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## Q. 1.1.12

Consider an FM signal

$f(t)=\cos \left[2 \pi f_{c} t+\beta_{1} \sin 2 \pi f_{1} t+\beta_{2} 2 \pi f_{2} t\right]$

The maximum deviation of the instantaneous frequency from the carrier frequency $f_{c}$ is

(a) $\beta_{1} f_{1}+\beta_{2} f_{2}$                                     (b) $\beta_{1} f_{2}+\beta_{2} f_{1}$

(c) $\beta_{1}+\beta_{2}$                                           (d) $f_{1}+f_{2}$

## Verified Solution

Maximum frequency deviation is

$=\frac{1}{2 \pi} \cdot\left|\frac{d \phi}{d t}\right|_{\max }$

φ = Phase deviation

\begin{aligned}&=\left.\frac{1}{2 \pi}\left\{\beta_{1} 2 \pi f_{1} \cos 2 \pi f_{1} t+\beta_{2} 2 \pi f_{2} \cos 2 \pi f_{2} t\right\}\right|_{\max } \\&=\frac{1}{2 \pi}\left(2 \pi \beta_{1} f_{1}+2 \pi \beta_{2} f_{2}\right) \\&=\beta f_{1}+\beta f_{2}\end{aligned}

Hence, the correct option is (a).