I am figuring out how to calculate average power for continuous periodic signals, I have learned what the rest of the equation means however I need some guidance on why the $(t)$ on the LHS is equal to $(2\pi t)$ on the RHS.
There are some holes in my understanding so please could this be explained in simple terms? I know it is something to do with $2\pi$ being the amount of radians required to do one full cycle of a circles circumference however it would help greatly if someone could make the connection here a bit clearer.
Continuous signals are mostly represented via voltage. The average power of a periodic voltage signal can be defined as: $$ P_{avg} = \frac{\int_{0}^{T} u(t)i(t) \mathrm{d}t}{\int_0^T\mathrm{d}t} = \frac{\int_{0}^{T} u(t)i(t) \mathrm{d}t}{T} $$ The voltage across an element is defined as $ u(t) = R(t)i(t) $. If we assume that we have a time invariant element and without lost of generality assume that $ R \equiv 1$ we get the average power as: $$P_{avg} = \frac{\int_{0}^{T} u^2(t) \mathrm{d}t}{T} $$ I assume that in your example the signal is a sine wave (even if it isn't by Fourier you can decompose it into sine waves) so the general expression for a sine singal would be: $$ u(t) = A\sin(\omega t + \phi) $$ In your example the phase $ \phi = 0 $ and the amplitude $ A = 1 $. The angular frequency $ \omega $ can be expressed in terms of period as $ \omega = 2\pi/T $, and because you have a $ 2\pi $ inside your function, then $ \omega = 2\pi $ and the period you have is then $ T = 1 $ so to just write it out: $$ P_°{avg} = \frac{1}{T} \int_{0}^{T} u^2(t) \mathrm{d}t = \frac{1}{T} \int_{0}^{T} A^2\sin^2(\omega t + \phi) \mathrm{d}t = |A = 1, \omega = 2\pi, \phi = 0, T =1| =$$ $$ = \frac{1}{1} \int_{0}^{1} \sin^2(2\pi t) \mathrm{d}t$$ I hope the answer is satisfactory.