Question #83928

1 Answer
Mar 25, 2017

Answer:

#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# 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 and 1# and no distortion is introduced in the AM wave when
#E_m <= 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.

#"Percent modulation"=E_m/E_Cxx100#

As shown in the figure, inserting given values we get
#75=E_m/12 xx100#
#=>E_m=75/100 xx12#
#=>E_m=9V" peak"#

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

Sometimes we express #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)#