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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.

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  • $\begingroup$ 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. $\endgroup$
    – abhilash
    Jun 16, 2023 at 18:13
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    $\begingroup$ related: this answer $\endgroup$
    – Matt L.
    Jun 17, 2023 at 8:35
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    $\begingroup$ @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. $\endgroup$
    – Matt L.
    Jun 17, 2023 at 16:41
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    $\begingroup$ @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. $\endgroup$
    – Matt L.
    Jun 18, 2023 at 16:47
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    $\begingroup$ 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. $\endgroup$
    – Matt L.
    Jun 18, 2023 at 16:49

1 Answer 1

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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.

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