GATE 2020 EC Q.47: solution and answer

Official answer: 2Source: Official answer key, GATE 2020 EC (IIT Delhi)

1. Read $m(t)$ from the figure: it is $0$ for $t<1\text{s}$, rises linearly from $0$ to $10\text{V}$ over $1\le t\le 3$, stays at $10\text{V}$ for $3\le t\le 6$, falls linearly from $10$ to $0\text{V}$ over $6\le t\le 7$, and is $0$ afterwards.

2. Slopes of $m(t)$: rising edge $\frac{dm}{dt}=\frac{10}{3-1}=5\text{V/s}$; flat part $0$; falling edge $\frac{dm}{dt}=-\frac{10}{7-6}=-10\text{V/s}$. The maximum value of $m(t)$ is $10\text{V}$.

3. The instantaneous angular frequency is the time derivative of the total phase. For PM the phase is $1000\pi t+K_{p}m(t)$, so $${\omega }_{PM}(t)=1000\pi +K_p\frac{dm}{dt}.$$

4. For FM the phase is $1000\pi t+K_f{\int }_{-\infty }^{t}m(\tau )d\tau$, and the derivative of the integral is $m(t)$, so $${\omega }_{FM}(t)=1000\pi +K_{f}m(t).$$

5. Taking $K_p>0$ and $K_f>0$ (they are deviation constants), the PM frequency is highest where $dm/dt$ is largest, which is the rising edge: $${\omega }_{PM,\max }=1000\pi +5K_p.$$ The falling edge with $-10\text{V/s}$ only lowers the frequency.

6. The FM frequency is highest where $m(t)$ is largest, which is $10\text{V}$: $${\omega }_{FM,\max }=1000\pi +10K_f.$$

7. Equate the two highest frequencies: $1000\pi +5K_p=1000\pi +10K_f\Rightarrow 5K_p=10K_f$, so $$\frac{K_p}{K_f}=2\text{s}.$$ The unit is consistent: $K_p$ is in rad/V and $K_f$ in rad/s/V, so the ratio is in seconds.