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Calculate an average power of a signal

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Given the following signal,

    ##v(t)=2\cos (2\pi f_{0}t)+4\cos (4\pi f_{0}t)+6\cos (6\pi f_{0}t)+8\cos (8\pi f_{0}t)##

    1. Calculate the signal's average power in the time domain.
    2. Calculate the signal's average power in the frequency domain.

    2. Relevant equations
    Didn't make a real attempt for a solution as yet, but I try to think of a shorter, easier way other than a straight calculation using the exact definition. If there's a hint according to the even form of the given cosines' amplitudes and frequencies that you could advise me of - it might make my way significantly easier here, I feel.

    Thanks in advance for any guidelines suggested!

    Adir.
     
  2. jcsd
  3. Oct 25, 2015 #2

    berkeman

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    Staff: Mentor

    The average power is zero.

    Oh wait, the average power is very high.

    Oh wait, what is missing that keeps us from calculating a power given the voltage only...? :smile:
     
  4. Oct 25, 2015 #3

    gneill

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    I think we have to presume that the "signal" is not necessarily electrical, but is a signal in the more general mathematical sense. In such a case, for a periodic function f(t) the power is defined to be (If my memory serves... it's been a while...):

    $$P = \frac{1}{T} \int_0^T |f(t)|^2 dt$$

    I think it gets more complicated when the signal is not periodic and you have to integrate using limits from -∞ to +∞.

    There may be a trick for finding the power of a sum of sinusoids involving summing the squares of the amplitudes and dividing by two... don't quote me on this :nb), as I say, it's been a while...
     
  5. Oct 26, 2015 #4
    Ok, so
    ##P_{av}=2^2+4^2+6^2+8^2=4+16+36+64=120\left [W \right ]##

    But why you decided to divide this sum by 2 if ##T## is unknown?
     
  6. Oct 26, 2015 #5

    gneill

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    T is the period of the periodic signal. In this case you can see that the lowest frequency is ##2 \pi f_o##, and all the other terms have frequencies that are multiples of this. That means the signal as a whole will repeat every ##1/f_o##, so that's its period.

    As to why the sum of squares is divided by two, that follows from the power integral for a cosine function. Do the integral for a single cosine; you can use the angular period in place of the time period since it covers the same domain for the function:
    $$\frac{1}{2 \pi} \int_0^{2 \pi} |cos(\theta)|^2 d \theta$$
     
  7. Oct 26, 2015 #6

    rude man

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    For the freq. domain, what did you learn about power for a Fourier series?
    For the time domain, it's (1/T)∫T v2(t) dt where T comprises an integer number of FUNDAMENTAL cycles.
    The assumption in problems like this is R = 1 Ω.
    Main thing is to get the same answer both ways! :smile:
     
  8. Oct 26, 2015 #7

    berkeman

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    Sorry, I don't get it. How can you ignore the load impedance when calculating the power of a voltage waveform supplied to that load?
     
  9. Oct 26, 2015 #8

    rude man

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    Yes, but in communications theory for example we talk of the power of a signal, assuming 1 ohm:
    http://mathworld.wolfram.com/AveragePower.html
    Note that v(t) was not specified to be a voltage, just a "signal".
     
  10. Oct 27, 2015 #9

    rude man

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    For sine waves, rms = peak/√2 so power = rms2 = peak2/2.
     
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