Answer to Question #104205 in Physics for DD

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 $E_{max}$, the maximum value of the envelope, and $E_{min}$, the minimum value of the envelope, the modulation index is

$$\mu=\frac{E_{max}-E_{min}}{E_{max}+E_{min}},\\ \space\\ E_{min}=E_{max}\frac{1-\mu}{1+\mu}=1.11\text{ V}.$$

The frequency present in the modulated output is

$$f=f_{carrier}+2\cdot f_{audio}=100+2\cdot20=140\text{ kHz}.$$