1

1 Fourier Series

Periodic Function A function is called “periodic” function of period p if f(x+np) = f(x), x, n >0, n: integer.

If both f and g are functions of period p, then af + bg is also a function of period p.

Fourier Series Period functions of period 2 can be represented in terms of trigonometric series as

1

0 sincos)(n

nn nxbnxaaxf (1)

If the series converges, it is called “Fourier series” of f(x), where a0, an, bn are called “Fourier coefficients”.

a0, an, bn are given by the “Euler formulas”:

,2,1,sin

cos)(

1;)(

2

10

ndx

nx

nxxf

b

adxxfa

n

n

(2)

Derivation of the Euler Formulas

Preliminary Orthogonality of trigonometric system

(1) nmmn

mnmxdxnx

0

coscos

Proof

mn

mndxxmnxmnmxdxnx

0)cos()cos(

2

1coscos

(2) nmmn

mnmxdxnx

0

sinsin

Proof

mn

mndxxmnxmnmxdxnx

0)cos()cos(

2

1sinsin

(3) 0cossin

mxdxnx

Proof 0)sin()sin(2

1cossin

dxxmnxmnmxdxnx

Now, integrating both sides of (1) from – to yields

0

1

0 2sincos)( adxnxbnxaadxxfn

nn

dxxfa )(

2

10

Likewise, multiplying both sides of (*) by cos nx and integrating both sides from – to yields

n

m

mm adxmxbmxanxnxanxdxxf

1

0 sincoscoscoscos)(

Likewise, n

m

mm bdxmxbmxanxnxanxdxxf

1

0 sincossinsinsin)( .

Clearly, a0 = 0 and

EX1

Odd function

“Kronecker Delta”

2

evenn

oddnnk

:0

:/4

xxx

kxf 5sin

5

13sin

3

1sin

4)(

.

NOTE The integral range does not need to be from – to , it can be “arbitrary” as long as it covers an entire

period, e.g., 0 to 2 and vice versa.

2 Functions of any period p = 2L

Assume g(v) is a function of period 2, then

1

0 sincos)(n

nn nvbnvaavg .

Now, let x = Lv/ (or v=x/L) (called “change of scale”), then let

nLx

Lgnx

Lgnvgvg 22)2()(

Let g(v) = f(x), then f(x+2nL) = f(x), i.e., periodic function of 2L.

1

0 sincos)(n

nnL

xnb

L

xnaaxf

with ,2,1,

sin

cos

)(1

;)(2

10

ndx

L

xn

L

xn

xfLb

adxxf

La

L

Ln

nL

L

(3)

a0=2k/4=k/2, bn=0

evenn

noddnnk

:0

12,:/2)1(

2/xv

Odd functions include only sine terms

Period 2L with respect to x

EX2

EX3 EX1 with period changed to 4

Even function

Even functions include only cosine terms

EX4 half-wave rectifier

Clearly, n=1 a1=0

3

21

2

1

2

1

2

2,:;0,:

n

E

nn

Eaevennaoddn nn

;

else

nEtdtntEbn

0

12/sinsin

/

0

3 Even, Odd Functions & Cosine, Sine Series

EVEN: fe(-x)=fe(x), bn=0,

1

0 cos)(n

neL

xnaaxf

with ,2,1,cos)(2

;)(1

000 ndx

L

xnxf

Ladxxf

La

L

en

L

e

ODD: fo(-x)=-fo(x), an=0,

1

sin)(n

noL

xnbxf

with ,2,1,sin)(

2

0 ndx

L

xnxf

Lb

L

on

.

{Even, Odd} functions become Fourier {Cosine, Sine} series, respectively. Thus, for any function f,

odd

n

n

even

n

noeL

xnb

L

xnaaxfxfxf

11

0 sincos)()()(

EX4b Consider w(t) = u(t)-(E/2)sin t, one can see that w(t) = (E/2)sin t, p = / becomes an even function.

It follows that ,2,1,2cossin2

2;sin

2

/

0

/

00 ntdtnt

Eatdt

Ea n

Consider f(x) = f1(x) + f2(x) where f1(x) = x, p=2 (f1 is an odd function), f2(x) = , then

Condition for Existence and Convergence

Existence L

Ldxxf )( (Absolutely integrable)

Convergence

EX5 Sawtooth wave

4

PARSEVAL’s IDENTITY states that

1

222

0

22)(

1

n

nn

L

Lbaadxxf

L if an and bn are the Fourier

coefficients corresponding to f(x) and if f(x) satisfies the Dirichlet conditions. For the proof, consider

dxxfL

xnb

L

xnaadxxf

L

Ln

nn

L

L)(sincos)(

1

0

2

, then use Euler’s formula.

Gibbs Phenomenon

The sinusoidal components of the signal that occur at

multiples of the fundamental frequency are called

harmonics. In general, for well-behaved (continuous)

periodic signals, a sufficiently large number of harmonics can

be used to approximate the signal reasonably well. For

periodic signals with discontinuities, however, such as a

periodic square wave, even a large number of harmonics will

not be sufficient to reproduce the square wave exactly,

resulting in the appearance of so-called “overshoot”. This

effect is known as Gibbs phenomenon and it manifests itself

in the form of ripples of increasing frequency and closer to

the transitions of the square signal. An illustration of Gibbs

phenomenon is shown in the figure on the right. The figure shows the result of adding 5, 9 and 13 harmonics.

The overshoots tend to shift toward the discontinuities, but their magnitude do not change much.

EX4c (Half-wave rectifier)

LLp

tL

Ltttu ,2,

00

0sin)(

ttt

Et

Etu

6cos

35

24cos

15

22cos

3

21sin

2)(

…(*)

LLp

tL

Ltttu ,2,

00

0cos)('

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Original

N=5

N=9

N=13

5

else

nEtdtntEatdtEa n

0

12/coscos;0cos

2

/

0

/

00

/

0

/

0

2

/

0

2/

0

)1(

)1cos(

)1(

)1cos(

)1sin()1sin(2

sincos

n

tn

n

tnE

dttntnE

tdtntEbn

1

2

1

2

1

2

2,:;0,:

2

n

nE

nn

Ebevennboddn nn

ttt

Et

Etu

6sin

35

64sin

15

42sin

3

22cos

2)('

EX Consider f(x) = x on the interval [-,] Fourier series:

1

1

sin)1(2

)(n

n

nxn

xf …(*)

From Parseval’s formula: 6

1

3

2

3

)(

3

114

1)1(2 2

12

233

12

2

1

21

nnn

n

nndxx

n

EX Consider f(x)=x,22

)()(22

xududuufxg

xx

Fourier series:

12

2

12

22

cos)1(2

3cos

)1(2

26)(

n

n

n

n

nxn

nxn

xg

Integrating term by term of (*) in the previous example yields

3cos

)1(2

2cos

)1(2

cos)1(2sin

)1(2)(

2

12

12

12

1

1

1

1

n

n

nn

n

n

xn

n

xn

x

nxnn

nxnn

nu

nnudu

nduuf

4 Application of Fourier Series to ODE for steady-state solution

Consider ODE when the input (external force) becomes periodic, e.g., my''+cy'+ky = r(t), where r(t) is periodic.

The idea is to represent r(t) in terms of Fourier series, and then solve the ODE for each term. For example,

Alternative Approach Using the transfer function Q(s) = (s2+0.05s+25)-1. For sinusoidal steady state response,

2

22

212

05.025

05.0252505.0)(

nn

njnnjnjnQ

;

nt

D

nnt

D

n

nejnQ

ny

nn

jnt

n sin05.0

cos254

)(Re4 2

22

Example

First Fourier series

Next solve

with

nth term becomes

6

The total solution is given by y = y1 + y3 + y5 + …

5 Application to PDE

(a) Heat Equation

Consider the heat conduction along a heat-conducting homogeneous rod with temperature distribution u(x, t).

Assumptions for one dimensional (1-D) case:

1. homogeneous rod with uniform cross section A and constant density ρ [kg/m3], specific heat [J/kgK],

thermal conductivity κ [W/mK]

2. insulated laterally so heat flows only in x-direction only

3. temperature is constant at all points of a cross section

The heat equation in this case is given by

2

2

22 );1()0,0(, cLxt

x

uc

t

u [m2/s] (for three dimensional (3-D) case: )0(,22

tuc

t

u)

with initial condition: u(x, 0) = f(x) (0 ≤ x ≤ L)…(2),

and boundary conditions : u(0, t) = 0, u(L, t) = 0, (t ≥ 0)… (3)

By method of separation of variables: u(x, t) = F(x)G(t)

tpc

tpc

xt CetGGpcGp

CtGG

CetGGpcGp

F

F

Gc

GGFcGF

22

22

)(

)(00

)(''

'')1(222

222

dependence-dependence-

2

2

)5(0),4(0'' 222 GpcGFpF

0sin,00)()0(0)()(,0)()0(sincos)()4()3(

pLALFFtGLFtGFpxBpxAxF

,2,1,sin)(,2,1, nL

xnxFn

L

npnpL n

,2,1,)(,0)3(2

2

neBtGL

cnGG

t

nnnnn

Eigenfunction ,2,1,sin)()(),(2

neL

xnBtGxFtxu

t

nnnnn

with corresponding eigenvalue n

dxL

xnxf

LB

L

xnBxfxue

L

xnBtxutxu

L

n

n

n

n

t

n

n

nn

0

1

)2(

11

sin)(2

sin)()0,(sin),(),(2

.

(b) Wave Equation

Consider a 1-D wave equation, which represents an up-down movement of a “thin” string.

);1()0,0(,2

22

2

2

Lxtx

uc

t

u

(for three dimensional (3-D) case: )0(,22

2

2

tuc

t

u)

with initial conditions (initial deflection, velocity): u(x, 0) = f(x) …(2a), ∂u(x,0)/∂t = g(x)…(2b), (0 ≤ x ≤ L)

and boundary conditions : u(0, t) = 0, u(L, t) = 0, (t ≥ 0)… (3)

By method of separation of variables: u(x, t) = F(x)G(t)

jpxjpx

pxpx

xt BeAexFFpFp

babaxxFF

BABeAexFFpFp

F

F

Gc

GGFcGF

)(''

0)(0''0

0)(''''

'')1(22

)3(

)3(22

dependence-dependence-

2

2

)5(0),4(0'' 222 GpcGFpF

0sin,00)()0(0)()(,0)()0(sincos)()4()3(

pLALFFtGLFtGFpxBpxAxF

,2,1,sin)(,2,1, nL

xnxFn

L

npnpL n

,2,1,sincos)(,0)3( 2 ntBtAtGL

cnGG nnnnnnn

A constant is required for

the RHS in order to satisfy

for all t and x.

Fourier sine series

7

Eigenfunction ,2,1,sinsincos),( nL

xntBtAtxu nnnnn

with corresponding eigenvalue n

11

sinsincos),(),(n

nnnn

n

nL

xntBtAtxutxu

dxL

xnxf

LA

L

xnAxfxu

L

n

n

n

a

0

1

)2(sin)(

2sin)()0,(

dxL

xnxg

LB

L

xnBxg

x

u L

nn

n

nn

t

b

0

10

)2(sin)(

2sin)(

6 Complex Fourier Series

Using Euler’s Formula, given by 2

sin;2

cosj

eex

eex

jxjxjxjx

, then

n

xL

nj

n

n

xL

nj

nn

n

xL

nj

nn

n

xL

njx

L

nj

n

xL

njx

L

nj

n

n

nn

ecejba

ejba

a

j

eeb

eeaax

L

nbx

L

naaxf

11

0

1

0

1

0

2222

22sincos)(

where c0 = a0, cn=(an-jbn)/2, c-n=cn*, n=1,2,…

L

L

xL

njL

L

L

L

nnn

L

Ldxexf

Ldx

L

xnxf

L

jdx

L

xnxf

L

jbacdxxf

Lac

)(2

1sin)(

2cos)(

2

1

2;)(

2

100

EX4c Half-wave rectifier with E=1. Note that L = .

/

0

)1()1(/

0

/

0 422sin

2dtee

jdte

j

eedttec tnjtnjtjn

tjtjtjn

n

/

0

)1(/

0

)1(/

0

)1()1(

)1()1(44 nj

e

nj

e

jdtee

jc

tnjtnjtnjtnj

n n: odd, n≠1 cn=0,

n = 1; 44

14

/

0

2

1

j

jdte

jc tj

; a1 = 0, b1 = ½

n:even 1

12;

1;

1

11

1

1

1

1

2

1

)1(

2

)1(

2

4 2002

nccaca

nnnnjnjjc nnnn

7 Fourier Transform

Consider a periodic function fL(x),

n

xL

nj

nL ecxf

)( , p=2L. Now, set

w

LLwww

L

nw nnn

1,, 1 .

Let L∞, assuming )(lim)( xfxf LL

is “absolutely integrable”, i.e.,

dxxfdxxfdxxf

b

baa)(P.V.)(lim)(lim

0

0

exists.

then w0, wnw (becomes “continuous”), the infinite sum of the Fourier series of fL(x) becomes infinite

integral as wdw in integrand:

n

xjwL

L

vjw

LL

n

xL

njL

L

vL

nj

LL

LL

nn edvevfw

edvevfL

xfxf )(2

lim)(2

1lim)(lim)(

dwedvevfedvevf

dwxf jwx

wf

jwvjwxjwv

)(ˆ

)(2

1)(

2)(

P.V. stands for Cauchy’s principal value

8

In engineering, Fourier Transform pair :

dwewfxffdxexfwff jwxjwx )(ˆ2

1)(]ˆ[)()(ˆ][ 1

FF

In mathematics, generally define

dwewfxfdxexfwf jwxjwx )(ˆ2

1)()(

2

1)(ˆ

Note also that conventionally variable pairs (x, w), (t (time), (frequency)) are used.

EX1 unit gate function

5.00

5.05.0

5.01

)()(

x

x

x

xrectxf

)2/( sinc2/

)2/sin()2/sin(2)(ˆ

2/2/2/2/5.0

5.0

5.0

5.0w

w

w

w

w

jw

ee

jw

ee

jw

edxewf

jwjwjwjwjwxjwx

.

EX2 0,0

0)(

aelse

xexf

ax

=e-axu(x), u(x) : unit step function

jwajwa

edxedxeewf

xjwaxjwajwxax

1

)()(ˆ

0

)(

0

)(

0.

Some Properties

Linearity : F[af + bg] = aF[f] + bF[g]

Symmetry or Duality : )(2)(ˆ)(ˆ)( wfxfwfxf

Proof

dxexfwfdxexfwfdwewfxf jwxjwxxwwxjwx )(ˆ)(2)(ˆ2

1)()(ˆ

2

1)( ,

.

Scaling :

a

wf

aaxfwfxf ˆ

||

1)()(ˆ)(

Proof a>0:

a

wf

advevf

aa

dvevfdxeaxfaxf ajwvajwv

axvjwx ˆ

||

1)(

1)()()( //

F

a<0, a=-b, b>0:

a

wf

advevf

bb

dvevfdxeaxfaxf ajwvbjwv

bxvjwx ˆ

||

1)(

1)()()( /)/(

F

EX1b gate function

2/0

2/5.0

2/1

)(

ax

ax

ax

a

xrectxf

)2/( sinc2/

)2/sin()2/sin(2)(ˆ

2/2/2/2/2/

2/

2/

2/waa

wa

waa

w

wa

jw

ee

jw

ee

jw

edxewf

jwajwajwajwaa

a

jwxa

a

jwx

Time shift : shift phase

00ˆ)()(ˆ)(

jwxewfxxfwfxf

Frequency shift : 0

modulation

ˆ)()(ˆ)( 0 wwfexfwfxfxjw

Special Cases

Delta Function :

1)(ˆ);()( dxexwfxxf jwx 1)( x

Furthermore,

dwex jwx

2

1]1[)( 1

F ;

Likewise, from duality )(2)(211)( wwx

dxew jwx

2

1]1[

2

1)( F

9

Harmonics : 0

)(2)(ˆ)( 000 wwdxedxeewfexf

xwwjjwxxjwxjw

Unit step function : f(x) = u(x) [= 1 (x≥0) and = 0 (x < 0)]

Let )(lim)(0

xuexu ax

a

,

2200000lim

1limlim)](lim[)(ˆ

wa

jwa

jwadweexuewu

aa

jwxax

a

ax

aF

Imaginary part : jwwa

jw

a

1lim

220

Real part : )(lim;tan220

1

22w

wa

a

a

wdw

wa

a

a

jwwwu /1)()(ˆ

Parseval’s Theorem :

dwwfdxxf

22)(ˆ

2

1)(

dwwfwfdwdvwvvfwfdwdvdxevfwf

dxdvevfdwewfdxxfxfdxxf

xwvj

xf

jvx

xf

jwx

)(ˆ)(ˆ2

1)()(ˆ

2

1)(ˆ

2

1)(ˆ

2

1)(ˆ

)(ˆ2

1)(ˆ

2

1)()()(

**)(*

)(

*

)(

*2

*

Significance Total energy in x (or time) domain = total energy in transform (or frequency) domain

Fourier transform of derivative : )(ˆ)(')(ˆ)( wfjwxfwfxf

Proof )(ˆ)()()(')]('[

0)(

wfjwdxexfjwexfdxexfxf jwx

xfx

jwxjwx

F

It follows that )(ˆ)(ˆ)]('[)('')(ˆ)( 2 wfwwfjwjwxfjwxfwfxf F

EX Consider 2

)( xxexf ; ][2

])'[(2

1)(ˆ)'(

2

1 2222 xxxx ejw

ewfexe FF

44)

2(

44)

2(

2

2

22

222

22

][w

z

wjwx

wwjwx

jwxxx edzeedxeedxeedxeee

F

Thus, 4

2

2

2][

2)(ˆ

w

x ejw

ejw

wf

F

Convolution

dppxgpfxgf )()())(*( gfgfwgxgwfxf ˆˆ*)(ˆ)(),(ˆ)(

)(ˆ)(ˆ)()(ˆ)()()()(

)()()()(]*[

)( wfwgdppfewgdppfedqeqgdppfdqeqg

dppfdxepxgdxedppxgpfgf

jwpjwpjwqpqjwqpx

jwxjwx

F

Duality of convolution : gffgwgxgwfxf ˆ*ˆ2

1)(ˆ)(),(ˆ)(

Exercise Prove this property.

EX Let f(x) = g(x) = rect(x), then for |x| > 1, f*g=0, -1< x<0, 1)(*5.0

5.0

xdpxgf

x

, 0<x<1,

xdpxgfx

1)(*

5.0

5.0. 321

1

0

0

1)1()1(]*[ IIIdxexdxexgf jwxjwx

F

10

w

w

jw

ee

jw

edxeI

jwjwjwxjwx sin

1

1

1

11

;2

0

1

0

1

0

12

11

w

e

jw

edxe

jwjw

xedxxeI

jwjwjwx

jwxjwx

2

1

0

1

0

1

03

11

w

e

jw

edxe

jwjw

xedxxeI

jwjwjwx

jwxjwx

2sinc

2sin

)2/(

1

2sin

4

2sin2

1

12

111]*[

2

2

2

2

2

2

2

22/2/

2222

ww

w

w

w

wj

w

eew

eeww

e

jw

e

w

e

jw

e

jw

eegf jwjwjwjw

jwjwjwjwjwjw

F

Using the “even” function property yields

2

2

2

1

0

2

1

0

1

0

1

0

1

0

)2/(sin4cos22cos2sin

2sin2sin2cos)1(2]*[

w

w

w

w

w

wxwxdx

ww

wxx

w

wxwxdxxgf

F

Alternatively, since )2/(sinc]*[),2/(sinc][ 2 wrectrectwrect FF .

8 Application of Fourier Transform to PDE

(a) Heat Equation Consider

)1(),0(,2

22

xt

x

uc

t

uwith initial condition: u(x, 0) = f(x) …(2),

Recall ][)](''[],[)]('[,)()]([)(ˆ 2 uwxuujwxudxexuxuwu jwxFFFFF

Take Fourier transform of (1) yields

uwcx

uc

t

uFFF

22

2

22

; ][][ u

tdxue

tdxe

t

u

t

u jwxjwxFF

ODEPDE

twuwct

twuuwc

t

u),(ˆ

),(ˆ 2222

FF …(3)

Take Fourier transform of (2) yields )(ˆ)0,(ˆ wfwu …(4)

From (3) twcCetwu

22

),(ˆ ; from (4) )(ˆ)0,(ˆ wfCwu twcewftwu

22

)(ˆ),(ˆ

Take inverse Fourier transform yields

dweewfdwetwutwutxu jwxtwcjwx 22

)(ˆ2

1),(ˆ

2

1)],(ˆ[),( 1

F

(b) Wave Equation Consider

);1(),0(,2

22

2

2

xt

x

uc

t

usubject to

with initial conditions : u(x, 0) = f(x) …(2a), ∂u(x,0)/∂t = 0…(2b),

and boundary conditions : u 0, ux 0, as |x|∞ (t ≥ 0)… (3)

where f(x) assumed to have Fourier transform

uwcucuu xxttttˆˆ 222 FF uwcutt

ˆˆ 22 cwtwBcwtwAtwu sin)(cos)(),(ˆ

For t=0, )(ˆ)()0,(ˆ wfwAwu , 0)()0,(ˆ wcwBwut , Thus 2

)(ˆcos)(ˆ),(ˆjcwtjcwt ee

wfcwtwftwu

)()(2

1)],(ˆ[),(

shiftfrequency 1 ctxfctxftwutxu F “d’Alembert’s solution”

9 Poisson’s Sum Formula

11

Consider a “periodic” impulse train with period p=2L,

k

kLxxs )2()( (Called comb, Ш (Shah) function).

Fourier series of s(x) is given by

LdxekLx

Ldxexs

Lc

L

L

xL

nj

k

L

L

xL

nj

n2

1)2(

2

1)(

2

1

and

kn

xL

nj

kLxeL

xs )2(2

1)(

Now, consider Fourier transform of s(x)

nnn

xL

nwj

jwx

n

xL

nj

L

nw

LL

nw

Ldxe

Ldxee

Lws )()(

2

2

2

1

2

1)(ˆ

)(

Let

kk

kLxgkLxxgxsxgxh )2()2(*)()(*)()( ”periodic version” of g(x)

FT of h(x) :

nnn L

nw

L

ng

LL

nwwg

LL

nw

Lwgwswgwh )()(ˆ)()(ˆ)()(ˆ)(ˆ)(ˆ)(ˆ

)(ˆ wg sampled at n/L. Take inverse FT of )(ˆ wh yields

n

xL

nj

k

eL

ng

LkLxg

)(ˆ2

1)2( Poisson’s Sum Formula

Fourier series vs Fourier transform

Since

n

xL

nj

n

xL

nj

n

k

eL

ng

LeckLxgxh

)(ˆ2

1)2()( , )(ˆ

2

1

L

ng

Lcn

.

For an example, consider a half-wave rectifier. First, let g(t) = sin t[u(t)-u(t-/)],

/

0

)()(/

0

/

0 2

1

2sin)(ˆ dtee

jdte

j

eedttewg twjtwjjwt

tjtjjwt

)(

1

)(

1

2

1

)()(2

1 /)(/)(/

0

)(/

0

)(

wj

e

wj

e

jwj

e

wj

e

j

wjwjtwjtwj

)()(2

1 2/)(2/)(2/)(

2/)(2/)(2/)(

wj

eee

wj

eee

j

wjwjwj

wjwjwj

)(

2/)sin(

)(

2/)sin(1 2/)(2/)(

w

we

w

we

j

wjwj

2/)(

2/)sin(

2/)(

2/)sin(

2

1 2/)(2/)(

w

we

w

we

j

wjwj

2

)(sinc

2

)(sinc

2

1 2/)(2/)( we

we

j

wjwj

Alternatively, let

)2/(

2

)2/(sin)(

trect

j

eetrectttg

tjtj

)()(2

2

2

wwjj

ee tjtj

F ;

2sinc

)2/( 2/ we

trect jw

F

Using gffg ˆ*ˆ2

1

yields

2

)(sinc

2

)(sinc

2

1)(ˆ 2/)(2/)( w

ew

ej

wg wjwj

Now, let

LkLtgkLttgtstgthkk

;)2()2(*)()(*)()(

12

2

)(sinc

2

)(sinc

2

1

2)(ˆ

2)(ˆ

2

1 2/)(2/)( ne

ne

jng

L

ng

Lc njnj

n

2

)1(sinc

2

)1(sinc

4

12

)1(

2

)1(

n

en

ej

nj

nj

n=0,

1

2/2/4

1

2sinc

2sinc

4

122

0

jj

jee

jc

jj

4

1

2

2sinc0 sinc

4

12

2

1j

ej

cj

, n: odd, n≠1 cn=0,

n:even: 1

11

)1(

2

)1(

2

4

1

2

)1(sinc

2

)1(sinc

4

12

2

)1(

2

)1(

nn

j

n

j

j

ne

ne

jc

nj

nj

n

Discrete (Sampled) Signal

Let fC(x) be a “continuous” signal, and fS(x) be a “sampled” signal with sampling period p=2L.

k

C

k

CCS kLxxfkLxxfxsxfxf )2()()2()()()()(

Using gffg ˆ*ˆ2

1

,

n

C

n

CCSL

nwwf

LL

nw

Lwfwswfwf )(*)(ˆ

2

1)(*)(ˆ

2

1)(ˆ*)(ˆ

2

1)(ˆ

which means )(ˆ wfC is repeated every /L, in other words, )(ˆ wfS is a “periodic” version of )(ˆ wfC . This is

called Discrete Time Fourier Transform (DTFT).

Summary

x (or time)

domain

Example Waveform Transform

(frequency)

domain

Example Spectrum

Continuous,

aperiodic

Continuous,

aperiodic

Continuous,

periodic

2L

Discrete (Fourier

series)

/L Discrete (sampled

signal)

2L

Continuous,

periodic (Periodic

version of Fourier

transform)

/L

10 Laplace Transform and Fourier Transform

Laplace to Fourier: Recall that Laplace transforms exist only for functions that satisfy the “growth restriction”

condition, which requires region of convergence (ROC) consideration (i.e., region of s where the transform

exists). Now, the conventional Laplace transform is defined for f(t), t≥0, as

0

)()( dtetfsF st,

which is sometimes called “unilateral” Laplace transform. Since f(t) is assumed to be 0 for t<0,

dtetfsF st)()( .(called “bilateral” Laplace transform)

Comparing to the Fourier transform for (t,) pair given by

dtetff tj )()(ˆ ,

13

it follows that by substituting s=j in F(s), provided that s=j (imaginary axis) is in the ROC, one can obtain

the Fourier transform, i.e., )()(ˆ jFf .

EX1 Consider f(t)=e-at, a>0, then F(s)=(s+a)-1. 1)()(ˆ

ajjFf .

EX2 Consider f(t) = sin t[u(t)-u(t-/)]. Note that w is used as the variable for frequency domain here.

/

2222

/2/2/2/

2/2/

2/)(2/)(

12

2

111

2

2/)(

2/cos

2/)(

2/cos

2

1

2

)(sinc

2

)(sinc

2

1)(ˆ

jwjwjwjw

jw

jwjw

wjwj

eww

e

ww

eee

w

wje

w

wje

j

we

we

jwf

)()(sin)(sin)(sin)()(sin)(

22

tut

sttuttutututsF LLLL

s

ess

2222

wjwj

ew

eww

jwFwf 1)()(ˆ222222

NOTE u(t)1/s, but )/(1)(ˆ ju ! This is because s > 0 (Re s > 0) for u(t).

Fourier to Laplace: Fourier transforms exist for functions that are “absolutely integrable”. Now, consider a

the Fourier transform of f(t) which satisfies f(t)=0 for t<0. It follows that

0

)()(ˆ dtetff tj .

This integral does not exist unless f(t) is absolutely integrable. Thus, an additional term is introduced to

guarantee the existence of integral, namely

)()()()(ˆ00

sFdtetfdteetff sttjt

.

Therefore, s=+j becomes complex and the ROC for s needs to be considered.

Inverse Laplace Transform

Recall the growth restriction dictates that Laplace transform F(s) exists if |f(t)|<Mekt, for t>0, ∃k,M. Can

choose >k such that 0)( ttkMe . Therefore, for f(t)=0, t<0,

ddefeetfe

tfe

jtjt

t

)]([

0)(

2

1)(

F

.

Moving the et term to the right hand side yields

j

j

sF

sstjs

jtjt dsdefej

ddeefeetf

)(

00)(

2

1)(

2

1)( .

Thus,

j

j

stdsesFj

tf

)(

2

1)( ”Bromwich integral”

where kk

sRemax , sk denotes the kth pole of F(s), i.e., > Maximum of Real part of poles.

Brief Introduction of Poles Let F(s)=A(s)/B(s), and sk denote a pole of F(s), then B(sk)=0.

(Recall eat(s-a)-1, and a becomes a pole)

EX 326

)(2

3/1

1

2/1

1

6/1

22

1)(

2

23

ttt eeetf

sssssssF

Thus, s>1 is the ROC, and it follows that > 1.

14

11 Fourier Cosine and Sine Transform

Recall that

0

0

00

)(

)sin)(cos)(

sinsincoscos)(1

)(cos2)(cos,0)(sin)(cos)(1

)(sin)(cos)(2

1

)(2

1)(

2

1)(

dwwxwBwxwA

dwdvwvwxwvwxvf

dwvxwdwvxwdwvxwdwdvvxwvf

dwdvvxwjvxwvf

dwdvevfdwdvevfexf vxjwjwvjwx

where

wvdvvfwBwvdvvfwA sin)(

1)(;cos)(

1)(

If f(x) is an “even” function, B(w)=0.

00

cos)()(;cos)(2

)( wxdwwAxfwvdvvfwA

If f(x) is an “odd” function, A(w)=0.

00

sin)()(;sin)(2

)( wxdwwBxfwvdvvfwB

Let ),(ˆ2)( wfwA c

then

00

cos)(ˆ2)(cos)()(ˆ wxdwwfxfwxdxxfwf cc

c

F

Let ),(ˆ2)( wfwB s

then

00

sin)(ˆ2)(sin)()(ˆ wxdwwfxfwxdxxfwf ss

s

F

Note that in mathamatics:

00

cos)(ˆ2)(cos)(

2)(ˆ wxdwwfxfwxdxxfwf cc

c

F,

00

sin)(ˆ2)(sin)(

2)(ˆ wxdwwfxfwxdxxfwf ss

s

F.

Some Examples

EX1

ax

axkxf

0

0)(

)( sincsinsin

cos)(ˆ

00

wakaw

wak

w

wxkwxdxkwf

aa

c .

waw

k

w

wxkwxdxkwf

aa

s cos1cos

sin)(ˆ

00

.

EX2 f(x) = e-kx, x > 0, k > 0

“Fourier Integral”

“Fourier Cosine Integral”

“Fourier Sine Integral”

“Fourier Cosine Transform (FCT)”

“Fourier Sine Transform (FST)”

15

0 2

2

0

2

00

0

coscos

sinsin

cos)(ˆ

wxdxew

k

w

wxke

wxdxew

k

w

wxewxdxewf

kxkx

kxkx

kx

c

2222

2

0202

2

/1

/cos)(ˆcos1

wk

k

wk

wkwxdxewf

w

kwxdxe

w

k kx

c

kx

.

0,0,cos2

cos)(ˆ2)(

0 220

kxdw

wk

wxkwxdwwfexf c

kx

.

It follows that 0,0,2

cos

0 22

kxek

dwwk

wx kx… (*)

Likewise,

0 2

2

0

2

00

0

sinsin1

coscos

sin)(ˆ

wxdxew

k

w

wxke

w

wxdxew

k

w

wxewxdxewf

kxkx

kxkx

kx

s

2222002

2

/1

/1sin)(ˆ1

sin1wk

w

wk

wwxdxewf

wwxdxe

w

k kx

s

kx

0,0,sin2

sin)(ˆ2)(

0 220

kxdw

wk

wxwwxdwwfexf s

kx

It follows that 0,0,2

sin

0 22

kxedwwk

wxw kx… (**)

Equations (*), (**) are called “Laplace Integrals”

Cosine transform of derivative : )0()(ˆ)(')(ˆ)( fwfwxfwfxf sc

Proof )0()(ˆsin)(cos)(cos)(')]('[

0)(

00fwfwwxdxxfwwxxfwxdxxfxf s

xfx

c

F

Sine transform of derivative : )(ˆ)(')(ˆ)( wfwxfwfxf cs

Proof )(ˆcos)(sin)(sin)(')]('[

0)(

00wfwwxdxxfwwxxfwxdxxfxf c

xfx

s

F

It follows that

)0(')]([)0(')]('[)](''[ 2 fxfwfxfwxf csc FFF

)0()]([)]('[)](''[ 2 wfxfwxfwxf scs FFF .

Application to PDE

Consider the heat equation given by:

)1()0,0(,2

22

xt

x

uc

t

uwith conditions u(x, 0) = f(x) …(2), u(0,t) = 0…(3)

Recall ][)](''[],[)]('[,)()]([)(ˆ 2 uwxuujwxudxexuxuwu jwxFFFFF

Take Fourier sine transform of (1) yields

ss

ssss

s uwct

uuwctwuuwc

x

uc

t

u

t

uˆ

ˆˆ),0(

ˆ 222222

2

22

FFF …(4)

16

Take Fourier sine transform of (2) yields )(ˆ)0,(ˆ wfwu ss …(5)

From (4) twc

s Cetwu22

),(ˆ ; from (5) )(ˆ)0,(ˆ wfCwu ss twc

ss ewftwu22

)(ˆ),(ˆ

Take inverse Fourier sine transform yields

00

1sin)(ˆ2

sin),(ˆ2

)],(ˆ[),(22

wxdwewfwxdwtwutwutxu twc

ss

F .

12 Practice Problems

f(x) or f(t) [For periodic functions, only one period is shown.] ODE

1

3

5

7

9

11

-1 0

1

1

0

1

-1 1

-2 0

1

-1 21

-1

0

1

-11

-1

-20

1

-1 21

-1

-2 0

1

-1 21

2

4

6

8

10

12

13

14

-1 0

1

1

0

1

-1 1

-2 0

1

-1 21

-1

0

1

-1 1

-1

-2 0

1

-1 21

f(x) = x2 (-1<x<1), p = 2

f(x) = x|x| (-1<x<1), p = 2

f(x) = e-|x| (-1<x<1), p = 2

a. y'+y = f(t)

b. y'+2y = f(t)

c. y'+3y = f(t)

d. y'+4y = f(t)

e. y''+3y'+2y = f(t)

f. y''+6y'+5y = f(t)

g. y''+5y'+4y = f(t)

h. y''+5y'+6y = f(t)

i. y''+4y'+4y = f(t)

Question

1. Find the Fourier series of f(x) (or f(t)), then find the “steady-state” solution for the ODE.

2. Find the Fourier transform of f(x) assuming that p∞.

3. Compare the Fourier transform found in prob. 2 with the Fourier series found in prob. 1.

4. Find the Fourier transform of the functions given above when they consist of only “positive” half, then compare it

with the corresponding Laplace transform.

5. Prove gffgwgxgwfxf ˆ*ˆ2

1)(ˆ)(),(ˆ)(

.

6. Find the Fourier cosine transforms of:

(a) –x + a, 0 < x < a (b) 2axe

(a > 0)

8. Show that

7. Find the Fourier sine transforms of:

(a) –x + a, 0 < x < a (b) x2axe

(a > 0)

a.

b.

17

c.

d.

9. Solve the following integral equations; (i.e., find f(x))

(a) awewxdxxf 4/

0

2

cos)(

(b) 2/

0

2

sin)( wwewxdxxf

(c) awjwx edxexf 4/2

)(

(d) wadxexf jwx sinc)(