Message signal and carrier signal are given as

$m=\cos(2 \pi \cdot f_{m} \cdot t) $

$c=A_{c}\cdot \cos(2 \pi \cdot f_{c} \cdot t) $

I was asked to find modulation index $a$ so that ratio of the power in the sidebands to the total power in the transmitted signal will be $\frac{1}{3}$

My attempt

We know that, in conventional AM, total power of transmitted signal $P_{u}= \frac{Ac^2}{2} + \frac{Ac^2}{2}\cdot a^2 P_{m} $ where $P_{m} $ is message signal's power.

I think that power in sidebands is related to $\frac{Ac^2}{2}\cdot a^2 P_{m}$ term

$\frac{Ac^2}{2}\cdot a^2 P_{m} = \frac{1}{3}P_{u}$

Submitting $P_{u}$

$\frac{Ac^2}{2}\cdot a^2 P_{m} = \frac{1}{3}(\frac{Ac^2}{2}(1 + a^2 P_{m}))$

$ a^2 P_{m} = \frac{1}{3}(1 + a^2 P_{m})$

$ a^2 = \frac{1}{2P_{m}}$

It seems false because modulation index might have been a numerical value.

  • 1
    $\begingroup$ I think what I did may not be false but less. I have found$P_{m}$ becomes $\frac{1}{2}$ submitting $0$ into autocorrelation function of $cos(2\pi*fm*t)$ $\endgroup$ – Uygar Uçar Jun 8 '18 at 21:21

this is John BG jgb2012@sky.com

In AM modulation, as shown in


the modulation index is the ratio between the amplitude of the modulated and the amplitude of the carrier.

So the sought modulation index $a=\frac{1}{Ac}$

modulating signal power: $\frac{1}{2}$

carrier power: $\frac{Ac^2}{2}$

side bands to total power ratio: $\frac{1}{(1+\frac{Ac^2}{2})}$

solving: $\frac{1}{(1+\frac{Ac^2}{2})} =\frac{1}{3}$ gives $Ac=2$

so the answer is $a=\frac{1}{2}$

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  • $\begingroup$ Thanks for your answer. How did you find modulating signal power as $\frac{1}{2}$? Sorry if simple . $\endgroup$ – Uygar Uçar Jun 9 '18 at 8:56
  • $\begingroup$ Carlson's Communication Systems, pages 31 and 33, energy and power definitions. How do I upload a picture that shows you such definitions? Also, I wrote the expressions in plain MATLAB but I see it's been edited to LaTEX. Is it ok to leave the MATLAB expressions, or do people have preference for the LaTEX format? $\endgroup$ – John Bofarull Jun 10 '18 at 1:01
  • $\begingroup$ There's a copy of Carlson's literature reference here: een.iust.ac.ir/profs/Beheshti/communication%20system/… $\endgroup$ – John Bofarull Jun 10 '18 at 1:02
  • $\begingroup$ Okay. I understand now. I think it is okay to leave MATLAB code here but you need to use it in code format. (After selecting all code, you need to click on bracket symbol over your answer, fifth symbol from left ) $\endgroup$ – Uygar Uçar Jun 10 '18 at 15:22
  • $\begingroup$ Uygar, 1/2W per ohm. Carlson and other Signal Processing literature references use this signal and power and energy definitions, tacitly assuming 'per ohm'. May ask you what else should I add to my answer so that you consider marking my answer as accepted answer? $\endgroup$ – John Bofarull Jun 10 '18 at 17:20

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