# Question #83928

Mar 25, 2017

$9 V \text{ peak}$

#### Explanation:

From the text of the question it is clear that this pertains to Amplitude Modulation.

Modulation index $m$ is a dimensionless quantity and is defined as the ratio of amplitudes of the modulating signal and of carrier. mathematically it can be stated as

$m = {E}_{m} / {E}_{C}$

modulation index $m$ takes values between $0 \mathmr{and} 1$ and no distortion is introduced in the AM wave when
${E}_{m} \le {E}_{C}$.

However, for ${E}_{m} > {E}_{C}$, $m$ becomes greater than $1$.
This will distort the the AM signal and is called over modulation.

Sometimes, this modulation index is expressed as a percentage and called as percentage modulation.

$\text{Percent modulation} = {E}_{m} / {E}_{C} \times 100$

As shown in the figure, inserting given values we get
$75 = {E}_{m} / 12 \times 100$
$\implies {E}_{m} = \frac{75}{100} \times 12$
$\implies {E}_{m} = 9 V \text{ peak}$

-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Sometimes we express $E m \mathmr{and} E c$ in terms of ${E}_{\max} \mathmr{and} {E}_{\min}$ as seen on a screen of a CRO.

As shown in the figure
${E}_{m} = \frac{{E}_{\max} - {E}_{\min}}{2}$
and ${E}_{C} = {E}_{\max} - {E}_{m}$
Using expression for ${E}_{m}$ we get
${E}_{C} = \frac{{E}_{\max} + {E}_{\min}}{2}$

these give
$m = \frac{{E}_{\max} - {E}_{\min}}{{E}_{\max} + {E}_{\min}}$