In AM modulation, the ratio of signal amplitude to carrier wave amplitude is called modulation index as m. When m>1 i.e when signal amplitude exceeds the carrier wave amplitude the modulated wave cancels out as shown below:

enter image description here

Why does the modulated wave becomes zero(cancels out) when m>1? How can this be explained graphically or mathematically?

  • \$\begingroup\$ \$y(t) = (1+m(t))c(t)\$. If \$m(t)=-1\$, then the modulated signal is 0. \$\endgroup\$
    – The Photon
    Commented May 22, 2018 at 16:47
  • \$\begingroup\$ @ThePhoton Sure, but m can be -1.5, then the whole carrier gets inverted and non null. So far, though, I haven't met anything like it, and thank goodness. \$\endgroup\$ Commented May 22, 2018 at 16:51
  • \$\begingroup\$ @aconcernedcitizen, I think OP's example is showing a case with m ~ 1.1 or something, so there's an extended period where \$1+m(t)\approx 0\$. \$\endgroup\$
    – The Photon
    Commented May 22, 2018 at 16:52
  • \$\begingroup\$ @ThePhoton Sure, but that only works if the input, m(t), is amplitude limited, otherwise it will happily go overboard. For example, 1+2*sin(t) will create a phase inversion and be null around the intersection point, in rest... Babel. :-) \$\endgroup\$ Commented May 22, 2018 at 16:57
  • 1
    \$\begingroup\$ @ThePhoton Yes, I understand, but the picture shows a flat segment for the part where the signal is overmodulated with negative m, yet the positive m peaks just fine, so there must be some limiting, or the picture is drawn wrong. You can verify this with any plotting program: (1+1.1*sin(x))*sin(10*x) will show non-null segment at -m, and non-limited peak at +m. To me, the picture is drawn wrong: -m is limited, +m isn't, otherwise, the responses would look like there are in my answer. \$\endgroup\$ Commented May 22, 2018 at 17:07

1 Answer 1


That will only happen if the modulating signal is limited at the input, i.e. if its amplitude will never exceed the maximum modulation index m=1. If so, then you get what's in the pictured. Or if you consider the strict mathematical formula, as @ThePhoton says in his comment (just added). Else, the carrier will try to accomodate the excess by phase inversion:


The first (bottom) trace shows a modulation of 0.5, for the middle one m=1, and the top one m=1.5. Note the phase inversion that happens at ~0.4s. Fromhere on, the carrier will only change in amplitude, having the same inversion. Usually, the signal is limited to less than m=1, for good reasons.

Here's the FFT of each modulation:


The black trace is with m=0.5, it shows two peaks, symmetric over 10Hz (the carrier), 1Hz apart (input). The blue trace shows the same two peaks, twice as big, for m=1 (of course...).

Corection: For m=-1.5, there are even larger sidelobes. But the detection can not be done now because the peaks suggest the modulation input is folded where the negative m is.

In addition, heres how it looks when the input is limited to \$\pm\$1, but m=1.5:


  • \$\begingroup\$ Is there still distortion for m=1.5? Will there be always distortion when m>1? \$\endgroup\$
    – user16307
    Commented May 22, 2018 at 17:09
  • \$\begingroup\$ For m>1 the negative peaks of the modulated signal will be clipped right? \$\endgroup\$
    – user16307
    Commented May 22, 2018 at 17:14
  • \$\begingroup\$ @user16307 They will only be clipped if there is a clipper active (which, normally, is). Otherise, purely mathematically, you can have (1+1000*sin(x))*sin(10*x), for example, but with the efect showsn in the FFT picture. \$\endgroup\$ Commented May 22, 2018 at 17:15
  • \$\begingroup\$ Im a bit confused about your FFT. If you modulate the 10Hz carrier with a signal which has freq. fs = 1Hz, the resulting wave is composed if 3 frequencies: fc, (fc+fs) and (fc-fs). But in your blue and black FFT spectrum I cannot see any 10Hz carrier component. (?) \$\endgroup\$
    – user16307
    Commented May 22, 2018 at 17:19
  • \$\begingroup\$ @user16307 I corrected the answer, the carrier is always there, it just overlaps with the others (the red trace is the last one). I've also shown what happens when the signal is overmodulated, but limited -- harmonics. \$\endgroup\$ Commented May 22, 2018 at 17:24

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