6
$\begingroup$

I am struggling a lot in finding the reference (book or, even better, academic paper) that states that:

$u(t)$ $t^{α} ↔^{FT} f^{-(α+1)}$

as I found here An Interesting Fourier Transform - 1/f Noise in the formula in the first box of the link.

I searched a lot, but I did not find any reference, and they are very important to me.

$\endgroup$
10
  • 3
    $\begingroup$ (+1) It looks like you left out U(t) on the time domain side, but nice link! I bumped into that pair long ago, also without U(t), so I will see if I can find the paper and post the reference in a comment. $\endgroup$ – Ed V Apr 6 at 12:11
  • 2
    $\begingroup$ Thank you. I have found a lot of paper about 1/f noise but no one that deal with corrispondence between frequency power law and time power law $\endgroup$ – Mardy Owens Apr 6 at 12:30
  • 4
    $\begingroup$ I see answers have appeared, that I upvoted, but I finally found the reference I remembered. My hardcopy is long gone, but Smit and Walg, Chromatographia, 9(10), 1976, pp. 483-489, give their expression 19 as -2 ln|t| <=> 1/f. This is supposedly a Fourier transform pair, and they simply stated it as such, but it seems highly unlikely now. Sorry for the initial wrong reference! $\endgroup$ – Ed V Apr 6 at 15:13
  • 3
    $\begingroup$ This expression is somehow consistent with the similarity (to the limit) of $t^\alpha$ as $\alpha \to 0$ and the logarithm, as suggested here What is the logarithm of a kilometer? Is it a dimensionless number? $\endgroup$ – Laurent Duval Apr 6 at 15:47
  • 3
    $\begingroup$ @Ed V You gave me a nice incentive to looking a Smit papers again, I have read some for works in chromatography $\endgroup$ – Laurent Duval Apr 6 at 16:14
7
$\begingroup$

You will find a "proof" of principle in math-stackexchange: Fourier transform of power function $t^\alpha$ (the title lacks the mention of the unit step function $u(t)$). At least formally, the Fourier transform of one-sided power law turns into a Laplace transform, and looks a lot like the Gamma function (the one related to the factorial).
You can find a similar description on page 276 of the chapter Appendix A. Mathematical Background, Power law. What I find troubling in this "simple proof" is that the convergence and well-behaved properties of the power-law functions are not satisfied. They are not integrable in the traditional sense, so it should be treated with tempered distributions or generalized functions, and Schwartz functions decaying faster than powers.

You will find a clean demonstration in Terence Tao blog: 245C, Notes 3: Distributions. 3. Tempered distributions, around Equation 9. In dimension $d$ ($d=1$ for you), the formula reads:

$$\widehat{u(t)t^\alpha}(f) = \frac{\pi^{-(d-\alpha)/2}\Gamma((d-\alpha)/2)}{\pi^{-\alpha/2}\Gamma(\alpha/2)}|f|^{-(d-\alpha)}$$

[Nota: this is a follow-up of question: Can I simplify $x=\frac{\ln(|fft((ifft(\sqrt{(f^{-5/3})}))^{2})|)}{\ln(f)}$]

$\endgroup$
2
  • 2
    $\begingroup$ Thank you. If you find other references, it's really important for me! $\endgroup$ – Mardy Owens Apr 6 at 13:43
  • 3
    $\begingroup$ Sure, I am keeping that in mind $\endgroup$ – Laurent Duval Apr 6 at 14:06
5
$\begingroup$

I haven't found a reference but it appears to be related to the derivative/integral properties of the FT: $n^{th}$ derivative in frequency corresponds to multiplication with $t^n$ in the time domain (divded by $i^n$) . If you start with FT of the unit step function $U(\omega) = \mathscr{F} \{u(t)\}$ then we can simply write

$$\mathscr{F} \{u(t)\cdot t^n\} = i^n\frac{d^n}{d\omega^n}\cdot U(\omega)$$

For n < 0 you can use either use integration or differentiation in time.

$\endgroup$
2
  • 2
    $\begingroup$ Shouldn't the power of $i$ be $n$? $\endgroup$ – Laurent Duval Apr 6 at 21:17
  • 2
    $\begingroup$ yes, thanks. I fixed it $\endgroup$ – Hilmar Apr 6 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.