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A message signal of frequency $10 \mathrm{kHz}$ and peak value of $10 \mathrm{~V}$ is used to moderate a carrier of frequency $1 \mathrm{MHz}$ and peak voltage $20 \mathrm{~V}$. The modulation index and side bands produced are
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The correct answer is:
$0.5$ and $1010 \mathrm{kHz}, 990 \mathrm{kHz}$
Frequency of message signal, $f_{m}=10 \mathrm{kHz}$
Carrier frequency, $f_{c}=1 \mathrm{MHz}=1000 \mathrm{kHz}$
Peak value of message signal, $V_{m}=10 \mathrm{~V}$
Peak value of carrier signal, $V_{c}=20 \mathrm{~V}$
$\therefore$ Modulation index, $\mu=\frac{V_{m}}{V_{c}}=\frac{10}{20}=0.5$
Upper side band frequency,
$$
f_{\text {USB }}=f_{c}+f_{m}=1000+10
$$
$$
=1010 \mathrm{kHz}
$$
Lower side band frequency,
$$
\begin{aligned}
f_{\text {LSB }} &=f_{c}-f_{m} \\
&=1000-10=990 \mathrm{kHz}
\end{aligned}
$$
Carrier frequency, $f_{c}=1 \mathrm{MHz}=1000 \mathrm{kHz}$
Peak value of message signal, $V_{m}=10 \mathrm{~V}$
Peak value of carrier signal, $V_{c}=20 \mathrm{~V}$
$\therefore$ Modulation index, $\mu=\frac{V_{m}}{V_{c}}=\frac{10}{20}=0.5$
Upper side band frequency,
$$
f_{\text {USB }}=f_{c}+f_{m}=1000+10
$$
$$
=1010 \mathrm{kHz}
$$
Lower side band frequency,
$$
\begin{aligned}
f_{\text {LSB }} &=f_{c}-f_{m} \\
&=1000-10=990 \mathrm{kHz}
\end{aligned}
$$
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