Answer to Question #104595 in Physics for Amra Musharraf

Question

20 kHz audio signal is amplitude modulated over a 100 kHz carrier frequency. The peak amplitude of carrier wave is 10 V. If the modulation index is 0.8, find out the frequencies present in the modulated output. Also calculate the ratio of maximum and the minimum amplitude of the envelope of modulated signal.

Answer 1

For the amplitude modulated signal

$$S=A_0(1+m\cdot cos(\Omega t))cos(\omega_0t)$$

initial condition

$A_M=10 V; m=0.8; \Omega=20kHz;\omega_0=100 kHz$

frequencies present in the modulated output

$\omega_1=\omega_0+\Omega; \omega_2=\omega_0-\Omega$

$\omega_1=100+20=120 kHz; \omega_2=100-20=80 kHz$

ratio of maximum and the minimum amplitude

$k=\frac{1+0.8}{1-0.8}=5$