${\omega }_c=4\pi \times 10^3$ rad/s: the copies occupy $0\le |\omega |\le 4\pi \times 10^3+4\pi \times 10^3=8\pi \times 10^3$ rad/s.

4. Bottom branch, ${\omega }_c=12\pi \times 10^3$ rad/s: the copies occupy $12\pi \times 10^3-4\pi \times 10^3\le |\omega |\le 12\pi \times 10^3+4\pi \times 10^3$, that is $8\pi \times 10^3\le |\omega |\le 16\pi \times 10^3$ rad/s.

5. The adder adds the two spectra (the Fourier transform is linear). Together they fill $0\le |\omega |\le 16\pi \times 10^3$ rad/s with no gap, so $y(t)$ is a low-pass signal whose highest frequency is ${\omega }_{\max }=16\pi \times 10^3$ rad/s.

6. Nyquist sampling frequency for a low-pass signal: $${\omega }_s=2{\omega }_{\max }=2\times 16\pi \times 10^3=32\pi \times 10^3\text{rad/s}$$

7. This is option (D).