Talk:Convolution

message: ...in the article is good. But it would be even better if the resultant convoluted function was shown. It can be a little bit hard to image in ones brain what the integral of the product of the two shown functions look like as the two functions slide over each other. I am not new to convolution I just have not used it for seven years or so and went here to see and quickly recap what it is all about. And for such a use case of the article a good figure is very powerfull. -- Slaunger 14:15, 24 October 2007 (UTC)
replies: text: I Agree that the visualization figure is very powerfull, but am reverting to previous version because I believe that now the Visual explanation of convolution figure is clutterring the article and is essentially a helper to the text. It was cleaner before. This is my opinion, if anyone else disagrees we could discuss. --D1ma5ad (talk) 22:28, 2 March 2008 (UTC)
replies: text:  Ok.  I can deal with the revert.  (For reference purposes, this was the edit in question.)  I think the image really needs to go, though, since it overwhelms the lead.  Perhaps someone should write an "explanation" section, and include this image as an accompanying visual aid. siℓℓy rabbit (talk) 21:51, 17 July 2008 (UTC)

text: I have to agree with the first comment, I can think of no reason not to include the convolved function. It would only take a few more inches of real estate and no additional explanation.   —Preceding unsigned comment added by 98.167.177.9 (talk) 03:29, 18 December 2008 (UTC)

text: It's the same function g in all three displays.  The lower part shows the reflected function.  SławomirBiały 10:50, 15 October 2015 (UTC)
thread_title:  The visualization figure 

message: (H * δ') * 1 = (H' * δ) * 1 = (δ * δ) * 1 = δ * 1 = 1
H * (δ' * 1) = H * (δ * 1') = H * (δ * 0) = H * 0 = 0
where H  represents heaviside's step function whose derivative is dirac's delta function
** sorry for the form in which I am presenting this, but I am not well familiarized with input of math equations
* hello, this article seems to be about convolution of functions. If you want to consider distributions, then you must impose another assumption, for example: One of the distributions must have a compact support for associativity to hold. In your example, certainly only Dirac has compact support. 78.128.195.197 (talk) 21:37, 10 March 2009 (UTC)
thread_title:  Note to associativity 

message: Some additional references I plan to use in my improvements of the article are:
* .
* .
* .
* .
I plan to add more to this list as I progress. siℓℓy rabbit (talk) 15:07, 16 July 2008 (UTC)
thread_title: References

message: I have added a short and rough section on translation invariance.  Obviously more should be said, since this is "the" most important property of convolutions.  Hopefully we can iterate to get this new addition of mine into a more reasonable shape.  siℓℓy rabbit (talk) 16:12, 24 July 2008 (UTC)
thread_title:  Translation invariance 

message: I am planning on adding a section to this article with some intuitive conceptions of the convolution (I am using Professor David Jerison's "bank account" analogy) with regards to its usage in solving ordinary differential equations with rest initial conditions. This intuition is primarily useful for introductory undergraduate differential equations, and as far as I can tell, not so much for advanced mathematics. This is possibly more relevant to the article superposition principle or Green's function and I would not have any objections if it moved over there, but it would be nice to see some a more layman explanation for this. — Edward Z. Yang(Talk) 22:10, 23 April 2009 (UTC)
replies: text: Ed, I reverted the long intuition example section.  If this kind of example can be sourced, maybe we can use it; but I didn't have the impression that it was particularly helpful or intuitive, being basically a bunch of integral math and such, just like the rest of the article.  Convolution is not so tricky that we need to resort to bank deposits to explain it, I think.  Dicklyon (talk) 05:18, 24 April 2009 (UTC)
replies: text: Source is here, albeit for a different example. This also works, although I am not sure if the link will always be available there. While the convolution may not be tricky (it is a relatively simple mathematical operation), many people will find the existence of a concrete example and/or usage of the term to help them understand what a convolution is, and why they might care. I have not restored the section. — Edward Z. Yang(Talk) 13:54, 24 April 2009 (UTC)
replies: text: Those don't support you novel approach of making a continuous compound interest problem out of it, which is itself an unfamiliar concept to anyone who is likely to have trouble with the math.  And they're not exactly WP:RS.  Dicklyon (talk) 15:44, 24 April 2009 (UTC)
replies: text: Sure. We can remove the analogy from it, and then reframe the section as a proof of the fundamental solution theorem. I would call them reputable sources under the self-publishing articles, but if that's not sufficient I can also find alternative sources. — Edward Z. Yang(Talk) 16:45, 24 April 2009 (UTC)
replies: text: In an area with so much published, there's probably no reason to resort to self-published material.  Dicklyon (talk) 19:37, 24 April 2009 (UTC)
replies: text: It seems to me that because it's so elementary, you're not going to find peer-reviewed material appropriate for this article. Would a textbook be acceptable, in your eyes? — Edward Z. Yang(Talk) 22:12, 24 April 2009 (UTC)
thread_title:  Intuition 

message: Translation invariant operators on Lp spaces are by definition operators that commute with all translations.  This is completely standard terminology: see any good book on Fourier analysis (E.g. Stein and Weiss, Duoandikotxea, Grafakos, etc.)  71.182.236.76 (talk) 12:44, 31 October 2009 (UTC)
thread_title:  Explanation of revert 

message: Two positive integrable and infinitely differentiable functions may have a nowhere continuous convolution.
That is there are infinitely differentiable (even entire) probability densities with a nowhere continuous convolution. 
If this happens when then the functions are badly unbounded.
I have constructed examples in the paper below and wrote a small correction to wikipedia some time ago, which has been "corrected". So I have added a line to the differentiation subtitle and a reference to my paper. I hope this will not appear too selfish.
 MR1654158 (99m:60026)
Uludağ, A. Muhammed(F-GREN-FM)
On possible deterioration of smoothness under the operation of convolution. (English summary)
J. Math. Anal. Appl. 227 (1998), no. 2, 335--358.
As the author states in the introduction: ``It is known that, as a rule, the operation of convolution improves smoothness. In this paper, the above statement is looked at critically. An example is constructed of two probability densities which are restrictions to $\bold R$ of entire functions, though their convolution possesses an infinite essential supremum on each interval.  —Preceding unsigned comment added by 134.206.80.237 (talk) 17:52, 4 March 2010 (UTC)
thread_title:  differentiation and convolution 

message: The starting animations don't mention that you have to reflect one of the functions (e.g. filter) for that visualisation to work. (Of course the signal is symmetric, so it's not wrong...) Tinos (talk) 02:03, 24 July 2010 (UTC)
replies: text: Put a message on the talk page of the guy who made it.  You're right; g(t) should be g(tau - t), and (f*g)(t) should be (f*g)(tau).  Dicklyon (talk) 03:51, 24 July 2010 (UTC)
replies: text: I don't think he has a talk page. He's provided the code so I can reproduce it. I think he's right with (f*g)(t), though? (The asterisk still needs fixing, though.)Tinos (talk) 06:31, 24 July 2010 (UTC)
replies: text: Actually, I think we're both wrong.  You need a variable of integration running opposite directions in f and g, so maybe they should be called f(tau) and g(t - tau), with result (f*g)(t).  But then there's also the fact that as plotted on the same x axis as the result, the f(tau) is really the same as f(t).  It's hard to do this in a way that's not somewhat misleading or confusing. Dicklyon (talk) 06:50, 24 July 2010 (UTC)

text: I've made the animations. However I just discovered Brian actually has a talk page on Wiki Commons! I've notified him (I think).Tinos (talk) 07:13, 25 July 2010 (UTC)

text: I've updated the article. I tried synchronising the animations but Firefox opens the first one before the second one! Anyone know how to get Wikipedia to sync them up? I guess it's not important, but would be kind of cool. Feel free to suggest improvements to the animations.Tinos (talk) 10:23, 26 July 2010 (UTC)
replies: text: Also, g(t) is simply g(t). I believe that it should never be linked to g(-t) or even g(-tau), as this simply leads to absolute and utter confusion. A different variable should be used instead, such as p(-t) = g(t). We know in advance that if g(t) were a causal function, then it's ok to write p(-t) = g(t) to generate a time-reversed function, and it is not ok to write g(-t) = g(t), since we all that that if g(t) were a causal function, then g(-t) would be zero. KorgBoy (talk) 04:27, 9 December 2018 (UTC)
thread_title:  Misleading Animations 

message: The first reference given (Bracewell) may be a great book, but it seems to be assuming from the very beginning that the reader is acquainted with a general idea of a transform. The WP lemma transform (mathematics) redirects to Integral transform. But what is a transform in general? Nobody seems to care to give a definition. --84.177.81.254 (talk) 17:35, 25 May 2011 (UTC)
replies: text: Think of it as a function whose input values and output values are not numbers, instead they are functions.  Some times the phrased is used for other things as well, such as the input and output values are vectors. (Such as linear transforms in linear algebra).   Thenub314 (talk) 04:55, 26 May 2011 (UTC)
replies: text:  Thanks! So the elements of a lambda algebra must be transforms, too, I think. --84.177.86.232 (talk) 17:37, 26 May 2011 (UTC)
thread_title:  transforms 

message: 
Every time I refer to this article that thing catches me by surprise and I jump ten feet in the air.  Why can't we convolve a nice butterfly?  — Preceding unsigned comment added by 69.242.232.143 (talk) 03:18, 10 July 2011 (UTC)
replies: text: I also find the image not quite to my taste.  It needs to be too large in order for the effects of the convolution to be visible.  A tiny part of the image (showing the hairs on the legs) could be much smaller and yet illustrate the sharpening effect more clearly.  Sławomir Biały (talk) 14:33, 10 July 2011 (UTC)
replies: text: Yes, it is too big. I think someone should crop off a leg; ideally the original artist. Tomgg (talk) 09:56, 9 August 2011 (UTC)

text: I agree. The same was done on Depth of Field (see Talk:Depth_of_field#Image_of_wolf_spider) for Wikipedia:Accessibility reasons (citing arachnophobia). —Ben FrantzDale (talk) 16:30, 9 August 2011 (UTC)

text: Another request for this - I can't finish the article because of that damn thing!  — Preceding unsigned comment added by 141.163.156.59 (talk) 12:21, 24 November 2011 (UTC)
replies: text: Done! Sławomir Biały (talk) 12:26, 24 November 2011 (UTC)
replies: text: No one has requested that the image be removed.  Rather, people would like it to be replaced.  I think we should keep it in the article until a better one comes along.  Yaris678 (talk) 12:51, 24 November 2011 (UTC)
replies: text: I disagree.  Many readers have already complained that the image is a significant impediment to reading the article.  It should therefore be removed until someone produces a satisfactory replacement.  It has been here for over four months already, despite objections.  We shouldn't wait indefinitely for someone to make an alternative.  Sławomir Biały (talk) 13:05, 24 November 2011 (UTC)
thread_title:  Does it really have to be a spider? 

message: Another lousy wikipedia article written by self-claimed experts to wank on about how they edited a wikipedia article.
The first paragraph is not bad.
The second paragraph, about the use of convolution in other groups is total wiki bullshit.
Put that crap down lower in the article.  I don't care you have a Ph.D and that you are correct.  Are you writing for more jackholes or for people that actually want to learn?
Similarly, get rid of stupid animations that have no ability to stop and let the person read.
DEMAND ATTENTION DEMAND ATTENTION DEMAND ATTENTION.
I made an animation, and regardless of how distracting it is, I will force you to read it.
In summation, typical crappy wikipedia article.
68.106.47.124 (talk) 17:35, 6 March 2012 (UTC)
replies: text: The lead of the article is supposed to provide a capsule summary of the article.  It isn't necessarily supposed to be the easiest part.  Since much of the article does focus on generalizations of the convolution, the second paragraph is appropriate and roughly proportional to the prominence of this topic in the article.  It's true that there should be more detail in the earlier sections on things that a larger group of readers might be interested in, like applications to image processing.  These can then also be mentioned in the lead, which has ample room for expansion.  At this point, I have no opinion about the animation. I seem to recall lobbying for its removal in the past whilst other editors have favored it's inclusion.  I think it should probably stay, at least pending a wider discussion. Sławomir Biały (talk) 18:53, 6 March 2012 (UTC)

text: I don't understand the ambivalence about the .gif animation.  I think it is clearly a positive feature of the article, and an excellent example of what computer-based learning can offer that hardcover books cannot.  A pause button would also be a nice, but minor, addition.  Perhaps someday Wikipedia will also figure out how to write articles on multiple levels... high school, undergrad, & grad.  Good things take time.--Bob K (talk) 12:27, 10 March 2013 (UTC)
thread_title:  Start off easy, then push the reader over a cliff 

message: The lead images to this article are not good.  They are so wide that they completely dominate the lead, squashing all of the text aside.  Moreover, the new captions I think are worse.  They are much less succinct that the original captions, and dwell on things like the difference between cross-correlation and convolution, which the images don't even illustrate.  If someone wants to write an examples section that includes these (or other) images, along with detailed explanations, that might be better.  Sławomir Biały (talk) 01:50, 12 March 2013 (UTC)
replies: text: I think the .gif "movies" are not helpful at their current location.  They would be more helpful with the formulas (Definition section) that they serve to illustrate.  The reason I brought up cross-correlation is so that I could equate g(t-\tau) to g(\tau-t), which is easier to grasp.  It was awkward, I agree, but maybe less awkward than g(t-\tau).  That was the hope.  I also thought it important to say how  t is defined.  But all those details became cumbersome.  I think it is better now. --Bob K (talk) 04:25, 12 March 2013 (UTC)
thread_title:  Lead images 

message: This section does not cover discrete convolution. Petergans (talk) 07:35, 18 May 2013 (UTC)
replies: text: I see Mark Viking has put in a link to a description of the method you are trying to push here. I've no problems with what they've done. Do you disagree with that paper? In particular equation (4)? Dmcq (talk) 10:05, 18 May 2013 (UTC)

text: There's a bit less to say about the domain of definition of the discrete convolution in Z.  The typical application is for absolutely summable sequences, although it can be defined in other \ell^p spaces.  I've added some discussion of this to the section on the discrete convolution.  Sławomir Biały (talk) 11:53, 18 May 2013 (UTC)
replies: text: It looks a bit funny because practically the same condition is in there twice, under the discrete convolution for the sequences and once under Domain of definition for the integrable functions. The same thing happens with Young's inequality in there twice for measures. Dmcq (talk) 12:22, 18 May 2013 (UTC)
replies: text: Yes, and it probably should appear yet again in the section on groups.  In some sense, this is in increasing level of mathematical sophistication.  It could probably be made more harmonious though.  I've removed it for now.  It's probably best not to dwell on these things.  Sławomir Biały (talk) 13:06, 18 May 2013 (UTC)
thread_title:  Domain of definition 

message: The sentence "They are valued for smoothing spectroscopic curves wile preserving the shape and phase of the peaks" is rubbish. See Numerical smoothing and differentiation#Signal distortion and noise reduction for details. Petergans (talk) 08:58, 25 May 2013 (UTC)
replies: text: That is cited to a paper published in IEEE Signal Processing magazine and written by a person with some fairly impressive credentials in signal processing. Perhaps you could point to a better source and explain what it says different thanks. Dmcq (talk) 09:38, 25 May 2013 (UTC)
replies: text: It's rubbish because spectroscopic curves don't have phase. The statement applies only to those electrical signals that do have phase. It lacks generality and has been quoted out of context. Petergans (talk) 18:53, 25 May 2013 (UTC)
replies: text: You're right; spectroscopic curves don't have phase.  Only in case of laser (or equivalent) illumination or in active RF "spectroscopy" are there phases.  I'll have to read the paper to determine the signal processing context.  — Arthur Rubin  (talk) 19:16, 25 May 2013 (UTC)

text:  Thanks for your comments. But you are confusing how spectrograms are produced with how they are filtered. Power spectrograms are the modulus squared of, for instance, the short-time Fourier transform of a signal and so do not retain the phase of the original signal. But the spectrogram is also a real valued integrable function and as such can be decomposed into Fourier components, with each component having its own phase. An S-G filter is an FIR filter, and like any filter, can potentially alter the phases of those components. Symmetric S-G filters, however, have zero phase and so do not alter the phases of those components. I apologize that my summary in the article was a bit too terse and thus confusing, but in filtering a spectrogram function, the concept of phase definitely exists. If you are looking for the source of "while preserving the shape and height of the peaks", that is on page 111 of the cited reference, top of the second column. The assertion "Symmetric S-G filters have zero phase" is on page 116, first column. --Mark viking (talk) 20:07, 25 May 2013 (UTC)
replies: text: I just had a another look and it looks like I didn't check the words here against there properly. I just grabbed words from here and then looked for something that corresponded there and found "Savitzky and Golay were interested in smoothing noisy data obtained from chemical spectrum analyzers, and they demonstrated that least squares smoothing reduces noise while maintaining the shape and height of waveform peaks (in their case, Gaussian shaped spectral peaks)" which seemed fine to me. I think maybe I read preserving the phase of the peaks as simply meaning they weren't shifted but it would be simpler to just say that. Dmcq (talk) 21:14, 25 May 2013 (UTC)
replies: text: Unfortunately, "maintaining the shape and height" is not, strictly speaking, correct. It is an approximation that is useful iff the convoluting function is chosen carefully. The SG filter has the following conservation properties. i) peak positions ii) peak symmetry iii} peak area. There are proofs in Appendix 7 of my book. Conservation of area (integral of convoluted function) is what is relevant here. If a SG filter is an integrable function then convolution#integration also proves it. Smoothing of a Gaussian, for example, reduces the height and increases the half-width. "Maintaining shape" is ambiguous, it could be taken to mean either height and half-width, or shape function.  I don't know about conservation of shape function. The literature on the SG filter is extensive , so maybe it has been proved. If so, a citation would be needed. What is certain is that the height and half-width  of the original shape function are not conserved. Petergans (talk) 08:06, 26 May 2013 (UTC)
replies: text: I took it as meaning a tendency rather than an absolute. The height of a peak tends to be maintained for instance compared to a blurring filter like an average or compared to an edge detection filter. I'm not sure what you mean by conserving peak area over and above that the coefficients add to 1 or peak position given that some peaks will disappear - I guess you have some special definition of those. Dmcq (talk) 10:14, 26 May 2013 (UTC)
replies: text: On further thought, SG-smoothing does not conserve shape function. In the FT co-domain the SG-smoothing filter is a sum of cosines, so whatever the FT of the shape function its form will not be conserved when mutiplied by the SG-FT. Let's have some clear and unambiguous wording, please. Petergans (talk) 10:48, 26 May 2013 (UTC)
replies: text: If the points follow a low degree polynomial the shape will be conserved. Dmcq (talk) 11:35, 26 May 2013 (UTC)
thread_title:  Applications 

message: I have now finished editing Savitzky–Golay filter for smoothing and differentiation. Please have a look at it.
The theory in this article (Convolution) is all about "pure" mathematics and has inadequate background on an important application such as the S–G filter. For example, on what basis can convolution be used to obtain derivatives of a function? I ask that you guys look again at the "applied" mathematics aspects of convolution. The S-G filter deserves to have a proper background treatment here. Petergans (talk) 08:27, 1 August 2013 (UTC)
thread_title:  Applied mathematics 

message: I think that the formula for differentiation is wrong and misses the boundary term:
\frac{d}{dx}(f * g)(x) = (f'  * g )(x) + f(0) g(x) = (f * g')(x) + f(x) g(0) 
I'm aware that the term missing in MathWorld as well. But just quoting from somewhere is no proof.
When taking the derivative, there is an "x" at the integral boundaries and at the integrands, which makes two terms. One may also compute the derivative explicitly from infinitesimal differences. Third way is to go via a Laplace transform, where the boundary terms show up explicitly:
\mathcal{L}_s[h'] = \int_0^\infty e^{-s x} h'(x) dx  = -h(0) - s \hat h(s) 
\mathcal{L}_s[(f * g)'] = -(f * g)(0) - s\widehat{f * g}(s) 
= - s\hat f(s) \hat g(s)
= \mathcal{L}_s[f'] \hat g(s) + f(0) g(s) 
= \mathcal{L}_s[(f'*g)(x) + f(0) g(x)]
since  (f * g)(0) = 0 .  — Preceding unsigned comment added by Felix0411 (talk • contribs) 15:11, 3 December 2013 (UTC)
replies: text: There is no boundary term. The reason you're getting one is that you aren't taking the convolution of f and g.  You've smuggled a heaviside function into the mix (notice that the lower limit of integration is zero, not negative infinity).  Sławomir Biały (talk) 15:52, 3 December 2013 (UTC)
replies: text: Upps, I didn't notice that. My version comes up naturally if f,g are response functions respecting causality. As far as I know the one-sided definition is called a "convolution" as well, sometimes with the attribute "one-sided". Should the article warn the reader about the different definitions?Felix0411 (talk) 16:42, 4 December 2013 (UTC)
replies: text: It seems like a good idea to mention the one-sided convolution as a variation of the standard (two-sided) one.  I wouldn't phrase it as a warning to the reader though.  Sławomir Biały (talk) 01:23, 6 December 2013 (UTC)
thread_title:  Differentiation: boundary term? 

message: I'm thinking that time inversion might need to be explained properly, or clarified. I'm thinking ---- (asking) --- since when is a function g(-t) a mirror image of g(t) along the vertical axis? (except for the case when somebody tells us that the function g(t) is an EVEN function). So, I'm thinking, if we talk about time-reversal and use this symbol g(-t) or g(-tau) then we need to explain a bit more....... such as to assume that g(t) is an EVEN function to begin with, right? KorgBoy (talk) 21:09, 9 August 2018 (UTC)
replies: text: Huh?  "Inversion" is used nowhere in the article, and it's not clear what problem you have with it.  –Deacon Vorbis (carbon • videos) 21:22, 9 August 2018 (UTC)
replies: text:  I meant 'time reversal'. See the article diagram, where it has the red-coloured saw-tooth? And then the diagrams below forms the mirror image of it, which people refer to as 'flipping' or time-reversal or mirroring about the vertical y-axis. At the moment, the problem I have with that is....... they begin with a function ...such as g(t), and then they appear to assume that the mirror imaged version is g(-t), which doesn't seem to be right. KorgBoy (talk) 21:31, 9 August 2018 (UTC)
replies: text: I still don't really understand what your concern is, but the diagram is reasonably accurate.  The definition of convolution involves reflecting one of the functions, so this is what you'd expect to see.  –Deacon Vorbis (carbon • videos) 21:45, 9 August 2018 (UTC)
replies: text: Ok. Well, you know how they start off with two functions, eg x(t) and g(t), and then they want to convolve them. One procedure involves mirror-imaging (reflecting) one of these functions ( such as g(t) ) about the vertical axis. The procedures often magically result in a function g(-t)..... or, using a different dummy variable g(-tau). However, if we consider the actual original function g(t) and purposly replace t with -t, then the actual version of g(-t) or g(-tau) is certainly not going to be a mirror-image of g(t). It just appears that those people that teach this mirroring approach don't tell you the full story, which leads to confusion. It appears to me that the procedure should be something more like ..... generate a different function f(-t) which will be defined as f(-t) = g(t), and the convolution is actually integral of x(tau).f(t-tau)dtau. It is not integral of x(tau).g(t-tau)dtau. KorgBoy (talk) 00:49, 10 August 2018 (UTC)
replies: text: The graph of g(-t) most definitely is the reflection of the graph of g(t) about the -axis.  The diagram reflects this fact (pun intended).  I'm afraid I really still don't understand your confusion (everything you're talking about in the article is more or less kosher), and this is really getting into WP:NOTFORUM territory.  If you still have questions, you can always ask at the Math ref desk.  –Deacon Vorbis (carbon • videos) 01:34, 10 August 2018 (UTC)
replies: text: Thanks! Actually, I don't have to ask the maths guys, since I'm only pointing out something very simple. For example, a function....defined by g(t) = sin(t) for t greater or equal to zero, and is equal to zero for t < 0. Now, if you plug in t = 0.2 into it... you get approximately 0.2. And if you plug in t = -0.2, you get 0. So clearly, a mirror imaging (about the vertical y axis) process is not merely handled by magically making 't' become '-t'. KorgBoy (talk) 06:59, 10 August 2018 (UTC)

text: The "reflection" of g(t) is h(t) = g(-t).  E.g., h(0.2) = g(-0.2), and h(-0.2) = g(0.2).  Sleep on it.  --Bob K (talk) 13:56, 10 August 2018 (UTC)

text: It is all clear now. The issue is due to the missing pieces of information that is always conveniently left-out in many procedures that involve the 'reversal' or 'mirror image' step. In the article page, you see two functions f(t) and g(t). And while there is a quick mention about "reversed and shifted", such quick mention of it is pretty much pointless, and is the life-story of many teachings with big holes in it. The article 'magically' transitions to a function such as g(-t), which the diagrams show is a mirror image of g(t). However, obviously, if the function g(t) is a causal function, where g(t) is zero for t < 0, then g(-t) is actually zero for negative values of '-t'. That is, g(-t) is generally not going to be g(+t) if g(t) is causal. So obviously, this magical g(-t) mirror image function seen in the article page is not as straight forward as the article makes it out to be. Instead, this mirror image g(-t) can only be properly understood by understanding that g(-t) is not actually g(t). Instead, a second function should be defined, such as p(-t) = g(t). So, in the end, the convolution will involve sliding p(-t) over f(t). That is, I believe that the procedures that show the sliding of g(-t) over f(t) are totally misleading --- because g(-t) is not really g(t) at all. The all-important step is the need to purposely form a new function p(-t), and define this function p(-t) as g(t). And just forget about g(-t) entirely. Everything will then be very clear from there. KorgBoy (talk) 23:31, 8 December 2018 (UTC)
replies: text:  It is indeed very clear to me Bob. It really is. You mentioned "No, it would end up sliding p(t) [which is g(-t)] over f(t)". Yes, p(t) would indeed be equal to g(-t) here, because if g(t) is causal, then g(-t) is zero for positive values of t, and p(t) is correctly equal to zero for positive values of t, since p(t) is g(-t). This basically indicates that p(-t) really describes the reflection of g(t) around the vertical axis. And, no, we have to slide p(-t) - the mirror image of g(t) - over f(t). And, once again, it is mathematically incorrect to state that g(-t) is the reflection of g(t) - except for the case where g(t) just so-happens to be symmetrical about the vertical axis. KorgBoy (talk) 10:42, 15 April 2019 (UTC)
thread_title:  Time inversion 

message: Quite a number of years ago I ran across a book online that talked about 2 dimensional digital image convolution implemented using FFT that could locate a pattern within a larger image...  In this case it was locating the image of George Washington on a US Federal Reserve Note.
http://www.dspguide.com/ch24/6.htm
Can reference to this application of "pattern recognition" of this math function (Convolution) be made in the article? Oldspammer (talk) 07:59, 2 September 2018 (UTC)
replies: text: Thank you.  I added it to External links section. --Bob K (talk) 17:06, 4 January 2019 (UTC)
thread_title:  Pattern recognition application mention 

message: In view of the recent thrashing between plain html vs LaTex, let's consider the compromise of transitioning to use of template .
Related links:
* Talk:Discrete-time_Fourier_transform#Inline_math_style_changes
* Aliasing#Sampling_sinusoidal_functions  (example)
--Bob K (talk) 23:13, 30 November 2018 (UTC)
thread_title:  Inline math style changes 

message: In explaining convolution using FFT, the article says Convolutions of the type defined above. Presumably that means non-circular, but as it comes right after the explanation of circular convolution, that isn't so obvious. Gah4 (talk) 21:26, 16 July 2021 (UTC)
thread_title:  ''Convolutions of the type defined above'' 

message: The following Wikimedia Commons file used on this page or its Wikidata item has been nominated for deletion:
* 
Participate in the deletion discussion at the . —Community Tech bot (talk) 20:25, 11 September 2021 (UTC)
thread_title:  A Commons file used on this page or its Wikidata item has been nominated for deletion 

message: The current definition on the page is:
: In mathematics (in particular, functional analysis), convolution is a mathematical operation on two functions (f and g) that produces a third function that expresses how the shape of one is modified by the other.
I think this is a bad definition since the idea of a convolution has nothing to do with function "shape", even from an intuitive perspective. And function "shape" is not a precisely defined concept to begin with. I think my definition which just explains that the operation can represent different things depending on context is much better. For example, the functions might be PDFs in which case convolution is related to adding two independeng RVs. Or the context might be signal processing which is a totally different application and context. Or it could be convolution of a function by a dirac delta, in which case the result is simply the function evaluated at a single point and really has nothing to do with shape. DMH43 (talk) 15:27, 26 December 2023 (UTC)
thread_title:  Definition of convolution 

message: * It is unusual to see actual computer code in an article.  I suspect there is a policy about that.
* Thank you for the /* reverse the signal */  code comment, but why only one comment?
* I think it needs an explanation of why one would want to use this program.  What is your motivation?
* In my experience, the usual motivation for circular convolution is to exploit the efficiency of the formula:
: x_{_N} * y\ =\ \scriptstyle{\rm DFT}^{-1} \displaystyle  \left[\scriptstyle{\rm DFT} \displaystyle \{x\}\cdot \scriptstyle{\rm DFT} \displaystyle \{y\}\right],
: implemented with a Fast Fourier transform algorithm.  (see  Eq.4c at Convolution_theorem#Periodic_convolution)
: And only the non-circular part of the output sequence is actually retained.
Bob K (talk) 13:10, 21 May 2024 (UTC)
replies: text: An article like this should not have code. It doesn't help with an understanding of the concepts except for a very few readers. Those special cases can use Google to find code. Johnuniq (talk) 01:38, 22 May 2024 (UTC)
thread_title:  Code for circular convolution (added 20-May-2024) 

message: I'm having difficulties trying to find the time reversal property, and people got it wrong too.
If
:y(t) = x(t)*h(t)
then
:y(-t) = x(-t)*h(-t)
This is the property.
If you look for the property you can find like in this presentation
, that they got it wrong, stating that
:y(-t) = x(t)*h(-t)=x(-t)*h(t)
I would like to propose to add the property so there is no more doubt in this topic. Bouchet07 (talk) 11:27, 25 May 2024 (UTC)
thread_title:  Add time reversal property