Answer to Question #112050 in Electric Circuits for EEFDD

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

$μ=\frac{​Emax​−Emin​​}{Emax​+Emin}$

applying componendo and dividendo

$\frac{\mu+1}{\mu-1}=\frac{E_{max}}{E_{min}}$

$\frac{E_{max}}{E_{min}}=1.11$

The frequency present in the modulated output is $f=f_{carrier}​+2⋅f_{audio}​=100+2⋅20=140 kHz.$