How to Calculate Percentage Modulation from RMS Voltage Before and After Modulation?

24 9

#1 21668169 19 May 2013 04:17

Ozai Ozuma Ozai Ozuma

Anonymous

Post #1
21668169 19 May 2013 04:17

I've encountered a problem. The question says that
"The R.M.S voltage of a carrier wave is 5V before modulation and 5.9V after modulation. Determine the percentage of modulation."

I've solved and obtained the answer as 1.25 but I;m not sure about that.
Please answer ASAP.
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#2 21668170 19 May 2013 10:07

Steve Lawson Steve Lawson

Anonymous

Post #2
21668170 19 May 2013 10:07

OK, it's been a very, verrrry long time since I last tinkered with such things, but since no one else is giving this a go, and you wanted this ASAP, I thought I would throw in my two cents:

To refresh my memory I googled "percent modulation formula" and got (from http://www.cliftonlaboratories.com/am_modulation.htm):

Percent Modulation = 100 x (Emax - Emin) / (Emax + Emin)

Emax(rms) = 5.9V
Emin(rms) = 5 - (5.9 - 5) = 4.1V

∴ Percent Modulation = 100 x (5.9 - 4.1) / (5.9 + 4.1) = 18%

Not positive if that is correct, though.
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#3 21668171 21 May 2013 05:04

David Adams David Adams

Anonymous

Post #3
21668171 21 May 2013 05:04

Usually that solution referred to the max and min values but not in the RMS values. There is one for RMS but I can't remember it. The problem is that you have to analyze is now that there is an audio component to the waveform. I don't think yours will work because how would you measure the RMS min voltage? You assumed that the 5.9 volts was the max.

So after a little memory work

M=modulation index = Vmodulating signal / V un-modulated carrier.

For me there is some missing information. In reality the Average value of the carrier does not change.
#4 21668172 21 May 2013 08:21

Steve Lawson Steve Lawson

Anonymous

Post #4
21668172 21 May 2013 08:21

The "Sq root of 2"s cancel out:

Percent Modulation = 100 x (Emax/√2 – Emin/√2) / (Emax/√2 + Emin/√2)

Percent Modulation = 100 x ((Emax – Emin)/√2) / ((Emax + Emin)/√2)

Percent Modulation = 100 x (Emax – Emin) / (Emax + Emin) * (√2/√2)
#5 21668173 22 May 2013 11:52

David Adams David Adams

Anonymous

Post #5
21668173 22 May 2013 11:52

Okay - ya got me as I was too lazy to work it out - '
He is still missing the Min value - needs a scope - but I suppose we could assume power levels - lets see
E unmodulated - 5 volts - assuming a 50 ohm load then Punmod = V^2 / R = 0.5 watts
Then the modulated power would be 5.9^2 / 50 = 0.6962 watts

Now I gotta dig
P(total) = Pc (1+m^2/2)
so M = sqrt ( 2 * ( (Pt/Pc) - 1))
therefore M = .886 = 88.6%

Whew !!!

But of course I did not solve this in time to get his homework solution handed in on time
#6 21668174 22 May 2013 18:28

Steve Lawson Steve Lawson

Anonymous

Post #6
21668174 22 May 2013 18:28

David, wouldn't the modulated wave have a much greater amplitude if it were modulated at 88.6%?

And, the Min value _is_ available: 5 volts is the RMS amplitude of the [unmodulated] carrier wave, so if the modulated wave has a max RMS amplitude of 5.9V, then 5 - (5.9 - 5) or 4.1V would be the minimum RMS amplitude (assuming the modulating wave is at a constant amplitude, at least for a cycle or two).

Is that correct, or am I full of beans?
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#7 21668175 25 May 2013 00:45

Donald Dahl Donald Dahl

Anonymous

Post #7
21668175 25 May 2013 00:45

Hi Steve,

I'll first admit that this is not my area of expertise. When I first read the question, I naively assumed the calculation would be

*100 x (5.9 - 5) / 5 = 18%*,

so I was curious when I saw that this agreed with your solution.

Let A = 5V, B = 5.9V

Using your formula and your definitions for Emax and Emin gives

Percent Modulation = 100 x [B - (A - (B - A) ) ] / [B + (A - (B - A) ) ]

= 100 x [B - (2A - B) ] / [B + (2A - B) ] = 100 x [2B - 2A] / 2A

= 100 x 2(B - A) / 2A = 100 x (B - A ) / A

= *100 x (5.9 - 5) / 5 = 18%*

So, assuming your interpretation was correct, we could have calculated the result directly from the values given in the problem.
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#8 21668176 25 May 2013 01:55

Steve Lawson Steve Lawson

Anonymous

Post #8
21668176 25 May 2013 01:55

Amazing!
#9 21668177 07 Jun 2013 23:26

Dishan Pangan Dishan Pangan

Anonymous

Post #9
21668177 07 Jun 2013 23:26

I also went through this at school before so i can relate : ) Just want to give a little hint. When you solve for percentage of modulation, ideally it ranges from 0 to 100% . So with your initial solution you got 1.25 or 125%. This is over-modulation, which causes signal distortion. You already have a clue that this is incorrect (because it goes beyond 100%).
#10 21668178 08 Jun 2013 11:31

Steve Lawson Steve Lawson

Anonymous

Post #10
21668178 08 Jun 2013 11:31

Huh...I assumed he meant 1.25% (especially due to the wording, "Determine the percentage of modulation")

But, I suppose you could be right, it could be 125% and I agree, that would be over modulated, and 5.9V after modulation is definitely not over modulated.
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Topic summary

The discussion addresses how to calculate the percentage modulation of an AM signal given the RMS voltage of the carrier before (5V) and after modulation (5.9V). Initial attempts to calculate the modulation index yielded incorrect or unclear results, such as 1.25 or 125%, which would indicate over-modulation. A commonly referenced formula for percent modulation is 100 × (Emax - Emin) / (Emax + Emin), where Emax and Emin are the maximum and minimum envelope voltages. However, this formula typically applies to peak voltages, not RMS values. One approach assumed the minimum RMS voltage as 5 - (5.9 - 5) = 4.1V, leading to an 18% modulation index. Another method used power relations: with a 50-ohm load, unmodulated power is calculated as V²/R, and total power after modulation as (5.9)²/50, then applying the formula M = sqrt[2 × (Pt/Pc - 1)] to find a modulation index of approximately 88.6%, which seems too high for the given voltages. The consensus favors the simpler calculation of percent modulation as 100 × (5.9 - 5) / 5 = 18%, consistent with the assumption that the RMS voltage increase corresponds directly to modulation depth. It is noted that modulation percentage should ideally range from 0 to 100%, and values exceeding this indicate over-modulation and distortion. The discussion highlights the importance of distinguishing between RMS and peak voltages in modulation calculations and the need for additional waveform information (e.g., minimum voltage or scope measurements) for precise determination.
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How to Calculate Percentage Modulation from RMS Voltage Before and After Modulation?

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