Chapter 6 Integral Transforms - YUctaps.yu.edu.jo/physics/Courses/Phys601/PDF/4_Phys601... · 2014....

1

© Dr. Nidal M. Ershaidat

Phys. 601: Mathematical Physics

Physics Department

Yarmouk University

Chapter 6 Integral Transforms

© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 4 Integral Transforms

2

Overview

1. Integral Transforms - Fourier

2. Development of the Fourier Integral

3. Fourier Transform – Inverse Theorem

4. Fourier Transform of Derivatives

5. Convolution Theorem

6.Momentum Representation

7. Transfer Functions

8. Laplace Transforms

9. Laplace Transform of Derivatives

10.Other Properties

11.Convolution (Faltungs) Theorem

12.Inverse Laplace Transform

Integral Transforms


4

In integrals of the form

g(αααα) is called the (integral) transform of f(t) by

the Kernel* K(αααα,t).The nature of the kernel defines the type of the

transform.

The variables αααα and t are called conjugate

variables. For example: frequency and time are

conjugate variables. It is also the case of

wavevector and position (k and x)

Introduction

(((( )))) (((( )))) (((( ))))dttKtfgb

a,αααα====αααα ∫∫∫∫ 1

* German word for nucleus


5

The inverse transform is defined by:

The importance of the integral transform

appears by looking carefully at equations 1 and 2.

Some problems are difficult to solve in their

original representations or in their domains.

The idea is to map the problem in another domain,

solving it in the new domain and then by choosing

the appropriate domain and using the inverse

transform the solution in the original domain is

mapped back!

Inverse Transform

(((( )))) (((( )))) (((( )))) αααααααααααα==== −−−−∫∫∫∫ dtKgtfb

a,1 2


6

Inverse Transform

The procedure is summarized schematically in

Fig. 6-1:

Figure 6-1: Schematic integral transform

2

Fourier Analysis


8

Fourier series are a basic tool for solving Fourier series are a basic tool for solving

ordinary differential equations (ODEs) and ordinary differential equations (ODEs) and

partial differential equations (PDEs) with partial differential equations (PDEs) with

periodic boundary conditions. Fourier periodic boundary conditions. Fourier

integrals for integrals for nonperiodicnonperiodic phenomena are phenomena are

developed in this chapter. The common developed in this chapter. The common

name for the field is name for the field is Fourier analysisFourier analysis..

Fourier Series and Integrals

Appendix 6-1: Fourier Series

The precursor of the transforms were the Fourier series to express functions in finite intervals. Later the Fourier transform was developed to remove the requirement of finite intervals.


10

Domains of Application

The Fourier transform is of fundamental

importance in a broad range of

applications, including both ordinary and

partial differential equations, probability,

quantum mechanics, waves, diffraction

and interferometry, signal and image

processing, and control theory, etc ...


11

Fourier Transform

The appropriate kernel is simply eiωωωωt and

its real part (cos ω ω ω ωt) or its imaginary part

(sinωωωωt).

Because these kernels are the functions

used to describe waves, Fourier

transforms appear frequently in studies

of waves and the extraction of

information from waves, particularly

when phase information is involved

(diffraction for example).


12

Domains of Application – Examples1

In optics, the diffraction pattern is the Fourier transform of the "obstacle" responsible of "diffracting" the waves.

The Fraunhofer diffraction pattern is the Fourier transform of the amplitude leaving the diffracting aperture.

In quantum mechanics the physical origin of the Fourier transforms is the

duality wave-matter, i.e. the wave nature of matter and our description of matter in terms of waves (Section 6).


13


The output of a stellar interferometer, for instance, involves a Fourier transform of the brightness across a stellar disk.

The electron distribution in an atom may be obtained from a Fourier transform of the amplitude of scattered X rays.

Electron scattering experiments were used, back in the 60's of the last century, in order to define the shape of a nucleus. The diffraction pattern" is used in order to define the "structure" of the scattering nucleus. By using an inverse Fourier transform.

3


14


In image processing, Fourier transform is a crucial tool. The input is a "spatial" (real) image which is decomposed into its sine and cosine components.

The output of the transformation, i.e. the result of applying the Fourier transform, is the image in the Fourier or frequency domain. In the Fourier domain image, each point represents a particular frequency contained in the spatial domain image.


15

The Fourier transform of a Gaussian function

is:

In order to calculate (3) we complete the

square in the exponent as follows:

This yields:

Example 1 - FOURIER TRANSFORM OF GAUSSIAN

22 tae

−−−−

(((( ))))(((( )))) (((( ))))� dteeg

tK

ti

tf

ta

∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−αααα

ωωωω−−−−

ππππ====ωωωω

,

22

2

1�� 3

2

22

2

222

42 aa

itatita


ωωωω−−−−−−−−====ωωωω++++−−−− 4

(((( )))) tdeeg ai

taa ′′′′

ππππ====ωωωω ∫∫∫∫

∞∞∞∞++++

∞∞∞∞−−−−

ωωωω−−−−′′′′−−−−


2

22

2

2

24

2

1 5


16

A simple change of variable (shift of origin),

which is a Gaussian, but in the (Fourier) ωωωω space.

Example 1 - FOURIER TRANSFORM OF GAUSSIAN

tddta

itt ′′′′====⇒⇒⇒⇒

ωωωω++++====′′′′

22

6

(((( ))))

(((( ))))22

44

4exp2

1

2

1

2

1 2222222

aa

dea

edteeg ataa

ωωωω−−−−====

ξξξξππππ

====ππππ

====ωωωω

ππππ

∞∞∞∞++++

∞∞∞∞−−−−

ξξξξ−−−−ωωωω−−−−∞∞∞∞++++

∞∞∞∞−−−−

−−−−ωωωω−−−− ∫∫∫∫∫∫∫∫��

The bigger a is, that is, the narrower the original

Gaussian is, the wider is its Fourier transform

~ .22 4ai

eωωωω−−−−

22 tae−−−−

gives:


17

This is justified by an application of Cauchy’s

theorem to the rectangle with vertices −T , T , T

+iω2a2 , −T + iω2a2 for T → ∞, noting that the

integrand has no singularities in this region

and that the integrals over the sides from ±±±±T to

± ± ± ±T + iω2a2 become negligible for T→∞.

Shifting the Origin


18

7

8

9

Laplace Transform

Hankel Transform

Mellin Transform

Other useful KernelsThree other useful kernels each giving rise to a particular transform are

(((( )))) (((( ))))∫∫∫∫∞∞∞∞ αααα−−−−====αααα0

dtetfg t

(((( )))) (((( )))) (((( ))))∫∫∫∫∞∞∞∞

αααα====αααα0

dttJttfg n

(((( )))) (((( ))))∫∫∫∫∞∞∞∞ −−−−αααα====αααα0

1dtttfg

te

αααα−−−−

(((( ))))tJt n αααα

1−−−−ααααt

Kernel Transform


19

Mellin and Laplace

We have seen Mellin transform for e-t.

(((( )))) (((( )))) (((( ))))!10

1 −−−−αααα====ααααΓΓΓΓ========αααα ∫∫∫∫∞∞∞∞ −−−−αααα−−−− dtteg t

(((( ))))10

!++++

∞∞∞∞ αααα−−−−

αααα========αααα ∫∫∫∫ n

tn ndtetg

Laplace transform of tn is

9

10

4

LinearityLinearity


21

Linear OperatorAll the previous integral transforms (Fourier, Laplace, Hankel and Mellin) are linear and we can write

(((( )))) (((( ))))tfg L====αααα

(((( )))) (((( ))))αααα==== −−−− gtf 1L

An inverse operator is expected to exist such that

11

12

In general, the determination of the inverse transform is the main problem in using integral transform.


22

Overview










10.Other Properties



22-- Development of Development of

the Fourier Integralthe Fourier Integral


24

Fourier Series & Fourier Integral

Fourier series are useful in representing certain functions

(1) over a limited range [0, 2 π π π π], [−L,L], and so on, or

(2) for the infinite interval (− ∞ ∞ ∞ ∞,∞∞∞∞), if the function is periodic.

Fourier transform is a generalization of Fourier series representing a nonperiodic function over the infinite range. Physically this means resolving a single pulse or wave packet into sinusoidal waves.


25

Deriving Fourier Integral

For a piecewise regular function, f(x) satisfying the Dirichlet conditions defined in the interval

[-L,L], we can write:

We start from the definition of the coefficients of Fourier series.

an, and bn, the Fourier coefficients are given

by:

(((( )))) ∑∑∑∑∑∑∑∑∞∞∞∞

====

∞∞∞∞

====

++++++++====11

0 sincos2

n

n

n

n xnbxnaa

xf

(((( )))) ,cos1∫∫∫∫

++++

−−−−

ππππ====

L

L

n dtL

tntf

La (((( )))) .sin

1∫∫∫∫

++++

−−−−

ππππ====

L

L

n dtL

tntf

Lb 14

13

5


26

Step 1

(((( )))) (((( )))) (((( ))))

(((( )))) ,sinsin1

coscos1

2

1

1

1

∫∫∫∫∑∑∑∑

∫∫∫∫∑∑∑∑∫∫∫∫++++

−−−−

∞∞∞∞

====

++++

−−−−

∞∞∞∞

====

++++

−−−−

ππππππππ++++

ππππππππ++++====

L

Ln

L

Ln

L

L

dtL

tntf

L

xn

L

dtL

tntf

L

xn

Ldttf

Lxf

The resulting Fourier series is:

using the trigonometric identity:

(((( )))) ββββαααα++++ββββαααα====ββββ−−−−αααα sinsincoscoscos

15-1

we have:

(((( )))) (((( )))) (((( )))) (((( ))))∑∑∑∑ ∫∫∫∫∫∫∫∫∞∞∞∞

====

++++

−−−−

++++

−−−−

−−−−ππππ

++++====1

cos1

2

1

n

L

L

L

L

dtxtL

ntf

Ldttf

Lxf 15-2


27

Step 2

The next step is to let L approach ∞∞∞∞, transforming the interval [-L,L] into [-∞∞∞∞,∞∞∞∞]. We also define a new variable ωωωω as

.,, ∞∞∞∞→→→→ωωωω∆∆∆∆====ππππ

ωωωω====ππππ

LwithLL

n

17

Then we have:

(((( )))) (((( )))) (((( ))))∑∑∑∑ ∫∫∫∫∞∞∞∞

====

∞∞∞∞++++

∞∞∞∞−−−−

−−−−ωωωωωωωω∆∆∆∆ππππ

→→→→1

,cos1

n

dtxttfxf 16

or

(((( )))) (((( )))) (((( ))))∫∫∫∫∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

−−−−ωωωωωωωωππππ

==== ,cos1

0

dtxttfdxf


28

Fourier Integral

Note the term a0 has vanished assuming that

Eq. 17 is taken as the Fourier integralFourier integral, under the

following conditions:

(((( ))))∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

dttf

1) f(x) is piecewise differentiable

2) f(x) is piecewise continuous

3) f(x) is absolutely integrable, i.e.is finite

exists.

(((( ))))∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

dxxf

Fourier IntegralFourier Integral

Exponential FormExponential Form


30

Fourier Integral TheoremEq. 17 can be written as:

using the fact that:

20(((( )))) (((( ))))∫∫∫∫∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

ωωωω∞∞∞∞++++

∞∞∞∞−−−−

ωωωω−−−− ωωωωππππ

==== ,2

1dtetfdexf tixi

19(((( )))) (((( )))) (((( )))) 0sin2

1====−−−−ωωωωωωωω

ππππ==== ∫∫∫∫∫∫∫∫

∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

dtxttfdxf

because sin ωωωω(t-x) is an odd function of ωωωω.

18(((( )))) (((( )))) (((( )))) ,cos2

1∫∫∫∫∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

−−−−ωωωωωωωωππππ

==== dtxttfdxf

Eq. 20 is called the Fourier integral.© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 4 Integral Transforms

31

The variable ωωωωThe variable The variable ωωωωωωωω introduced here is an introduced here is an arbitrary mathematical variable. arbitrary mathematical variable.

In many physical problems, however, it In many physical problems, however, it

corresponds to the angular frequency corresponds to the angular frequency ωωωωωωωω. .

We may then interpret We may then interpret Eq. 18Eq. 18 or or Eq. 20Eq. 20 as a as a

representation of representation of ff((xx)) in terms of a in terms of a

distribution of infinitely long sinusoidal distribution of infinitely long sinusoidal

wave trains of angular frequency wave trains of angular frequency ωωωωωωωω, in , in which this frequency is a continuous which this frequency is a continuous

variable.variable.

6

Important ApplicationImportant Application

Derivation of Dirac Delta Derivation of Dirac Delta

FunctionFunction


33

A Useful Representation of δδδδUsing the Fourier integral we can define Dirac Using the Fourier integral we can define Dirac

delta function asdelta function as

Appendix 6Appendix 6--33

21(((( )))) (((( )))) (((( ))))∫∫∫∫∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

−−−−ωωωω−−−−∞∞∞∞++++

∞∞∞∞−−−−

−−−−ωωωω ωωωωππππ

====ωωωωππππ

====−−−−δδδδ dedext txixti2

1

2

1


34

Overview










10.Other Properties



33-- Fourier TransformFourier Transform

Inverse TheoremInverse Theorem


36

Using the Exponential Transform

Let us define Let us define gg((ωωωωωωωω)), the Fourier transform of the , the Fourier transform of the function function ff((tt)), by, by

Exponential TransformExponential Transform

22(((( )))) (((( )))) .2

1∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

ωωωω

ππππ≡≡≡≡ωωωω dtetfg ti

Then, from Then, from Eq. 20Eq. 20, we have the inverse relation,, we have the inverse relation,

23(((( )))) (((( ))))∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

ωωωω−−−− ωωωωωωωωππππ

==== .2

1degtf ti


37

Important Remarks�� Eq. 22Eq. 22 and and Eq. 23Eq. 23 are almost symmetrical, are almost symmetrical,

differing in the sign of differing in the sign of ii..

�� In physics we are more interested in the In physics we are more interested in the Fourier transform (Fourier transform (Eq. 22Eq. 22 and and Eq. 23Eq. 23) rather than ) rather than Eq. 20Eq. 20 (Fourier integral)(Fourier integral)

�� The symmetry is a matter of choice or The symmetry is a matter of choice or

convenience. This factor is sometimes replaced convenience. This factor is sometimes replaced

by by 11 in one equation and the entire factor in one equation and the entire factor ½½ππππππππin the other.in the other.

ππππ21

ππππ21

7

38

Eq. 23bEq. 23b may be interpreted as an expansion of may be interpreted as an expansion of

ff((rr)) in a continuum of plane wave in a continuum of plane wave

eigenfunctions; eigenfunctions; gg((kk)) then becomes the then becomes the

amplitude of the wave amplitude of the wave exp( exp( -- ii k k . . rr))..

The 3D Form - PhysicsMoving the Fourier transform pair (Fourier Moving the Fourier transform pair (Fourier

transform and its inverse) to threetransform and its inverse) to three--

dimensional space, the pair becomes:dimensional space, the pair becomes:

The integrals are over all space.The integrals are over all space.

23a(((( ))))(((( ))))

(((( )))) ,2

1 323 ∫∫∫∫

⋅⋅⋅⋅

ππππ==== rderfkg rki

��

23b(((( ))))(((( ))))

(((( )))) .2

1 323 ∫∫∫∫

⋅⋅⋅⋅−−−−

ππππ==== kdekgrf rki

��

Exercise: Verify Exercise: Verify Eq. 23aEq. 23a and and Eq. 23bEq. 23b by substituting the by substituting the

leftleft--hand side of one equation into the integrand of the hand side of one equation into the integrand of the

other equation and using the threeother equation and using the three--dimensional delta dimensional delta

function.function.

Fourier Cosine and Fourier Cosine and

Sine TransformsSine Transforms


40

Case of Even and Odd Functions

If If ff((xx)) is even then is even then Eq. 22Eq. 22 and and Eq. 23Eq. 23, respectively, , respectively,

can be written as:can be written as:

24(((( )))) (((( )))) ,cos2

0

∫∫∫∫∞∞∞∞++++

ωωωωππππ

====ωωωω dtttfg cc

25(((( )))) (((( ))))∫∫∫∫∞∞∞∞++++

ωωωωωωωωωωωωππππ

====0

.cos2

dxgtf cc

If If ff((xx)) is odd then is odd then Eq. 22Eq. 22 and and Eq. 23Eq. 23, respectively, , respectively, can be written as:can be written as:

26(((( )))) (((( )))) ,sin2

0

∫∫∫∫∞∞∞∞++++

ωωωωππππ

====ωωωω dtttfg ss

27(((( )))) (((( ))))∫∫∫∫∞∞∞∞++++


====0

.sin2

dxgtf ss© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 4 Integral Transforms

41

ParityThe pair, Eq. 24 and Eq. 25 are called the Fourier cosine transforms. The other pair, Eq. 26 and Eq. 27 are the sine Fourier sine transforms.

The Fourier cosine transforms and the Fourier sine transforms each involve only positive values (and zero)

The parity of f(x) is used to establish the transforms; but once the transforms are

established, the behavior of the functions f and gfor negative argument is irrelevant.

The transform equations impose a definite parity:

even for the Fourier cosine transform and odd for even for the Fourier cosine transform and odd for

the Fourier sine transformthe Fourier sine transform..


42

Physical Meaning

In In Eq. 27Eq. 27, , ff((xx)) is being described is being described by a continuum by a continuum

of sine waves. The amplitude of of sine waves. The amplitude of sinsinωωωωωωωωxx is given is given by by , in which , in which ggss((ωωωωωωωω)) is the Fourier sine is the Fourier sine transform of transform of ff((xx))..

(((( )))) (((( ))))∫∫∫∫∞∞∞∞++++


====0

.sin2

dxgtf ss

(((( )))) (((( ))))ωωωωππππ sg2

In the following example the important In the following example the important

application of the Fourier transform in the application of the Fourier transform in the

resolution of a finite pulse of sinusoidal wavesresolution of a finite pulse of sinusoidal waves, ,

is detailed.is detailed.


43

Imagine that an infinite wave train sinω0t is clipped by Kerr cell or saturable dye cell shutters (Fig. 6-2) so that we have

In Fig. 6-2, N, which represents the number of cycles of the

wave train, is equal to 5.

Example 2 – FINITE WAVE TRAIN

(((( ))))

ωωωω

ππππ>>>>

ωωωω

ππππ

8


44

f(t) is odd, thus

Integrating, we find the amplitude function:

Amplitude in the Fourier Space

29(((( )))) .sinsin20

0

0∫∫∫∫ωωωωππππ

ωωωωωωωωππππ

====ωωωω

N

s dtttg

30(((( ))))(((( ))))(((( ))))

(((( ))))(((( ))))(((( ))))

(((( )))).

2

sin

2

sin2

0

00

0

00

ωωωω++++ωωωω

ωωωωππππωωωω++++ωωωω−−−−

ωωωω−−−−ωωωω

ωωωωππππωωωω−−−−ωωωω


NNgs

Dependence of g(ωωωω) on frequency.

For large ωωωω and ω ≈ω ≈ω ≈ω ≈ ωωωω0, the first term will dominate because of the denominator (ωωωω-ωωωω0).


45

Fig. 6-3 shows the first term.

Single-Slit Diffraction Pattern

This is the amplitude curve for the single-slit diffraction pattern!


46

For large N, g(ωωωω) may also be interpreted as a Dirac Delta distribution.

The contributions outside the central maximum being small in this case,

N vs. Spread in frequency (∆ω∆ω∆ω∆ω)

N

0ωωωω====ωωωω∆∆∆∆

can be taken as a good measure of the spread in frequency of our wave pulse.

The larger N is, the smaller is the frequency spread.

For small N, the spread is large and the secondary maxima become more important.

31


47

If we are dealing with em waves we have

∆∆∆∆E represents an uncertainty in the energy of our pulse. There is also an uncertainty in the time

because our wave of N cycles requires 2ππππN/ωωωω0seconds to pass.

Uncertainty PrincipleUncertainty Principle

ωωωω∆∆∆∆====∆∆∆∆⇒⇒⇒⇒ωωωω==== �� EE 32

0

2

ωωωω

ππππ====∆∆∆∆

Nt

The product hN

Nh

NtE ====

ωωωω⋅⋅⋅⋅

ωωωω====

ωωωω

ππππ⋅⋅⋅⋅ωωωω∆∆∆∆====∆∆∆∆⋅⋅⋅⋅∆∆∆∆

0

0

0

2�

The Heisenberg uncertainty principle states that:

which is clearly satisfied in this example.

,42 ππππ

====≥≥≥≥∆∆∆∆⋅⋅⋅⋅∆∆∆∆h

tE�

34

35

33Taking


48

Overview










10.Other Properties



Fourier Transform Fourier Transform

of Derivativesof Derivatives

9


50

Starting from the definition of the Fourier Starting from the definition of the Fourier

transform for transform for ff((xx))

and for and for dfdf//dxdx

Integrating by parts:Integrating by parts:

Transform of the 1st Derivative

(((( )))) (((( ))))∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

ωωωω

ππππ====ωωωω dxexfg xi

2

1

(((( )))) (((( ))))xfvdvdx

xdf

eiduuexixi

====⇒⇒⇒⇒====

ωωωω====⇒⇒⇒⇒==== ωωωωωωωω

(((( ))))(((( ))))

.2

11 ∫∫∫∫

∞∞∞∞++++

∞∞∞∞−−−−

ωωωω

ππππ====ωωωω dxe

dx

xdfg xi

36

37


51

Differentiation becomes multiplication

We haveWe have

We use the fact that, apart from some cases, We use the fact that, apart from some cases,

ff((xx)) must vanish as must vanish as xx→→±∞±∞±∞±∞±∞±∞±∞±∞ in order for the in order for the Fourier transform of Fourier transform of ff((xx)) to exist. The first term to exist. The first term

of the of the rhsrhs vanishes and we obtain:vanishes and we obtain:

(((( )))) (((( )))) (((( )))) (((( ))))

(((( ))))

.2

1

2

11

�� ωωωω

∞∞∞∞++++

∞∞∞∞−−−−

ωωωω∞∞∞∞++++

∞∞∞∞−−−−

ωωωω ∫∫∫∫ππππωωωω−−−−


g

xixi dxexfiexfg

(((( )))) (((( )))).1 ωωωωωωωω−−−−====ωωωω gigi.e.i.e. the transform of the derivative is (the transform of the derivative is (−−iiωωωωωωωω) ) times the transform of the original function.times the transform of the original function.

38

39


52

ConsiderConsider

Integrating by parts:Integrating by parts:

Transform of the 2nd Derivative

(((( )))) (((( )))) (((( )))) (((( )))) (((( ))))ωωωωωωωω−−−−====ωωωωωωωω−−−−====ωωωω gigig 212

(((( ))))(((( ))))

∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

ωωωω

ππππ====ωωωω dxe

dx

xfdg xi

2

2

22

1

(((( )))) (((( ))))dx

xdfvdv

dx

xfd

eiduuexixi

====⇒⇒⇒⇒====

ωωωω====⇒⇒⇒⇒==== ωωωωωωωω

2

2

(((( ))))(((( )))) (((( ))))

(((( ))))

.2

1

2

1

10

2

�� ωωωω

∞∞∞∞++++

∞∞∞∞−−−−

ωωωω

≡≡≡≡

∞∞∞∞++++

∞∞∞∞−−−−

ωωωω ∫∫∫∫ππππωωωω−−−−


g

xixi dxedx

xdfie

dx

xdfg

40

41


53

This may readily be generalized to the This may readily be generalized to the nnthth

derivative to yieldderivative to yield

provided all the integrated parts vanish as provided all the integrated parts vanish as

xx →±∞→±∞→±∞→±∞→±∞→±∞→±∞→±∞. .

This is the power of the Fourier transform, the This is the power of the Fourier transform, the

reason it is so useful in solving (partial) reason it is so useful in solving (partial)

differential equations. The operation of differential equations. The operation of

differentiation has been replaced by a differentiation has been replaced by a

multiplication in multiplication in ωωωωωωωω--space.space.

Generalization – nth Derivative

(((( )))) (((( )))) (((( )))) ,ωωωωωωωω−−−−====ωωωω gig nn 42

An important An important

ExampleExample

Heat Flow PDEHeat Flow PDE


55

kk is the thermal conductivity (MKS unit = is the thermal conductivity (MKS unit = W.mW.m--11.K.K--11))

The law of heat conduction, also known as Fourier's

law, states that the time rate of heat transfer time rate of heat transfer qqthrough a material is proportional to the negative

gradient in the temperature (T) and to the area, at right angles to that gradient, through which the heat is flowing.

Fourier's Law – Heat ConductionIn heat transfer, conduction is the transfer of heat energy by microscopic diffusion and collisions of particles or quasi-particles within a body due to a temperature gradient.

43Tkq ∇∇∇∇−−−−====��

The differential form of Fourier's law, in which we look The differential form of Fourier's law, in which we look

at the flow rates or fluxes of energy locally, is given byat the flow rates or fluxes of energy locally, is given by

10


56

a is a constant called the thermal diffusivity, also known as Fourier constant.

where cP is the heat capacity at constant pressure.

a is related to the thermal conductivity of a material of density ρ ρ ρ ρ by the simple relation:

In order to find the temperature field ψψψψ(x,t)(Temperature at x at instant t) for a given system, one has to solve the heat equation (which can be derived from Fourier's law)

Heat Equation

(((( )))),

,2

22

xa

t

tx

∂∂∂∂

ψψψψ∂∂∂∂====

∂∂∂∂

ψψψψ∂∂∂∂

ρρρρ====

Pc

ka

44


57

Defining the Fourier transform of Defining the Fourier transform of ψψψψψψψψ((xx,,tt)) asas

Integrating we obtain

The previous equation is an ODE for the Fourier transform ΨΨΨΨ of ψψψψ in the time variable t.

Taking the Fourier transform of both sides of Eq. 44we get

Temperature Field

(((( )))) (((( )))) .,, 22 tat

tωωωωΨΨΨΨωωωω−−−−====

∂∂∂∂

ωωωωΨΨΨΨ∂∂∂∂

(((( )))) (((( ))))∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

ωωωωψψψψππππ

====ωωωωΨΨΨΨ dxetxt xi,2

1,

(((( )))) Clntat,ln ++++ωωωω−−−−====ωωωωΨΨΨΨ 22 oror (((( )))) taeCt22

, ωωωω−−−−====ωωωωΨΨΨΨ

45

46


58

The integration constant C may still depend on ω and, in general, is determined by initial conditions.

The Integration Constant

In fact, C=ΨΨΨΨ(ωωωω,0) is the initial spatial distribution of ΨΨΨΨ, so it is given by the transform (in x) of the initial distribution of ψψψψ, namely, ψψψψ(x, 0).

Putting this solution back into our inverse Fourier transform, this yields

(((( )))) (((( ))))(((( ))))

.2

1,

,

22

∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−


ωωωωΨΨΨΨ

ωωωω−−−− ωωωωωωωωππππ

====ψψψψ deeCtx xi

t

a t

�� 47


59

Taking ψψψψ(ωωωω,0) = δδδδ(ωωωω,0), C is ωωωω-independent.

δ δ δ δ Function Initial Temperature Distribution

*ψψψψ(x,t) is the inverse Fourier transform of

C exp (-a2ωωωω2t).

(((( )))) ,4

exp2

,2

2

−−−−

ππππ====ψψψψ

ta

x

a

Ctx

Integrating Eq. 47 by completing the square as we did in Example 1 (where we calculated the Fourier transform of a Gaussian*), we get

48


60

Overview










10.Other Properties



5 5 -- Convolution Convolution

TheoremTheorem

11


62

Convolution - DefinitionConvolution is used to solve differential Convolution is used to solve differential

equations, to normalize momentum wave equations, to normalize momentum wave

functions (next section), and to investigate functions (next section), and to investigate

transfer functions.transfer functions.

(((( )))) (((( ))))∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−


==== dyyxfygg*f2

1

Let us consider two functions Let us consider two functions ff((xx)) and and gg((xx)) with with

Fourier transforms Fourier transforms FF((tt)) and and GG((tt)), respectively. , respectively. We define the operationWe define the operation

as the convolution of the two functions as the convolution of the two functions ff and and gg over over

the interval the interval ((−− ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞,,∞∞∞∞∞∞∞∞)). Some authors use the German . Some authors use the German word word FaltungFaltung (which means folding) instead of (which means folding) instead of

convolution.convolution.

49


63

Convolution - UseThis form of an integral appears in probability This form of an integral appears in probability

theory in the determination of the probability theory in the determination of the probability

density of two random, independent variablesdensity of two random, independent variables


64

Convolution – Graphical Illustration

For For ff((yy) = ) = ee−−yy , , ff((yy)) and and ff((xx −− yy)) are plotted in are plotted in Fig. 6Fig. 6--44. Clearly, . Clearly, ff ((yy)) and and ff ((xx −− yy)) are mirror images of are mirror images of each other in relation to the vertical line each other in relation to the vertical line yy = = xx/2/2, ,

i.e.i.e., we could generate , we could generate ff((xx−−yy)) by folding over by folding over ff((yy))on the line on the line yy = = xx/2/2..


65

Back to the Electrostatic Analog

The solution of Poisson's equation (The solution of Poisson's equation (Chapter 3 Eq. Chapter 3 Eq.

109109), ), i.e.i.e.

can be written as

(((( )))) (((( )))) .4

12

21

2

01 ∫∫∫∫∫∫∫∫∫∫∫∫ ττττ−−−−

ρρρρ

εεεεππππ−−−−====ψψψψ d

rr

rr ��

��

(((( )))) (((( ))))(((( ))))�

(((( ))))

.2

4

2

12

1

210

21

21

2

∫∫∫∫∫∫∫∫∫∫∫∫ ττττ

−−−−

ππππ

εεεεππππ−−−−ρρρρ

ππππ====ψψψψ

−−−−

−−−−

drrrr

rrf

rg ��

��

��

�

which we may interpret as the convolution of a

charge distribution and a weighting function, 1

210

2

4−−−−

−−−−

ππππ

εεεεππππ−−−− rr

��

(((( ))))2r�

ρρρρ

50

51


66

Fourier Transform and Convolution

(((( )))) (((( )))) (((( )))) (((( )))) (((( ))))

(((( )))) (((( ))))[[[[ ]]]]

(((( ))))

(((( )))) (((( )))) gfdtetGtF

dtedyeygtF

dydtetFygdyyxfyg

xti

xti

tG

yti

tyxi

*

2

1

2

1

2

≡≡≡≡====

ππππ====

ππππ====−−−−

∫∫∫∫

∫∫∫∫ ∫∫∫∫

∫∫∫∫ ∫∫∫∫∫∫∫∫

∞∞∞∞++++

∞∞∞∞−−−−

−−−−

∞∞∞∞++++

∞∞∞∞−−−−

−−−−

ππππ

∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

−−−−−−−−∞∞∞∞++++

∞∞∞∞−−−−

��

Let's transform Let's transform Eq. 36Eq. 36 by introducing the Fourier by introducing the Fourier

transforms transforms

52

This result may be interpreted as follows: The Fourier inverse transform of a product of Fourier transforms is

the convolution of the original functions, f ∗∗∗∗ g.© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 4 Integral Transforms

67

Fourier Transform and Convolution

(((( )))) (((( )))) (((( )))) (((( ))))∫∫∫∫∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

−−−−==== dyygyfdttGtF

For the special case For the special case xx = 0= 0, , Eq. 52Eq. 52 givesgives

53

The minus sign in The minus sign in −−yy suggests that suggests that

modifications be tried. We now do this with modifications be tried. We now do this with gg∗∗∗∗∗∗∗∗instead of instead of gg using a different technique.using a different technique.

12

Parseval'sParseval's RelationRelationUnitarity of Fourier TransformUnitarity of Fourier Transform


69

Eq. 52Eq. 52 and the corresponding sine and cosine and the corresponding sine and cosine

convolutions are often labeled convolutions are often labeled ParsevalParseval’’ss relations relations

by analogy with by analogy with ParsevalParseval’’ss theorem for Fourier theorem for Fourier

series (series (Arfken Chapter 19Arfken Chapter 19-- 6th6th , Chapter 20 in the 7, Chapter 20 in the 7thth editionedition).).

relates the product of the function relates the product of the function ff and and gg** to to

their respective Fourier transforms (their respective Fourier transforms (FF and and GG**) ) (in the transform (Fourier) space)(in the transform (Fourier) space)

Parseval's Relation

54

The The Parseval'sParseval's relationrelation

(((( )))) (((( )))) (((( )))) (((( ))))∫∫∫∫∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

====ωωωωωωωωωωωω dttgtfdGF **


70

21’(((( )))) (((( )))) (((( ))))∫∫∫∫∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

ωωωω−−−−−−−−

∞∞∞∞++++

∞∞∞∞−−−−

−−−−ωωωω

ππππ====

ππππ====−−−−ωωωωδδδδ dtedtex xtixti

2

1

2

1

Using delta function representation, we can writeUsing delta function representation, we can write

we havewe have

Derivation of the Parseval's Relation

55

Integrating over Integrating over tt and usingand using

(((( )))) (((( )))) (((( )))) (((( ))))∫∫∫∫ ∫∫∫∫ ∫∫∫∫∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

ωωωω−−−−∞∞∞∞++++

∞∞∞∞−−−−ππππ

⋅⋅⋅⋅ωωωωωωωωππππ

==== dtdxexGdeFdttgtf txiti **

2

1

2

1

56

(((( )))) (((( )))) (((( )))) (((( )))) (((( ))))

(((( )))) (((( )))) ,*

**

∫∫∫∫

∫∫∫∫ ∫∫∫∫∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

ωωωωωωωωωωωω====

ωωωωωωωω−−−−δδδδωωωω====

dGF

ddxxxGFdttgtf


71

Special case f(t) = g(t)In the very important case where In the very important case where ff((tt) = ) = gg((tt)), the , the integrals on both sides of integrals on both sides of Eq. 56Eq. 56 are nothing else are nothing else

but normalization integrals.but normalization integrals.

This important relation guarantees that if This important relation guarantees that if ff((tt)) is is

normalized in the normalized in the ""tt--spacespace"", then its Fourier , then its Fourier

transform transform FF((ωωωωωωωω)) (in the transform (frequency) (in the transform (frequency) space) is normalized too!space) is normalized too!

57(((( )))) (((( )))) (((( )))) (((( ))))∫∫∫∫∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

ωωωωωωωωωωωω==== dFFdttftf **

This is what we call the unitarity of Fourier This is what we call the unitarity of Fourier

transform which has a large importance in transform which has a large importance in

quantum physics.quantum physics.


72

Unitarity of Fourier Transform

It may be shown that the Fourier It may be shown that the Fourier

transform is a unitary operation (in the transform is a unitary operation (in the

Hilbert space Hilbert space LL22, square , square integrableintegrable

functions). The functions). The Parseval'sParseval's relation is a relation is a

reflection of this unitary property.reflection of this unitary property.


73

ApplicationsIn In FraunhoferFraunhofer diffraction optics the diffraction diffraction optics the diffraction

pattern (amplitude) appears as the transformpattern (amplitude) appears as the transform

of the function describing the aperture. of the function describing the aperture.

With intensity proportional to the square of the With intensity proportional to the square of the

amplitude the amplitude the ParsevalParseval relation implies that relation implies that

the energy passing through the aperture the energy passing through the aperture

seems to be somewhere in the diffraction seems to be somewhere in the diffraction

patternpattern——a statement of the conservation of a statement of the conservation of

energy. energy.

Parseval’sParseval’s relations may be developed relations may be developed

independently of the inverse Fourier transform independently of the inverse Fourier transform

and then used rigorously to derive the inverse and then used rigorously to derive the inverse

transform. transform.

13


76

A rectangular pulse is described by

a) The Fourier exponential transform is

Example 3 – Single Slit Diffraction

(((( ))))

>>>>

14


82

Examples - Thermodynamics

In thermodynamics temperature In thermodynamics temperature TT and entropy and entropy

SS are conjugate variables. Pressure are conjugate variables. Pressure ((PP)) and and

volume volume ((VV)) are also conjugate variables. The are also conjugate variables. The

pair pair ((TT,,SS)) or the pair or the pair ((PP,,VV)) are used to define all are used to define all the properties of a thermodynamics system the properties of a thermodynamics system

such as the internal energy.such as the internal energy.

In fact all thermodynamic potentials are In fact all thermodynamic potentials are

expressed in terms of conjugate variables. expressed in terms of conjugate variables.

In statistical physics, pairs of extensive and In statistical physics, pairs of extensive and

intensive properties of a given system form intensive properties of a given system form

pairs of conjugate variables pairs of conjugate variables


83

Important Consequences in Physics

The duality leads naturally to an uncertainty The duality leads naturally to an uncertainty

principle in physics called the Heisenberg principle in physics called the Heisenberg

uncertainty principle.uncertainty principle.

84

59

3)3) The expectation value, The expectation value, i.e.i.e. the average position of the the average position of the

particle along the particle along the xx--axisaxis isis

2)2) ψψψψψψψψ((xx) ) dxdx is normalized (total probabilities = is normalized (total probabilities = 11))

(((( )))) (((( )))) (((( )))) 1* ====ψψψψψψψψ==== ∫∫∫∫∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

dxxxdxxP 58 (((( )))) (((( ))))∫∫∫∫∞∞∞∞

∞∞∞∞−−−−

ψψψψψψψψ==== dxxxxx *

Real SpaceIn this section we shall start with the usual space In this section we shall start with the usual space

distribution and derive the corresponding momentum distribution and derive the corresponding momentum

distribution. distribution.

For the one dimensional case our wave function For the one dimensional case our wave function ψψψψψψψψ((xx))has the following properties:has the following properties:

1)1) ψψψψψψψψ**((xx)) ψψψψψψψψ((xx) ) dxdx is the probabilityis the probability of finding the of finding the quantum system between quantum system between xx and and xx++dxdx..


85

2)2) gg((pp)) is normalized (total probabilities = is normalized (total probabilities = 11))

(((( )))) (((( )))) .1* ====∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

dppgpg 60

3)3) The expectation value, The expectation value, i.e.i.e. the average the average momentum of the particle ismomentum of the particle is

(((( )))) (((( )))) .*∫∫∫∫∞∞∞∞

∞∞∞∞−−−−

==== dxpgppgp 61

Momentum SpaceWe want a function We want a function gg((pp)) that will give the same that will give the same information about the momentum:information about the momentum:

1)1) gg**((pp)) gg((pp) ) dpdp is the probability is the probability that the particle has that the particle has

a momentum between a momentum between pp and and pp++dpdp..


86

(((( )))) (((( )))) ,2

1∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

−−−−ψψψψππππ

==== dxexpg xpi �

�

63

64

Momentum (Fourier) SpaceSuch a function is given by Fourier transform of Such a function is given by Fourier transform of

the space function the space function ψψψψψψψψ((xx)),, i.e.i.e.

(((( )))) (((( )))) .2

1 **∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

ψψψψππππ

==== dxexpg xpi �

�

The corresponding 3D momentum function isThe corresponding 3D momentum function is

(((( ))))(((( ))))

(((( )))) .rderpg pri∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

⋅⋅⋅⋅−−−−ψψψψππππ

==== 323

2

1 ��

�

�

62

Parseval'sParseval's relation guarantees the normalization of relation guarantees the normalization of

gg((pp)) if if ψψψψψψψψ((xx)) is normalized.is normalized.


87

Checking property (3) means showing thatChecking property (3) means showing that

We replace the momentum functions by Fourier We replace the momentum functions by Fourier

transformed space functions, and the first integral transformed space functions, and the first integral

becomesbecomes

where where ppxx is the momentum operator in the space is the momentum operator in the space

representation,representation,

Expectation Values

(((( )))) (((( )))) (((( ))))�

(((( ))))∫∫∫∫∫∫∫∫∞∞∞∞

∞∞∞∞−−−−

∞∞∞∞

∞∞∞∞−−−−

ψψψψψψψψ======== dxxdx

d

ixdxpgppgp

xp

�**65

(((( )))) (((( )))) (((( )))) .2

1 *∫∫∫∫ ∫∫∫∫ ∫∫∫∫

∞∞∞∞

∞∞∞∞−−−−

′′′′−−−−−−−− ′′′′ψψψψ′′′′ψψψψππππ

dxxddpxxep xxpi �

�66

15


88

Substituting into Substituting into Eq. 66Eq. 66 and integrating by parts, and integrating by parts,

holding holding xx′′ and and pp constant, we obtainconstant, we obtain

Expectation ValuesUsing the plane wave identityUsing the plane wave identity

(((( )))) (((( )))) ,

−−−−====

′′′′−−−−−−−−′′′′−−−−−−−− �� xxpixxpi eidx

dep 67

Here Here pp is a constant, not an operator.is a constant, not an operator.

(((( ))))[[[[ ]]]] (((( )))) (((( ))))∫∫∫∫ ∫∫∫∫ ∫∫∫∫∞∞∞∞

∞∞∞∞−−−−

∞∞∞∞++++

∞∞∞∞−−−−

′′′′−−−−−−−− ′′′′ψψψψ′′′′ψψψψ⋅⋅⋅⋅ππππ

==== .2

1 * dxxdxdx

d

ixdpep xxpi�

�

�68

Here we assume Here we assume ψψψψψψψψ((xx)) vanishes as vanishes as xx → ±∞→ ±∞→ ±∞→ ±∞→ ±∞→ ±∞→ ±∞→ ±∞, , eliminating the integrated part. Using the Dirac eliminating the integrated part. Using the Dirac

delta function, delta function, Eq. 68Eq. 68 reduces to reduces to Eq. 65Eq. 65 to verify our to verify our momentum representation (momentum representation (qedqed))..


89

Overview










10.Other Properties




90

Self Reading

7- Fourier Transfer Functions

Arfken 6th Edition– Section 15.7, pages: 961-964Arfken 7th Edition– Section 20.5, pages: 997-1002See both sections.


91

Overview








8.Laplace Transforms9. Laplace Transform of Derivatives

10.Other Properties



88-- Laplace Laplace

TransformsTransforms


93

For instance, For instance, FF((tt)) may diverge exponentially for may diverge exponentially for

large large tt. However, if there is some constant. However, if there is some constant ss00such thatsuch that

Definition

The Laplace transform The Laplace transform ff((ss)) or or LL of a function of a function FF((tt))is defined byis defined by

The integral need not exist!The integral need not exist!

(((( )))) (((( )))){{{{ }}}} (((( )))) (((( )))) .lim00

∫∫∫∫∫∫∫∫∞∞∞∞

−−−−−−−−

∞∞∞∞→→→→============ dttFedttFetFsf ts

ats

aL 69

(((( )))) ,0

∫∫∫∫∞∞∞∞

dttF

(((( )))) ,0 MtFe ts ≤≤≤≤−−−− 70

16


94

where where MM is a positive constant for is a positive constant for

sufficiently large sufficiently large tt, , tt > > tt00, the Laplace , the Laplace

transform (transform (Eq. 46Eq. 46) will exist for ) will exist for ss > > ss00; ; FF((tt)) is is

said to be of exponential order. said to be of exponential order.

Counterexample

As a counterexample, As a counterexample, does not satisfy does not satisfy

the condition given by the condition given by Eq. 47Eq. 47 and is not of and is not of

exponential order. does not exist.exponential order. does not exist.

(((( ))))2tetF ====

{{{{ }}}}2teL


95

"Failures" exist

The Laplace transform may also fail to exist The Laplace transform may also fail to exist

because of a sufficiently strong singularity because of a sufficiently strong singularity

in the function in the function FF((tt)) as as tt →→00; that is,; that is,

71∫∫∫∫∞∞∞∞

−−−−

0

dttents

diverges at the origin for diverges at the origin for nn ≤≤≤≤≤≤≤≤ −1−1. The Laplace . The Laplace transform transform LL{{ttnn}} does not exist for does not exist for nn ≤≤≤≤≤≤≤≤ −−11..


96

The operation denoted LL by is linear.

Linearity

Since, for two functions Since, for two functions FF((tt)) and and GG((tt)), for , for

which the integrals existwhich the integrals exist

(((( )))) (((( )))){{{{ }}}} (((( )))){{{{ }}}} (((( )))){{{{ }}}}tGbtFatGbtFa LLL ++++====++++ 72

Elementary Elementary

FunctionsFunctions


98

Self Reading

Elementary Functions

Arfken 6th Edition –pages: 965-966Arfken 7th Edition–pages: 1008-1010

Inverse TransformInverse Transform

17


100

Definition

The Laplace transform of The Laplace transform of FF((tt)) is (is (Eq. 69Eq. 69))

The inverse is, by definition,The inverse is, by definition,

(((( )))) (((( )))){{{{ }}}} (((( )))) .0

∫∫∫∫∞∞∞∞

−−−−======== dttFetFsf tsL

(((( )))) (((( )))){{{{ }}}} (((( )))) .0

1

∫∫∫∫∞∞∞∞

−−−− ======== dssfesftF tsL 73

This inverse transform is not unique.This inverse transform is not unique.


101

where N(t) is a null function, indicating that

Two functions F1(t) and F2(t) may have the same

transform, f(s). However, in this case F1(t) – F2(t)

= N(t)

For all positive t0. (Lerch's theorem)

Unicity of L-1 – Lerch's Theorem

74(((( )))) ,00

0

====∫∫∫∫t

dttN

In physics, we need L-1 to be unique and therefore we take N(t) = 0.


102

(1) A table of transforms can be built up and

used to carry out the inverse transformation,

exactly as a table of logarithms can be used to

look up antilogarithms,

Several Methods exist:

2) A general technique for L−1 using the

calculus of residues, (Arfken in Section 15.12 ),

Determination of L-1

3) Numerical inversion.

Partial Fraction Partial Fraction

ExpansionExpansion


104

Utilization of a table of transforms (or inverse

transforms) is facilitated by expanding f(s) in partial fractions.

Often, the Laplace transform f(s) occurs in the form of a

fraction g(s)/h(s), where g(s) and h(s) are polynomials

with no common factors, g(s) being of lower degree

than h(s).

Using Tables

If the factors of h(s) are all linear and distinct, then by the method of partial fractions we may write

75(((( ))))n

n

as

c

as

c

as

csf

−−−−++++++++

−−−−++++

−−−−==== �

2

2

1

1

where the ci are independent of s. The ai are the roots

of h(s). © Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 4 Integral Transforms

105

Partial Fraction ExpansionIf any one of the roots, say, a1, is multiple

(occurring m times), then f(s) has the form

76(((( )))) ∑∑∑∑====

−−−−

−−−−====

−−−−++++++++

−−−−++++

−−−−====

n

i i

i

n

nmm

as

c

as

c

as

c

as

csf

2

1,

2

1,2

1

,1�

Finally, if one of the factors is quadratic,

(s2 + ps + q), then the numerator, instead of being a simple constant, will have the form (See Example 3)

77qsps

bsa

++++++++

++++2

18


106

Let

The partial fraction method consists in writing the previous fraction as the sum of two fractions. (Note the degrees of the polynomials in the numerators in the rhs)

The lhs is developed and like powers of s are

equated, i.e.

Example 3 – Partial Fraction Expansion

(((( )))) (((( ))))222

kss

ksf

++++==== 78

80

(((( )))) (((( ))))22 ksbsa

s

csf

++++

++++++++==== 79

(((( )))) (((( ))))bsasksck ++++++++++++==== 222

.;0;,0 02212 skkcsbsac ============++++⇒⇒⇒⇒

which gives, for s=0, c = 1, b = 0 and a = -1,© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 4 Integral Transforms

107

We finally have

F(t) is obtained by calculating the Laplace

transform of the two fractions in the rhs.

and consequently:

Example 3 – Partial Fraction Expansion

(((( )))) (((( )))) ,1

22 ks

s

ssf

++++−−−−==== 81

{{{{ }}}}s

L1

1 ==== 82{{{{ }}}} 22cos kss

ktL++++

====and

(((( )))){{{{ }}}} (((( )))) ktkss

ssf cos1

122

111 −−−−====

++++−−−−

==== −−−−−−−−−−−− LLL 83

We have (See Arfken, Elementary Functions, pages 965-966).

99-- Laplace Transform Laplace Transform

of Derivativesof Derivatives


109

Self Reading

9- Laplace Transform of Derivatives

2) Arfken 6th Edition –Section 15.9 pages: 971-9787th Edition – Section 20.8 pages: 1016-1038.

1) Lecture by Dr. Raed AlmomaniSee

Last Lecture

��

ةةةةررررييييخخخخألألألألاااا ةةةةلقلقلقلقححححللللاااا

© Dr. Nidal M. Ershaidat

Chapter 4

Integral Transforms

Appendix 4-1

Fourier Series

Introduction

© Dr. Nidal M. Ershaidat Phys. 601 Chapter 6: Integral Transforms - Appendix 6-2 Calculating Fourier Series 3

Piecewise regular functionsDefinition:

A piecewise regular function is a function

f(x) which has a finite number of

discontinuities and a finite number of

extrema values over a single interval.

0

Example: (((( ))))

====

++++∈∈∈∈====

nxfor

nnxforxxf

0

1{,}

2 3 4 5 61


0

The sawtooth function

Another Example

(((( ))))

ππππ


The following integral identities (for m,n#0):

represent the orthogonality relationships of the sine and cosine functions.

Orthogonality

(((( )))) (((( ))))∫∫∫∫ππππ++++

δδδδππππ====

20

0

x

x

nmdxxnsinxmsin

(((( )))) (((( ))))∫∫∫∫ππππ++++

δδδδππππ====

20

0

x

x

nmdxxncosxmcos

(((( )))) (((( ))))∫∫∫∫ππππ++++

====

20

0

0

x

x

dxxncosxmsin

(((( ))))∫∫∫∫ππππ++++

====

20

0

0

x

x

dxxmsin (((( ))))∫∫∫∫ππππ++++

====

20

0

0

x

x

dxxmcos

δ δ δ δmn = Kronecker symbol

Harmonic Analysis


The computation and study of Fourier series is

known as harmonic analysis and is extremely

useful as a way to break up an arbitrary

periodic function into a set of simple terms

that can be plugged in, solved individually, and

then recombined to obtain the solution to the

original problem or an approximation to it to

whatever accuracy is desired or practical.

Computation of Fourier Series

Calculation of

Fourier Coefficients


For this purpose we integrate both sides of the Fourier expansion, i.e.:

Computing a0

According to the orthogonality relations we have:

which gives:

Coefficients of Fourier Series

(((( ))))∫∫∫∫ππππ++++

ππππ====

2

0

0

0

1x

x

dxxfa

(((( )))) ∑∑∑∑ ∫∫∫∫∑∑∑∑ ∫∫∫∫∫∫∫∫∫∫∫∫∞∞∞∞

====

ππππ++++∞∞∞∞

====

ππππ++++ππππ++++ππππ++++

++++++++====1

2

1

22

0

2 0

0

0

0

0

0

0

02 n

x

x

n

n

x

x

n

x

x

x

x

dxxninsbdxxncosadxa

dxxf

(((( )))) 00

2

22

0

0

aa

dxxf

x

x

ππππ====ππππ××××====∫∫∫∫ππππ++++

��0

��0


For this purpose we first multiply both sides of the Fourier expansion by cos mx and integrate

over the period [x0,x0+2ππππ], i.e.:

Calculation of an

(((( )))) mn

nmn

x

x

aaxdxmcosxf ∑∑∑∑∫∫∫∫∞∞∞∞

====

ππππ++++

====δδδδ====ππππ 1

20

0

1


(((( ))))

∑∑∑∑ ∫∫∫∫

∑∑∑∑ ∫∫∫∫∫∫∫∫∫∫∫∫

∞∞∞∞

====

ππππ++++

∞∞∞∞

====


++++++++====

1

2

1

22

0

2

0

0

0

0

0

0

0

02

n

x

x

n

n

x

x

n

x

x

x

x

dxxninsxmcosb

dxxncosxmcosaxdxmcosa

xdxmcosxf

�� 0

�� 0

��

mnδδδδππππ


��

mnδδδδππππ

(((( )))) mn

nmm

x

x

bbxdxmsinxf ∑∑∑∑∫∫∫∫∞∞∞∞

====

ππππ++++

====δδδδ====ππππ 1

20

0

1

Now we first multiply both sides of the Fourier expansion by sinmx and integrate over the

period [x0,x0+2ππππ] , i.e.:


Calculation of bn

(((( ))))

∑∑∑∑ ∫∫∫∫

∑∑∑∑ ∫∫∫∫∫∫∫∫∫∫∫∫∞∞∞∞

====

ππππ++++

∞∞∞∞

====


++++

++++====

1

2

1

22

0

2

0

0

0

0

0

0

0

0

inssin

cossinsin2

sin

n

x

x

n

n

x

x

n

x

x

x

x

dxnxmxb

dxnxmxamxdxa

mxdxxf

�� 0

�� 0


The orthogonality relations for a periodic

function on the interval [-ππππ, ππππ] become:

Orthogonality

(((( )))) (((( ))))∫∫∫∫ππππ++++

ππππ−−−−

δδδδππππ==== nmdxxnsinxmsin

(((( )))) (((( ))))∫∫∫∫ππππ++++


δδδδππππ==== nmdxxncosxmcos

(((( )))) (((( ))))∫∫∫∫ππππ++++


==== 0dxxncosxmsin

(((( ))))∫∫∫∫ππππ++++


==== 0dxxmsin (((( ))))∫∫∫∫ππππ++++


==== 0dxxmcos

δδδδmn = Kronecker symbol


The FS coefficients are thus given by the following relations:

n = 1, 2, 3, ∞∞∞∞. Note that we distinguish the coefficient of the constant term a0 by writing it in a special form in order to preserve symmetry with the definitions of an and bn.

Coefficients of Fourier Series

(((( ))))∫∫∫∫ππππ++++

ππππ−−−−ππππ

==== dxxfa1

0

(((( )))) (((( ))))∫∫∫∫ππππ++++


==== dxxnxfan cos1

(((( )))) (((( ))))∫∫∫∫ππππ++++


==== dxxnxfbn sin1


A Fourier series converges to the function equal

to the original function at points of continuity or

to the average of the two limits at points of

discontinuity)

f

If f satisfies the Dirichlet conditions.

Convergence of Fourier Series

(((( )))) (((( ))))

(((( )))) (((( ))))

ππππππππ−−−−====

++++

ππππ


Using the definitions of the coefficients, the Fourier series is written as:

General Form of Fourier Series

(((( )))) (((( )))) (((( ))))

(((( )))) ]

[

∫∫∫∫

∫∫∫∫∑∑∑∑∫∫∫∫ππππ++++

ππππ++++∞∞∞∞

====

ππππ++++

++++++++ππππ

====

2

2

1

2

0

0

0

0

0

02

1

x

x

x

xn

x

x

dunucosufxnsin

dunucosufxncosduufxf

Application


Example: Find the Fourier expansion for the periodic function:

Example

(((( ))))

ππππ≤≤≤≤≤≤≤≤

≤≤≤≤≤≤≤≤ππππ−−−−====

xforx

xforxf

0

00

2ππππ 3ππππππππ0-2ππππ -ππππ-3ππππ


Let’s compute the Fourier coefficients

Solution1

(((( ))))

ππππ≤≤≤≤≤≤≤≤

≤≤≤≤≤≤≤≤ππππ−−−−====

xforx

xforxf

0

00

(((( ))))∫∫∫∫ππππ++++


==== dxxfa1

02

10

1

0

0ππππ

====ππππ

++++ππππ

==== ∫∫∫∫∫∫∫∫ππππ++++


dxxdx

(((( )))) (((( ))))∫∫∫∫ππππ++++


==== dxxncosxfan1

(((( ))))∫∫∫∫ππππ++++

ππππ====

0

1dxxnsinx

(((( ))))∫∫∫∫ππππ++++

ππππ====

0

1dxxncosx

(((( )))) (((( ))))∫∫∫∫ππππ++++


==== dxxnsinxfbn1


Solution2

(((( )))) (((( )))) (((( ))))[[[[ ]]]] (((( ))))[[[[ ]]]]1110110

−−−−−−−−====−−−−ππππ====∫∫∫∫ππππ++++

n

ncosncos

ndxxnsin

n

�� vu

We need to compute the integral

xu ==== (((( )))) (((( )))) (((( ))))xnsinn

dxxncosvdxxncosdv1

========⇒⇒⇒⇒==== ∫∫∫∫,

(((( )))) (((( )))) (((( ))))

−−−−

ππππ====

ππππ==== ∫∫∫∫∫∫∫∫

ππππ++++ππππππππ++++

000

111dxxnsin

nxnsin

n

xdxxncosxan

where we used the integration by parts:

��

∫∫∫∫−−−− duv��

∫∫∫∫ dvu

(((( )))) 00

====

π

xnsinn

x


Thus we have:

If n is even

If n is odd

an

(((( )))) (((( ))))[[[[ ]]]]nnn

dxxncosxa 1111

2

0

−−−−−−−−ππππ

====ππππ

==== ∫∫∫∫ππππ++++

ππππ

====2

2

0

n

an


bnNow we compute bn , We proceed as for an, i.e.

(((( ))))∫∫∫∫ππππ++++

ππππ====

0

1dxxnsinxbn

�� vu

(((( )))) (((( ))))∫∫∫∫ππππ++++ππππ

ππππ++++

ππππ−−−−====

00

1dxxncos

nxncos

n

x

��

∫∫∫∫−−−− duv��

∫∫∫∫ dvu

(((( )))) (((( )))) (((( ))))nnn

ncosnn 1

11++++

−−−−====

−−−−−−−−====

ππππ−−−−====

(((( ))))

ππππ++++ ∫∫∫∫

ππππ++++

0

11dxxncos

n(((( ))))

ππππ

−−−−

ππππ====

0

1xncos

n

x


The Fourier series for the function

Solution

(((( ))))

ππππ≤≤≤≤≤≤≤≤

≤≤≤≤≤≤≤≤ππππ−−−−====

xforx

xforxf

0

00

2ππππ 3ππππππππ0-2ππππ -ππππ-3ππππ

(((( ))))(((( )))) (((( ))))

nxsinn

nxcosn

xfn n

nn

∑∑∑∑ ∑∑∑∑∞∞∞∞

====

∞∞∞∞

====

++++−−−−

++++−−−−−−−−

++++ππππ

====1 1

1

2

111

4

][

is


The sawtooth function

Another Example

(((( ))))

ππππ


Solution4and the integrals

(((( )))) (((( )))) (((( ))))

(((( )))) 1

000

1

1

++++

ππππ−−−−ππππ−−−−ππππ−−−−


====ππππππππ

−−−−====

++++

−−−−==== ∫∫∫∫∫∫∫∫

n

nncos

n

dxxncosn

xncosn

xdxxnsinx

(((( )))) (((( )))) (((( ))))

(((( )))) 1000

1

1

++++

ππππππππππππ


====ππππππππ

−−−−====

++++

−−−−==== ∫∫∫∫∫∫∫∫

n

nncos

n

dxxncosn

xncosn

xdxxnsinx


Useful Integrals

(((( )))) 00

====∫∫∫∫ππππ++++

dxxncos

(((( )))) 00

====∫∫∫∫ππππ−−−−

dxxncos

(((( )))) (((( ))))(((( ))))nn

dxxnsin 111

0

−−−−−−−−====∫∫∫∫ππππ++++

(((( )))) (((( ))))(((( ))))1110

−−−−−−−−====∫∫∫∫ππππ−−−−

n

ndxxnsin

(((( )))) (((( ))))(((( ))))nn

dxxncosx 111

2

0

−−−−−−−−====∫∫∫∫ππππ−−−−

(((( )))) (((( ))))(((( ))))1112

0

−−−−−−−−====∫∫∫∫ππππ++++

n

ndxxncosx

(((( )))) (((( ))))n

dxxnsinxn ππππ

−−−−====++++


∫∫∫∫1

0

1

(((( )))) (((( ))))n

dxxnsinxn ππππ

−−−−====++++

ππππ

∫∫∫∫1

0

1


(((( )))) (((( )))) (((( )))) (((( ))))∫∫∫∫∫∫∫∫ππππ++++



++++ππππ++++ππππ

====

0

0

11dxxncosxdxxncosx

an

(((( )))) (((( ))))∫∫∫∫ππππ++++


==== dxxncosxfan1

(((( )))) (((( ))))

(((( )))) (((( )))) (((( ))))∫∫∫∫∫∫∫∫

∫∫∫∫∫∫∫∫ππππ++++ππππ++++

ππππ−−−−ππππ−−−−


++++ππππ

++++ππππππππ

++++ππππ

====

00

00

11

11

dxxncosdxxncosx

dxxncosdxxncosx

(((( ))))(((( )))) (((( ))))(((( ))))0

1110

11122

−−−−−−−−−−−−

ππππ++++++++

−−−−−−−−

ππππ====

nn

nn

0====


(((( )))) (((( )))) (((( )))) (((( ))))∫∫∫∫∫∫∫∫ππππ++++



++++ππππ++++ππππ

====

0

0

11dxxnsinxdxxnsinx

bn

(((( )))) (((( ))))

(((( )))) (((( )))) (((( ))))∫∫∫∫∫∫∫∫

∫∫∫∫∫∫∫∫ππππ++++ππππ++++

ππππ−−−−ππππ−−−−


++++ππππ

++++ππππππππ

++++ππππ

====

00

00

11

11

dxxnsindxxnsinx

dxxnsindxxnsinx

(((( )))) (((( )))) (((( )))) (((( ))))nnnn

nnnn 11111111

++++++++++++++++−−−−

++++−−−−

++++−−−−

++++−−−−

====

(((( )))) 114 ++++

−−−−====n

n

(((( )))) (((( ))))∫∫∫∫ππππ++++


==== dxxnsinxfbn1

Now we compute bn, We proceed as for an, i.e.


The Fourier series for the function

The Series

(((( ))))(((( ))))

∑∑∑∑∞∞∞∞

====

++++−−−−

++++ππππ====1

11

4n

n

xnsinn

xf

is

(((( ))))

ππππ


Using the definitions:

Substituting in the definition of Fourier series

we get:

Complex Form of Fourier Series

(((( )))) ,2

sini

eexn

xnixni −−−−−−−−==== (((( )))) ,

2cos

xnixni eexn

−−−−++++====

(((( )))) ∑∑∑∑∑∑∑∑∞∞∞∞

====

∞∞∞∞

====

++++++++====11

0 sincos2 n

n

n

n xnbxnaa

xf

(((( )))) [[[[ ]]]] [[[[ ]]]]∑∑∑∑∑∑∑∑∞∞∞∞

====

−−−−∞∞∞∞

====

−−−− −−−−++++++++++++====11

0

222 n

xnixnin

n

xnixnin eei

bee

aaxf

[[[[ ]]]] [[[[ ]]]]∑∑∑∑∑∑∑∑∞∞∞∞

====

−−−−∞∞∞∞

====

++++++++−−−−++++====11

0

2

1

2

1

2 n

xni

nn

n

xni

nn ebiaebiaa


can be written in the form

The previous equation

The limit n = - ∞∞∞∞ takes into account the terms e-inx

Complex Form of Fourier Series

(((( )))) [[[[ ]]]] [[[[ ]]]]∑∑∑∑∑∑∑∑∞∞∞∞

====

−−−−∞∞∞∞

====

−−−− −−−−++++++++++++====11

0

222 n

xnixnin

n

xnixnin eei

bee

aaxf

[[[[ ]]]] [[[[ ]]]]∑∑∑∑∑∑∑∑∞∞∞∞

====

−−−−∞∞∞∞

====

++++++++−−−−++++====11

0

2

1

2

1

2 n

xni

nn

n

xni

nn ebiaebiaa

(((( )))) ∑∑∑∑∞∞∞∞

∞∞∞∞−−−−====

====n

xni

n ecxf


For n > 0, we have

In order to find cn we multiply both sides of the

equation:

by and integrate over the interval [x0 ,x0+ππππ], i.e.xmie −−−−

The Complex Coefficients

2

0

0

ac ====

(((( )))) ∑∑∑∑∞∞∞∞

∞∞∞∞−−−−====

====n

xni

n ecxf

nnn biac −−−−====*cbiac nnnn ====++++====−−−−

(((( )))) (((( ))))∑∑∑∑ ∫∫∫∫∫∫∫∫∞∞∞∞

∞∞∞∞−−−−====

ππππ++++

−−−−

ππππ++++

−−−− ====n

x

x

xmni

n

x

x

xmi dxecdxexf

22 0

0

0

0

∑∞

∞−=

δπ=

n

mnnc2

(((( ))))∫∫∫∫ππππ++++

−−−−

ππππ====⇒⇒⇒⇒

20

02

1x

x

xmi

m dxexfc

Some Properties and

Uses of Fourier Series


The problem, as we have seen, is to determine

the number of coefficients one should calculate

in order to get as close as possible to the

shape of the periodic function f(x).

The rate of convergence of the series gives an

idea when to stop.

Convergence


Convergence - Properties

1- If the function f(x) is discontinuous , then

some of the Fourier coefficients will vary as

1/n. The convergence in gene

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