A short introduction to distribution theory

Sven NordeboSchool of Computer Science, Physics and Mathematics

Linnaeus University

8 september 2010

1 The classical Fourier integral

Consider the Fourier transform pairX() = F{x(t)} =

x(t)eit dt

x(t) = F1{X()} = 12pi

X()eit d,(1)

where the integrals are assumed to exist in the usual sense, i.e., as Riemann integrals,or as Lebesgue integrals [3]. If e.g., the function x(t) is absolutely integrable, i.e.,if x(t) L1 and |x(t)| dt < exists finitely, then the Fourier transform X() =F{x(t)} exists in the sense of the integral in (1), and X() is a continuous andbounded function, i.e., |X()| < A where A is a constant, see e.g., [3]. Similarly,if the function X() L1, then the inverse Fourier transform x(t) = F1{X()}exists in the sense of the integral in (1), and it is a continuous and bounded function.Alternatively, if x(t) L2, i.e., if |x(t)|2 dt

Define now the linear functional

x(t), (t) =

x(t)(t) dt (3)

which should be interpreted as a linear mapping from the space of testing functions(t) to the real (or complex) numbers.

Theorem: The linear functional (3) has the following symmetry properties withrespect to the differential and the Fourier transform, respectively.

x(t), (t) = x(t), (t), (4)Fx(t), () = x(),F(t). (5)

Proof: The proof of (4) is as follows

x(t), (t) =

x(t)(t) dt = [x(t)(t)]

x(t)(t) dt = x(t), (t).(6)

The proof of (5) is as follows

Fx(t), () =

(

x(t)eit dt)() d

=

(

x()eit d)(t) dt

=

x()

(

(t)eit dt)

d = x(),F(t). (7)

2 Definition of tempered distributions

The space S of testing functions of rapid descent [4] (also called the Schwartz class)consists of all functions (t) which are infinitely differentiable (infinitely smooth),i.e., all the derivatives (k)(t) exist and are continuous for any order k 0, andare such that, as |t| , (t) and all its derivatives (k)(t) decay faster than anypower of 1/|t|. In other words, for any order k 0 and integer power m 0, thereis a constant ckm such that |(k)(t)| ckm 1|t|m for all t.

The space S of tempered distributions [4] (also called distributions of slowgrowth) consists of all continuous linear functionals on the space S of testing func-tions of rapid descent. Hence, a tempered distribution x(t) is a rule that assigns anumber x(t), (t) to each (t) in S, in such a way that the following conditionsare fulfilled:

Linearity: If 1(t) and 2(t) are in S and if is a number, then

x(t), 1(t) + 2(t) = x(t), 1(t)+ x(t), 2(t) (8)x(t), 1(t) = x(t), 1(t). (9)

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Continuity: If the sequence n(t) of testing functions in S converges to zeroas n, then x(t), n(t) also converges to zero. Formally, the linear functionalx(t), (t) is continuous if the following implication hold

limn

max

3.2 Limits of distributions

A sequence xn(t) of distributions is said to converge to the distribution x(t) if

limnxn(t), (t) = x(t), (t) (18)

for all testing functions (t) S. In this case, we write limn

xn(t) = x(t). The same

definition is made with an indexed set of distributions where e.g., lim0

x(t) = x(t)

means that lim0x(t), (t) = x(t), (t) for all testing functions (t) S.

Example: The distribution x(t) =12pi

et2

22 (Gaussian pulse) has zero mean,

variance 2 and integralx(t) dt = 1, and hence

lim0x(t), (t) = lim

0

x(t)(t) dt = (0)

x(t) = (0) = (t), (t).(19)

It is therefore concluded that lim0

x(t) = (t).

Example: The rectangular pulse xT (t) =1T

(u(t + T2) u(t T

2)) has integral

xT (t) dt = 1, and it has the same property as above that

limT0xT (t), (t) = lim

T0

xT(t) dt = limT0

1

T

T/2T/2

(t) dt = (0) = (t), (t),(20)

and hence that limT0

xT (t) = (t). Obviously, we can construct many distribution

(pulse) sequences with this property.

3.3 Delay, reflection, symmetry and periodic distributions

The delay x(t ) of a distribution x(t) is defined by

x(t ), (t) = x(t), (t+ ) (21)

for all testing functions (t) S. Note that this property is shared by (3) by asimple substitution of variables.

The reflection x(t) of a distribution x(t) is defined by

x(t), (t) = x(t), (t) (22)

for all testing functions (t) S. Note that this property is shared by (3) by asimple substitution of variables. A distribution is said to be even if x(t) = x(t),and odd if x(t) = x(t).

A periodic distribution x(t) = x(t+ T ) is defined by the property that

x(t), (t) = x(t+ T ), (t) = x(t), (t T ) (23)

for all testing functions (t) S.

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Example: The delay of a delta distribution is defined by

(t ), (t) = (t), (t+ ) = () (24)

for all (t) S.

Example: The impulse train is defined by

m=(tmT ), (t) =

m=

(tmT ), (t)

=

m=(t), (t+mT ) =

m=

(mT ) (25)

for all (t) S.

3.4 Multiplication of distributions

The multiplication of two distributions x(t) and y(t) is in general not defined. Ho-wever, if (at least) one of the distributions is also a function g(t) of slow growth, thenthe product g(t)x(t) is well defined. A function g(t) of slow growth is infinitely dif-ferentiable (infinitely smooth), i.e., the derivatives g(k)(t) exist for all orders k 0,and they grow slower than some polynomial, i.e., |g(k)(t)| Ck(1 + |t|)Nk for all t,where the constant Ck and the polynomial order Nk depend on k. It can be shownthat the product of a testing function (t) S and a function g(t) of slow growthis also a testing function, i.e., g(t)(t) S, see e.g., [4]. Hence, the product of adistribution x(t) and a function g(t) of slow growth (e.g., a polynomial) is definedas the distribution g(t)x(t) by the following property

g(t)x(t), (t) = x(t), g(t)(t), (26)

for all testing functions (t) S.

Example: If g(t) is a function of slow growth, then g(t)(t) = g(0)(t).

To prove this, consider the following equalities

g(t)(t), (t) = (t), g(t)(t) = g(0)(0) = g(0)(t), (t), (27)

for all testing functions (t) S. Hence g(t)(t) = g(0)(t), by the definition ofequality of distributions.

Example: The product of the delta distribution (t) with itself, i.e., (t)(t)can not be defined as a distribution. Hence, the notation 2(t) has no meaningfulinterpretation.

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3.5 Derivatives of distributions

The derivative of a distribution x(t) S is a distribution x(t) S defined byx(t), (t) = x(t), (t) (28)

for all testing functions (t) S. Note that the right hand side of (28) is welldefined as a distribution, and that the definition is chosen in accordance to theclassical property (4). By induction, it follows from the definition (28) that all thederivatives of x(t) exist as distributions, i.e., x(k)(t) S for all integer orders k 1.In particular, the derivatives x(k)(t) are defined by

x(k)(t), (t) = (1)kx(t), (k)(t) (29)for all testing functions (t) S.

Example: The derivatives of the delta distribution are defined as follows

(t), (t) = (t), (t) = (0) (30)and

(k)(t), (t) = (1)k(t), (k)(t) = (1)k(k)(0). (31)

Example: The distributional derivative of the unit step function u(t) and thesign function sgn(t) are given by

u(t) = (t) (32)

sgn(t) = 2(t). (33)

To prove this, let (t) be an arbitrary testing function in S. Then

u(t), (t) = u(t), (t) = 0

(t) dt = [(t)]0 = (0) = (t), (t)(34)

which shows that u(t) = (t). The derivative of the sign function follows immedia-tely by writing sgn(t) = 2u(t) 1.

3.6 Fourier transform of distributions

It can be shown that if (t) is a testing function of rapid descent, then the classicalFourier transform F(t) defined in (1) exists and generates a new function () =F(t) which is also a testing function of rapid descent, see e.g., [4]. Hence, if (t) Sthen F(t) S. Based on this property, the Fourier transform of a distributionx(t) S is a distribution Fx(t) S defined by

Fx(t), () = x(),F(t) (35)for all testing functions (t) S. Note that the right hand side of (35) is well defi-ned as a distribution, and that the definition is chosen in accordance to the classical

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property (5).

Example: F1 = 2pi().

To prove this, let () = F(t) where (t) S.

F1, () = 1,F(t) = 1,() =

() d = 2pi(0) = 2pi(), (),(36)

which shows that F1 = 2pi().

Example: F(t) = 1.

To prove this, let () = F(t) where (t) S.

F(t), () = (),F(t) = (),() = (0) =

(t) dt = 1, (),(37)

which shows that F(t) = 1.

3.7 Fouriertransform of a derivative

Let x(t) be a distribution in S and X() = Fx(t). The Fourier transform of aderivative is given by

Fx(t) = iX(), (38)and hence, by induction

Fx(k)(t) = (i)kX() (39)where k 1.

Proof: The proof is based on the corresponding properties of the testing func-tions, together with the definition of the Fourier transform. Hence, let (t) be anarbitrary testing function in S. Then

Fx(t), () = x(),F(t) = x(), ddF(t) = x(),F{it(t)}

= Fx(t),i() = iX(), (), (40)

which shows that Fx(t) = iX().

3.8 Derivative of the Fourier transform

Let x(t) be a distribution in S and X() = Fx(t). The Fourier transform of tx(t)is given by

Ftx(t) = i dd

X(), (41)

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and hence, by induction

Ftkx(t) = ik dk

dkX() (42)

where k 1.

Proof: Let (t) be an arbitrary testing function in S and () = F(t). Then

Ftx(t), () = x(),F(t) = x(), () = ix(), i()= ix(),F d

dt(t) = iFx(t), d

d() = i d

dFx(t), ()

= i dd

X(), () (43)

which shows that Ftx(t) = i ddX().

3.9 Inversion of the Fourier transform

If x(t) is a distribution in S , then

FFx(t) = 2pix(t). (44)It follows from this property that the Fourier transform F is an invertible operator,i.e., the inverse Fourier transform F1 exists.

Proof: Let (t) be an arbitrary testing function in S. Then FF(t) = 2pi(t).Hence,

FFx(t), (t) = Fx(),F() = x(t),FF(t) = x(t), 2pi(t)= 2pix(t), (t) (45)

which shows that FFx(t) = 2pix(t).

3.10 Even and odd symmetries

If x(t) is a distribution in S , then the following symmetry properties holdx(t) even x(t) oddx(t) odd x(t) evenx(t) even Fx(t) evenx(t) odd Fx(t) odd.

(46)

Proof: Suppose that x(t) is an even distribution, i.e., x(t) = x(t) and let (t)be an arbitrary testing function in S. Then

x(t), (t) = x(t), (t) = x(t), ddt(t) = x(t),(t)

= x(t), (t) = x(t), (t) = x(t), (t), (47)

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which shows that x(t) is odd.Consider next the Fourier transform, and let X() = Fx(t) and () = F(t).

Then

X(), () = X(), () = Fx(t), ()= x(),F(t) = x(),() = x(),()

= x(),F(t) = Fx(t), () = X(), (), (48)

which shows that X() = Fx(t) is even. The proofs for odd distributions are similar.

3.11 The step and sign functions

Based on the distribution theory introduced above, we are now ready to establishthe following important Fourier transforms:

Fsgn(t) = 2i

(49)

Fu(t) = 1i

+ pi() (50)

F 1pit

= isgn(), (51)

where the distributions 1i

and 1pit

should be interpreted in the sense of Cauchy prin-cipal value integrals.

Proof: By taking the Fourier transform of the distributional identity

sgn(t) = 2(t) (52)

we obtainiFsgn(t) = 2, (53)

which has the general distributional solution

Fsgn(t) = 2i

+ A() (54)

where A is a constant. Now, since sgn(t) is an odd distribution, the distributionFsgn(t) must also be odd, and since () is an even distribution, it is concludedthat A = 0. By writing u(t) = 1

2(sgn(t) + 1) it follows immediately that Fu(t) =

1i

+ pi(). Finally, the Fourier transform of 1pit

is obtained as an inverse Fouriertransform based on the formula FFx(t) = 2pix(t). Hence

FFsgn(t) = F 2i

= 2pisgn(t), (55)

which shows that F 1pit

= isgn().

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3.12 Time shift and modulation

Time shift and modulation are important operations. Let x(t) be a distribution andX() = Fx(t). Then

Fei0tx(t) = X( 0) (modulation) (56)Fx(t t0) = eit0X() (time shift). (57)

Proof: The proof is based on the corresponding time shift and modulation pro-perties of the testing functions, together with the definition of the Fourier transform.Hence, let (t) be an arbitrary testing function in S and () = F(t). Then

Fei0tx(t), () = ei0x(),F(t) = x(), ei0() = x(),F(t+ 0)= Fx(t), ( + 0) = X(), ( + 0) = X( 0), (), (58)

which shows that Fei0tx(t) = X( 0).Similarly,

Fx(t t0), () = x( t0),F(t) = x( t0),() = x(),( + t0)= x(),Feit0t(t) = Fx(t), eit0() = eit0X(), (), (59)

which shows that Fx(t t0) = eit0X().

3.13 Convolution of distributions

The convolution of two distributions x(t) and y(t) is in general not defined. However,if (at least) one of the distributions has a Fourier transform which is also a functionof slow growth, then the product Fx(t)Fy(t) is well defined as a distribution. Inthis case, the convolution x(t) y(t) is a distribution defined by F{x(t) y(t)} =Fx(t)Fy(t), or

x(t) y(t) = F1{X()Y ()}, (60)where X() = Fx(t) and Y () = Fy(t). Note that the definition (60) is in agree-ment with the corresponding properties of the classical Fourier transform (1).

In general, if both distributions x(t) and y(t) have Fourier transforms X() andY () which are also ordinary functions, and which can be multiplied together toyield a well defined distribution X()Y (), then the convolution x(t) y(t) is welldefined by (60).

Example: The Fourier transform of the delta distribution is F(t) = 1, whichis a function of slow growth. Hence, the convolution of a distribution x(t) with thedelta distribution is given by

x(t) (t) = F1{Fx(t)F(t)} = F1{Fx(t)} = x(t). (61)

Hence, the delta distribution is a unity operator under convolution. Note e.g., that(t) (t) = (t).

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Example: The following distributional identity is valid

1

pit 1pit

= F1{F 1pitF 1pit} = F1{(isgn())2} = F1{1} = (t) (62)

where the function (sgn())2 = 1 almost everywhere (except possibly at = 0), andhence (sgn())2 = 1 in the distributional sense.

3.14 Fourier transform of limits

If xn(t) is a sequence of distributions in S that converges to x(t) S , i.e., if

limn

xn(t) = x(t), then limn

Fxn(t) = Fx(t). The same property holds for anyindexed set of distributions. Hence, if lim

a0xa(t) = x(t), then lim

a0Fxa(t) = Fx(t).

This property shows that the Fourier transform F is a continuous mapping from S to S .

Proof: Let (t) be an arbitrary testing function in S. Then

limnFxn(t), () = lim

nxn(),F(t) = x(),F(t) = Fx(t), () (63)

which shows that limn

Fxn(t) = Fx(t).

Example: Define the distribution xa(t) by the ordinary function

xa(t) =

{eat t > 0eat t < 0 (64)

where a > 0. The distributional Fourier transform Xa() = Fxa(t) exists also as aclassical Fourier integral, and is given by

Xa() =2i

(a i)(a+ i) . (65)

Since lima0+

xa(t) = sgn(t), it is concluded that

lima0+

Fxa(t) = lima0+

2i(a i)(a+ i) = Fsgn(t) =

2

i, (66)

where the limits should be interpreted in the distributional sense, and the distribu-tion 1

icorresponds to a principal value integral.

Example: Define the distribution xa(t) by the ordinary function

xa(t) = eatu(t) (67)

where a > 0. The distributional Fourier transform Xa() = Fxa(t) exists also as aclassical Fourier integral, and is given by

Xa() =1

a+ i. (68)

Since lima0+

xa(t) = u(t), it is concluded that

lima0+

Fxa(t) = lima0+

1

a+ i= Fu(t) = 1

i+ pi(), (69)

where the limits should be interpreted in the distributional sense, and the distribu-tion 1

icorresponds to a principal value integral.

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4 Relation to the classical Laplace transform

The Region Of Convergence (ROC) of the classical Laplace transform

X(s) = Lx(t) =

x(t)est dt (70)

is the (open) region in the complex s-plane where the integral converges absolutely.Hence, if s = + i ROC, then

|x(t)est| dt =

|x(t)|et dt 0. The Laplace transform is given by

X(s) =2s

(a s)(a+ s) , (75)

where the region of convergence ROC is given by a < Re s < a. Since s = i ROC, the Fourier transform is given by

X() = Fx(t) = X(s)|s=i = 2i(a i)(a+ i) . (76)

Example: Let the function x(t) be given by

x(t) = eatu(t) (77)

where a > 0. The Laplace transform is given by

X(s) =1

a+ s, (78)

where the region of convergence ROC is given by Re s > a. Since s = i ROC,the Fourier transform is given by

X() = Fx(t) = X(s)|s=i = 1a+ i

. (79)

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Example: Let x(t) be given by the sign function

x(t) = sgn(t). (80)

The classical Laplace transform does not exist in this case. To see this, let s = +i,and consider the integral

|x(t)est| dt =

|sgn(t)|et dt =

et dt =, (81)

which is divergent for any value of .

Example: Let x(t) be given by the unit step function

x(t) = u(t). (82)

The Laplace transform is given by

X(s) =1

s, (83)

where the region of convergence ROC is given by Re s > 0. Since s = i / ROC, weare not allowed to draw the conclusion (73) about the relation between the Fouriertransform and the Laplace transform. In fact, the Fourier transform does not existin the classical sense (1), but it can be interpreted as a distribution. Hence, in thiscase, we have that

X() = Fx(t) = 1i

+ pi() 6= X(s)|s=i = 1i. (84)

We have, however,

X(s) =

u(t)est dt =

u(t)eteit dt = Fetu(t) = 1 + i

=1

s(85)

where s = + i, and where the Fourier integral exists in the classical sense (1) for > 0. Since lim

0+etu(t) = u(t) in the distributional sense, the following relation

between the classical Laplace transform and the distributional Fourier transform isvalid

lim0+

X(s)|s=+i = lim0+

1

+ i= lim

0+Fetu(t) = Fu(t) = 1

i+ pi() = X()

(86)where the limits should be interpreted in the distributional sense, and the distribu-tion 1

icorresponds to a principal value integral.

5 Hilbert transform relationships

Hilbert transform relationships have many important applications in physics, te-lecommunication, etc., see e.g., [1]. Here, we summarize briefly some importantproperties.

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5.1 Application of the Hilbert transform in the time domain

Let X() = Fx(t) where x(t) is a distribution. When the product isgn()X() iswell defined as a distribution, it defines the Hilbert transform in the time domainby

1

pit x(t) = F1{isgn()X()}. (87)

In telecommunication applications, the Hilbert transform has the important pro-perty of shifting the phase of a signal by pi/2 radians without changing its envelope.This is a property that can be used e.g., to obtain the quadrature components of asignal in a telecommunication system, see e.g., [2].

Example: Let 0 be the carrier frequency, and let (t) be an ordinary functionwith band limited spectrum () = F(t) where () = 0 for || < B, and whereB < 0. The Hilbert transform has the following phase shifting properties

1

pit cos0t = sin0t (88)

1

pit ((t) cos0t) = (t) sin0t. (89)

Proof: The proof can be outlined as follows

1

pit cos0t = F1{isgn()(pi( 0) + pi( + 0))}

= F1{ipi( 0) + ipi( + 0)} = F1{2pi2i( 0) 2pi

2i( + 0)}

=1

2iei0t 1

2iei0t = sin0t. (90)

and

1

pit ((t) cos0t) = F1{isgn()(1

2( 0) + 1

2( + 0))}

= F1{ i2

( 0) + i2

( + 0)} = F1{ 12i

( 0) 12i

( + 0)}

=1

2i(t)ei0t 1

2i(t)ei0t = (t) sin0t. (91)

5.2 Application of the Hilbert transform in the frequencydomain

Let X() = Fx(t) where x(t) is a distribution. When the product sgn(t)x(t) is welldefined as a distribution, it defines the Hilbert transform in the frequency domainby

F{sgn(t)x(t)} = 12pi

2

iX() (92)

or, equivalently1

piX() = F{isgn(t)x(t)}. (93)

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An important application of the Hilbert transform in the frequency domain, iswith the fundamental physical properties of a causal linear system. A linear andtime invariant system is said to be causal if the impulse response x(t) is supportedin [0,). In other words, if x(t) is the impulse response of a causal linear system,then x(t) = 0 for t < 0. For distributions, this means that x(t), (t) = 0 for alltesting functions (t) which are supported in t < 0, see [4].

An arbitrary distribution x(t) can be decomposed in its even and odd parts xe(t)and xo(t), respectively, where

x(t) = xe(t) + xo(t) (94)

and

xe(t) =x(t) + x(t)

2(95)

xo(t) =x(t) x(t)

2. (96)

If X() = Fx(t) = ReX() + i ImX() and if x(t) is real, it can be shown thatReX() is even and ImX() is odd. Hence, for real distributions x(t) the followingrelationships hold

ReX() = Fxe(t) (97)i ImX() = Fxo(t). (98)

Suppose now that x(t) is casual with x(t) = 0 for t < 0. It is then readily verifiedthat

sgn(t)x(t) = x(t) (99)

sgn(t)x(t) = x(t), (100)and hence that

xe(t) = sgn(t)xo(t) (101)

xo(t) = sgn(t)xe(t) (102)

where it has been assumed that the product sgn(t)x(t) is well defined as a distribu-tion.

A situation of practical interest is when x(t) contains a delta function C(t)supported at the origin t = 0. In this case, the relations (101) and (102) above mustbe modified as follows

xe(t) = sgn(t)xo(t) + C(t) (103)

xo(t) = sgn(t)(xe(t) C(t)) (104)where the distribution C(t) is associated with the even part since (t) is even.

By taking the Fourier transform of (101) and (102) (assuming that C = 0) thefollowing Hilbert transform relationships are now obtained

ReX() =1

2pi

2

i i ImX() (105)

i ImX() =1

2pi

2

i ReX() (106)

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or

ReX() =1

pi ImX() =

1

piImX( ) d (107)

ImX() = 1pi ReX() =

1

piReX( ) d (108)

where the integral (when it exists) should be interpreted as a Cauchy principal value.In conclusion, for a causal linear system where x(t) = 0 for t < 0, the real part

ReX() of the frequency function X() is given explicitly by the Hilbert transformof its imaginary part ImX(), and vice versa.

Referenser

[1] F. W. King. Hilbert transforms vol. III. Cambridge University Press, 2009.

[2] J. G. Proakis. Digital Communications. McGraw-Hill, third edition, 1995.

[3] W. Rudin. Real and Complex Analysis. McGraw-Hill, New York, 1987.

[4] A. H. Zemanian. Distribution theory and transform analysis. Dover Publications,New York, 1965.

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