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If we take the average of two quantities a and b we have to sum the two and then divide by 2.
If we evaluate over 0 to T we may miss some of the information that is contained in the waveform before t=0. To account for this symmetrical limits allow looking at both the minus and plus axis values. But when we do that we end up with too much quantity so we have to divide by 2.
So if we want to integrate but dont want to miss all that for -t we would have to integrate from -T to T and then let T go to infinity, and then we would have twice the quantity we needed so we'd have to divide by 2 (to get the actual average).
So we could use limits -T/2 and T/2 and just divide by the 'usual' T. It probably also works to use limits of -T to T and then divide by 2T.
Another way of looking at it is that for two quantities a and b the average is (a+b)/2, where a is the stuff to the left of 0 and b is the stuff to the right of 0. The sum a+b is too much because that's just the sum and not the average. Dividing by 2 gives us the average.
One of the ideas here is to be able to include non periodic signals as well as the periodic ones.
Well for the integral equation you've shown there with different limits and different multipliers outside the integration, i think it becomes obvious that they have to be the same because on the left we are looking at two periods of the same thing as we are looking at on the right, except on the right we are looking at only one period so the quantity under the integral on the right has to be one half of the quantity under the integral on the left, which means to maintain equality we have to multiply the left side integral only by 1/2 which is what is being done. Here because of the period T we have to assume the waves are periodic.
We only have to look at a simple wave to see how this works. If we have half the wave on the left of zero and half on the right, we have one period of the wave. If we have one full period on the left of zero and one full period on the right we have two periods of the wave. It makes sense that the area for two halves of the period is one half of the area of two full periods. So in the equation since we are looking at two full periods vs two half periods we should see twice the area on the left as on the right unless we multiply the left by 1/2 (which is done already).
But this is a little elementary (if you graph the periodic function it becomes so very obvious that two areas are equal to twice one area of the same thing) so i think maybe what you meant to ask was what if we take the limits as T goes to infinity. In this case i think we should consider a non periodic function that extends over the whole x axis and that could be a function like:
y=sin(t)+sin(pi*t)
This function is not periodic yet it extends over all t, so performing the required operations should show the same result on both sides of the equation. However, the equation we are talking about is usually shown with the limits of -T/2 to T/2 so that would be the thing to stick with.
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