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AM amplitude modulation-

  1. Nov 17, 2013 #1
    AM amplitude modulation---

    Hi everyone, I am learning AM recently, but I have some question that keep troubling me..

    1. We know that AM wave is produced by the multiplication of carrier wave and modulation wave. But why does the equation of AM wave is not a direct multiplication of Vm×Vc but consist of a constant? Can anyone explain the general concept of the AM wave equation?

    Why it is
    Vam = [Ec + Emcos(ωmt)]cos(ωct)

    but not
    Vam = [Eccos(ωct)] [Emcos(ωmt)]

    2. In single sideband AM, why does the cancelling of one sideband doesn't affect the information in the wave? And what is the effect to the wave pattern after the AM wave has been transform from double sideband (DSBFC) to single sideband (SSBFC)?

    Last edited: Nov 17, 2013
  2. jcsd
  3. Nov 17, 2013 #2
    Hi! :smile:

    The amplitude of the carrier is EC. This is the base amplitude of the carrier. when it is modulated with a modulating signal eM, It's amplitude is changing according to the instantaneous value of eM.

    Have a look at the waveforms:


    Vam = [Ec + Emcos(ωmt)]cos(ωct)

    In this equation,
    [Ec + Emcos(ωmt)] refers to the instantaneous amplitude of the am wave, which is, you know, changing with time...
    cos(ωct) is the frequency component of the wave.

    Compare with v = Vm sin (ωt + [itex]\phi[/itex])

    If it was Vam = [Eccos(ωct)] [Emcos(ωmt)]

    That means [Eccos(ωct)] is the amplitude. In this case, the amplitude goes negative in one half cycle. It is meaningless. So the main amplitude (with zero modulating signal) is EM. But the instantaneous value of the modulating signal ADDS to this main amplitude EM. So that's why you have that equation! Have a closer and more detailed look at that waveform, it will be clear to you!
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