Editing Dirac comb (section)

== Fourier transform ==
The [[continuous Fourier transform|Fourier transform]] of a Dirac comb is also a Dirac comb. For the Fourier transform <math>\mathcal{F}</math> expressed in [[Fourier transform#Other conventions|frequency domain]] (Hz) the Dirac comb <math>\operatorname{\text{Ш}}_{T}</math> of period <math>T</math> transforms into a rescaled Dirac comb of period <math>1/T,</math> i.e. for
:<math>\mathcal{F}\left[ f \right](\xi)= \int_{-\infty}^{\infty} dt f(t) e^{- 2 \pi i\xi t}, </math> 
:<math>\mathcal{F}\left[ \operatorname{\text{Ш}}_{T} \right](\xi) = \frac{1}{T} \sum_{k=-\infty}^{\infty} \delta(\xi-k \frac{1}{T}) = \frac{1}{T} \operatorname{\text{Ш}}_{\ \frac{1}{T}}(\xi) ~</math>
is proportional to another Dirac comb, but with period <math>1/T</math> in frequency domain (radian/s). The Dirac comb <math>\operatorname{\text{Ш}}</math> of unit period <math>T=1</math> is thus an [[continuous Fourier transform#Eigenfunctions|eigenfunction]] of <math>\mathcal{F}</math> to the [[eigenvalue]] <math>1.</math>

This result can be established{{sfn|Bracewell|1986}} by considering the respective Fourier transforms <math>S_{\tau}(\xi)=\mathcal{F}[s_{\tau}](\xi)</math> of the family of functions <math>s_{\tau}(x)</math> defined by

:<math>s_{\tau}(x) = \tau^{-1} e^{-\pi \tau^2 x^2} \sum_{n=-\infty}^{\infty} e^{-\pi \tau^{-2} ( x-n)^{2} }.</math>

Since <math>s_{\tau}(x)</math> is a convergent series of [[Gaussian function|Gaussian functions]], and Gaussians [[Fourier transform#Square-integrable functions, one-dimensional|transform]] into [[Normal distribution#Fourier transform and characteristic function|Gaussians]], each of their respective Fourier transforms <math>S_\tau(\xi)</math> also results in a series of Gaussians, and explicit calculation establishes that

:<math>S_{\tau}(\xi) = \tau^{-1}  \sum_{m=-\infty}^{\infty}  e^{-\pi \tau^2 m^2} e^{-\pi \tau^{-2} ( \xi-m)^{2} }.</math>

The functions <math>s_{\tau}(x)</math> and <math>S_\tau(\xi)</math> are thus each resembling a periodic function consisting of a series of equidistant Gaussian spikes <math>\tau^{-1} e^{-\pi \tau^{-2} ( x-n)^{2} }</math> and <math>\tau^{-1} e^{-\pi \tau^{-2} ( \xi-m)^{2} }</math> whose respective  "heights" (pre-factors)  are determined by slowly decreasing Gaussian envelope functions which drop to zero at infinity.  Note that in the limit <math>\tau \rightarrow 0</math> each Gaussian spike becomes an infinitely sharp [[Dirac delta function|Dirac impulse]] centered respectively at <math>x=n</math> and <math>\xi=m</math> for each respective <math>n</math> and <math>m</math>, and hence also all pre-factors <math> e^{-\pi \tau^2 m^2}</math> in <math>S_{\tau}(\xi)</math> eventually become indistinguishable from <math> e^{-\pi \tau^2 \xi^2}</math>. Therefore  the functions <math>s_{\tau}(x)</math> and their respective Fourier transforms <math>S_{\tau}(\xi)</math> converge to the same function and this limit function is a series of infinite equidistant Gaussian spikes, each spike being multiplied by the same pre-factor of one, i.e., the Dirac comb for unit period:

:<math>\lim_{\tau \rightarrow 0} s_{\tau}(x) = \operatorname{\text{Ш}}({x}),</math>   and   <math>\lim_{\tau \rightarrow 0} S_{\tau}(\xi) = \operatorname{\text{Ш}}({\xi}).</math>

Since <math>S_{\tau}=\mathcal{F}[s_{\tau}]</math>, we obtain in this limit the result to be demonstrated:
:<math>\mathcal{F}[\operatorname{\text{Ш}}]= \operatorname{\text{Ш}}.</math>
The corresponding result for period <math>T</math> can be found by exploiting the [[Fourier transform#Time scaling|scaling property]] of the [[Fourier transform]],
:<math>\mathcal{F}[\operatorname{\text{Ш}}_T]= \frac{1}{T} \operatorname{\text{Ш}}_{\frac{1}{T}}.</math>

Another manner to establish that the Dirac comb transforms into another Dirac comb starts by examining [[Fourier transform#Fourier transform for periodic functions|continuous Fourier transforms of periodic functions]] in general, and then specialises to the case of the Dirac comb. In order to also show that the specific rule depends on the [[Fourier transform#Other conventions|convention]] for the Fourier transform, this will be shown using angular frequency with <math>\omega=2\pi \xi :</math> for any periodic function <math>f(t)=f(t+T)</math> its Fourier transform
:<math>\mathcal{F}\left[ f \right](\omega)=F(\omega) = \int_{-\infty}^{\infty} dt f(t) e^{-i\omega t} </math>  obeys:
:<math>F(\omega) (1 - e^{i \omega T}) = 0</math>
because Fourier transforming <math>f(t)</math> and <math>f(t+T)</math> leads to <math>F(\omega)</math> and <math>F(\omega) e^{i \omega T}.</math> This equation implies that <math>F(\omega)=0</math> nearly everywhere with the only possible exceptions lying at <math>\omega= k \omega_0,</math> with <math>\omega_0=2\pi / T</math> and <math>k \in \mathbb{Z}.</math> When evaluating the Fourier transform at <math>F(k \omega_0)</math> the corresponding Fourier series expression times a corresponding delta function results. For the special case of the Fourier transform of the Dirac comb, the Fourier series integral over a single period covers only the Dirac function at the origin and thus gives <math>1/T</math> for each <math>k.</math> This can be summarised by interpreting the Dirac comb as a limit of the [[Dirichlet kernel#Relation to the periodic delta function|Dirichlet kernel]] such that, at the positions <math>\omega= k \omega_0,</math> all exponentials in the sum <math> \sum\nolimits_{m=-\infty}^{\infty} e^{\pm i \omega m T} </math> point into the same direction and add constructively. In other words, the [[Fourier transform#Fourier transform for periodic functions|continuous Fourier transform of periodic functions]] leads to
:<math>F(\omega)= 2 \pi \sum_{k=-\infty}^{\infty} c_k \delta(\omega-k\omega_0) </math> with <math>\omega_0=2 \pi/T,</math>
and
:<math>c_k = \frac{1}{T} \int_{-T/2 }^{+T/2} dt f(t) e^{-i 2 \pi k t/T}.</math>
The [[#Fourier series|Fourier series]] coefficients <math>c_k=1/T</math> for all <math>k</math> when <math>f \rightarrow \operatorname{\text{Ш}}_{T}</math>, i.e. 
:<math>\mathcal{F}\left[ \operatorname{\text{Ш}}_{T} \right](\omega) = \frac{2 \pi}{T} \sum_{k=-\infty}^{\infty} \delta(\omega-k \frac{2 \pi}{T})</math>
is another Dirac comb, but with period <math>2 \pi/T</math> in angular frequency domain (radian/s).

As mentioned, the specific rule depends on the [[Fourier transform#Other conventions|convention]] for the used Fourier transform. Indeed, when using the [[Dirac delta function#Scaling and symmetry|scaling property]] of the Dirac delta function, the above may be re-expressed in ordinary frequency domain (Hz) and one obtains again:
<math display="block">\operatorname{\text{Ш}}_{\ T}(t) \stackrel{\mathcal{F}}{\longleftrightarrow} \frac{1}{T} \operatorname{\text{Ш}}_{\ \frac{1}{T}}(\xi) = \sum_{n=-\infty}^{\infty}\!\! e^{-i 2\pi \xi n T},</math>

such that the unit period Dirac comb transforms to itself:
<math display="block">\operatorname{\text{Ш}}\ \!(t) \stackrel{\mathcal{F}}{\longleftrightarrow} \operatorname{\text{Ш}}\ \!(\xi).</math>

Finally, the Dirac comb is also an [[Fourier transform#Eigenfunctions|eigenfunction]] of the unitary continuous Fourier transform in [[Fourier transform#Other conventions|angular frequency]] space to the eigenvalue 1 when <math>T=\sqrt{2 \pi}</math> because for the unitary Fourier transform 
:<math>\mathcal{F}\left[ f \right](\omega)=F(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} dt f(t) e^{-i\omega t}, </math>
the above may be re-expressed as 
<math display="block">\operatorname{\text{Ш}}_{\ T}(t) \stackrel{\mathcal{F}}{\longleftrightarrow} \frac{\sqrt{2\pi}}{T} \operatorname{\text{Ш}}_{\ \frac{2\pi}{T}}(\omega) = \frac{1}{\sqrt{2\pi}}\sum_{n=-\infty}^{\infty} \!\!e^{-i\omega nT}.</math>