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The equation below states the following:

P = G * v^2(t)

W = Integral (P) over all time

Paverage = W/(time period)

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Can someone explain the time period (1/2t)? Why is it not 1/T (T = period of the time varying signal)?

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  • \$\begingroup\$ You're mixing up two concepts- the given formula is true for any signal. You can integrate over a single cycle (or an integral number of cycles) of a periodic signal and get the same answer. \$\endgroup\$ – Spehro Pefhany May 2 '17 at 17:10
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This formula applies to any (power or energy) signal, not just a periodic signal.

You are correct that for a periodic signal you could change the limits of integration to \$t_0\$ and \$t_0+T\$ (for any value of \$t_0\$ you like), change the time period factor to \$1/T\$, and not have to take a limit.

But not every signal is periodic.

Can someone explain the time period (1/2t)? Why is it not 1/T (T = period of the time varying signal)?

Actually, it's \$1/(2\tau)\$. This is because the limits of integration are \$-\tau\$ and \$\tau\$, so the total time period being integrated over is \$2\tau\$.

It's not \$T\$ because no period \$T\$ has been defined as part of the problem, and we haven't assumed the signal is periodic.

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The 2 has to do with the conversion from Peak Voltage to RMS Voltage .

\$ V_{rms} ≈ 0.707 V_{pk} = \frac{1}{\sqrt{2}} V_{pk}\$ thus

\$V^2_{rms}=\frac{V^2_{pk}}{2}\$

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  • \$\begingroup\$ The 2 is because the limits are from \$-\tau\$ to \$\tau\$, so the total durationof the integration is \$2\tau\$. \$\endgroup\$ – The Photon May 2 '17 at 16:56
  • \$\begingroup\$ that makes sense , but as τ goes to ±∞ but , what is 2*±∞ or ∞/2? It's an academic formula with very little practical use. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 2 '17 at 17:07
  • \$\begingroup\$ Whenever you deal with infinities, you have to deal with limits rather than try to treat infinity as just another number. And \$\lim_{t\to\infty}1/(2t)\$ is very well defined. \$\endgroup\$ – The Photon May 2 '17 at 17:19

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