Question

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

Answer 1

The frequency present in the modulated output is f = 140 kHz

Explanation:

Given data:

• Frequency of audio signal = 20 kHz
• Carrier frequency = 100 kHz
• Peak amplitude of carrier wave = 10 V
• Modulation index = 0.8

Solution:

Let's assume

E(max) = the maximum value of the envelope

E(min) = the minimum value of the envelope,

The modulation index is

μ =  E  (max)  − E(min)  /  E( max)  + E( min)

E(min)   = E(max)  1−μ  / 1+μ   = 1.11 V

Now calculate frequency present in the modulated output is

f = f  carrier  + 2⋅f  audio  

​f = 100 + 2.200

f = 140 kHz

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