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Power of a sine wave (electronics engineering)

  1. Jun 15, 2009 #1
    This is obviously an electronics question.
    In communication systems, to calculate the power of a sine wave, the formula below is used
    Power (Sine Wave) = 1/2 * (peak amplitude)^2

    This formula is apparently a standard electronics formula.
    I'm trying to understand where it comes from. How is it possible to get a power figure from a voltage only? There is no information on period, current, or resistance. Can some one please help explain the logic behind this formula for me?

    Maybe its a simple trick, to me they've probably simply integrated the sign wave, but the limits must cancel out... i Don't know.... help please..

  2. jcsd
  3. Jun 15, 2009 #2
    priscared, I'm pretty sure the definition makes the assumption of "1" unit of resistance - where the unit is whatever makes sense to produce the power units you choose. I've asked this question several times, and that is the only answer I've seen.
  4. Jun 16, 2009 #3
    yeah... normalised with a 1 Ohm resistor.
  5. Jun 21, 2009 #4
    >> I'm trying to understand where it comes from.

    It is derived from the http://en.wikipedia.org/wiki/Poynting_vector" [Broken].

    >> How is it possible to get a power figure from a voltage only?

    P = UI = U*(U/R) = U2/R
    Last edited by a moderator: May 4, 2017
  6. Jun 21, 2009 #5
    Have you ever tried Fourier Transform of a pure sine?
  7. Jun 21, 2009 #6
    what does that have to do with anything trambolin? I mean I know rayleighs equality... but we are wondering about the units here, right?
  8. Jun 21, 2009 #7
    I meant power of a infinite signal in terms of 2-norm, Plancherel, parseval etc. If everthing is normalized as you did for 1 ohm resistance... but anyway, nevermind, maybe you are right.
  9. Jun 22, 2009 #8
    hmmm... indeed the poynting vector may have some relevance. But if some1 could point me to a meaningful explanation i would be impressed. I am assuming the assumptions is resistance = 1. And thats how the formula exists.

    The rule almost seems like an anomaly, every1 uses it without thinking about its origin...
  10. Jun 23, 2009 #9
  11. Jun 23, 2009 #10
    " A signal's instantaneous power is
    defined to be its square, as if it were a voltage or current passing through a 1 Ω resistor. "

    THanks for the link

    This rule is obviously just an arbitrary convention. It's obviously the definition of "instantaneous power".

    BTw i can't see how this has anything to do with Fourier transforms.
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