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CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

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Question : 18 of 27
Marks: +1, -0
A signal of low frequency fmf_m is to be transmitted using a carrier wave of frequency fcf_c. Derive the expression for the amplitude modulated wave and deduce expressions for the lower and upper sidebands produced. Hence, obtain the expression for modulation index.
Solution:  
Derive expression for amplitude modulated wave.
Deducing expression for lower and upper side bands.
Obtaining expression for modulation index.
Let a carrier wave be given by
c(t)=Acsinωctc(t)=A_c \sin \omega_c t where ωc=2πfc\omega_c = 2 \pi f_c
And signal wave be
m(t)sinωmtm(t) \sin \omega_m t where ωm=2πfm\omega_m = 2 \pi f_m
The modulated signal is
cm(t)=(Ac+Amsinωmt)sinωctc_m(t)=(A_c + A_m \sin \omega_m t) \sin \omega_c t
cm(t)=Ac(1+AmAcsinωmt)sinωctc_m(t)=A_c \left(1+\frac{A_m}{A_c} \sin \omega_m t\right) \sin \omega_c t
cm(t)=Acsinωct+μAc2cos(ωcωm)tc_m(t)=A_c \sin \omega_c t + \mu \frac{A_c}{2} \cos (\omega_c - \omega_m) t μAc2cos(ωc+ωm)t- \mu \frac{A_c}{2} \cos (\omega_c + \omega_m) t
The modulation index μ=AmAc\mu = \frac{A_m}{A_c}
Lower frequency band ωcωm\omega_c - \omega_m
Upper frequency band ωc+ωm\omega_c + \omega_m
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