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A message signal is used to modulated a carrier signal of frequency $5 \mathrm{MHz}$ and peak voltage of $40 \mathrm{~V}$. In the process, two side-bands are produced seperated by $40 \mathrm{kHz}$. If the modulation index is 0.75 , then the peak voltage and frequency of the messages signal, respectively are
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The correct answer is:
$30 \mathrm{~V} ; 20 \mathrm{kHz}$
Given, frequency of carrier signal, $f=5 \mathrm{MHz}$
and peak voltage, $V_c=40 \mathrm{~V}$
Modulation index, $\mu=0.75$
$\therefore \quad \mu=\frac{V_m}{V_c} \quad \begin{array}{r}{\left[\because V_m \text { is peak voltage }\right.} \\ \text { of message signal }]\end{array}$
$\begin{aligned} 0.75 & =\frac{V_m}{40} \\ \Rightarrow \quad V_m & =40 \times 0.75 \\ V_m & =30 \mathrm{~V}\end{aligned}$
Since, difference of frequencies of two side bands is equal to the band width ( $2 \mathrm{fm})$.
i.e., Band width $=40 \mathrm{kHz}$
$2 f_{\mathrm{m}}=40 \mathrm{kHz} \Rightarrow f_{\mathrm{m}}=20 \mathrm{kHz}$
Hence, the peak voltage and frequencey will be $30 \mathrm{~V}$ and $20 \mathrm{kHz}$.
and peak voltage, $V_c=40 \mathrm{~V}$
Modulation index, $\mu=0.75$
$\therefore \quad \mu=\frac{V_m}{V_c} \quad \begin{array}{r}{\left[\because V_m \text { is peak voltage }\right.} \\ \text { of message signal }]\end{array}$
$\begin{aligned} 0.75 & =\frac{V_m}{40} \\ \Rightarrow \quad V_m & =40 \times 0.75 \\ V_m & =30 \mathrm{~V}\end{aligned}$
Since, difference of frequencies of two side bands is equal to the band width ( $2 \mathrm{fm})$.
i.e., Band width $=40 \mathrm{kHz}$
$2 f_{\mathrm{m}}=40 \mathrm{kHz} \Rightarrow f_{\mathrm{m}}=20 \mathrm{kHz}$
Hence, the peak voltage and frequencey will be $30 \mathrm{~V}$ and $20 \mathrm{kHz}$.
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