DIGITAL PROCESSING OF SPEECH AND IMAGE SIGNALS

Lecture

DIGITAL PROCESSING

OF

SPEECH AND IMAGE SIGNALS

RWTH Aachen, WS 2006/7

Prof. Dr.-Ing. H. Ney, Dr.rer.nat. R. SchluterLehrstuhl fur Informatik 6

RWTH Aachen

1. System Theory and Fourier Transform

2. Discrete Time Systems

3. Spectral Analysis

4. Fourier Transform and Image Processing

5. LPC Analysis

6. Wavelets

7. Coding

8. Image Segmentation and Contour-Finding

Completions: L. Welling, A. Eiden; April 1997Completions: J. Dahmen, F. Hilger, S. Koepke; Mai 2000Completions: F. Hilger, D. Keysers; Juli 2001Translation: M. Popovic, R. Schluter; April 2003Corrections: D. Stein; October 2006

Literature:

A. V. Oppenheim, R. W. Schafer: Discrete Time Signal Processing,Prentice Hall, Englewood Cliffs, NJ, 1989.

A. Papoulis: Signal Analysis, McGraw-Hill, New York, NY, 1977. A. Papoulis: The Fourier Integral and its Applications, McGraw-HillClassic Textbook Reissue Series, McGraw-Hill, New York, NY, 1987.

W. K. Pratt: Digital Image Processing, Wiley & Sons Inc, New York,NY, 1991.

Further reading:

T. K. Moon, W. C. Stirling: Mathematical Methods and Algorithmsfor Signal Processing. Prentice Hall, Upper Saddle River, NJ, 2000.

J. R. Deller, J. G. Proakis, J. H. L. Hansen: Discrete-Time Processingof Speech Signals, Macmillan Publishing Company, New York, NY,1993.

W. H. Press, S. A. Teukolsky, W. T. Vetterling, B. P. Flannery: Nu-merical Recipes in C, Cambridge Univ. Press, Cambridge, 1992.

L. Rabiner, B. H. Juang: Fundamentals of Speech Recognition, Pren-tice Hall, Englewood Cliffs, NJ, 1993.

T. Lehmann, W. Oberschelp, E. Pelikan, R. Repges: Bildverarbeitungfur die Medizin, Springer Verlag, Berlin, 1997.

L. Berg: Lineare Gleichungssysteme mit Bandstruktur, VEB DeutscherVerlag der Wissenschaften, Berlin, 1986.

Contents

1 System Theory and Fourier Transform 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Linear time-invariant Systems . . . . . . . . . . . . . . . . 11

1.3 Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . 16

1.4 Properties of the Fourier Transform . . . . . . . . . . . . . 25

1.5 Parseval Theorem . . . . . . . . . . . . . . . . . . . . . . . 33

1.6 Autocorrelation Function . . . . . . . . . . . . . . . . . . . 34

1.7 Existence of the Fourier Transform . . . . . . . . . . . . . 35

1.8 -Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

1.9 Motivation for Fourier Series . . . . . . . . . . . . . . . . . 41

1.10 Time Duration and Band Width . . . . . . . . . . . . . . . 45

2 Discrete Time Systems 51

2.1 Motivation and Goal . . . . . . . . . . . . . . . . . . . . . 52

2.2 Digital Simulation using Discrete Time Systems . . . . . . 53

2.3 Examples of Discrete Time Systems . . . . . . . . . . . . . 56

2.4 Sampling Theorem (Nyquist Theorem) and Reconstruction 61

2.5 Logarithmic Scale and dB . . . . . . . . . . . . . . . . . . 70

2.6 Quantization . . . . . . . . . . . . . . . . . . . . . . . . . 72

2.7 Fourier Transform and zTransform . . . . . . . . . . . . . 74

2.8 System Representation and Examples . . . . . . . . . . . . 78

2.9 Discrete Time Signal Fourier Transform Theorem . . . . . 88

2.10 Discrete Fourier Transform: DFT . . . . . . . . . . . . . . 90

2.11 DFT as Matrix Operation . . . . . . . . . . . . . . . . . . 98

2.12 From Continuous Fourier Transform to Matrix Representa-tion of Discrete Fourier Transform . . . . . . . . . . . . . . 102

2.13 Frequency Resolution and Zero Padding . . . . . . . . . . 104

2.14 Finite Convolution . . . . . . . . . . . . . . . . . . . . . . 105

2.15 Fast Fourier Transform (FFT) . . . . . . . . . . . . . . . . 108

2.16 FFT Implementation . . . . . . . . . . . . . . . . . . . . . 118

i

2.17 Cyclic Matrices and Fourier Transform . . . . . . . . . . . 124

3 Spectral analysis 1313.1 Features for Speech Recognition . . . . . . . . . . . . . . . 1323.2 Short Time Analysis and Windowing . . . . . . . . . . . . 1353.3 Autocorrelation Function and Power Spectral Density . . . 1593.4 Spectrograms . . . . . . . . . . . . . . . . . . . . . . . . . 1653.5 Filter Bank Analysis . . . . . . . . . . . . . . . . . . . . . 1683.6 Mel-frequency scale . . . . . . . . . . . . . . . . . . . . . . 1713.7 Cepstrum . . . . . . . . . . . . . . . . . . . . . . . . . . . 1733.8 Statistical Interpretation of the Cepstrum Transformation 1833.9 Energy in acoustic Vector . . . . . . . . . . . . . . . . . . 185

4 Fourier Transform and Image Processing 1874.1 Spatial Frequencies and Fourier Transform for Images . . . 1884.2 Discrete Fourier Transform for Images . . . . . . . . . . . 1964.3 Fourier Transform in Computer Tomography . . . . . . . . 1974.4 Fourier Transform and RST Invariance . . . . . . . . . . . 199

5 LPC Analysis 2075.1 Principle of LPC Analysis . . . . . . . . . . . . . . . . . . 2085.2 LPC: Covariance Method . . . . . . . . . . . . . . . . . . . 2125.3 LPC: Autocorrelation Method . . . . . . . . . . . . . . . . 2135.4 LPC: Interpretation in Frequency Domain . . . . . . . . . 2165.5 LPC: Generative Model . . . . . . . . . . . . . . . . . . . 2215.6 LPC: Alternative Representations . . . . . . . . . . . . . . 223

6 Outlook: Wavelet Transform 2256.1 Motivation: from Fourier to Wavelet Transform . . . . . . 2266.2 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . 2276.3 Discrete Wavelet Transform . . . . . . . . . . . . . . . . . 229

7 Coding (appendix available as separate document) 233

8 Image Segmentation and Contour-Finding 237

ii

List of Figures

1.1 Oscillograms of three time functions composed as sum of 20partial oscillations. a) n = 0, b) n =

2 , c) n statistical. 3

1.2 Amplitude spectrum of a time function composed as sum of20 partial tones. . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 from left to right: original photo, low-pass and high-passfiltered version . . . . . . . . . . . . . . . . . . . . . . . . 3

1.4 Phase manipulation for portion of a speech signal (vowel o)sampled at 8kHz, 25ms analysis window (200 samples), 512point FFT . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.5 Phase manipulation for portion of a speech signal (consonantn) sampled at 8kHz, 25ms analysis window (200 samples),512 point FFT . . . . . . . . . . . . . . . . . . . . . . . . 5

1.6 Phase manipulation for a Heavisidefunction (stepfunction) 6

1.7 Schematic representation of the physiological mechanism ofspeech production . . . . . . . . . . . . . . . . . . . . . . . 8

2.1 Digital photo . . . . . . . . . . . . . . . . . . . . . . . . . 58

2.2 Gradient image . . . . . . . . . . . . . . . . . . . . . . . . 58

2.3 Several real cases of Laplace Operator subtraction from orig-inal image. a) Original image b) Original image minusLaplace Operator (negative values are set to 0 and valuesabove the grey scale are set to the highest grade of grey) . 60

2.4 Ideal reconstruction of a band-limited signal (from Oppen-heim, Schafer)a) original signal b) sampled signal c) reconstructed signal 64

iii

2.5 Sampling of band-limited signal with different sampling rates:b) sampling rate higher than Nyquist rate - exact reconstruc-tion possiblec) sampling rate equal to Nyquist rate - exact reconstructionpossibled) sampling rate smaller than Nyquist rate - aliasing - exactreconstruction not possible . . . . . . . . . . . . . . . . . . 65

2.6 Amplitude spectrum of the voiceless phoneme s from theword ist . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

2.7 Logarithmic amplitude spectrum of the phoneme s . . . 71

2.8 Amplitude spectrum of the voiced phoneme ae from theword Ah . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

2.9 Logarithmic amplitude spectrum of the phoneme ae . . . 71

2.10 Amplitude spectrum of a speech pause . . . . . . . . . . . 71

2.11 Logarithmic amplitude spectrum of a speech pause . . . . 71

2.12 Hanning window . . . . . . . . . . . . . . . . . . . . . . . 103

2.13 Example of a linear convolution of two finite length signals:a) two signals;b) signal x[n-k] for different values of n:i) n < 0, no overlap with h[k], therefore convolution y[n] =0ii) n between 0 and Nh +Nx 2, convolution 6= 0iii) n > Nh +Nx 2, no overlap with h[k], convolution y[n]= 0c) resulting convolution y[n]. . . . . . . . . . . . . . . . . . 106

2.14 Flow diagram for decomposition of one N -DFT to two N/2DFTs with N = 8 . . . . . . . . . . . . . . . . . . . . . . . 110

2.15 Flow diagram of an 8pointFFT using Butterfly operations. 111

2.16 Flow diagram of an 8pointFFT using Butterfly operations. 120

2.17 Input and output arrays of an FFT. a) The input array con-tains N (N is power of 2) complex input values in one realarray of the length 2N . with alternating real and imagi-nary parts. b) The output array contains complex Fourierspectrum at N frequency values. Again alternating real andimaginary parts. The array begins with the zero-frequencyand then goes up to the highest frequency followed withvalues for the negative frequencies. . . . . . . . . . . . . . 122

iv

3.1 Example for the application of the Discrete Fourier Trans-form (DFT). . . . . . . . . . . . . . . . . . . . . . . . . . . 138

3.2 a) signal v[n]; b) DFT-spectrum V [k]; c) Fourier spectrumV (ej). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

3.3 a) signal v[n]; b) DFT-spectrum V [k]; c) Fourier spectrumV (ej). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

3.4 a) DFT of length N = 64; b) DFT of length N = 128; c)Fourier spectrum V (ej). . . . . . . . . . . . . . . . . . . . 151

3.5 Influence of the window function:above: speech signal (vowel a); central: 512 point FFTusing rectangle window; below: 512 point FFT using Ham-ming window . . . . . . . . . . . . . . . . . . . . . . . . . 158

3.6 Fourier Transform of a voiced speech segment:a) signal progression, b) high resolution Fourier Transform,c) low resolution Fourier Transform with short Hammingwindow (50 sampled values), d) low resolution Fourier Trans-form using autocorrelation function (19 coefficients), e) lowresolution Fourier Transform using autocorrelation function(13 coefficients) . . . . . . . . . . . . . . . . . . . . . . . . 162

3.7 Signal progression and autocorrelation function of voiced(left) and unvoiced (right) speech segment . . . . . . . . . 163

3.8 Temporal progression of speech signal and four autocorrela-tion coefficients . . . . . . . . . . . . . . . . . . . . . . . . 164

3.9 a) wide-band spectrogram: short time window, high timeresolution (vertical lines), no frequency resolution; for voicedsignals provides information on formant structure b) narrow-band spectrogram: long time window, no time resolution,high frequency resolution (horizontal lines); for voiced sig-nals provides information on fundamental frequency (pitch) 166

3.10 Wide-band and narrow-band spectrogram and speech am-plitude for the sentence Every salt breeze comes from thesea. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

3.11 Above: logarithmized power spectrum of a spoken vowel(schematic).Below: corresponding cepstrum (inverse Fouriertransformof the logarithmized power spectrum). . . . . . . . . . . . 177

v

3.12 Cepstral smoothing: speech signal (vowel a), windowedspeech signal (Hamming window), spectrum obtained fromthe whole cepstrum (blue) and smoothed spectrum obtainedfrom the first 13 cepstral coefficients (red). . . . . . . . . . 178

3.13 Homomorph analysis of a speech segment: signal progres-sion, homomorph smoothed spectrum using 13 and 19 cep-stral coefficients . . . . . . . . . . . . . . . . . . . . . . . . 179

4.1 TVimage (analog) . . . . . . . . . . . . . . . . . . . . . . 1934.2 Digitized TVimage . . . . . . . . . . . . . . . . . . . . . . 1934.3 Amplitude spectrum of Figure 4.2 . . . . . . . . . . . . . . 1934.4 Low-pass filtered . . . . . . . . . . . . . . . . . . . . . . . 1934.5 High-pass filtered . . . . . . . . . . . . . . . . . . . . . . . 1944.6 High-pass enhancement . . . . . . . . . . . . . . . . . . . . 194

5.1 LPCanalysis of one speech segmenta) signal progression, b) prediction error (K=12), c) LPCspectrum with K=12 coefficients, d) spectrum of the predic-tion error (K=12), e) LPCspectrum with K=18 coefficients 219

5.2 LPCSpectra for different prediction orders K . . . . . . . 220

List of Tables

2.1 Fourier transform pairs . . . . . . . . . . . . . . . . . . . . 872.2 Fourier transform Theorems . . . . . . . . . . . . . . . . . 88

Chapter 1

System Theory and FourierTransform

Overview:1.1 Introduction

1.2 Linear time-invariant Systems

1.3 Fourier Transform

1.4 Properties of Fourier Transform

1.5 Parseval Theorem

1.6 Autocorrelation Function

1.7 Existence of the Fourier Transform

1.8 Function

1.9 Fourier Series

1.10 Duration and Band Width

Digital Processing of Speech and Image Signals WS 2006/2007

1

1.1 Introduction

What distinguishes the Fourier Transform (FT) fromother transformations?

1. Mathematical property of linear time-invariant systems:

FT decomposes the time signal into eigenfunctionseigenfunctions keep their form by passing thelinear time-invariant system

A x = x

Magnitude of FT: shift invariant2. Physical observation:

Human ear produces sort of FT, essentially only magnitude of FT(strictly speaking: short-time FT)

Example:Time functions with different evolution can sound equally.The human ear either senses sense phase differences of partialtones of the complete sound of stationary processes very weakly,or does not sense them at all.

Fourier transform in speech processing:

Calculation of the spectral components of speech Basic method for obtaining observations (features) forspeech recognition


2

0

0

= 0

0

0

= pi/2

0

0

random

Figure 1.1: Oscillograms of three timefunctions composed as sum of 20 partialoscillations. a) n = 0, b) n =

pi2, c) n

statistical.

0

Figure 1.2: Amplitude spectrum of a timefunction composed as sum of 20 partialtones.

Figure 1.3: from left to right: original photo, low-pass and high-pass filtered version


3

amplitude spectrum

0

original signal

0

0

inverse FT for phase (f) = 0

0

0

Inverse FT for phase (f) = pi2

0

0

Inverse FT for random phase (f)

0

0

Figure 1.4: Phase manipulation for portion of a speech signal (vowel o) sampled at 8kHz,25ms analysis window (200 samples), 512 point FFT


4

amplitude spectrum

0

original signal

0

0


0

0

inverse FT for phase (f) = pi2

0

0

inverse FT for random phase (f)

0

0

Figure 1.5: Phase manipulation for portion of a speech signal (consonant n) sampled at8kHz, 25ms analysis window (200 samples), 512 point FFT


5

amplitude spectrum

0

original signal

0


0

0

inverse FT for phase (f) = pi2

0

0

inverse FT for random phase (f)

0

0

Figure 1.6: Phase manipulation for a Heavisidefunction (stepfunction)


6

Why Fourier?

Roughly:

Production, description and algorithmic operations on signals (func-tions or measurement curves over the time axis) can be described verywell in Fourier domain (frequency domain).

Deeper reason:

Production, description and algorithmic operations on signals are largelybased on linear time-invariant (LTI) operations.

Fourier Transform: simple representation of LTI-operations (later:convolution theorem)

Why continuous?

Real world is continuous Computer (digital = time discrete = sampled)model of the real world


7

t AT

glottal

pulsesvocal

tract

filter

radiation

from lips

and nose

t

T

speech [a:]a)

b)

0 4 0 4 0 4 0 4

1/T

|E(f)| [dB] |V(f)| [dB][a:] |A(f)| [dB] |S(f)| [dB]

* * =

[a:]

MUSCLEFORCE

LUNGVOLUME

PHARYNXCAVITY

NASALCAVITY

MOUTHCAVITY

TONGUEHUMP

VELUM

VOCALCORDS

TRACHEA ANDBRONCHI

LARYNXTUBE

NOSEOUTPUT

MOUTHOUTPUT

Figure 1.7: Schematic representation of the physiological mechanism of speech production


8

signal (speech, image)

feature extraction(signal analysis)

feature vector(pattern vector)

(pattern)comparison

decision

reference data(vectors, features)

Examples:

Spoken language Written numbers (letters) Cell recognition (red blood cells)


9

Examples of applications of Fourier Transform:

Electrical switchgears Recognition and coding

Speech and general acoustic signals

Image signals

Time series analysis: Astronomical measurement curves

Stock-market course

. . .

Computer tomography Solving differential equations Description of image production in optical systems


10

1.2 Linear time-invariant Systems

Example:

speech production

electrical systems

S

h(t)input signal

x(t)output signal

y(t)

symbolic:

{t y(t)} = S {t x(t)}simplified:

y(t) = S {x(t)} Note: the complete time domain of the function is important, notindividual positions in time t.

more exact:

y = S {x}LTISystem: (LTI = Linear Time-Invariant)

Linear:Additive:

S {x1 + x2} = S {x1}+ S {x2}

Homogeneous:

S { x} = S {x} , IR

Time-invariant:{t y(t t0)} = S {t x(t t0)} , t0 IR


11

Mathematical theorem:

Linearity and time invariance result in convolution representation Output signal y(t) of LTI system S with input signal x(t):

y(t) =

x(t ) h() d

=

x() h(t ) d

= x(t) h(t)

h: impulse response of the system S

1/

e (t)

t t

x (t)

i

system response h (t) to excitation e (t):h(t) = S {e(t)}

signal x(t) is represented as sum of amplitude weighted and timeshifted elementary functions e(t):

x(t) = lim0

[i

x(i) e (t i) ]


12

Hence the following holds for the output signal y(t):

y(t) = S {x(t)}

= S

{lim0

i

x(i) e(t i) }

= lim0

[S

{i

x(i) e (t i) } ]

additivity:

= lim0

[i

S { x(i) e(t i) }]

homogeneity (for x(i) and ):

= lim0

[i

x(i) S { e (t i) } ]

time invariance:

= lim0

[i

x(i) h (t i) ]

limiting case 0 :

di

h(t) h(t)result:

y(t) =

x() h(t ) d = x(t) h(t)

h(t): impulse response of the system


13

Examples of LTI-operations:

Oscillatory systems (electrical or mechanical) withexternal excitation:

x(t) h() y(t)

y(t) =

h(t ) x() d

y(t) + 2y(t) + 2y(t) = x(t)

, : parameters depending on the oscillatory system

More general electrical engineering systems:high-pass, low-pass, band-pass

Sliding average value:

x(t) S y(t) := x(t)

x(t) =1

T

+T/2T/2

x(t+ ) d

Differentiator:x(t) S y(t) := x(t)

Comb filter: hypothesized period Tx(t) S y(t) := x(t) x(t T )

In general: linear differential equations with coefficients ck and dlk

cky(k)(t) =

l

dlx(l)(t)

[ + further constraints ]


14

Example of a non-linear system:

system: y(t) = x2(t)

x(t) = A cos(t)

= y(t) = A2 cos2(t) = A2

2(1 + cos(2t))

frequency doubling


15

1.3 Fourier Transform

Sinusoidal oscillation:

x(t) = A sin ( t + )

amplitude A

phase / null phase

angular frequency = 2 f

j2 = 1, j C

cos

sin

Im

Re

1

1

complex representation: ej = cos + j sin , IR

cos =ej + ej

2and sin =

ej ej2j

dimension:

DIM() DIM(t) = 1

DIM() =1

DIM(t)=

1

[sec]= [Hz]


16

LTI-System

y(t) =

x(t )h()d = x(t) h(t)

Determine the following specific input signal:x(t) = A ej(t+)

For this input signal the output signal becomes:

y(t) =

A ej((t)+)h()d

= A ej(t+)

h()ejd

H() = F {h()}

= x(t) H()

Definition of the Fourier transform:

H() =

h()ejd = F {h()} = F { h()}

( decomposition into ej)

H() is called transfer function of the system

Remark about x(t) = A ej(t+):

The shape of the input signal x(t), i.e. its frequency (eigenfunc-tion) remains invariant

Amplitude (intensity) and phase (time shift) are depending onH() (eigenvalue)

( analogy to the problem of eigenvalues in linear algebra)


17

Remarks

FT is complex:H() = Re {H()} + j Im {H()} = |H()| ej()

Amplitude (spectrum):

|H()| =Re {H()}2 + Im {H()}2

Phase (spectrum):

() =

arctan

(Im {H()}Re {H()}

)Re {H()} > 0

arctan

(Im {H()}Re {H()}

)+ Re {H()} < 0

2Re {H()} = 0,Im {H()} > 0

2

Re {H()} = 0,Im {H()} < 0


18

Examples of Fourier transforms:

1. Rectangle function

h(t) = rect(t

T) =

{1, |t| T/20, |t| > T/2

H() =

h(t)ejtdt =

T2

T2

ejtdt =1

j[ej

T2 ej T2

]

=2

sin(

T

2) =

T sin(T

2)

T

2

(here: Im {H()} = 0)

t

h(t)

H()


19

2. Double-sided exponential

h(t) = e|t| with > 0

H() =

h(t)ejtdt

=

0

e(+j)tdt+

0

e(j)tdt

=

[e(+j)t

(+ j) +e(j)t

( j)]0

= 0 + 0 1(+ j) 1

( j)=

j + + j2 + 2

=2

2 + 2

Imaginary part equals 0 Infinite spectrum No zeros

h(t)

t

H( )

If h(t) is symmetric (i.e. h(t) = h(t)), imaginary parts drop awayand the real part is sufficient


20

3. Damped oscillations

h(t) = e|t| cos(t) with > 0

H() =

h(t)ejtdt

=

0

e(+j)t cos(t)dt+

0

e(j)t cos(t)dt

=

0

e(+j)tejt + ejt

2dt+

0

e(j)tejt + ejt

2dt

= . . . (elementary calculation)

=

2 + ( )2 +

2 + ( + )2

Limiting case:

H()|= = 1

+

2 + (2)2

= tends towards or if tends towards 0

H( )

| |

h(t)

t


21

4. Modulated rectangle function (truncated cosine)

h(t) =

{cos( t), |t| T/2

0, |t| > T/2

H() =

h(t)ejtdt

=

T2

T2

cos( t)ejtdt

= . . . (elementary calculation)

=T

2

sin(( ) T

2

)( ) T

2

+

sin

(( + )

T

2

)( + )

T

2

| |

h(t)

t

H( )

h(t)

t

| |

H( )


22

Fourier Transform pairs (u = /2)

Rectangle function

1

-1/2 1/2

-1/2 1/2

Triangle function

1

Exponential function

e-|x|

Gaussian function

e-x2

Unit impulse

(x)

1

Sinc function

Squared sinc function

sin(piu)piu

1

2+(2piu)22

piu2

e-pi

1


23

Inverse Fouriertransform

H() =

h(t)ejtdt

assumption: h(t) =1

2

H()ejtd

with: H() =

h()ejd

inserting H() in h(t):

h(t) =1

2lim

,T

TT

h() ej(t) d

d=

1

2lim

limT

TT

ej(t) d h() d

= lim

limT

1

TT

sin ((t ))t h() d

= lim

1

sin ((t ))t h() d

= h(t)

due to:

lim

1

sin(t)

th(t) dt = h(0)

formal expression:

h(t) =

12

ej(t) d

= (t )

h() d

( distribution theory, see there for stronger proof)Digital Processing of Speech and Image Signals WS 2006/2007

24

1.4 Properties of the Fourier Transform

Symmetry

H() =

h(t) ejt dt = F {h(t)}

h(t) =1

2

H() ejt d = F1 {H()}

F 2{h(t)} = F{H()} = 2h(t)

F1 F{h(t)} = F1{H()} = h(t)

Time domain and frequency domain are correlated symmetrically. Properties of FT are valid in both domains, especially the convolutiontheorem (see later).


25

Theorems for the Fourier transform

H() =

ejt h(t) dt

consider the equation:

H() = F {h(t)}more exact:

{ H()} = F {t h(t)}

1. Linearity: integral operator is linear

2. Inverse scaling, similarity principle:

h(t) ejt dt =1

||

h() ej d

F{h(t)} = 1|| H(

), IR\{0}

Note:

Absolute value, because integral boundaries are swapped for < 0.

3. Shift: h(t t0)

h(t t0) ejt dt = ejt0

h(t t0) ej(tt0) dt

= ejt0

h() ej d


26

= F{h(t t0)} = ejt0H() t0 IR

with H() = F{h(t)}

important:

| F{h(t t0)} | = | F{h(t)} | , because

|ejt0| = |eju| = | cosu j sinu|=

cos2 u + sin2u

= 1

4. Symmetry and antisymmetry:

h(t) = h(t) results in Im{H()} = 0

h(t) = h(t) results in Re{H()} = 0

5. Complex conjugation: suppose that h(t) is a complex function

h(t) ejt dt =

h(t) ejt dt

=

h(t) ejt dt = H()

F{h(t)} = H() = F{h(t)}

Special case: h(t) is real, so h(t) = h(t)

= H() = H() = | H() | = | H() | = | H() |


27

6. Differentiation:

dh

dt=

t

12

H() ejt d

=

1

2

H() j ejt d

F{dh(t)dt

} = j F{h(t)}

Interpretation: differentiation = enhancement of high frequencies(due to the multiplication with )

7. Integration:

F{t

h()d} = 1

jF{h(t)}

Proof: similar to differentiation or inversion

8. Modulation principle:

F{h(t) cos(0t)} =

h(t) cos(0t) e

jt dt

=1

2

h(t) ej0t ejt dt +

h(t) ej0t ejt dt

=

1

2

h(t) ej(0)t dt +

h(t) ej(+0)t dt

=

1

2[ H( 0) + H( + 0) ]

and similarly

F{ h(t) sin(0t) } = 12j

[ H( 0) H( + 0) ]


28

h(t), H() x(t) X()

y(t) Y()

Convolution theorem

Convolution in time domain corresponds to multiplication in frequencydomain

Time domain: y(t) = x(t) h(t) =

x(t ) h() d

Frequency domain:

Y () =

ejt

h() x(t ) d dt

=

h()

x(t ) ejt dt d

=

h() X() ej d (shifting)

= X()

h() ej d

= X() H()


29

Likewise, multiplication in time domain corresponds to convolution infrequency domain (note the factor 12):

Time domain: y(t) = a(t) b(t)Frequency domain:

Y () =

a(t) b(t) ejt dt

=

a(t)1

2

B()ejt ejt d dt

=1

2

B()

a(t)ej()t dt d

=1

2

A( ) B()d

=1

2A() B()

Motivation for the Fourier transform:FT gives the simplest representation of the system operation, be-cause every LTI-System can be interpreted as convolution of the inputsignal x(t) and the impulse response of the system h(t). Convolutioncan be then efficiently calculated using FT and convolution theorem.

Mathematical: eigenfunctions


30

Example: Oscillator with excitation

x(t) Oscillator y(t)

y(t) + 2 y(t) + 2 y(t) = x(t)

x(t) =1

2

+

X()ejtd

y(t) =1

2

+

Y ()ejtd

y(t) =1

2

+

Y ()j ejtd

y(t) =1

2

+

Y ()[2] ejtd

+

[2 + 2j + 2]Y ()ejtd =+

X()ejtd

+

{[2 + 2j + 2] Y ()X()}

=0

ejtd = 0 t

In this way we obtain the transfer function of an oscillator:

H() =Y ()

X()=

1

2 + 2j + 2


31

h(t) =1

2

+

H()ejtd

(can be given explicitly)

y(t) =

+

x(t) h(t )d

Note:y(t) does not contain the component which corresponds to the homoge-neous differential equation of the oscillator.

FourierTransform

Convolution with h(t)

Multiplication with H() = F{h(t)}

Inverse FourierTransform

x(t)

X()

y(t)

Y()


32

1.5 Parseval Theorem

Convolution theorem:

F1 {H() X()} =

h(t) x( t) dt

()1

2

H() X() ej d = (h x) ()

We make two special assumptions:

i) x(t) := h(t), then: X() = H()ii) = 0

Inserting in () results in:

1

2

H()H() d =

h(t)h(t) dt

1

2

|H()|2 d =

|h(t)|2 dt = E

Energy E in time domain = Energy E in frequency domain(up to the factor

1

2; aid: use normalization factor

12

for both

directions of Fourier Transform)

Physical aspect: energy conservation Mathematical aspect: unitary (orthogonal) representation in vectorspace

|H()|2 is called power spectral density.


33

1.6 Autocorrelation Function

Autocorrelation function

Autocorrelation function of time continuoussignal or function h(t) is defined as:

R(t) =

h() h(t+ )d

The following equation is valid:R(t) = h(t) h(t) which results in R(t) = R(t)

Fourier transform gives: (Wiener-Khinchin Theorem)F{R(t)} = H() H() = |H()|2

Thus: Fourier transform connects autocorrelationfunction R(t) and power spectral density |H()|2

|H()|2 =

R(t) ejt dt =

R(t) cos(t) dt

Remark:autocorrelation is a special case of the cross correlation between sig-nals x() and h(t)

Ch,x =

h() x(t+ )d


34

1.7 Existence of the Fourier Transform

Conditions for h(t) for the existence of the Fourier transform

H() =

ejt h(t) dt , h(t) =1

2

ejt H() d

When are those equations valid?

Sufficient conditions:

1. h(t) is absolutely integrable:

|h(t)|dt

Impulse response:

y(t) =

h(t )() d

= h(t) (t)

= h(t)

Consequence:

h(t) 1

(t) dt = 1

A function like (t) does not exist. But it is possible to define thefunctional for each function t h(t):

[t h(t)] (h) := h(0)

1.8 -Function

Starting point: definition of the -function as a boundary case ofa function (t):

lim0

+

f(t) (t) dt = f(0) (1.1)

Possible realizations of (t)

a) (t) =

1

2t [,+]

0 otherwise

b) (t) =1

2 + t2


36

c) (t) =1

sin (t/)

t

d) (t) =122

et2

22

During inversion of the Fourier transform we have formally ob-tained:

(t) =1

2

+

ejt d = lim

1

sin (t)

t(1.2)

Fourier transform F{(t)}:

F{(t)} =+

ejt(t) dt

due to (1.1) the following holds:

F{(t)} = ejt|t=0 = 1

Another derivation using (1.2):

(t) =1

2

+

ejt F{(t)} d general

=1

2

+

ejt d according to (1.2)

Comparison results in:

F{(t)} = 1


37

From this we obtain the following equations:

From symmetry property:

F{1} = 2 ()

From shifting theorem:

F{ej0t 1} = 2 ( 0)

cos (0 t) =1

2

[ej0t + ej0t

]=

1

2

+

( 0) ejt d ++

( + 0) ejt d

=

1

2

+

[ ( 0) + ( + 0) ] ejt d

F{ cos (0 t) } = [ ( 0) + ( + 0) ]

Note: another derivation:

consider damped oscillations

1

2e|t| cos (0t)

in the limit 0 .


38

Comb function

define comb function (pulse train, sequence of -impulses):

x(t) =+

n=(t nT )

Fourier transform of comb function:

X() =

+

x(t) ejt dt

=

+

+n=

(t nT ) ejt dt

=+

n=

+

(t nT ) ejt dt

=+

n=ejnT

= . . . (see Papoulis 1962, p. 44)

=2

T

+n=

( n2T)

in words:

-impulse sequence with period T in time domain

produces

-impulse sequence with period 1T in frequency domain(i.e. 2T in -frequency domain)

comb function is transformed to comb function


39

Comb function

cos(0t)

sin(0t)

-2pi-4pi-6pi 2pi 4pi 6piT T T T T T

T 3T-T-3T 6T-6T

(t-nT)n=-

12j((-0)+(+0))

((-0)+(+0))12

(-n2pi/T)n=-

2piT

0 0

0

0


40

1.9 Motivation for Fourier Series

x : IR IRt x(t)

Consider a periodical function x with period T :

x(t) = x(t+ T ) for each t IRthen also x(t) = x(t+ kT ) for k Z

Examples:

Constant function:x0(t) = A0

Harmonic oscillator:

x1(t) = A1 cos (2

Tt + 1) , A1 > 0

All higher harmonic:

xn(t) = An cos (n2

Tt + n) , An > 0

therefore

x(t) =n=0

An cos (n 0 t + n) with 0 =2

T, An 0

is periodical with period T = 20

Another notation:

x(t) =

n=Bn e

j n 0 t where Bn is a complex number


41

Line spectrum representation

Real measured signal has always a widespread spectrum.

Reasons:

Strictly periodical signal (almost) never exists Period can fluctuate

Wave form within one period can fluctuate

Only a finite section of the signal is analyzed(window function)

Only a strictly periodical signal has a sharp line spectrum

Remarks:

Fourier series are actually not strictly related to periodical functions:a finite interval of IR is sufficient (the signal is then interpreted asinfinitely prolonged).

By transition from the finite interval to the complete real axis theFourier series becomes Fourier integral.


42

Calculation of Fourier coefficients:

Consider a periodical function x(t) with period T = 20 approach:

x(t) =+

n=an e

j n0 t a C

multiplication with ej m0 t where m IN and integration over oneperiod result in:

+T/2T/2

x(t) ej m 0 t dt =+

n=an

+T/2T/2

ej (nm) 0 t dt

Due to orthogonality holds:+T/2T/2

ej (nm) 0 t dt ={T if n = m0 if n 6= m

Then:T/2

T/2

x(t) ej m 0 t dt = am T

Result:

an =1

T

+T/2T/2

x(t) ej n 0 t dt

=1

T

+T/2T/2

x(t) cos (n 0 t) dt j 1T

+T/2T/2

x(t) sin (n 0 t) dt


43

Spectrum of a periodical function

If x(t) is periodical with the period T = 20 , then

x(t) =+

n=an e

j n0 t, an C

The Fourier transform X() is:

X() = F{x(t)}

=+

n=an F{ej n 0 t}

= 2( n0)

= 2 +

n=an ( n0)

Note:This derivation is formal, because the Fourier integral does notexist in the usual sense;strict derivation within the scope of distribution theory.

In words:a periodic function with the period T has a Fourier transform in theform of a line spectrum with the distance 0 =

2T between the com-

ponents.


44

1.10 Time Duration and Band Width

1. Similarity principle:

F{h(t)} = 1|| H(

)

t

h( t)0

2. Special case: h(t) with

Im {H()} = 0 ( h(t) symmetrical )and

Re {H()} 0

h(t) has maximum for t = 0:

h(t) =1

2

H() cos(t) d 12

H() d = h(0)

define:

T =1

h(0)

h(t) dt

B =1

H(0)

H() d

from

T =H(0)

h(0)and B = 2

h(0)

H(0)

follows

T B = 2


46

3. In general: normalized impulse h(t) IR with

h2(t) dt = 1, h(t) IR

T 2 :=

h2(t) t2 dt

B2 :=

|H() |2 2 d = 2

[h(t)]2 dt

Results in uncertainty relation:

T B

2

Proof: Cauchy-Schwarz inequality

| xT y | ||x|| ||y||

[ t h(t) ] h(t) dt

2

=1

4

[ t h(t)]2 dt

= T 2

[ h(t) ]2 dt

B2

2From: partial integration

u(t) v(t) dt = u(t) v(t)

u(t) v(t) dt

[ h(t) h(t) ]

u(t)

tv(t)

dt =1

2h(t)2 t

1

2h2(t) 1 dt

[ h(t) h(t) ] t dt = 0 12


47

Equality sign is valid for linear dependency:

h(t) = t h(t)dh

h= t dt

log(h) = 12 t2 + const., > 0

= Optimum T B =

2for Gauss impulse

h(t) =2

e12 t2

Variance: 2 =1

Quantum Physics: similar statement about position and impulseof a particle


48

4. Finite positive signal

g(t)

{ 0 0 t T= 0 t < 0 or T < t

0 T

g(t)

t

The following is valid for the amplitude spectrum |G()|:

|G()| =

+

g(t) ejtdt

+

|g(t)| |ejt|dt

=

+

|g(t)| dt

because g(t) 0= G(0)

Define the band width B as:

|G(B)|2 = G2(0)

2and |G(B)|2 |G()|2 for || < B

Then:

T B 2


49

Proof:

The following inequalities are valid:

a2 + b2 (a b)2

2 a, b IR

| sin|+ | cos| 1 IR

For the Fourier-Transform of g(t) holds:

Re{G()} =T0

g(t) cost dt

Im{G()} = T0

g(t) sint dt

For 0 t 2holds: cost 0, sint 0

and therefore: cost+ sint = | cost|+ | sint| 1

Re{G()} Im{G()} =T0

g(t) [cost+ sint] dt

T0

g(t) 1 dt

= G(0)

|G()|2 = Re2{G()}+ Im2{G()}

[Re{G()} Im{G()}]2

2

12G2(0) |G(B)|2


50

Chapter 2

Discrete Time Systems

Overview:2.1 Motivation and Goal

2.2 Digital Simulation using Discrete Time Systems

2.3 Examples of Discrete Time Systems

2.4 Sampling Theorem and Reconstruction

2.5 Logarithmic Scale and dB

2.6 Quantization

2.7 Fourier Transform and zTransform

2.8 System Representation and Examples

2.9 Discrete Time Signal Fourier Transform Theorems

2.10 Discrete Fourier Transform (DFT)

2.11 DFT as Matrix Operation

2.12 From continuous FT to Matrix Representation of DFT

2.13 Frequency Resolution and Zero Padding

2.14 Finite Convolution

2.15 Fast Fourier Transform (FFT)

2.16 FFT Implementation


51

2.1 Motivation and Goal

If we want to process a continuous time signal x(t) with a computer, wehave to sample it at discrete equidistant time points

tn = n TSwhere TS is called sampling period.

Terminology:time discrete is often called digital, where this adjective often(but not always) denotes the amplitude quantization,i.e. the quantization of the value x(n TS).

Advantages of digital processing in comparison to analog components:

independent of analog components and technical difficulties with re-spect to their realization;

in principle arbitrary high accuracy; also non-linear methods are possible,in principle even every mathematical method.


52

2.2 Digital Simulation using Discrete Time Systems

Task definition:

Given:Analog system with input signal x(t) and output signal y(t);Sampling with sampling period TS

Wanted:Discrete System with input signal x[n] and output signal y[n], suchthat

x[n] = x(nTS)

results in

y[n] = y(nTS)

For which signals is such a digital simulation possible? The sampling theorem gives (most of) the answer.


53

LTI System (analog to continuous time case):

Linearity:Homogeneity:

S { x[n]} = S {x[n]}

Additivity:

S {x1[n] + x2[n]} = S {x1[n]} + S {x2[n]}

Shift invariance:S {x[n n0]} = y[n n0], n0 whole number


54

Representation of an LTI System as discrete convolution:

Unit impulse:

[n] =

{1, n = 00, n 6= 0

The signal x[n] is represented with amplitude weighted and time shiftedunit impulses [n]. The system reacts on [n] with h[n]:

h[n] = S {[n]}Input signal:

x[n] =

k=x[k] [n k]

Output signal:

y[n] = S

{ k=

x[k] [n k]}

Additivity

=

k=S { x[k] [n k] }

Homogeneity

=

k=x[k] S { [n k] }

Time invariance

=

k=x[k] h[n k]

Input signal x[n] and output signal y[n] of a discrete time LTI system arelinked through discrete convolution.h[n] is called impulse response like in continuous time case.


55

2.3 Examples of Discrete Time Systems

Difference calculation:y[n] = x[n] x[n n0]

1-2-1-averaging:y[n] = 0.5 x[n 1] + x[n] + 0.5 x[n+ 1]

sliding window averaging (smoothing)

y[n] =1

2M + 1

Mk=M

x[n k]

weighted averaging: instead of constant weight

h[n] =1

2M + 1

arbitrary weights can be used:

y[n] =M

k=Mh[k] x[n k]

Note: the only difference from general case isfinite length of the convolution kernel h[n].

First order difference equation:(recursive averaging, averaging with memory)

y[n] y[n 1] = x[n]

(Digital) resonator (second order difference equation)y[n] y[n 1] y[n 2] = x[n]

Image processing:Gradient calculation and image enhancement(Roberts Operator, Laplace Operator)


56

Roberts Cross Operator

gray values x[i, j]

i i+1

j

j+1

2

|x[i, j]|2 = (x[i, j] x[i+ 1, j + 1])2 + (x[i, j + 1] x[i+ 1, j])2

Note: non-linear operation

simplified:

|x[i, j]| = |x[i, j] x[i+ 1, j + 1]|+ |x[i, j + 1] x[i+ 1, j]|


57

Figure 2.1: Digital photo

Figure 2.2: Gradient image


58

Laplace Operator discrete approximation of the second derivation

2x[i, j] = 2ix[i, j] +2jx[i, j]= x[i+ 1, j] 2x[i, j] + x[i 1, j] +

x[i, j + 1] 2x[i, j] + x[i, j 1]= x[i+ 1, j] + x[i 1, j] + x[i, j + 1] + x[i, j 1] 4x[i, j]

1 -2 1

1

1

1

1

1

1-4

-2 i-1 i i+1

j+1jj-1

Image enhancement:

y[i, j] = x[i, j]2x[i, j]= h[i, j] x[i, j]


59

Figure 2.3: Several real cases of Laplace Operator subtraction from original image. a)Original image b) Original image minus Laplace Operator (negative values are set to 0and values above the grey scale are set to the highest grade of grey)


60

2.4 Sampling Theorem (Nyquist Theorem) and Re-

construction

The following will be analyzed and derived respectively:

How should we choose the sampling period TS, if we want to represent acontinuous signal x(t) with its sample values x(nTS) so that the signal x(t)can be exactly reconstructed from its sample values?

Fourier transform of the continuous time signal x(t):

X() = F { x(t) } =

x(t) ejt dt

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

X() ejt d (2.1)

Signal x(t) has limited bandwidth with upper limit B, which means:X() = 0 for all || BNote: X(B) = 0

X() in domain B < < B can be represented as Fourier Series:

X() =

n=an exp(jn

B) (2.2)

The coefficients an are given by:

an =1

2B

BB

X() exp(jn

B) d (2.3)

Comparison of the equations (2.1) and (2.3) shows that the coefficientsan are given by the values of the inverse Fourier transform of x(t) atpoints

tn =n

B(2.4)

The band limitation of X() has to be considered for the integrationlimits in (2.1). Result:

an = x(n

B)

B(2.5)


61

Inserting Eq. (2.5) into Eq. (2.2) and then in Eq. (2.1) results in:

x(t) =1

2

BB

B

n=

x(n

B) exp(jn

B) exp(jt) d

After swapping summation and integration and subsequent integra-tion:

x(t) =

n=x(n

B)sin(B (t n

B))

B (t nB

)

Reconstruction of the signal x(t) from sample values is possible ifequidistant sample values x(

n

B) = x(n Ts) have the distance TS

TS =

B(2.6)

The sampling period TS corresponds to the sampling frequency S:

S =2

TS

Equation (2.6) shows that if the sampling frequency isS := 2 B

the original signal x(t) can be reconstructed exactly.

In the Fourier series representation of X() in equation (2.2), theperiod 2 B has been supposed.B is the highest frequency component of the signal x(t).


62

Since X() is equal to zero for || B, the period 2 B can besubstituted with every period 2 B where B B. The previousderivation is also valid for this B.

WhenB =

TS

then:

x(t) =

n=x(nTS)

sin( (t nTS)/TS) (t nTS)/TS

(reconstruction formula)

Note: limt0sin(t)t = 1 (lHopitals rule)

The condition B B results in:

TS B

(2.7)

for the sampling period TS and in:

S 2 B (2.8)for the sampling frequency S.

The equations (2.7) and (2.8) are denoted as sampling theorem. Thesampling frequency has to be at least twice as high as the upper limitfrequency of the signal B where X() = 0 for || B. If and onlyif this condition is satisfied, an exact reconstruction (without approx-imation) of a continuous signal x(t) from its sample values x(nTS) ispossible.

Note: The sampling frequency S = 2 B is also calledNyquist frequency.


63

tx(t)

t

xs(t)

xr(t)

T

T

a)

b)

c)

Figure 2.4: Ideal reconstruction of a band-limited signal (from Oppenheim, Schafer)a) original signal b) sampled signal c) reconstructed signal


64

X()

a)

b)

-S S

. . . . . .

XS1() S > 2,

XS2()

c)

, S = 2

-S S

. . . . . .

(Nyquist rate)

XS3()

, S < 2 (aliasing)

. . . . . .

SS

d)

Figure 2.5: Sampling of band-limited signal with different sampling rates:b) sampling rate higher than Nyquist rate - exact reconstruction possiblec) sampling rate equal to Nyquist rate - exact reconstruction possibled) sampling rate smaller than Nyquist rate - aliasing - exact reconstruction not possible


65

Another proof using delta- and comb-function:

Sampling of the continuous signal x(t) with S =2TS

Band limitation: X() = 0 for || B(always possible: analog to low-pass with T () = 0 for || B)

Sampling procedureMultiplication of a function with a comb-function in time domain

xs(t) = Ts x(t) +

n=(t nTs)

results in a convolution with a comb-function in frequency domain:

Xs() = Ts 12

X() 2Ts

+n=

( 2n

Ts

)

=

+

X()+

n=

( 2n

Ts

)d

=+

n=X

( n2

Ts

)

= sampled signal has periodical Fourier spectrum(Analogy to Fourier series: periodical signal has line spectrum, i.e.discrete spectrum)

No overlap if:

B S B2B S


66

In so-called digital simulation, the signal x(t) is represented by itssampled values x(n TS) measured at equidistant time points withdistance TS. With a proper sampling period TS an exact reconstruc-tion of the signal x(t) from the sampled values x(n TS) is possible.

If it is possible to exactly reconstruct the signal x(t) from the sampledvalues x(nTS), then it is possible to perform a discrete time processingof the sampled values x(n TS) on a computer, which is equivalent tothe continuous time processing of the signal x(t) (digital simulation).

Continuous time processing:

y(t) =

x() h(t ) d

Discrete time processing: Sampling period TS

x[n] := x(nTS)

y(nTS) =

k=x(kTS) h(nTS kTS) TS, h[n] = h(nTS)

y[n] =

k=x[k] h[n k]

As a result of the convolution theorem (convolution in time domaincorresponds to multiplication in frequency domain), the band limitedinput signal gives an also band limited output signal which is exactlydetermined by its sampled values.


67

Important:

In the domain || < S/2 the Fourier transform of a continuous timesignal x(t) is identical with the Fouriertransform of the correspondingsampled discrete time signal x(nTS):

X() =

x(t) exp(jt) dt

for || S/2 is identical to

TS XS() = TS

n=x(nTS) exp(jTSn)

= TS

n=x(nTS) exp(j2

Sn)

Inverse Fourier transform of discrete time signal:

x(nTS) =1

S

S/2S/2

XS() exp(jTSn) d

One period:

S2

S2

2S

The Fourier transform of a discrete time signal is periodic in withthe period 2/TS = S.

The Fourier transform of a discrete time signal iscontinuous in .


68

Frequency normalization

Define the normalized frequency N :

N : = 2

S

Definition: ( now denotes a normalized frequency)

Fourier transform of discrete time signal x[n]:

X(ej) =+

n=x[n] exp(jn)

Note the notation X(ej).

Inverse Fourier transform of discrete time signal x[n]:

x[n] =1

2

X(ej) exp(jn) d


69

2.5 Logarithmic Scale and dB

Why?

large dynamic range for the amplitude values of a signal

x(t) = A cos t

A := amplitude(pressure, velocity, inclination, current, voltage, ... )linear variable

A0 := reference amplitudepredefined value for calibration

dB := decibel

A[dB] 20 lg AA0

, lg log10

= 10 lg A2

A20, A2 = quadratic variable = energy, intensity

because of 210 = 1024 = 103:

1 bit more =

{factor 2 for amplitude = 6 dB= factor 4 for intensity

3 dB = factor 2 for intensity


70

01

2

3

4

5

6

7

0 1000 2000 3000 4000 5000 6000 7000 8000

A

f / Hz

Phonem: s

Figure 2.6: Amplitude spectrum of thevoiceless phoneme s from the wordist

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 1000 2000 3000 4000 5000 6000 7000 8000

log

A

f / Hz

Phonem: s

Figure 2.7: Logarithmic amplitude spec-trum of the phoneme s

0

2

4

6

8

10

12

0 1000 2000 3000 4000 5000 6000 7000 8000

A

f / Hz

Phonem: ae

Figure 2.8: Amplitude spectrum of thevoiced phoneme ae from the wordAh

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 1000 2000 3000 4000 5000 6000 7000 8000

log

A

f / Hz

Phonem: ae

Figure 2.9: Logarithmic amplitude spec-trum of the phoneme ae

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1000 2000 3000 4000 5000 6000 7000 8000

A

f / Hz

Pause

Figure 2.10: Amplitude spectrum of aspeech pause

-3

-2.5

-2

-1.5

-1

-0.5

0

0 1000 2000 3000 4000 5000 6000 7000 8000

log

A

f / Hz

Pause

Figure 2.11: Logarithmic amplitudespectrum of a speech pause


71

2.6 Quantization

Uniform quantization

-XMAX XMAX

Quantisation: x = Q(x)

B bits correspond to 2B quantisation levels

Boundaries: x0, x1, . . . , xk, . . . , xK where K = 2B

Width of one quantisation level using uniform quantisation:

= 2 XMAX2B

Quantisation error:

2e =

+

(x x)2 p(x) dx =Kk=1

xkxk1

(x xk)2 p(x) dx

for uniform quantisation:

a) xk xk1 = = const(k)b) xk =

12(xk1 + xk)

uniform distribution with p(x) = const(x) results in:

2e =k

212 1K

=212

=X2MAX3 22B


72

signal-to-noise ratio in dB (general definition):

SNR[dB] := 10 lg2x2n

2x = power of the signal x2n = power of the noise n

SNR = signal-to-noise ratio

signal-to-quantisation noise ratio (special case):

SNR[dB] := 10 lg2x2e

2e = power of the noise caused by quantisation errors

uniform quantisation using B bits:

SNR[dB] = 6.02 B + 4.77 20 lg XMAXx

if signal amplitude has Gaussian distribution, only 0.064% of sampleshave amplitude greater than 4x:

SNR[dB] = 6.02 B 7.2 for XMAX = 4x


73

2.7 Fourier Transform and zTransform

Transfer function and Fourier transform

Eigenfunctions of discrete linear time invariant systems (analog to timecontinuous case):

x[n] = ej n < n < ( is dimensionless here)

Proof:

x[n] = ej n

y[n] =

k=h[k] ej (nk)

= ej n

k=h[k] ej k

Define: H(ej ) =

k=h[k] ej k

Remark:The Fourier transform of a discrete time signal is already introduced asFourier series during the derivation of sampling theorem and reconstruc-tion formula (equation (2.2)).

Result: y[n] = ej n H(ej )


74

ztransform:

Fourier transform of a discrete time signal: x[n]

X(ej) =+

n=x[n] ejn

periodic in

is normalized frequency, thence:

<

X is evaluated on the unit circle (ej)

Generalization: X is evaluated for any complex values z. That results in ztransform:

X(z) =+

n=x[n] zn

Reasons for ztransform1. analytically simpler, function theory methods are applicable

2. better handling of convergence problem:

convergence of finite signal, i.e. x[n] = 0 for each n > N0

convergence of infinite signal depends on z

Inverse ztransform:

x[n] =1

2j

X(z) zn1 dz

formally: z = ej dz = jzd

x[n] =1

2

20

X(ej) ejn d


75

Example of Fourier transform and ztransform:

Truncated geometric series

x[n] =

{an 0 n N 10 otherwise

ztransform

X(z) =N1n=0

an zn =N1n=0

(a z1)n =1 (a z1)N1 a z1

=1

zN1zN aNz a

Fourier transform

ztransform results in Fourier transformation using substitution

z = ej

X(ej) =1 aN ejN1 a ej

special case for a = 1 (discrete time rectangle):

= exp

(j(N 1)

2

) sin(N2

)sin(2

)


76

Proof for the ztransform inversion

Statement:

x[k] =1

2j

X(z) zk1 dz

Cauchy integration rule1

2j

zkdz =

{1 k = 10 k 6= 1

1

2j

X(z) zk1dz =

1

2j

n

x[n] zn+k1dz

=n

x[n]1

2j

zn+k1dz

6= 0 only for n = k= x[k]

Fourier:z = ej = dz = j ej d

Then:

x[n] =1

2j

+

X(ej) (ej)n1 j ejd

Integration path is unit circle because of ej

=1

2

+

X(ej) ejn d


77

2.8 System Representation and Examples

Example 1: Difference calculation

Difference equationy[n] = x[n] x[n n0], n0 integral number

Fourier transform gives:

n=y[n] ejn =

n=

x[n] ejn

n=x[n n0] ejn

Y (ej) = X(ej)

n=

(x[n] ejn ejn0

)= X(ej) ejn0 X(ej)

Then follows:H(ej) =

Y (ej)

X(ej)

= 1 ejn0

|H(ej)|2 = (1 cos(n0))2 + sin2(n0)= 1 2cos(n0) + cos2(n0) + sin2(n0)= 2 (1 cos(n0))

|H(ei)|2

0

1

2

3

4

5

n0


78

Example 2: First order difference equation

Delay

y[n]x[n]

+

y[n-1]

x[n] + y[n 1] = y[n] y[n] y[n 1] = x[n]

Method 1: Estimation of transfer function H(ej)from impulse response h[n]:

From the Eq. above with y[n] = h[n] and x[n] = [n] follows:h[n] = [n] + h[n 1]

= [n] + [n 1] + 2 [n 2] + =

{n, n 00, otherwise

Fourier spectrum/transfer function H(ej)

H(ej) =+

k=h[k] ejk

=+k=0

k ejk

=+k=0

( ej

)k=

1

1 ej for || < 1


79

Method 2: Estimation of transfer function H(ej) usingFourier transform of difference equation:

Difference equation:y[n] y[n 1] = x[n]

Fouriertransform:Y (ej) ej Y (ej) = X(ej)

Result:

H(ej) =Y (ej)

X(ej)

=1

1 ej


80

Example 3: Linear difference equations (with constant coefficients)

Difference equation:

y[n] =Ii=0

b[i] x[n i]Jj=1

a[j] y[n j]

z-transform:

Y (z) = X(z)Ii=0

b[i]zi Y (z)Jj=1

a[j]zj

Result:

H(z) =Y (z)

X(z)

=

Ii=0

b[i] zi

1 +Jj=1

a[j] zj

=+

n=h[n] zn

Using the definition of H(z) we can optain the impulse response as afunction of the coefficients of the difference equation in the above term.

Remark:

If we factorise denominator and numerator polynoms into linear fac-tors, we can obtain a zero-pole-representation of a discrete time LTIsystem:

H(i) =Ii=1(z vi)Jj=1(z wj)

with zeros vi C and poles wj C.


81

in general:

h[n] has infinite number of non-zero values

= IIRfilter: Infinite Impulse Response

but if: a[j] 0 jh[n] identical to zero outside of a finite interval

h[n] =

{b[n] n = 0, . . . , I0 otherwise

= FIRfilter: Finite Impulse Response


82

Example 4:Impulse response as truncated geometric series

h[n] =

{an 0 n M a IR0 otherwise

H(z) =Mn=0

an zn =1 aM+1 z(M+1)

1 a z1

system operation:

y[n] =

k=h[k] x[n k]

=Mk=0

akx[n k]

or as difference equation (recursively)

y[n] a y[n 1] = x[n] aM+1x[nM 1]


83

For this example we consider the zero-pole-representation:

H(z) =1 (za)(M+1)1 (za)1

> 0

Zeros: zk = a ej 2pikM+1 k = 0, 1, . . . ,MPole: z0 = a

(cancelled by zero z0 = a)

M=11

Re

Im

a


84

Example 5:Fibonacci numbers

Difference equation:

n 0 h[n+ 2] = h[n+ 1] + h[n]h[0] = h[1] = 1

n < 0 h[n] = 0

H(z) =

n=h[n]zn

= 1 + z1 +n=0

h[n+ 2]z(n+2)

= 1 + z1 +n=0

h[n+ 1]z(n+2) +n=0

h[n]z(n+2)

= 1 + z1 + z1n=0

h[n+ 1]z(n+1) + z2n=0

h[n]zn

= 1 + z1 (1 +n=1

h[n]zn) H(z)

+ z2n=0

h[n]zn H(z)

= 1 + z1H(z) + z2H(z)

H(z)(1 z1 z2) = 1H(z) =

1

1 z1 z2=

15

( a1 az1

b

1 bz1)

where a =1 +

5

2and b =

152


85

For a and b the following holds:

1

1 (az )=

n=0

( az

)n=

n=0

anzn

That results in:

H(z) =n=0

15

(an+1 bn+1 ) zn

!=

n=0

h[n] zn

h[n] =15

(an+1 bn+1 )


86

Table 2.1: Fourier transform pairs

signal Fouriertransform

1. [n] 1

2. [n n0] ejn0

3. 1 ( < n

2.9 Discrete Time Signal Fourier Transform Theo-

rem

Basically there is no difference between FT theorem for the continuoustime and the discrete time case because summation has the same proper-ties as integration. Only differentiation and difference calculation are notcompletely analog, because it is not possible to form a derivative in thediscrete time case.

Table 2.2: Fourier transform Theorems

signal Fouriertransformx[n], y[n] X(ej), Y (ej)

1. ax[n] + by[n] aX(ej) + bY (ej)

2. x[n nd], ejndX(ej)nd is integral number

3. ej0nx[n] X(ej(0))

4. x[n] X(ej)X(ej) if x[n] is real

5. nx[n] jdX(ej)

d

6. x[n] y[n] X(ej)Y (ej)

7. x[n]y[n]1

2

pipi

X(ej)Y (ej())d

8. x[n] x[n 1] (1 ej)X(ej)|1 ej|2 = 2(1 cos)

Parseval theorem

9.

n=

|x[n]|2 = 12

pipi

|X(ej)|2d

10.

n=

x[n]y[n] =1

2

pipi

X(ej)Y (ej)d


88

Example 1 corresponding to Theorem 5:

X(ej) =+

k=x[k] ejk

d

dX(ej) =

d

d

(+

k=x[k] ejk

)

=+

k=

d

d

(x[k] ejk

)=

+k=

x[k] (jk) ejk

j dd

X(ej) =+

k=k x[k] ejk

F {n x[n]} = j dd

F {x[n]}

Example 2 corresponding to Theorem 8:

F {x[n] x[n 1]} =+

k=x[k] ejk

+k=

x[k 1] ejk

=+

k=x[k] ejk

+k=

x[k] ejk ej

= X(ej)(1 ej)

= |F {x[n] x[n 1]} |2 = |F {x[n]} |2 |1 ej|2= |F {x[n]} |2 2 (1 cos())


89

2.10 Discrete Fourier Transform: DFT

The Fourier transform for discrete time signals and systems has been ex-plained on the previous pages. For discrete time signals with finite lengththere is also another Fourier representation called Discrete Fourier Trans-form (DFT).

The DFT plays a central role in digital signal processing.

Decisive reasons:

fast algorithms exist for DFT calculation(Fast Fourier Transform, FFT).

discrete frequencies k can be better represented in the computer thancontinuous frequencies .


90

Assume a discrete time signal x[n] with finite length (see also chapter 3.2on page 135):

x[n] =

{x[n] 0 n N 10 otherwise

Note: For a continuous time signal it is impossible in the strict sense to beband-limited and time-limited (truncation effect = Windowing).

The discrete time signal Fourier transform for x[n] is:

X(ej) =N1n=0

x[n] exp(jn)

is a continuous variable. The period is 2. Frequency discretisationis made by sampling along the frequency axis.

The Fourier transform X(ej) is evaluated at

k =2

Nk where k = 0, 1, . . . , N 1

Define:X[k] : = X(ej)| = k

N=8

Re

ImC


91

Discrete Fourier Transform (DFT):

X[k] =N1n=0

x[n] exp(j 2N

k n), k = 0, 1, . . . , N 1

Inverse DFT:

x[n] =1

N

N1k=0

X[k] exp(j2

Nk n), n = 0, 1, . . . , N 1

Remark:

This equation can be proven by inserting the equation for X[k] in theequation for x[n] and using the orthogonality:

1

N

N1n=0

exp

(j2

Nkn

)=

{1 k = m N, m is integral number0 otherwise

Note:

Consider the analogy between inverse DFT (above) and inverseFourier transform of discrete time signal:

x[n] =1

2

20

X(ej) ejn d

Under the given conditions the integral is equal to the sum(without approximation!).


92

Remarks:

DFT coefficients X[k] are not an approximation of the discrete timesignal Fourier transform X(ej). On the contrary:

X[k] = X(ej)| = k Number of the coefficients X[k] depends on the signal length N . Afiner sampling of the discrete time signal Fourier transform is possibleby appending zeros to the signal x[n] (ZeroPadding).

x[n]

nN-1


93

Interpretation of Fourier coefficients

Fourier transform X(ej) of the time discrete signal x[n]

|X(e )|

pipi

j

Evaluation at N discrete sampling points

k =2

Nk

yields the DFT coefficients X[k].

At first k lies in the domain k = N2+ 1, . . . , 0, . . . ,

N

2.

|X(e )|, |X[k]|

pipi

j

-N/2+1 N/21 20-1 k


94

Because of the periodicity of X(ej) the coefficients X[k] can also beobtained by shifting the sampling points with negative frequency intothe positive frequency domain (by one period).

Then k = 0, . . . , N/2, . . . , N 1.

X[k] =N1n=0

x[n] exp(j 2N

k n)

|X(e )|, |X[k]|

pipi

j

1 20 N-1 k

Interpretation of coefficients for general signal x[n]:k = 0 f = 0

1 k N2 1 0 < f < fS

2

k =N

2 fS

2N

2+ 1 k N 1 fS

2< f < 0


95

Symmetric relations by real signals:

For DFT coefficients X[k] of a real signal x[n] the following holds:X[k] = X[N k]

Re(X[k]) = Re(X[N k])Im(X[k]) = Im(X[N k])

For the amplitude spectrum |X[k] | the following holds:|X[k] |2 = Re2{X[k]} + Im2{X[k]}

= |X[N k] |2


96

Realization of DFT:

/* PI = 3.14159265358979 */

/* x: input signal */

/* N: length of input signal */

/* Xre, Xim: real and imaginary part of DFT coefficients */

void dft (int N, float x[], float Xre[], float Xim[]) {

int n, k;

float SumRe, SumIm;

for (k=0; k

2.11 DFT as Matrix Operation

Notation with unit roots

X[k] =N1n=0

x[n] exp (2jN

k n)

=N1n=0

x[n] W knN

where WN := exp (2jN

)

N=12

W =10N

W 1N

W 2NW 3N

Periodicity of WNunit root:

exp (j k) = exp (j2Nk) = (WN)

k

WN := exp (j2N)


98

Note:

1. W rN = Wr mod NN

2. W kNN = (WNN )

k = 1k = 1 k Z

3. W 2N = [exp (2j

N)]2 = exp (2j

N2)

= exp ( 2jN/2

) = WN/2 N even

4. WN/2N = exp (

2j

N

N

2) = exp (j) = 1

5. Wr+N/2N = W

N/2N W

rN = W rN


99

DFT as matrix multiplication

X[k] =N1n=0

x[n] exp (2jN

k n)

=N1n=0

W knN x[n]

=N1n=0

{WN}kn x[n]

with the matrix {WN} and the matrix elements:

{WN}kn := W knN

Inversion:

x[n] =1

N

N1k=0

X[k] exp (2j

Nk n)

=1

N

N1k=0

(W1N )kn X[k]

=1

N

N1k=0

{W1N }kn X[k]

Therefore for the matrix {WN}1 holds:

{WN}1 := 1N{W1N }


100

DFT matrix operation: properties

DFT: invertible linear mapping

N complex signal values N complex Fourier components

N real signal values N2complex Fourier components

(due to symmetry)

in words:DFT causes no information loss in the signal.

Parseval theorem for DFTgeneral Fourier:

N1n=0

|x[n]|2 = 12

+|X(ej)|2d

special DFT: (recalculate for yourself!)

N1n=0

|x[n]|2 = 1N

N1k=0

|X[k]|2

in words:

Disregarding the factor1

N, the DFT is a norm conserving (= energy

conserving) transformation (mathematical terminology: unitary).


101

2.12 From Continuous Fourier Transform to Matrix

Representation of Discrete Fourier Transform

Assumption: band-limited signal x(t)

Fourier transform of the continuous time signal x(t):

X() = F{x(t)} =

x(t) ejtdt (2.9)

For the exact reconstruction (without approximation) of the continuoustime signal from sampled values, the samples x[n] = x(n Ts) must havethe distance of at most

Ts =

B

(sampling theorem).

This results in the Fourier transform of the discrete time signal x[n]:

X(ej) =

x[n] ejn (2.10)

where is frequency normalised on T s

Functions (2.9) and (2.10) agree in interval [S/2,+S/2] = [B,+B].X

()||

BB SS


102

00.2

0.4

0.6

0.8

1

0 N-10

0.2

0.4

0.6

0.8

1

0 N-1

Figure 2.12: Hanning window

The signal x[n] is further decomposed by applying a window function w[n](windowing):

w[n] =

{. . . n = 0, . . . , N 10 otherwise

Windowed signal y[n]:y[n] = w[n] x[n]

can be analyzed using Fourier transform or DFT.

Y (ejk) =N1n=0

y[n] ejk

DFT:

k =2k

Nwhere k = 0, . . . , N 1

Y [k] =N1n=0

y[n] e2piNkn

Matrix representation:

K = N

Y [0]...

Y [k]...

Y [K 1]

=

...

e2pijNnk

...

y[0]...

y[n]...

y[N 1]


103

2.13 Frequency Resolution and Zero Padding

Task: signal x[n] with finite length N is given.

Wanted: Fourier transform X(ejk) at

k =2

Kk, where k = 0, 1, . . . , K 1 and K > N

Inserting the definitions:

X(ejk) =N1n=0

x[n] exp (2jK

k n)

=K1n=0

x[n] exp (2jK

k n)

where x[n] =

{x[n] n = 0, . . . , N 10 n = N, . . . ,K 1

i.e. Zero Padding (appending zeros).

Matrix representation:

X[0]...

...X[K 1]

=

W nkK

x[0]...

...x[N 1]

0...0

n = 0

n = N 1n = N

n = K 1

Note:Zero Padding does not introduce any additional information into the sig-nal. This is only a trick so that DFT and particularly FFT (Fast FourierTransform) can be performed with a

higher frequency resolution .


104

2.14 Finite Convolution

Input signal and convolution kernel have finite duration

Consider finite convolution:

Impulse response: h[n] 0 for n 6 {0, 1, 2, . . . , Nh 1} Input signal: x[n] 0 for n 6 {0, 1, 2, . . . , Nx 1} Output signal:

y[n] =

k=h[k] x[n k]

=

Nh1k=0

h[k] x[n k]

k

k

h[k]

x[-k]

N-1

-(N-1)x

h

0

n=0

Altogether: Nx +Nh 1 positions with overlap Therefore only Nx + Nh 1 values of output signal can be differentfrom zero:

y[n] =

0 n > Nx +Nh 2. . . n = 0, 1, . . . , Nx +Nh 20 n < 0


105

k0 Nh-1=12

1

k0 Nx-1=4

x[k]1 0.8

0.60.4

0.2

k0-Nx

x[n-k] n=-1,

k0

x[n-k] n=m (m>0 & m Nh +Nx 2, no overlap with h[k], convolution y[n] = 0c) resulting convolution y[n].


106

Finite convolution using DFT

Convolution theorem:

y[n] =

k=h[k] x[n k]

Fourier:

Y (ej) = H(ej) X(ej), 0 2Also valid for sample frequencies:

k :=2

Nk, k = 0, . . . , N 1 for any N

Notation: Y [k] = H[k] X[k]

Question: How to choose the length N of the DFT ? Reminder: different lengths

x[n]: Nx non-zero values

h[n]: Nh non-zero values

y[n]: Ny = Nx +Nh 1 non-zero values Answer:

The convolution theorem is certainly correct for any N > 0.

If we want to calculate the output signal completely from Y [k],we have to know Y [k] for

at least N = Nx +Nh 1frequency values k = 0, 1, . . . , N 1.

In words: for the DFT length N must be valid:

N Nx + Nh 1Method: Zero Padding, i.e. appending zeros.

Note: The FFT will be introduced on the next pages. A comparison ofcosts for realization of the finite convolution by DFT and FFT can befound at the end of paragraph 2.15.


107

2.15 Fast Fourier Transform (FFT)

Principle of FFT:Calculation of the DFT can be done by successive decomposition intosmaller DFT calculations. In this way, the number of elementary oper-ations (multiplications and additions) is dramatically reduced:

FFT:

N 2 N2ldN operations

N = 1024 :2 NldN

=2 102410

= 200 factor of velocity gain

The matrix is decomposed into a product of sparse matrices, therefore Nwith lot of prime factors is convenient (not necessarily only powers of two).

Terminology for different variants of FFT:

in time in frequency in place: yes/no radix 2 radix 4 decomposition to prime factors instead of N = 2n

History

1965 Cooley and Tukey1942 Danielson and Lanczos1905 Runge1805 Gauss


108

Algorithms which are based on a decomposition of the signal x[n] are calleddecimationintime algorithms.

The case N = 2 is considered in the following.

X[k] =N1n=0

x[n] exp(j 2N

k n) where k = 0, 1, . . . , N 1

=N1n=0

x[n] W nkN where WnkN = exp(j

2

Nk n)

Decomposition of the sum over n into the sums over even and odd n:

X[k] =

N/21r=0

x[2r] W 2rkN +

N/21r=0

x[2r + 1] W(2r+1)kN

=

N/21r=0

x[2r] (W 2N)rk + W kN

N/21r=0

x[2r + 1] (W 2N)rk

Because of

W 2N = exp(2j2

N) = exp(j 2

N/2) = WN/2

for k = 0, . . . , N 1 holds:

X[k] =

N/21r=0

x[2r] W rkN/2 + WkN

N/21r=0

x[2r + 1] (WN/2)rk

= G[k] + W kN H[k]

Each of the two sums corresponds to the DFT with the length N/2.The first sum is a N/2DFT of the even indexed signal values x[n],the second sum is a N/2DFT of the odd indexed values.

The DFT of the length N can be obtained by getting the two N/2DFTs together, with the factor W kN .


109

Complexity:The complexity O(N 2) of one-dimensional FT can be reduced by adequateresorting values from two FTs with length N2 and complexity O(2 (N2 )2) =N2

2 . By successive application of this resorting the complexity can be re-duced to O(N logN).The case N = 23 = 8 is considered in the following.

X[4] can be obtained from H[4] and G[4] according to previous equa-tion.

Because of the DFTlength N2 = 4:H[4] = H[0] and G[4] = G[0]

And then:

X[4] = G[0] + W 4N H[0]

The values X[5], X[6] and X[7] can be obtained analogously.

Flow diagram for decomposition of one N -DFT into two N/2DFTs:

x[n] X[k]

N/2-point

DFT

N/2-point

DFT

x[0]

x[2]

x[4]

x[6]

x[1]

x[3]

x[5]

x[7]

G[0]

G[1]

G[2]

G[3]

H[0]

H[1]

H[2]

H[3]

X[0]

X[1]

X[2]

X[3]

X[4]

X[5]

X[6]

X[7]

W0N

W1N

W2N

W3N

W4N

W5N

W6N

W7N

Figure 2.14: Flow diagram for decomposition of one N -DFT to two N/2DFTs withN = 8


110

Further analogous decomposition, until only DFTs with the lengthN = 2 remain (so called Butterfly Operation)

Resulting flow diagram of the FFT:

x[n] X[k]x[0]

x[4]

x[2]

x[6]

x[1]

x[5]

x[3]

x[7]

W0N

W0N

W0N

W0N

-1

-1

-1

-1

W0N

W2N

-1

-1

-1

-1

W0N

W1N

W2N

W3N

X[0]

X[1]

X[2]

X[3]

X[4]

X[5]

X[6]

X[7]

W0N

W2N

-1

-1

-1

-1

Figure 2.15: Flow diagram of an 8pointFFT using Butterfly operations.


111

Complexity reduction

Number of complex multiplications in FFTis N/2 ldN .

Comparison:Direct application of the DFT definition needsN 2 complex multiplications.

Example: N = 1024 = 210

N 2

N/2 ldN 200

Complexity reduction by factor 200

FFT with the base 2 is not minimal according to number of addi-tions, FFT with the base 4 can be better.


112

Matrix representation of the FFT principle

The complex Fourier matrix can be decomposed into the product ofr = ldN matrices, each of them having only two non-zero elements ineach column.

The following graph shows the decomposition of the Fourier matrix inthe case of inverse transformation.

w corresponds to W1N

X = |wnk|X = T3 T2 T1 TS xThis is how the decomposition into r + 1 = 4 matrices looks like(w4 = 1, w8 = 1):

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 w w2 w3 w4 w5 w6 w7 1 w w2 w3 w4 w5 w6 w7

1 w2 w4 w6 w8 w10 w12 w14 1 w2 w4 w6 1 w2 w4 w6

1 w3 w6 w9 w12 w15 w18 w21 1 w3 w6 w w4 w7 w2 w5

|wnk| = 1 w4 w8 w12 w16 w20 w24 w28 = 1 w4 1 w4 1 w4 1 w41 w5 w10 w15 w20 w25 w30 w35 1 w5 w2 w7 w4 w w6 w3

1 w6 w12 w18 w24 w30 w36 w42 1 w6 w4 w2 1 w6 w4 w2

1 w7 w14 w21 w28 w35 w42 w49 1 w7 w6 w5 w4 w3 w2 w


113

Signal flow diagram

The calculation operations which correspond to the matrix representationof FFT can be showed in a signal flow diagram.

T3 T2 T1 TS1 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 00 1 0 0 0 w 0 0 0 1 0 w2 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 1 0 0 00 0 1 0 0 0 w2 0 1 0 -1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 00 0 0 1 0 0 0 w3 0 1 0 w2 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 1 01 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 00 1 0 0 0 w 0 0 0 0 0 0 0 1 0 w2 0 0 0 0 1 -1 0 0 0 0 0 0 0 1 0 00 0 1 0 0 0 w2 0 0 0 0 0 1 01 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 00 0 0 1 0 0 0 w3 0 0 0 0 0 1 0 w2 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 1

TS T1 T2 T3

-1

-1

-1

-1

-1

-1

-1

2

2

-1

2

3

1

1

1

1

X[0]

X[1]

X[2]

X[3]

X[4]

X[5]

X[6]

X[7]

x[0]

x[1]

x[2]

x[3]

x[4]

x[5]

x[6]

x[7]


114

Matrices T1, T2 and T3 contain exactly two non-zero elements in eachrow.

Non-zero elements are realizing the Butterfly Operation.

Matrix T1: step width of the Butterfly Operation is 1Matrix T2: step width of the Butterfly Operation is 2Matrix T3: step width of the Butterfly Operation is 4

step widths can be found: in signal flow diagram

distance between the non-zero elements in T1, T2 and T3


115

Butterfly Operation

Signal flow diagram and matrix representation of the FFT are basedon the following basic operation:

Xm-1[p]

Xm-1[q]

Xm[p]

Xm[q]-1

WrN

For two input values Xm1[p] and Xm1[q] this operation producestwo output values Xm[p] and Xm[q]. The output values are thereby alinear combination of the input values.

Because of the flow graph, the operation is calledButterfly Operation.

Xm[p] = Xm1[p] +W rN Xm1[q]Xm[q] = Xm1[p]W rN Xm1[q]

[Xm[p]Xm[q]

]=

[1 W rN1 W rN

] [Xm1[p]Xm1[q]

]


116

Bit Reversal

The matrix representation of the FFT uses a sorting matrix, i.e. thesignal which is to be transformed is at first resorted.

Example for N = 8:

n Binary representation Reversed n

0 000 000 01 001 100 42 010 010 23 011 110 64 100 001 15 101 101 56 110 011 37 111 111 7

Bit Reversal is a necessary part of the FFTAlgorithm.

Bit Reversal for N = 23


117

2.16 FFT Implementation

Fortran versionC adapted from: Oppenheim, Schafer p. 608

C SUBROUTINE FFT DecimationInTime (X, ld n) **********************************

C ****************************************************************************

PARAMETER PI = 3.14159265358979

PARAMETER N max = 2048

COMPLEX X(N max) ! array for input AND output

COMPLEX Temp ! temporary storage

COMPLEX W uni ! root of unity

COMPLEX W pow ! powers of W uni

INTEGER N, ld N, ip, iq, iqbeq, j, k, i exp, istp

N = 2**ld n

IF (N.GT.N max) STOP

C BIT Reversed Sorting *********************************************************

j = 1

DO i = 1, N-1

IF (i.LT.j) THEN ! swap X(j) and X(i)

Temp = X(j)

X(j) = X(i)

X(i) = Temp

ENDIF

k = N/2

DO WHILE (k.LT.j)

j = j - k

k = k / 2

ENDDO

j = j + k

ENDDO

C End of Bit Reversed Sorting **************************************************

C FFT Butterfly Operations *****************************************************

DO i=1, ld N

i exp = 2**i ! exponent

istp = i exp/2 ! stepsize

W pow = (1.0,0.0)

W uni = CMPLX (COS (PI/FLOAT(istp)), -SIN(PI/FLOAT(istp)))

DO ipbeg = 1, istp

DO ip = ipbeg, N, i exp

iq = ip + istp

Temp = X(iq) * W pow

X(iq) = X(iq) - Temp

X(ip) = X(iq) + Temp

ENDDO

W pow = W pow * W uni

ENDDO

ENDDO

C End of FFT Butterfly Operations **********************************************

RETURN

END


118

Explanations about Fortran Program

Two program parts:

1. Bit Reversal

2. Butterfly Operations

3 loops with variables i, ipbeg, ip are controlling the execution ofthe Butterfly operations

outer loop i:i specifies the level of the FFT

With exception of the first level, Butterfly operations are nested.Therefore two loops are used for the Butterfly operations withinone level.

middle loop, ipbeg:ipbeg: goes over the nested Butterfly operationsi=1: ipbeg=1i=2: ipbeg=1,2i=3: ipbeg=1,2,3,4iqbeg: specifies the sequence of starting points for inner loop

inner loop, ip:ip: specifies the first element of the Butterfly operationistp: step width of the Butterfly operationiq=ip+istp: specifies the second element for Butterfly operationinner loop is started once per nesting


119

x[0]

x[4]

x[2]

x[6]

x[1]

x[5]

x[3]

x[7]

W0N

W0N

W0N

W0N

-1

-1

-1

-1

W0N

W2N

-1

-1

-1

-1

W0N

W1N

W2N

W3N

X[0]

X[1]

X[2]

X[3]

X[4]

X[5]

X[6]

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