I’m not sure on this solution!

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.

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.

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.

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)

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

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?

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.

Amazing!

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%).

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.