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A carrier wave of peak voltage $60 \mathrm{~V}$ is used to transmit a message signal. Then the peak voltage of the modulating signal in order to have a modulation index of $90 \%$ is
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The correct answer is:
$54 \mathrm{~V}$
Modulation index,
$\mu=\frac{\text { Amplitude of message } A_M}{\text { Amplitude of carrier } A_C}$
here, $\quad \mu=90 \%=\frac{9}{10}$ and $A_C=60 \mathrm{~V}$
$\therefore \quad \frac{9}{10}=\frac{A_M}{60}$
$\Rightarrow \quad A_M=\frac{9 \times 60}{10}=54 \mathrm{~V}$
So peak voltage of modulating signal is $54 \mathrm{~V}$.
$\mu=\frac{\text { Amplitude of message } A_M}{\text { Amplitude of carrier } A_C}$
here, $\quad \mu=90 \%=\frac{9}{10}$ and $A_C=60 \mathrm{~V}$
$\therefore \quad \frac{9}{10}=\frac{A_M}{60}$
$\Rightarrow \quad A_M=\frac{9 \times 60}{10}=54 \mathrm{~V}$
So peak voltage of modulating signal is $54 \mathrm{~V}$.
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