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In a proof for the Wiener-Khintchine theorem (See p. 572-573)

Screenshot from the referenced page of text book.

I have seen the following operation being done:

\begin{align*} S_{xx}(f) &= \lim_{T \rightarrow \infty} \int_{-2T}^{2T} Rxx(\tau) e^{-j 2\pi f \tau} (1 - \frac{|\tau|}{2T})d\tau \\ &= \lim_{T \rightarrow \infty} \int_{-\infty}^{\infty} Rxx(\tau) e^{-j 2\pi f \tau} q_T(\tau)d\tau \\ &= \int_{-\infty}^{\infty} Rxx(\tau) e^{-j 2\pi f \tau} d\tau {\qquad\qquad\qquad\qquad\qquad\square}\\ \end{align*}

The argument given for this proof is that the function $(1 - \frac{|\tau|}{2T})$ will be approximately equal to 1 if $T$ goes to infinity and $\tau$ is a finite value, thus giving the final result.

I would argue that this does not immediately hold: we are integrating $\tau$ from $-2T$ to $2T$, and at both of these bounds the value of $(1 - \frac{|\tau|}{2T})$ would be 0: we are essentially finding the Fourier Transform of $Rxx(\tau)$ windowed with an infinitely large triangular window, but the value of this function will not be constant over the domain as we take the limit since it must be equal to 0 at the integration bounds. In my mind, this would highly distort the shape, and thus impact the result of this integral.

While an alternative approach would be possible (e.g. saying this windowing has the effect of convolving with an infinitely thin sinc^2 function in the frequency domain which is approximately equal to a delta distribution, thus not affecting the spectrum), I am wondering if there is some reasoning for this step in time-domain specifically.

This was also partially mentioned in the response to this question.

Finn Heijink
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  • At a time move only one variable, either you process the limit -OR- you process the integration. Don't do both at the same instant. – abhilash Jun 16 '23 at 18:13
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    related: this answer – Matt L. Jun 17 '23 at 08:35
  • "[...] but the value of this function will not be constant over the domain as we take the limit [...]". Yes, in the limit it would, because $\lim_{T\to\infty}(1-|\tau|/(2T))=1$. And the bounds where the triangle would be zero are shifted towards infinity. – Matt L. Jun 17 '23 at 13:36
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    @V.V.T: I know that one, but the OP doesn't seem to want to take the route via the frequency domain, but wants to see the proof directly in the time domain; at least that's what I understood from the post. And my remark about the limit was just concerning the OP's statement that I quoted in my previous comment. – Matt L. Jun 17 '23 at 16:41
  • @MattL. thank you for the response. I am not too familiar with improper integrals hence my confusion, but would this not cause issues considering that infinity is still a part of the domain of the integral? How do we differentiate between $\lim_{T \rightarrow \infty} 2T$ and $\infty$ itself, considering that we made the simplification of the integral by using $\infty > 2T$ when defining $q_T(\tau)$? – Finn Heijink Jun 18 '23 at 14:15
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    @FinnHeijink: The authors used infinite limits because they defined $q_T(\tau)$ to be zero for $|\tau|>2T$. That's why the second equation after the line $S_{xx}(f)=\ldots$ in your post is wrong, because in that case the triangular function would become negative for $|\tau|>2T$, when it actually should be zero. – Matt L. Jun 18 '23 at 16:47
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    I could write up an answer, but I'm afraid I couldn't add much more than what you already read in the other answer you linked to at the end of your question. I.e., formulate a sufficient condition such that the Wiener-Khinchin theorem is satisfied. – Matt L. Jun 18 '23 at 16:49
  • @MattL. you are completely right, thanks for letting me know. I think the proofs mentioned earlier using the Inverse Fourier Transform will have to suffice in that case, thanks for all the help. – Finn Heijink Jun 18 '23 at 19:34
  • @FinnHeijink: Also check out the proof I linked to in my first comment. It avoids that kind of trouble. – Matt L. Jun 18 '23 at 19:39
  • @MattL. I know this is very late but I thought I would still let you know, the time-domain approach could probably work if we consider the input signals to be absolutely integrable and smooth (which we can do if we use applicable sequences which approach the input signal of interest), since we can use the dominated convergence theorem to move the limit of T inside of the integral and end up with the known result. I am still trying to learn more and more about the mathematics involved in signal processing so please let me know if you agree or disagree with the method. – Finn Heijink Oct 13 '23 at 15:27
  • @FinnHeijink: I think it would probably be best if you added your proof to your question, so the community can check it and give you feedback. – Matt L. Oct 14 '23 at 08:48

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Since we are dealing with convergence in the distributional sense when working with non-L1 functions, we can handle the functions inside of the integral as Schwarz-class functions: we can use the dominated convergence theorem due to the absolute convergence, meaning the limit can be placed inside of the integral. This yields:

$\lim_{T \rightarrow \infty} \int_{-\infty}^{\infty} Rxx(\tau) e^{-j 2\pi f \tau} q_T(\tau)d\tau = \int_{-\infty}^{\infty} \lim_{T \rightarrow \infty} Rxx(\tau) e^{-j 2\pi f \tau} q_T(\tau)d\tau = \int_{-\infty}^{\infty}Rxx(\tau) e^{-j 2\pi f \tau}d\tau = \mathscr{F}\{R_{xx}(\tau)\}$

Which concludes the proof.

Finn Heijink
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