Question

A message signal of frequency $10kHz$ and peak voltage $10$ volts is to modulate a carrier of frequency $1MHz$ and peak voltage of $20$, what are modulation index and side bands produced?
(A) $0.5,$ $101kHz,$ $99kHz$
(B) $0.5,$ $1010kHz,$ $990kHz$
(C) $2,$ $1010kHz,$ $990kHz$
(D) $0.6,$ $101kHz,$ $99kHz$

Answer 1

In order to solve this question, consider the equations of carrier and modulating signal and find the AM signal using them. Express the value of AM in terms of carrier signal. From there, you can find the modulating signal. IF you do some manipulation, you can see that the AM signal has three frequencies which includes the side bands. From there, find the side bands.

Complete step by step answer:
The instantaneous carrier signal is given ${V_c} = {A_c}\sin 2\pi {f_c}t$, where ${A_C}$ is peak voltage of the carrier wave.
The modulating signal is given as ${V_m} = {A_m}\sin 2\pi {f_m}t$, where ${A_m}$ is peak voltage of modulating voltage.
The AM signal is given as ${V_{AM}} = {A_{AM}}\sin 2\pi {f_c}t$, where ${A_{AM}}$ is the amplitude of the AM signal which is given as ${A_{AM}} = {A_c} + C{A_m}\sin 2\pi {f_m}t$. $C$ is the modulating constant and its value lies between $0 - 1$. If nothing is mentioned, we take $C = 1$.
Substituting the value of ${A_{AM}}$ in the above equation, we get,

$${V_{AM}} = \left( {{A_c} + {A_m}\sin 2\pi {f_m}t} \right)\sin 2\pi {f_c}t \\\ \Rightarrow {V_{AM}} = {A_c}\left( {1 + \dfrac{{{A_m}}}{{{A_c}}}\sin 2\pi {f_m}t} \right)\sin 2\pi {f_c}t \\\ \Rightarrow {V_{AM}} = {A_c}\left( {1 + m\sin 2\pi {f_m}t} \right)\sin 2\pi {f_c}t \\\$$

Here, $m = \dfrac{{{A_m}}}{{{A_c}}}$ is the modulation index.
Now, let us do some manipulations,
${V_{AM}} = {A_c}\sin 2\pi {f_c}t + m{A_c}\left( {\sin 2\pi {f_m}t} \right)\left( {\sin 2\pi {f_c}t} \right) \\\ \Rightarrow {V_{AM}} = {A_c}\sin 2\pi {f_c}t + \dfrac{{m{A_c}}}{2}\left( {2\sin \left( {2\pi {f_m}t} \right)\sin \left( {2\pi {f_c}t} \right)} \right) \\\ \Rightarrow {V_{AM}} = {A_c}\sin 2\pi {f_c}t + \dfrac{{m{A_c}}}{2}\left( {\cos 2\pi \left( {{f_c} - {f_m}} \right)t - \cos 2\pi \left( {{f_c} + {f_m}} \right)t} \right) \\\ \therefore {V_{AM}} = {A_c}\sin 2\pi {f_c}t + \dfrac{{m{A_c}}}{2}\cos 2\pi \left( {{f_c} - {f_m}} \right)t - \dfrac{{m{A_c}}}{2}\cos 2\pi \left( {{f_c} + {f_m}} \right)t \\\$
As you can see, the AM signal is a combination of carrier signal, a lower sideband signal frequency and a higher side band signal frequency.
Therefore, we have the side bands having frequencies as ${f_c} - {f_m}$ and ${f_c} + {f_m}$.
So, as we have been given in the question that ${A_m} = 10V$ and ${A_c} = 20V$, the modulation index will be $m = \dfrac{{{A_m}}}{{{A_c}}} = \dfrac{{10}}{{20}} = 0.5$.
The sidebands will have values $1000 + 10 = 1010kHz$ and $1000 - 10 = 990kHz$.
Hence, the modulation index and side bands produced are $0.5,$ $1010kHz,$ $990kHz$.

So, the correct answer is “Option B”.

Note:
Remember that the modulation index is defined as the ratio of the modulating signal peak voltage and carrier signal peak voltage. Also remember that the AM signal has a combination of three frequencies namely carrier signal, lower side band signal and upper side band signal frequencies.