Question #115073

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.

Expert's answer

As per the given question,

Frequency of the audio signal (fc)=20kHz(f_c)=20kHz

Frequency of the signal (fs)=100kHz(f_s)=100kHz

Amplitude of the carrier signal =10V=10V

Modulation index =0.8=0.8

modulating frequency fm=fs×10V0.8=100kHz×100.8=1250kHzf_m=\dfrac{f_s\times 10V}{0.8}=\dfrac{100kHz\times 10}{0.8}=1250kHz

Lower band frequency =fs+fc=120kHz=f_s+f_c =120kHz

upper band frequency =fsfc=100kHz20kHz=80kHz=f_s-f_c=100kHz-20kHz=80kHz

Hence the required ratio=120kHz80kHz=3:2=\dfrac{120kHz}{80kHz} =3:2



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