A message signal is super-imposed with a carr | vedclass

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(D) The standard form of an amplitude-modulated signal is given by:$C_m(t) = A_c \sin(\omega_c t) + \frac{\mu A_c}{2} \sin((\omega_c + \omega_m)t) - \

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$\frac{2 A_2}{A_1}, \frac{\omega_3-\omega_2}{2}$

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(D) The standard form of an amplitude-modulated signal is given by:$C_m(t) = A_c \sin(\omega_c t) + \frac{\mu A_c}{2} \sin((\omega_c + \omega_m)t) - \frac{\mu A_c}{2} \sin((\omega_c - \omega_m)t)$.Comparing this with the given equation $C_m(t) = A_1 \sin(\omega_1 t) + A_2 \sin(\omega_2 t) - A_2 \sin(\omega_3 t)$:We identify the carrier amplitude $A_c = A_1$ and carrier frequency $\omega_c = \omega_1$.The sideband frequencies are $\omega_c + \omega_m = \omega_3$ and $\omega_c - \omega_m = \omega_2$.Subtracting these two equations: $(\omega_c + \omega_m) - (\omega_c - \omega_m) = \omega_3 - \omega_2$,which gives $2\omega_m = \omega_3 - \omega_2$,so $\omega_m = \frac{\omega_3 - \omega_2}{2}$.Comparing the amplitudes of the sidebands: $\frac{\mu A_c}{2} = A_2$.Substituting $A_c = A_1$,we get $\frac{\mu A_1}{2} = A_2$,which implies $\mu = \frac{2 A_2}{A_1}$.Thus,the modulation index is $\frac{2 A_2}{A_1}$ and the angular frequency of the message signal is $\frac{\omega_3 - \omega_2}{2}$.

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