# Talk:Dirac comb

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## Shah function

This is the same as the so-called Shah function, right? -- Pgabolde 20:23, 26 August 2005 (UTC)

I think so - I don't know if

${\displaystyle \sum \delta (t-nT)}$

is the Shah function or whether it needs T=1. Anyway, I created a redirect for Shah function to here. PAR 21:27, 26 August 2005 (UTC)

Thanks, that's what I meant to do. -- Pgabolde 15:41, 27 August 2005 (UTC)

## i changed the symbol for the Dirac comb...

... to ΔT(t) or ${\displaystyle \Delta _{T}(t)\ }$ to differentiate it from the "nascent" impulse functions in the Dirac delta function article which have the same symbol with subscript as the former symbol for the Dirac comb. two qualitatively different functions, same symbol --- bad. r b-j 05:44, 8 June 2006 (UTC)

oh, and i forgot to mention that i looked up all of the articles linked to Dirac comb and fixed any reference there so we're consistent. r b-j 05:45, 8 June 2006 (UTC)

## Scaling property

Is the "scaling property" mentioned somewhere in the literature? I don't see the point of mentioning it here or how it can be of immediate used, even if it appears to be correct. In fact, they way it is presented now does not even mention the Dirac comb function explicitly. A more interesting scaling property is the following:

${\displaystyle \Delta _{T}(t)=\sum _{k=-\infty }^{\infty }\delta (t-kT)=\sum _{k=-\infty }^{\infty }\delta (T[t/T-k])={\frac {1}{|T|}}\sum _{k=-\infty }^{\infty }\delta (t/T-k)={\frac {1}{|T|}}\Delta _{1}(t/T)}$

which means that any Dirac comb function ${\displaystyle \Delta _{T}}$ can be obtained by an appropriate scaling of the normalized Dirac comb function ${\displaystyle \Delta _{1}}$. This property is useful since it can simplify certain derivations, and we only need to remember the Fourier transform of ${\displaystyle \Delta _{1}}$ from which the Fourier transform of ${\displaystyle \Delta _{T}}$ easily can be derived. --KYN 21:02, 2 August 2006 (UTC)

knock yourself out. there is a mention of this scaling at Dirac_delta_function#Delta_function_of_more_complicated_arguments. r b-j 20:59, 2 August 2006 (UTC)

Yes, I see it, but can you please tell me what information the "Scaling property" section of the article provides about the comb function? --KYN 21:02, 2 August 2006 (UTC)

## The shah symbol

The article refers to Bracewell and he uses the shah-symbol ${\displaystyle III(t)}$ to denote the Dirac comb function. If Bracewell is the base reference, why not stick to that symbol and why not call it the shah function, instead of using ${\displaystyle \Delta }$ and mainly call it the Dirac comb function? In principle, it doesn't matter much to me which notation and names are used as long as they can be motivated. If we are going to use ${\displaystyle \Delta }$ and Dirac comb function, I would like to see a reference which establishes this notation and name. --KYN 21:00, 2 August 2006 (UTC)

Bracewell isn't the only reference about the Dirac comb. Oppenheim and Schafer is an authorative reference and they call it s(t). your "shah" function is just three letters: "I" and what we should really get is the cyrillic symbol. r b-j 21:05, 2 August 2006 (UTC)

OK, I agree that the shah symbol may appear somewhat shaky through the pseudo-latex renderer, even though it may be possible to make the wiki-guys to implement a proper shah symbol. Why not stick to s(t) and add Oppenheim and Schafer to the reference list? Where does the ${\displaystyle \Delta }$ symbol comes from? --KYN 22:36, 2 August 2006 (UTC)

The only way I've found to make the symbol is to use three upright 'I' letters:
${\displaystyle \mathrm {III} _{T}(t)}$
However, the hinting causes these 'I's to have odd spacing. --Tweenk (talk) 01:36, 10 June 2013 (UTC)

## Fourier transform

Why ${\displaystyle \quad {1 \over T}\sum _{k=-\infty }^{\infty }\delta \left(f-{k \over T}\right)\quad =\sum _{n=-\infty }^{\infty }e^{-i2\pi fnT}}$? Thank you. --Abdull 18:52, 12 June 2007 (UTC)

Let x(t) be periodic, so that x(t) = x(t-T) for all t. Then the Fourier series for x(t) is
${\displaystyle x(t)=\sum _{n=-\infty }^{\infty }c_{n}e^{i2\pi nf_{0}t}\ }$
where
${\displaystyle f_{0}\equiv {\frac {1}{T}}\ }$
and
${\displaystyle c_{n}=f_{0}\int _{t_{0}}^{t_{0}+T}x(t)e^{-i2\pi nf_{0}t}dt\ }$
and where
${\displaystyle -\infty can be anything.
Now set
${\displaystyle x(t)={\frac {1}{T}}\sum _{k=-\infty }^{\infty }\delta (t-kT)\ }$
and solve for cn. You'll find that all cn = 1 so that
${\displaystyle x(t)=\sum _{n=-\infty }^{\infty }e^{i2\pi nf_{0}t}\ }$ .
Then ask yourself: What is the Fourier Transform of ${\displaystyle e^{i2\pi nf_{0}t}\ }$ ?
207.190.198.135 20:05, 12 June 2007 (UTC)
I found this very useful, but personally took me a little time to digest so want to clarify a little more.
First look at the function:
${\displaystyle x(f)=\sum _{k=-\infty }^{\infty }\delta (f-{\frac {k}{T}})}$
obviously it is a periodic function with period 1/T (as it is a sum of evenly spaced delta functions, each 1/T apart in f-space). Next take the Fourier series of x(f) by the ansatz of the form:
${\displaystyle x(f)=\sum _{n=-\infty }^{\infty }c_{n}e^{i2\pi nfT}}$
where the coefficients ${\displaystyle c_{n}}$ determined by the standard Fourier trick (of multiplying and then integrating orthogonal functions since ${\displaystyle \int _{f_{0}}^{f_{0}+1/T}e^{i2\pi mfT}e^{-i2\pi nfT}df={\frac {1}{T}}\delta _{mn}}$) to get
${\displaystyle c_{n}={\frac {1}{T}}\int _{f_{0}}^{f_{0}+1/T}x(t)e^{-i2\pi nfT}df}$ or
${\displaystyle c_{n}={\frac {1}{T}}\int _{f_{0}}^{f_{0}+1/T}\left(\sum _{k=-\infty }^{\infty }\delta (f-{\frac {k}{T}})\right)e^{-i2\pi nfT}df}$
Remembering that ${\displaystyle e^{-i2\pi nfT}}$ is periodic over 1/T, we can choose a simple range with ${\displaystyle f_{0}=-1/2T}$, so the only value of ${\displaystyle k/T}$ that is in the range we are integrating (-1/2T,1/2T) over is with ${\displaystyle k=0}$, giving us ${\displaystyle c_{n}={\frac {1}{T}}}$
Therefore we have shown that ${\displaystyle \sum _{k=-\infty }^{\infty }\delta (f-{\frac {k}{T}})={\frac {1}{T}}\sum _{n=-\infty }^{\infty }e^{i2\pi nfT}}$ which is the desired eqn. 24.58.159.5 (talk) 07:44, 16 September 2008 (UTC)

Since you have no problem with:

${\displaystyle 1=\int _{-{\frac {1}{2T}}}^{\frac {1}{2T}}\delta (f)e^{-i2\pi nfT}df=\int _{-\infty }^{\infty }\delta (f)e^{-i2\pi nfT}df,}$

I assume you're OK with the transform pair:   ${\displaystyle 1\ {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\ \delta (f).}$

Then just write out the Poisson summation formula (row 602, Relationship between functions and their Fourier Transforms):

${\displaystyle \sum _{k=-\infty }^{\infty }F\left(\nu +{\frac {k}{T}}\right)=T\sum _{n=-\infty }^{\infty }f(nT)\ e^{-i2\pi nT\nu }}$

explicitly for the case  ${\displaystyle F(\nu )=\delta (\nu ).\quad \quad {\mbox{QED}}\,}$

The proof of the Poisson summation formula is worth knowing.

--Bob K (talk) 17:19, 16 September 2008 (UTC)

--User:rdsomma I changed the variables in this section, the left hand sides were given as a function of t and the right hand sides were given as a function of ${\displaystyle f}$ or ${\displaystyle \omega }$. Also, the (unitary) Fourier transform of the comb was incorrect. The period is ${\displaystyle 2\pi /T}$ , not ${\displaystyle {\sqrt {2\pi }}/T}$. — Preceding unsigned comment added by Rdsomma (talkcontribs) 19:37, 23 October 2013 (UTC)

## Bed of nails?

Should the article say something about what Bracewell calls the "bed of nails" function? This and its natural higher-dimensional analogs are very important in crystallography. It's surprising that, as far as I can tell, Wikipedia lacks an article on them. Sławomir Biały (talk) 14:41, 4 July 2009 (UTC)

## A fact

It can be shown that

${\displaystyle \delta (\sin x)=\sum _{n=-\infty }^{\infty }\delta (t-n\pi ),}$

in case if someone is interested... Zitterbewegung Talk 23:42, 28 October 2009 (UTC)

This seems to beg the question a bit, depending on what one means by the composition of δ with a smooth function. This identity is a trivial consequence of the usual definition (see Dirac delta function) although it is possible that there are other ways of defining the composition in which the statement becomes more interesting (I don't know). Sławomir Biały (talk) 17:06, 21 November 2009 (UTC)

## Directional statistics

Can someone help here - The Dirac comb of period 2π is to circular statistics what the Dirac delta function is to linear statistics. E.g see wrapped distribution. The links in this article to circular statistics articles are constantly being removed because the relationship is not clear to some editors from reading this article. I have added some information in the intro, but any other ideas would be welcome. PAR (talk) 09:48, 18 November 2010 (UTC)

It would help if the article gave a proper indication of the supposed relationship to Directional statistics.... there was nothing at all there before this latest edit. It would be good at least to have a section headed "use in directional statistics", or some such thing, even if it does little more than point to the article you mentioned. And there isn't an immediate analogue with the Dirac delta function as that can be treated as equivalent to a probability distribution in it own right, whereas a Dirac comb cannot ... because it integrates to infinity and so is not equivalent to a probability distribution. Melcombe (talk) 15:12, 25 November 2010 (UTC)
First there was the probability nav box, which was removed. So I put back the nav box along with a short explanation. I will expand the explanation, and include the nav box.
Directional (or wrapped, or circular) probability distributions are never integrated over the real number line (from negative infinity to positive infinity), they are always integrated over the unit circle (an interval of length 2π), as explained in the wrapped distribution article. Since all wrapped distributions are periodic with period 2π the limits of integration are irrelevant as long as the length of the interval is 2π. As a result, the Dirac comb of course integrates to unity. In linear statistics, the variable x often extends from -infinity to infinity. In directional statistics, the angle is in some interval of length 2π In other words the angles θ and θ+2π are considered identical. The analog of x in linear statistics is usually taken to be z=exp(θ), which has a period of 2π. PAR (talk) 22:14, 25 November 2010 (UTC)

## Deletion of Scaling property section.

I have to fully agree with the IP who deleted the Scaling property section. It was contributed long ago with one of many edits by an editor who often thinks he's making the article better and isn't. Undoing his "improvements" over the project's scope would be a full-time job. The equation could make sense, but as it was, is just incorrect. One quick look at the period of the the periodic function on the two sides of the equation would immediately reveal the error. 71.169.185.162 (talk) 03:59, 27 December 2011 (UTC)

I restored the scaling property, expressed properly, with justification. PAR (talk) 04:33, 27 December 2011 (UTC)

## Schwartz distribution?

Now this needs source or proof. Is it really a Schwartz distribution? — Preceding unsigned comment added by 92.52.10.154 (talk) 12:17, 26 December 2014 (UTC)

Added "citation needed" tag. -- intgr [talk] 13:07, 26 December 2014 (UTC)

## Terms need defining

The very first equation already introduces ${\displaystyle \delta }$, ${\displaystyle t}$ and ${\displaystyle T}$ defining only ${\displaystyle T}$. ${\displaystyle \delta }$ and ${\displaystyle t}$ should be defined as well. ${\displaystyle t}$, by implication is time of course, and ${\displaystyle \delta }$ one presumes is the Dirac delta function, but this is not a page in the middle of text on signal processing, rather a standalone page that should, in manner of such pages introduce terms as needed, not assume understanding. Of course I know the challenge is always where to draw the line and some level of assumed knowledge is inescapable (we're not going to define ${\displaystyle \Sigma }$ or ${\displaystyle k}$ in this context for example). But ${\displaystyle \delta }$ and ${\displaystyle t}$ here are not in that category, IMO. --Bwechner (talk) 04:11, 20 September 2016 (UTC)