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CBSE Class 12 Physics 2019 Delhi Set 1 Paper

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Question : 24 of 27
Marks: +1, -0
(a) If AA and BB represent the maximum and minimum amplitudes of an amplitude modulated wave, write the expression for the modulation index in terms of AA and BB.
(b) A message signal of frequency 20 kHz20\,\mathrm{kHz} and peak voltage 10 V10\,\mathrm{V} is used to modulate a carrier of frequency 2 MHz2\,\mathrm{MHz} and peak voltage of 15 V15\,\mathrm{V}. Calculate the modulation index. Why the modulation index is generally kept less than one?
Solution:  
(a) Obtaining the expression for modulation index in terms of AA and BB
(b) Calculation of μ\mu
Reason
We are given that
A=Ac+AmA = A_c + A_m
 and B=Ac−Am\text{ and } B = A_c - A_m
Ac=A+B2A_c = \frac{A+B}{2}
Am=A−B2A_m = \frac{A-B}{2}
∴μ=AmAc\therefore \mu = \frac{A_m}{A_c}
=A−BA+B= \frac{A-B}{A+B}
(b) We have
μ=AmAc\mu = \frac{A_m}{A_c}
=1015=23= \frac{10}{15} = \frac{2}{3}
μ\mu is kept less than one to avoid distortion.
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