# PSDs and Parseval's theorem

OK, I have been banging my head for quite a while trying to make sense of this simple equation:

The $1/T$ is what is causing me all the grief. I assume that the square of the Fourier transform of the signal $X$ is a finite function. So won't the $T$ in the denominator mean that the PSD, $S_x$, will always tend toward zero? If $X$ does increase to infinity with time, can someone explain this to me? I am thinking that $X$ is a measure of the frequency content of the random time signal, and that shouldn't change appreciably with the time taken to measure it.

• Can you give a source for that equation? – Matt L. Sep 19 '17 at 16:12

## 1 Answer

That equation is indeed simple, but it's also wrong. So your doubts are completely justified. What is probably meant is something like

$$S_X(f)=\lim_{T\rightarrow\infty}\frac{1}{T}\left|X_T(f)\right|^2\tag{1}$$

with

$$X_T(f)=\int_{-T/2}^{T/2}x(t)e^{-j2\pi f t}dt\tag{2}$$

This definition of the power spectrum is used for functions $x(t)$ with finite power (but infinite energy) for which the Fourier integral does not exist.

• @ChrisDavidson: Concerning your (deleted) answer, note that $\lim_{T\rightarrow\infty}|X_T(f)|^2$ does not exist because for power signals the Fourier integral does not exist. – Matt L. Sep 19 '17 at 19:37