Question

Write the formula for band-width in amplitude modulation.

Answer 1

We know that Amplitude modulation is a technique in electronic communication which is used to modify the amplitude of the carrier waves. Now, bandwidth is the difference between highest and lowest frequencies of the signal. Using this concept we will find the bandwidth of amplitude modulation.

Formula used: $s\left( t \right)=\left[ {{A}_{c}}+{{A}_{m}}\cos \left( 2\pi {{f}_{m}}t \right) \right]\cos \left( 2\pi {{f}_{c}}t \right)$,$BW={{f}_{\max }}-{{f}_{\min }}$

Complete step-by-step answer:
Before finding the bandwidth of Amplitude modulated waves we will understand what is Amplitude modulation (AM).
So, Amplitude modulation is a technique in electronic communication which is used to modify the amplitude of the carrier waves. It is generally used in Radio waves or radio signals. The equation of amplitude modulated wave can be given as,
$s\left( t \right)=\left[ {{A}_{c}}+{{A}_{m}}\cos \left( 2\pi {{f}_{m}}t \right) \right]\cos \left( 2\pi {{f}_{c}}t \right)$ ………………..(i)
Where, ${{A}_{c}}$ is amplitude of carrier wave, ${{A}_{m}}$is amplitude of modulated wave, ${{f}_{c}}$is the frequency of carrier wave and ${{f}_{m}}$ is the frequency of modulated wave.
Now, bandwidth can is the difference between highest and lowest frequencies of any wave and it can be given mathematically as,
$BW={{f}_{\max }}-{{f}_{\min }}$ ………………(ii)
Where, ${{f}_{\max }}$is maximum frequency and ${{f}_{\min }}$is minimum frequency.
Now, considering the equation of amplitude modulated wave, $s\left( t \right)=\left[ {{A}_{c}}+{{A}_{m}}\cos \left( 2\pi {{f}_{m}}t \right) \right]\cos \left( 2\pi {{f}_{c}}t \right)$
It can also be written as,
$\Rightarrow s\left( t \right)=\left[ {{A}_{c}}\cos \left( 2\pi {{f}_{c}}t \right)+{{A}_{m}}\cos \left( 2\pi {{f}_{m}}t \right)\cos \left( 2\pi {{f}_{c}}t \right) \right]$
Now, multiplying and dividing the term with $2{{A}_{c}}$, we will get,
$\Rightarrow s\left( t \right)=\left[ {{A}_{c}}\cos \left( 2\pi {{f}_{c}}t \right)+\dfrac{2{{A}_{c}}}{2{{A}_{c}}}{{A}_{m}}\cos \left( 2\pi {{f}_{m}}t \right)\cos \left( 2\pi {{f}_{c}}t \right) \right]$
$\Rightarrow s\left( t \right)={{A}_{c}}\cos \left( 2\pi {{f}_{c}}t \right)+\dfrac{{{A}_{c}}\mu }{2}\left[ 2\cos \left( 2\pi {{f}_{m}}t \right)\cos \left( 2\pi {{f}_{c}}t \right) \right]$
Where, $\mu =\dfrac{{{A}_{m}}}{{{A}_{c}}}$ . Hence from the equation it can be said that amplitude modulated waves have three frequencies those are, carrier frequency ${{f}_{c}}$, upper sideband frequency ${{f}_{c}}+{{f}_{m}}$ and lower sideband frequency ${{f}_{c}}-{{f}_{m}}$,
Here, ${{f}_{\max }}={{f}_{c}}+{{f}_{m}}$ and ${{f}_{\min }}={{f}_{c}}-{{f}_{m}}$, on substituting in the equation (ii) we will get,
$BW={{f}_{c}}+{{f}_{m}}-\left( {{f}_{c}}-{{f}_{m}} \right)$
$\Rightarrow BW=2{{f}_{m}}$
Thus, we can say that the bandwidth of amplitude modulated waves is twice the frequency of modulation of the wave.

Note: Many times, students might consider modulation index as ratio of amplitude of carrier wave to that of amplitude of modulated wave instead of ratio of modulated wave to that of amplitude of carrier wave and solve the problem based on that but that is wrong and students must take care about formulas.