Last night in class we talked about the average power for any periodic function. enter image description here

I'm just confused of why the one halves are in the bounds for this equation. Since we are letting tau approach infinity the bounds just approach negative infinity and infinity respectively. If anyone can give an explination on why they put 1/2 I would appreciate it.

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    \$\begingroup\$ That makes the entire integral be over a time period of τ, rather than 2τ. You could define it either way, it makes no difference. \$\endgroup\$ – Hearth Feb 8 at 18:02
  • \$\begingroup\$ The first starts at any time and averages one T relating cycle, while the 2nd averages over all time before and after T/2 so for random power not periodic cycles, like last month’s electric bill and next month /2 months and so on. kWh / hrs/mo. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 8 at 18:12

The bounds are not actually important mathematically, but they help with meaning.

You could have written \$P=\lim_{\tau\to\infty}\frac{1}{2\tau}\int_{-\tau}^\tau p(t)dt\$ and gotten the same results. However, if you consider all of the different power functions:

  • Finite bound: \$P=\frac{1}{\tau}\int_{T}^{T+\tau} p(t)dt\$
  • One sided infinite: \$P=\lim_{\tau\to\infty}\frac{1}{\tau}\int_{0}^\tau p(t)dt\$
  • Two sided infinite: \$P=\lim_{\tau\to\infty}\frac{1}{\tau}\int_{\frac{-\tau}{2}}^\frac{\tau}{2} p(t)dt\$

We see that \$\tau\$ has the same meaning of "period" or size of integration window in all of these cases. This is convenient for looking for parallels between the different equations. If we integrated the two sided infinite equation from \$-\tau\$ to \$\tau\$, it would obscure the connection between these different forms.

You'll also see this happen later with many transforms, like the Fourier transform. We could change the way we define things to avoid some annoying \$2\pi\$ terms that arise, but that would then obscure the connection between the Fourier transform of some data and its original form.


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