Talk:Discrete-time Fourier transform

The $$X(e^{i\omega})$$ notation
This article uses the notation $$ X(e^{i\omega}) $$ for the resulting function, already in the definition of the transform, without providing a rationale. I know that this notation is in common use, in particular in the signal processing community. But as far as I can see, it is not a necessary concept in the defintion of the transform. I would rather see an defintion like


 * $$X'(\omega) = \sum_{n=-\infty}^{\infty} x[n] \,e^{-i \omega n}\,$$

which, after providing the rationale, later is rephrased as


 * $$X(e^{i \omega}) = \sum_{n=-\infty}^{\infty} x[n] \,e^{-i \omega n}\,$$

and pointing out that X and X' are two related, but different, functions.

This distinction between X and X' (as they are defined here) is very important. For example, in the relation


 * $$x[n] \cdot y[n] \, \Leftrightarrow \frac{1}{2 \pi} X(e^{i \omega}) * Y(e^{i \omega}) $$

it is important to realize that the convolution operation should be made on the X' and Y' functions rather than on the X and Y functions. This is not obviously clear from the current notation used in these formulas. --KYN 21:00, 31 August 2006 (UTC)


 * The rationale can be found at Discrete-time_Fourier_transform. But I am a member of the signal processing community who never liked the $$X(e^{i \omega})\,$$ notation.  The $$X_T(f)\,$$ notation, used in the article, is better, because it can represent both sides of the equality:


 * $$\sum_{k = -\infty}^{\infty} X(f - {k f_s}) = \sum_{n=-\infty}^{\infty} T\cdot x(nT) \cdot e^{- i 2\pi f n T}\,$$


 * Therefore I would lean toward defining the DTFT in terms of $$f\,$$ or $$\omega\,$$ and save the $$X(e^{i \omega})\,$$ notation for the Z-transform intro.
 * --Bob K 00:26, 1 September 2006 (UTC)


 * I would lean towards $$X(\omega)$$ myself as well. Note that this has the added benefit of being more similar to the other Fourier transform articles. —Steven G. Johnson 04:14, 1 September 2006 (UTC)


 * That's an obvious but very important point. I believe that the $$X(e^{i \omega})\,$$ notation obscures the fact that it's just a normal Fourier transform of $$x_T(t)\,$$.  I believe it's because of that confusion that Rbj insists that  the DTFT is subsequent to a proof of the sampling/reconstruction theorem.  But as we all know, one can sample a signal and do a DTFT with the samples without ever proving the sampling theorem.  And having done that, the sampling theorem becomes rather obvious and trivial to prove, compared to the  bloated proof now offered by Wikipedia.  --Bob K 14:41, 1 September 2006 (UTC)

In Discrete-time_Fourier_transform we learn that the resulting function of the transform is periodic, but I don't see that as a sufficient rationale for writing $$X(e^{i\omega})$$. There must be something more than this? Why else has this rather peculiar notation become so widespread? It can't be just to constantly remind the user about the periodicity of X? --KYN 11:17, 2 September 2006 (UTC)


 * We also learn that it "emphasizes the relationship to the Z-transform". The symbol $$X\,$$ is overloaded, thus ambiguous.  At various times, it can represent any of these functions:


 * $$X_1(\omega) = \mathcal{F}\{x(t)\}\,$$
 * $$X_2(\omega) = \mathcal{F}\{x_T(t)\}\,$$
 * $$X_3(z) = \sum_{n=-\infty}^{\infty} x[n] \,z^{-n}$$


 * We can remove one of the ambiguities by noting that:


 * $$X_2(\omega) = X_3(e^{i \omega})\,$$


 * $$X_2\,$$ is a special case of $$X_3\,$$, so we simply dispense with the 2nd definition. Many narrowly focussed DSP articles have no use for $$X_1(\omega)\,$$ either, leaving no ambiguity at all.  I believe that is the rationale.  However, a coherent sequence of encyclopedia articles has to build from the ground up.  So I don't believe that rationale is appropriate for us.  We could avoid overloading if we really wanted to, just as I have done above with subscripts, but that would be non-standard.  So we have to rely on context and prose to make our articles unambiguous.  That is the standard, as far as I can tell.
 * --Bob K 15:21, 2 September 2006 (UTC)


 * I guess another possible rationale, depending on your feelings about the existence of $$\mathcal{F}\{x_T(t)\}\,$$, is that $$X(e^{i \omega})\,$$ seems to circumvent the whole debate over use of the Dirac comb function. But if that is the rationale, I would attribute it to the mathematical community, not the signal processing community.  --Bob K 15:36, 2 September 2006 (UTC)


 * I think the definition with exp(i omega) is really a z transform, and we should just not use it except for that. Go back to X(omega) throughout.  At present we have an incompatible mixture, so something must be done. Dicklyon 04:11, 6 December 2006 (UTC)

And the table...
The table uses X(omega), not the form defined in the lead. And some of the entries have periodic DTFTs while some do not. That is, the ones with exp(i omega) in them are obviously periodic, but the the ones like delta(omega+a) are not; they seem to just be copied from regular Fourier transforms. These need to be fixed. Dicklyon 04:09, 6 December 2006 (UTC)

Oh, I missed this hack:


 * $$\omega \!$$ is a real number in $$(-\pi,\ \pi)$$, representing continuous angular frequency (in radians per sample).
 * The remainder of the transform $$(|\omega| > \pi \,)$$ is defined by: $$X(\omega + 2\pi k) = X(\omega)\,$$

Is this enough to make the table always correct? Or are there some entries where an infinite sum is really needed? Probably the sinc-squared needs to be patched up at least, or W limited to 1, otherwise there's overlap that needs to be summed up. And what about a general exp(an) for complex a? Dicklyon 04:33, 6 December 2006 (UTC)

The table is still very screwed up and inconsistent. For example, the factors in front of the delta functions for the exp and cos are not consistent. One just got changed, but it didn't bring them any closer to agreement. Anyone want to take this on? And maybe the section on sampling should be made consistent with the X(omega) definition instead of its own X(f)? Dicklyon 19:32, 14 December 2006 (UTC) OK, I made some changes. Someone should check me. Dicklyon 05:03, 18 December 2006 (UTC)

Alejo2083, thanks for checking me; you may be right that the one I deleted as absurd is not; I missed the n in the denominator. Anyway, I searched back and found that you're the source of the table, so I'd appreciate your comments on the rest of the changes, and also a source please, since I don't want to have to do the math to check all these. Dicklyon 00:27, 19 December 2006 (UTC) I checked a few in Matlab to see how they look, which is where it became clear that the sinc-square and sinc both needed a factor of 1; and I noticed the W condition for the one I thought was absurd, which is fine. Dicklyon 01:15, 19 December 2006 (UTC)


 * no problem :-) I added that line in the table because I made an exam about Z-transform and I had to use such a transform several times, but it was not in any table. So I calculated it in general, using generic parameters, and I published it on wikipedia, hoping it could be useful for somebody else. If you think the table is growing too big, we might consider moving it to its own page, keeping on this page just the main transforms such as sine and cosine, sinc, and a few more. Alessio Damato 12:41, 19 December 2006 (UTC)

Symmetry Properties
The new table entries have errors. For instance this equality:


 * $$X_R(e^{i \omega}) = Re\{x[n]\}\!$$

One side is time domain and the other is frequency domain.

And this:


 * $$\angle X(e^{i \omega}) = \angle X(e^{-i \omega}) \!$$

For instance, consider $$x[n] = \delta [n - k] \!$$ and $$X(e^{i \omega}) = e^{-i k \omega} \!$$ :


 * $$\angle X(e^{i \omega}) = -k\omega \!$$
 * $$\angle X(e^{-i \omega}) = k\omega \!$$

So:  $$-k\omega \ \stackrel{\mathrm{???}}{=}\ \ k\omega\!$$

I haven't checked everything. Those just leaped out at me. --Bob K 04:17, 22 February 2007 (UTC)

I am replacing a discussion with this synopsis: The following text (or something similar) should accompany : If we were to decompose the real and imaginary parts of a complex function into their even and odd parts, we would have four components. And there is a one-to-one mapping between the components of a complex time function and the components of its complex frequency transform. The mapping is illustrated in the figure. --Bob K (talk) 01:54, 17 December 2018 (UTC)

notation
In DTFT the frequency variable is commonly symbolized by $$\Omega \!$$ and not $$\omega \!$$. $$\omega \!$$ is used in continuous Fourier Transform. 132.69.231.117 19:10, 26 February 2007 (UTC)


 * So are you saying that
 * $$\Omega \!$$ is most common (Google appears to disagree with that), or
 * $$\Omega \!$$ should be mentioned as another commonly used notation?


 * --Bob K 13:44, 27 February 2007 (UTC)


 * Most of the ones I checked used the small omega, but I also saw some variations on that. Probably not worth noting.  Dicklyon (talk) 02:18, 8 October 2009 (UTC)

Impulses in DTFT of u[n]
After I reverted an anon's uncommented addition of delta functions, User:Vaamarnath added back the impulses in the table, still unsourced, but this time I checked. Sure enough, that's pretty much what all sources say. This puzzles me, since the expression 1/(1 - exp(i omega)) already has infinities at all the frequencies where the impulses were added (the denominator being zero whenever omega is a multiple of 2 pi). But I don't understand distributions and functions with infinities in them well enough to know exactly what's going on here, and none of the sources appear to comment on it. I suppose the infinities there don't really do anything, since they're approached with opposite signs from the two sides; is that why the delta functions are needed? Dicklyon (talk) 02:18, 8 October 2009 (UTC)


 * I suppose it's the same reason (whatever that is) that:
 * $$\mathcal{F}\bigl\{u(t)\bigr\} = \sqrt{\frac{\pi}{2}} \left( \frac{1}{i \pi \omega} + \delta(\omega)\right)$$
 * also has a double singularity at zero.
 * --Bob K (talk) 11:22, 9 October 2009 (UTC)


 * Also note that without the delta in this inverse transform:
 * $$\mathcal{F}^{-1}\{2 \mathrm{u}(f)\} = \delta(t) + j\cdot {1 \over \pi t} $$
 * an analytic signal would have no real part.
 * --Bob K (talk) 11:46, 9 October 2009 (UTC)

Formula for multiplication in time
The use of the convolution symbol on the titular line in the table is somewhat unclear on what type of convolution must be taken; while it's obvious that it's a continuous convolution, it's less (immediately) obvious that it must also be a periodic one.

This wouldn't be a problem since you could just click the handy link to the convolution article, but the current wikipedia article on convolution doesn't seem to define a periodic convolution anywhere, and the circular convolution article just states one (admittedly useful) property of the (normal) convolution of periodic signals while apparently leaving the actual convolution definition the same.

I mean it's obvious what should be done; just don't periodize H in the second formula (from the circular convolution article) and you're set, but what's needed is an explicit definition that states that.

I admit I don't really like the clutter of the new formula over the old one, but this should be more mathematically accurate and will have to do unless someone adds to the convolution article or has a better idea. 128.12.178.64 (talk) 22:50, 14 March 2011 (UTC)


 * Good point. And thank you... I think this was a missing link for me.  I have revised the intro to Circular convolution to clarify what it means in terms of two functions that are periodic to begin with.  And of course such functions can always be expressed as periodic summations, so circular convolution always has an equivalent representation in terms of conventional convolution.  I'm not sure, however, if this affects what you have done to the DTFT article.  It seems pretty clear now.  Reverting would not necessarily be an improvement.
 * --Bob K (talk) 03:27, 16 March 2011 (UTC)

Alternate notation
The alternate notation (z-transform like) is used in the properties table, but it doesn't seem to say so, so the results are confusing. We should clarify. Dicklyon (talk) 19:41, 12 January 2013 (UTC)

Rambling, sub-standard, misplaced paragraph
The following paragraph was inserted as a "Definition" on Aug 21. Besides the obvious flaws, it is not a definition, and it seems more concerned with the sampling theorem than with the subject of this article. Discrete-Time Fourier Transform is related to Nyquist–Shannon sampling theorem. Continuous Fourier Transform (CFT) decomposite a non-periodic signal and gives its spectrum ratios/relations (due to integral over infinite period, unlike Fourier Series directly giving coeffiecients. The Inverse Continuous Fourier Transform is the Fourier coefficent expansion form for CFT). For digital processing such as using computer, discrete data is required. Shannon's original proof in Nyquist–Shannon sampling theorem logically showed that under proper sampling condition the original Continuous Fourier Transform X(ω) could be recovered by calculating its Discrete-time Fourier transform (the maximum unique frequency component DTFT represents cannot exceed the sampling frequency, and due to aliasing the maximum unique frequency is limited under half of the sampling frequency (Nyquist frequency), therefore the spectrum becomes periodic), and therefore its Fourier coefficient expansion form can be recovered, and the original signal. As long as the sampling frequency Fs >= 2B which is the Nyquist frequency (B is the bandwidth limit of the signal), using discrete samples to unambiguiously recover the original signal is achievable. Discrete Fourier transform is the counterpart of DTFT dealing with discrete sampling and reconstruction of periodic signal. --Bob K (talk) 12:47, 24 August 2017 (UTC)

Side note explaining equation 2 has mistake
Should use $$e^{-i 2\pi f T n}$$ rather than delta function. — Preceding unsigned comment added by 128.179.187.178 (talk) 10:28, 23 November 2018 (UTC)


 * You are thinking of equation 3:
 * $$\sum_{n=-\infty}^{\infty} x[n] \cdot \delta(t-nT) \quad\stackrel{\mathcal{F}}{\Longleftrightarrow}\quad \sum_{k=-\infty}^{\infty} X\left(f - k/T\right).

$$
 * Eq 2 equates two frequency domain functions via the Poisson Summation formula. Eq 3 relates a time-domain and freq domain function via the Fourier transform.
 * --Bob K (talk) 15:54, 23 November 2018 (UTC)

Inline math style changes
Generally, large-scale style changes shouldn't be done without some discussion to establish consensus. Personally, I think the use of math tags for math inside sentences is a bit of mess, and that the template Template:math is generally better for this (it's also a smaller change from what was there in this case). Anyone want to work in that direction? Dicklyon (talk) 21:14, 25 November 2018 (UTC)


 * Thank you for the template link. It has the advantages you've mentioned, and is very "professional" looking.  I don't mind the changes from straight html to LaTex, in fact I like it, but it's admittedly not as slick as the template.  I support your idea.  But I am not as motivated as user Fvultier.
 * --Bob K (talk) 15:17, 26 November 2018 (UTC)
 * It sure does get tiring... Dicklyon (talk) 05:25, 27 November 2018 (UTC)
 * I'm about done converting it that way. Open to further suggestions. Dicklyon (talk) 06:13, 28 November 2018 (UTC)
 * I agree that the mvar template is the way to go. Using normal text for inline formulas looks ugly and it is difficult for the reader to distinguish formulas and text quickly. I noticed that the mvar template does not produce the exactly same results as the latex formulas. Fvultier (talk) 10:46, 28 November 2018 (UTC)
 * The math and mvar templates simply choose a compatible font/size/style for otherwise html text (compatible with the text around, but with serif font). It's not powerful like LaTeX equations, but often the best choice for math in sentences. Dicklyon (talk) 04:55, 29 November 2018 (UTC)


 * It's the serif font that looks "professional" to me ( $fs$ vs fs). LaTex is lovely by itself, but contrasts with surrounding plain text.   isn't really needed.  It's just a shortcut for

when you want the entire expression italicized; i.e. $N$ = $N$.
 * Aliasing has a lot of inline math, which I just converted from LaTex to

, for the experience. And I think it's a significant improvement. My browser "zoom" was inadvertently set at 90%. Consequently, I ended up using a lot of &nbsp to avoid crowding. After discovering my problem, I did a comparison (with and without), and decided that the extra whitespace is still a slight improvement. So I left it that way, but I expect someone will remove it all someday.
 * --Bob K (talk) 15:06, 29 November 2018 (UTC)


 * A couple of tricks I discovered are: (1) how to create the "definition" symbol, and (2) how to change $fN$ into $fN$ in this formula: $fN( f ) Δ = | f – N fs|.$
 * --Bob K (talk) 15:51, 29 November 2018 (UTC)