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D AUBECHIES F ILTERS

Patrick J. Van Fleet

Center for Applied MathematicsUniversity of St. Thomas

St. Paul, MN USA

Joint Mathematical Meetings, 8 January 2007

8 JANUARY 2007 (S ESSION 2) DAUBECHIES F ILTERS JMM M INICOURSE #5 1 / 5

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R EVIEW T HE H AA R WAVELET T RANSFORM

In Session 1, we constructed the orthogonal Haar Wavelet Transform :

W 8 =

h 1 h 0 0 0 0 0 0 00 0 h 1 h 0 0 0 0 00 0 0 0 h 1 h 0 0 00 0 0 0 0 0 h 1 h 0

g 1 g 0 0 0 0 0 0 00 0 g 1 g 0 0 0 0 00 0 0 0 g 1 g 0 0 00 0 0 0 0 0 g 1 g 0

where

h = ( h 0 , h 1 ) = ( √2/ 2, √2/ 2), g = ( g 0 , g 1) = ( √2/ 2, −√2/ 2)


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L OWPASS /H IGHPASS F ILTERS L OWPASS F ILTERS

h = ( h 0 , h 1) = ( √2/ 2, √2/ 2) is called a lowpass lter .Lowpass lters tend to preserve non-oscillatory data and either

dampen or annihilate oscillatory data.Lowpass lters are best analyzed using a Fourier series.


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Pretending we’re engineers for the moment, let’s “form” theFourier series

H (ω) =1

k = 0

h k e ik ω = √22

+ √22

e i ω

We wish to plot |H (ω)| so simplifying gives

H (ω) = √22

+ √22

e i ω

=√22

e i ω/ 2 e −i ω/ 2 + e i ω/ 2

= √2e i ω/ 2

cos (ω/ 2)So

|H (ω)| = √2cos (ω/ 2)


L O ASS /H G ASS F S L O ASS F S

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H (ω) =1

k = 0

h k e ik ω = √22

+ √22

e i ω


H (ω) = √22

+ √22

e i ω

=√22

e i ω/ 2 e −i ω/ 2 + e i ω/ 2

= √2e i ω/ 2

cos (ω/ 2)So

|H (ω)| = √2cos (ω/ 2)



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Plotting |H (ω)| = √2cos (ω/ 2)



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Suppose h = ( h n , . . . , h m ) with Fourier series

H (ω) =m

k = n

h k e ik ω

We will say h is a lowpass lter if

H (0) =m

k = n

h k = √2

and

H (π) =m

k = n

h k e ik π =m

k = n

h k (−1)k = 0



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L OW SS /H G SS F S L OW SS F S

Suppose h = ( h n , . . . , h m ) with Fourier series

H (ω) =m

k = n

h k e ik ω

We will say h is a lowpass lter if

H (0) =m

k = n

h k = √2

and

H (π) =m

k = n

h k e ik π =m

k = n

h k (−1)k = 0


L OWPASS /H IGHPASS F ILTERS H IGHPASS F ILTERS

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If we write down the Fourier series for g = ( √22 , −

√22 ), we have

G (ω) =√22 −

√22 e −i ω

Simplifying gives

G (ω) =

√22 −

√22 e −

i ω

=√22

e −i ω/ 2 e i ω/ 2 −e −i ω/ 2)

= √2ie −i ω/ 2 sin (ω/ 2)

So

|G (ω)| = √2|sin (ω/ 2)|



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√22 ), we have

G (ω) =√22 −

√22 e −i ω

Simplifying gives

G (ω) =

√22 −

√22 e −

i ω

=√22

e −i ω/ 2 e i ω/ 2 −e −i ω/ 2)

= √2ie −i ω/ 2 sin (ω/ 2)

So

|G (ω)| = √2|sin (ω/ 2)|



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Plotting |G (ω)| = √2|sin (ω/ 2)|



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Suppose g = ( g n , . . . , g m ) with Fourier series

G (ω) =

m

k = n g k e

ik ω

We will say g is a highpass lter if

G (0) =m

k = n

h k = 0

and

G (π) =m

k = n h k e ik π =

m

k = n h k (−1)k = √2

Highpass lters tend to preserve oscillatory data and eitherdampen or annihilate non-oscillatory data.



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G (ω) =

m

k = n g k e ik ω


G (0) =m

k = n h k = 0

and

G (π) =m

k = n h k e ik π =

m

k = n h k (−1)k = √2




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G (ω) =

m

k = n g k e ik ω


G (0) =m

k = n h k = 0

and

G (π) =

m

k = n h k e ik π =

m

k = n h k (−1)k = √2



O RTHOGONAL F ILTERS D AUBECHIES 4-T AP F ILTERS

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We can build longer orthogonal lters h , g .Orthogonal lters are those that give rise to an orthogonal wavelettransformation W N (N even).It turns out length 3 doesn’t work, so we will try length 4.Let lowpass h = ( h 0 , h 1 , h 2 , h 3 ) and highpass g = ( g 0 , g 1 , g 2 , g 3).



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We form the matrix

W 8 =

h 3 h 2 h 1 h 0 0 0 0 00 0 h 3 h 2 h 1 h 0 0 00 0 0 0 h 3 h 2 h 1 h 0h 1 h 0 0 0 0 0 h 3 h 2

g 3 g 2 g 1 g 0 0 0 0 00 0 g 3 g 2 g 1 g 0 0 00 0 0 0 g 3 g 2 g 1 g 0g 1 g 0 0 0 0 0 g 3 g 2



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The order of the lters is not important - it turns out reectionswork as well.The reverse order I’ve used reects the fact that the wavelet matrixwas rst motivated using convolution (see Section 5.1 of the text).Note the “wrapping of rows”. This makes it easy to set uporthogonality conditions for W 8 , but is not practical in applications.

In this case, the fourth and eighth elements of the transformeddata will be built using elements from the beginning and end of theinput vector.This makes sense if the data are periodic ...

But data are not typically periodic!We can periodize input data, but then we need more from ourlters ... that’s coming!



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The order of the lters is not important - it turns out reectionswork as well.The reverse order I’ve used reects the fact that the wavelet matrixwas rst motivated using convolution (see Section 5.1 of the text).Note the “wrapping of rows”. This makes it easy to set uporthogonality conditions for W 8 , but is not practical in applications.





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The order of the lters is not important - it turns out reectionswork as well.

The reverse order I’ve used reects the fact that the wavelet matrixwas rst motivated using convolution (see Section 5.1 of the text).Note the “wrapping of rows”. This makes it easy to set uporthogonality conditions for W 8 , but is not practical in applications.





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Orthogonality Conditions

h 20 + h

21 + h

22 + h

23 = 1

h 0h 2 + h 1h 3 = 0

g 20 + g 21 + g 22 + g 23 = 1

g 0g 2 + g 1g 3 = 0

h 0g 0 + h 1g 1 + h 2g 2 + h 3g 3 = 0h 0g 2 + h 1g 3 = 0

h 2g 0 + h 3g 1 = 0



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and Lowpass/Highpass Conditions

h 0 + h 1 + h 2 + h 3 = √2h 0 −h 1 + h 2 −h 3 = 0

g 0 + g 1 + g 2 + g 3 = 0g 0 −g 1 + g 2 −g 3 = √2



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Look at the 5th equation:

h 20 + h

21 + h

22 + h

23 = 1

h 0h 2 + h 1h 3 = 0

g 20 + g 21 + g 22 + g 23 = 1

g 0g 2 + g 1g 3 = 0

h 0g 0 + h 1g 1 + h 2g 2 + h 3g 3 = 0h 0g 2 + h 1g 3 = 0

h 2g 0 + h 3g 1 = 0

Given h 0 , h 1 , h 2 , h 3 , can you nd g 0 , g 1 , g 2 , g 3 to satisfy it?



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How about

g 0 = h 3 , g 1 =

−h 2 , g 2 = h 1 , g 3 =

−h 0

Then

h 0g 0 + h 1g 1 + h 2g 2 + h 3g 3 = h 0 (h 3)+ h 1(−h 2)+ h 2(h 1)+ h 3(−h 0) = 0

Moreover, if h 0 + h 1 + h 2 + h 3 = √2 and h 0 −h 1 + h 2 −h 3 = 0, then

g 0 + g 1 + g 2 + g 3 = h 3 −h 2 + h 1 −h 0 = 0

and

g 0 −g 1 + g 2 −g 3 = h 3 −(−h 2) + h 1 −(−h 0) = √2

g is highpass!



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How about

g 0 = h 3 , g 1 =

−h 2 , g 2 = h 1 , g 3 =

−h 0

Then



g 0 + g 1 + g 2 + g 3 = h 3 −h 2 + h 1 −h 0 = 0

and

g 0 −g 1 + g 2 −g 3 = h 3 −(−h 2) + h 1 −(−h 0) = √2

g is highpass!



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There’s more good news ...

The orthogonality conditions reduce toh 20 + h 21 + h 22 + h 23 = 1

h 0h 2 + h 1h 3 = 0

Add to that the lowpass conditions:h 0 + h 1 + h 2 + h 3 = √2h 0 −h 1 + h 2 −h 3 = 0

and we have a (quadratic) system to solve.The moral of the story is we can build g from h .


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h 0h 2 + h 1h 3 = 0





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The bad news (for mathematicians) is that the system has aninnite number of solutions.The good news (for engineers) is that the system has an innitenumber of solutions.

We can add a condition!The condition Ingrid Daubechies added:“Flatten” H (ω) at ω = π .Set H (π) = 0!



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ORTHOGONAL

FILTERS

DAUBECHIES

4-TAP

FILTERS

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Differentiating

H (ω) = h 0 + h 1 e i ω + h 2 e 2i ω + h 3 e 3i ω

givesH (ω) = ih 1 e i ω + 2ih ,e 2i ω + 3ih 3 e 3i ω

Plugging in ω = π and simplifying gives the condition

h 1 −2h 2 + 3h 3 = 0



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So we want to solveh 20 + h 21 + h 22 + h 23 = 1

h 0h 2 + h 1h 3 = 0h 0 + h 1 + h 2 + h 3 = √2h 0 −h 1 + h 2 −h 3 = 0

h 1 −2h 2 + 3h 3 = 0



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Let’s look at the notebook

DaubechiesFilters.nb


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