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Power and energy signals

  1. Sep 23, 2011 #1
    Hello PF,

    I have trouble understanding these 2 types of signals. I mean definition is clear, math is also mostly clear, but intuition isn't.

    Can somebody explain to me, why signals that have finite energy have 0 average power. Physical explanation if you may.

    I learned that average power in AC circuits is power dissipated in Joule heating, but I cannot relate that to signals.
  2. jcsd
  3. Sep 24, 2011 #2
    To add: I have trouble understanding the concept of average power.
  4. Sep 24, 2011 #3
    A periodic signal repeats itself every period: g(t+ T0) = g(t).

    In a sinusoid for example, g(t) = sin(t), intuitively you can say that in one period, half the time g(t) varies from 0 to 1, and half the time g(t) varies from 0 to -1. In other words, g(t) varies from 1 to -1 in one period. And since g(t) is periodic, you can observe that on average g(t) is zero over the whole time interval. This kind of leads to the notion that the average of just the g(t) is not useful.

    So instead, we take the average of g(t)^2 (definition of a power signal). This makes any negative value of g(t) positive as to avoid any cancellation with positive value of g(t).

    If we take g(t) = sin(t), then g(t)^2 = sin(x)^2

    To the get intuitive feel for it, we notice that sin(x)^2 fluctuates evenly between 0 and 1 in one period. So on average sin(x)^2 is (0+1)/2 = 1/2. It means that the function spends most of the time around 1/2.

    Hence, the power{sin(t)} = 1/2 (average of sin(t)^2 per period). To obtain the average of a sin(t) from the power, take the square root of power{sin(t)}= 1/sqrt(2) = 0.707 (root mean square or RMS).

    If that kind of made sense, an energy signal is defined to approach zero as time goes to infinity. Something like f(t)= e^-t is an energy signal, which is not periodic, so its power is not defined. And likewise, a power signal doesn't approach zero as time goes to infinity, so its energy is not defined.
    Last edited: Sep 24, 2011
  5. Sep 24, 2011 #4
    Thank you for your in depth reply. I think I understand what you stated in your post. In any case you gave me a lot of things to work with in order to understand. Some tricks that you showed me are not in textbooks. I appreciate that very much.

    One more thing tho, do you by not defined, mean infinity?
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