GATE 2020 EC Q.42: solution and answer

Official answer: BSource: Official answer key, GATE 2020 EC (IIT Delhi)

1. The demodulator multiplies $x(t)$ by a local carrier that is 40 Hz away from $f_c$:
   $$v(t)=4\cos (1000\pi t)\cos (2\pi f_{c}t)\cos (2\pi (f_c+40)t).$$

2. Apply $\cos A\cos B=\frac{1}{2}[\cos (A-B)+\cos (A+B)]$ to the two carrier factors:
   $$\cos (2\pi f_{c}t)\cos (2\pi (f_c+40)t)=\frac{1}{2}\left[\cos (80\pi t)+\cos (2\pi (2f_c+40)t)\right].$$

3. The second term is centred near $2f_c\approx 2\text{MHz}$. It lies far above the 510 Hz cut-off of the ideal LPF, so it is removed, together with its sidebands at about $2\text{MHz}\pm 500\text{Hz}$. After the filter, only the low-frequency part of the product remains:
   $$4\cos (1000\pi t)\cdot \frac{1}{2}\cos (80\pi t)=2\cos (1000\pi t)\cos (80\pi t).$$

4. Expand the product of cosines:
   $$2\cos (1000\pi t)\cos (80\pi t)=\cos (1080\pi t)+\cos (920\pi t).$$

5. The two terms have frequencies $\frac{1080\pi }{2\pi }=540\text{Hz}$ and $\frac{920\pi }{2\pi }=460\text{Hz}$.

6. The LPF has unit gain for $|f|\le 510\text{Hz}$ and zero gain beyond. The 460 Hz term lies inside the passband and is passed. The 540 Hz term lies outside and is blocked.

7. Hence $y(t)=\cos (920\pi t)$, which is option (B).