0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

$T$ 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.

$\endgroup$
  • $\begingroup$ what i am asking about is the value of T itself, if it goes to infinity, that includes numbers that aren't multiples of the Periodic time, so why is it even there ? $\endgroup$ – bigworld12 Feb 18 '18 at 23:34
  • $\begingroup$ @bigworld12 : I don't understand it. Why is this a problem? As I said, the limit is well-defined, say, for sinusoidal processes. $\endgroup$ – akhmeteli Feb 19 '18 at 0:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.