# Average Power Equation

The equation below states the following:

P = G * v^2(t)

W = Integral (P) over all time

Paverage = W/(time period)

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

• 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. – Spehro Pefhany May 2 '17 at 17:10

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.

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

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

• The 2 is because the limits are from $-\tau$ to $\tau$, so the total durationof the integration is $2\tau$. – The Photon May 2 '17 at 16:56
• that makes sense , but as τ goes to ±∞ but , what is 2*±∞ or ∞/2? It's an academic formula with very little practical use. – Tony Stewart Sunnyskyguy EE75 May 2 '17 at 17:07
• 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. – The Photon May 2 '17 at 17:19