GATE 2024 IN Q.48: solution and answer

Official answer: ASource: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. The received signal $y(t)=\cos \left(2\pi f_{1}t+\beta \cos (2\pi f_{2}t)\right)$ is an angle-modulated carrier. Its instantaneous phase is ${\theta }_i(t)=2\pi f_{1}t+\beta \cos (2\pi f_{2}t)$, so $\beta$ is the peak phase deviation in radians.

2. Differentiate the phase to get the instantaneous frequency: $$f_i(t)=\frac{1}{2\pi }\frac{d{\theta }_i}{dt}=f_1-\beta f_2\sin (2\pi f_{2}t)$$ The peak deviation from the carrier is therefore $\Delta f=\beta f_2$, which gives $\beta =\frac{\Delta f}{f_2}$.

3. For frequency modulation with sensitivity $k_f=500\text{Hz/V}$, the frequency deviation is set by the peak message amplitude. Here $A_m=0.5\text{V}$, since $m(t)=0.5\sin (2\pi f_{2}t)$.

4. Frequency deviation: $$\Delta f=k_f{A}_m=500\times 0.5=250\text{Hz}$$ so the instantaneous frequency swings $\pm 250$ Hz about $f_1=40$ kHz.

5. Modulation index: $$\beta =\frac{\Delta f}{f_2}=\frac{250}{1000}=0.25$$

6. Check: with $\beta =0.25$, the peak of $\beta f_2\sin (2\pi f_{2}t)$ is $0.25\times 1000=250$ Hz, which agrees with $k_f{A}_m$. The sign of the sinusoidal term depends on how $m(t)$ and the phase term are paired, but the peak deviation does not.

7. Result: $\beta =0.25$ and $\Delta f=\pm 250\text{Hz}$, which is option (A).