Understanding the Definition of Average Power in Sinusoidal Functions

[jakey]

I am slightly confused by the definition of average power if the power function $p(t)$ is sinusoidal. Why is it that only one period is considered?

I mean I know that it simplifies calculations but if we assume that the period of $p(t)$ is $T$ and I compute the average power over $[0,\sqrt{2}T]$, I do not get the same result had I computed the average power over $[0,T].$

Case in point: If $p(t)=\frac{1}{2}(1+\cos(2x))$ then the average power is not the same for both cases mentioned...

 

[sophiecentaur]

If you want to know the instantaneous power then you would use VI. For a mean power value, you could choose any period you liked, to integrate over, but you would need to state that period (absolute phase intervals). It seem perfectly reasonable to me to choose a single cycle (or any integral number) because it's the most likely thing that anyone else would do. Any other period would be arbitrary and could introduce a massive range of possible outcomes (as you seem to be finding).

 

[jakey]

so are you saying that this is simply a definition? I'm sorry but I still can't seem to understand it...so the average power based on this definition, then, is merely an approximation of the "real mean"?

 

[sophiecentaur]

Think how you'd tackle a 1kW electric heater. If you took the first 10ms of one cycle, the average power would be somewhat less. Would that make sense?

 

[DrewD]

so are you saying that this is simply a definition? I'm sorry but I still can't seem to understand it...so the average power based on this definition, then, is merely an approximation of the "real mean"?

The "real" mean depends on how long you are averaging over the power. Averaging over one period gives the long-time average. If you were given a power function and asked the average power over a certain period of time, then you would average over just that time. If the period (of the power function) is short, you are probably interested in the average over many periods rather than one small random interval. The average over one period approximates this quite well.