Talk:Convolution/Archive 3

Archive 1

Archive 2

Archive 3

Archive 4

Applications

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)

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)

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)

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)

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)

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)

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)

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)

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)

If the points follow a low degree polynomial the shape will be conserved. Dmcq (talk) 11:35, 26 May 2013 (UTC)

Applied mathematics

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)

Differentiation: boundary term?

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:

${\mathcal {L}}_{s}[h']=\int _{0}^{\infty }e^{-sx}h'(x)dx=-h(0)-s{\hat {h}}(s)$

${\mathcal {L}}_{s}[(f*g)']=-(f*g)(0)-s{\widehat {f*g}}(s)=-s{\hat {f}}(s){\hat {g}}(s)={\mathcal {L}}_{s}[f']{\hat {g}}(s)+f(0)g(s)={\mathcal {L}}_{s}[(f'*g)(x)+f(0)g(x)]$ since $(f*g)(0)=0$. — Preceding unsigned comment added by Felix0411 (talk • contribs) 15:11, 3 December 2013 (UTC)

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)

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)

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)

Error with visual explanation?

The second part of the figure illustrates the step 2 in the algorithm, "Reflect one of the functions: g(tao)→g(-tao)" However, the corresponding part of the figure showing the reflected filter is incorrectly labeled as g(t - tao) instead of g(-tao) Richard☺Decal (talk) 00:12, 27 February 2015 (UTC)

My interpretation is that the 2nd part of the figure illustrates both steps 2 and 3. Perhaps they should be labeled 2a and 2b. Would that fix it for you?

--Bob K (talk) 12:31, 27 February 2015 (UTC)

I have updated the image so that the steps shown correspond to those in the article text. In particular, the first reflection of g does not include the time offset. I don't know what the post below is about, since the image under discussion has no saw-tooth in it. Sławomir Biały (talk) 13:08, 7 May 2015 (UTC)

The second saw-tooth needs to be reflected. It is correct in the consequential summation operations under i, but it is wrong. I dont agree with the relabling, each operation is a different case. — Preceding unsigned comment added by 85.199.231.18 (talk) 10:14, 7 May 2015 (UTC)

The visualization figure

...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)

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)

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)

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)

The integral of their product is the area of the yellow region. or The integral of their product is the area of the triangle. —Preceding unsigned comment added by 58.107.79.95 (talk) 12:34, 4 February 2010 (UTC)

The first red triangle (function g) should be the other way round, as in the lower part of the diagram where f and g overlap. — Preceding unsigned comment added by 134.76.90.227 (talk) 09:58, 15 October 2015 (UTC)

It's the same function g in all three displays. The lower part shows the reflected function. Sławomir
Biały 10:50, 15 October 2015 (UTC)

Misleading intuition of convolution as area of overlap?

Several of the visuals and some of the text indicate that the value of the convolution/cross-correlation/auto-correlation is equivalent to the shaded region of overlap. This doesn't seem to be generally true, and it's unclear to me if this is a mistake, or if the text should indicate that this is only true for special cases. Consider the bottom-right example in the intro graphic (the auto-correlation of a triangular shape). We can define the curve

$g(x)={\begin{cases}1-x,&0<x<1\\0,&otherwise\end{cases}}$

The area of overlap is the integral of the minimum of the curve and the lagged curve, and this is not the same as the integral of the product of the curve and the lagged curve. — Preceding unsigned comment added by Drummist180 (talk • contribs) 00:22, 2 April 2018 (UTC)

Good point. So I clarified the figure caption.

--Bob K (talk) 14:58, 2 April 2018 (UTC)

convolute and correlate?

In discussion of an other article, there is the question about the name for the result of doing a convolution. That is, convolution is the operation, in the same way that addition and division are operations. In those cases, the results are sum and quotient. Is the result of convolution a convolute? Also, for correlation is it correlate? Gah4 (talk) 23:11, 17 July 2018 (UTC)

No.

--Bob K (talk) 01:32, 18 July 2018 (UTC)

Then do you happen to know what it is? Otherwise, how do you know it isn't? Gah4 (talk) 01:46, 18 July 2018 (UTC)

It's just a convolution and a correlation, analogous to the 4th sentence at Fourier transform: "The term Fourier transform refers to both the frequency domain representation and the mathematical operation that associates the frequency domain representation to a function of time."

The definition of convolution at https://www.google.com/search?q=convolution is "a function derived from two given functions by integration that expresses how the shape of one is modified by the other". Thus one can say "The function f*g is a convolution.  And the process of computing f*g is also called convolution."

--Bob K (talk) 20:13, 19 July 2018 (UTC)

Well, that is the other way around. Operations are the ones ending in "ion". I believe some say "Fourier Transformation" for the operation, or "Fourier Transform Operation", and then some might shorten it leaving out the "operation" part. But convolution and correlation, with the ion ending, are the operations. I believe that correlate is sometimes used in statistics. The one that started this is derivative vs. differentiation. Sometimes the operation is taking the derivative, and I believe there is also taking the Fourier Transform for its operation. Gah4 (talk) 21:21, 19 July 2018 (UTC)

A noun is needed here. Convolution is a noun. Your suggestion, convolute, is a verb.

--Bob K (talk) 11:59, 20 July 2018 (UTC)

It seems that correlate is used as a noun in statistics, but maybe not convolute. Gah4 (talk) 14:18, 20 July 2018 (UTC)

Inline math style changes

In view of the recent thrashing between plain html vs LaTex, let's consider the compromise of transitioning to use of template {{math}}.
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)

Misleading Animations

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)

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)

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)

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)

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)

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)

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)

Why the time inversion?

The article doesn't explain why g is reversed. What is the point of time inverting it? Egriffin 17:46, 28 October 2007 (UTC)

What do you mean by "the point"? Convolution is defined with time inversion, and as such, happens to have many useful applications. If you don't perform time-inversion, you have cross-correlation instead; which also has lots of useful applications. Oli Filth^(talk) 17:56, 28 October 2007 (UTC)

Or why g instead of f? If you look at it, it makes no difference, since the variable of integration could just as well run the other way, and gets integrated out. In the result, you'll find that if either f or g is shifted to later, then their convolution shifts to later. For this to work this way, the integral needs to measure how they align against each other in opposite order. But think of the variable of integration as some "sideways" dimension, not time, and there's not no "time reversal" to bother you. Or think in terms of the PDF of the sum of two independent random variables: their PDFs convolve, as you can work out, but there is no time involved and no reversal except relative inside the integral. Dicklyon (talk) 16:27, 26 February 2008 (UTC)

It helps to think of $\int _{-\infty }^{\infty }f(\tau )g(t-\tau )\,d\tau$ as a weighted average of the function $g(\tau )$:

• • up to the moment "t", if $f(\tau )$ happens to be zero for all negative values of $\tau$, or
  • centered around the moment "t", if $f(\tau )$ happens to be symmetrical around $\tau =0$.

The weighting coefficient, $f(\tau ),$ for a positive value of $\tau ,$ is the weight applied to the value of function $g$ that occurred $\tau$ units (e.g. "seconds") prior to the moment "t". You may either infer that from the formula, or you may define  $f(\tau )$ that way and infer (i.e. derive) the formula from that definition.

Maybe your point is that something like this needs to be stated in the article, not here.

--Bob K (talk) 12:03, 1 May 2008 (UTC)

This question worth considering in the article, IMO. I do not know about other fields, but convolution is a very important thing in the theory of abstract systems: you apply some input u(t) and the theory computes output y(t). The relationship between y(t) and u(t), the system, is usually given implicitly, by a differential equation, which is easily resolved into the from y = Hu in Laplace domain. The inverse Laplace brings solution back to time domain and it is a convolution between impulse response h(t) and u(t). Taking into account that any function f(t) can be represented as a sum of impulses that are 0 everywhere besides t, where impulse is f, you can understand: why convolution. Though impulse response, h(t), depends on system, it usually looks like a single smoothed triangle, close to 0+ on t-axis, since the reference impulse, δ(x), is applied at time 0. If input impulse is applied at different time, τ, the argument to the h function is reduced by that amount, h(t-τ). The output y at moment t is the sum of all such impulse responses (every is scaled by amplitude of input, u(τ)). This explains both integration over τ ∈ [0,t] (input is zero before 0 and future, time > t, must not affect y(t)) and multiplication of u(τ) by some `time inverse' h(t-τ). Look at the illustration of two-impulse u:

This can be put the other way around: τ is the time difference between impulse time t-τ and current time t, thus its contribution into y(t) is u(t-τ)·h(τ). Surely, this helps and must be explained in the article. Additionally, in this picture, no signal is reversed in time. In this view, convolution has nothing to do with correlation. The reference to correlation is misleading, IMO. --Javalenok (talk) 20:23, 28 July 2010 (UTC)

I have realized that, in discrete domain, it is even easier to explain. Take a difference relation, e.g. x[k+1] = Ax[k] + Bu[k], where x is a state and u is the input vector. Then, x[2] = Ax[1] + Bu[1] = A(Ax[0] + Bu[0]) + Bu[1] = A²x[0] + ABu[0] + Bu[1], x[3] = A³x[0] + A²Bu[0] + ABu[1] + Bu[u],

$x_{k}=A^{k}x_{0}+\sum _{i=0}^{k-1}{A^{k-1-i}}Bu_{i}$. 

The convolution seems to come up every time you solve a non-homogeneous differential equation. That is, dx/dt = ax(t) + bu(t) has a solution

$x(t)=x(0)e^{at}+\int _{0}^{t}{e^{a(t-\tau )}bu(\tau )}d\tau$. 

Herein, h[k] = A^kB and h(t) = e^atb are the impulse responses. --Javalenok (talk) 06:57, 1 August 2010 (UTC)

I think the clearest distillation of this post is: the contribution at time t of an impulse u occurring at time t−τ is u(t−τ)h(τ). However, I also agree with Dicklyon that the main point seems to be the symmetry of the convolution. All other considerations are too application-specific to be a compelling answer to "why" the time inversion happens. The most familiar case of convolution is the product of two polynomials (or series):

$(\sum a_{n}x^{n})(\sum b_{n}x^{n})=\sum _{n}\left(\sum _{k=0}^{n}a_{k}b_{n-k}\right)x^{n}.$

Here there's no deep reason "why" it's k in one and n−k in the other: it just works out that way so that the total homogeneity on each term in n. The Fourier series of a function, as a series in the variable x = e^iθ, obeys the same relation (with a few trivial changes), and again there is no additional need to look for an explanation of "why". Sławomir Biały (talk) 14:40, 1 August 2010 (UTC)

The question of 'why the time inversion?' is because the mathematical formula for the continuous time convolution has a part that involves $g(t-\tau )$. The formula is not straight forward because there is an important piece of information that is often left out, which confuses a lot of people. The term $g(-\tau )$ is actually quite misleading. It should always be mentioned that an intermediate function such as p(-t) first needs to be formed, where p(-t) is purposely defined to be the mirror image of g(t) along the vertical axis. And once this mirror function p(-t) is defined, we then completely forget about the original function g(t), and we form a "brand new" function g(-t) = p(-t). Here, g(-t) should no longer be associated with the original g(t) function, since there is no direct connection - which is clearly understood by taking a hypothetical example case of the original g(t) function being causal, or equal to zero for t < 0, implying g(-t) = 0 when '-t' is negative. Obviously, the 'actual' g(-t) function is nothing like the newly "fabricated" g(-t) function. Hence we do not associate the newly constructed g(-t) function with the original g(t). The last step is to change the variable 't' to 'tau', leading to $g(-\tau )$, which then leads to $g(t-\tau )$ for the sliding procedure. The main point is that g(-t) is not directly related to the original function g(t), even though the same alphabetic letter 'g' is used for both of these functions. That's the catch. And, in my opinion, it's very poor form to teach people that g(-t) is the reflection of g(t) when we all know it is not how it is. Also, for discrete-time convolution, the 'time reversal' step merely provides a 'graphical way' to implement the discrete summation formula. KorgBoy (talk) 07:16, 2 January 2019 (UTC)

Pattern recognition application mention

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)

Thank you. I added it to External links section. --Bob K (talk) 17:06, 4 January 2019 (UTC)

Time inversion

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)

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)

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)

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)

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)

The graph of $g(-t)$ most definitely is the reflection of the graph of $g(t)$ about the y-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)

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)

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)

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)

I'm glad it's all clear now, if it really is. But statements like: "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" make me think that you should actually look at an image of the computer screen reflected in a mirror. And I don't know what to make out of "such as p(-t) = g(t). So, in the end, the convolution will involve sliding p(-t) over f(t)." No, it would end up sliding p(t) [which is g(-t)] over f(t).
--Bob K (talk) 21:31, 4 January 2019 (UTC)

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)

Okay, this is a warning to both KorgBoy and to Deacon Vorbis that what's going on here has been noticed:

• KorgBoy, it is inappropriate and potentially a violation of Wikipedia guidelines to go back and modify talk page posts that you have previously made. It is possible to modify them but it must be made clear what you have done. See REDACT for both the policy detail and how to do it right. Please go back and conform your recent talk page modifications to that standard and follow it in the future.
• Decon Vorbis, there's nothing in TPG that gives you the right to revert what KorgBoy has done and you've clearly violated 3RR. TPG probably would, under the "format correction" allowance, let you modify his posts to add the redlining that he's not been doing or you could post a note to the discussionsaying that he's done it and giving diffs to warn others if you think that's needed, but you don't just get to revert them and you sure don't get to edit war over them. If KorgBoy continues to modify his posts without following REDACT, report it to ANI.

I'm not reporting anyone to administrators or enforcement noticeboards at this point, but I'm watching this page and both parties' need to stop what they're doing. Regards, TransporterMan (TALK) 15:59, 23 April 2019 (UTC)

asterisk operator

I understand where the editor who removed the word "operator"[1] is coming from, since I admittedly never heard a mathematician call it by that name. However, I am somewhat chagrined about the loss of precision. This is not an idiosyncratic choice of words, but the official Unicode name of the character (U+2217). This is the only term that I'm aware of that unambiguously distinguishes it from all other types of asterisks. Providing that name is relevant in the context of the article. I'll therefore reinsert it with a mention of Unicode, which hopefully makes it clearer. — Sebastian 17:07, 22 June 2010 (UTC)

This is strangely over-specific. It is not written in stone that the convolution needs to be the "asterisk operator", many authors simply use an asterisk or conventional five-pointed star, not to mention the fact that the unicode glyph need not be used at all. I will adjust the text accordingly. Sławomir Biały (talk) 18:28, 22 June 2010 (UTC)

Your latest edit, in which you removed any mention of Unicode and even the link to the pertinent article about the symbol appears to be an emotional overreaction. I consider this an edit war, and I will not continue on that path. Instead, I ask you politely to undo at least the excess of your reversion.

You do have a point that there are more characters possible. That invites the comparison with multiplication. Not only does that article have a section on multiplication#Notation and terminology, but we even have an article on the multiplication sign itself. Obviously, the asterisk operator isn't as important as to warrant its own article, but at least is should be mentioned and referred to in the one article that describes its use, don't you think so? — Sebastian 19:18, 22 June 2010 (UTC)

How about if we just add one sentence like this: "Unicode defines the "[[asterisk operator]]" (U+2217) for this."? — Sebastian 19:30, 22 June 2010 (UTC)

There is List_of_mathematical_symbols_by_subject describing many mathematical symbols and their uses. It does look like (U+2217) is an asterisk operator, not specifically a convolution operator. A common use for asterisk, to indicate a footnote, is raised more like a superscript. I suspect that is what distinguishes the operator form. Well, also, mathematical operators have different typographical spacing than text operators. Gah4 (talk) 18:49, 7 June 2019 (UTC)

You might do well to familiarize yourself with the consensus policy. You were WP:BOLD, and made an edit. I the made a change to your original edit, because I felt that the word "operator" was a poor choice in a mathematical context. You changed this back to mandate the use of unicode. I changed your subsequent edit because it imposes an artificial standard on the typography of the convolution operation. (E.g., the glyph $\star$ is perhaps equally common in the literature.) What I have not seen demonstrated here is that any mention of Unicode is at all appropriate in the article. Does Unicode actually mandate that this character must be used for convolution? If so, can we please see a reference to that effect? Sławomir Biały (talk) 19:42, 22 June 2010 (UTC)

9 years after the fact, I know: With all the discussion on how to deal with "asterisk" no one has mentioned that, if the diagram shown and the cross-correlation article are correct, that asterisk is used for convolution, and star used for cross-correlation. To at-least be consistent, I've removed the "or star" reference with the expectation that someone with more knowledge/passion will correct me. Ch'marr (talk) 15:28, 7 June 2019 (UTC)

I have a question related to that, regarding the first line of the definition: What is ƒ∗g supposed to be? Is that actually used in literature for the convolution (never saw that before) or is it just supposed to be a placeholder for asterix or star? If the latter is case, it might be better to simply explicitly write both notation rather than having such a placeholder construction, which in doubt might just be irritating to readers.--Kmhkmh (talk) 20:53, 22 June 2010 (UTC)

Sounds like your set-up doesn't display this Unicode character properly — i see ∗ as a slightly larger asterisk, but i'm guessing it appears as an empty box to you? The lack of universal support for the full Unicode set may be another reason for not being too dogmatic about its use. I notice the Table of mathematical symbols only gives the standard asterisk (*) in the HTML column of the entry for convolution. I don't claim to understand the relationship between Unicode and HTML. Qwfp (talk) 22:01, 22 June 2010 (UTC)

Yes I'm seeing an empty box, while the other asterixs are fine. Possibly just my setup, however since I'm using a fairly standard installation (default xp + default firefox) this problem might occur for a larger number of readers, hence it might be a not a good idea to use that unicode charachter in the article.--Kmhkmh (talk) 22:25, 22 June 2010 (UTC)

Hmm, I'm using XP & Firefox too, and i've just tried a few different fonts (Arial, Tahoma, Candara) and its ok in all. Qwfp (talk) 07:06, 23 June 2010 (UTC)

Ah sorry actually I used opera 10.53, when i looked at it. I checked with my firefox 3.5.9 and there it worked. But I don't quite get why we would need to use different asterixs throughout the text anyway, I mean why don't we use * only, since it is used throughout most of the article anyway and is displayed correctly on any system (due to being ascii).--Kmhkmh (talk) 09:17, 23 June 2010 (UTC)

As far as I know, symbols for multiplication related operations have been confused since Fortran started using the asterisk, commonly called star, for multiplication. Well, dot product and cross product (small centered dot and cross, respectively) usually work out right. Anything that looks slightly like an asterisk is probably within range for convolution. Usually, but maybe not always, you can tell from context whether convolution or cross-correlation is meant. Gah4 (talk) 18:27, 7 June 2019 (UTC)

Introductory definition

The introduction sounds like it might be the same as the functional inner product. It's extremely unclear. — Preceding unsigned comment added by Edgewalker81 (talk • contribs) 19:46, 4 January 2018 (UTC)

The initial definition doesn't mention that one of the function is reversed:

...ing the area overlap between the two functions as a function of the amount that one of the original functions is translated.

Shouldn't it read "translated and reversed"?

Joelthelion (talk) 14:31, 15 September 2014 (UTC)

Or maybe the figure is incorrect? — Preceding unsigned comment added by 198.161.2.212 (talk) 18:07, 22 January 2016 (UTC)

Think you are right. The red line in the figure (function g) should be the same in convolution and cross-correlation. Because - if you convolve f and g and later you correlate f and g(rotated 180 degrees) - you should get the same result.   — Preceding unsigned comment added by 2001:4643:EBFE:0:3923:18BE:E93E:A96 (talk) 14:02, 19 November 2016 (UTC) 

If we assume that the two top rows are labeled correctly (with f's and g's), then the 3 red traces should all look identical, as they did in the Oct 7 version. Then what's wrong is the $(f\star g)$ picture. It should look like $(f*g)$ because f(t) is symmetrical. Alternatively, it should be re-labeled $(g\star f).$

--Bob K (talk) 15:40, 19 November 2016 (UTC)

I expanded the figure to depict all the combinations of convolution, correlation, and order of operation (e.g. f*g and g*f). That should help.

--Bob K (talk) 19:04, 19 November 2016 (UTC)

Thank you for correcting it Bob K.  — Preceding unsigned comment added by 2001:4643:EBFE:0:C916:D6C0:D6BD:B173 (talk) 10:06, 20 November 2016 (UTC) 

I don't know much about math (hence wanting to know what convolution is) but found the introductory definition lacks a brief description of what it is. Sentence 1 says vaguely that it is "a mathematical operation on two functions". Sentence 2 talks about how the term refers to both the result and to the process of computing it. Sentence 3 says how it is similar or different from cross-correlation (a concept the reader may also know little about at this point). Sentence 4 is very technical "For continuous functions, the cross-correlation operator is the adjoint of the convolution operator." At this point I gave up looking for a brief description and started looking at the diagram...

I found this on this Wolfram MathWorld page. Their first sentence is: "A convolution is an integral that expresses the amount of overlap of one function g as it is shifted over another function f. It therefore "blends" one function with another."

The key for me was the shifting or 'sliding' of one function over the other. I think that should be mentioned at some point in the introductory section. I would try editing it but thought it might be better to leave it to a proper mathematician do it.

--Billtubbs (talk) 15:40, 8 June 2019 (UTC)