Question

For an amplitude modulated wave, the maximum amplitude is found to be $10V$ while the minimum amplitude is found to be $2V$ , Find the modulation index, $\mu$ . What would be the value of $\mu$ if the minimum amplitude is zero volt?

Answer 1

In order to answer this question, first we will rewrite the given facts and then to find the modulation index for an amplitude modulated wave, we will apply its formula in the terms of maximum and minimum amplitudes, i.e. $\mu = \dfrac{{{E_{\max }} - {E_{\min }}}}{{{E_{\max }} + {E_{\min }}}}$.

Complete step by step answer:
Given that, the maximum amplitude is, ${E_{\max }} = 10V$ , and
The minimum amplitude is, ${E_{\min }} = 2V$
And we have to find the modulation index, $\mu = ?$ :
So, we will apply the formula to find the modulation index in the terms of maximum and minimum amplitudes:
$\mu = \dfrac{{{E_{\max }} - {E_{\min }}}}{{{E_{\max }} + {E_{\min }}}}$
Now, we will substitute the given values of maximum and minimum amplitudes-
$\Rightarrow \mu = \dfrac{{10 - 2}}{{10 + 2}} = \dfrac{8}{{12}} = 0.67$
Therefore, the required value of the modulation index is 0.67.

Again, as per the question: If the minimum amplitude is $0V$ and the maximum amplitude remains same as $10V$ , then the modulation index will be: (Again we will apply the same upper formula of modulation index)
$\mu = \dfrac{{{E_{\max }} - {E_{\min }}}}{{{E_{\max }} - {E_{\min }}}}$
This time, we will substitute 0 volt instead of minimum amplitude:-
$\therefore \mu = \dfrac{{10 - 0}}{{10 + 0}} = 1$

Hence, the required modulation index is one.

Note: A integer between 0 and 1 should be used as the modulation index. When m is bigger than 1, the modulated waveform suffers from significant distortion. This phenomenon occurs when ${V_m}$ exceeds ${V_c}$ , and it is also referred to as overmodulation. When ${V_m}$ equals ${V_c}$ and m equals 1, we have the optimum situation.