Question #83928

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Question #83928

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Answer 1 · 2017-03-25T07:09:56

$9 V peak$ $9V" peak"$

Explanation:

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

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Modulation index $m$ $m$ 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

$m = \frac{E_{m}}{E_{C}}$ $m=E_m/E_C$

modulation index $m$ $m$ takes values between $0 and 1$ $0 and 1$ and no distortion is introduced in the AM wave when
$E_{m} \leq E_{C}$ $E_m <= E_C$ .

However, for $E_{m} > E_{C}$ $E_m > E_C$ , $m$ $m$ becomes greater than $1$ $1$ .
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.

$Percent modulation = \frac{E_{m}}{E_{C}} \times 100$ $"Percent modulation"=E_m/E_Cxx100$

As shown in the figure, inserting given values we get
$75 = \frac{E_{m}}{12} \times 100$ $75=E_m/12 xx100$
$\Rightarrow E_{m} = \frac{75}{100} \times 12$ $=>E_m=75/100 xx12$
$\Rightarrow E_{m} = 9 V peak$ $=>E_m=9V" peak"$

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

Sometimes we express $E m and E c$ $Em and Ec$ in terms of $E_max and E_min$ as seen on a screen of a CRO.

As shown in the figure
$E_m=(E_max-E_min)/2$
and $E_C=E_max-E_m$
Using expression for $E_m$ we get
$E_C=(E_max+E_min)/2$

these give
$m=(E_max-E_min)/(E_max+E_min)$

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Question #83928

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