Talk:Gibbs phenomenon

Long formulas

Added title. —Nils von Barth (nbarth) (talk) 22:36, 3 April 2009 (UTC)

Teorth, try to keep formulas as short as possible. For example, on one of my computers, when I viewed the page in its previous state, the formula for the sum could not fit alongside the pictures and was pushed below them, making an ugly enormous gap in the page. The same holds for the formulas in the "Formal mathematical description of the phenomenon" — try to make them look aesthetic even with a narrow window.

Oh, and I think the half error was on the other side — please check my calculations. Gadykozma 13:13, 28 Oct 2004 (UTC)

OK, I've split the longest equation into three pieces; feel free to truncate the others as necessary. Regarding all the factors of 2, I think I've fixed them all now... Terry 18:05, 28 Oct 2004 (UTC)

Overshoot in the limit?

Added title. —Nils von Barth (nbarth) (talk) 22:36, 3 April 2009 (UTC)

Doesn't the overshoot remain (though with zero width) in the infinite limit? If so, then the arguments in the second paragraph should be changed since they are false.74.98.54.54 19:12, 12 May 2007 (UTC)

No the overshoot does not exist at all in that case. "An infinite sum of continuous functions can be discontinuous" is correctly stated. Cuddlyable3 10:16, 14 May 2007 (UTC)

The fact that "An infinite sum of continuous functions can be discontinuous" is correct and irrelevant, and more over it doesn't follow from this fact that the infinite sum "does not exhibit the Gibbs phenomenon". In fact, the described phenomenon is a property of the sequence of partial sums and not of it's (pointwise) limit, so there no meaning to the statement that the limit function "does not . It doesn't belong in this article, and if it does, it cannot be phrased by assigning a true or false value to the statement that the limit function "exhibits the Gibbs phenomenon" because this sentence does not accept one of these two values. Hadaso (talk) 09:24, 9 December 2007 (UTC)

Hadaso is correct. I'm changing the wording. JamesCrook (talk) 22:03, 31 October 2010 (UTC)

Square Wave Definition

The Description section defines the square wave function this way: "More precisely, this is the function f which equals π / 4 between 2nπ and (2n + 1)π and − π / 4 between 2(n + 1)π and 2(n + 2)π for every integer n; thus this square wave has a jump discontinuity of height π / 2 at every integer multiple of π."

I don't think that's quite right. Shouldn't it be "... π / 4 between 2nπ and (2n + 1)π and − π / 4 between (2n + 1)π and 2(n + 1)π" ? DRE 15:49, 13 March 2007 (UTC)

Looks like it. I've changed it. Michael Hardy 21:52, 13 March 2007 (UTC)

Pointwise limit of the partial sums

I have removed the following sentence from the article's intro paragraph

“ The pointwise limit of the partial sums of a Fourier series does not exhibit an overshoot near a jump discontinuity. ”

It does not make sense to me for a couple reasons:

1. Pointwise limit with respect to what?
2. The partial sums do exhibit an overshoot near a discontinuity, so it seems to be a contradiction.

Mr. PIM (talk) 11:51, 29 July 2008 (UTC)

Well it is a consequence of Dirichlet's theorem that the Fourier series for a piecewise C^1 function converge to the function :everywhere except jump discontinuities where it converges to the average value on either side of the jump. By:

1. limit of the partial sums I assume he means $g(x)=\lim _{N\to \infty }S_{N}f(x)$
2. The partial sums do exhibit an overshoot, but g(x) does not for examples like the square wave of sawtooth wave.

I may put this back with a little more explanation. Thenub314 (talk) 06:25, 12 September 2008 (UTC)

Explanation of overshoot

I am not sure the following sentence is the correct description of the problem:

The overshoot is a consequence of trying to approximate a discontinuous function with a partial (i.e. finite) sum of continuous functions.

I thought, but my memory is often wrong, that wavelets did not suffer from this phenomenon because they were localized. Am I correct about this? Thenub314 (talk) 12:24, 12 September 2008 (UTC)

You are right, that sentence is completely misleading. That continuous functions cannot converge uniformly to discontinuous functions is obvious, but that has nothing to do with Gibbs phenomenon. Just take the example when you approximate the jump by straight line segments of increasing slope, there is no overshoot then. Gibbs phenomenon means that there is not even uniform convergence on any interval a<x<a+h, with a the discontinuity point. But at that interval the limit function is continuous. So Gibbs phenomenon is a property of the particular choice of continuous functions you use, it holds for trigonometric polynomials, but not wavelets or piecewise linear functions. 129.16.29.31 (talk) 06:15, 15 September 2008 (UTC)

The example of the straight line segments of increasing slopes seem to indicate to me that the Gibbs phenomenon is a result of approximating jump discontinuities with continuous functions having no jump discontinuity (the example of the straight line segments has jump discontinuities). Mr. PIM (talk) 16:25, 23 September 2008 (UTC)

You could do it easily well with smooth versions of these functions. Gibbs phenomenon does not result from attempting to approximate a discontinuous function by continuous (or even smooth functions) functions. Thenub314 (talk) 16:48, 23 September 2008 (UTC)

Show me an example. Mr. PIM (talk) 17:55, 23 September 2008 (UTC)

Ok, take any positive real valued smooth function φ(x) supported in the interval [0,1]. Let me know if you want to see an example of such a function.

Define: $\psi (x)={\frac {\int _{-\infty }^{x}\varphi (t)\,dt}{\int _{-\infty }^{\infty }\varphi (t)\,dt}}$. Some properties of ψ:

• ψ is smooth;
• ψ(x)=1 for x> 1;
• ψ(x)=0 for x<0;
• 0≤ψ(x)≤1 for all x.

For n=1, 2, 3, … define $\psi _{n}(x)=\psi (n\cdot x)$, then

• ψ_n is smooth;
• ψ_n(x)=1 for x> 1/n;
• ψ_n(x)=0 for x<0;
• 0≤ψ_n(x)≤1 for all x.

Hence $H(x)=\lim _{n\to \infty }\psi _{n}(x)={\begin{cases}0&{\text{for }}x\leq 0\\1&{\text{for }}x>0\end{cases}}$.

So, there is a sequence of smooth functions that converges to a function with a jump discontinuity. Thenub314 (talk) 18:32, 23 September 2008 (UTC)

Hi Thenub314 – this is a good question, and a much-confused point (I’ve been spending the past week reading and thinking about this in the context of ringing artifacts), and is much easier to understand from the POV of signal processing, namely viewing Fourier sums as a brick-wall low-pass filter, i.e., convolution by sinc.

I’ve written this up at Gibbs phenomenon#Signal processing explanation, but briefly, undershoot and overshoot are caused because you’re convolving with sinc, which takes on negative values. Thus output values are affine combinations of input, not convex combinations, and thus can fall outside the range of the original function.

This does not occur if you use a Gaussian as low-pass filter, for instance (visually, it smooths the discontinuity, without over- or under-shooting). Wavelets have negative values, so they will have ringing.

For added reference, see nascent delta function and mollifier.

—Nils von Barth (nbarth) (talk) 23:10, 3 April 2009 (UTC)

(Corrected statement on wavelets. —Nils von Barth (nbarth) (talk) 03:20, 12 April 2009 (UTC))

As for wavelets, I don’t know if ringing happens with them – if you convolve with a wavelet, then yes, it has negative values, so you can get undershoot and overshoot, but if you take a wavelet transform and low-pass filter it, I don’t know – it may depend on the wavelet: I’m not very familiar with wavelet transforms, and don’t know what the (analog of the?) impulse/step response would be. However, with basis functions with compact support, any ringing will be local, as you say.

As a datum, I’ve written a small comparison of wavelet transforms vs. Fourier transforms here:

Discrete wavelet transform#Comparison with Fourier transform

using Haar wavelets, and there low-pass filtering yields no ringing.

—Nils von Barth (nbarth) (talk) 03:20, 12 April 2009 (UTC)

It appears that wavelet transforms can exhibit ringing, as discussed in JPEG 2000, and demonstrated at Baseline JPEG and JPEG2000 Artifacts Illustrated.

—Nils von Barth (nbarth) (talk) 03:30, 12 April 2009 (UTC)

Very nice! thanks Nils. I should have known all this I suppose, but it has been a long time since I have seriously done any signal processing. Now of course the question is, the overshoot/undershoot for the FT is rather well understood, do you get the same percentage via other methods. That is does the amount of overshoot depend only on the fact that you have negative values (which I doubt) or does it depend on the relative strength of negative terms in your convolution filter? Thanks again. Thenub314 (talk) 08:33, 18 April 2009 (UTC)

History

The history section is badly cited and seems a bit off of what I believe are the referenced articles. For one, there is in fact a 1898 letter to the editor by Michelson on the topic in the October 6th, 1898 edition of Nature, but it makes no reference to any graphing machine. Instead it points to the seeming existence of a tangent line to an isolated point in the Fourier expansion of the sawtooth function which to Michelson seems ridiculous. Further, applying the label "paper" to these articles seems a little misleading because the entire back and forth discussion between Michelson, Gibbs, and Love is in the shape of more informal letters to the editor. I hesitate change the section entirely because I'm hoping that someone can get a solid cite for that graphing machine or possibly point to some other papers by them that might have been floating around at this time in some other source. —Preceding unsigned comment added by Twb5 (talk • contribs) 20:50, 8 November 2010 (UTC)

Animation not working?

The animation of the square wave does not work unless you actually go into the link http://upload.wikimedia.org/wikipedia/commons/f/f8/SquareWave.gif Not sure why this is. Nice gif by the way though. Handsofftibet (talk) 12:51, 5 June 2012 (UTC)

About the constant

Throughout the article, the constant "0.089490..." is used.
I believe that the value given is problematic.
See: http://www.wolframalpha.com/input/?i=%28Si%28Pi%29+-+Pi%2F2%29+%2F+Pi
which produces: 0.0894898722...

Apparently, fifth digit to the right of the decimal point was rounded up. But the notation used implies that the fifth and sixth digit are respectively 9 and 0 (rather than 8 and 9). — Preceding unsigned comment added by Inspect All Information (talk • contribs) 12:47, 17 July 2014 (UTC)

You can't be serious. Sixth digit to the right of the decimal point was rounded up, not fifth. 0.0894898722 then becomes 0.089490. Really? It is school level material. 109.252.90.66 (talk) 15:21, 3 February 2021 (UTC)

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Clarification needed

The section History contains this sentence:

"Inspired by correspondence in Nature between Michelson and A. E. H. Love about the convergence of the Fourier series of the square wave function, J. Willard Gibbs published a note in 1898 pointing out the important distinction between the limit of the graphs of the partial sums of the Fourier series of a sawtooth wave and the graph of the limit of those partial sums."

But what exactly does the phrase "the limit of the graphs" means?

It *might* have meant "the graph of the function that is the limit of the functions in the graphs", but this does not appear to be the meaning, since the text makes a distinction between the two things.

So: What does this phrase mean? 2601:200:C000:1A0:20BE:A512:2183:1640 (talk) 18:40, 22 June 2022 (UTC)

I removed edit that added too much to lede.

User Goodphy (talk · contribs) made an edit which added the following text to the lede's first paragraph:

For a point of discontinuity, as more Fourier series constituents are taken, the first overshoot in the oscillatory behavior around the point approaches ~ 9 % of the discontinuity jump and the oscillation does not disappear but gets closer to the point so that the integral of the oscillation approaches to zero (i.e., zero energy in the oscillation). At the discontinuity point, the Fourier series gives the average of the function's both side limits toward the discontinuity. The Fourier series, if it exists, may uniformly converge to a continuous function while it may pointwisely converge to a function with discontinuities.

I ended up deleting it cause I thought it was too much details for the lede's first paragraph. But it doesn't seem to be a bad explanation. So maybe that user or someone else can try to incorporate it or part of it in another manner somewhere else in the article. Em3rgent0rdr (talk) 22:22, 29 June 2023 (UTC)

Needs a proof

The Gibbs phenomenon is about overshooting the correct value as a point of discontinuity is reached. Not so much about "oscillations". So overshooting would be better language to use for this.

But also: The overshooting phenomenon has a short proof that ought to be included in the article.

I hope someone knowledgeable about this subject can provide it here.