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I am studying the Wiener-Khinchin Theorem.

But, I am wondering why the Dirichlet condition, which says that the autocorrelation function of WSS should be absolutely integrable, is sufficient for the existence of the Fourier transform.

And why the autocorrelation function $R(\tau)=E[x(t)x(t+\tau)]$ of WSS is absolutely integrable $\int_{-\infty}^{\infty}|R(\tau)|d\tau \lt \infty$ ?

Is it the case that it is not actually integrable, but it is just assumed that the autocorrelation function of the WSS process is absolutely integrable?

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The Dirichlet conditions are sufficient conditions to guarantee the existence and convergence of the Fourier series or the Fourier transform. A proof can be found here.

The autocorrelation function of a WSS proceess is not necessarily absolutely integrable per se. In the Wiener-Khinchin Theorem, this assumption is made so that the power spectrum can be defined via the Fourier transform.

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Thank you. By the way, if the autocorrelation function is absolutely integrable (or, square integrable), then does that indicate finite energy of a signal? In this case, do I have to see energy spectral density because power spectral density goes to zero as interval $T$ goes to $\infty$? – Jon Lee Oct 8 '12 at 15:00
The autocorrelation function being absolutely integrable (or, square integrable) does not mean the random process is finite energy. Actually, a nontrivial stationary random process certainly cannot have finite energy, e.g. white noise.For the definition of Winer-Khinchin Theorem, there is no $T$ involved. – chaohuang Oct 9 '12 at 2:00
I was wondering if you were saying Fourier transform pair of PSD and autocorrelation function for the definition of Wiener-Khinchin Theorem. Here ( ) is a proof involved with interval $T$ that goes to $\infty$. – Jon Lee Oct 9 '12 at 5:33
Yes, but the Wiener-Khinchin theorem is really a definition rather than a theorem. The proof is to show that the definition through limit ($T \rightarrow \infty$) is equivalent to the one through the Fourier transform of the autocorrelation function. – chaohuang Oct 9 '12 at 5:49
So, I was wondering if I have to consider energy or power because power spectral density goes to zero as $T$ goes to $\infty$. – Jon Lee Oct 9 '12 at 15:04

There is a fundamental mistake in both answers. It is not necessary to assume the autocovariance function is integrable. For a non-ergodic stationary process, it will not be integrable. The reason why Wiener gets credit for this theorem, instead of physicists like Schuster or Einstein, is that he was able to rigorously make sense of its Fourier transform anyway, in a new way, which he called «Generalised Harmonic Analysis», instead of the usual notion of the Fourier transform as given by the integral you write down. (In fact, he even anticipated Laurent Schwartz's notion of a distribution in his work on this.) So the Wiener-Khintchine theorem states that as long as the original process $f$ is stationary and has an auto-covariance function at all, then in this new sense of Fourier transform (which works even when the Dirichlet conditions are not satisfied), the power spectral density function (which can have infinities since it is the derivative of a function which is not differentiable, and so only makes rigorous sense as a distribution) is the Fourier transform of the auto-covariance function.

==About power and finite energy signals==

If the signal has finite energy, the power is zero, as follows from your formulas below.

But only a transient signal can have finite energy. The probability of sampling a transient signal from a stationary process is zero. Transient is the exact complete opposite of stationary.

A simple unit square wave for one-cycle only has finite energy, but zero power, and as you can calculate its sample auto-covariance function easily, you see it is zero. It has to be, since the Wiener-Khintchine theorem says the Fourier transform of the auto-covariance is the power spectral density and we just saw the power is zero.

Summarizing: finite energy (which means transient) ==> zero power ==> zero sample auto-covariance function.

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Now, about power. Power is already per unit time, so you don't go on and divide it again by time when, as here, the process is stationary, so that in some sense, the power is constant. (Its statistical expectation over all possible signals is constant.) So the formula in the other answer, below, is really for the total power of the signal, not any sort of time-averaged power.On the right hand side, you are dividing by T, and that makes it be per unit time, which is what gives it the dimensions of power. To get time-averaged power, you'd have to divide by T again, which is useless. – José Sanchez Nov 6 '12 at 0:25
Thank for your answer, Sanchez. But if the estimate of power spectral density had infinity, is the estimate still correct? Isn't it inconsistent? I am also wondering the existence.. – Jon Lee Dec 10 '12 at 18:19

This is about power and energy of signals.. I thought the power spectral density may approach zero when interval $T$ goes to $\infty$. Maybe I need to understand this compared with Wiener-Khinchin theorem... Please help me.

Total energy of signal $x(t)$: ${E_\infty } = \mathop {\lim }\limits_{T \to \infty } \int_{ - T}^T {{{\left| {x\left( t \right)} \right|}^2}dt} = \int_{ - \infty }^\infty {{{\left| {x\left( t \right)} \right|}^2}dt}$

time-averaged power: ${P_\infty } = \mathop {\lim }\limits_{T \to \infty } \frac{1} {{2T}}\int_{ - T}^T {{{\left| {x\left( t \right)} \right|}^2}dt}$

if signal energy is finite, ${P_\infty } = \mathop {\lim }\limits_{T \to \infty } \frac{{{E_\infty }}} {{2T}} = 0$.

For some infinite energy, $P_\infty \gt 0$. For example, $x(t)=4$ has infinite energy, but average power $P_\infty=16$.

When it comes to the Wiener-Khinchin theorem, it is assumed the autocorrelation function $r(\tau)$ is absolutely integrable. $\int_{ - \infty }^\infty {\left| {r\left( \tau \right)} \right|d\tau } < \infty $.

I was confusing this integrability compared with the time-averaged power spectral density in the Wiener-Khinchin theorem.

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So you can see that the "time-averaged power" $P_{\infty}=0$ for finite energy signals as $T \rightarrow \infty$, but I still don't see why the time-averaged power spectral density goes to zero when $r(\tau) \in \mathbf{L}^1(\mathbb{R})$. Maybe you confuse the "time-averaged power" with the autocorrelation function. – chaohuang Oct 10 '12 at 5:30

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