Question

A 1000kHz carrier wave is modulated by an audio signal of frequency range $100 - 5000\,{\text{Hz}}$. Then the width of channel in kHz is
A. 5
B. 10
C. 20
D. 50

Answer 1

Use the formula for the bandwidth of the channel. This formula gives the relation between the width of the channel and the maximum frequency of the audio signal. Convert the obtained frequency for the width of the channel into kHz from Hz.

Formula used:
The width of the channel is given by
$\beta = 2{f_m}$ …… (1)
Here, $\beta$ is the width of the channel and ${f_m}$ is the maximum frequency of the audio signal.

Complete step by step answer:
We have given that the frequency of the carrier wave Is $1000\,{\text{kHz}}$.
The range of frequency of the audio signal is $100 - 5000\,{\text{Hz}}$.
Hence, the minimum frequency of the audio signal is $100\,{\text{Hz}}$ and the maximum frequency of the audio signal is $5000\,{\text{Hz}}$.
${f_m} = 5000\,{\text{Hz}}$
We can determine the width of the channel using equation (1).
Substitute $5000\,{\text{Hz}}$ for ${f_m}$ in equation (1).
$\beta = 2\left( {5000\,{\text{Hz}}} \right)$
$\Rightarrow \beta = 10000\,{\text{Hz}}$
Hence, the width of the channel is $10000\,{\text{Hz}}$.
But we need to determine the width of the channel in kilohertz.
Convert the unit of width of channel in kilohertz.
$\Rightarrow \beta = \left( {10000\,{\text{Hz}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{kHz}}}}{{1\,{\text{Hz}}}}} \right)$
$\Rightarrow \beta = 10\,{\text{kHz}}$
Therefore, the width of the channel in kHz is $10\,{\text{kHz}}$.

Hence, the correct option is B.

Note:
In the question, the width of the channel is asked in the unit kHz. After substituting the maximum frequency in the formula, the obtained value of the width of the channel is in hertz. Don’t forget to convert it in kilohertz. Also the students should always remember that we have to substitute the maximum value of frequency from the given frequency range of the audible signal and not the minimum frequency or the mean of frequency range.