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When the signal is modulated onto the carrier in the electromagnetic spectrum, that signal occupies the small portion of the spectrum surrounding the carrier frequency. It also cause sidebands to be generated at frequencies above and below the carrier frequency.

But how and why are those sidebands generated in AM and FM and why are there so many sidebands generated in FM while just two are generated in AM ? Please provide a practical example, as I already know how they are generated mathematically.

What I know is, in the time domain, when the original signal is put into the carrier signal, it is actually multiplied with the carrier signal which means that in frequency-domain the original signal is convolved with the carrier signal. Those two Sidebands in AM are actually the Fourier transform of the carrier signal.

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    $\begingroup$ If no sidebands are generated, how can one tell the difference between a modulated carrier and an unmodulated carrier? $\endgroup$
    – Dilip Sarwate
    May 4, 2012 at 18:51
  • $\begingroup$ @Effected By 'sidebands' are you referring to the base-band spectrum, or are you referring to the rolling off side lobes? $\endgroup$
    – Spacey
    May 4, 2012 at 18:57
  • $\begingroup$ by sidebands i mean the generated freq. which is equal to the difference and the sum of Carrier and signal frequency $\endgroup$
    – Sufiyan Ghori
    May 4, 2012 at 19:29
  • $\begingroup$ If you know the mathematical representation of AM and FM signals, then you can also calculate their spectrum using the Fourier transform. This illustrates what the sidebands look like for each type of modulation. $\endgroup$
    – Jason R
    May 4, 2012 at 20:32
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    $\begingroup$ By definition there can only be, at most, two sidebands, one on one side of the carrier and one on the other. The sidebands, as hotpaw2 explains, are where the actual information is, and their width is proportional to the maximum amount of information the channel can carry. $\endgroup$
    – Daniel R Hicks
    May 4, 2012 at 21:46

3 Answers 3

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Carrying information requires bandwidth.

For a given S/N ratio, modulating a signal to carry more information will thus expand its bandwidth. Call the addition bandwidth "side bands". If you don't add side bands to a fixed frequency carrier, you can't expand its bandwidth, and thus you can't transmit any information (other than the presence of a constant carrier).

For AM, AM is not PM (phase modulation). Any additional bandwidth (as required to carry information in the modulating signal) on one side of the carrier will usually have a different phase (change of phase with respect to time from any reference point) from the carrier. To neutralize this phase difference, AM modulation has to add some additional matching bandwidth on the opposite side of the carrier to carry a signal that will exactly cancel any phase shift of the spectrum the first side, so that AM doesn't become PM.

With FM, modulating a carrier changes the signal frequency to new frequencies. You can also call those additional new frequencies so generated "side bands".

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  • $\begingroup$ What about VSB and SSB modulation? $\endgroup$
    – CyberMen
    May 7, 2012 at 13:59
  • $\begingroup$ SSB allows the phase to shift or be modulated, thus not requiring a redundant (in terms of information content) opposite sideband to cancel phase shift. $\endgroup$
    – hotpaw2
    May 7, 2012 at 14:15
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I interpret the question as follows: If we modulate a carrier with a pure tone using AM, we get a single set of sidebands, but if we modulate with phase modulation, we get an infinite number of sidebands, spaced at the modulation frequency. Why?

It is easy to see why amplitude modulation at a single frequency gives exactly two sidebands. Simply multiply out the expression for AM:

y(t) = (1 + m cos(Ω t)) exp(i ω t)

y(t) = (1 + (m/2) ( exp(i Ω t) + exp(-i Ω t) )) exp(i ω t)

Here we see that we get sidebands offset by the modulation frequency Ω from the carrier frequency ω.

Now, phase modulation. I refer you to this animation (generated by this matlab script) of the phasor diagram:

Phasor animation of phase modulation

As seen in the animation, the higher order sidebands are necessary to keep the amplitude of the resultant phasor (in red) constant and thus produce pure phase modulation. You can see how each pair of higher order sidebands is needed to correct the deviation from a circular arc introduced by the lower order sidebands.

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  • $\begingroup$ Try out some really high modulation depths with the matlab script -- it's mesmerizing! $\endgroup$
    – nibot
    Jul 24, 2012 at 12:05
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Mixing audio with a carrier is exactly the same as mixing an incoming signal with a local oscillator to get an intermediate frequency. In both cases you result with the original frequencies, the sum of the frequencies, and the difference between the two frequencies. Whenever you mix frequencies, this results. When two people sing together, harmonics result. If the difference between their notes is in the audible range, you will hear it. I have heard quartets sing, and a deep bass note emerge that wasn't sung or played.

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    $\begingroup$ When two people sing together, harmonics result? and if only person is singing, harmonics are nowhere to be found? $\endgroup$
    – Dilip Sarwate
    Jul 22, 2015 at 12:50
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    $\begingroup$ Really interesting how these sidebands somehow emerge in FM almost unexpectedly and not as intuitive as just adding sine waves together. I suppose it’s very hard to think in the Frequency domain as an entire dimension that is convolutedly interrelated with the more familiar time & amplitude domain. I suppose that’s why they call it the “convolution”? $\endgroup$
    – TrinitronX
    Dec 6, 2018 at 3:21

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