In mathematics, a Dirac comb (also known as sha function, impulse train or sampling function) is a periodic function with the formulafor some given period
T
\delta
\delta
The symbol
\operatorname{Ш
Because the Dirac comb function is periodic, it can be represented as a Fourier series based on the Dirichlet kernel:
The Dirac comb function allows one to represent both continuous and discrete phenomena, such as sampling and aliasing, in a single framework of continuous Fourier analysis on tempered distributions, without any reference to Fourier series. The Fourier transform of a Dirac comb is another Dirac comb. Owing to the Convolution Theorem on tempered distributions which turns out to be the Poisson summation formula, in signal processing, the Dirac comb allows modelling sampling by multiplication with it, but it also allows modelling periodization by convolution with it.[2]
The Dirac comb can be constructed in two ways, either by using the comb operator (performing sampling) applied to the function that is constantly
1
\delta
In signal processing, this property on one hand allows sampling a function
f(t)
\operatorname{Ш
f(t)
\operatorname{Ш
The scaling property of the Dirac comb follows from the properties of the Dirac delta function. Since
\delta(t)=
1 | |
a |
\delta\left(
t | |
a |
\right)
a
a
\operatorname{Ш
See also: Dirichlet kernel. It is clear that
\operatorname{Ш
T
All Fourier coefficients are 1/T resulting in
When the period is one unit, this simplifies to
Remark: Most rigorously, Riemann or Lebesgue integration over any products including a Dirac delta function yields zero. For this reason, the integration above (Fourier series coefficients determination) must be understood "in the generalized functions sense". It means that, instead of using the characteristic function of an interval applied to the Dirac comb, one uses a so-called Lighthill unitary function as cutout function, see, p.62, Theorem 22 for details.
The Fourier transform of a Dirac comb is also a Dirac comb. For the Fourier transform
l{F}
\operatorname{Ш
T
1/T,
l{F}\left[f\right](\xi)=
infty | |
\int | |
-infty |
dtf(t)e-,
l{F}\left[\operatorname{Ш
1/T
\operatorname{Ш
T=1
l{F}
1.
This result can be established by considering the respective Fourier transforms
S\tau(\xi)=l{F}[s\tau](\xi)
s\tau(x)
s\tau(x)=\tau-1
-\pi\tau2x2 | |
e |
infty | |
\sum | |
n=-infty |
-\pi\tau-2(x-n)2 | |
e |
.
Since
s\tau(x)
S\tau(\xi)
S\tau(\xi)=\tau-1
infty | |
\sum | |
m=-infty |
-\pi\tau2m2 | |
e |
-\pi\tau-2(\xi-m)2 | |
e |
.
The functions
s\tau(x)
S\tau(\xi)
\tau-1
-\pi\tau-2(x-n)2 | |
e |
\tau-1
-\pi\tau-2(\xi-m)2 | |
e |
\tau → 0
x=n
\xi=m
n
m
-\pi\tau2m2 | |
e |
S\tau(\xi)
-\pi\tau2\xi2 | |
e |
s\tau(x)
S\tau(\xi)
\lim\taus\tau(x)=\operatorname{Ш
\lim\tauS\tau(\xi)=\operatorname{Ш
Since
S\tau=l{F}[s\tau]
l{F}[\operatorname{Ш
T
l{F}[\operatorname{Ш
Another manner to establish that the Dirac comb transforms into another Dirac comb starts by examining 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 convention for the Fourier transform, this will be shown using angular frequency with
\omega=2\pi\xi:
f(t)=f(t+T)
l{F}\left[f\right](\omega)=F(\omega)=
infty | |
\int | |
-infty |
dtf(t)e-i\omega
F(\omega)(1-ei)=0
f(t)
f(t+T)
F(\omega)
F(\omega)ei.
F(\omega)=0
\omega=k\omega0,
\omega0=2\pi/T
k\inZ.
F(k\omega0)
1/T
k.
\omega=k\omega0,
infty | |
\sum\nolimits | |
m=-infty |
e\pm
F(\omega)=2\pi
infty | |
\sum | |
k=-infty |
ck\delta(\omega-k\omega0)
\omega0=2\pi/T,
ck=
1 | |
T |
+T/2 | |
\int | |
-T/2 |
dtf(t)e-i.
ck=1/T
k
f → \operatorname{Ш
l{F}\left[\operatorname{Ш
2\pi/T
As mentioned, the specific rule depends on the convention for the used Fourier transform. Indeed, when using the scaling property of the Dirac delta function, the above may be re-expressed in ordinary frequency domain (Hz) and one obtains again:
such that the unit period Dirac comb transforms to itself:
Finally, the Dirac comb is also an eigenfunction of the unitary continuous Fourier transform in angular frequency space to the eigenvalue 1 when
T=\sqrt{2\pi}
l{F}\left[f\right](\omega)=F(\omega)=
1 | |
\sqrt{2\pi |
Multiplying any function by a Dirac comb transforms it into a train of impulses with integrals equal to the value of the function at the nodes of the comb. This operation is frequently used to represent sampling.
Due to the self-transforming property of the Dirac comb and the convolution theorem, this corresponds to convolution with the Dirac comb in the frequency domain.
Since convolution with a delta function
\delta(t-kT)
kT
(\operatorname{Ш
This leads to a natural formulation of the Nyquist–Shannon sampling theorem. If the spectrum of the function
x
(-B,B)
\tfrac{1}{2B}
\operatorname{Ш
1 | \Pi\left( | |
2B |
f | |
2B |
\right)(2B\operatorname{Ш
In time domain, this "multiplication with the rect function" is equivalent to "convolution with the sinc function" (p.33-34). Hence, it restores the original function from its samples. This is known as the Whittaker–Shannon interpolation formula.
Remark: Most rigorously, multiplication of the rect function with a generalized function, such as the Dirac comb, fails. This is due to undetermined outcomes of the multiplication product at the interval boundaries. As a workaround, one uses a Lighthill unitary function instead of the rect function. It is smooth at the interval boundaries, hence it yields determined multiplication products everywhere, see, p.62, Theorem 22 for details.
In directional statistics, the Dirac comb of period
2\pi
In linear statistics, the random variable
(x)
x
-infty
+infty
(\theta)
\theta
2\pi
2\pi
2\pi