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Questions about a problem in AM

  1. Jan 29, 2014 #1
    I may post a few questions here and there in the upcoming days in this thread (all related to AM). I begin with the first one:


    and its solution:


    Relevant equations

    I have a two questions, both related to understanding the core of the AM process here:

    1. What exactly is the peak envelope power? If it equals 32 kW, then why do I need to square the whole (2Ac) term? If Am=Ac (equal amplitudes of both message signal and carrier signal), then why do I need that square operation? I feel like that's a purely basic understanding I lack to completely understand what's going on here.

    2. Sx=1/2
    Sx is supposed to be the average power of the message signal. Assuming only the amplitude of the carrier signal is known now (i.e Am=Ac=8kW) - How it's possible to calculate that 1/2 value?

  2. jcsd
  3. Jan 29, 2014 #2

    rude man

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    Homework Helper
    Gold Member

    Power goes as the square of amplitude.
    You can't write " ... i.e Am=Ac=8kW ... "

    Amplitude is V/m, power is E*H = poynting vector magnitude = (V^2)/Z with Z = 377 ohms for a plane wave in vacuo.
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