Integral Transform

Integral TransformIntegral Transform

Dongsup Kim

Department of Biosystems, KAISTFall, 2004

Integral TransformsIntegral Transforms Integral transform:

– K(,t): Kernel– Mapping a function f(t) in t-space into another function g() in -

space– Fourier, Wavelet, Z-transform, Laplace, Hilbert, Radon, etc

An original problem is difficult to solve in the original coordinate. Often, the transform of the problem can be solved more easily. Then, the inverse transform returns the solution from the

transform coordinates to the original system.

dttftKgb

a )(),()(

Problem in

Transform space

Original

problem

Solution in

Transform space

Solution of

original problem

Integral transform

Relatively easy solution

Difficult solution

Inverse transform

Fourier Transform, IFourier Transform, I A physical process can be described either in the time

domain by a function of time t, h(t), or in the frequency domain as a function of frequency f, H(f)

h(t) and H(f) are two different representations of the same process.

One goes back and forth between these two representations by means of the Fourier transform,

– Dirac delta function:

Using angular frequency =2f,

dfefHth

dtethfH

ift

ift

2

2

)()(

)()(

deHth

dtethH

ti

ti

)(2

1)(

)()(

n

n

ixtnn

ndte

x

nxxxx

2

1sin)( ),(lim)(

)()()( afxfax

Examples, IExamples, I FT of Dirac delta function:

FT of cos(0t)

t

(t)

1)( 2 dtet ift

0{cos( )}tFcos(0t)

t

Examples, IIExamples, II FT of exp(2if0t)=exp(i0t)

Sum

F {exp(i0t)}

exp(i0t)

t

t Re

Im

FT

Properties, IIProperties, II Correspondence between symmetries in the two

domains:

Scaling and shifting

odd and real is )( odd andimaginary is )(

even andimaginary is )(even andimaginary is )(

odd andimaginary is )( odd and real is )(

even and real is )( even and real is )(

)()( odd is )(

)()( even is )(

*)]([)( imaginary is )(

*)]([)( real is )(

fHth

fHth

fHth

fHth

fHfHth

fHfHth

fHfHth

fHfHth

shiftingfrequency : )()(

shifting time: )()(

scalingfrequency : )()(||

1

scaling time: )(||

1)(

ansformFourier tr: )()(

02

20

0

0

ffHeth

efHtth

bfHb

th

b

a

fH

aath

fHth

tif

ift

ScalingScaling

t

t

t

Shortpulse

Medium-lengthpulse

Longpulse

)(||

1)(

a

fH

aath

Properties, IIIProperties, III With two functions h(t) and g(t), and their FT H(f) and

G(f), the convolution, g*h, is defined by

Convolution theorem: the FT of the convolution is the product of the individual FTs.

The correlation, Corr(g,h)

Correlation theorem (for two real functions, g and h):

Autocorrelation, Wiener-Khinchin theorem:

Parseval’s theorem:

dthghg )()(*

)()(* fHfGhg

dhtghgCorr )()(),(

)(*)(),( fHfGhgCorr

2|)(|)(*)(),( fGgHfGggCorr

dffHdtth

22)()(power total

Properties, IVProperties, IV FT of derivatives;

in general

application: partial differential equation; wave equation

)(2)(2)(

)()(

)()(

22

21

2

fifHdtethifeth

dtedt

tdhfH

dtethfH

iftift

ift

ift

)()2()(

)( 2 fHifdtedt

thdfH nift

n

n

n

)()0,( ,),(1),(

2

2

22

2

xfxyt

txy

vx

txy

)()(),(

)()0,()( ,)(),(

),(1),()2(

22

22

2

2

22

vtxfdeeFtxy

dxexfYFeFtY

t

tY

vtYi

xitvi

xitvi

Sampling theorem, ISampling theorem, I Suppose function h(t) is sampled at evenly spaced

intervals in time;

– 1/: Sampling rate For any sampling interval , there is a special frequency

fc, called Nyquist critical frequency, given by

– ex: critical sampling of a sine wave of Nyquist critical frequency is two sample points per cycle.

A function f is “bandwidth limited” if its Fourier transform is 0 outside of a finite interval [-L, L]

Sampling Theorem: If a continuous function h(t), sampled at an interval , is bandwidth limited to frequency smaller than fc, i.e., H(f)=0 for all |f|>fc, then the function h(t) is completely determined by its samples hn.

,...2,1,0,1,2..., ),( nnhhn

2/1cf

n

cn nt

ntfhth

)(

)](2sin[)(

Sampling theorem, IISampling theorem, II For bandwidth limited signals, such as music in concert

hall, sampling theorem tells us that the entire information content of the signal can be recorded by sampling rate -1 equal to twice the maximum frequency pass by the amplifier.

For the function that is not bandwidth limited to less then the Nyquist critical

frequency, frequency component that lies outside of the frequency range, -fc

< f < fc is spuriously moved into that range

(aliasing).– Demo applet

http://www.cs.brown.edu/exploratories/freeSoftware/repository/edu/brown/cs/exploratories/applets/nyquist/nyquist_limit_java_plugin.html





Discrete Fourier transform, IDiscrete Fourier transform, I Suppose we have N consecutive sampled values

where is the sampling interval, and assume N is even. With N numbers of input, we can produce no more than

N independent number of outputs. Therefore, we seek estimates only at the discrete values;

Then, discrete Fourier Transform (DFT) is

; DFT maps N complex numbers (the hk’s) into N complex numbers (the Hn’s).

;trapezoidal approximation of the integral

1 ,...,2 ,1 ,0 , ),( Nkktthh kkk

2,...

2 ,

NNn

N

nfn

n

N

k

Niknk

N

k

tifk

tifn HehehdtethfH knn

1

0

/21

0

22)()(

Discrete Fourier transform, IIDiscrete Fourier transform, II

It’s periodic in n, with period N; H-n = HN-n, n=1,2,…

With this conversion, one lets the n in Hn vary from 0 to N-1. Then n and k vary exactly over the same range.

With this convention, – zero frequency n=0

– positive frequencies, 0 < f < fc 1 n N/2-1

– negative frequencies, –fc < f < 0 N/2+1 n N-1

– the value n = N/2 both f = fc and f = -fc

The DFT has symmetry properties almost exactly the same as the continuous Fourier transform.

1

0

/2N

k

Niknkn ehH

1

0

/)(21

0

/2/)(21

0

/)(2N

kn

Nnikk

N

k

NikNNnikk

N

k

NnNikknN HeheehehH

Discrete Fourier transform, IIIDiscrete Fourier transform, III The discrete inverse Fourier transform is

– Proof:

Parseval’s theorem:

– Proof?

1

0

/21 N

n

Niknnk eH

Nh

Nekj

e

eekj

hehN

eehN

eHN

N

n

Nnkji

Nkji

kjiN

n

Nnkji

k

N

j

N

n

Nnkjij

N

n

NiknN

j

Nijnj

N

n

Niknn

1

0

/)(2

/)(2

)(21

0

/)(2

1

0

1

0

/)(21

0

/21

0

/21

0

/2

, if

01

1 , if

111

1

0

21

0

2 1 N

nn

N

kk H

Nh

Fast Fourier Transform (FFT), IFast Fourier Transform (FFT), I DFT appears to be an O(N2) process. Danielson and Lanczos; DFT of length N can be rewritten as

the sum of two DFT of length N/2.

We can do the same reduction of Hk0 to the transform of its

N/4 even-numbered input data and N/4 odd-numbered data. For N = 2R, we can continue applying the reduction until we

subdivide the data into the transforms of length 1. For every pattern of log2N number of 0’s and 1’s, there is one-

point transformation that is just one of the input number hn

)( , /2101/20

12/

012

)2//(2/212/

02

)2//(2

12/

012

/)12(212/

02

/)2(2

1

0

/2

Nik

kkk

Nikk

N

kk

NijkNikN

jj

Nijk

N

jj

NkjiN

jj

Nkji

N

jj

Nijkk

eWHWHHeH

heehe

hehe

heH

nhH nk somefor 001...1001

Fast Fourier Transform (FFT), IIFast Fourier Transform (FFT), II For N=8

Since WN/2 = -1, Hk0 and Hk

1 have period N/2,

Diagrammatically (butterfly),

There are N/2 butterflies for this stage of the FFT, and each butterfly requires one multiplication

12/,...,1,0for , 102

10 NkHWHHHWHH kk

kN/kkk

kk

)12/,...,1,0(

101

100

NkHWHH

HWHH

kk

kk

kk

kk

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

)]()[()()(

)()(

74

32

54

164

22

44

0

74

32

54

164

22

44

0

76

54

32

166

44

22

0

77

66

55

44

33

22

10

7

0

hWhWhWhWhWhWhWh

hWhWhWhWhWhWhWh

hWhWhWhWhWhWhWh

hWhWhWhWhWhWhWhhWH

kkkkkkk

kkkkkkk

kkkkkkk

kkkkkkk

jj

jkk

Fast Fourier Transform (FFT), Fast Fourier Transform (FFT), IIIIII

So far,

The splitting of {Hk} into two half-size DFTs can be repeated on Hk

0 and Hk1 themselves,

713

303

13

612

202

12

511

101

11

4100

010

313

303

03

212

202

02

111

101

01

010

000

00

HHWHH

HHWHH

HHWHH

HHWHH

HHWHH

HHWHH

HHWHH

HHWHH

k

14/,...,1,0for ,

,112101

4112101

0120004

012000

NkHWHHHWHH

HWHHHWHH

kk

kN/kkk

kk

kk

kN/kkk

kk

Fast Fourier Transform (FFT), Fast Fourier Transform (FFT), IIIIII

So far,

– {Hk00} is the N/4-point DFT of {h0, h4,…, hN-4},




– Note that there is a reversal of the last two digits in the binary expansions of the indices j in {hj}.

713

303

13

111

2101

111

612

202

12

110

0100

110

511

101

11

111

2101

101

410

000

10

110

0100

100

313

303

03

011

2001

011

212

202

02

010

0000

010

111

101

01

011

2001

001

010

000

00

010

0000

000

HHWHHHWHH

HHWHHHWHH

HHWHHHWHH

HHWHHHWHH

HHWHHHWHH

HHWHHHWHH

HHWHHHWHH

HHWHHHWHH

Fast Fourier Transform (FFT), Fast Fourier Transform (FFT), IVIV

Fast Fourier Transform (FFT), VFast Fourier Transform (FFT), V If we continue with this process of halving

the order of the DFTS, then after R=log2N stages, we reach where we are performing N one-point DFTs.

– One-point DFT of the number hj is just the identity hj hj

– Since the reversal of the order of the bits will continue, all bits in the binary expansion of j will be arranged in reverse order.

– Therefore, to begin the FFT, one must first rearrange {hj} so it is listed in bit reverse order.

For each of the log2N stages, there are N/2 multiplications, hence there are (N/2)log2N multiplications needed for FFT.

– Much less time than the (N-1)2 multiplications needed for a direct DFT calculation.

– When N=1024, FFT=5120 multiplication, DFT=1,046,529 savings by a factor of almost 200.

Fast Fourier Transform (FFT), Fast Fourier Transform (FFT), VIVI

Integral Transform

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