Say that the average power of a signal i(t) is G.
What is the average power of i(t)cos(ωt)?
I think it might be 0.5G, but I don't understand why.
Can you help please explain it?
Thank you.
Hi,
This question is a little ill posed. On the one hand you are asking about something that sounds common but on the other hand you are asking a question that is very uncommon. Let me try to explain a few points here to help clear this up...
The average power of a current all by itself is zero. That's because a current with no voltage produces no power. This makes us assume that you are probably asking about the one ohm energy content of the signal. That would mean that the resistance is assumed to be 1 ohm and since we know the current we can calculate the instantaneous power and so we can find the average power.
The second point is that it is a sinusoidal wave which has an average of zero, but when we deal with a sinusoid we dont usually average over the entire period but over 1/2 the total period. Thus if the frequency is f, we dont average over the time period 1/f (which would give us zero which doesnt do us any good) but instead we average over 1/2f, which is half the period. This gives us the average because power from a sinusoid is generated in both halves of the sine wave. Doing it this way we would find that the average is i^2/2 for some peak value represented by i, which would be a constant. But you stated that i is not a constant and is really an unspecified function of time.
The third point, and probably the most critical, is that you are stating that i(t) is not constant. That means that i(t) is a function of time. This really throws the problem out the window because there's no way to know what the average is of a wave represented by a function that is not known or not specified.
However, since you have specified the average of i(t) as G, what we need to prove then is that the integral of the square of a function i(t) times the integral of the square of the function cos(wt) is equal to the integral of the square of the two functions multiplied together over the period T/2 (and each integral scaled by 2/T or 2/Ta):
(2/Ta)*Integral i(t)^2 dt * (2/T)*Integral cos^2(wt) dt = (2/T)*Integral ((i(t)*cos(wt))^2) dt (all over the period 0 to T/2 or Ta/2)
or in Latex:
[LATEX]\frac{2}{Ta}\int^{Ta/2}_{0} i(t)^2\,\, dt * \frac{2}{T}\int^{T/2}_{0} cos^2(wt) \,\,dt = \frac{2}{T}\int^{T/2}_{0} (i(t)*cos(wt))^2 \,\,dt[/LATEX]
I dont think this is true in general unless some constraint is imposed on the function i(t) such as maybe it has to be the same frequency as cos(wt). If there is some constraint on i(t) you should let us know now. See if you can find at least one function for i(t) that doesnt work in the above.
This assumes i understand your question correctly.
Interesting, it looks like it works for sinusoidal signals so far when the signals are harmonically related.