Mumbai University > Electronics > Sem 4 > Priniciples of Communication Engineering

Given :

Equation of FM wave, efm(t) = 20 $cos(16 \pi 10^6 t + 25sin(2 \pi 10^3 t))$

Solution :

Standard equation of an FM wave is given as,

efm(t) = $A_c . cos(2 \pi f_c t + \beta sin ( 2 \pi f_m t))$

By comparing the above two equations, we get,

Amplitude of the carrier signal, $A_c$ = 20 V

Frequency of carrier signal , $f_c$ = 8 x $10^6$ Hz = 8 MHz

Frequency of the message signal $f_m$ = 1 x $10^3$ Hz = 1KHz

Modulation Index , $\beta$ = 25

But, $ \beta f = \frac{\Delta f}{f m} $

i.e. $ \Delta f=\beta \mathrm{fm} $

$\Delta f$ = 25 x 1KHz = 25KHz

Frequency deviation , $ \Delta f$ is 25KHz

Power of FM wave is $P_c = \frac{Ac^2}{2R}$

Assume, $ R=1 \Omega $ and substituting the value of $A_c$

$P_c$ = $\frac{(20)^2}{2(1)}$ = 200 W

Power of FM wave is 200 Watts.