Given a multi-tone amplitude modulated wave as follows :

s(t) = Ac[1+2kasin(w1) + kasin(w2)] cos(wct)

where wc is the carrier frequency and w1 and w2 are the modulating frequency components. My question is : what is the largest value of amplitude sensitivity ka for which envelope detection can be performed on s(t) to recover the message signal without distortion ?

Method 1 (calculate effective modulation index)

Effective modulation index for multi-tone AM signal is given by \$u_{eff}=\sqrt{u_1^2+u_2^2}\$ (ueff = sqrt(u12 + u22) where u1 and u2 represent modulation index for single tone AM signal and sqrt stands for square-root. By definition, u = maximum value of message signal after modulation.So u1 = 2Ka and u2 = Ka. In the case of single tone AM wave, for envelope detector to work without any distortion, the condition is u<=1. Analogously for multi-tone AM wave the same condition becomes ueff<=1.

4ka2 + ka2 <= 1


Method 2

Here again we take analogy from single-tone AM for envelope detection condition. However, instead of ueff<=1, we do

u1 + u2 <=1

2ka + ka <= 1

ka <= 0.33

Please tell me which of one of these two methods is correct ?

(Sorry for poor formatting. I am a novice to MathJax )

  • \$\begingroup\$ Apart from the answer, can someone please tell me how to format my question correctly ? Like I couldn't get square root symbol to come on screen, etc. \$\endgroup\$ Commented Jun 26, 2022 at 14:14
  • 1
    \$\begingroup\$ it's probably easier to use mathjax for the more complicated equations, I've done one for you as an example \$\endgroup\$
    – Neil_UK
    Commented Jun 26, 2022 at 14:52
  • \$\begingroup\$ Thanks a lot @Neil \$\endgroup\$ Commented Jun 27, 2022 at 8:49
  • \$\begingroup\$ note how the curly brackets are used, to group things for the sub _ operator, or group terms under the sqrt operator \$\endgroup\$
    – Neil_UK
    Commented Jun 27, 2022 at 10:25

1 Answer 1


Do not take the modulation index to mess this, because you already have the time domain expression for the modulated output s(t).

I guess you only forgot to insert t. I mean you planned to write this:

s(t) = Ac[1+2kasin(w1t) + kasin(w2t)] cos(wct)

Distortion (see NOTE1) occurs if the amplitude term goes momentarily to negative. To prevent it means you keep 1+2kasin(w1t) + kasin(w2t) non-negative.

The worst case occurs when the sines happen to be at the same time =-1.

You want 1-2ka- ka >0 i.e. ka < 1/3 as you have already written.

NOTE1: The distortion gets its birth in the common envelope detector(= diode detector). Synchronous detector (=mixing the signal back down) doesn't have that problem.

  • \$\begingroup\$ Thanks a lot for your answer ! \$\endgroup\$ Commented Jun 27, 2022 at 10:33

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