The modulation index used in amplitude modula | vedclass

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(B) Initially, the modulation index $\mu_1 = 0.5$ and the message signal amplitude $A_m = 10 \,V$.Using the formula for modulation index, $\mu = \frac

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Reduced by $7.5 \,V$

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(B) Initially, the modulation index $\mu_1 = 0.5$ and the message signal amplitude $A_m = 10 \,V$.Using the formula for modulation index, $\mu = \frac{A_m}{A_c}$, we find the initial carrier amplitude $A_{c_1} = \frac{A_m}{\mu_1} = \frac{10}{0.5} = 20 \,V$.When the modulation index is changed to $\mu_2 = 0.8$ while keeping $A_m = 10 \,V$ constant, the new carrier amplitude is $A_{c_2} = \frac{A_m}{\mu_2} = \frac{10}{0.8} = 12.5 \,V$.The change in carrier amplitude is $\Delta A_c = A_{c_1} - A_{c_2} = 20 \,V - 12.5 \,V = 7.5 \,V$.Since the carrier amplitude decreased from $20 \,V$ to $12.5 \,V$, the peak voltage of the carrier signal must be reduced by $7.5 \,V$.

The modulation index used in amplitude modulation of a message signal of peak voltage $10 \,V$ is $0.5$. By how much should the peak voltage of the carrier signal be changed, so as to obtain the modulation index of $0.8$, keeping the peak voltage of the message signal the same as before?

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