Question

A carrier wave of peak voltage $15\,V$ is used to transmit a message signal. Find the peak voltage of the modulating signal in order to have a modulation index of $60\,\% $.

Answer 1

Hint: Use the formula of the modulation index given below. Rearrange this formula based on the need of the question. Substitute the known values in the rearranged formula, the obtained result gives the peak voltage of the modulating signal.

Useful formula:
The formula of the modulation index is given by
$m = \dfrac{{PM}}{{PC}}$
Where $m$ is the modulation index, $PM$ is the peak value of the modulating signal and $PC$ is the peak value of the carrier signal.

Complete step by step solution:
It is given that
The peak voltage of the carrier wave, $PC = 15\,V$
The modulation index, $m = 60\,\% $
Modulation index is the ratio of the peak value of the modulating signal to the peak value of the carrier signal. Using the formula of the modulation index,
$m = \dfrac{{PM}}{{PC}}$
Rearranging the formula, based on the needed answer.
$PM = m \times PC$
Substituting the known values in the above formula.
$PM = 60\,\% \times 15$
The percentage is also written as $\dfrac{1}{{100}}$ . So
$PM = \dfrac{{60}}{{100}} \times 15$
By performing the further simplification in the above step.
$PM = 0.6 \times 15$
By multiplying the terms in the right hand side of the above equation,
$PM = 9\,V$

Thus the peak voltage of the modulating signal is obtained as $9\,V$.

Note: The modulating signal is the signal that we want to transmit like radio, TV signal etc. But these waves have less frequency. In order to transmit it with a greater frequency, the carrier signal is used to carry the modulating signal in high frequency.