Question #104595
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.
1
Expert's answer
2020-03-10T11:33:35-0400

For the amplitude modulated signal


S=A0(1+mcos(Ωt))cos(ω0t)S=A_0(1+m\cdot cos(\Omega t))cos(\omega_0t)

initial condition

AM=10V;m=0.8;Ω=20kHz;ω0=100kHzA_M=10 V; m=0.8; \Omega=20kHz;\omega_0=100 kHz

frequencies present in the modulated output

ω1=ω0+Ω;ω2=ω0Ω\omega_1=\omega_0+\Omega; \omega_2=\omega_0-\Omega

ω1=100+20=120kHz;ω2=10020=80kHz\omega_1=100+20=120 kHz; \omega_2=100-20=80 kHz

ratio of maximum and the minimum amplitude

k=1+0.810.8=5k=\frac{1+0.8}{1-0.8}=5


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