Question

A message signal of frequency $100MHz$ and peak voltage $100V$ is used to execute amplitude modulation on a carrier wave of frequency $300GHz$ and peak voltage $400V$. The modulation index and difference between the side band frequency are:
A.)$4;1\times {{10}^{8}}Hz$
B.)$0.25;1\times {{10}^{8}}Hz$
C.)$4;2\times {{10}^{8}}Hz$
D.)$0.25;2\times {{10}^{8}}Hz$

Answer 1

Hint: Modulation index is the ratio of the amplitude of the message signal to the amplitude of the carrier signal. The difference between the side band frequencies can be found out by writing each side band frequency in terms of the message signal frequency and carrier wave frequency.

Formula used:

Modulation index ${{m}_{a}}$ in amplitude modulation is given by

${{m}_{a}}=\dfrac{{{A}_{m}}}{{{A}_{c}}}$
where ${{A}_{m}},{{A}_{c}}$ are the amplitudes of the message wave and carrier wave respectively.

The frequencies ${{f}_{1}}$ and ${{f}_{2}}$ of the side bands in amplitude modulation is given by

${{f}_{1}}={{f}_{c}}+{{f}_{m}}$
${{f}_{2}}={{f}_{c}}-{{f}_{m}}$

Where ${{f}_{c}},{{f}_{m}}$ are the frequencies of the carrier and message waves respectively.

Complete step by step answer:
During amplitude modulation, a message wave is superimposed over a carrier wave. As explained in the hint, the modulation index in amplitude modulation is the ratio of the amplitudes of the message wave to the amplitude of the carrier wave. Hence,
Modulation index ${{m}_{a}}$ in amplitude modulation is given by

${{m}_{a}}=\dfrac{{{A}_{m}}}{{{A}_{c}}}$ --(1)

where ${{A}_{m}},{{A}_{c}}$ are the amplitudes of the message wave and carrier wave respectively.

Now, the peak voltage of the wave can be considered as the amplitude of the wave.
Therefore, according to the question,

the amplitude of the message wave is ${{A}_{m}}=100V$
the amplitude of the carrier wave is ${{A}_{c}}=400V$
Hence, using (1), the modulation index ${{m}_{a}}$ will be
$\dfrac{{{A}_{m}}}{{{A}_{c}}}=\dfrac{100V}{400V}=\dfrac{1}{4}=0.25$
Hence, the modulation index of the amplitude modulation is $0.25$.

The frequencies ${{f}_{1}}$ and ${{f}_{2}}$ of the side bands in amplitude modulation is given by

${{f}_{1}}={{f}_{c}}+{{f}_{m}}$ --(2)
${{f}_{2}}={{f}_{c}}-{{f}_{m}}$ --(3)

Where ${{f}_{c}},{{f}_{m}}$ are the frequencies of the carrier and message waves respectively.

Now, according to the question the frequency of the message wave is ${{f}_{m}}=100MHz=100\times {{10}^{6}}Hz={{10}^{8}}Hz$ $\left( \because 1MHz={{10}^{6}}Hz \right)$ --(4)

The frequency of the carrier wave is ${{f}_{c}}=300GHz=300\times {{10}^{9}}=3\times {{10}^{11}}Hz$ $\left( \because 1GHz={{10}^{9}}Hz \right)$ --(5)

Now, the difference $\Delta f$ between the side band frequencies will be

$\Delta f=\left| {{f}_{1}}-{{f}_{2}} \right|$

Using (2) and (3), we get,

$\Delta f=\left| {{f}_{c}}+{{f}_{m}}-\left( {{f}_{c}}-{{f}_{m}} \right) \right|=\left| {{f}_{c}}+{{f}_{m}}-{{f}_{c}}+{{f}_{m}} \right|=\left| 2{{f}_{m}} \right|=2{{f}_{m}}$ $\left( \because {{f}_{m}}>0 \right)$ --(6)
Now using (4) and (6), we get,
$\Delta f=2\times 1\times {{10}^{8}}Hz=2\times {{10}^{8}}Hz$.

Hence the difference in the frequencies of the two sidebands are $2\times {{10}^{8}}Hz$.
Hence, the correct option is D) $0.25;2\times {{10}^{8}}Hz$.

Note: Students often get confused between the carrier wave and the message wave. This leads to a lot of silly errors. A good way to remember the difference between the two is to remember that the carrier wave carries the message signal and hence must have a much greater frequency, so that it can travel easily through the medium without much attenuation. Since, the attenuation of a wave is dependent directly on its wavelength, the attenuation (loss in energy and signal) is less when the wavelength is low (or the frequency is high). Therefore, the carrier waved does not suffer much attenuation in the medium and can travel efficiently.