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WAVELETS: Theory and Applications
An Introduction
Willy Hereman
Dept. of Mathematical and Computer Sciences
Colorado School of Mines
Golden, Colorado
USA
Ph.D. Program, Department of Physics
University of Antwerp
Antwerp, Belgium
December 4-15, 2000
1
.
Introduction to WAVELETS: Theory and Applications
LECTURE NOTESHOMEWORK ASSIGNMENT & EXAM
BOOKS ON WAVELETS
Prof. Willy Hereman
Dept. of Mathematical and Computer Sciences
Colorado School of Mines
Golden, Colorado
USA
Honors Course–Numerical Algorithms
Department of Applied Mathematics
University of Stellenbosch
Stellenbosch, South Africa
March 9-23, 2001
Wed.: 10-10:50, Fri.: 10-10:50, 11-11:50
2
Topics—Outline
• Introduction: History, motivation, applications
• Key ideas and definitions
• Haar and Daubechies wavelets
• Scaling and wavelet functions
• Multi-resolution analysis
• Wavelet analysis: decomposition and reconstruction
• Fast Fourier Transform (FFT) versus Fast Wavelet Transform (FWT)
• Vanishing moments, smoothness, approximation
• Low and high pass filters
• Quadrature Mirror Filters (QMF)
• Construction of Daubechies’ wavelets
• Construction of scaling and wavelet functions
• Selected applications
• Software demonstration
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• INTRODUCTION (Andre Weideman)
• BACKGROUND AND EXPERIENCE WITH WAVELETS
* 1993- Work with Gregory Beylkin at CU - Boulder.
* 1993 - ’94 Sabbatical year devoted to wavelets and applications.
* 1993 - Short course in Ghent, Belgium (my alma mater).
* 1994
- Work on coiflets (with Monzon and Beylkin),
- work on Dubuc-Deslauriers’ subdivision scheme and wavelets,
- work on Battle-Lemarie spline based wavelets.
* Course on wavelets at CSM-Golden, CO (1995).
* Short course on wavelets in Antwerp. Fall 2000 (Also, May 1995).
* 1998 Paper on coiflets.
* Currently wavelet issues related to applications (facial recognition, fin-gerprints, etc.).
* Belgian connection with Ingrid Daubechies and Wim Sweldens.
• PLAN AT STELLENBOSCH
* Gentle introduction to wavelets (concepts).
* American style: ask questions.
* Media
o Lecture notes in library.
o Copies of slides in library.
o Reference books in library.
o Review papers in office.
o Wavelet Explorer (Mathematica).
• WHO HAS EXPERIENCE WITH WAVELETS?
* o Theory.
o Applications? Which?
* Who has heard about wavelets and where, what context?
* Relation to field of study.
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* Background in mathematics.
* Reasons to take course?
• GRADE FOR WAVELET SHORT COURSE
* One assignment 4 or 5 problems.
* One examination question (open book).
• INFO ABOUT WAVELETS
o long list of books (copy in library).
o thousands of papers.
o two dozen good review papers.
o Wavelet Digest (Wavelets for Kids) http://www.wavelet.org/wavelet.
o Electronic resources.
o Wavelet Software.
o Info on WEB (see course Weideman).
DEMO OF WAVELET EXPLORER
To get to Wavelet Explorer.
- Start Mathematica.
- Click on Help.
- Click on Add-ons.
- Click on Wavelet Explorer.
To get intro to Wavelet ExplorerFrom wavelet Explorer Pick Fundamentals of Wavelets
To use it in your own notebook in Mathematica
In [1] Needs [“Wavelets‘ Wavelets‘”].
In [2] ? *aub*.
DaubechiesFilter[n] (brief description)
In [3] ??DaubechiesFilter[n] (for full description with all options)
In [4] DaubechiesFilter[3].
For filtertaps of Daubechies averaging filter with 3 vanishing moments.
5
GENERAL IDEAS ABOUT WAVELETS
* Arena (setting) Mathematics behind data, signal, image analysis or process-ing (Engr.)and numerical analysis (Mathematics/Applied Mathematics).
* Wavelets is at interface of engineering and mathematics (compare with math-ematical physics, mathematical biology, mathematical finance, ....)
* Counterpart to Fourier analysis
FFT ←→ FWT (or discrete DWF)Fast Fourier transform Fast wavelet transform
Great Discovery of Jean-Baptiste Fourier (1768-1830).
A periodic signal (sound, function) can be decomposed in harmonics (sinesor cosines, or complex exponentials).
* Shortcomings of Fourier analysis (weaknesses)
- sines and cosines wiggle (extend) infinitely far,
- does not handle gaps or localized signals well, or faulty data (too manycoefficients needed to make cancellations to zero),
- a transient sharp “blip” needs many Fourier components. Huge numberof non-zero coefficients,
- does not handle fast variations (highly oscillatory signal) well,
- trouble at discontinuities (overshoot, undershoot, Gibbs phenomena).
* Strong points of Fourier analysis
- functions are analytic and simple,
- simple (sines and cosines) orthogonal functions,
- fast FFT is n log n algorithm - due to trick invented in 1965 by J. Tuckeyand J. Cooley (same trick applies for wavelets),
- few coefficients in series are needed to represent a smooth periodic func-tion,
- very applicable recursive algorithms,
- caused a revolution in scientific computing,
- continuous and discrete transforms are available,
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→ Try to keep what’s good and try to improve on bad properties =⇒Wavelets.
* Key requirements
o a few functions (no longer sines/cosines) should give good approximation,
o orthogonal functions,
o we want localization in both time (via translations) and (space) frequency(wavelength) via dilations and contractions,
o we want fast, recursive algorithm,
o zoom in - zoom out property,
o something highly applicable,
o mathematically on solid foundation (requires new mathematical con-cepts, ideas, ....construction),
o continuous and discrete transforms.
* Cost:
=⇒ no longer analytical functions,
=⇒ not as simple as Fourier series or Fourier transform.
WHAT IS A WAVELET?
“a little wiggle, ripple”
French: ondelette
How many types of wavelets are there? - About 2 dozen practical types.(Daubechies’ wavelets, Meyer wavelets, biorthogonal wavelets, Battle-Lemariewavelets, ...)
HOW DOES ONE GET TO WAVELET BASIS?
In Fourier analysis:Take sinusoids (or complex exponentials) as basis functions and then study theproperties of the Fourier series (or Fourier integral)
In wavelet analysis:One poses the desired properties and then derives the resulting basis functions.
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[For instance, multiresolution is a key property, orthogonality also, and perhapssymmetry, vanishing moments property].
Key idea:You want to construct a basis for IL2(IR) (= collection/space of all square inte-grable functions)
f ∈ IL2(IR) if∫ +∞
−∞f(x)2dx <∞
CONCEPTUAL ANALOGIES (OTHER PERSPECTIVES)
1. The game of averages and differences and vanishing moments
Student #1 has X1 amount of money in bank accountStudent #2 has X2 amount of money in bank account
You can either give X1, X2 itself or give the
average
S1 =X1 +X2
2⇐⇒ X1 = S1 +D1
difference
D1 =X1 −X2
2⇐⇒ X2 = S2 −D1
Advantage ifX1 = X2 =⇒ D1 = 0
.If X1 = X2 follow the same constant trend
then simple “differences” creates a zero student number
“he same amount in account” −→ difference = zero, −→ average = how mucheach has.
Question: Is it possible to generalise this simple idea? YES.
Look at it from a “signal data” processing point of view
8
X1 X2︸ ︷︷ ︸ −→ X3 X4︸ ︷︷ ︸ X5 . . .
move over in steps of two.
for averages •(1
2
)= h0 •
(1
2
)= h1
for differences •(1
2
)= g0 •
(1
2
)= g1
S1 = 12X1 + 1
2X2 = h0X1 + h1X2
D1 = 12X1 − 1
2X2 = g0X1 + g1X2
Choices
(1
2,1
2
)= h0, h1(
1
2,−1
2
)= (g0, g1)
or (1, 1) = (h0, h1)
(1,−1) = (g0, g1)
or
(1√2,
1√2
)= (h0, h1)(
1√2,− 1√
2
)= (g0, g1) Haar case
Similarly S2 = h0X3 + h1X4
D2 = g0X3 + g1X4, etc.
Leaping ahead
If X1, X2, X3, X4, . . . lie on line can we find h0, h1 and g0, g1 so that D1, D2
etc. are zero? No.
We need four numbers
h0, h1, h2, h3 for averaging (low pass filter)g0, g1, g2, g3 for differencing so that (high pass filter)
D1 = g0X1 + g1X2 + g2X3 + g4X4 = 0
if X1, X2, X3, X4, . . . are on line
Also, D2 = 0 etc.
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If Xi are on parabola we will need
(h0, . . . , h5) for averaging(g0, . . . , g5) for differencing, so
that g0X1 + . . .+ g5X6 = D1 = 0
if Xi follow quadratic trend
a+ bX + cX2
↓ ↓ ↓
.constant linear and quadratic combined
We will need six numbers.
Etc. can be done for xn also.
If Xi follow on a+ bX + cX2 + . . .+ fXn
then you need 2n “magical” numbers with special properties so that
D1 = 0 ,D2 = 0 , etc.
Examples of data following trends
* Pictures: cone and ball on tiled floor. Show and explain image compres-sion.
* Cartoon: how are the Disney cartoons made? Game for kids: picturebook with slightly different (shifted figures) on frozen background.
* Video: TV broadcasting. Transmit only the differences.−→ Life is full of redundancy!
2. Vanishing moments
If you could construct a wavelet function ψ(x) which corresponds to “dif-ferencing” then you could require that ψ(x) annihilates constant, linear,quadratic, cubic, ..... trends . . . up to degree say xM−1
so∫x`ψ(x) dx = 0 for ` = 0, 1, 2, . . .M − 1︸ ︷︷ ︸
M moments
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Since a function f(x) can be written in McLaurin series
f(x) = f(0)︸ ︷︷ ︸constant
+ f ′(0)x︸ ︷︷ ︸linear
+
quadraticf ′′(0)
2x2
︸ ︷︷ ︸ + . . .+f (M−1)
(M − 1)!xM−1 + . . .
degree M − 1
you would have
∫f(x)ψ(x)dx = 0 + 0 + 0 . . .+ 0 +
∫ xM
M !ψ(x)dx+ . . .︸ ︷︷ ︸
highly oscillating part
−→ Wavelets would be sensitive to fast oscillating pieces of f(x) and anni-hilate slower varying pieces of f(x).
3. Fingerprint story
Popularized wavelets(FBI competition for fingerprint compression)
(Old systems for identification was based on facial features. Bertillion 1884).
- 37 million fingerprint cards for criminals (some in duplicate),
- 250 million fingerprint cards in total (job applications, immigration,background checks for employment ..., weapon purchases, ... ),
- 60 km of “thin” cardboard,
- each card has 10 rolled imprints of fingers, 2 unrolled thumb prints, twofull hands,
- cataloguing is several million cards behind,
- request for identity check may take between 100 - 140 days,
- enormous office building full of these cards.
−→ Need for automation and electronic storage (via scanning), saves space.Allows for on-line verification (in shortest possible time, while suspect isbeing held).Should be accessible from every police car (digitizing tablet, modem, ...).
- FBI has approximately 350 employees to handle the verification requests,cataloguing, removing duplicates, etc. ...
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- Automatic system: IAFIS (Automatic Fingerprint Identification Sys-tem).
* cards are digitized and stored on 2 computers in a nuclear attackproof building (cellar 20 m deep in Clarksburg, West Virginia),
* old card collection is moved to Federicksburg,
* scanning all cards took 5 years,
* cost: $640 million,
* every police car has a scanner and they can fully automatically checkthe enormous database.
- FBI get 45,000 requests per day for verification.
- FBI adds 5,000 fingerprints to database per day.
- Average checking time (for routine stuff, like for application for weaponspermit) takes 30 seconds.
- Computer database is 40 terabytes.1 byte = 8 bids (zero or one)40 × 1012 bytes (= 40,000 PC with harddrive of 10 gigabytes). Evencompressed!
How does it get that large?
Each square inch fingerprint image is broken into 500 × 500 grid of smallboxes, called pixels.
Each pixel is given a gray-scale from 0 to 255
↓ ↓white black
−→ darkness of one pixel requires 8 bits = 1 byte.
Start multiplying: 1 whole card ∼= 10 MB.
(Mega byte = 106 bytes
Giga byte = 109 bytes
Tera byte = 1012 bytes)
=⇒ 2500 Tera byte = 2500 1012
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250 million cards ∼= 2500106 106︸︷︷︸MB∼= 2500× 102 × 10GB︸ ︷︷ ︸
harddrive∼= 250, 000PC’s. (quarter million PC’s.)
For ≈ 37 million criminals alone
=⇒ 37,000 PC’s. 1,000 cards fill up one PC!
Impossible without serious data compression for storage and transmission(via JPEG (Joint Photographic Experts Group) or something else.
−→ FBI competition won by Ron Coifman (Yale) and Bradley and Brislawn,at Los Alamos National Laboratory (biorthogonal wavelets).
DATA COMPRESSION:
Sending 1 card without compression via 56K modem would take≈ 20 minutes
Standard compression ratios 3/1 or 2/1Wavelet compression ratios 15/1 or 20/1 (5% kept)
−→ with 20/1 compression
transmission time = 1 minute.
Space savings also factor 15 or 20−→ 1/20 of space
(For cirminals’ card storage −→ less than 2,000 PC’s)
Wavelet algorithms closer to home:
* wavelet chip in modems (Aware Corp.)
* JPEG 2000 standard for internet pictures (you see the details come in).
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APPLICATIONS OF WAVELETS
Signal and Data processing:
- analysis (eg. detection of edges, detection of faults, abrupt changes,defects like “cracks” in materials, . . . ),
- compression (reduction of storage space, fast transmission, as in fin-gerprint application),
- smoothing (attenuation of noise, denoising),
- synthesis (reconstruction after compression or/and smoothening, orother types of modification).
Types of signals
1D : sound, speech, time-series (e.g. history of a stock in financial markets),
2D : pictures, maps, . . . ,
3D : spatial diffusion (e.g. of heat of body, dopamine injected in brain, . . . ).
SAMPLE APPLICATIONS
−→ Reconstruction of Johannes Brahms’ early recording (1889) of First Hun-garian Dance (story in Scientific American 1993).
Music speech
−→ Medical applications: analysis of
– MRI scan,
– electro cardiogram,
– ultrasonic images,
– mammograms (tumor detection),
– various other modern scanning, visualization methods in medicine.
Medical imaging
−→ Military applications:formation of images based on partial radar data (distinguish ambulancefrom tank).
Military imaging
−→ High speed data transmission:
- single pictures (come gradually in focus),
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- video (explain principle).
Digital communication
−→ Denoising technology:(project for ADA Technologies: detection of metals like vanadium, mer-cury, lead in smoke, laser spectroscopy −→ least squares method andwavelets).
Data analysis
−→ Mathematical, engineering applications:fast PDE solvers, Ax = b solvers, numerical analysis
FFT vs FWT
−→ Geophysical applicationsseismic methods for detection of rock formations, pockets of gas, detec-tion of oil fields, . . . .
CSMMassive data
−→ Finance:analysis of time-series (prediction of behaviour of stocks, other financialintruments . . . ) .
Money matters
−→ Metereology:analysis of weather data, climate, temperature in oceans,. . . .
Massive data
SHORT HISTORY“Wavelets come from a synthesis of ideas”.
• First wavelet: Haar 1910 (box functions).
• Engineering experiments in 1960’s −→ 1980’s.
• Experiments with windowed Fourier Transforms.
• 1980: Jean Morlet (geophysicist)- wavelets for analysis of seismic data.
• 1980 Alex Grossmann: theoretical study of wavelets and wavelet trans-form.
• 1985: Yves Meyer (harmonic analyst)
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- connections with other fields in maths,
- connection with singular integral operators.
• 1986 - 1988: Ingrid Daubechies (VUB colleague)
- frames (generalized basis for Hilbert space),
- 1988 - major breakthrough,
- “Daubechies” family of orthonormal wavelets with compact support,
- connection with coherent states.
- Inspired by work of Mallat and Meyer on multi-resolution analysisand applications in decomposition and reconstruction of images (com-puter graphics).
- “Ten lectures on Wavelets”.
• Beyond 1988: exponential growth in theoretical developments.(5,000 papers in Science Citation Index).Applications in physics (coherent states), quantum field theory, electricalengineering, computer science, statistics, etc., etc., . . . .
“SYNTHESIS OF IDEAS”
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SCALING AND WAVELET FUNCTIONS
* Wavelet is little ripple, a little wave.
ψ(x), not symmetric, such that∫ψ(x)dx = 0, and such that(admissibility condition)
ψ(x) will have special properties.
* Start with “mother” wavelet, concentrated, say, in interval [0,1].
* Generate other wavelet by moving the mother wavelet left or right in unitsteps, and by dilating or compressing the mother wavelet by repeated factorsof 2.
Compare with images (zoom into a picture)
dilate (stretch)zoom out−→ lower resolution details of picture,
compress (squeeze)zoom in−→ higher resolution details of picture.
Look at TV-screen:large screen (blurred image) expanded imagesmall screen (sharper image) compressed image.
Definition (Mathematical)
Let IL2IR be the Hilbert space of square integrable functions on real line.A wavelet is a funtion ψ ∈ IL2(IR) such that
2j/2ψ(2jx− k) j, k ∈ ZZ
is an orthonormal basis of IL2(IR)
There are two families in the game:
ϕ(x) scaling function −→ ϕj,k(x)↓father where ϕj,k(x) = 2j/2ϕ(2jx− k) = 2j/2ϕ(2j(x− k/2j))
↑ ↑2 parameter family kids (boys)
ψ(x) wavelet function −→ ψj,k(x) −→ kids(girls)↓mother where ψj,k(x) = 2j/2ψ(2jx− k) = 2j/2ψ(2j(x− k/2j))
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↑2 parameter family
Haar wavelet Daubechies wavelet
(DAUB4) 2 vanishing momentsPicture motherPicture fatherPictures of scaling and wavelet functions
Key features (desired properties).
* The basis should be orthonormal “vectors” (here functions) should be mutu-ally orthogonal, and properly normalized
−→ fast algorithms to compute coefficients.
* The basis functions (wavelets) should have vanishing moments−→ sparsity (lots of zeros) everywhere.
* The algorithms should be recursive. True due to the multi-resolution prop-erties of the basis (always devide or multiply by factor 2).
* There should be no need for the functions themselves
so, the family {2j/2ψ(2jx− k)}j,k is not needed.
Instead one uses “filter coefficients or taps” for averaging, differencing (youwant to maintain a constant norm, independent of j).
* Normalization
If ψj,k(x) = cψ(2jx− k)Require ‖ψj,k(x)‖2 = ‖ψ0,0(x)‖2 = ‖ψ‖2∫ψj,k(x)
2dx = c2∫ψ2(2jx− k)dx, set z = 2jx− k
= c22−j∫ψ2(z)dz = c22−j‖ψ‖2 = 1.‖ψ‖2
so we need c = 2j/2.
Note: Daubechies wavelets
If ψ is a Daubechies wavelet with M vanishing moments that generates an or-thonormal basis of IL2(IR) then
ψ is supported on [0, 2M − 1] or [−M + 1,M ] if translated to leftover M − 1
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ϕ is supported on [0, 2M − 1].
Vanishing moments
2 vanishing moments means∫ψ(x) dx = 0,
∫xψ(x) dx = 0
4 vanishing moments∫x` ψ(x) dx = 0 for ` = 0, 1, 2, 3
M vanishing moments∫x` ψ(x) dx = 0 ` = 0, 1, 2, . . . ,M − 1.
Wavelets with vanishing moments will annihilate (ignore, or be blind to) certaintrends; linear (including const), quadratic, cubic, quartic, quintic, etc. and onlybe sensitive ( pick out) higher degree oscillations.
Expanding and contracting wavelets.Compare with music.ψ quarter note at middle C (where your start)1V 2ψ(x2) half note a lower octave (stretch)
2ψ(4x) 16th note at 2 octaves higher (contract)
THE REFINEMENT OR TWO-SCALE DIFFERENCE EQUATION
Haar caseϕ(x) = characteristic function on (0,1)
ϕ(x) =
1 , 0 ≤ x < 10 , elsewhere (x < 0 or x ≥ 1).
Observe ∫ 1
0ϕ(x)dx = 1 =
∫ 1
0ϕ2(x)dx.
Also, for scaling function
ϕ(x) = ϕ(2x) + ϕ(2x− 1).
Refinement equation for Haar case.
Other names: dilation equation; two-scale difference equation.
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Verify
0 ≤ x < 1/2 1 = 1 + 0
1/2 ≤ x ≤ 1 1 = 0 + 1
x < 0 or x ≥ 1 0 = 0 + 0.
Also, for wavelet function
ψ(x) = ϕ(2x)− ϕ(2x− 1).
Note: wavelet function can be defined in terms of scaling function.
IN GENERAL (DAUBECHIES WAVELET AND OTHERS)
ϕ(x) =√
2L−1∑k=0
hkϕ(2x− k)
ψ(x) =√
2L−1∑k=0
gkϕ(2x− k)
Equivalently
ϕ(x/2) =√
2L−1∑k=0
hkϕ(x− k), similar for ψ(x/2)
L = 2 (length of filter)
For Haar case
ϕ(x) = ϕ(2x) + ϕ(2x− 1)
=√
2(
1√2ϕ(2x) + 1√
2ϕ(2x− 1))
ψ(x) = ϕ(2x)− ϕ(2x− 1)
= 1√2
(1√2ϕ(2x)− 1√
2ϕ(2x− 1))
−→ (h0, h1) =(
1√2 ,
1√2
)= H
−→ (g0, g1) =(
1√2 ,−
1√2
)= G
Filters for averaging and differencing.
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In general
H = (h0, h1, h2, . . . , hL−1) low pass filterG = (g0, g1, g2, . . . , gL−1) high pass filter.
Always come in pairs
Note: As vectors−→H ⊥
−→G
i.e. ϕ⊥ψ as functionsH −→ G reverse the order and switch every other sign
H = (h0, h1, . . . , hL−1) −→ G = (hL−1,−hL−2, . . . ,−h0).
Reverse (of refinement equation)
ϕ(2x) =1
2(ϕ(x) + ψ(x))
ϕ(2x− 1) =1
2(ϕ(x)− ψ(x))
Later, general:
ϕ(2x− `) =∑k
(a`−2kϕ(x− k) + b`−kψ(x− k))
Remarks
1. In general there is no (explicit) analytical form for ϕ(x) or ψ(x)
2. ϕ(x) is evaluated via the refinement equation at dyadic points x = k2j
j, k ∈ ZZ
−→ (Assignment plus old notes. Eigenvalue problem).
Example:
Set x = 1/2 ϕ(1/2) =
0︷ ︸︸ ︷ϕ(1) +
1︷ ︸︸ ︷ϕ(0) = 1
Also ϕ(−1/2) = ϕ(−1) + ϕ(−2)= 0 + 0 = 0
So, if ϕ is known at integers then ϕ is known at the halvesRecursive!
ϕ(1/4) = ϕ(1/2) + ϕ(−1/2) = 1
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General ϕ(2j−1n) = ϕ(2jn) + ϕ(2jn− 1)
3. To be able to use {ϕ(x − k)} k ∈ ZZ to approximate even simple functionwe assumethat ϕ(x) and its translates form a partition of unity.
∑k∈ZZ
ϕ(x− k) = 1, ∀x ∈ IR
Proof in Fourier domain
ϕ(0) = 1 =⇒∑k
ϕ(x− k) = 1.
More general:
f(t) =∑nfn(t) where fn(t) = ϕ(t− n)f(t) −→ localized function.
For Haar case:ϕ(x) = ϕ(2x) + ϕ(2x− 1)
4. Refinement equations were known in context of splines.
Examples
* Piecewise linear spline (hat function)
N2 =
x , 0 ≤ x < 12− x , 1 ≤ x < 20 , elsewhere
Satisfies
N2(x) =1
2N2(2x) +N2(2x− 1) +
1
2N2(2x− 2)
* Piecewise quadratic spline
N3(x) =
1/2x2 , 0 ≤ x ≤ 13/4− (x− 3/2)2 , 1 ≤ x < 21/2(x− 3)2 , 2 ≤ x ≤ 30 , elsewhere
Satisfies
N3(x) =1
4N3(2x) +
3
4N3(2x− 1) +
3
4N3(2x− 2) +
1
4N3(2x− 3)
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5. In general
ϕ(x) =√
2L−1∑k=0
hkϕ(2x− k)∫ϕ(x)dx=1=⇒
L−1∑k=0
hk =√
2 linear
ψ(x) =√
2∑gkϕ(2x− k)
∫ψ(x)dx=0=⇒
L−1∑k=0
gk = 0 linear
Proof - Integrate both sides
1 =∫ϕ(x)dx =
√2L−1∑k=0
hk∫ϕ
z︷ ︸︸ ︷(2x− k) dx
=
√2
2
L−1∑k=0
hk∫ϕ(z)dz︸ ︷︷ ︸
1
=1√2
L−1∑k=0
hk
−→L−1∑k=0
hk =√
2
Verify for Haar case 1√2 + 1√
2 = 2√2 =√
2∫ψ(x)dx = 0 =⇒ ∑
k gk = 0
Similar proof for∑k gk = 0
Verify for Haar case 1√2 −
1√2 = 0
6. Orthogonality properties
ϕ(x) ⊥ ϕ(x− k), ∀k ∈ ZZ0 (k 6= 0)
Inner product in IL2(IR)
< f, g >=∫ +∞
−∞f(t) g(t) dt
So < ϕ,ϕn >=< ϕ,ϕ(x− n) >= δno
One can show thatϕ(x)⊥ϕ(x− n), if n 6= 0
andϕ2(x) normalized (n = 0)
∫ϕ2(x)dx = 1
leads to (No proof here - proof in extended notes)
23
(i)L−1∑k=0
h2k = 1 ←− quadratic (n = 0) case above
(ii)
L−1−2m∑`=0
h2m+` h` = 0 m = 1, 2, . . . ,L
2− 1 (n 6= 0 )above
Show in detail via shifts of twoh0 h1 h2 h3 . . .
h0 h1 . . .
Verify for Haar case
(i) (1√2)2 + (
1√2)2 =
1
2+
1
2= 1
L = 2 m = 0 - no condition of type (ii)
ϕ ⊥ ψ (h0, h1) ⊥ (g0, g1) = (h1,−h0)
so, h0h1 − h1h0 = 0.
For
L = 8
h0h2 + h1h3 + h2h4 + h3h5 + h4h6 + h5h7 = 0h0h4 + h1h5 + h2h6 + h3h7 = 0h0h6 + h1h7 = 0
7. Admissibility Condition ∫ψ(x)dx = 0
ψ(x) = h0ϕ(2x)− h1ϕ(2x− 1)
∫ψ(x)dx = 0 =
h1
2
1︷ ︸︸ ︷∫ϕ(z)dz − h0
2
1︷ ︸︸ ︷∫ϕ(z)dx
=⇒ h0 = h1
In general,
L−1∑k=0
h2k =L−1∑k=0
h2k+1 =1√2, and also
L−1∑k=0
(−1)khk = 0 (H ⊥ G)
24
REPRESENTING FUNCTIONS IN A WAVELET BASIS
Multi-resolution
For Haar case (1910)
{ψj,k(x)}k,j∈ZZ is a fully orthonormal basis for IL2(IR).
So for f(x) ∈ IL2(IR)
f(x) =+∞∑j=−∞
+∞∑k=−∞
ajkψj,k(x)
=+∞∑j=−∞
+∞∑k=−∞
ajk2j/2ψ(2jx− k)
↑ wavelets
In practice, Haar also used
f(x) =+∞∑
k=−∞ckϕ(x− k) +
+∞∑k=−∞
+∞∑j=0
djkψj,k(x)
↓ ↓translates at scale adding weighted
j = 0 wavelets at different
scales, starting from j = 0
and allowing translates also.
Effect of j
x←→ 2jx space←→ scale
t←→ 2jt time←→ frequency (via dilation)
Effect of k localization in space (or time)
(via translation)
25
RESOLUTION - MULTI-RESOLUTION
A few examples
1. Our number system has multi-resolution.
Compare with computation of circumference circle2 radius = π
π = 3.1415926535
j’aime a faire connaitre le nombre utile aux sages.
3 = 3. 1100 −→ what scaling function shows at a given resolution
3.1 = 3. 1100 + 1.
1
10︸ ︷︷ ︸ −→ better resolution
wavelet contribution - encodes the detail
3.14 = 3. 1100 + 1.
1
10︸ ︷︷ ︸+ 4.1
102︸ ︷︷ ︸+ . . .
Other direction: stretching (adding more and more detail), the scalingfunction too much we end up seeing nothing at all. Represent 3.14159 in10 = 101 or 100 = 102 scales.
2. Picture of Dali’s painting.
“Gala naked watching the sea”.
(Salvador Dali, 1976, Dali Museum, Figueres, Spain).
Zoom in: Look at detail (close-up) −→ Gala (Dali’s wife).
Zoom out: Look at coarse picture (from afar) −→ Abraham Lincoln.
Dali was aware of images at different scales of resolution.
26
Basic notions (First for Haar case)Inspiration for multi-resolution properties.
1. Consider the scaling function ϕ and the associated 2 parameter scaling family
ϕj,k(x) = 2j/2ϕ(2jx− k) j, k ∈ ZZ
For fixed j : {ϕj,k(x)}k∈ZZ spans a subspace Vj of IL2(IR).
More precise: We define the subspace −→ closure.
Vj = spank∈ZZ
{ϕj,k(x)
}j fixed
and ϕj,k(x) is a basis for Vj, the space of piecewise constant functions (withdyastic points k
2j ).
Example:
If f(x) ∈ V0 then f(x) =∑k∈ZZ
ak
ϕ(x−k)︷ ︸︸ ︷ϕ0,k(x)
V0 = space of piecewise constant functions over integers.
V1 = space of ... over halves = larger space than V0,
because the functions ϕ1,k(x) =√
2ϕ(2x− k)=√
2ϕ(2(x− k/2))
are narrower and translated in smaller steps.
can represent finer detail
So V0 ⊂ V1
In general . . . ⊂ V−2 ⊂ V−1 ⊂ V0 ⊂ V1 ⊂ V2 ⊂ . . .
Vj ⊂ Vj+1 −→ Question: what “lies” between them?
2. Consider the wavelet function ψ and the associated 2 parameter family
ψj,k(x) = 2j/2ψ(2jx− k) j, k ∈ ZZ.
For fixed j : {ψj,k(x)}k∈ZZ spans Wj
27
{ψj,k}k∈ZZ is basis for Wj, again with piecewise functions on
dyadic grid {k/2j}is a fully orthonormal basis.
3. {ψj,k}(j,k∈ZZ
basis for IL2IR. This basis is complete.
It is an infinite basis in terms of wavelet functions.
4. Practical alternative
Take the set{ϕj0,k(x) = j0/2ϕ(2j0x− k),
ψj,k = 2j/2ψ(2jx− k)}j≥0, k∈ZZ.
which also forms a basis for IL2(IR).
In practice, one quite often takes j0 = 0
and therefore takes{ϕ(x), ϕ(x± 1), ϕ(x± 2), . . . , . . .ψ(x), ψ(x± 1), ψ(x± 2), . . .√
2ψ(2x± 1),√
2ψ(2x± 2), . . ....
2j/2ψ(2jx−k)} as the basis
Note: The set {ϕj0,k(x)}k∈ZZ spans the same subspace as {ψj,k(x)}j≤j0, k∈ZZ.
5. Visualization of Haar’s multi-resolution
V0 coarse
V1 finer V0 ⊂ V1 ⊂ V2
V2 still finer...
Vj by having j sufficiently large the subspace Vj is quite large.
Using the wavelets ψj,k(x) takes care of the differences.
The precise relationship between Vj,Wj and IL2(IR) is given via the Multires-olution Analysis (MRA).
28
6. Beyond the Haar case
The same construction is possible, the same concepts and ideas hold for manytypes of wavelet families (Daubechies wavelets in particular).
Advantage: We will be able to use more regular (smoother) functions ϕ andψ at the cost of a more elaborate construction and loss of analytic or closedforms, longer filters, etc.
Multi-resolution analysis (Meyer and Mallat - 1986)Multi-resolution analysis of IL2(IR) is an increasing sequence of closed (nested)
subspaces
{Vj}j∈ZZ of IL2(IR),
. . . ⊂ V−2 ⊂ V−1 ⊂ V0 ⊂ V1 ⊂ V2 ⊂ . . . ⊂ Vj ⊂ Vj+1 ⊂ . . .
Such that
(i)⋂k∈ZZ
Vj = {0} or limj→−∞
Vj = {0} (no overlap, separate spaces)
(ii)⋃k∈ZZ
Vj = IL2(IR) or limj→∞
Vj = IL2(IR)
⋃j
Vj is dense in IL2(IR)
(iii) For any f ∈ IL2(IR), for any j ∈ ZZ
f(x) ∈ Vj iff f(2x) ∈ Vj+1(Also f(x) ∈ V0 ⇐⇒ f(2jx) ∈ Vj
)
(iv) For any f ∈ IL2(IR), for any k ∈ ZZ
f(x) ∈ V0 iff f(x− k) ∈ V0
(v) ∃ϕ ∈ V0 such that
{ϕ(x− k)}k∈ZZ is an orthonormal basis of V0
.
Notes:
(1) ϕ is the scaling function.
(2) ψ will come as a by-product.
29
(3) Riesz basis: A countable set {fn} of a Hilbert space IL2(IR) is a Riesz basisif ∀f ∈ IL2(IR) it can be uniquely written as
f =∑n∈ IN
cnfn and ∃A,B > 0 such that
A‖f‖2 ≤∑n|cn|2 ≤ B‖f‖2
Now define Wj as the orthogonal complement of Vj in Vj+1 ,i.e. Vj ⊕Wj = Vj+1, j ∈ ZZ.
Then IL2(IR) is represented as the direct sum
IL2(IR) =⊕j∈ZZ
Wj
In practice
1. There is a coarsest scale “n” and
V−n ⊂ . . . ⊂ V−1 ⊂ V0 ⊂ V1 ⊂ . . .
−→ IL2(IR) = V−n ⊕Wj,j≥−n low resolution
2. There is a finite number of scales, select j = 0 to be the coarsest scale
V0 ⊂ V1 ⊂ . . . ⊂ Vn, V0 ⊂ IL2(IR)
↓ ↓coarsest finest
3. In numerical implementations the subspace V0 is finite dimensional.
Be aware of other choices in the literature where ϕj,k(x) = 2−j/2ϕ(2−jx− k)Then
Vj = Vj+1 ⊕Wj+1 j −→ −jand
⊃ V−2 ⊃ V−1 ⊃ V0 ⊃ V1 ⊃ . . .
30
EXPLANATION OF MULTI-RESOLUTION ANALYSISHaar basis on [0,1]
V0 is spanned by ϕ(x)
W0 is spanned by ψ(x)
V0⊥W0 and V1 = V0 ⊕W0
V1 = space of all piecewise constant functions on half intervals.
Basis for V1 consists of√
2ϕ(2x) and√
2ϕ(2x− 1) −→√
2ϕ(2x− k), k = 0, 1.
√2ϕ(2x) =
√2
2(ϕ(x) + ψ(x))
√2ϕ(2x− 1) =
√2
2(ϕ(x)− ψ(x))
↑ ↑∈ V0 ∈ W0
Since V0 ⊂ V1 −→ ϕ(x) can be written in terms of√
2ϕ(2x) and√
2ϕ(2x− 1)
Indeed
ϕ(x) =1√2
(√2ϕ(2x) +
√2ϕ(2x− 1)
)Sine
W0 ⊂ V1 −→ ψ(x) = ϕ(2x)− ϕ(2x− 1)
=1√2
(√2ϕ(2x)−
√2ϕ(2x− 1)
)
Basis for W1 consists of√
2ψ(2x) and√
2ψ(2x− 1) −→√
2ψ(2x− k), k = 0, 1.
V2 is spanned by 2ϕ(22x− k) = 2ϕ(4x− k), k = 0, 1, 2, 3−→ space of piecewise functions on quarter interval.
W2 is spanned by 2ψ(4x− k), k = 0, 1, 2, 3
V2 = V1 ⊕W1 = V0 ⊕W0 ⊕W1
31
So, a function f ∈ V2 can be written in terms of
ϕ(x), ψ(x), 2ψ(2x) and√
2ψ(2x− 1)
But V2 is also spanned by
2ϕ(4x), 2ϕ(4x− 1), 2ϕ(4x− 2) and 2ϕ(4x− 3).
In general,
Vj is spanned by 2j/2ϕ(2jx− k) for fixed j and k = 0, 1, . . . , 2j − 1
Wj is spanned by 2j/2ψ(2jx− k), and k = 0, 1, . . . , 2j − 1
Vj+1 = Vj ⊕Wj
= Vj−1 ⊕Wj−1 ⊕Wj
= · · ·= · · ·
Vj+1 = V0 ⊕W0 ⊕W1 ⊕ · · · ⊕Wj
and the translates of wavelets on the right are also translates of scaling functionson the left.
Next step, move from [0,1] to IR by allowing all translations (no restrictions onindex k).
A basis for IL2IR consists of ϕ(x− k)k∈ZZ together with
ψj,k = 2j/2ψ(2jx− k) with j ≥ 0, k ∈ ZZ
Another basis for IL2IR consists of
ψj,k(x) = 2j/2ψ(2jx− k) ∀j ∈ ZZ , ∀k ∈ ZZ.
Orthogonality Issues
1. For fixed j Vj⊥Wj∫ϕj,kψj,`dx = 0 ∀k, `
2. Since Wj ⊂ Vj+1 and Wj+1⊥Vj+1∫ψj,k(x)ψj+1,`dx = 0
=⇒ Wj⊥Wj+1
Then Wj+1 ⊂ Vj+2 and Wj+2⊥Vj+2 =⇒ Wj+1⊥Wj+2
All Wj are mutually orthogonal.
32
3. Since Vj ⊂ Vj+1 and Wj+1⊥Vj+1∫ψp,kψq,`dx = 0 if p 6= q
=⇒ Vj⊥Wj+1∫ϕj,kψj+1,`dx = 0.
4. It is not true that Vj⊥Vj+1, but ϕ⊥ its translates.
Representing a function in a wavelet basis
There are 5 different approaches (points of view).Each one leads to a key interpretation, to key properties and to an algorithm
to perform wavelet decomposition.
Approaches:
1. inner product method,
2. matrix method (change of basis),
3. convolution method,
4. FWT method,
5. projection matrices method.
Example to illustrate the methods
Take f(x) = 9ϕ(4x) + 1.ϕ(4x− 1) + 2ϕ(4x− 2) + 0.ϕ(4x− 3).
Rewrite in basis 2j/2ϕ(2jx− k) on V2
here j = 2 , k = 0, 1, 2, 3
f(x) =9
22ϕ(4x− 0)︸ ︷︷ ︸+1
22ϕ(4x− 1)︸ ︷︷ ︸+1 2ϕ(4x− 2)︸ ︷︷ ︸+0ϕ(4x− 3)︸ ︷︷ ︸
=9
2ϕ2,0(x) +
1
2ϕ2,1(x) + 1ϕ2,2(x) + 0ϕ2,3(x)
basis vectors for V2
We want to write
f(x) = aϕ(x) + bψ(x) + c√
2ψ(2x) + d√
2ψ(2x− 1)
↓ ↓ ↓ ↘ ↙∈ V2 ∈ V0 ∈ W0 ∈ W1
33
There are 5 different methods to compute a, b, c and d.
METHOD #1:Inner product method (not efficient)
a = < f(x), ϕ(x) >=9
2< ϕ2,0, ϕ > +
1
2< ϕ2,1′ϕ > +1 < ϕ2,2′ϕ > +0
=9
2< 2ϕ(4x), ϕ(x) > +
1
2< 2ϕ(4x− 1), ϕ(x) > + < 2ϕ(4x− 2), ϕ(x) >
=9
22.
1
4+
1
2.2.
1
4+ 2.
1
4+ 0
=9
4+
1
4+
2
4= 3.
Similarlyb = < f(x), ψ(x) >= · · · = 2
c = < f(x),√
2ψ(2x) >= · · · = 4/√
2
d = < f(x),√
2ψ(2x− 1) >= · · · = 1/√
2
−→ (x) = 3.ϕ(x) + 2ψ(x) +4√2.√
2ψ(2x) +1√2.√
2ψ(x− 1).
So,
f(x) = 3ϕ(x) + 2ψ(x) +4√2
√2ψ2x+
1√2
√2ψ(2x− 1)
f (2)(x) = f (0) + g(0) + g(1)
↓ ↓ ↓ ↓∈ V2 ∈ V0 ∈ W0 ∈ W1
1 coeff 1 coeff 2 coeffs
In general g(j) is made up with functions on mesh 12j .
Full decomposition
f (n)(x) = f (0) + g(0) + g(1) + g(2) + · · ·+ g(n−1)
# coefficients 1 + 1 + 2 + 4 + 8 + · · ·+ 2n−1︸ ︷︷ ︸2n coefficients
After J steps of decomposition
f (n)(x) = f (n−J)(x) + g(n−J)(x) + · · ·+ g(n−1)(x)
# coefficients 2n−J + · · · 2n−2 + 2n−1 = 2n coefficients
34
METHOD #2: Matrix Method
Old basis B = {2ϕ(4x), 2ϕ(4x− 1), 2ϕ(4x− 2), 2ϕ(4x− 3)}New basis B′ = {ϕ(x), ψ(x),
√2ψ(2x),
√2ψ(2x− 1)}
9/21/210
=
1/2 1/2 1/
√2 0
1/2 1/2 −1/√
2 0
1/2 −1/2 0 1/√
2
1/2 −1/2 0 −1/√
2
a
b
c
d
⇐⇒→f= R
→b
↑ ↑ R = reconstruction matrixold coordinates If matrix with as columns the coordinates
of the new basis vectors in terms of theold basis.
Column #1 : 1/21/21/21/2
because
ϕ(x) =1
22ϕ(4x) +
1
22ϕ(4x− 1) +
1
22ϕ(4x− 2) +
1
22ϕ(4x− 3)
Similarly, for columns 2, 3 and 4
ψ(x) =1
22ϕ(4x) +
1
22ϕ(4x− 1)− 1
22ϕ(4x− 2)− 1
22ϕ(4x− 3)
√2ψ(2x) =
1√22ϕ(4x)− 1√
22ϕ(4x− 1) + 0.2ϕ(4x− 2) + 0.2ϕ(4x− 3)
√2ψ(2x− 1) = 0.2ϕ(4x)− 0.2ϕ(4x− 1) +
√22ϕ(4x− 2)− 1√
22ϕ(4x− 3)
→b= D
→f D = decomposition matrix
We need→b= R−1
→f= R>
→f
because the columns of R are orthonormal vectors.
35
−→
a
b
c
d
=
1/2 1/2 1/2 1/21/2 1/2 −1/2 −1/2
1/√
2 −1/√
2 0 0
0 0 1/√
2 −1/√
2
9/21/210
=
32
4/√
2
1/√
2
.
So, f (2)(x) = f(x) = 3ϕ(x) + 2ψ(x) + 4√2
√2ψ(2x) + 1√
2
√2ψ(2x− 1)
= 3ϕ(x) + 2ψ(x) + 4ψ(2x) + 1ψ(2x− 1)
Picture; 9ψ(4x) + ϕ(4x− 1) + 2ϕ(4x− 2) + 0ϕ(4x− 3)
METHOD #3: Convolution Method
For Haar case H = ( 1√2 ,
1√2), G = ( 1√
2 , −1√2)
→s
(n) H−→ →s
(n−1) H−→ →s
(n−2) · · · −→ →s
(0)
G ↓ G ↓ G ↓ ↓ G
→d
(n−1) →d
(n−2) →d
(n−3) →d
(0)︸ ︷︷ ︸set aside
stop at level
zero.
→s(0)
. . . . .→d
(0)
. . . . .→d
(1)
. . . . .→d
(2)
. . . . ....
. . . . .→d
(n−1)
.
→s
(2)= 9/2 1/2 1 0
H−→→s(1)= 5/
√2 1/
√2
H−→→s(0)= 3
↓ G ↓ G
→d
(1)= 4/
√2 1/
√2 d(0) = 2
→s
(0)
. . .→d
(0)
. . .→d
(1)
36
Assemble the result:
9/21/210
D
=⇒
3. . . . .
2. . . . .
4/√
2√2
In general
→sn=
...
...
...
...
...
2n data −→
→s
(0)
. . . .→d
(0)
. . . .→d
(1)
. . . .→d
(2)
. . . .......
. . . .→dn−1
1
1
2
4
8
...
2n−1
2n data
Convolution9/2 1/2 1 0 −→ 5/
√2 1/
√2
↘ ↙ ↘↙ ↘ ↙Average: 5/
√2 1/
√2 Average: 3
Difference: 4/√
2 1/√
2 Difference: 2
37
METHOD #4: Fast Wavelet Transform
Decompose the matrix D into a product of 3 sparser matrices.
Decompose at level 1: V2 = V1 ⊕W1
h0 h1 −→g0 g1 −→
1√2
1√2 0 0
1√2 −
1√2 0 0
0 0 1√2
1√2
0 0 1√2 −
1√2
9/21/210
=
5/√
2
4/√
2
1/√
2
1/√
2
− low frequency− high frequency− low frequency− high frequency
or M1→f=
→b1
butterfly (structured) matrix
Bring high frequencies to bottom via a permutation1 0 0 00 0 1 00 1 0 00 0 0 1
5/√
2
4/√
2
1/√
2
1/√
2
=
5/√
2
1/√
2
4/√
2
1/√
2
− low frequencies− low frequencies− high frequencies− high frequencies
or M2→b1=
→b2
very sparse
Decompose at level 2: V1 = V0 ⊕W0
1√2
1√2 0 0
1√2 −
1√2 0 0
0 0 1 00 0 0 1
5/√
2
1/√
2
4/√
2
1/√
2
=
32
4/√
2
1/√
2
or M3
→b2=
→b
So
M1→f =
→b1
M2→b1
↙=
→b2
M3→b2
↙=
→b
M3M2M1
→f =
→b
= D→f =
→b
D = M3M2M1
38
Count number of operations
For
D→f=
→b −→∼ N 2 operations (not counting additions)
D is a dense N ×N matrix 12 multiplications
For
M1→f =
→b1
M2→b1 =
→b2
M3→b2 =
→b
8 multiplicationsno cost, only shuffling4 multiplications
∼ 12 multiplications
In general N log2N algorithm instead of N 2algorithm.
Compare with FFT Tukey, Cooley (1965)(See slides)
* Same idea: write dense matrix as product of sparser “butterfly” and permu-tation matrices.
* Computation count N log2N versus N 2
If N = 1024 N 2 = 1, 048576
N log2N = 10240.
• Show M1 for Daubechies case (DAUB4)
METHOD #5 Projection matrices Method
See e.g. Daubechies’ Ten Lectures on Wavelets.
• Comparison with FFT.
QUADRATURE MIRROR FILTERS(A look in the Fourier domain)
The Fourier transform.
There are 3 equivalent choices, depending on where one puts the “2π”.
39
Each choice exists of two versions, depending on whether the minus is put inthe exponent of the direct or the inverse transform.
So, we have
1. (Ff)(ω) = f(ω) def=1√2π
∫ +∞
−∞f(x)e−iωxdx
→ ω is in radians per second (1 radian = 12π circle)
2. f(ω) def=∫ +∞
−∞e−2πiωtdt
→ ω in Hertz
3. f(ω) =∫ +∞
−∞f(x)e−iωxdx
→ in radians.
For these notes:
Direct Transformation: f(ω) = (Ff)(ω) =∫ +∞
−∞e−2πiωtf(t) dt
Inverse Transformation: f(t) = (F−1f)(t) =∫ +∞
−∞e2πiωtf(ω) dω
1. Apply the Fourier Transformation to the refinement equation
(in real space)
ϕ(x) =√
2L−1∑k=0
hkϕ(2x− k)
ϕ(ω) =∫ +∞
−∞e−2πiωxϕ(x)dx
=√
2L−1∑k=0
hk∫ +∞
−∞e−2πiωxϕ(2x− k)dx
Set u = 2x− k −→ x =u+ k
2−→ dx =
du
2
=
√2
2
L−1∑k=0
hk∫ +∞
−∞e−2πiω[u+k
2 ]ϕ(u)du
=1√2
L−1∑k=0
hk∫ +∞
−∞e−2πiω
2 ke−2πiω2 uϕ(u)du
40
=1√2
L−1∑k=0
hke−2πiω
2 k∫ +∞
−∞e−2πi(ω
2 )uϕ(u)du
=1√2
L−1∑k−0
hke−2πiω
2 kϕ
(ω
2
)
Define
m0(ω) =1√2
L−1∑k=0
hke−2πikω
Then
ϕ(ω) = m0
(ω
2
)ϕ
(ω
2
)
or equivalently,
ϕ(2ω) = m0(ω)ϕ(ω) refinement equation in Fourier domain!
Notes: m0(ω) is a trigonometric polynomial
m0(ω + 1) = m0(ω)
If we know m0(ω) =⇒ we know hk.
2. Keep refining
Since
ϕ(ω) = m0
(ω
2
)ϕ
(ω
2
)
then ω → ω/2 ϕ(ω2
)= m0
(ω4
)ϕ(ω4
)
then ω → ω/2 ϕ(ω4
)= m0
(ω8
)ϕ(ω8
)
We get
ϕ(ω) = m0
(ω
2
)m0
(ω
4
)m0
(ω
8
)· · ·m0
(ω
32
)ϕ
(ω
32
)...
Keep going ω −→ ω/2
ϕ(ω) =N∏j=1
m0
(ω
2j
)ϕ
(ω
2N
)
41
Take limit for N −→∞
ϕ(ω) =∞∏j=1
m0
(ω
2j
)ϕ(0)
Butϕ(0) =
∫e−2πi0tϕ(t)dt =
∫ϕ(t)dt = 1
So,
ϕ(ω) =∞∏j=1
m0
(ω
2j
)
Infinite refinement of the two-scale difference relation in the Fourier domain.
Similarly,ψ(ω) = m1(ω/2)ϕ(ω/2)
or ψ(2ω) = m1(ω)ϕ(ω)
with m1(ω) def=1√2
L−1∑k=0
gke−2πikω
Also, by continuing the refinement process
ψ(ω) = m1(ω/2)ϕ(ω/2)
= m1(ω/2)m0(ω/4)ϕ(ω/4)
= m1(ω/2)m0(ω/4)m0(ω/8)ϕ(ω/8)
...
=⇒ ψ(ω) = m1
(ω
2
) ∞∏j=2
m0
(ω
2j
)
Since G ⊥ H so gk = (−1)khL−1−k
one has m1(ω) = e−2πi(1−L)ωm0(ω + 1
2
)
42
3. Change of notation
Set z = e−2πiω
Then m0(ω) −→ H(z) =√
2m0(ω(z))
H(z) =L−1∑k=0
hkzk
And
m1(ω)→ G(z) =√
2m1(ω(z))
G(z) =L−1∑k=0
gkzk
Also, m1(ω) is related to m0(ω + 1/2) =⇒ G(z) = −zL−1H(−1z)
DERIVATION OF THE QMF CONDITION
(Quadrature mirror filters)
Tools needed
Parseval’s Identity
< f, g >=∫ +∞
−∞f(x)g(x)dx =
∫ +∞
−∞f(ω)g(ω)dω =< f, g >
Consequence (Plancherel’s formula)
‖ f ‖2 =∫ffdx =
∫|f |2dx
=∫f(ω)f(ω)dω =
∫|f(ω)|2dω =‖ f ‖2
(F is an isometric isomorphism)
First, we prove that the following statements are equivalent
(i)
< ϕ(x− k), ϕ(x− `) >=∫ +∞
−∞ϕ(x− k)ϕ(x− `)dx = δkl
i.e. ϕ ⊥ ϕ translates.
43
(ii)
ϕ(ω) satisfies∫ +∞
−∞e−2πikω|ϕ(ω)|2dω = δk0
(iii)+∞∑
k=−∞|ϕ(ω + k)|2 = 1
Proof: Define the function K(ω) =+∞∑
k=−∞|ϕ(ω + k)|2
Then
(1) K(ω) is 1 periodic
K(ω + 1) =∑k∈ZZ| ϕ(ω + k + 1) |2=
∑`∈ZZ| ϕ(ω + `) |2= K(ω)
(2) ∫ 1
0K(ω)dω =
∫ 1
0
∑k
| ϕ(ω + k) |2 dω
=∑k
∫ 1
0| ϕ(ω + k) |2 dω
Set z = ω + k
=∑k
∫ k+1
k| ϕ(z) |2 dz =
∫ +∞
−∞| ϕ(z) |2 dz
Plancherel=∫ +∞
−∞| ϕ(z) |2 dz = 1.
So, K(ω) is 1 periodic and integrable (integrates to 1)
So, K(ω) has a Fourier series
K(ω) =+∞∑
n=−∞cne
2πinω
Where cn =∫ 1
0e−2πinωK(ω)dω
=∫ 1
0e−2πinω
+∞∑k=−∞
| ϕ(ω + k) |2 dω
44
Set ω + k = z
=+∞∑
k=−∞
∫ k+1
ke−2πin(z−k) | ˆϕ(z) |2 dz
=+∞∑
k=−∞
∫ k+1
ke−2πinz | ϕ(z) |2 dz
=∫ +∞
−∞e−2πinzϕ(z)ϕ(z)dz
=∫ +∞
−∞ϕ(z − n)ϕ(z)dz
Parseval=∫ +∞
−∞ϕ(x− n)ϕ(x)dx = δno
So, cn = δn0
Hence, K(ω) =+∞∑
n=−∞δnoe
2πinω = 1
=⇒+∞∑
k=−∞| ϕ(ω + k)2 |= 1.
Consequences
(1) If ϕ(0) = 1
Then from+∞∑
k=−∞| ϕ(ω + k) |2=|=⇒ ϕ(k) = δk0
One can show that ϕ(k) = δk0 ⇐⇒+∞∑
k=−∞| ϕ(ω + k) |2= 1.
(2)+∞∑
k=−∞ϕ(x− k) = 1 Partition of unity
Proof: Similar to the above by setting
P (x) =∑k
ϕ(x− k)
showing that P (x+ 1) = P (x) and P (x) integrable
Compute cn =∫ 1
0e−2πinx∑
k
ϕ(x− k)dx
= · · · = δn0 , etc.
45
(3)ϕ(k) = δk0 ⇐⇒
∑k∈ZZ
ϕ(x− k) = 1 ∀x
Compare with Poisson summation formula
+∞∑n=−∞
f(n) =+∞∑
k=−∞f(k)
Now we can show that
| m0(ω) |2 + | m0(ω + 1/2) |2= 1 QMF condition
or, equivalently,
H(z)H(1/z) +H(−z)H(−1/z) = 2
Proof
use ϕ(2ω) = m0(ω)ϕ(ω) (1)
DenoteK(ω) =
∑k
| ϕ(ω + k) |2= 1
K(2ω) =∑k
| ϕ(2ω + k) |2= 1
(1)=
∑k
| ϕ(ω +k
2) |2| m0(ω +
k
2) |2
Split in even and odd terms
=∑
` (k=2`)| ϕ(ω + `) |2| m0(ω + `) |2
+∑
` (k=2`+1)| ϕ(ω + `+
1
2) |2| m0(ω + `+
1
2) |2
m0(ω) is 1-periodic
=∑`
| m0(ω) |2| ϕ(ω + `) |2 +∑`
| m0(ω +1
2) |2| ϕ(ω + `+
1
2) |2
= | m0(ω) |2∑`
| ϕ(ω + `)︸ ︷︷ ︸1
|2 + | m0
(ω +
1
2
)|2∑`
| ϕ(ω + `+
1
2
)︸ ︷︷ ︸
1
|2
= | m0(ω) |2 + | m0(ω +1
2) |2= 1 QMF
46
Rewrite in terms of H(z)
H(z)H(1/z) +H(−z)H(−1/z) = 2
Proof
m0(ω) =1√2H(z) since z = e−2πiω
m0(ω) =1√2
∑hke−2πikω =
1√2
∑k
hke2πiωk
= 1√2H(1/z)
m0(ω + 1/2) =1√2
∑k
hke−2πik(ω+1/2)
=1√2
∑k
hk(−1)ke−2πikω
=1√2
∑k
hk(−z)k =1√2H(−z)
m0(ω + 1/2) = 1√2H(−1/z)
Som0(ω)m0(ω) +m0(ω + 1/2)m0(ω + 1/2) = 1
−→ H(z)√2
H(1/z)√2
+H(−z)√
2
H(−1/z)√2
= 1
orH(z)H(1/z) +H(−z)H(−1/z) = 2
Expresses that ϕ ⊥ ϕ translates.
Similarly,ϕ ⊥ ψ G(z)H(1/z) +G(−z)H(−1/z) = 0
ψ ⊥ ψ G(z)G(1/z) +G(−z)G(−1/z) = 2
For reversal H ←→ G
G(z) = −zL−1H(−1/z)
47
Vanishing moments for ψ(x)
Theorem. The following statements are equivalent.
(i) The wavelet function ψ has M vanishing moments
i.e.∫xmψ(x)dx = 0 , m = 0, 1, 2, · · · ,M − 1
(ii) The filter coefficients hk satisfyL−1∑k=0
(−1)kkmhk = 0,
m = 0, 1, 2, · · · ,M − 1
(iii) H(z) has z = −1 as a zero with multiplicity M
i.e. H(−1) = H ′(−1) = · · · = H(M−1)(−1) = 0
or H(m)(−1) = 0 , m = 0, 1, 2, · · · ,M − 1
Hence, H(z) =√
2(1+z2 )MQ(z)
(iv) The matrix which determines the values of ϕ(x) at integers has eigenval-ues 1, 1
2 , (12)
2, · · · , (12)M−1
(v) The Fourier transform of ψ , i.e. ψ(ω), has a M -fold zero at ω = 0
(vi) The Fourier transform of ϕ, i.e. ϕ(ω), has a M -fold zero at ω = k, exceptat 0 where ϕ(0) = 1.
Proof (some samples, to get the flavour)
ψ(ω) =∫ +∞
−∞ψ(x)e−2πiωxdx
ψ(0) =∫ +∞
−∞ψ(x)dx = 0 −→ ψ(0) = 0
ψ′(0) =dψ
dω|ω=0 =
∫ +∞
−∞ψ(x)(−2πix)e−2πiωxdx
So,∫xψ(x)dx = 0 ←→ ψ′(0) = 0, etc.
ψ(ω) = m1(ω/2)ϕ(ω/2) −→ ψ(0) = 0 =⇒ m1(0) = 0 −→ H(−1) = 0, etc.
and also ϕ(0) = 1.
48
Example: DAUB6. (3 vanishing moments)
∫xmψ(x)dx = 0 m = 0, 1, 2
So, H(z) =√
2
(1 + z
2
)3Q(z) (1)
On the other hand H(z) =5∑
k=0hkz
k = h0 + h1z + · · ·+ h5z5 (2)
−→ Q(z) is degree 2
So, Q(z) = a0 + a1z + a2z2
H(1) =√
2Q(1) =5∑
k=0hk =
√2 ( due to normalization)
So, Q(1) = 1 =⇒ a0 + a1 + a2 = 1
Put H(z) =√
2
(1 + z
2
)3Q(z) =
√2
(1 + z
2
)3(a0 + a1z + a2z
2) (3)
into QMF
and take coeff of z0, z or 1z , z
2 or 1z2
=⇒
10(a2
0 + a21 + a2
2) + 15(a0a1 + a1a2) + 6a0a2 = 163(a2
0 + a21 + a2
2) + 8(a0a1 + a1a2) + 10a0a2 = 0a0a1 + a1a2 + 6a0a2 = 0
=⇒
a2
0 + a21 + a2
2 = 19/4a0a1 + a1a2 = −9/4a0a2 = 3/8
Solvea0 = 1
4
(1 +√
10 +√
5 + 2√
10)
a1 = 14
(2− 2
√10)
= 12
(1−√
10)
a2 = 14
(1 +√
10−√
5 + 2√
10)
Substitute in (3), compare with (2)
49
=⇒
h0 = 1
16√
2
(1 +√
10 +√
5 + 2√
10)
......
h5 = 116√
2
(1 +√
10−√
5 + 2√
10)
DAUBECHIES WAVELETS: SOLVING THE QMF
We will solve
| m0(ω) |2 + | m0(ω + 1/2) |2= 1 QMF
Set| m0(ω) |2= M0(ω)
then| m0(ω + 1/2) |2= M0(ω + 1/2)
QMF : M0(ω) +M0(ω + 1/2) = 1
For wavelets with M vanishing moments we have
m0(ω) =
1 + e−2πiω
2
M L(ω)
→ trigonometric polynomial in e−2πiω
M0(ω) =
1 + e−2πiω
2
1 + e2πiω
2
M IL(ω)
whereIL(ω) = L(ω)L(ω)
= |L(ω)|2
= cos2M(πω)IL(ω)
→ is trigonometric polynomial in e−2πiω
→ can be written in sin2 πω
→ say P (sin2 πω)
So we will assume the form
M0(ω) = cos2M(πω)P (sin2 πω)Then M0(ω + 1/2) = sin2M(πω)P (cos2 πω)
50
Set y = sin2 πω M0(ω) = (1− y)MP (y)Then 1− y = cos2 πω M0(ω + 1/2) = yMP (1− y)
So we need to find P (y) such that
(1− y)MP (y) + yMP (1− y) = 1
Note: Doing this all for z = e−2πiω would give
(z + 1)2M
(4z)M︸ ︷︷ ︸(1−y)M
P (y)︷ ︸︸ ︷Q(z)Q(1/z) + (−1)M
(z − 1)2M
(4z)M︸ ︷︷ ︸yM
Q(−z)Q(−1/z)︸ ︷︷ ︸P (1−y)
= 1
Step 1: Find P (y)
Most general solution
P (y) =M−1∑k=0
M + k + 1
k
yk + yMR
(1
2− y
)
where R(y) is an odd polynomial chosen so that P (y) ≥ 0 for 0 ≤ y ≤ 1
Let’s consider the case R = 0⇐⇒ minimum length IL where L = 2M
(If R 6= 0 then M > 2L , more freedom).
Argument: P (y) polynomial in y
P (y) ∼ (1− y)−M Taylor∞∑k=0
M + k + 1
k
yk−→ must be truncated at degree M − 1
P (1− y) ∼ y−M
Candidate
P (y) =M−1∑k=0
M + k + 1
k
yk
or Q(z)Q(1/z) =M−1∑k=0
(−1)k(1− z)2k(4z)−k(M + k − 1)!
k!(M − 1)!
51
Example: DAUB6 case
LHSQ(z) = a0 + a1z + a2z
2
Q(
1z
)= a0 + a1
z + a2
z2
Q(z)Q
(1
z
)= a2
0 + a21 + a2
2 + (a0a1 + a1a2)
(z +
1
z
)+ a0a2
(z2 +
1
z2
)
RHS
2∑k=0
(−1)k(1− z)2k
(4z)k
2 + k
k
=
2
0
−3
1
(1− z)2
4z+
4
2
(1− z)4
(4z)2
=19
4− 9
4
(z +
1
z
)+
3
8
(z2 +
1
z2
)
=⇒
a2
0 + a21 + a2
2 = 19/4a0a1 + a1a2 = −9/4a0a2 = 3/8
a0 + a1 + a2 = 1 since Q(1) = 1
Solve a0 = 1
4
(1 +√
10 +√
5 + 2√
10)
a1 = 12
(1−√
10)
a2 = 14
(1 +√
10−√
5 + 2√
10)
−→ H(z) =√
2(
1+z2
)3 (a0 + a1z + a2z
2)
≡ h0 + h1z + · · ·+ h5z5
−→
h0 = 116√
2
(1 +√
10 +√
5 + 2√
10)
......
......
h5 = 116√
2
(1 +√
10−√
5 + 2√
10)
52
GENERAL POLYNOMIAL SOLUTION OF QMF
(1− y)MP (y) + yMP (1− y) = 1
Use Theorem of Bezout (gives existence and uniqueness of polynomial solu-tions).
Theorem: If p1(x) and p2(x) are two polynomials of degree n1 and n2 with nocommon zero, then there exists two unique polynomials q1(x) and q2(x) of degree≤ n2 − 1 and ≤ n1 − 1 so that
p1(x)q1(x) + p2(x)q2(x) = 1
Apply Bezout’s theorem
p1(y) = (1− y)M degreeMp2(y) = yM degreeM
∃ q1(y) and q2(y) so that
(1− y)Mq1(y) + yMq2(y) = 1
Now substitute 1− y for y
−→ yMq1(1− y) + (1− y)Mq2(1− y) = 1
Since q1(y) and q2(y) are unique =⇒ q1(y) = q2(1− y)So, (1− y)Mq1(y) + yMq1(1− y) = 1
and q1(y) is of degree ≤M − 1
So P (y) = PM(y) (the polynomial we sought)
Exists and is unique
−→ PM(y) =M−1∑k=0
M + k − 1
k
ykStep 2 How do we get Q(z) or m0(ω) now that we have M0(ω) =| m0(ω) |2
or P (y) = Q(z)Q(1/z)
=⇒ Use spectral Factorization Theorem of Riesz
53
Let A(ξ) be a positive trigonometric polynomial invariant under the symmetryξ −→ −ξ
then A(ξ) is of the form
A(ξ) =M∑m=0
am cos(mξ) , am ∈ IR
and there then exists a trigonometric polynomial B(ξ) of degree M , i.e.
B(ξ) =M∑m=0
bmeimξ , bm ∈ IR
such that | B(ξ) |2= A(ξ)
B(ξ) is “root”.
In practice: Factor A(ξ) and investigate the complex (or real) root structure.
Example: DAUB 6
A(ξ) = Q(z)Q(1/z) =38 −
94z + 19
4 z2 − 9
4z3 + 3
8z4
z2
4 rootsz1
z1
2 inside unit circle|z1| = 0.325406
Re(z1) = 0.28725
z−11
z1−1
2 outside unit circle
Q(z) = cte(z2 − 2zRe(z1)+ | z1 |2
)
Q(1) = 1 −→ Q(z) = 1.8818688(z2 − 0.57450z + 0.1058894)
Then H(z) =√
2(1+z2 )3Q(z) = h0 + h1z + . . .+ h5z
5
−→ h1, h2, · · · , h5 values
Doing so one gets the numerical values of the DAUB6 filter taps.For the roots outside the unit circle in the Gauss plane one gets
54
h0 = 116√
2
(1 +√
10 +√
5 + 2√
10)∼= 0.3326705529500825 · · ·
h1 = 116√
2
(5 +√
10 + 3√
5 + 2√
10)∼= 0.8068915093110924 · · ·
h2 = 18√
2
(5−√
10 +√
5 + 2√
10)∼= 0.4598775021184914 · · ·
h3 = 18√
2
(5−√
10−√
5 + 2√
10)∼= −0.1350110200102545 · · ·
h4 = 116√
2
(5 +√
10− 3√
5 + 2√
10)∼= −0.0854412738820267 · · ·
h5 = 116√
2
(1 +√
10−√
5 + 2√
10)∼= 0.03522629188570 · · ·
Using the roots inside the unit circle one gets the filter in reversed order.
Construction of scaling and wavelet filters
Eigenvalue-Eigenvector Problem
Example: DAUB6 (L = 6)
ϕ(x) =√
2l−1∑k−0
hkϕ(2x− k)
Assume that ϕ is bounded to an interval [a, b].
Then
[a, b]matching←→
[a
2,b
2
],
[a+ 1
2,b+ 1
2
], · · · ,
[a+ L+ 1
2,b+ L− 1
2
]
So
a ≡ a2 −→ a = 0
b ≡ b+L−12 −→ 6 2b = 6 b+ L− 1
b = L− 1
So ϕ(x) lives on interval [0, L− 1]Same for ϕ(x)
Supp ϕ = Supp ψ = [0, L− 1]
55
For L = 6 Set ck =√
2hkϕ(0) = c0ϕ(0) + 0 (Since ϕ(−1) = 0, etc.)ϕ(1) = c0ϕ(2) + c1ϕ(1) + c2ϕ(0) + 0ϕ(2) = c0ϕ(4) + c1ϕ(3) + c2ϕ(2) + c3ϕ(1) + c4ϕ(0)ϕ(3) = c0ϕ(6) + c1ϕ(5) + c2ϕ(4) + c3ϕ(3) + c4ϕ(2) + c5ϕ(1)ϕ(4) = c3ϕ(5) + c4ϕ(4) + c5ϕ(3)ϕ(5) = c5ϕ(5)
Since all ci 6= 0, i = 0, · · · , 5 =⇒ ϕ(0) = ϕ(5) = 0
ϕ(0) = 0ϕ(1) = c0ϕ(2) + c1ϕ(1)ϕ(2) = c0ϕ(4) + c1ϕ(3) + c2ϕ(2) + c3ϕ(1)ϕ(3) = c2ϕ(4) + c3ϕ(3) + c4ϕ(2) + c5ϕ(1)ϕ(4) = c4ϕ(4) + c5ϕ(3)ϕ(5) = 0
Write non-trivial equations in matrix form.
ϕ(1)ϕ(2)ϕ(3)ϕ(4)
=
c1 c0 0 0c3 c2 c1 c0c5 c4 c3 c20 0 c5 c4
ϕ(1)ϕ(2)ϕ(3)ϕ(4)
→ϕ= A
→ϕ
Eigenvalue - Eigenvector Problem
λ→v= A
→v where λ = 1
→v unknown.
Due to partition of unity∑vi = 1, so,
ϕ(1) + ϕ(2) + ϕ(3) + ϕ(4) = 1 (∗)
Eigenvalues of matrix A are 14 ,
12 , 1,
1−√
108
56
If you use
c0 = 116
(1 +√
10 +√
5 + 2√
10)
c1 = 116
(5 +√
10 + 3√
5 + 2√
10)
c2 = 18
(5−√
10 +√
5 + 2√
10)
c3 = 18
(5−√
10−√
5 + 2√
10)
Using (∗) you get for the eigenvector components
ϕ(1) = 1.28634ϕ(2) = −0.385837ϕ(3) = 0.0952675ϕ(4) = 0.00423435
−→allows one to computeϕ , ψ at dyadicpoints . . .
57
Reference Books on Wavelets available at Engr. Library:
• Ingrid Daubechies, Ten Lectures on Wavelets, CBMS-NSF Regional Confer-ence Series in Applied Mathematics, vol. 61, SIAM, Philadelphia, 1992
• Michael Frazier, An Introduction to Wavelets through Linear Algebra, SpringerVerlag, New York, 1998
• Eugene Hernandez and Guido Weiss, A First Course on Wavelets, CRC Press,Boca Raton, Florida, 1996
• Gerald Kaiser, A Friendly Guide to Wavelets, Birkhauser Press, Boston, 1994
• Yves Meyer (transl: Robert D. Ryan), Wavelets, Algorithms and Applica-tions, SIAM, Philadelphia, 1993
• Yves Nievergelt, Wavelets Made Easy, Birkhauser, Boston, 1999
Review Papers available from Instructor’s Office (220)
• Adhemar Bultheel, “Learning to swim in a sea of wavelets,” Bull. Belg.Math. Soc. vol. 2 (1995) 1-44.
• Pieter W. Hemker, Tom H. Koornwinder, and Nico M. Temme, “Wavelets:Mathematical preliminaries,” from: T.H. Koornwinder (ed.), Wavelets: AnElementary Treatment of Theory and Applications, World Scientific, Singa-pore, 1993, pp. 13-26.
• Peter Meuller and Brani Vidakovic, “Wavelets for Kids: tutorial paper for thestatistical community,” (with Mathematical notebook, part B), ISDS, DukeUniversity. FTP to ftp.isds.duk.edu in directory /pub/brani/wav4Kids[A-B].ps.Z.uu
• Peter Nacken, “Image compression using wavelets,” from: T.H. Koornwinder(ed.), Wavelets: An Elementary Treatment of Theory and Applications,World Scientific, Singapore, 1993, pp. 81-92.
• Adri B. Olde Daalhuis, “Computing with Daubechies’ wavelets,” from: T.H.Koornwinder (ed.), Wavelets: An Elementary Treatment of Theory and Ap-plications, World Scientific, Singapore, 1993, pp. 93-105.
58
• T.K. Sarkar, et al. “A tutorial on wavelets from an electrical engineeringperspective: Part 1: Discrete wavelet techniques”, IEEE Antennas and Prop-agation Magazine, vol. 40, no.5, 1998, 49-70.
• T.K. Sarkar, et al. “A tutorial on wavelets from an electrical engineeringperspective: Part 2: “The continuous case,” IEEE Antennas and PropagationMagazine, vol. 40, no. 6, 1998, 36-49.
• Nico M. Temme, “Wavelets: First steps,” from: T.H. Koornwinder (ed.),Wavelets: An Elementary Treatment of Theory and Applications, WorldScientific, Singapore, 1993, pp. 1-12.
Books available at Instructor’s Office 220 Engr. Bldg.:
• C. Sidney Burrus, Ramesh A. Gopinath and Haitao Guo, Introduction toWavelets and Wavelet Transforms: A Primer, Prentice Hall, Upper SaddleRiver, New Jersey, 1998
• Charles K. Chui, Wavelets: A Mathematical Tool for Signal Analysis, SIAM,Philadelphia, 1997
• Ingrid Daubechies, Ten Lectures on Wavelets, CBMS-NSF Regional Confer-ence Series in Applied Mathematics, vol. 61, SIAM, Philadelphia, 1992
• Barbara Burke Hubbard, The World According to Wavelets: The Story of aMathematical Technique in the Making (2nd edition), A.K. Peters, Natick,Massachusetts, 1998
• Charles K. Chui, Wavelets: A Mathematical Tool for Signal Analysis, SIAM,Philadelphia, 1997
• Stephane Mallat, A Wavelet Tour of Signal Processing (2nd edition), Aca-demic Press, New York, 1999
• Yves Meyer (transl: Robert D. Ryan), Wavelets, Algorithms and Applica-tions, SIAM, Philadelphia, 1993
• Yves Nievergelt, Wavelets Made Easy, Birkhauser, Boston, 1999
• Gilbert Strang and Truong Nguyen, Wavelets and Filter Banks, Wellesley-Cambridge Press, Wellesley, Massachusetts, 1996
• M. Victor Wickerhauser, Wavelet Packet Laboratory, A.K. Peters, Wellesley,Massachusetts, 1994
59
TW Honns: Numeriese Algoritmes Huiswerk # 4(Vir 3/4/01) 2001
INSTRUCTIONS:
(i) General instructions (a) and (b) from the previous assignments apply.
(ii) The computational part of the assignment should be done with software ofyour choice.
Recommended: MATLAB, Mathematica, Fortran or C.
Task 1: Wavelet Decomposition & Reconstruction with Haar Wavelets
Consider a function f(x), that is piecewise constant on quarter-intervals:
f(x) =
2, 0 ≤ x < 14
0, 14 ≤ x < 1
2
6, 12 ≤ x < 3
4
4, 34 ≤ x < 1
(1)
(i) Express f(x) in terms of the scaling function φ2,0(x) = 2φ(4x) and its translatesφ2,1(x) = 2φ(4x− 1), φ2,2(x) = 2φ(4x− 2), φ2,3(x) = 2φ(4x− 3).
Let ~fB denote the coordinate vector of f(x) with respect to the (old) basis B,made up with the above 4 functions belonging to the scaling family φj,k(x) =
2j2φ(2jx− k). What is the explicit form of ~fB?
(ii) Express f(x) in terms of the wavelet (new) basis B′ = {φ0,0(x) = φ(x), ψ0,0(x) =ψ(x), ψ1,0(x) =
√2ψ(2x), ψ1,1(x) =
√2ψ(2x− 1)}.
Let ~fB′ denote the coordinate vector of f(x) with respect to the (new) basis B′,made up with the above 4 functions. What is the explicit form of ~fB′?
(iii) Find the explicit form of the transformation matrix D such that ~fB′ = D~fB.
The matrix D is called the decomposition matrix.
(iv) Find the explicit form of the transformation matrix R such that ~fB = R~fB′.
The matrix R is called the reconstruction matrix.
(v) Show that the matrices D and R are both orthogonal matrices.
(vi) Show that D can be written as the product of three matrices, i.e. D =D2P1D1, where D1 is the matrix for averaging and differencing at level 1, D2
60
is the matrix for averaging and differencing at level 2, and P1 is a permutationmatrix which brings averages together (at the top) and differences together (atthe bottom).
(vii) Verify your analytic work on the computer. Define the matrices D1,D2 andP1 and compute their product to obtain D. Compute the matrix R explicitly.Verify the formulas in (iii) and (iv).
(viii) Compute ~fB′ via repeated convolutions with the low-pass filterH = (h0, h1) =( 1√
2 ,1√2) and the high-pass filter G = (g0, g1) = (h1,−h0) = ( 1√
2 ,−1√2).
How may times can you apply each filter if you start with the data defined byf(x)?
(ix) If there are 4 data entries (the four values of f(x)) then all the matrices aresize 4× 4.
Generalize the above problem to sizes 8× 8 and 16× 16.
For instance, for the 8× 8 case, B = {2√
2φ(8x− k), k = 0, 1, 2, ..., 7} andB′ = {φ(x), ψ(x),
√2ψ(2x),
√2ψ(2x− 1), 2ψ(4x), 2ψ(4x− 1), 2ψ(4x− 2), 2ψ(4x−
3)}.Set up the matrices D and R for each case.
(x) Show that for the 8×8 case, the matrix D can be written as D = D3P2D2P1D1.
The explicit forms of all the decomposition and permutation matrices should bein your computer program.
(xi) Show that for the 16 × 16 case, the matrix D can be written as D =D4P3D3P2D2P1D1. The explicit forms of all the decomposition and permutationmatrices should be in your computer program.
(xii) Apply the Haar decomposition and reconstruction algorithm to the followingdata sets
~fB = 2√
2[1, 3, 0, 5, 7, 4, 2, 10]T (2)
and~fB = [4, 20, 0, 8, 4, 12, 12, 16, 24, 28, 4, 8, 0, 12, 36, 24]T (3)
Determine ~fB′ for both cases. Have the computer do the computations!
Task 2: Construction of Daubechies Filter(s) (DAUB4) with 2 vanishingmoments
This task involves the computation of the Daubechies filter(s) (DAUB4), H andG, each with 4 filter taps, corresponding to wavelets with two vanishing moments
61
(M = 2).
The goal is to find the real numbers (preferably 16 digits beyond the decimalpoint) h0, h1, h2 and h3 such that
(i) G(g0, g1, g2, g3) is a high-pass filter which annihilates both constant and lineartrends (or signals). Thus, we require two vanishing moments for ψ : i.e.∫ ∞
−∞ψ(x)dx =
∫ ∞−∞
xψ(x)dx = 0. (4)
(ii) H(h0, h1, h2, h3) is normalized in the usual way:∫ ∞−∞
φ(x)dx = 1; (5)
(iii) The scaling function φ(x) is orthogonal to its translates, i.e.∫ ∞−∞
φ(x)φ(x− n)dx = δn0, (6)
where the Kronecker δn0 equals 1 when n = 0 and zero otherwise.
First, express the conditions in (i), (ii) and (iii) in terms of the filter taps of thelow-pass filter H. Second, solve that system of five(*) equations for h0, h1, h2 andh3. How many solutions?
(*) Remarks:
(a) The quadratic condition coming from n = 0 in (iii) is actually redundant,but can and should be used to verify the results. So, one has to solve four (threelinear, one quadratic) equations.
(b) The filter taps can be expressed in closed (irrational) form. Compare theclosed forms with the numerical values. Do they match?
Task 3: Construction of the DAUB4 Scaling and Wavelets Functions
This task involves the construction of the scaling and wavelets functions, φ(x) andψ(x), for the Daubechies wavelets with two vanishing moments (DAUB4), wherethe low-pass and high-pass filters, H and G, have 4 filter taps. The computationscan be done by hand or (easier) with a symbolic manipulation program such asMathematica or Maple.
One of the solutions obtained in Task 2 reads
h0 = 0.4829629131445343
h1 = 0.836516303737808
h2 = 0.2241438680420134
h3 = -0.1294095225512604
62
(i) Use the exact, closed form irrational representation corresponding to the abovenumerical values of the filter H = {h0, h1, h2, h3}, of length L = 4, and the two-scale difference (refinement or dilation) equations for φ(x) and ψ(x) :
φ(x) =√
2L−1∑k=0
hkφ(2x− k) =L−1∑k=0
ckφ(2x− k) with ck =√
2hk, (7)
ψ(x) =√
2L−1∑k=0
gkφ(2x− k) =L−1∑k=0
dkφ(2x− k) with dk =√
2gk, (8)
to determine the values of the scaling function φ(x) at the integers. To do this,go through the following steps:
(a) Prove that the scaling function φ(x) is compactly supported on the interval[0,L-1] = [0,3].
(b) Prove that φ(0) = φ(3) = 0.
(c) Set up the eigenvalue-eigenvector problem A~v = λ~v to determine φ(1) andφ(2).
What is the matrix A? What is the vector ~v? What are the eigenvalues of A? Doyou notice anything special about these eigenvalues? What are the eigenvectors ofA? Which eigenvalue and eigenvector will allow you to determine φ(1) and φ(2)?
(d) Use the resolution of the identity (also called partition of unity) property ofthe scaling function, ∑
k∈ZZφ(x− k) = 1 (9)
to normalize φ(1) and φ(2). Compute the closed forms of the irrational numbersφ(1) and φ(2)?
(ii) What is the compact support of the wavelet function ψ(x)? Why?
(iii) Again, use the refinement equations for φ(x) and ψ(x) to compute the valuesof these functions at the so-called diadic (dyadic) rational (x = k 2−j), k and j
integer). Once the values of φ(x) and ψ(x) at integers are computed, (7) and (8)determine the values at the halves, quarters, eights, etc. In general, by repeatedapplications of the refinement equations, you can obtain the values of φ(x) andψ(x) at diadic points x = k 2−j for k = ...,−2,−1, 0, 1, 2, ...; j = 1, 2, 3, .... Fordecent resolution of the pictures, it suffices to run this iteration process till j = 4or 5. Determine explicitly the (closed-form irrational or numerical) values of φ(x)and ψ(x) at x = 0, 1
32 ,316 ,
12 ,
58 ,
32 ,
114 , and 3.
(iv) Graph the various iterations of φ(x) and ψ(x). Make sure that the horizontaland vertical axes are properly labelled.
63
Task 4: Scaling and Wavelet Functions via a Cascade Algorithm
Consider the following cascade (iterative) scheme based on the refinement equationfor φ(x) :
φ(i+1)(x) =√
2L−1∑k=0
hk φ(i)(2x− k), i ≥ 0, (10)
where the initial φ(0)(x) must be given. We use the characteristic function on theunit interval
φ(0)(x) =
1, 0 ≤ x < 1,
0, elsewhere.(11)
(i) For the Haar case (L = 2), take h0 = h1 = 1√2 , and implement the cascade
algorithm to compute the values of φ(1)(x), φ(4)(x) and φ12)(x).
(ii) Make pictures of these three scaling functions over their intervals of support.For a decent resolution of the graphs, use steps of 1
64 . Do not go beyond theboundaries of the supporting intervals.
Note: If you use Mathematica in order to efficiently compute the values of φ(i)(x),use a construction like
phi[0,x_] := 0 /; x<0.0
phi[0,x_] := 0 /; x>=1.0
phi[0,x_] := 1 /; (x<1.0 && x>=0.0)
h0=1/Sqrt[2];
h1=1/Sqrt[2];
phi[i_,x_] := phi[i,x] =
N[Sqrt[2]*(h0*phi[i-1,2*x]+h1*phi[i-1,2*x-1]),8] /; i >= 1
(iii) Still for the Haar case, use the relation (8) to compute the values of ψ(12)(x).
(iv) Make a picture of this wavelet function.
(v) What is the function corresponding to φ(∞)(x)? Does it have an analyticalform? What is it?
(vi) What is the function corresponding to ψ(∞)(x)? Is there an analytical form?What is it?
64
(vii) Repeat steps (i)-(iv) for the case of Daubechies wavelets (DAUB4) with twovanishing moments, where
h0 = 0.4829629131445343,
h1 = 0.836516303737808,
h2 = 0.2241438680420134,
h3 = -0.1294095225512604.
(viii) Answer the corresponding questions for the DAUB4 case: Can you still use(11) as the starting point? [Hint: Experiment with different initial conditions].What could one use instead?
(ix) Assuming that (10) has a ‘fixed point’ for any choice of a low-pass filter,prove that the cascade algorithm always converges to the true scaling functioncorresponding to that filter.
(x) Prove or disprove that the ‘fixed point’ φ(x) depends on the initial φ(0)(x).
[Hint: for (ix) and (x) work with the equivalent of (10) in the Fourier domain].
65
TW Honns: Numeriese Algoritmes Exam Question # 4(Vir Vrydae 6/4/01) 2001
Question 4: Wavelets: solving the Quadrature Mirror Filter condition
Based on the Quadrature Mirror Filter (QMF) condition it is possible to computeFinite Impulse Response (FIR) low-pass and high-pass filters, (h0, h1, h2, ..., hL−1)and (g0, g1, g2, ..., gL−1), for the Daubechies wavelets. The QMF condition reads
H(z)H(1
z) +H(−z)H(−1
z) = 2, ∀z, (12)
for
H(z) =L−1∑k=0
hkzk, (13)
where L is the length of the filter and hk are the low-pass filter taps.
The conditionH(1) =
√2 (14)
normalizes the filter (and the scaling function).
For a wavelet ψ(x) with M vanishing moments, i.e.∫ ∞−∞
xmψ(x)dx = 0, m = 0, 1, 2, ...,M − 1, (15)
the polynomial H(z) and its derivatives up to order M−1 must be zero at z = −1.Thus,
H(−1) = H ′(−1) = H ′′(−1) = ...... = H(M−1)(−1) = 0. (16)
(i) Use the normalization condition H(1) =√
2 and QMF (12) to show thatH(−1) = 0.What does the latter condition imply in terms of the filters (h0, h1, h2,..., hL−1)and (g0, g1, g2,..., gL−1)? What is relevance in signal processing?
(ii) Show that (14) and (16) imply that H(z) is of the form
H(z) =√
2
(1 + z
2
)MQ(z), (17)
where Q(z) is polynomial and Q(1) = 1.
(iii) Show that for a FIR filter with M vanishing moments Q(z) has degree M−1.
(iv) Consider the Haar case (L = 2,M = 1). What is Q(z)? Determine H(z) andthe filter coefficients from (17) by comparing with (13).
66
(v) If the QMF condition is rewritten in terms of Q(z), one can show (see LectureNotes, no need to prove this) that Q(z) must satisfy
Q(z)Q(1
z) =
M−1∑k=0
(−1)k(M − 1 + k)!
k!(M − 1)!(1− z)2k(4z)−k. (18)
What does (18) reduce to for the Haar case?
(vi) Substitute (13) directly into (12) to compute the conditions for h0, h1 for theHaar case. How many conditions do you get? How many are trivial? How manyare nontrivial? What do these conditions express? Compute h0 and h1 explicitly,using (14) to normalize.
(vii) Consider DAUB4, where L = 4,M = 2, corresponding to compactly sup-ported wavelets with two vanishing moments. What is the degree of Q(z)? Use(18) and Q(1) = 1 to determine Q(z) explicitly. Then, continuing with one ofthe (two possible) solutions, use (17) to compute H(z). Finally, use (13) to findclosed-form irrational expressions for the filter coefficients h0, h1, h2, and h3.
(viii) Substitute (13) directly into (12) to compute the conditions for the hk forthe DAUB4 low-pass filter (L = 4). How many conditions do you get? How manyare trivial? How many are nontrivial? What do these conditions actually express?Verify that low-pass filter (h0, h1, h2, h3) computed in (vii) satisfies all conditions,including (14).
67
Books, Review Papers, Info about Wavelets
Compiled by Willy Hereman—–Updated: February 9, 2001
Books on Wavelets
1. Ali N. Akansu and Richard A. Haddad, Multi-resolution Signal Decompo-sition: Transforms, Subbands, and Wavelets, Academic Press, New York,1992
2. Ali N. Akansu and M. Medley (eds.), Wavelet, Subband and Block Trans-forms in Communications and Multimedia, Kluwer Academic Press, Nor-wood, Massachusetts, 1998
3. Ali N. Akansu and Mark J.T. Smith, Subband and Wavelet Transforms,Design and Applications, Kluwer Academic Press, Boston, 1995
4. Akram Aldroubi and Michael Unser (eds.), Wavelets in Medicine and Biology,CRC Press, Boca Raton, Florida, 1996
5. Anestis Antoniadis and G. Oppenheim (eds.), Wavelets and Statistics, SpringerVerlag, Berlin, 1995
6. M. Barlaud (ed.), Wavelets in Image Communication, Elsevier, Dublin, 1995
7. John J. Benedetto and Michael W. Frazier (eds.), Wavelets: Mathematicsand Applications, CRC Press, Boca Raton, Florida, 1993
8. Andrew Bruce and Hong-Ye Gao, Applied Wavelet Analysis with S-Plus,Spinger Verlag, Berlin, New York, 1996
9. C. Sidney Burrus, Ramesh A. Gopinath and Haitao Guo, Introduction toWavelets and Wavelet Transforms: A Primer, Prentice Hall, Upper SaddleRiver, New Jersey, 1998
10. Y.T. Chan, Wavelet Basics, Kluwer Academic Publishers, Norwood, Mas-sachusetts, 1995
11. Charles K. Chui, An Introduction to Wavelets. Vol. 1 series: Wavelet Anal-ysis and its Applications, Academic Press, New York, 1992
12. Charles K. Chui (ed.), Wavelets: A Tutorial in Theory and Applications.Vol. 2 series: Wavelet Analysis and its Applications, Academic Press, NewYork, 1992
68
13. Charles K. Chui, Laura Montefusco and Luigia Puccio (eds.), Wavelets: The-ory, Algorithms and Applications. Vol. 5 series: Wavelet Analysis and itsApplications, Academic Press, Orlando, Florida, 1995
14. Charles K. Chui, Wavelets: A Mathematical Tool for Signal Analysis, SIAM,Philadelphia, 1997
15. Albert Cohen and Robert Ryan, Wavelets and Multiscale Signal Processing,Chapman & Hall, London, 1995
16. J.M. Combes, A. Grossmann and Ph. Tchamitchian (eds.), Wavelets: Time-Frequency Methods and Phase Space, Proc. Int. Conf. Marseille, France,December 1987, Springer Verlag, New York, 1990
17. Wolfgang Dahmen, Andrew Kurdila and Peter Oswald (eds.), MultiscaleWavelet Methods for Partial Differential Equations. Vol. 6. series: WaveletAnalysis and Applications, Academic Press, San Diego, 1997
18. Ingrid Daubechies, Ten Lectures on Wavelets, CBMS-NSF Regional Confer-ence Series in Applied Mathematics, 61, SIAM, Philadelphia, 1992
19. I. Daubechies, S. Mallat and A. Willsky (eds.), Special Issue on WaveletTransforms and Multiresolution Signal Analysis, IEEE Transactions on In-formation Theory 38 1992
20. Ingrid Daubechies (ed.), Different Perspectives on Wavelets, Proc. of Sym-posia in Applied Mathematics 47, Lecture Notes of Short Course, San Anto-nio, Texas, January 1993, AMS, Providence, Rhode Island, 1993
21. Guy David, Wavelets and Singular Integrals on Curves and Surfaces, LectureNotes Nankai Institute of Mathematics, Tianjin, People Republic of China,June 1988, Springer Verlag, Berlin, 1991
22. Marie Farge, J.C.R. Hunt and J.C. Vassilicos (eds.), Wavelets, Fractals andFourier Transforms, Clarendon Press (Oxford University Press), Oxford, U.K.,1993
23. F.J. Fliege, Multirate Digital Signal Processing: Multirate Systems, FilterBanks and Wavelets, Wiley & Sons, New York, 1994
24. Efi Foufoula-Georgiou and Praveen Kumar (eds.), Wavelets in Geophysics,Academic Press, San Diego, 1994
69
25. Michael Frazier, An Introduction to Wavelets through Linear Algebra, SpringerVerlag, New York, 1998
26. Eugene Hernandez and Guido Weiss, A First Course on Wavelets, CRC Press,Boca Raton, Florida, 1996
27. Matthius Holschneider, Wavelets: An Analysis Tool, Clarendon Press (Ox-ford University Press), Oxford, U.K., 1995
28. Barbara Burke Hubbard, The World According to Wavelets: The Story of aMathematical Technique in the Making (2nd edition), A.K. Peters, Natick,Massachusetts, 1998
29. S. Jaffard and Yves Meyer, Wavelet Methods for Pointwise Regularity andLocal Oscillations of Functions, AMS, Providence, Rhode Island, 1996
30. Kurt Jetter and Florencio I. Utreras (eds.), Multivariate Approximation:From CAGD to Wavelets, Proc. Int. Workshop, Santiago, Chili, September1992, World Scientific, Singapore, 1993
31. Jean-Pierre Kahane and P.G. Lemarie-Rieusset, Fourier Series and Wavelets,Gorden & Breach, London, 1995
32. Gerald Kaiser, A Friendly Guide to Wavelets, Birkhauser Press, Boston, 1994
33. T.H. Koornwinder (ed.), Wavelets: An Elementary Treatment of Theory andApplications, World Scientific, Singapore, 1993
34. Andrew F. Laine (ed.), Mathematical Imaging: Wavelet Applications in Sig-nal and Image Processing, Proc. Meeting, San Diego, CA, July 1993, SPIE2034, Bellingham, Washington, 1994
35. Andrew F. Laine, Wavelet Theory and Applications, Kluwer Academic Pub-lishers, Dordrecht, The Netherlands 1995
36. P.G. Lemarie (ed.), Les Ondelettes en 1989, Lecture Notes in Mathematics,1438, Springer Verlag, Berlin, 1990
37. LeMehaute, Laurent and L.L. Schumaker (eds.), Wavelets, Images and Sur-face Fitting, A.K. Peters, Wellesley, Massachusetts, 1994
38. Alfred K. Louis, Wavelets: Theory and Applications, John Wiley, New York;Teubner, Germany, February 1997
70
39. Stephane Mallat, A Wavelet Tour of Signal Processing (2nd edition), Aca-demic Press, New York, 1999
40. Peter R. Massopust, Fractal Functions, Fractal Surfaces and Wavelets, Aca-demic Press, Orlando, Florida, 1994
41. Yves Meyer, Ondelettes et Operateurs I, II et III, Hermann, Paris, 1990
42. Yves Meyer, Wavelets and Applications, Research Notes in Applied Mathe-matics, 20, Proc. Int. Conf. Marseille, France, May 1989, Masson, Milan andSpringer Verlag, Berlin, 1991
43. Yves Meyer, Wavelets and Operators. Vol. 1, Cambridge Studies in AdvancedMathematics 37, Cambridge University Press, Cambridge, 1992
44. Yves Meyer, Les Ondelettes, Algorithmes et Applications, Armand Colin,1992
45. Yves Meyer (transl: Robert D. Ryan), Wavelets, Algorithms and Applica-tions, SIAM, Philadelphia, 1993
46. Yves Meyer, Wavelets and Operators. Vol. 2: Calderon-Zygmund Operatorsand Multilinear Operators, Cambridge Studies in Advanced Mathematics 48,Yves Meyer, Cambridge University Press, Cambridge, 1995
47. Yves Meyer and S. Roques (eds.), Progress in Wavelet Analysis and Applica-tions, Proc. Int. Conf. on Wavelets and Applications, Toulouse, France, June1992.
48. Michel Misiti, Yves Misiti, Georges Oppenheim and Jean-Michel Poggi, WaveletToolbox User’s Guide, The MathWorks Inc., Natick, Massachusetts, 1996
49. Marius Mitrea, Clifford Wavelets, Singular Integrals and Hardy Spaces, Lec-ture Notes in Mathematics 1575, Springer Verlag, Berlin, 1994
50. Rodolphe L. Motard and Babu Joseph (eds.), Wavelet Applications in Chem-ical Engineering, Kluwer Academic Publishers, Boston, 1994
51. David E. Newland, An Introduction to Random Vibrations, Spectral andWavelet Analysis (3rd edition), John Wiley, New York; Longman, U.K., 1993
52. Yves Nievergelt, Wavelets Made Easy, Birkhauser, Boston, 1999
71
53. R. Todd Ogden, Essential Wavelets for Statistical Applications and DataAnalysis, Birkhauser Press, Boston, 1996
54. Mary B. Ruskai, Gregory Beylkin, Ronald Coifman and Ingrid Daubechies,Wavelets and Their Applications, Stephane Mallat, Yves Meyer and LouiseRaphael, Jones and Bartlett, Boston, Massachusetts, 1992
55. Larry L. Schumaker and Glenn Webb (eds.), Recent Advances in WaveletAnalysis, Wavelet Analysis and its Applications 3, Academic Press, NewYork, 1993
56. Eric J. Stollnitz, Tony D. DeRose and David H. Salesin, Morgan Kaufmann,San Francisco, 1996
57. Gilbert Strang and Truong Nguyen, Wavelets and Filter Banks, Wellesley-Cambridge Press, Wellesley, Massachusetts, 1996
58. Harold H. Szu (ed.), Wavelet Applications, Proc. Meeting Orland, Florida,April 1994, SPIE 2242, Bellingham, Washington, 1994
59. C. Taswell, Handbook of Wavelet Transform Algorithms, Birkhauser Press,Boston, 1996
60. Anthony Teolis, Computational Signal Processing with Wavelets, BirkhauserPress, Boston, 1996
61. Ahmed H. Tewfik, Wavelets and Multiscale Signal Processing Techniques:Theory and Applications, 1997
62. Bruno Torresani, Analyse Continue par Ondelettes, Savoirs Actuels, CNRSEditions, Paris, 1995
63. J.C. van den Berg (ed.), Wavelets in Physics, Cambridge University Press,Cambridge, 1998
64. Martin Vetterli and J. Kovacevic, Wavelets and Subband Coding, PrenticeHall, Des Moines, Iowa, 1995
65. Gilbert G. Walter, Wavelets and Other Orthogonal Systems with Applica-tions, CRC Press, Boca Raton, Florida, 1994
66. Mladen Victor Wickerhauser, Adapted Wavelet Analysis from Theory to Soft-ware, A.K. Peters, Wellesley, Massachusetts, 1994
72
67. M. Victor Wickerhauser, Wavelet Packet Laboratory, A.K. Peters, Wellesley,Massachusetts, 1994
68. Gregory W. Wornell, Signal Processing with Fractals: A Wavelet-based Ap-proach, Prentice Hall, Upper Saddle River, New York, 1996
69. P. Wojtaszczyk, A Mathematical Introduction to Wavelets, London Mathe-matical Society Graduate Texts, London, U.K., 1997
70. Randy K. Young, Wavelet Theory and Its Applications, Series in Engineer-ing and Computer Science. VLSI, computer architecture and digital signalprocessing, Kluwer Academic Publishers, Dordrecht, 1993
Some Review Papers on Wavelets
1. Marc A. Berger, “Lectures on Wavelets and Lectures on Products of RandomMatrices,” School of Mathematics, Georgia Institute of Technology, Atlanta,GA 30332 (1991) 72 pages
2. Gregory Beylkin, “Wavelets and Fast Numerical Algorithm,” Proc. Symposiain Appl. Math. 47 (1993) 89-177
3. Gregory Beylkin, “Wavelets, Multiresolution Analysis and Fast NumericalAlgorithm,” Draft of INRIA Lectures, 1991
4. Adhemar Bultheel, “Learning to swim in a sea of wavelets,” Bull. Belg. Math.Soc. 2 (1995) 1-44.
5. Ingrid Daubechies, “Orthonormal Bases of Compactly Supported Wavelets,”Comm. Pure Appl. Math. 41 (1988) 909-996
6. Christopher E. Heil and David F. Walnut, “Continuous and Discrete WaveletTransforms, SIAM Review 31 (1989) 628-666
7. Bjorn Jawerth and Wim Sweldens, “An Overview of Wavelet Based Multires-olution Analyses,” SIAM Review 36 (1994) 377-412
8. Don Lancaster, “Hardware Hacker: Understanding transforms, video com-pression secrets, ... and more wavelet breakthroughs,” Radio-ElectronicsMagazine, July 1991, 68-?.
9. Stephane G. Mallat, “A Theory for Multiresolution Signal Decomposition:The Wavelet Representation,” IEEE Trans. Patt. Anal. Mach. Intell. 11(1989) 674-693
73
10. Stephane G. Mallat, “Multiresolution Approximations and Wavelet Orthonor-mal Bases of L2(R), Trans. AMS 315 (1989) 69-87
11. Stephane G. Mallat, “Multifrequency Channel Decomposition of Images andWavelet Model, IEEE Trans. Acoust. Speech Signal Process. 37 (1989) 2091-2110
12. Peter Meuller and Brani Vidakovic, Wavelets for Kids: tutorial paper for thestatistical community (with Mathematical notebook, part B), ISDS, DukeUniversity. FTP to ftp.isds.duk.edu in directory /pub/brani/wav4Kids[A-B].ps.Z.uu
13. Yves Meyer, Book Reviews, Bulletin (New Series) of the AMS 28 (1993)350-360. Review of the books “An introduction to wavelets,” by C. Chui,and “Ten Lectures on Wavelets,” by I. Daubechies
14. Olivier Rioul and Martin Vetterli, “Wavelets and Signal Processing,” IEEESignal Proc. Mag. 8 (1991) 14-38
15. Alistair C.H. Rowe and Paul C. Abbott, “Daubechies wavelets and Mathe-matica,” Computers in Physics 9(6) (1995) 635-648
16. Gilbert Strang, “Wavelet Transforms Versus Fourier Transforms,” Bulletin(New Series) AMS 28 (1993) 288-305
17. Gilbert Strang, “Wavelets and Dilation Equations: A Brief Introduction,”SIAM Review 31 (1989) 614-627
18. Gilbert Strang, “Wavelets,” American Scientist 82 (1994) 250-255
19. Martin Vetterli, “Filter Banks Allowing Perfect Reconstruction,” Signal Pro-cessing 10 (1986) 219-244
20. Martin Vetterli and Cormac Herley, “Wavelets and Filter Banks: Theory andDesign,” IEEE Trans. Acoust. Speech Signal Process. 40 (1992) 2207-2232
Additional Information about Wavelets
• Journal: Applied and Computational Harmonic Analysis: Time-Frequencyand Time-Scaled Analysis, Wavelets, Numerical Algorithms and Applications(ACHA), C.K. Chui, R.R. Coifman and I. Daubechies (eds.), Academic Press,First issue: Nov.-Dec. 1993
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• Wavelet Digest, editor: Wim Sweldens, Lucent Technologies, Bell Laborato-ries,http://wim.sweldens.com. Add yourself via email: [email protected], URL:http://www.wavelet.org/
• Video Tape: Wavelets Making Waves in Mathematics and Engineering, In-grid Daubechies, MAA Invited Lectures, Baltimore, Maryland, January 1992,AMS-Selected Lectures in Mathematics
• Wavelets on World Wide Web
• Special Issue(s) on wavelets of IEEE Transactions on Information Theory(March 1992)
• Literature Survey by Stefan Pittner, Institute for Applied and NumericalMathematics, Technical University Vienna, Austria, 700 entries, 220 pages,cost $20, 1993, Contact: [email protected]
• Journal: Fourier Analysis and Applications, CRC Press, Boca Raton, Florida(1995-present)
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