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Page 1: Fourier transform

Fourier Transform

SOLO HERMELIN

Updated: 22.07.07Run This

Page 2: Fourier transform

Fourier Transform

dttjtftfF exp:F

SOLO

Jean Baptiste JosephFourier

1768 - 1830

F (ω) is known as Fourier Integral or Fourier Transformand is in general complex

jAFjFF expImRe

Using the identities

tdtj

2exp

we can find the Inverse Fourier Transform Ftf -1F

002

1

2exp

2expexp

2exp

tftfdtfdd

tjf

dtjdjf

dtjF

2exp:

dtjFFtf -1F

002

1

tftfdtf

If f (t) is continuous at t, i.e. f (t-0) = f (t+0)

This is true if (sufficient not necessary)f (t) and f ’ (t) are piecewise continue in every finite interval1

2 and converge, i.e. f (t) is absolute integrable in (-∞,∞)

dttf

Page 3: Fourier transform

Fourier TransformSOLO

tf-1F

F FProperties of Fourier Transform

Linearity 1

221122112211 exp: FFdttjtftftftf

F

Symmetry 2 tF

-1FF f2

tFdttjtFfdt

tjtFfd

tjFtft

F

exp22

exp2

exp

Proof:

Conjugate Functions3 tf *

-1FF *F

Proof:

tfd

tjFd

tjFtf ****

2exp

2exp 1-F

Page 4: Fourier transform

Fourier Transform

a

Faa

d

ajfdttjtaftaf

ta 1

expexp:F

FjdttjjtfFd

ddttjtftfF nn

n

n

FF expexp:

SOLO

tf-1F

F FProperties of Fourier Transform

Scaling4

Derivatives5

Proof:

taf-1F

F

a

Fa

1

Proof:

Corollary: for a = -1 tf

-1FF F

tftj n-1F

F

Fd

dn

n

tftd

dn

n

-1FF Fj n

Fj

dtjjFtf

td

ddtjFFtf nn

n

n1-1- FF

2

exp2

exp

Page 5: Fourier transform

Fourier TransformSOLO

tf-1F

F FProperties of Fourier Transform

Convolution6

Proof:

212121

212121

expexpexp

expexpexp:

FFFdjfdduujufjf

ddttjtfjfdtdtfftjdtff

ut

F

tftf 21-1F

F 21 * FF

dtfftftf 2121 :*-1F

F 21 FF

The animations above graphically illustrate the convolution of two rectangle functions (left) and two Gaussians (right). In the plots, the green curve shows the convolution of the blue and red curves as a function of t, the position indicated by the vertical green line. The gray region indicates the product as a function of g (τ) f (t-τ) , so its area as a function of t is precisely the convolution.

http://mathworld.wolfram.com/Convolution.html

Page 6: Fourier transform

Fourier TransformSOLO

tf-1F

F FProperties of Fourier Transform

dFFdttftf 2*

12*

1 2

1Parseval’s Formula7

Proof:

dttjtfF exp11

22exp

2exp 2

*

112*

2*

12*

1

dFF

ddttjtfFdt

dtjFtfdttftf

22exp

2exp 21122121

dFF

ddttjtfFdt

dtjFtfdttftf

2exp*

2

*

2

dtjFtf

dttjtfFdttjtfF expexp 1111

dFFdFFdttftf 212121 2

1

2

1

Page 7: Fourier transform

Signal Duration and BandwidthSOLO

tf-1F

F FRelationships from Parseval’s Formula

dFFdttftf 2*

12*

1 2

1Parseval’s Formula7

Choose tstjtftf m 21

,2,1,0

2

12

22

ndd

Sddttst

m

mm

tftj n-1F

F

Fd

dn

n

and use 5a

Choose n

n

td

tsdtftf 21 and use 5b tf

td

dn

n

-1FF Fj n

,2,1,02

1 22

2

ndSdttd

tsd mn

n

Choosec

,2,1,0,,2,1,0

2*

mndd

SdS

jdt

td

tsdtstj

m

mn

n

n

nmm

n

n

td

tsdtf 1

tstjtf m2

Page 8: Fourier transform

Fourier TransformSOLO

tf-1F

F FProperties of Fourier Transform

Modulation9

Shifting: for any a real 8

Proof:

ttf 0cos -1F

F 002

1 FF

Proof:

tjtjt 000 expexp2

1cos

atf -1F

F ajF exp tajtf exp-1F

F aF

Fajdajfdttjatfatfat

expexpexp:F

aFdttajtfdttjtajtftajtf

expexpexp:expF

use shifting property with a=±ω0

Page 9: Fourier transform

atf -1F

F ajF exp

Fourier TransformSOLO tf

-1FF FProperties of Fourier Transform (Summary)

Linearity 1 221122112211 exp: FFdttjtftftftf

F

Symmetry 2

tF-1F

F f2

Conjugate Functions3 tf *

-1FF *F

Scaling4 taf-1F

F

a

Fa

1

Derivatives5 tftj n-1F

F

Fd

dn

n

tftd

dn

n

-1FF Fj n

Convolution6

tftf 21-1F

F 21 * FF

dtfftftf 2121 :*-1F

F 21 FF

dFFdttftf 2*

12*

1 2

1

Parseval’s Formula7

Shifting: for any a real 8 tajtf exp

-1FF aF

Modulation9 ttf 0cos -1F

F 002

1 FF

dFFdFFdttftf 212121 2

1

2

1

Page 10: Fourier transform

Fourier TransformSOLO

dttjtftfF exp:F

2exp

dtjFFtf 1-F

-1FF

dttjtfF exp

* - complex conjugate

dttjtfF exp**

imaginarytf

realtf

0Re

0Im

tf

tf

tftf

tftf

*

*

*

*

FF

FF

realtf *FF

imaginarytf *FF

Therefore

Fourier Transform of Real or Imaginary Functions

Page 11: Fourier transform

Fourier TransformSOLO

realtf *FF

imaginarytf *FF

realtf

FF

FF

ImIm

ReRe

imaginarytf

FF

FF

ImIm

ReRe

dttjtftfFjFF exp:ImRe F

Fdttjtfdttjtftf expexpF

tftftftf eveneven 5.0: tftftftf oddodd 5.0:

realtf

FjFFtftftftf

FFFtftftftf

evenodd

eveneven

Im5.05.0:

Re5.05.0:

F

F

FFFtftftftf

FjFFtftftftf

evenodd

eveneven

Re5.05.0:

Im5.05.0:

F

F imaginarytf

Fourier Transform of Real or Imaginary Functions (continue – 1)

Page 12: Fourier transform

Fourier Transform

FRe

FIm

Real & Even

t

tfIm

tfRe

Real & Even

SOLO

FRe

FIm

Imaginary & Odd

t

tfIm

tfRe

Real & Odd

FRe

FIm

Imag. & Even

t

tfIm

tfRe

Imag. &Even

FRe

FIm

Real & Odd

t

tfIm

tfRe

Imag. & Odd

realtf

tftftftf even 5.0:

tftftftf even 5.0:

imaginarytf

FF

FFtftf even

ReRe

5.0FF

*FF

tftftftf odd 5.0:

FjFj

FFtftf even

ImIm

5.0FF

realtf *FF

*FF

tftftftf odd 5.0:

FjFj

FFtftf even

ImIm

5.0FF

imaginarytf *FF

FF

FFtftf even

ReRe

5.0FF

Fourier Transform of Real or Imaginary Functions (continue – 2)

Page 13: Fourier transform

Fourier TransformSOLO

dtttftfjdtttftf

dttjttftfdttjtftfF

oddevenoddeven

oddeven

sincos

sincosexp:F

tftftftf eveneven 5.0: tftftftf oddodd 5.0: tftftf oddeven

0000

0

cos2coscoscoscoscos dtttfdtttfdfdtttfdtttfdtttf eveneven

f

eveneven

t

eveneven

even

0coscoscoscoscos000

0

dtttfdfdtttfdtttfdtttf odd

f

oddodd

t

oddodd

odd

0sinsinsinsinsin000

0

dtttfdfdtttfdtttfdtttf even

f

eveneven

t

eveneven

even

0000

0

sin2sinsinsinsinsin dtttfdtttfdfdtttfdtttfdtttf oddodd

f

oddodd

t

oddodd

odd

Therefore

00

sin2cos2exp: dtttfjdtttfdttjtftfF oddeven F

0

cos25.0 dtttfFFF eveneven

0

sin25.0 dtttfjFFF oddodd

Odd and Even Parts

Page 14: Fourier transform

Fourier Transform

2cosImsinRe

2sinImcosRe

2sincosImRe

2exp:

dtFtFj

dtFtF

dtjtFjF

dtjFFtf -1F

SOLO

00: ttfCausal

Causal Functions

A causal functions is a equal zero for negative t

tftftftf eveneven 5.0: tftftftf oddodd 5.0:

Since

and 0 ttf we have 022 ttftftf oddeven

realtf

2sinImcosRe

dtFtFFtf -1F

causalrealtf &

02

sinIm42

cosRe4

2sinIm22

2cosRe22

00

td

tFd

tF

dtFtf

dtFtftf oddeven

FF

FF

ImIm

ReRe

0sinIm2

cosRe2

00

tdtFdtFtf

Page 15: Fourier transform

Fourier TransformSOLO

00: ttfCausalReal & Causal Functions 022 ttftftf oddeven

causalrealtf & 0sinIm2

cosRe2

00

tdtFdtFtf

dttjttfFjFF sincosImRe

dtduttuuFdttdutuuFdtttfF

cossinIm

2cossinIm

2cosRe

00

dvdtttvvFdttdvtvvFdtttfF

sincosRe

2sincosRe

2sinIm

00

Therefore

dtduttuuFF

cossinIm2

Re0

dtdvttvvFF

sincosRe2

Im0

But also

dtduttuuFF

coscosRe2

Re0

dtdvttvvFF

sinsinRe2

Im0

Real & Causal Functions

Real & Causal Functions

FF

FF

ImIm

ReRe

Page 16: Fourier transform

Fourier TransformSOLO

2sin

2cos

dtFj

dtFtftftf oddeven

tftftf

tf eveneven

2

: tftftf

tf oddodd

2

:

tfd

tFtftf

tf eveneven

2cos

2

tfd

tFjtftf

tf oddodd

2sin

2 FjFF ImRe

2sinRe

2sinIm

2cosIm

2cosReImRe

dtFj

dtF

dtFj

dtFtfjtftf

2sinIm

2cosReRe

dtF

dtFtf

2sinRe

2cosImIm

dtF

dtFtf

2sin

2cos

dtFj

dtFtftftftftf oddevenoddeven

Page 17: Fourier transform

Fourier TransformSOLO

Examples of Fourier Transform

Page 18: Fourier transform

Fourier TransformSOLO

Examples of Fourier Transform

Page 19: Fourier transform

Fourier TransformSOLO

Examples of Fourier Transform

Page 20: Fourier transform

Fourier TransformSOLO

Examples of Fourier Transform

Page 21: Fourier transform

Fourier TransformSOLO

Examples of Fourier Transform

Page 22: Fourier transform

Fourier TransformSOLO

Examples of Fourier Transform

Page 23: Fourier transform

Fourier Transform

2

10

2

11

t

tt

2

12

1

t

t

Rectangle1

t

tt

t

t

2/1

2/

2/1

lim

Limiter

tt sgnlimlim20

0

tlim

t

2/1

2/1

SOLO

Special Symbols

10

11

t

ttt

11

t

t

Triangle1

00

01

t

ttH

0

tH

t

Heavisideunit step

1

01

01sgn

t

tt

0

tsgn

t

Signum1

1

0

ttd

dlim

t

2/1

Area = 1td

d

Page 24: Fourier transform

Fourier Transform

00

0

0

2/1lim

2/1

2/

2/1

limlimlim:000 t

t

t

t

t

tt

t

td

dt

td

dt

SOLO

Special Symbols

0

ttd

dlim

t

2/1

Area = 1td

d

δ (t) function

Since tt sgn2

1limlim

0

we have also

δ (t) function is defined as:

ttd

dt sgn

2

1

0

ttd

dlim

t

2/1

Area = 1

0 t

tArea = 1

0

t

tt

t

t

2/1

2/

2/1

lim

Limiter

tt sgnlimlim20

0

tlim

t

2/1

2/1

Page 25: Fourier transform

Fourier TransformSOLO

Special Symbols

Properties of δ (t) function

0

ttd

dlim

t

2/1

Area = 1

0 t

tArea = 1

0

tt δ (t) is a even function: 2

00

0

0

2/1lim

0 t

t

t

tt

1

3

002

1

ffdtttfdtttfuu

Proof:

002

1

002

1lim

2

1lim

sgn2

1limsgn

2

1limsgnlim

2

1lim

0

0

sgn2

1

ff

fTfTffTfTftfdtfdTfTf

tfdtttftdtfdtttf

T

T

TT

T

TT

T

TT

T

TT

tdt

dtT

TT

4 Fourier Transform 10exp2

10exp

2

1exp

jjdttjtt F

Page 26: Fourier transform

Fourier Transform

xLn

x

xJ

xAi

x

x

x

x

xx

2exp

1lim

11lim

1lim

sin1

lim

4exp

2

1lim

lim

lim1

2

0

/10

0

0

2

0

1

0

220

SOLO

Special Symbols

δ (t) function The δ (t) function can be defined as the following limit as ε→0

Ai is the Airry function,

0

3

3cos

1dttx

txAi

dxnjxJ n sinexp2

1

Friedrich WilhelmBessel

1784 - 1846

Edmond NicolasLaguerre

1834 - 1886

Jn (x) is the Bessel function of the first kind,

and Ln (x) is the Laguerre polynomial of arbitrary positive order.

Page 27: Fourier transform

Fourier TransformSOLO

Special Symbols

δ (t) function The δ (t) function can be defined also by the limit n→∞

x

xn

xn

2

1sin

2

1sin

2

1lim

tnsincn

tnn

xnnx

n

n

n

lim

lim

explim 22

2/10

2/11

x

xx

x

xxsinc

sin

Page 28: Fourier transform

Fourier TransformSOLOδ (t) function

0

ttd

dlim

t

/1Area = 1

0 t

tArea = 1

0

2/2/

20

2

1

:

02

t

tet

tfj

Use

It’s Fourier Transform is

ff

ff

ffj

edtedtetf

tffjtffjtfj

0

0

2/

2/0

22/

2/

22 sin

2

11 0

0

For any function φ (t), defined at t=0- and t=0+, we have

002

11lim

1lim

1lim

1lim

1limlim

0

2/0

2/

00

0

2/

2

0

2/

0

2

0

2/

2/

2

00

000

tttt

dttedttedttedttt tfjtfjtfj

tt

0lim

Page 29: Fourier transform

Fourier Transform

20

2

1

:

02

t

tet

tfj

SOLOδ (t) function

ff

fff

0

0sin

tt

0lim

1sin

limlim0

0

00

ff

fff

0

0.25

1.00

89.0

3 dBf

1

4 dBf

2

nnf

1

0 f1

0 f0f

0.50

0.75

ff

ff

0

0sin

fdet tfj 2

Page 30: Fourier transform

Fourier Transform

2/0

2/1

:

0

0

fff

fffffS f

SOLOδ (f) function

Define:0f

f

f/1 Area =1

f

0ff Area = 1

0 f

2/0 ff 2/0 ff 0f

In the time domain we obtain:

tfj

ff

ff

tfjff

ff

tfjtfjff e

tf

tf

tj

e

ffde

ffdefSts 0

0

0

0

0

2

2/

2/

22/

2/

22 sin

2

11

For any function Φ (f), defined at f=f0- and f=f 0+ , we have

00

2/0

2/

0

2/0

2/

0

2/

2/00

2

11lim

1lim

1lim

1lim

1limlim

0

0

0

0

0

0

0

0

0

0

fffff

fff

dfff

dfff

dfff

dfffS

f

fff

ff

ff

f

fff

ff

ff

ff

fff

ff

00

lim fffS ff

00

:lim fffS ff

tfj

ff

ets 02

0lim

tdeff tffj 020

Page 31: Fourier transform

Fourier Transform

N

NnN Tntftf :

SOLOfN (f) N-Periodic Extension of a function f (t)

DefineN- extension of f (t)

0

N

NnN Tntftf

t

0 T1

T1 + T2 > T

-T2

tf

t

T

Page 32: Fourier transform

Fourier Transform

N

NnN Tntt :

SOLOδN (f) function

Define

Let find the Fourier transform of δN (f)

Tf

TNf

ee

j

j

ee

eee

eee

e

ee

etdenTttdetf

TfjTfj

TNfjTNfj

TfjTfjTfj

TNfjTNfjTfj

Tfj

NTfjTNfj

N

Nn

TnfjN

Nn

tfjtfjNN

sin

21

2sin2

2

1

1

2

12

2

12

2

12

2

12

2

1222

222

We can see that

,2,1,0

sin

12sin

sin

1212sin

kf

Tf

TNf

kTf

NkTNf

T

kf NN

N-extension of δ (t)

Page 33: Fourier transform

Fourier TransformSOLOδN (f) function (continue – 1)

N

NnN Tntt :

N

Nn

TnfjN e

Tf

TNff

2

sin

12sin

NT 0

fN12 N

TN 12

1

Tf

10 TN 12

1

T2

1

T2

1

T

1

f

0-NT T 2T-T

tN

t

δN (t) is a periodic function with a time period of T .

ΔN (f) is a periodic function with a frequency period of f0 = 1/T .

Tn

n

TTnj

edfedff

N

Nn

nn

N

Nn

T

T

TnfjN

Nn

T

T

TnfjT

T

N

1sin1

2

0001

2/1

2/1

22/1

2/1

22/1

2/1

Page 34: Fourier transform

Fourier TransformSOLOδN (f) function (continue – 2)

NT 0

fN12 N

TN 12

1

Tf

10 TN 12

1

T2

1

T2

1

T

1

f

0-NT T 2T-T

tN

t

When N → ∞ the peak goes to infinity and the null-to-null bandwidth goes to zero.This resembles to a delta function. To prove that this is the case let compute:

ΔN (f) is a periodic function with a frequency period of f0 = 1/T , with peak amplitude of(2 N+1) and null-to-null bandwidth of 2/ [(2N+1) T].

Tdff

T

T

N

12/1

2/1

01

limlim2/1

2/1

22/1

2/1

T

dffedfffN

Nn

T

T

Tnfj

N

T

T

NN

Page 35: Fourier transform

Fourier Transform

0

m Tmf

Tf

11

Tf

10

T

1

f

0 nTT 2T-T

n

Tnt

t

T

1

T

1

T

2

T

2T

m

SOLOδN (f) function (continue – 3)

Tdff

T

T

N

12/1

2/1

01

limlim2/1

2/1

22/1

2/1

T

dffedfffN

Nn

T

T

Tnfj

N

T

T

NN

Therefore

m

T

T

NN T

mf

Tdfff 1

lim:2/1

2/1

Page 36: Fourier transform

Fourier TransformSOLOδN (f) function (continue – 4)

Let compute the convolution between f (t) and δN (f)

tfTntfdTntfdTntfttf N

N

Nn

N

Nn

N

NnN

:

Therefore ttftf NN Using this relation the Fourier Transform of fN (t) is given by

Tf

TfNfFffFfF NN

sin

12sin Tntfttftf

N

NnNN

If N → ∞ then

mm

m

T

mf

T

mF

TT

mffF

T

T

mf

TfFffFfF

11

1

m

tT

mj

m

n

eT

mF

T

T

mf

T

mF

T

Tntftf

2

1

1

1F

Page 37: Fourier transform

Fourier TransformSOLOδN (f) function (continue – 4)

m T

mf

T

mF

TfF 1

m

tT

mj

n

eT

mF

TTntftf

21

f∞ (t) is a periodic function with a time period of T .

F∞ (f) is a periodic function with a frequency period of f0 = 1/T .

We obtained the Fourier Series description of a periodic function

tdetf

TT

mF

Taeatf

tT

mj

mm

tT

mj

m

22 11

If we define

2/0

2/0

Tt

Tttftf

2/

2/

20

T

T

tfj tdetffF

then

2/

2/

22 1 T

T

tT

mj

mm

tT

mj

m tdetfT

aeatf

Page 38: Fourier transform

Fourier Transform

xxxf

SOLO

Simple Fourier Series

0cos1

cos1

dxxnxdxxnxfan

nn

xn

n

xnx

dxxnxdxxnxfb

n

n

1

0

2

0

12

sincos2

sin1

sin1

http://en.wikipedia.org/wiki/Fourier_series

Square wave equation

xNN

xxxfSN 1sin1

13sin

3

1sin

Sawtooth wave equation

http://en.wikipedia.org/wiki/Sawtooth_wave

http://en.wikipedia.org/wiki/Square_wave

xnn

xxxfSn

N sin1

22sinsin21

http://mathworld.wolfram.com/FourierSeries.html

Page 39: Fourier transform

Fourier Transform

122

sin

2sin

8

kTriangle k

xkkxf

SOLO

Simple Fourier Series

Triangular wave equation

Page 40: Fourier transform

Fourier Transform

122

sin

2sin

8k

Triangle k

xkkxf

SOLO

Simple Fourier Series

Triangular wave equation

http://mathworld.wolfram.com/FourierSeries.html

Page 41: Fourier transform

SignalsSOLO

Signal Duration and Bandwidth

then

tdetsfS tfi 2

fdefSts tfi 2

t

t2

t

2ts

ff

f2

2fS

2/1

2

22

:

tdts

tdtstt

t

tdts

tdtst

t2

2

:

Signal Duration Signal Median

2/1

2

2224

:

fdfS

fdfSff

f

fdfS

fdfSf

f2

22

:

Signal Bandwidth Frequency Median

Fourier

Page 42: Fourier transform

Signals

fdefSts tfi 2

SOLO

Signal Duration and Bandwidth (continue – 1)

dffSfSdfdesfS

dfdesfSdfdefSsdss

tfi

tfitfi

2

22

fdefSts tfi 2

fdefSfitd

tsdts tfi 22'

dffSfSfdfdesfSfi

dfdesfSfidfdefSfsidss

tfi

tfitfi

222

22

2'2

'2'2''

dffSds 22

Parseval Theorem

From

From

dffSfdtts2222

4'

Page 43: Fourier transform

Signals

dffS

fdfd

fSdfS

i

dffS

fdtdetstfS

dffS

tdfdefStst

dffS

tdtstst

tdts

tdtst

t

fifi

22

2

2

2

22

2

2:

SOLO

Signal Duration and Bandwidth

tdetsfS tfi 2

fdefSts tfi 2Fourier

tdetstifd

fSd tfi 22

fdefSfitd

tsd tfi 22

tdts

tdtd

tsdtsi

tdts

tdfdefSfts

tdts

fdtdetsfSf

tdts

fdfSfSf

fdfS

fdfSf

f

fifi

22

2

2

2

22

2 2222

:

Page 44: Fourier transform

Signals

dffSfdttstdttsdttstdtts

222222

2

2 4'4

1

dffSdts22

SOLO

Signal Duration and Bandwidth (continue – 1)

0&0 ftChange time and frequency scale to get

From Schwarz Inequality:

dttgdttfdttgtf22

Choose tstd

tsdtgtsttf ':&

dttsdttstdttstst22

''we obtain

dttstst 'Integrate by parts

sv

dtstsdu

dtsdv

stu '

'

dttststdttsstdttstst '' 2

0

2

dttsdttstst 2

2

1'

dffSfdtts2222

4'

dffS

dffSf

dtts

dttst

dtts

dffSf

dtts

dttst

2

222

2

2

2

222

2

244

4

1

assume 0lim

tstt

Page 45: Fourier transform

SignalsSOLO

Signal Duration and Bandwidth (continue – 2)

22

2

222

2

24

4

1

ft

dffS

dffSf

dtts

dttst

Finally we obtain ft 2

1

0&0 ftChange time and frequency scale to get

Since Schwarz Inequality: becomes an equalityif and only if g (t) = k f (t), then for:

dttgdttfdttgtf22

tftsteAttd

sdtgeAts tt 222:

22

we have ft 2

1

Page 46: Fourier transform

Laplace’s Transform

sdej

jdej

fdet tsf

js

tjf

js

tfj

2

1

2

1 2:

:

2:

:

2

f

ts dtetftfsF0

L

SOLO

Laplace L-Transform

Laplace’s Transform

To find the Inverse Laplace’s Transform (L -1) we use:

00

dsdefdsedefdsesFj

j

tsj

j

tssj

j

ts

tfdtf

j

j

ts dsesFj

tf2

1

For a signal f (t) we define the Laplace’s Transform (L)

Pierre-Simon Laplace1749-1827

0

2 dtfj tfj2

Page 47: Fourier transform

Laplace’s Transform

C2

f

a

0t

00

t

js s - plane

SOLO

Laplace L-Transform (continue – 1)

The Inverse Laplace’s Transform (L -1) is given by:

j

j

ts dsesFj

tf2

1

Using Jordan’s Lemma (see “Complex Variables” presentation or the end of this one)

Jordan’s Lemma GeneralizationIf |F (z)| ≤ M/Rk for z = R e iθ where k > 0 and M are constants, then

for Γ a semicircle arc of radius R, and center at origin:

00lim

mzdzFe zm

R

where Γ is the semicircle, in the left part of z plane.

x

y

R

we can write

j

j

tstsf

f

dsesFj

dsesFj

sFtf

2

1

2

11-L

dsesFj

dsesFj

dsesFj

sFtf ts

C

tsj

j

ts

2

1

2

1

2

1

0

1-L

If the F (s) has no poles for σ > σf+, according to Cauchy’s Theoremwe can use a closed infinite region to the left of σf+, to obtain

Page 48: Fourier transform

Laplace’s TransformSOLO

Properties of Laplace L-Transform

s - Domaint - Domain

tf

f

st sdtetfsF Re0

1 if

M

iii zsFc maxRe

1

Linearity

M

iii tfc

1

3 000 1121 nnnn ffsfssFs Differentiation n

n

td

tfd

4

t

tdf

ss

sF 0

lim1Integration

t

df

5 s

sFReal DefiniteIntegration

t

df0

t

ddf0 0

2s

sF

2

a

sF

a

1Scaling taf

Page 49: Fourier transform

Laplace’s TransformSOLO

Properties of Laplace L-Transform (continue – 1)

s - Domaint - Domain

tf

f

st sdtetfsF Re0

6 n

n

sd

sFdMuliplicity by tn tft n

7

0

dssFDivision by t t

tf

8 sFe sTime shifting tutf

9 asF Complex Translations

tfe ta

10 sHsF Convolutiont - plane

0

dthfthtf

11

j

j

dsHFj

sHsFj

2

1

2

1Convolution s - plane

thtf

Page 50: Fourier transform

Laplace’s TransformSOLO

Properties of Laplace L-Transform (continue – 2)

s - Domaint - Domain

tf

f

st sdtetfsF Re0

12 Initial Value Theorem sFstfst

limlim0

13 Final Value Theorem sFstfst 0limlim

14 Parseval’s Theorem

j

j

j

j

ts

j

j

ts

dssGsFj

dsdtetgsFj

dttgdsesFj

dttgtf

2

1

2

1

2

1

0

00

Page 51: Fourier transform

tf

0n

T Tntt

0

*

n

T TntTnfttftf

tf *

tfT t

f

ts dtetftfsF0

L

SOLO

Sampling and z-Transform

0

1

1

00sT

n

sTn

n

T eeTnttsS LL

0

00**

1

1

2

1

f

j

j

tsT

n

sTn

n

de

Fj

ttf

eTnfTntTnf

tfsF

L

LL

tse

ofPoleststs

FofPoles

tsts

n

nsT

e

FResd

e

F

j

e

FResd

e

F

j

eTnf

sF

1

1

0

*

112

1

112

1

2

1

Poles of

Tse 1

1

Poles of

F

planes

Tnsn

2

j

j

0s

Laplace Transforms

The signal f (t) is sampled at a time period T.

12

R

R

Poles of

Tse 1

1

Poles of

F

plane

Tnsn

2

j

j

0s

Z Transform

Page 52: Fourier transform

Fourier Transform

tf

0n

T Tntt

0

*

n

T TntTnfttftf

tf *

tfT t

SOLO

Sampling and z-Transform (continue – 1)

nnTse

nts

T

njs

T

njs

e

ofPolests

T

njsF

TeT

Tn

jsF

T

njsF

eT

njs

e

FRessF

ts

n

ts

212

lim

2

1

2

lim1

1

2

21

1

*

Poles of

F

j

0s

T

2

T

2

T

2

Poles of

*F plane

js

The signal f (t) is sampled at a time period T.

The poles of are given by tse 1

1

T

njsnjTsee n

njTs 221 2

n T

njsF

TsF

21*

Page 53: Fourier transform

Fourier TransformSOLO

F F-1

frequency-B/2 B/2B

F F-1

-B/2 B/2

B

1/Ts-1/Ts frequency

Sample

Sampling a function at an interval Ts (in time domain)

Anti-aliasing filters is used to enforce band-limited assumption.

causes it to be replicated at 1/ Ts intervals in the other (frequency) domain.

Sampling and z-Transform (continue – 2)

Bandlimited Continuous Time Signal

1/B sec

ampl

itud

e

time (sec)

-0.4

-0.2

0.2

0

0.4

0.6

0.8

1

0 5 10 15-15 -10 -5

Discrete-Time (Sampled) Signal

ampl

itud

e

sample

-0.4

-0.2

0.2

0

0.4

0.6

0.8

1

0 10 20-20 -10

Page 54: Fourier transform

Fourier Transform

tf

0n

T Tntt

0

*

n

T TntTnfttftf

tf *

tfT t

SOLO

Sampling and z-Transform (continue – 3)

0z

planez

Poles of

zF

C

The signal f (t) is sampled at a time period T.

The z-Transform is defined as:

iF

iF

iiF

Ts

FofPoles

T

F

n

n

ze

ze

F

zTnf

zFsFtf

1

0*

1

lim:Z

00

02

1 1

n

RzndzzzFjTnf

fCC

n

Page 55: Fourier transform

Fourier TransformSOLO

Sampling and z-Transform (continue – 4)

0

* 21

n

nsT

n

eTnfT

njsF

TsF

We found

The δ (t) function we have:

1

dtt fdtttf

The following series is a periodic function: n

Tnttd :

therefore it can be developed in a Fourier series:

n

n

n T

tnjCTnttd 2exp:

where: T

dtT

tnjt

TC

T

T

n

12exp

12/

2/

Therefore we obtain the following identity:

nn

TntTT

tnj 2exp

Second Way

Page 56: Fourier transform

Fourier Transform

dttjtftfF 2exp:2 F

0

* 21

n

nsT

n

eTnfT

njsF

TsF

dtjFFtf 2exp2:2-1F

SOLOSampling and z-Transform (continue – 5)

We found

Using the definition of the Fourier Transform and it’s inverse:

we obtain

dTnjFTnf 2exp2

0

111

0

* exp2exp2expnn

n sTndTnjFsTTnfsF

111

* 2exp22 dTnjFjsFn

nn T

nF

Td

T

n

TFjsF 2

1122 111

*

We recovered (with –n instead of n)

n T

njsF

TsF

21*

Second Way (continue)

Making use of the identity: with 1/T instead of T

and ν - ν 1 instead of t we obtain:

nn T

n

TTnj 11

12exp

nn

TntTT

tnj 2exp

Page 57: Fourier transform

Claude Elwood Shannon 1916 – 2001

http://en.wikipedia.org/wiki/Claude_E._Shannon

Fourier TransformSOLO

Henry Nyquist1889 - 1976

http://en.wikipedia.org/wiki/Harry_Nyquist

Nyquist-Shannon Sampling Theorem

The sampling theorem was implied by the work of Harry Nyquist in 1928 ("Certain topics in telegraph transmission theory"), in which he showed that up to 2B independent pulse samples could be sent through a system of bandwidth B; but he did not explicitly consider the problem of sampling and reconstruction of continuous signals. About the same time, Karl Küpfmüller showed a similar result, and discussed the sinc-function impulse response of a band-limiting filter, via its integral, the step response Integralsinus; this band-limiting and reconstruction filter that is so central to the sampling theorem is sometimes referred to as a Küpfmüller filter (but seldom so in English).

The sampling theorem, essentially a dual of Nyquist's result, was proved by Claude E. Shannon in 1949 ("Communication in the presence of noise"). V. A. Kotelnikov published similar results in 1933 ("On the transmission capacity of the 'ether' and of cables in electrical communications", translation from the Russian), as did the mathematician E. T. Whittaker in 1915 ("Expansions of the Interpolation-Theory", "Theorie der Kardinalfunktionen"), J. M. Whittaker in 1935 ("Interpolatory function theory"), and Gabor in 1946 ("Theory of communication").

http://en.wikipedia.org/wiki/Nyquist-Shannon_sampling_theorem

Page 58: Fourier transform

Fourier Transform

Poles of

S

j

0s

plane

js

fB2

SOLO

Nyquist-Shannon Sampling Theorem (continue – 1)

• Signal can be recovered if Fourier spectrum of the sampling signal do not overlap.

sB

2/sB2/sBf

A

fB

2/fB2/fB

f

A

sTA /

sT/1

Poles of

S

j

0s

sT

2Poles of

*Splane

js

fB2sT

2

sT

2

• Start with a band limited signal s (t) 2

0 fBfforfS

• Sample s (t) at a time period Ts, replicates spectrum every 1/Ts Hz.

k sTkfjSfS

12*

fjs 2

nsTnttsts *

k sTjksSsS

2*

L-1

L

FF-1

Page 59: Fourier transform

Fourier Transform

2

1

2

B

T

B

s

SOLO

Nyquist-Shannon Sampling Theorem (continue – 2)

• Signal can be recovered if Fourier spectrum of the sampling signal do not overlap.

BB

Ts

22

1

(Nyquist Sampling Rate)

• Complex signal band-limited to B/2 Hz requires B complex samples/second, or 2 B real samples/seconds (twice the highest frequency)

sB

2/sB2/sBf

A

fB

2/fB2/fB

f

A

sTA /

sT/1

• Start with a band-limited signal f (t) 2

0 fBfforfF • Sample f (t) at a time period Ts,

replicates spectrum every 1/Ts Hz.

Nyquist-Shannon Sampling Theorem:

fB

2/fB2/fB

f

sTA /

sT/1

BandlimitedFilter

Page 60: Fourier transform

Fourier TransformSOLO

The Discrete Time Fourier Transform (DTFT)

sB

2/sB2/sBf

A

fB

2/fB2/fB

f

A

sTA /

sT/1

• Start with a band limited signal s (t) 2

0 fBfforfS

• Sample s (t) at a time period Ts, replicates spectrum every 1/Ts Hz.

k sTkfSfS

1*

nss

ns

TntTns

Tnttsts

*

tdetsfS tfj 2

fdefSts tfj 2F

F-1

Continuous Fourier Transform

F

F-1

sT

ts*

tsq

t ts

Discretization of a Continuous Signal

fdefSTnts sTnfjs

2

n

nf

fj

s

Tf

n

TnfjsDTFT

ss

s

s eTnseTnsfS

2

1

2:

DTFT provides an approximation of the continuous-time Fourier transform.

Discrete Time Fourier Transform (DTFT)Define

Page 61: Fourier transform

Fourier TransformSOLO

The Discrete Time Fourier Transform (DTFT) (continue-1)

• Signal can be recovered if Fourier spectrum of the sampling signal do not overlap.

Discretization of a Continuous Signal

fdefSTnts sTnfjs

2

DTFT-

1

DTFT

Discrete Time Fourier Transform (DTFT)

n

nf

fj

s

Tf

n

TnfjsDTFT

ss

s

s eTnseTnsfS

2

1

2:

We can see that

n

DTFTnkj

nf

fj

sn

nf

fkfj

ssDTFT fSeeTnseTnsfkfS ss

s

1

222

The Discrete Time Fourier Transform SDTFT (fs) is periodic with period fs.Let compute

ns

sn

nmnm

ss

f

fs

nmf

fj

s

n

f

f

nmf

fj

s

f

f n

nmf

fj

s

f

f

mf

fj

DTFT

TmsTnm

nmfTns

fnm

j

eTns

fdeTnsdfeTnsdfefS

s

s

s

s

s

s

s

s

s

s

s

s

1sin

2

10

2/

2/

2

2/

2/

22/

2/

22/

2/

2

n

TnfjsDTFT

seTnsfS 2:

s

s

s

T

T

nTfjDTFTss dfefSTTns

2/1

2/1

2

Page 62: Fourier transform

Fourier TransformSOLO

The Discrete Time Fourier Transform (DTFT) (continue-2)

Normalization of the frequency

DTFT-1

DTFT

n

TnfjsDTFT

seTnsfS 2:

s

s

s

T

T

nTfjDTFTss dfefSTTns

2/1

2/1

2

2/1,2/1

2/1,2/1

:

*

*

f

TTf

Tff

ss

s

n

nfjDTFT ensfS *2* :

DTFT-1

DTFT

2/1

2/1

*2 ** dfefSns nfjDTFT

Example 1,,1,002 NneAns nfj

1*

0

0

*

*

**

**

*2

*21

0

*2*

0

0

0

00

00

0

0

0

*sin

*sin

1

1

Nffj

ffj

Nffj

ffjffj

NffjNffj

ffj

NffjN

n

nffjDTFT

eff

NffA

e

e

ee

eeA

e

eAeAfS

0.50.40.30.20.1-0.1 0-0.2-0.3-0.4-0.5

N = 20

20

15

10

5

0

13.2 dB

f0 = 0.25|SDTFT(f*)|

Normalized Frequency

Page 63: Fourier transform

Fourier TransformSOLO

The Discrete Time Fourier Transform (DTFT) (continue-3)

n

nfjDTFT ensfS *2* :

DTFT-1

DTFT

2/1

2/1

*2 ** dfefSns nfjDTFT

Example

22&8,,00

21,,10,902

nn

nens

nfj

0

5

10

15

0 50 100 150 200 250 300

Δf

SPECTRUM OF 12-SAMPLE PULSE

DTFT

Sig

nal a

mpl

itud

e

27&4,,00

26,,10,302

nn

nens

nfj

0

10

20

30

0 50 100 150 200 250 300

Δf /2

SPECTRUM OF 24-SAMPLE PULSE

Sign

al a

mpl

itude

DTFT

0

0.5

1

0 5 10 15 20 25 30

12-SAMPLE PULSE

Signal sample

Sign

al am

plitu

de

0

0.5

1

0 5 10 15 20 25 30

24-SAMPLE PULSE

Signal sample

Sign

al am

plitu

de

Frequency Resolution Increases with Observation Time N Ts

DTFT

DTFT

Page 64: Fourier transform

Fourier Transform

1

0

2

:N

n

nkN

j

sDFT eTnskS

SOLO

The Discrete Fourier Transform (DFT)

Assume a periodic sequence, sampled at a time period Ts, such that s (n Ts) = s [(n+kN) Ts]

The Discrete Fourier Transform (DFT) requires an input function that is discrete and whose non-zero values have a limited (finite) duration.

Unlike the Discrete-time Fourier transform (DTFT), it only evaluates enough frequency components to reconstruct the finite segment that was analyzed. Its inverse transform cannot reproduce the entire time domain, unless the input happens to be periodic (forever). Therefore it is often said that the DFT is a transform for Fourier analysis of finite-domain discrete-time functions

For the sequence s (0), s (Ts),…,s [(N-1) Ts] we define the Discrete Fourier Transform:

ts*

sT sT tsT

N Ts N Ts N Ts

Page 65: Fourier transform

Fourier Transform

1

0

1

0

2

:N

n

nks

N

n

nkN

j

sDFT WTnseTnskS

SOLO

The Discrete Fourier Transform (DFT) (continue – 1)

For the sequence s (0), s (Ts),…,s [(N-1) Ts] we define the Discrete Fourier Transform:

where is a primitive N'th root of unityand is periodic

Nj

eW2

:

7

8

2

j

e

0

8

2

j

e

1

8

2

j

e

2

8

2

j

e3

8

2

j

e

4

8

2

j

e

5

8

2

j

e 6

8

2

j

e

1

n

Nm

Nj

n

Nj

Nmn

Nj

Nmn WeeeW

1

222

N

N

N s

s

s

s

s

s

W

NNNNNNN

NNNNNNN

NN

NN

NN

S

DFT

DFT

DFT

DFT

DFT

TNs

TNs

Ts

Ts

Ts

WWWWW

WWWWW

WWWWW

WWWWW

WWWWW

NS

NS

S

S

S

1

2

2

1

0

1

2

2

1

0

1121211101

1222221202

1222221202

1121211101

1020201000

NNN sWS NW is a Vandermonde type of Matrix

Page 66: Fourier transform

Fourier TransformSOLO

The Discrete Fourier Transform (DFT) (continue – 2)

nNmn WW

NH

NN IN

WW1

Nj

eW2

1

2* WeW N

j

1121211101

1222221202

1222221202

1121211101

1020201000

NNNNNNN

NNNNNNN

NN

NN

NN

N

WWWWW

WWWWW

WWWWW

WWWWW

WWWWW

W

1112121110

2122222120

2122222120

1112121110

0102020100

*

NNNNNNN

NNNNNNN

NN

NN

NN

TN

HN

WWWWW

WWWWW

WWWWW

WWWWW

WWWWW

WW

Let multiply those two matrices

mkN

mkW

W

W

WW

WWWWWWWWWW

mk

mk

N

mk

NmkN

j

jmk

mNNkmjjkmkmkmk

HNN

01

1

1

1

1

1

0

111100

,

Where IN is the NxN identity matrix

Page 67: Fourier transform

Fourier Transform

1

0

1

0

2

:N

n

nks

N

n

nkN

j

sDFT WTnseTnskS

SOLO

The Discrete Fourier Transform (DFT) (continue – 3)

For the sequence s (0), s (Ts),…,s [(N-1) Ts] we defined the Discrete Fourier Transform:

7

8

2

j

e

0

8

2

j

e

1

8

2

j

e

2

8

2

j

e3

8

2

j

e

4

8

2

j

e

5

8

2

j

e 6

8

2

j

e

1

NNN sWS NW is a Vandermonde type of Matrix

We found that

NH

NN IN

WW1

Where IN is the NxN identity matrix

Therefore the Inverse Discrete Fourier Transform (IDFT) is

NH

NN SWN

s1

1

0

21

0

11 N

n

nkN

j

DFT

N

k

nkDFTs ekS

NWkS

NTns

D.F.T.

I.D.F.T.

Page 68: Fourier transform

Fourier TransformSOLO

The Discrete Fourier Transform (DFT) (continue – 4)

Second way to find the Inverse Discrete Fourier Transform (IDFT). Let compute:

1

0

1

0

21

0

1

0

21

0

2 N

n

N

k

rnkN

j

s

N

k

N

n

rnkN

j

s

N

k

rkN

j

DFT eTnseTnsekS

Nmrn

NmrnN

rnN

jrnN

rnjrn

rnN

rnN

rn

rnN

rnN

jrnN

rnjrn

rnN

rn

rnN

jrnN

rnjrn

e

e

e

e

ern

Nj

rnj

rnN

j

Nrn

Nj

N

k

rnkN

j

0cossin

cossin

sin

sin

cossin

cossin

sin

sin

2sin

2cos1

2sin2cos1

1

1

1

1

2

2

2

2

1

0

2

,2,1,01

0

2

mTmNrsNekS s

N

k

rkN

j

DFT

Page 69: Fourier transform

Fourier Transform

1

0

1

0

2

:N

n

nks

N

n

nkN

j

sDFT WTnseTnskS

SOLO

The Discrete Fourier Transform (DFT) (continue – 1)

For the sequence s (0), s (Ts),…,s [(N-1) Ts] we define the Discrete Fourier Transform:

where is a primitive N'th root of unityand is periodic

Nj

eW2

:

7

8

2

j

e

0

8

2

j

e

1

8

2

j

e

2

8

2

j

e3

8

2

j

e

4

8

2

j

e

5

8

2

j

e 6

8

2

j

e

1

n

Nm

Nj

n

Nj

Nmn

Nj

Nmn WeeeW

1

222

s

s

s

s

s

NN

NN

NN

NN

DFT

DFT

DFT

DFT

DFT

TNs

TNs

Ts

Ts

Ts

WWWWW

WWWWW

WWWWW

WWWWW

WWWWW

NS

NS

S

S

S

1

2

2

1

0

1

2

2

1

0

12210

23320

23420

12210

00000

Page 70: Fourier transform

Fourier TransformSOLO

The Discrete Fourier Transform (DFT) (continue – 5)

The DFT ant Inverse DFT (IDFT) are given by

1

0

21 N

k

nkN

j

DFTs ekSN

Tns

1

0

2

:N

n

nkN

j

sDFT eTnskS

IDFT

DFT

with the periodic properties

,2,1,0

m

TnsTmNns ss

,2,1,0

m

kSNmkS DFTDFT

The sequence s (0), s (Ts),…,s [(N-1) Ts] can be interpreted to be a sequence of finitelength, given for r = 0, 1,…,N-1, and zero otherwise or a periodic sequence, defined for all r.

ts*

sT sT tsT

N Ts N Ts N Ts

ts*

sT tN Ts

Page 71: Fourier transform

Fourier Transform

1

0

2

:N

n

nkN

j

sDFT eTnskS

SOLO

The Discrete Fourier Transform (DFT) (continue – 6)

The DFT ant Inverse DFT (IDFT) are given by

1

0

21 N

k

nkN

j

DFTs ekSN

Tns

IDFT

DFT

n

nfjDTFT ensfS *2* :

2/1

2/1

*2 ** dfefSns nfjDTFT

IDTFT

DTFT

The DTFT ant Inverse DTFT (IDTFT) where given by

We can see that DFT is a sampled version of DTFT by tacking:

2/1,2/1

2/1,2/1

1,,1,0

*

*

f

TTf

NkTN

kf

N

kfTf

ss

ss

1,,1,0:1

0

2

NkfSeTnskSsTN

kfDTFT

N

n

nkN

j

sDFT

Page 72: Fourier transform

Fourier TransformSOLO

The Discrete Fourier Transform (DFT) (continue –7)

We can see that DFT is a sampled version of DTFT :

1,,1,0:1

0

2

NkfSeTnskSsTN

kfDTFT

N

n

nkN

j

sDFT

10 12 14 16 18 0 2 4 6 8 k

f0 0.1 0.2 0.3 0.4 0.5-0.1-0.2-0.3-0.4-0.50

5

10

15

20

25

s [n]= exp (j 2 π f0 n), 0 ≤ n ≤ 19N = 20

f0 = 0.25

|SD

TF

T (

f)| a

nd

|SD

TF (

k)|

10 12 14 16 18 0 2 4 6 8 k

f0 0.1 0.2 0.3 0.4 0.5-0.1-0.2-0.3-0.4-0.50

5

10

15

20

25s [n]= exp (j 2 π f0 n), 0 ≤ n ≤ 19

N = 20

f0 = 0.275

|SD

TF

T (

f)| a

nd

|SD

FT (

k)|

By changing f0 from 0.25 to 0.275 we move |SDTFT (f)| to the right, and since the samplingpoints didn’t change, we obtain different |SDFT (k)| values.

Page 73: Fourier transform

Fourier TransformSOLO

The Discrete Fourier Transform (DFT) (continue – 8)

We can see that DFT is a sampled version of DTFT :

1,,1,0:1

0

2

NkfSeTnskSsTN

kfDTFT

N

n

nkN

j

sDFT

10 12 14 16 18 0 2 4 6 8 k

f0 0.1 0.2 0.3 0.4 0.5-0.1-0.2-0.3-0.4-0.50

5

10

15

20

25

s [n]= exp (j 2 π f0 n), 0 ≤ n ≤ 19N = 20

f0 = 0.25

|SD

TF

T (

f)| a

nd

|SD

TF (

k)|

10 12 14 16 18 0 2 4 6 8 k

f0 0.1 0.2 0.3 0.4 0.5-0.1-0.2-0.3-0.4-0.50

5

10

15

20

25s [n]= exp (j 2 π f0 n), 0 ≤ n ≤ 59

N = 60

f0 = 0.25

|SD

TF

T (

f)| a

nd

|SD

FT (

k)|

Increase sampling density from N=20 to N=60.

Page 74: Fourier transform

SOLOProperties of The Discrete Fourier Transform (DFT) (continue – 9)

mns mkN

j

DFT ekS2

Linearity 1 nsns 2211

Shift of a Sequence2

3

4

5

Periodic Convolution

6

7

Conjugate

8

9

IDFTDFT

1

0

2

:N

n

nkN

j

DFT enskS

1

0

21 N

k

nkN

j

DFT ekSN

ns

kSkS DFTDFT 2211

nsns 21 , Periodic Sequence(Period N)

kSkS DFTDFT 21 , DFT(Period N)

nlN

jens

2 lkSDFT

1

021

N

m

mnsms kSkS DFTDFT 21

nsns 21

1

021

1 N

lDFTDFT lkSlS

N

ns kSDFT

ns kSDFT

Real & Imaginary nsRe

nsImj

2/kSkSkS DFTDFTeven

2/kSkSkS DFTDFTodd

Page 75: Fourier transform

SOLOProperties of The Discrete Fourier Transform (DFT) (continue – 10)

2/: nsnsnseven kSDFTReEven Part10

11

12 Symmetric Proprties)only when s (n) is real(

Parseval’s Formula

IDFTDFT

1

0

2

:N

n

nkN

j

DFT enskS

1

0

21 N

k

nkN

j

DFT ekSN

ns

nsns 21 , Periodic Sequence(Period N)

kSkS DFTDFT 21 , DFT(Period N)

lkSDFT

kSkS

kSkS

kSmkSm

kSkS

kSkS

DFTDFT

DFTDFT

DFTDFT

DFTDFT

DFTDFT

II

ReRe

Odd Part 2/: nsnsnsodd

Page 76: Fourier transform

Fourier TransformSOLO

Fast Fourier Transform (FFT)

John Wilder Tukey 1915 – 2000

http://en.wikipedia.org/wiki/John_Tukey

James W. Cooley1926 -

http://www.ieee.org/portal/pages/about/awards/bios/2002kilby.html

The Cooley-Tukey algorithm, is the most common fast Fourier transform (FFT) algorithm. It re-expresses the discrete Fourier transform (DFT) of an arbitrary composite size N = N1N2 in terms of smaller DFTs of sizes N1 and N2, recursively, in order to reduce the computation time to O(N log N) for highly-composite N (smooth numbers).

FFTs became popular after J. W. Cooley of IBM and John W. Tukey of Princeton published a paper in 1965 reinventing the algorithm (first invented by Gauss) and describing how to perform it conveniently on a computer

Page 77: Fourier transform

Fourier TransformSOLO

Fast Fourier Transform (FFT)

The radix-2 DIT Algorithm

The radix-2 decimation-in-time (DIT) FFT is the simplest and most common form of the Cooley-Tukey algorithm, although highly optimized Cooley-Tukey implementations typically use other forms of the algorithm as described below. Radix-2 DIT divides a DFT of size N into two interleaved DFTs (hence the name "radix-2") of size N/2 with each recursive stage.

1

0

1

0

2

:N

n

nks

N

n

nkN

j

sDFT WTnseTnskS

For the sequence s (0), s (Ts),…,s [(N-1) Ts] we define the Discrete Fourier Transform:

1,1, 22/12

*2

jNjevenN

NNj

Nj

eWeWWeWeW

Suppose N is a power of 2; i.e. N=2L (L is integer). Since N is a even integer, let compute SDFT (k) by separate s (nTs) into two (N/2)-point sequences consisting of the even-numberedpoints (n=2r) and odd numbered points (n=2r+1).

12/

0

212/

0

2

12/

0

1212/

0

2

122

122

N

n

kr

Nk

N

N

n

kr

N

N

n

krN

N

n

krNDFT

WrsWWrs

WrsWrskS

Page 78: Fourier transform

Fourier TransformSOLO

Fast Fourier Transform (FFT)

The radix-2 DIT Algorithm (continue – 1)

2/2/

2222

NN

jN

j

N WeeW

We divided the N-point DFT into two N/2-points DFTs.

kH

N

n

krN

kN

kG

N

n

krN

N

n

krN

kN

N

n

krNDFT

WrsWWrs

WrsWWrskS

12/

02/

12/

02/

12/

0

212/

0

2

122

122

Since

N/2 pointsDFT

N/2 pointsDFT

0s

2s 4s

2/Ns

1s

3s

5s

12/ Ns

ΣWNN/2

WNN/2+1

WNN/2+2

WNN-1

WNN/2-1

WN0

WN1

WN2

Σ

Σ

Σ

Σ

Σ

Σ

Σ

2/NSDFT

12/ NSDFT

22/ NSDFT

1NSDFT

1DFTS 2DFTS

0DFTS

12/ NSDFT

1G

2G

0G

12/ NG

1H

2H

0H

12/ NH

Page 79: Fourier transform

Fourier TransformSOLO

Fast Fourier Transform (FFT)

The radix-2 DIT Algorithm (continue – 2)

We divided the N-point DFT into two N/2-points DFTs.

4 pointsDFT

4 pointsDFT

0s

2s

4s

6s

1s

3s

5s

7s

ΣW84

W85

W86

W87

W83

Σ

Σ

Σ

Σ

Σ

Σ

Σ

4DFTS

5DFTS

6DFTS

7DFTS

1DFTS

2DFTS

0DFTS

3DFTS

1G

2G

0G

3G

1H

2H

0H

3H

W80

W81

W82

Reduction of an 8-points FFT to two4-points FFTs

2 pointsDFT

0s

2s

4s

6s

Σ

Σ

Σ

Σ

W40

W41

W42

2 pointsDFT

1G

2G

0G

3GW43

1DFTS

0DFTSΣ

Σ 1s

0s

A 2-points FFT

Reduction of an 4-points FFT to two2-points FFTs

Page 80: Fourier transform

Fourier TransformSOLO

Fast Fourier Transform (FFT)

The radix-2 DIT Algorithm (continue – 3)

ΣW84

W85

W86

W87

W83

Σ

Σ

Σ

Σ

Σ

Σ

Σ

4DFTS

5DFTS

6DFTS

7DFTS

1DFTS

2DFTS

0DFTS

3DFTS

1G

2G

0G

3G

1H

2H

0H

3H

W80

W81

W82

2 pointsDFT

Σ

Σ

Σ

Σ

W40

W41

W42

2 pointsDFT

W43

2 pointsDFT

Σ

Σ

Σ

Σ

W40

W41

W42

2 pointsDFT

W43

0s

2s

4s

6s

1s

3s

5s

7s

1st Stage 2nd Stage 3th Stage

Flow Diagram for an 8-points FFT

Page 81: Fourier transform

Fourier TransformSOLO

Fast Fourier Transform (FFT)

The radix-2 DIT Algorithm (continue – 2) kkj

kN

N

jNk

N eeW 12

2/

We divided the N-point DFT into two N/2-points DFTs.

12/

01

2/12/

0

2/ 2/2/N

n

knN

NkN

N

n

NnkN

knNDFT WWNnsnsWNnsWnskS

k

Since N/2 is an even integer (N=2L)

N/2 pointsDFT

N/2 pointsDFT

Σ

Σ

Σ

Σ

2DFTS 4DFTS

0DFTS

2NSDFT

1G

2G

0G

12/ NG

1H

2H

0H

12/ NH

1DFTS 3DFTS 5DFTS

1NSDFT

WN0

Σ

Σ

Σ

Σ

WN1

WN2

WNN/2-1

1s 2s

0s

12/ Ns

2/Ns 12/ Ns 22/ Ns

1Ns

tgofFFTN

N

n

nlN

WW

N

N

n

nlN

ng

DFT WngWNnsnslkSNN

L

2/

12/

02/

2

12/

0

2 2/2

2/2

thofFFTN

N

n

nlN

WW

N

N

n

nlN

nh

nNDFT WnhWWNnsnslkS

NN

L

2/

12/

02/

2

12/

0

2 2/2

2/12

Page 82: Fourier transform

Fourier TransformSOLO

Fast Fourier Transform (FFT)

The radix-2 DIT Algorithm (continue – 3)

We divided the N-point DFT into two N/2-points DFTs.

4 pointsDFT

4 pointsDFT

Σ

Σ

Σ

Σ

2DFTS

4DFTS

0DFTS

6DFTS

1G

2G

0G

3G

1H

2H

0H

3H

1DFTS

3DFTS

5DFTS

7DFTS

W80

Σ

Σ

Σ

Σ

W81

W82

W83

1s

2s

0s

3s

4s

5s

6s

7s

2- pointsDFT 2DFTS

1DFTS

0DFTS

1'G

0H

0'G

1H

Σ

Σ

Σ

Σ

1G

2G

0G

3G

2- pointsDFT 3DFTS

W40

W41

Reduction of an 8-points FFT to two4-points FFTs

Reduction of an 4-points FFT to two2-points FFTs

1DFTS

0DFTSΣ

Σ 1s

0s

A 2-points FFT(Butterfly)

Page 83: Fourier transform

Fourier TransformSOLO

Fast Fourier Transform (FFT)

The radix-2 DIT Algorithm (continue – 4)

Σ

Σ

Σ

Σ

W80

Σ

Σ

Σ

Σ

W81

W82

W83

1s

2s

0s

3s

4s

5s

6s

7s

2DFTS

4DFTS

0DFTS

6DFTS

1DFTS

3DFTS

5DFTS

7DFTS

Σ

Σ

Σ

Σ

W40

W41

Σ

Σ

Σ

Σ

W40

W41

Σ

Σ

Σ

Σ

Σ

Σ

Σ

Σ

1st Stage 2nd Stage 3th Stage

Flow Diagram for an 8-points FFT

Page 84: Fourier transform

Fourier Transform

1,,1,0:1

0

2

NkeTnskS

N

n

nkN

j

sDFT

86424648

162566425624

321024160102464

6440963844096160

1281638489616384384

SOLO

Fast Fourier Transform (FFT)

Arithmetic Operations for a Radix FFT versus DFT

For N = 2L we have L stages of Radix FFT and:

For N-point DFT we have:

For each row we have N complex additions and N complex multiplications, therefore for the N rows we have

Number of complex additions DFT = Number of complex multiplications DFT = NxN=N2

Number of complex additions FFT =N L=N log2 N

Number of complex additions FFT =N/2 (multiplications per stage) x L -1 =N/2 log2 (N/2)

Operation

Complex additions Complex multiplications

DFT DFTFFT FFTN=2L

Approximate number of Complex Arithmetic Operations Required for 2L-point DFT and FFT computations

Page 85: Fourier transform

SOLO Complex Variables

Laurent’s Series (1843)

Power Series

If f (z) is analytic inside and on the boundary of the ringshaped region R bounded by two concentric circles C1 andC2 with center at z0 and respective radii r1 and r2 (r1 > r2),

then for all z in R:

Pierre Alphonse Laurent1813 - 1854

C1

x

y

RC2R2

R1

z0

z

z'

r

P1

P0

z'

1 00

0

nn

n

n

n

nzz

azzazf

,2,1,0''

'

2

1

2

1

0

nzdzz

zf

ia

C

nn

,2,1,0''

'

2

1

1

1

0

nzd

zz

zf

ia

C

nn

Proof:

Since z is inside R we have R1 <|z-z0|=r < R2 , and |z’-z0|= R1 on C1 and R2 on C2.

Start with the Cauchy’s Integral Formula:

212

0

1

1

01

''

''

'

''

'

''

'

''

'

''

'

'

0

CCC

P

P

P

PC

dzzz

zfdz

zz

zfzfdzdz

zz

zfdz

zz

zfdz

zz

zfdz

zz

zfzf

Page 86: Fourier transform

SOLO Complex Variables

Laurent’s Series (continue - 1)

Power SeriesPierre Alphonse Laurent

1813 - 1854

C1

x

y

RC2R2

R1

z0

z z'

r

Proof (continue – 1):

Since z and z’ are inside R we have R1 >|z-z0|=r >R2, |z’-z0|=R1.

From Cauchy’s Integral Formula:

21

''

''

'

'

CC

dzzz

zfdz

zz

zfzf

Use the identity:

11

1

1 12n

n

For I integral:

nn

zz

zz

zz

zzzz

zz

zz

zz

zz

zz

zzzzzz 0

0

0

0

1

0

0

0

0

0

0

00 '

'1

1

''1

'

1

'1

1

'

1

'

1

n

n

n

R

C

n

n

n

zs

C

n

za

C

za

C

Rzzzazzzazazzzz

zdzf

i

zz

zzzz

zdzf

izz

zz

zdzf

izz

zdzf

i

n

n

0000100

0

0

1

0

0

02

00

2

0

2

01

2

00

2

''

''

2

'

''

2

1

'

''

2

1

'

''

2

1

1

''

'

2

1

C

zdzz

zf

i

We have:

n

n

n

C

n

n

n R

r

rR

MRdR

rRR

Mr

zzzz

zdzfzzR

11

1

2

0

1

110

0

2''

''

20

where |f (z)|<M in R and r/R1< 1, therefore: 0

n

nR

Page 87: Fourier transform

SOLO Complex Variables

Laurent’s Series (continue - 2)

Power SeriesPierre Alphonse Laurent

1813 - 1854

C1

x

y

RC2R2

R1

z0

z z'r

Proof (continue – 1):

Since z and z’ are inside R we have R1 >|z-z0|=r > R2, |z’-z0|=R2.

From Cauchy’s Integral Formula:

21

''

''

'

'

CC

dzzz

zfdz

zz

zfzf

Use the identity:

11

1

1 12n

n

For II integral:

nn

zz

zz

zz

zzzz

zz

zz

zz

zz

zz

zzzzzz 0

0

0

0

1

0

0

0

0

0

0

00

'

'1

1''1

1'

1

11

'

1

n

n

n

R

C

n

n

n

za

C

n

za

CC

Rzzzazzzazzzz

zdzfzz

i

zzzz

zdzf

izzzz

zdzf

izdzf

i

n

n

1

001

1

001

0

0

1

0

1

00

2

0

0

0

01

0

01

00

'

'''

2

1

1

'

''

2

11

'

''

2

1''

2

1

C

zdzz

zf

i'

'

'

2

1

We have:

n

n

n

C

n

n

n r

R

rR

RMdR

rRr

MR

zzzz

zdzfzzR

2

2

2

2

0

2

2

2

0

0

2'

'''

2

1

0

where |f (z)|<M in R and R2/r< 1, therefore: 0

n

nR Return to Table of Contents

Page 88: Fourier transform

Z2 TransformC1

x

y

R

C2r2 z0

z

r

z'

C

r1

SOLO

Z-Transform Two Sided

n

nzTnfzF

Example 1

TnaTnf

C

n dzzzFj

Tnf 1

2

1

1

0

0

0/1

/

01

1

n k

T

T

Tk

TT

nT

n

T

T

nT

n

nT

n

nTn

nazaz

az

a

z

a

z

z

a

nza

z

az

a

z

azazF

Page 89: Fourier transform

Z2 TransformSOLO

Z-Transform Two Sided

Example 2

gg

ff

rz

r

rr

gg rzr

gfgf rrzrr

C

dz

GFj

TngTnf

1

2

1Z

gfgf

fg

gf

rrzrr

nrrz

nrzr

0&/

0&/

C

rz

r

C n

n

n

n

n

rrC

nn

n

n

dz

GFj

dz

TngFj

zTngdzFj

zTngTnfTngTnf

gg

nf

nf

11

1

2

1

2

1

2

1

00

Z

Page 90: Fourier transform

Z2 TransformSOLO

Z-Transform Two Sided

Example 2 (continue – 1)

zban

ba

zba

z

b

zb

z

a

aResd

b

zb

z

a

aj

zban

z

ba

z

baResd

z

baj

ba

TT

TT

TT

CT

T

T

T

b

z

bb

z

T

T

T

T

C

TT

TTTTa

b

za

TT

TnTn

T

TT

T

TT

&01111

1

2

1

&01

1

1

11

1

1

1

11

2

1

Z

C

dz

GFj

TngTnf

1

2

1Z

gfgf

fg

gf

rrzrr

nrrz

nrzr

0/

0/

Page 91: Fourier transform

Z TransformSOLO

Properties of Z-Transform Functions

Z - Domaink - Domain

kf

ffk

k rzrzkfzF0

1

ii ff

M

iii rzrzFc minmax

1

Linearity

M

iii kfc

1

2 ,2,10 kkfmkf zFz mShifting

mkf

m

k

kmm zkfzFz1

mkf

m

k

kmm zkfzFz1

1kf 0fzFz

3 Scaling kfak

ffk

krazrazakfzaF

0

11

Page 92: Fourier transform

Z TransformSOLO

Properties of Z-Transform Functions (continue – 1)

4 Periodic Sequence kf

1111 ffN

N

rzrzFz

z

N = number of units in a period

Rf1- ,+ = radiuses of convergence in F(1) (z)

F(1) (z) = Z -Transform of the first period

5 Multiplication by k kfk

ff rzrzd

zFdz

6 Convolution

0

:m

mkhmfkhkf hfhf rrzrrzHzF ,min,max

7 Initial Value zFfz

lim0

8 Final Value existsfifzFzkfzk

1limlim1

Z - Domaink - Domain

kf

ffk

k rzrzkfzF0

Page 93: Fourier transform

Z TransformSOLO

Properties of Z-Transform Functions (continue – 2)

9 Complex Conjugate kf * ff rzrzF **

10 Product khkf hfhf

C

rrzrrz

zdzHzF

j,1,

2

1 1

12 Correlation

1,1,2

1 11

0

krrzrr

z

zdzzHzF

jkmhmfkhkf hfhf

C

k

m

11 Parceval’s Theorem

hfhf

Ck

rrzrrz

zdzHzF

jkhkf ,1,

2

1 1

0

Z - Domaink - Domain

kf

ffk

k rzrzkfzF0

Page 94: Fourier transform

Z TransformSOLO

Table of Z-Transform Functions

Z - Domaink - Domain

kf f

k

k RzzkfzF

0

1

mkf 110 11 mfzfzfzFz mm 2

mkf zFz m3

kfkfkf 1: 01 fzzFz 4

kfkfkfkf 122:2 1021 2 fzfzzzFz 5

kf3 2130331 23 fzfzzfzzzzFz 6

Page 95: Fourier transform

L2 Transform

0 aetf ta

SOLO

Laplace Transform Two Sided

j

j

j

j

ts

ts

j

j

tts

gg

tftf

dsGFj

ddtetgFj

dtetgdeFj

dtetgtftgtf

2

1

2

1

2

1

00

2L

Hence

ff

gf

j

j

ts

tfor

tfor

dsGFj

dtetgtftgtf

0

0

2

12L

Example 1

srealasreala

tastastastaststata

asasas

e

as

edtedtedteee

11

0

0

0

0

2L

0&1

0&1

2

tsrealaas

tasrealase

f

fta

L

aa sreal

simag

taetf

t

1

Page 96: Fourier transform

L2 Transform

0 aetf ta

SOLO

Laplace Transform Two Sided

Example 2

ab

dbsa

ee

fg

j

j

tbta

112L

0&1

0&1

2

tsrealaas

tasrealase

f

fta

L

0 betg tb

Find the two sided Laplace transform of f (t) g (t)

0&1

0&1

2

tsrealbbs

tbsrealbse

f

ftb

L

ba

dbsa

ee

gf

j

j

tbta

112L

basbsaRes

a

111

basbsaRes

a

111

C1

b a

0t

00

t

C2

ba

0t

00

t

Page 97: Fourier transform

SOLO

References

A. Papoulis, “The Fourier Integral and its Applications”, McGraw Hill, 1962

R.N. Bracewell, “The Fourier Transform and its Applications”, McGraw Hill, 1965, 1978

J.W. Goodman,“Introduction to Fourier Optics”, McGraw Hill, 1968

H. Stark, Ed. “Applications of Optical Fourier Transform”, Academic Press, 1982

A. Papoulis, “Systems and Transforms with Applications in Optics”, McGraw Hill, 1968

Fourier Transform

Athanasios Papoulis1921 - 2002 Ronald N. Bracewell

1921 -

Joseph W. GoodmanWilliam Ayer Professor, Emeritus

Packard 352Department of Electrical Engineering

Stanford UniversityStanford, CA 94305

Email :[email protected]

Page 98: Fourier transform

April 9, 2023 98

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

Page 99: Fourier transform

Raymond Paley1907 - 1933

Norbert Wiener1894 - 1964

Paley – Wiener Condition

A necessary and Sufficient condition for a square-integrablefunction A (ω) ≥ 0 to be the Fourier spectrum of a causal functionis the convergence of the integral:

dA

21

ln

SOLO

Page 100: Fourier transform

The Mellin Transform

0

1: s

M exfsF

SOLO

Hjalmar Mellin1854 - 1933

Putting: tdexdex tt 11 sts ex

tdeefsF tst

M

We can see that the Mellin Transform of the function f (t) is identical to theBilateral Laplace Transform of f (e-t).

Page 101: Fourier transform

SOLO

Example

0

sindk

k

krLet compute:

x

y

R

A

B

C

D

E

F

G

H

Rx Rx

For this use the integral: 0ABCDEFGHA

zi

dzz

e

Since z = 0 is outside the region of integration

0

BCDEF

ziR xi

GHA

zi

R

xi

ABCDEFGHA

zi

dzz

edx

x

edz

z

edx

x

edz

z

e

00

0000

sin2

sin2

sinlim2limlimlim dk

k

rkidx

x

xidx

x

xidx

x

eedx

x

edx

x

eR

R

R xixi

R

R xi

RR

xi

R

idideideie

edz

z

e i

ii

eii

i

eiez

GHA

zi

00

1

0

0

00limlimlim

012

2

0

/2/2sin

0

sin

00

R

RRReRii

i

eRieRz

BCDEF

zi

eR

dedededeRieR

edz

z

e i

ii

Therefore: 0sin

20

idkk

rkidz

z

e

ABCDEFGHA

zi 2

sin

0

dkk

kr

Complex Variables

Page 102: Fourier transform

SOLO Complex Variables

Cauchy’s Theorem

C

x

y

R

Proof:

0C

dzzf

If f (z) is analytic with derivative f ‘ (z) which is continuous at all points insideand on a simple closed curve C, then:

yxviyxuzf ,, Since is analytic and has continuous first order derivative

y

ui

y

v

x

vi

x

u

zd

fdzf

iyzxz

'

y

u

x

v

y

v

x

u

& Cauchy - Riemann

0

00

RR

dydxy

v

x

uidydx

y

u

x

v

dyudxvidyvdxudyidxviudzzfCCCC

q.e.d.

Augustin Louis Cauchy ) 1789-1857(

Page 103: Fourier transform

SOLO Complex Variables

The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals

Jordan’s LemmaIf |F (z)| ≤ M/Rk for z = R e iθ where k > 0 and M are constants, then

where Γ is the semicircle arc of radius R, center at origin, in theupper part of z plane, and m is a positive constant.

0lim

zdzFe zmi

R

x

y

R

Proof:

0lim

zdzFe zmi

R

using:

q.e.d.

0

deRieRFezdzFe iieRmieRz

zmi i

i

2/

0

sin

1

0

sin

1

0

sin

0

sincos

00

2

dReR

MdRe

R

MdReRFe

deRieRFedeRieRFedeRieRFe

Rm

k

Rm

k

iRm

iiRmRmiiieRmiiieRmi ii

2/0/2sin for2/

1sin

/2

Rm

k

Rm

k

Rm

k

iieRmi eR

Mde

R

Mde

R

MdeRieRFe

i

1

2222/

0

/2

1

2/

0

sin

1

0

012

limlim0

Rm

kR

iieRmi

Re

R

MdeRieRFe

i

Marie Ennemond Camille Jordan1838 - 1922

Page 104: Fourier transform

SOLO Complex Variables

The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals

Jordan’s Lemma GeneralizationIf |F (z)| ≤ M/Rk for z = R e iθ where k > 0 and M are constants, then

for Γ a semicircle arc of radius R, and center at origin:

00lim

mzdzFe zmi

R

x

y

R

where Γ is the semicircle, in the upper part of z plane.

1

00lim

mzdzFe zmi

R

xy

R

where Γ is the semicircle, in the down part of z plane.

2

00lim

mzdzFe zm

R x

y

R

where Γ is the semicircle, in the right part of z plane.

3

00lim

mzdzFe zm

R

where Γ is the semicircle, in the left part of z plane.

4x

y

R

Page 105: Fourier transform

SOLO Complex Variables

The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals

Integral of the Type (Bromwwich-Wagner)

jc

jc

ts sdsFei2

1

The contour from c - i ∞ to c + i ∞ is called Bromwich Contour

Thomas Bromwich1875 - 1929

x

y

0t

R

c

x

y0t

R c

0

0

2

1

lim2

1

2

1

tzFeRes

tzFeReszdzF

i

sdsFesdsFei

sdsFei

tf

tz

planezRight

tz

planezLeft

ts

ic

ic

ts

R

ic

ic

ts

where Γ is the semicircle, in the right part of z plane, for t < 0.

where Γ is the semicircle, in the left part of z plane, for t > 0.

This integral is also the Inverse Laplace Transform.