What does the $T$ really represent in the signal power equation?

Question

In the signal power equation

$$ P(g(t)) = \lim _{T\to \infty }{\frac{\left(\int_{-\frac{T}{2}}^{\frac{T}{2}}g\left(t \right)^2dt\ \right)}{T}} $$ What does the $T$ really represent?

From my understanding, the target of the power equation is to get the average power throughout the signal, which is basically Energy / Time , so we divide the integral "Energy" over the time $T$.

but i don't seem to understand why are we taking the limit of $T$ to infinity at all?

e.g. if the signal was a sine wave $$g(t)=\sin(t)$$ then using This calculator in desmos i found out that we only get the average power as 0.5 (the correct answer) if $T$ was a multiple of the periodic time of the sine wave

so changing $T$ to other values than multiples of the periodic time gives wrong answers, so why are we taking the limit to infinity at all? or does $T$ has another meaning?

Edit 1:

Also, can we change the upper/lower bounds in the energy equation to $\int_0^T$?

i checked and they give the same value

Answer 1

$T$ is indeed time. For a sinusoidal signal, it does not matter if you have an integer number of periods if the number of periods is very large. If you have a million periods, then what you get over a part of a period is divided by a large value of $T$ and does not affect the value of the limit.

If you integrate from $-T/2$ to $T/2$, this may be OK as a definition, but for a signal that starts at the time moment of 0 you will get a different result.

Answer 2

so changing 𝑇 to other values than multiples of the periodic time gives wrong answers, so why are we taking the limit to infinity at all? or does 𝑇 has another meaning?

This is "false" for $T\rightarrow\infty$: $$\lim_{T\rightarrow\infty} \frac{1}{T}\int_{-T/2}^{T/2} \sin^2 t = \lim_{T\rightarrow\infty}\frac{T-\sin T}{2T} = 1/2 $$ so the formula gives consistent result with averaging the power over a periodic time (because $T\rightarrow\infty$ is secretly a periodic time for all periodic functions).

So then why are we taking $T$ to infinity.

The formula is given assuming your function $g$ is over all time, which it needs to be for a truly periodic function. Otherwise the Fourier transform won't look like delta functions at the specified frequency. In practice, no signal is like this but we still treat them as such for the purposes of finding the spectrum and also the average power because this treatment is more meaninful, i.e. it answers the questions we want answered.

Also, can we change the upper/lower bounds in the energy equation to $\int^𝑇_0$?

If for some reason your signal was defined only from $t=0$ to $t = T$ for some $T$, then changing the bounds would give you the correct average power. If you take $T$ to infinity again because your signal starts at $0$ and goes for all time, then it is still averaging over a periodic time so the average power is still correct.