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

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

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

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

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 ?

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