QJEEM22Phc512: JEE Main 2022 Physics - Class 12 JEEM22Phc512 Solution: Find the modulation index of an AM wave having 8   V…

In an Amplitude Modulated (AM) signal, the amplitude of the carrier signal is varied in accordance with the information being sent. The extent of this variation is given by the modulation index, often denoted by $\mu$ .

In the problem statement, we are given a peak-to-peak variation in the signal of 8V. This is essentially twice the amplitude of the modulating signal, because the peak-to-peak value represents the total variation from the minimum to the maximum. Thus, we find the amplitude of the modulating signal ( $A_m$ ) as follows:

$A_m = \frac{8V}{2} = 4V$

The maximum amplitude of the AM wave is 9V. This includes the amplitude of the carrier signal and the amplitude of the modulating signal. We therefore find the amplitude of the carrier signal ( $A_c$ ) as follows:

$A_c = 9V - A_m = 9V - 4V = 5V$

Finally, the modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier signal:

$\mu = \frac{A_m}{A_c} = \frac{4V}{5V} = 0.8$

Thus, the modulation index of the AM signal is 0.8, meaning that the amplitude of the carrier signal is varied by 80% of its original amplitude to encode the information being sent. The correct answer among the provided options is therefore Option A: 0.8.