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## Q. 1.1.1

The baseband signal m(t) shown in the figure is phasemodulated to generate the PM signal $\phi(t)=\cos \left(2 \pi f_{c} t+\right.$  km(t)). The time t on the x-axis in the figure is in milliseconds. If the carrier frequency is $f_{c}$ = 50 kHz and k = 10π, then the ratio of the minimum instantaneous frequency (in kHz) is to the maximum instantaneous frequency (in kHz) is _____ (rounded off to 2 decimal places). ## Verified Solution

\begin{aligned}\phi(t) &=\cos \left(2 \Pi f_{c} t+k m(t)\right) \\\theta_{i}(t) &=2 \Pi f_{c} t+k m(t) \\f_{i} &=\frac{1}{2 \Pi} \frac{d}{d t} \theta_{i}(t) \\&=\frac{1}{2 \Pi} \frac{d}{d t} \theta_{i}(t) \\&=\frac{1}{2 \Pi} \frac{d}{d t}\left[2 \Pi f_{c} t+k m(t)\right] \\&=\frac{1}{2 \Pi}\left[2 \Pi f_{c}+k\right] \frac{d}{d t} m(t) \\&=f_{c}+\frac{k}{2 \Pi} \frac{d}{d t} m(t) \\f_{c} &=50 KHz \\k &=10 \Pi\end{aligned}

\begin{aligned}&\therefore f_{\max }=50 k +\frac{10 \Pi}{2 \Pi} \cdot \frac{2}{1 ms }=60 k \left[\left.\frac{d m(t)}{d t}\right|_{\max }=2\right] \\&t_{\min }=50 k -\frac{10 \Pi}{2 \Pi} \cdot 1 \times 10^{3}=45 k \left[\left.\frac{d m(t)}{d t}\right|_{\max }=-1\right] \\&\frac{t_{\max }}{f_{\min }}=\frac{60 k }{45 k }=0.75\end{aligned}