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Conventional AM average power

  1. Mar 21, 2014 #1
    1. The problem statement, all variables and given/known data
    Trying to calculate Conventional AM average transmitted power of the modulated signal
    s(t) = A (1+ka m(t) ) cos(2∏f t)
    where
    A: carrier amplitude
    ka: amplitude sensitivity
    m(t) : message signal
    f: Carrier frequency

    my professor's lecture notes calculate the average power as
    0.5 A2( 1+< ka2 m2(t)>)
    where <.> is the time average.
    2. Relevant equations

    Average Power (for periodic Signals) = 1/T (∫0T x2(t) dt)

    3. The attempt at a solution
    average power for cos(2∏ft) =0.5
    but applying the Average power rule I can't get this integral to be equal to the formula my professor's got!
    Thanks in advance.
     
    Last edited: Mar 21, 2014
  2. jcsd
  3. Mar 22, 2014 #2

    rude man

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    Rewrite s(t) as A km(t)cos(wt) + Acos(wt), w = 2 pi f.
    Then do s^2(t). Hint: as you will recall, (a+b)^2 = a^2 + 2ab + b^2.

    So making the appropriate substitutions for a and b,
    avg{s(t)^2} = avg(a^2) + avg(2ab) + avg(b^2).
    Etc.
     
    Last edited: Mar 23, 2014
  4. Mar 23, 2014 #3
    so avg{s(t)^2} = avg{A^2 k^2 m(t)^2 cos^2(wt)} + avg{2A^2 k m(t)cos^2(wt)}+avg{A^2 cos^2(wt)}
    still who says avg{A^2 k^2 m(t)^2 cos^2(wt)} = avg{cos^2(wt)} * avg{A^2 k^2 m(t)}
     
  5. Mar 23, 2014 #4

    rude man

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    Excellent question! The answer is that f(t) is assumed to be slowly-varying compared to cos(wt), and that the avergage of f(t) = 0. This is certainly the case in "ordinary" AM where f(t) is limited to a few sinusoidal (zero-avg.) KHz whereas w = 2 pi*1 MHz typical.

    The prof should have stated this for you IMO.
     
  6. Mar 23, 2014 #5
    Well, he did state that assumption, although what I understand is that the assumption that the average of the message is 0 means we will neglect the term avg{2A^2 k m(t)cos^2(wt)}.
    However,my question is
    if ∫ ab is not equal to ∫a ∫ b
    then how come average(ab) = average(a) average(b)
    Thanks !
     
  7. Mar 23, 2014 #6

    rude man

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    Right.

    Are you referring to the third term 2ab or the term A^2 k^2 m^2(t) cos^2(wt)? I'll assume the latter since you seem to agree that the 2ab term disappears. The answer is that m(t) varies slowly compared to cos(wt) so m(t) is approx. constant over 1 cycle of cos(wt) and so that term average boils down to A^2 k^2 <m^2>/2.
     
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