Question

$1000kHz$carrier wave is amplitude modulated by the signal frequency $200Hz-4000Hz$. The channel width of this case is:
1. $8kHZ$
2. $7.6kHz$
3. $3.8kHz$
5. $400kHz$
6. $4kHz$

Answer 1

Hint:We use the formula for the bandwidth of the channel. This method gives the relation between the width of the channel and the maximum frequency of the signal. Convert obtained frequency for the width of the channel into kHz from Hz

Formula used:

Width of the channel is given as$\beta =2{{f}_{m}}\ldots \ldots (1)$
Here, $\beta$ is the width of the channel and ${{f}_{m}}$ is the maximum frequency of the given signal.


Complete answer:
We have given,
The frequency of the carrier wave = $1000kHz$
The range of frequency is$200Hz-4000Hz$
The minimum frequency is $200Hz$and the maximum frequency is $4000Hz$

${{f}_{m}}=4000Hz$
We can determine the width of the channel by using equation (1).
Substitute $4000Hz$ for ${{f}_{m}}$ in equation (1).
$\begin{align}
  & \beta =2(4000)Hz \\
 & \Rightarrow \beta =8000Hz \\
\end{align}$
So, the width of the channel is $8000Hz$.
But we have to find the width of the channel in kilohertz.
Convert the unit of width of the channel into kilohertz.
$\begin{align}
  & \Rightarrow \beta =\left( 8000Hz \right)\left( \frac{{{10}^{-3}}kHz}{1Hz} \right) \\
 & \Rightarrow \beta =8kHz \\
\end{align}$
So, the width of the channel is $8kHz$
Hence, the correct option is B.



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