Talk:Fourier transform/Archive 1

Untitled
Sounds like trying to tell someone what music sounds like, when there's a piano right in front of you. How 'bout a formula or two? Perhaps the Fourier transform? User:Kevin_baas -2003.05.06

I didn't understand the part about the basis functions. The article says that any normal function can be decomposed into an infinite series of sines and / or cosines. Does it mean that each sine term is a basis function for the original function? In other words (like we have n-dimensional vectors), is the function in an infinite-dimensional space? Forgive my lack of mathematical rigour; I'm just in high school! Gokul


 * Yes. You understand it correctly. Pay no attention to the gibberish below. -- Kevin Baas 21:33, 16 Jan 2004 (UTC)

The infinite-dimensional space is the set of all "quadratically integrable" functions, i.e., those satisfying


 * $$\int_0^{2\pi}\left|f(x)\right|^2\,dx<\infty.$$

The functions sin(nx), cos(nx) for an "orthogonal basis", but not a Hamel basis. That it is not a Hamel basis means that not every quadratically integrable function is a linear combination of finitely many basis functions. (Some people say that phrase is a redundancy--that "linear combination" by definition means just finitely many. If so, that's why redundancy is sometimes useful!)  Michael Hardy 20:31, 15 Nov 2003 (UTC)


 * What is why? The fact that they pointed out that it's a redundancy, i.e. that it is confusing insofar as one expects it to give new information, and one is puzzled trying to figure out what that information is, when really its a wild goose chase?  Because that's what someone means when they call something redundant.  They aren't merely pointing out the fact that its redundant, they are implying that this poses a problem.  You say that this "posing a problem" is usefull.  How? -- Kevin Baas 08:58, 14 Jan 2004 (UTC)


 * Mr. Baas, why the belligerence in your tone? Including the word "gibberish"? To say "a linear combination of finitely many basis functions" is an instance of how redundancy can sometimes make things clear where they might otherwise be confusing.  "that's what someone means when they call something redundant"??  No, I don't think that's what "redundant" means. Michael Hardy 01:29, 17 Jan 2004 (UTC)


 * I didn't say that that's what redundant means. Try reprocessing the syntax of that sentence. -- Kevin Baas 09:35, 17 Jan 2004 (UTC)

Regarding that last edit by Wile, I don't see how it improved the article. It seems to me have made it less readable. What do other people think? -- Kevin Baas 08:58, 14 Jan 2004 (UTC)


 * As the author of the edit in question, let me explain. The article was phrased rather strongly in signal processing terms -- time domain vs frequency domain and so on. There is nothing about the FT which requires interpretation in those terms. I tried to rephrase it in general terms, and to identify the signal processing interpretation as a special case. I also added the defn of the inverse transform. Wile E. Heresiarch 06:21, 15 Jan 2004 (UTC)

In reference to other edits since the one I made -- this link should yield the appropriate diff (at least until the page is changed again). The more recent edits seem problematic. The first paragraph says the FT "can be thought of" or "may be thought of" in a couple of ways: that is not appropriate for an encyclopedia article. First we have to say, definitely, what it is. Then we can talk about how that is interpreted or applied. Also, under the heading "Generalizations" some stuff has been cut out, including the definition of A and B, now orphaned in the article. I agree that section can be clarified but information has been lost. I'm inclined to revert some of these changes; I'll do so in a few days unless I hear otherwise. Wile E. Heresiarch 06:21, 15 Jan 2004 (UTC)


 * RE: "some information has been lost": Apparently, we have different interpretations of "information". My interpretation is based on information theory and semiotics.  You seem to take a modernist approach.  Perhaps you would be satisfied if this page was purely mathematical formulas?  For me, that would make the page have little or no information.


 * The article should be clearly readable for someone who does not know what a fourier transform is. It should not require the reader to be versed in higher-level mathematics, or, indeed, anything esoteric.  Ofcourse, there is a degree of neccessary knowledge; it can't be written for a first grader, but it shouldn't be written for a graduate either.  Rather, it should be written as if it was written by a graduate: in a common language.  Graduates are discouraged from writting esoterically.


 * I also the dispute the value of "generality" as you define it. Sure, if we want to be purely general, we'd give a mathematical formula without explanation.  Perhaps we'd use some code words like "linear operator", ect. but these would simply be links to formulas, and these links would be within statements written in the language of mathematics, i.e. logically precise, like "given x, we restrict the rotation of theta to the contour of z, and solve for y...".  This is not how good text books are written, and it is not how encyclopedia articles are written.  The introduction should be conceptual and visual.  I'm sure there are numerous applications for the fourier transform, just as there are numerous applications for addition besides counting apples (such as binary logic, compression, encryption, ect).  This doesn't mean that we shouldn't use apples to demonstrate the principle of addition.  For more on writting mathematics articles, see WikiProject Mathematics.


 * And I'm sorry the, fourier transform is principally used in signal processing. It is not "used" in pure mathematics, because pure mathematics doesn't "use" anything, rather it rejects all meaning and works in absurd logical manipulations (which is about as usefull as doing binary logic by hand, and amounts to the same in practice (see Bertrand Russell, Principia Mathematica).)  My contentions about pure mathematics aside, the point is that the article should be pedagogical.  Otherwise, it's nothing but a bunch of gibberish.  Your edits have made it less pedagogical. -- Kevin Baas 21:33, 16 Jan 2004 (UTC)


 * The Fourier transform has many applications to things other than signal processing, in pure mathematics, in physics, in engineering, in cryptology, in number theory, in combinatorics, in probability theory, in statistics, and in many other fields. See Dym & McKean's book to read about many of these. Michael Hardy 01:21, 17 Jan 2004 (UTC)


 * In physics, I assume you mean quantum physics, which operates in the frequency domain (it works with waves). In electrical engineering, same thing, in structural or mechanical engineering, vibrations, as frequency domain as you can get.  Number theory and pure mathematics are not applications, as previously discussed.  Probability theory and statistics (which are pretty much different section of the same thing), i don't know what you are refering to.  Perhaps the original signal is fourier transformed before statistics are done on it?  (By signal I mean set of data which constitues the information being manipulated/processed; the variables of the equation(s), if you will.)  Cryptology, sure. In the conventional sense, the symbol string is a signal, but oh well, I'm soft on this one as I am with probablity theory, I don't know much about the fourier transform's usage in it.   In any case, all this is irrelevant because I never contested the fact that the fourier transform is used in other fields.  Furthermore, it is besides the point of my argument.  That last paragraph was just a sidenote, completely peripherary to my argument.  I could have, and probably should have, left it out. -- Kevin Baas 09:35, 17 Jan 2004 (UTC)

It is to be regretted that discrete and other non-continuous Fourier transforms have been purged from this page. Michael Hardy 01:22, 17 Jan 2004 (UTC)


 * I think this page is about the fourier transform that is an integral transform. I agree that there should also be a page on fourier series and the discrete fourier transform.  I also think that there should be links to those pages on this page. -- Kevin Baas 09:35, 17 Jan 2004 (UTC)

Magnitude and phase shift
This part here, "The real parts of the resulting complex-valued function F represent the amplitudes of their respective frequencies (s), while the imaginary parts represent the phase shifts." I don't believe is correct.. Isn't it the case that the Absolution value of the complex value is the magnitude, i.e. Mag = sqrt(Re^2+Im^2) while the phase is given by arctan(im/re)?


 * You are correct, Mr. 202.72.131.230. Article fixed. -- Cyrius|&#9998 11:55, May 7, 2004 (UTC)

Redundancies with Continuous Fourier transform
I think that, originally, Fourier transform was meant to discuss the general idea of Fourier transforms, and continuous Fourier transform was meant to discuss that particular case. However, as it stands, Fourier transform is now mostly about the continuous Fourier transform, and there is a tremendous amount of redundancy. I would suggest merging a lot of the present material into continuous Fourier transform. &mdash;Steven G. Johnson 00:35, May 19, 2004 (UTC)


 * I agree there is a lot of redundant material. I guess it seems useful to have a certain amount of redundancy, but there is too much at present. So I'm in favor of doing some merging, if you are inclined to do so. Happy editing, Wile E. Heresiarch 15:49, 19 May 2004 (UTC)


 * I agree; but please keep a formulation of the Fourier Plancherel Theorem as a Theorem. And also keep links to Potryagin duality. CSTAR 16:10, 19 May 2004 (UTC)

=Plancherel vs. Parseval?=

The Wikipedia article on the Plancherel theorem and the one in MathWorld seem to be completely different. The Wikipedia one states a theorem about the Fourier transform being in L2. The MathWorld one states a specific identity that is a generalization of Parseval's theorem. The reference books I have on my desk don't even mention Plancherel, but I wonder if there aren't two completely different things, a Plancherel theorem about L2-ness and unitarity of the Fourier transform, and a Plancherel identity referenced on MathWorld. Could someone please look into this? (And who came first, Plancherel or Parseval?) And please give references; neither the Wikipedia nor the MathWorld articles cite anything for the Plancherel theorem. &mdash;Steven G. Johnson 21:42, May 19, 2004 (UTC)


 * There are references in the article on Pontryagin duality, but I'll put some in the Fourier analysis article specific to the case discussed there.CSTAR 22:03, 19 May 2004 (UTC)


 * References should go under Plancherel theorem, not (just) under Fourier.


 * I and (everybody I know) has always called the unitary property of Fourier as Plancherel's theorem, although Hormander in the reference I provide calls it Parseval (Yosida calls it Plancherel). I don't personally know Hormander :( The generalizations to non-commutative groups are called Plancherel theorems (see Dixmier, in reference list of Pontryagin duality). I will object if you remove Plancherel, but will keep quite if you leave in Parseval.  Somebody with historical knowledge of this might add something. It probably should not be hard to figure out. I wouldn't consider Mathworld authoritative. CSTAR 22:21, 19 May 2004 (UTC)


 * Everyone in the physics and engineering community that I know calls it Parseval. =)  Ditto for the books I have on my shelf: Oppenheim & Schafer (Discrete-time signal processing), Arfken &amp; Weber (Mathematical Methods for Physicists), Jackson (Classical Electrodynamics), ... and Google gives twice as many hits for "parseval theorem" compared to "plancherel theorem".  To be clear, I'm not proposing to remove the reference to Plancherel, but the situation with respect to Parseval needs to be clarified, especially if the latter terminology is more common.  (I don't regard Mathworld as authoritative, but the fact that they list a generalized version of Parseval under Plancherel is a clue for something to look for.)  &mdash;Steven G. Johnson 00:03, May 20, 2004 (UTC)


 * Note also that the Wikipedia entry for Plancherel theorem doesn't currently even mention unitarity, it only says that the Fourier spectrum is L2-integrable. So, Wikipedia isn't even consistent with itself on this. &mdash;Steven G. Johnson 00:06, May 20, 2004 (UTC)
 * OK I've fixed these CSTAR 02:26, 20 May 2004 (UTC)

The following books call it Plancherel:


 * Loomis, Abstract Harmonic Analysis (Parseval's is the corresponding identity for Fourier Series)
 * Yosida Functional Analysis
 * Dixmier, ''Les C*-algebres et leurs Representations '
 * Lang SL_2(R)

It seems to be Math vs Physics and engineering....

CSTAR 01:18, 20 May 2004 (UTC)

Update: I did some checking, and it seems that Parseval's theorem pre-dates Plancherel (and, in fact, predates Fourier transforms), but is also less general; it was a result about series that was used to prove "unitarity" of the Fourier series. I'm guessing that this historical precedence is why the name continues to be attached to all unitarity properties of Fourier transforms, at least in physics and engineering. &mdash;Steven G. Johnson 02:26, Jun 6, 2004 (UTC)
 * Remark on update: Plancherel theorem, in mathematics is used to refer in general -- even for non commutative groups -- to the unitary property of the decomposition of the left regular representation of a group into primary representations. Those representations which appear in the decomposition are called square-integrable (at least type I and unimodular groups). The work of Harish-Chandra to a large extent was devoted to finding specific forms for the abstract Fourier transform.

FT Properties editing
I've just edited the properties of the Fourier Transform as I found there was something wrong with the convolution theorem : to be coherent with the previous definition of the trasform, the factor $$\frac{1}{\sqrt{2\pi}}$$ must be added. I've also written something about the important derivation property..very useful in many fields of pure and applied science. --Giuscarl 20:22, 4 Jun 2004 (UTC)
 * You're right.CSTAR 23:34, 4 Jun 2004 (UTC)

New edit
Fourier transform is a decomposition into functions having a very specific form. Mentioning waves is misleading and should be changed..also sinusoidal should be replaced by something like complex exponential.

CSTAR 01:29, 11 Jun 2004 (UTC)


 * I agree that "waves" is a bit more vague than I would like; on the other hand, I can understand the editor's inclination to write something handwavy that gives a non-mathematical person some impression of what's going on; it's not the function of the introduction to be an exact formal definition. Regarding "sinusoidal", however, in what way is a decomposition into complex amplitudes times complex exponentials not a decomposition in terms of sinusoidal basis functions, and vice versa? &mdash;Steven G. Johnson 05:04, Jun 11, 2004 (UTC)


 * Sorry, my objection is to the word basis in the phrase sinusoidal basis functions.


 * These are not basis functions on L^2 (because they aren't in L^2). They are tempered distributions, but in what sense are they a basis in the space of tempered distributions? I suppose one could come up with a definition but it wouldn't be helpful as an introduction.


 * why use equivalent in the defining sentence? Why not just say the fourier transform represents a function as an integral transform complex exponentials (or if you really want of sinusoidal functions).  This is my understanding of what you;'re trying to say,  i.e., f = F(g) for some g. I dunno maybe you mean something else.


 * I'm not against "handwavy", but why does signal come in here?. OK I'll settle for something that would accurately express the decomposition. However, signal is a function in the time domain (At least that's the way I've always thought of it, like the signal on my radio) and decomposition into waves is not a "frequency" decomposition.  In short bringing up waves and wave propagation here seems like a bad idea. If you can tell me how this clarifies matters I'll shut up.CSTAR


 * First, please separate the discussion of sinusoidal basis functions from the edit about decomposing a signal. The latter I am ambivalent about.  As for the former, the usual meaning of "basis" that I am familiar with is that you represent elements of a vector space by a linear combination of things in that space times coefficients in the field of that space.  Furthermore, since part of the point of the article should be to define what it means to "transform" a function, I don't think it's useful to define the Fourier transform as an "integral transform" of something (especially since not all forms of the Fourier transform involve integrals). &mdash;Steven G. Johnson 23:04, Jun 11, 2004 (UTC)


 * ("sinusoidal" is understandable to a wider audience than "complex exponential", and it is perfectly correct. In fact, arguably more correct, since "complex exponential" on its face includes exponentials of arguments that are not purely imaginary.)

Rewrite of introductory sentences
How does this sound?


 * The Fourier transform, named for Jean Baptiste Joseph Fourier, is an integral transform which represent an almost arbitrary function on the real line in terms of complex exponentials. Equivalently, it represents a function on the real line as a continuous superposition of sine and cosine waves.

or some combination thereof.

CSTAR 16:26, 11 Jun 2004 (UTC)


 * That version only describes the continuous FT. &mdash;Steven G. Johnson 22:54, Jun 11, 2004 (UTC)


 * OK I'm confused. Do you also want to include discrete Fourier transforms on finite abelian groups in this introduction? In what sense are the basis elements (in this case they are basis elements) "sinusoidal" or "waves". CSTAR 23:10, 11 Jun 2004 (UTC)

Looks good
OK OK, it looks fine now. Thanks. CSTAR 23:12, 11 Jun 2004 (UTC)

F notation
I don't know where else to ask. What's the difference between $$\mathcal{F}$$ and $$F\,$$. Are they used correctly in this article? - Omegatron 01:56, Sep 19, 2004 (UTC)


 * Seems clear to me from the article. $$F$$ is the function whose Fourier transform is to be found, and $$\mathcal{F}(F)$$ is the transformed function, so that $$\mathcal{F}$$ is the transform itself, i.e., the mapping from one space of functions to another. Michael Hardy 22:25, 19 Sep 2004 (UTC)


 * If you know it, explain it in the integral transform article, as well. - Omegatron 02:02, Sep 19, 2004 (UTC)


 * The function $$F\,$$ is the Fourier transform of the function $$f\,$$. $$\mathcal{F}$$ is the Fourier transform operator.CSTAR 02:18, 19 Sep 2004 (UTC)


 * Ok. But when do you use the $$\mathcal{F}$$? $$\mathcal{F}(f(x))$$?  hmm is there an article for operator?  ...  Yes, but it doesnt mention the notation.  - Omegatron 02:28, Sep 19, 2004 (UTC)


 * Strictly speaking, $$(\mathcal{F} F)(t)$$ is the right way to parse the thing, and $$\mathcal{F}(f(x))$$ is a solecism, often used by engineers and sometimes used even by mathematicians. When you use $$\mathcal{F}$$ alone would include such things as when you say "the Fourier transform is a 90 &deg; rotation of the space of square-integrable functions".  I seem to recall that the article titled operator was something of an Augean stable, but that was months ago; I don't know if it's improved. Michael Hardy 22:30, 19 Sep 2004 (UTC)


 * f(x) denotes the value of f at x. f denotes a function x denotes a real number. $$\mathcal{F}(f(x))$$ would make sense only if $$\mathcal{F}$$ were defined for numbers, but it;s defiend for functions.CSTAR 02:34, 19 Sep 2004 (UTC)


 * Oh oh I see. So


 * $$F(\omega) = \mathcal{F}(f(t))$$

Yeah. If you're picky you will note that t has no real purpose in the above formula. It really should be within the scope of a binding operator such as $$ t \mapsto f(t) \quad $$.; this however by rules of lambda-calclus reduces to the term f.CSTAR 02:47, 19 Sep 2004 (UTC)


 * $$\mathcal{F}^{-1}(F)(t)$$ is used in the article. is this the correct way to say the above, are they both incorrect, or is the t just extraneous but it doesn't really matter? - Omegatron 02:54, Sep 19, 2004 (UTC)


 * The variable t occurs on the right hand side as well as the left hand side. The RHS is a term (in this case an integral of an exponential); You can't get rid of the t.  Try it. What would you get?CSTAR 03:12, 19 Sep 2004 (UTC)


 * In that case, I don't understand. I thought you meant that it was correct to use $$F(\omega) = \mathcal{F}(f)$$. - Omegatron 03:30, Sep 19, 2004 (UTC)


 * If you think of $$F(\omega) = \mathcal{F}(f)$$ as meaning two things: (a) $$F = \mathcal{F}(f)$$ and (b) a suggestion that the symbol $$\omega$$ is reserved to name an independent variable to name the argument of the function F, since $$\omega$$ is often thought of (for instance by physicist or engineers) as frequency.CSTAR 03:40, 19 Sep 2004 (UTC)


 * I thought I understood, but I guess not. The term "scope of a binding operator" would probably help.  You don't have to teach it to me if it's something I don't already know.  Just point me where to look.  Regardless, is the article notation right? - Omegatron 14:42, Sep 19, 2004 (UTC)


 * See variable-binding operator or some such thing. There is a link to this somewhere in wikipedia. CSTAR 15:36, 19 Sep 2004 (UTC)


 * It's at free variables and bound variables. Michael Hardy 22:14, 19 Sep 2004 (UTC)

I'm used to engineering notation, where $$X(f) = \mathcal{F}(x(t))$$, so you can use f for frequency and avoid confusing Fs. Then of course there's $$F(f) = \mathcal{F}(f(x))$$. :-) - Omegatron 02:40, Sep 19, 2004 (UTC)

In fact, I vote that we change f(t)->F(&omega;) into some other letter (I know you won't use X, but maybe g?), to avoid confusing newcomers to the Fourier transform. - Omegatron 02:42, Sep 19, 2004 (UTC)


 * I didn't make the choice of notation here. I'll leave your suggestion to somebody else.CSTAR 02:46, 19 Sep 2004 (UTC)

One thing the "engineering notation" does not allow for is the idea that functions have values. E.g., if f(x) = x3 for all values of x, then f(2) is the value of that function at 2, and is equal to 8. If you say f(&omega;) is the function to be transformed, and g(t) is the transformed function, then g(2) should be the value of the transformed function at the point t = 2. But if you use the "engineers' notation" and write $$\mathcal{F}(f(\omega)) = g(t)$$, then you cannot plug 2 into the left side. But watch this: $$(\mathcal{F} f)(2)$$ is the result of plugging 2 into the transformed function.

One thing to be said for the difficulties introduced by the engineers' notation that are avoided by the cleaner, simpler, but more abstract "mathematicians' notation", is that perhaps sometimes one ought not to be evaluating these functions pointwise anyway! But that's a slightly bigger can of worms than what I want to open at this moment ... Michael Hardy 22:38, 19 Sep 2004 (UTC)


 * All I meant by "engineering notation" was not using the letter f as a function, since it would too easily get confused with frequency as a variable in lowercase and script F for the fourier transform in uppercase, leading to protracted discussions about why there are two capital Fs and why one is script and the other is not and what it all means. :-)


 * *Reads the recommended articles on bound variables* - Omegatron 03:40, Sep 20, 2004 (UTC)


 * Let me drop in w/ my $0.02 -- I agree w/ Mike Hardy that conventional engineering notation (which puts function arguments in inappropriate places) is imprecise & misleading, and the usual "pure math" notation is superior. That said, it is certainly confusing & unnecessary to have different kinds of "F" running around; making technical distinctions based on font types is problematic IMHO. So, Omegatron, I'm not opposed to replacing f by g, likewise F by G, throughout the article. I'll consider doing that myself. Regards & happy editing, Wile E. Heresiarch 15:11, 20 Sep 2004 (UTC)


 * Check Continuous Fourier transform as well. - Omegatron 15:31, Sep 20, 2004 (UTC)

Assessment comment
Substituted at 21:15, 4 May 2016 (UTC)