GATE 2023 EE Q.63: solution and answer

Official answer: 80Source: Official answer key, GATE 2023 EE (IIT Kanpur)

1. Product-to-sum identity: $2\cos A\cos B=\cos (A+B)+\cos (A-B)$, so $$x(t)=\cos (240\pi t)+\cos (120\pi t)$$ The signal holds two tones, $f_1=120\text{Hz}$ and $f_2=60\text{Hz}$.

2. Nyquist check: $f_{\max }=120\text{Hz}$ needs $f_s\ge 240\text{Hz}$. Here $f_s=200\text{Hz}$, so the 120 Hz tone aliases.

3. Spectrum of the sampled signal: every tone $f_0$ reappears at $|k{f}_s\pm f_0|$ for integer $k$. For $f_0=60\text{Hz}$: $60,140,260,340,\dots \text{Hz}$. For $f_0=120\text{Hz}$: $80,120,280,320,\dots \text{Hz}$ (since $200-120=80$).

4. Ideal LPF, cut-off 100 Hz: it passes only the components below $100\text{Hz}$. Those are $60\text{Hz}$ and the alias at $80\text{Hz}$. The original $120\text{Hz}$ tone and every higher image are removed. No component sits exactly at 100 Hz, so the edge of the passband does not matter.

5. The filtered output is $\propto \cos (120\pi t)+\cos (160\pi t)$, so the highest frequency in it is $80\text{Hz}$.