Question

A signal of $5\,Hz$ frequency modulates a carrier of frequency $1\,MHz$ and peak voltage $25\,V$. If the amplitude at the sidebands of the amplitude modulated signal is $5\,V$, then the modulation index is:
A) $0.8$
B) $0.6$
C) $0.5$
D) $0.2$
E) $0.4$

Answer 1

Hint:Here we have to use the concept of modulation index.
The modulation index specifies how much the modulated component of the carrier signal changes about its un modulated level.
The FM modulation index is equal to the frequency deviation ratio of the modulating index; it can be found that there is no concept that contains the carrier frequency, which means that it is entirely independent of the carrier frequency.

Complete step by step solution:
Given,
Peak voltage of the carrier, ${V_c} = 25\,V$

Peak voltage of the modulated signal, ${V_m} = 2 \times 5 = 10\,V$

Modulation index $ = $ ?
Here peak voltage is equal to twice the amplitude of voltage.
We know that modulation index is given by:
$\mu = \dfrac{{{V_m}}}{{{v_c}}}$

Therefore, modulation index, $\mu = \dfrac{{{V_m}}}{{{V_c}}} = \dfrac{{10}}{{25}} = 0.4$

Hence, if the amplitude at the sidebands of the amplitude modulated signal is $5\,V$, then the modulation index is $0.4$.

Thus, option E is correct.

Additional information:Modulation is a method of varying one or more characteristics of a continuous waveform, called a carrier signal, with a modulating signal that usually includes information to be transmitted. The modulator is a modulating device.
If the modulation frequency is greater than the modulation index of $1$ i.e. more than $100$ percent of modulation, this induces over-modulation.
The carrier encounters ${180^ \circ }$ phase reversals where the carrier’s level will aim to be below zero. These phase reversals give rise to additional sidebands arising from phase modulation. These sidebands, induced by phase reversal, stretch to infinity. This will cause a significant amount of interference.

Note:Here we have to pay attention that the peak voltage is twice the amplitude of voltage. We do not multiply $2$ then the answer would be wrong.