Algorithms of Wavelets and Framelets - University of Albertabhan/notes.pdf · 2011-07-22 · Algorithms of Wavelets and Framelets Bin Han Department of Mathematical and Statistical

Algorithms of Wavelets and Framelets

Bin Han

Department of Mathematical and Statistical SciencesUniversity of Alberta, Canada

Present at Summer School onApplied and Computational Harmonic Analysis

Edmonton, Canada

July 29–31, 2011

Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 1 / 64

Outline of Tutorial: Part I

Algorithms for discrete wavelet/framelet transform

Perfect reconstruction, sparsity, stability

Multi-level fast wavelet transform

Oblique extension principle

Framelet transform for signals on bounded intervals


Notation

l(Z) for signals: all v = {v(k)}k∈Z : Z → C.l0(Z) for filters: all finitely supported sequencesu = {u(k)}k∈Z : Z → C on Z.For v = {v(k)}k∈Z ∈ l(Z), define

v ⋆(k) := v(−k), k ∈ Z,

v(ξ) :=∑

k∈Z

v(k)e−ikξ, ξ ∈ R.

Convolution u ∗ v and inner product:

[u ∗ v ](n) :=∑

k∈Z

u(k)v(n − k), n ∈ Z,

〈v , w〉 :=∑

k∈Z

v(k)w(k), v , w ∈ l2(Z)


Subdivision and Transition Operators

The subdivision operator Su : l(Z) → l(Z):

[Suv ](n) := 2∑

k∈Z

v(k)u(n − 2k), n ∈ Z

The transition operator Tu : l(Z) → l(Z) is

[Tuv ](n) := 2∑

k∈Z

v(k)u(k − 2n), n ∈ Z.


Subdivision and Transition Operators

The upsampling operator ↑d : l(Z) → l(Z):

[v ↑d](n) :=

{v(n/d), if n/d is an integer;

0, otherwise,n ∈ Z

the downsampling (or decimation) operator↓d : l(Z) → l(Z):

[v ↓d](n) := v(dn), n ∈ Z.

Subdivision and transition operators:

Suv = 2u ∗ (v ↑2) and Tuv = 2(u⋆ ∗ v)↓2.


Discrete Framelet Transform (DFrT): Decomposition

Let u0, . . . , us ∈ l0(Z) be filters for decomposition.

For data v ∈ l(Z), a 1-level framelet decomposition:

wℓ :=√

22 Tuℓ

v , ℓ = 0, . . . , s,

where wℓ are called framelet coefficients.

Grouping together, a framelet decompositionoperator W : l(Z) → (l(Z))1×(s+1):

Wv :=√

22 (Tu0

v , . . . , Tusv), v ∈ l(Z).


Reconstruction

Let u0, . . . , us ∈ l0(Z) be filters for reconstruction.

A one-level framelet reconstruction byV : (l(Z))1×(s+1) → l(Z):

V(w0, . . . , ws) :=

√2

2

s∑

ℓ=0

Suℓwℓ, w0, . . . , ws ∈ l(Z).

prefect reconstruction: VWv = v for any data v .

A filter bank ({u0, . . . , us}, {u0, . . . us}) has theperfect reconstruction (PR) if VW = Id l(Z).


Perfect Reconstruction (PR) Property

TheoremA filter bank ({u0, . . . , us}, {u0, . . . , us}) has the perfectreconstruction property, that is,

v = VWv =1

2

s∑

ℓ=0

SuℓTuℓ

v , ∀ v ∈ l(Z),

if and only if, for all ξ ∈ R,

u0(ξ) u0(ξ) + u1(ξ) u1(ξ) + · · ·+ us(ξ) us(ξ) = 1,

u0(ξ + π) u0(ξ) + u1(ξ + π) u1(ξ) + · · ·+ us(ξ + π) us(ξ) = 0.


Matrix form of Perfect Reconstruction

The perfect reconstruction (PR) condition can beequivalently rewritten into the following matrix form:

[ u0(ξ) · · · us(ξ)u0(ξ + π) · · · us(ξ + π)

] [u0(ξ) · · · us(ξ)

u0(ξ + π) · · · us(ξ + π)

]⋆

= I2,

where I2 denotes the 2× 2 identity matrix and A⋆ := AT.

a filter bank ({u0, . . . , us}, {u0, . . . , us}) has PR is calleda dual framelet filter bank.


Biorthogonal Wavelet Filter Bank

A dual framelet filter bank with s = 1 is called abiorthogonal wavelet filter bank, a nonredundant filterbank.

PropositionLet ({u0, . . . , us}, {u0, . . . , us}) be a dual framelet filterbank. Then the following statements are equivalent:

(i) W is onto or V is one-one;

(ii) VW = Id l(Z) and WV = Id(l(Z))1×(s+1) , that is, Vand W are inverse operators to each other;

(iii) s = 1.


Duality in l2(Z)

LemmaLet u ∈ l0(Z) be a finitely supported filter on Z. ThenSu : l2(Z) → l2(Z) is the adjoint operator ofTu : l2(Z) → l2(Z). More precisely,

〈Suv , w〉 = 〈v , Tuw〉, ∀ v , w ∈ l2(Z). (1)

The space (l2(Z))1×(s+1) has inner product:

〈(w0, . . . , ws), (w0, . . . , ws)〉 := 〈w0, w0〉+· · ·+〈ws , ws〉,and

‖(w0, . . . , ws)‖2(l2(Z))1×(s+1) := ‖w0‖2

l2(Z) + · · · + ‖ws‖2l2(Z).


Role of√

22

in DFrT

TheoremLet u0, . . . , us ∈ l0(Z). Then TFAE:

(i) ‖Wv‖2(l2(Z))1×(s+1) = ‖v‖2

l2(Z) for all v ∈ l2(Z), that is,

‖Tu0v‖2

l2(Z)+· · ·+‖Tusv‖2

l2(Z) = 2‖v‖2l2(Z), ∀ v ∈ l2(Z);

(ii) 〈Wv ,W v〉 = 〈v , v〉 for all v , v ∈ l2(Z);

(iii) the filter bank ({u0, . . . , us}, {u0, . . . , us}) has PR:[

u0(ξ) · · · us(ξ)u0(ξ + π) · · · us(ξ + π)

] [u0(ξ) · · · us(ξ)

u0(ξ + π) · · · us(ξ + π)

]⋆

= I2,

{u0, . . . , us} with PR is called a tight framelet filter bank.Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 12 / 64

Orthogonal Wavelet Filter Bank

A tight framelet filter bank with s = 1 is called anorthogonal wavelet filter bank.

PropositionLet {u0, . . . , us} be a tight framelet filter bank. Then thefollowing are equivalent:

1 W is an onto orthogonal mapping satisfying〈Wv ,W v〉 = 〈v , v〉 for all v , v ∈ l2(Z);

2 for all w0, . . . , ws , w0, . . . , ws ∈ l2(Z),

〈V(w0, . . . , ws),V(w0, . . . , ws)〉 = 〈(w0, . . . , ws), (w0, . . . , ws)〉;

3 s = 1.


Examples

To list a filter u = {u(k)}k∈Z with support [m, n],

u = {u(m), . . . , u(−1), u(0), u(1), . . . , u(n)}[m,n],

{u0, u1} is the Haar orthogonal wavelet filter bank:

u0 = {12, 1

2}[0,1], u1 = {12,−1

2}[0,1]. (2)

({u0, u1}, {u0, u1}) is a biorthogonal wavelet filterbank, where

u0 = {−18, 1

4, 3

4, 1

4,−1

8}[−2,2], u1 = {1

4,−1

2, 1

4}[0,2],

u0 = {14, 1

2, 1

4}[−1,1], u1 = {1

8, 1

4,−3

4, 1

4, 1

8}[−1,3].


Illustration: I

Apply the Haar orthogonal filter bank to

v = {1, 0,−1,−1,−4, 60, 58, 56}[0,7]

Note that

[Tu0v ](n) = v(2n)+v(2n+1), [Tu1

v ](n) = v(2n)−v(2n+1), n ∈ Z.

We have the wavelet coefficients:

w0 =√

22{1,−2, 56, 114}[0,3], w1 =

√2

2{1, 0,−64,−2}[0,3].

Note that

[Su0w0](2n) = w0(n), [Su0

w0](2n + 1) = w0(n), n ∈ Z

[Su1w1](2n) = w1(n), [Su1

w1](2n + 1) = −w1(n), n ∈ Z.


Illustration: II

Hence, we have√

22 Su0

w0 = 12{1, 1,−2,−2, 56, 56, 114, 114}[0,7],√

22 Su1

w1 = 12{1,−1, 0, 0,−64, 64,−2, 2}[0,7].

Clearly, we have the perfect reconstruction of v :

√2

2 Su0w0+

√2

2 Su1w1 = {1, 0,−1,−1,−4, 60, 58, 56}[0,7] = v

and the following energy-preserving identity

‖w0‖2l2(Z) + ‖w1‖2

l2(Z) = 161372

+ 41012

= 10119 = ‖v‖2l2(Z).


Diagram of 1-level DFrTs

input

√2u⋆

0

√2u⋆

1

√2u⋆

s

↓2

↓2

↓2

processing

processing

processing

↑2

↑2

↑2

√2u0

√2u0

√2u0

⊕ output

Figure: Diagram of a one-level discrete framelet transform using adual framelet filter bank ({u0, . . . , us}, (u0, . . . , us}).


Sparsity of DFrT

One key feature of DFrT is its sparse representationfor smooth or piecewise smooth signals.

It is desirable to have as many as possible negligibleframelet coefficients for smooth signals.

Smooth signals are modeled by polynomials. Letp : R → C be a polynomial: p(x) =

∑mn=0 pnx

n.

a polynomial sequence p|Z : Z → C by[p|Z](k) = p(k), k ∈ Z.

N0 := N ∪ {0}.


Polynomial Differentiation Operator

Polynomial differentiation operator:

p(x − i d

dξ

)f(ξ) :=

∞∑

n=0

pn

(x − i d

dξ

)nf(ξ).

p(x − i d

dξ

)f(ξ) =

∞∑

n=0

(−i)n

n!p(n)(x)f(n)(ξ)

=

∞∑

n=0

xn

n!p(n)

(−i d

dξ

)f(ξ).

Generalized product rule for differentiation:

p(x − i d

dξ

)(g(ξ)f(ξ)

)=

∞∑

n=0

(−i)n

n!g(n)(ξ)p(n)

(x − i d

dξ

)f(ξ).

[p(−i d

dξ

)(e ixξf(ξ))

]∣∣∣ξ=0

=[p(x − i d

dξ

)f(ξ)

]∣∣∣ξ=0

.


Convolution with Polynomials

LemmaLet u = {u(k)}k∈Z ∈ l0(Z). Then for any polynomialp ∈ Π, p ∗ u is a polynomial with deg(p ∗ u) 6 deg(p),

[p ∗ u](x) =∑

k∈Z

p(x − k)u(k) =[p(x − i d

dξ

)u(ξ)

]∣∣∣ξ=0

=∞∑

n=0

(−i)n

n!p(n)(x)u(n)(0) =

∞∑

n=0

xn

n!

[p(n)

(−i d

dξ

)u(ξ)

]∣∣∣ξ=0

.

Moreover, p ∗ (u ↑2) = [p(2·) ∗ u](2−1·),

p(n)∗u = [p∗u](n), p(·−y)∗u = [p∗u](·−y), ∀ y ∈ R.


Big O Notation

For smooth functions f and g, it is often convenient touse the following big O notation:

f(ξ) = g(ξ) + O(|ξ − ξ0|m), ξ → ξ0

to mean that the derivatives of f and g at ξ = ξ0 agreeto the orders up to m − 1:

f(n)(ξ0) = g(n)(ξ0), ∀ n = 0, . . . , m − 1.


Transition Operator Acting on Polynomials

TheoremLet u ∈ l0(Z). Then for any polynomial p ∈ Π,

Tup = 2[p ∗ u⋆](2·) = p(2·) ∗ u =∞∑

n=0

2(−i)n

n!p(n)(2·)u(n)(0),

where u is a finitely supported sequence on Z such thatû(ξ) = 2u(ξ/2) + O(|ξ|deg(p)+1), ξ → 0.

In particular, for any integer m ∈ N, TFAE:1 Tup = 0 for all polynomial sequences p ∈ Πm−1;2 Tuq = 0 for some q ∈ Π with deg(q) = m − 1;3 u(ξ) = O(|ξ|m) as ξ → 0;4 u(ξ) = (1 − e−iξ)mQ(ξ) for some 2π-periodic

trigonometric polynomial Q.Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 22 / 64

Vanishing Moments

We say that a filter u has m vanishing moments ifany of items (1)–(4) in Theorem holds.

Most framelet coefficients are zero for any inputsignal which is a polynomial to certain degree.

If u has m vanishing moments. For a signal v , if vagrees with some polynomial of degree less than mon the support of u(· − 2n), then [Tuv ](n) = 0.


Coset Sequences

For u = {u(k)}k∈Z and γ ∈ Z, we define the associatedcoset sequence u[γ] of u at the coset γ + 2Z to be

u[γ](ξ) :=∑

k∈Z

u(γ + 2k)e−ikξ,

that is,

u[γ] = u(γ + ·)↓2 = {u(γ + 2k)}k∈Z.


Subdivision Operator on Polynomials

Sup is not always a polynomial for p ∈ Π.

For example, for p = 1 and u = {1}[0,0], we have[Sup](2k) = 2 and [Sup](2k + 1) = 0 for all k ∈ Z.

LemmaLet u ∈ l0(Z) and q be a polynomial. TFAE:

(i)∑

k∈Zq(−1

2− k)u(1 + 2k) =

∑k∈Z

q(−k)u(2k),

that is, (q ∗ u[1])(−12) = (q ∗ u[0])(0);

(ii) [q(−i ddξ)(e

−iξ/2u[1](ξ))]|ξ=0 = [q(−i ddξ)u

[0](ξ)]|ξ=0;

(iii) [q(− i2

ddξ)u(ξ)]|ξ=π = 0.


Subdivision Operator on Polynomials

TheoremLet u = {u(k)}k∈Z. For m ∈ N, TFAE:

1 SuΠm−1 ⊆ Π;2 Suq ∈ Π for some q ∈ Π with deg(q) = m − 1;3 SuΠm−1 ⊆ Πm−1;4 u has m sum rules:

u(ξ + π) = O(|ξ|m), ξ → 0;

5 u(ξ) = (1 + e−iξ)mQ(ξ) for some 2π-periodic Q;

6 e−iξ/2u[1](ξ) = u[0](ξ) + O(|ξ|m), ξ → 0.

Moreover, Sup = 2−1p(2−1·) ∗ u.Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 26 / 64

Linear-phase Moments (LPM)

LemmaLet u ∈ l0(Z). Let p be a polynomial and definem := deg(p). For c ∈ R, p ∗ u = p(· − c) ⇐⇒ if u hasm + 1 linear-phase moments with phase c:

u(ξ) = e−icξ + O(|ξ|m+1), ξ → 0.

PropositionLet u ∈ l0(Z) and c ∈ R. Then u has m + 1 linear-phasemoments with phase c ⇐⇒ Tup = 2p(2 · +c) for allp ∈ Πm. Similarly, u has m + 1 sum rules and m + 1linear-phase moments with phase c ⇐⇒Sup = p(2−1(· − c)) for all p ∈ Πm.Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 27 / 64

Symmetry

We say that u = {u(k)}k∈Z : Z → C has symmetry if

u(2c − k) = ǫu(k), ∀ k ∈ Z

with 2c ∈ Z and ǫ ∈ {−1, 1}.u is symmetric if ǫ = 1; antisymmetric if ǫ = −1.c is the symmetry center of the filter u.A symmetry operator S to record the symmetry type:

[Su](ξ) :=u(ξ)

u(−ξ), ξ ∈ R.

[Su](ξ) = ǫe−i2cξ.


LPM and Symmetry

PropositionSuppose that u ∈ l0(Z) has m but not m + 1linear-phase moments with phase c ∈ R. If m > 1, thenthe phase c is uniquely determined by u through

c = i u′(0) =∑

k∈Z

u(k)k .

Moreover, if u has symmetry: u(2c0 − k) = u(k) for allk ∈ Z for some c0 ∈ 1

2Z, then c = c0 (that is, the phase

c agrees with the symmetry center c0 of u).


Example: B-spline filters

B-spline filter of order m: aBm(ξ) := 2−m(1 + e−iξ)m

aB4 (0) = 1, aB

4

′(0) = −2i , aB

4

′′(0) = −5, aB

4

′′′

(0) = 14i .

For p ∈ Π3,

[p ∗ aB4 ](x) = p(x) − 2p′(x) +

5

2p′′(x) − 7

3p′′′(x).

[TaB4p](x) = 2p(2x) + 4p′(2x) + 5p′′(2x) +

14

3p′′′(2x).

[SaB4p](x) = p(x/2) − p′(x/2) +

5

8p′′(x/2) − 7

24p′′′(x/2).


Multi-level Fast Framelet Transform: Decomposition

Let a be a primal low-pass filter and b1, . . . , bs beprimal high-pass filters for decomposition.

For a positive integer J , a J-level discrete frameletdecomposition is given by

vj−1 :=√

22 Tavj , wj−1;ℓ :=

√2

2 Tbℓvj ,

ℓ = 1, . . . , s, j = J , . . . , 1,

where vJ : Z → C is an input signal.

decomposition operator WJ : l(Z) → (l(Z))1×(sJ+1):

WJvJ := (wJ−1;1, . . . , wJ−1;s, . . . , w0;1, . . . , w0;s, v0),


Thresholding and Quantization

− ε

2ε

2− ε

2ε

2− q

2q2

Figure: The hard thresholding function, the soft thresholdingfunction, and the quantization function, respectively. Boththresholding and quantization operations are often used to processthe framelet coefficients in a discrete framelet transform.


Multi-level Reconstruction

Let a be a dual low-pass filter and b1, . . . , bs be dualhigh-pass filters for reconstruction.

a J-level discrete framelet reconstruction is

vj :=

√2

2Savj−1 +

√2

2

s∑

ℓ=1

Sbℓwj−1;ℓ, j = 1, . . . , J .

a J-level discrete reconstruction operatorVJ : (l(Z))1×(sJ+1) → l(Z) is defined by

VJ(wJ−1;1, . . . , wJ−1;s, . . . , w0;1, . . . , w0;s, v0) = vJ ,

A fast framelet transform with s = 1 is called a fastwavelet transform.


Diagram of Multi-level FFrT

input

√2a⋆

√2b⋆

1

√2b⋆

s

↓2

↓2

↓2

√2a⋆

√2b⋆

1

√2b⋆

s

↓2

↓2

↓2

processing

processing

processing

↑2

↑2

↑2

√2a

√2b1

√2bs

⊕processing

processing

↑2

↑2

↑2

√2a

√2b1

√2bs

⊕ output

Figure: Diagram of a two-level discrete framelet transform using adual framelet filter bank ({a; b1, . . . , bs}, (a; b1, . . . , bs}).


Stability

A multi-level discrete framelet transform employing adual framelet filter bank {a; b1, . . . , bs}, {a; b1, . . . , bs})has stability in the space l2(Z) if there exists C > 0 suchthat for J ∈ N0,

‖WJv‖(l2(Z))1×(sJ+1) 6 C‖v‖l2(Z), ∀ v ∈ l2(Z)

and

‖VJ ~w‖l2(Z) 6 C‖~w‖(l2(Z))1×(sJ+1), ∀ ~w ∈ (l2(Z))1×(sJ+1).


Stability

PropositionSuppose that a multi-level discrete framelet transformhas stability in the space l2(Z). Then all the waveletdecomposition operators must have uniform stability inspace l2(Z):

1C‖v‖l2(Z) 6 ‖WJv‖l2(Z) 6 C‖v‖l2(Z), ∀v ∈ l2(Z), J ∈ N.

Moreover, wavelet reconstruction operators have uniformstability: for all J ∈ N,

1C‖~w‖(l2(Z))1×(sJ+1) 6 ‖VJ ~w‖l2(Z) 6 C‖~w‖(l2(Z))1×(sJ+1) , ~w ∈ (l2(Z))1×(sJ+1)

if and only if s = 1.Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 36 / 64

Stability of Multi-level FFrT

TheoremLet ({a; b1, . . . , bs}, {a; b1, . . . , bs}) be a dual frameletfilter bank with a(0) = â(0) = 1. Define

ϕ(ξ) :=

∞∏

j=1

a(2−jξ), ϕ(ξ) :=

∞∏

j=1

â(2−jξ), ξ ∈ R.

Then a multi-level discrete framelet transform hasstability in the space l2(Z) ⇐⇒ ϕ, ϕ ∈ L2(R) and

b1(0) = · · · = bs(0) = b1(0) = · · · = bs(0) = 0.


Multi-level FFrT Again

vj =√

22Tavj+1 = · · · = (

√2

2)J−jTa∗(a↑2)∗···∗(a↑2J−j−1),2J−j v

wj ;ℓ =√

22Tbℓ

vj+1 = (√

22

)J−jTa∗(a↑2)∗···∗(a↑2J−j−2)∗(bℓ↑2J−j−1),2J−j v

wj ;k(k) = 〈v , 2(J−j)/2[a∗(a↑2)∗· · ·∗(a↑2J−j−2)∗(bℓ ↑2J−j−1)](·−2J−jk)〉Similarly, we have

VJ(0, . . . , 0, v0) = (√

22

)J−jSa∗(a↑2)∗···∗(a↑2J−1),2Jv0

VJ(0, . . . , 0, wj ;ℓ, 0, . . . , 0) = (√

22

)J−jSa∗(a↑2)∗···∗(a↑2J−j−2)∗(bℓ↑2J−j−1),2J−j wj ;ℓ.


Define filters aj and aj by

aj = a ∗ (a↑2) ∗ · · · ∗ (a↑2j−1)

andaj := a ∗ (a↑2) ∗ · · · ∗ (a↑2j−1).

Now a J-level discrete framelet transform employing({a; b1, . . . , bs}, {a; b1, . . . , bs}) is equivalent to

v =∑

k∈Z

〈v , 2J/2aJ(· − 2Jk)〉2J/2aJ(· − 2Jk) +

J−1∑

j=0

s∑

ℓ=1

∑

k∈Z

〈v , [2(J−j)/2aJ−j−1 ∗ (bℓ ↑2J−j−1)](· − 2J−jk)〉[2(J−j)/2aJ−j−1 ∗ (bℓ ↑2J−j−1)](· − 2J−jk).


Lack of Vanishing Moments

LemmaLet ({a; b1, . . . , bs}, {a; b1, . . . , bs}) be a dual frameletfilter bank. If all b1, . . . , bs have m vanishing momentsand all b1, . . . , bs have m vanishing moments, then

(i) the primal low-pass filter a must have m sum rules,that is, a(ξ + π) = O(|ξ|m) as ξ → 0;

(ii) the dual low-pass filter a must have m sum rules,that is, â(ξ + π) = O(|ξ|m) as ξ → 0;

(iii) aâ has m + m linear-phase moments with phase 0:

1 − a(ξ)â(ξ) = O(|ξ|m+m), ξ → 0.


Example of B-spline Filters

Let a(ξ) = aBm(ξ) = 2−m(1 + e−iξ)m and

â(ξ) = aBm(ξ) = 2−m(1 + e−iξ)m be two B-spline filters.

a(0)â(0) = 1,

[aâ]′(0) =i(m − m)

2,

[aâ]′′(0) =(m − m)2 + m + m

4.


Oblique Extension Principle

Theorem({a; b1, . . . , bs}, {a; b1, . . . , bs})Θ has the followinggeneralized perfect reconstruction property:

Θ ∗ v =1

2Sa(Θ ∗ Tav) +

1

2

s∑

ℓ=1

SbℓTbℓ

v , ∀ v ∈ l(Z),

⇐⇒ for all ξ ∈ R,

Θ(2ξ)a(ξ)â(ξ) + b1(ξ)b1(ξ) + · · ·+ bs(ξ)

bs(ξ) = Θ(ξ),

Θ(2ξ)a(ξ + π)â(ξ) + b1(ξ + π) b1(ξ) + · · · + bs(ξ + π) bs(ξ) = 0.


Vanishing Moments of OEP

LemmaLet ({a; b1, . . . , bs}, {a; b1, . . . , bs})Θ be an OEP-basedfilter bank. Suppose that all b1, . . . , bs have m vanishingmoments and all b1, . . . , bs have m vanishing moments,where m, m ∈ N. Then

Θ(ξ) − Θ(2ξ)a(ξ)â(ξ) = O(|ξ|m+m), ξ → 0.

If in addition Θ(0) 6= 0, then the primal low-pass filter amust have m sum rules and the dual low-pass filter amust have m sum rules.


OEP-based Tight Framelets

PropositionLet θ, a, b1, . . . , bs ∈ l0(Z). Then

‖θ∗Tav‖2l2(Z)+‖Tb1

v‖2l2(Z)+· · ·+‖Tbs

v‖2l2(Z) = 2‖θ∗v‖2

l2(Z),

⇐⇒ ({a; b1, . . . , bs}, {a; b1, . . . , bs})Θ has PR:[

b1(ξ) · · · bs(ξ)

b1(ξ + π) · · · bs(ξ + π)

][b1(ξ) · · · bs(ξ)

b1(ξ + π) · · · bs(ξ + π)

]⋆

=MΘ,a,

where Θ := θ ∗ θ⋆ and

MΘ,a :=

[Θ(ξ) − Θ(2ξ)|a(ξ)|2 −Θ(2ξ)a(ξ + π)a(ξ)

−Θ(2ξ)a(ξ)a(ξ + π) Θ(ξ + π) − Θ(2ξ)|a(ξ + π)|2]

.


OEP-based Dual Framelets

({a; b1, . . . , bs}, {a; b1, . . . , bs})Θ having PR iscalled an OEP-based dual framelet filter bank.

{a; b1, . . . , bs}Θ having PR is called an OEP-basedtight framelet filter bank.


Fejer-Riesz Lemma

LemmaLet Θ be a 2π-periodic trigonometric polynomial withreal coefficients (or with complex coefficients) such thatΘ(ξ) > 0 for all ξ ∈ R. Then there exists a 2π-periodictrigonometric polynomial θ with real coefficients (or withcomplex coefficients) such that |θ(ξ)|2 = Θ(ξ) for allξ ∈ R. Moreover, if Θ(0) 6= 0, then we can further

require θ(0) =√

Θ(0).


OEP-based Tight Framelet

LemmaLet {a; b1, . . . , bs}Θ be an OEP-based tight framelet

filter bank. If Θ(0) > 0, then Θ(ξ) > 0 for all ξ ∈ R andconsequently, there exists θ ∈ l0(Z) such that Θ = θ ∗ θ⋆

holds and θ(0) =√

Θ(0).


OEP with s = 1

TheoremLet ({a; b}, {a; b})Θ be an OEP-based dual framelet

filter bank with Θ(0) 6= 0. Then

Θ(2ξ)Θ(π) = Θ(ξ)Θ(ξ + π)

[â(ξ)

ˆb(ξ)

â(ξ + π)ˆb(ξ + π)

] [a(ξ) b(ξ)

a(ξ + π) b(ξ + π)

]⋆

=

[1 00 1

],


Continued...

Theoremwhere all filters are finitely supported and are given by

â(ξ) := â(ξ)Θ(2ξ)/Θ(ξ) andˆb(ξ) := ˆb(ξ)/Θ(ξ).

That is, ({a; b}, {a; b}) is a biorthogonal wavelet filterbank. Moreover,

b(ξ) = ce i(2n−1)ξâ(ξ + π),ˆb(ξ) = c−1e i(2n−1)ξa(ξ + π)


DFrT by OEP Filter Bank

({a; b1, . . . , bs}, {a; b1, . . . , bs})Θ.An OEP-based J-level discrete frameletdecomposition is exactly the same as the one before.For given low-pass framelet coefficients v0 andhigh-pass framelet coefficients wj ;ℓ, ℓ = 1, . . . , s andj = 0, . . . , J − 1, an OEP-based J-level discreteframelet reconstruction is

v0 := Θ ∗ v0,

vj :=

√2

2Savj−1 +

√2

2

s∑

ℓ=1

Sbℓwj−1;ℓ, j = 1, . . . , J ,

recover vJ from vJ via the relation vJ = Θ ∗ vJ .


Diagram of DFrT using OEP

input

√2a⋆

√2b⋆

1

√2b⋆

s

↓2

↓2

↓2

√2a⋆

√2b⋆

1

√2b⋆

s

↓2

↓2

↓2

processing

processing

processing

Θ ↑2

↑2

↑2

√2a

√2b1

√2bs

⊕

processing

processing

↑2

↑2

↑2

√2a

√2b1

√2bs

⊕ Θ−1 output

Figure: Diagram of a two-level discrete framelet transform using anOEP-based dual framelet filter bank({a; b1, . . . , bs}, (a; b1, . . . , bs})Θ.


Avoiding Deconvolution using OEP

input

multi-level discrete framelet transform using

({a1; b1,1, . . . , b1,s1}, {a1 ; b1,1, . . . , b1,s1

})Θ1


({a2; b2,1, . . . , b2,s2}, {a2; b2,1, . . . , b2,s1

})Θ2


({an ; bn,1, . . . , bn,s1}, {an ; bn,1, . . . , bn,s1

})Θn

Θ1

Θ2

Θn

⊕ output

Figure: Diagram of an n-branch multi-level discrete averagingframelet transform using dual framelet filter banks{({am; bm,1, . . . , bm,sm}, (am; bm,1, . . . , bm,sm})Θm

, m = 1, . . . , n.Therefore, deconvolution is completely avoided.


Signals on Intervals

Signals: v b = {v b(k)}N−1k=0 : [0, N − 1] ∩ Z → C.

Extend v b from interval [0, N − 1] to Z.

Tu(·−2m)v = [Tuv ](· − m), m ∈ Z.

extend v b from [0, N − 1] ∩ Z to a sequence v on Z

by any method that the reader prefers.

zero-padding: v(k) = 0 for k ∈ Z\[0, N − 1].

Must recover v b(0), . . . , v b(N − 1) exactly.

Keep all [SuℓTuℓ

v ](n), n = 0, . . . , N − 1.

fsupp(u) = [n−, n+] with n− 6 0 and n+ > 0.


12 [SuTuv ](n) =

∑k∈Z

[Tuv ](k)u(n − 2k),n =0, . . . , N − 1.

Record all the framelet coefficients:

[Tuv ](k), k = [−n+

2 ], . . . , [N−1−n−2 ].

we always have [−n+

2 ] 6 0 and [N−1−n−2 ] >

N2 − 1.

framelet coefficients {[Tuv ](k)}N2 −1

k=0 must berecorded.

extra: [Tuv ](k), k = [−n+

2], . . . ,−1 and

k = N2 , . . . , [N−1−n−

2 ].

Ideal situation can happen ⇐⇒ u vanishes outside[0, 1].


Periodic Extension

Proposition

Let u ∈ l1(Z) be a filter and v b = {v b(k)}N−1k=0 . Extend

v b into an N-periodic sequence v on Z as follows:

v(Nn + k) := v b(k), k = 0, . . . , N − 1, n ∈ Z.

Then the following properties hold:

(i) u ∗ v is an N-periodic sequence on Z;

(ii) Suv is a 2N-periodic sequence on Z;

(iii) If N is even, then Tuv is an N2 -periodic;

(iv) If N is odd, then Tuv is an N-periodic.


DFrT using Periodic Extension

A one-level discrete periodic framelet decomposition:

wbℓ =

{wb

ℓ (k) :=√

22

[Tuℓv ](k)

}N2−1

k=0, ℓ = 0, . . . , s,

where v is the N-periodic extension of v b.Grouping all framelet coefficients,

Wper (v b) = (wb0 , wb

1 , . . . , wbs )T.

a one-level discrete periodic framelet reconstruction:

v b = Vper (wb0 , . . . , wb

s ) :={

v b(k) :=√

22

s∑

ℓ=0

[Suℓwℓ](k)

}N−1

k=0,

VperWper = IN .Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 56 / 64

Tight Framelet Filter Bank

A tight framelet filter bank by Ron-Shen, where

u0 = {14, 1

2, 1

4}[−1,1],

u1 = {14,−1

2, 1

4}[−1,1],

u2 = {−√

24

, 0,−√

24}[−1,1].

A test input data:

v = {1, 0,−1,−1,−4, 60, 58, 56}[0,7]


Example: Tight Framelet Filter Bank

We extend v b to an 8-periodic sequence v on Z, given by

v = {. . . , 1, 0,−1,−1,−4, 60, 58, 56, 1, 0,−1,−1,−4, 60, 58, 56, 1, 0,−

Then all sequences Tu0v , Tu1

v , Tu2v are 4-periodic and

w0 =√

22Tu0

v =√

22{. . . , 29

2,−3

4, 51

4, 58, 29

2,−3

4, 51

4, 58,29

2,−3

4, 51

4, 58, . . .}

w1 =√

22Tu1

v =√

22{. . . , 27

2, 1

4, 67

4, 0, 27

2, 1

4, 67

4, 0,27

2, 1

4, 67

4, 0, . . .},

w2 =√

22Tu0

v = {. . . ,−14, 14,−59

4,−29,−14, 1

4,−59

4,−29, − 14, 1

4,−59

4,

It is also easy to check that√2

2 (Su0w0 + Su1

w1 + Su2w2) = v .


PropositionLet u ∈ l1(Z) such that u(2c − k) = ǫu(k). Extendv b = {v b(k)}N−1

k=0 , with both endpoints non-repeated(EN), into (2N − 2)-periodic v byv(k) = v b(2N − 2 − k).

(i) Then u⋆ ∗ v is (2N − 2)-periodic with

[u⋆ ∗ v ](−2c − k) = [u⋆ ∗ v ](2N − 2 − 2c − k) = ǫ[u⋆ ∗ v ](k),

and [−⌊c⌋, N − 1 − ⌈c⌉] is its control interval.

(ii) If c ∈ Z, then Tuv is (N − 1)-periodic with

[Tuv ](−c − k) = [Tuv ](N − 1− c − k) = ǫ[Tuv ](k),

and [⌈−c2⌉, ⌊N−1−c

2 ⌋] is a control interval of Tuv .


Proposition

Let u ∈ l1(Z) with ǫ ∈ {−1, 1} and c ∈ 12Z. Extend v b,

with both endpoints repeated (ER), into 2N-periodic vby v(k) = v b(2N − 1 − k).

(i) Then u⋆ ∗ v is 2N-periodic with

[u⋆ ∗v ](−1−2c−k) = [u⋆ ∗v ](2N−1−2c −k) = ǫ[u⋆ ∗v ](k),

and [−⌊12 + c⌋, N − ⌈1

2 + c⌉] is its control interval.

(ii) If c − 12 ∈ Z, then Tuv is an N-periodic sequence:

[Tuv ](−12−c−k) = [Tuv ](N− 1

2−c−k) = ǫ[Tuv ](k),

and [⌈−14− c

2⌉, ⌊N

2− 1

4− c

2⌋] is its control interval.


Table 1: Endpoint Nonrepeated (EN)

filter u u⋆ ∗ v with v extended by EN Tuv with v extended by EN

c = 0ǫ = 1

(2N − 2)-periodic,symmetric about 0 and N − 1,a control interval [0, N − 1].

(N − 1)-periodic,

symmetric about 0 and N−12

,

a control interval [0, N2

− 1].

c = 0ǫ = −1

(2N − 2)-periodic,antisymmetric about 0 and N − 1,a control interval [0, N − 1],[u⋆ ∗ v](0) = [u⋆ ∗ v](N − 1) = 0.

(N − 1)-periodic,

antisymmetric about 0 and N−12

,


− 1],

[Tuv](0) = 0.

c = 1ǫ = 1

(2N − 2)-periodic,symmetric about −1 and N − 2,a control interval [−1, N − 2].

(N − 1)-periodic,

symmetric about − 12

and N2

− 1,


− 1].

c = 1ǫ = −1

(2N − 2)-periodic,antisymmetric about −1 and N − 2,a control interval [−1, N − 2],[u⋆ ∗ v](−1) = [u⋆ ∗ v](N − 2) = 0.

(N − 1)-periodic,

antisymmetric about − 12

and N2

− 1,


− 1],

[Tuv]( N2

− 1) = 0.

The decomposition filter u has the symmetrySu(ξ) = ǫe−i2cξ, where ǫ ∈ {−1, 1} and c ∈ {0, 1}. v isa symmetric extension with both endpoints non-repeated(EN) of an input signal v b = {v b(k)}N−1

k=0 .


Table 2: Endpoints Repeated (ER)

filter u u⋆ ∗ v with v extended by ER Tuv with v extended by ER

c = 12

ǫ = 1

2N-periodic,symmetric about −1 and N − 1,a control interval [−1, N − 1].

N-periodic,

symmetric about − 12

and N−12

,


− 1].

c = 12

ǫ = −1

2N-periodic,antisymmetric about −1 and N − 1,a control interval [−1, N − 1],[u⋆ ∗ v](−1) = [u⋆ ∗ v](N − 1) = 0.

N-periodic,

antisymmetric about − 12

and N−12

,


− 1]

.

c =− 1

2ǫ = 1

2N-periodic,symmetric about 0 and N,a control interval [0, N].

N-periodic,symmetric about 0 and N

2,


].

c =− 1

2ǫ = −1

2N-periodic,antisymmetric about 0 and N,a control interval [0, N],[u⋆ ∗ v](0) = [u⋆ ∗ v](N) = 0.

N-periodic,antisymmetric about 0 and N

2,


],

[Tuv](0) = [Tuv]( N2

) = 0.

The decomposition filter u has the symmetrySu(ξ) = ǫe−i2cξ, where ǫ ∈ {−1, 1} and c ∈ {−1

2,12}. v

is a symmetric extension with both endpoints repeated(ER) of an input signal v b = {v b(k)}N−1

k=0 .


Example: Biorthogonal Wavelet Filter Bank

Since Su0 = 1 and Su1 = e−i2ξ, extend v b by bothendpoints non-repeated (EN):

v = {. . . , 58, 60,−4,−1,−1, 0, 1, 0,−1,−1,−4, 60, 58, 56, 58, 60,−4,−

Then Tu0v is 7-periodic and is symmetric about 0, 7/2:

w0 =√

22Tu0

v =√

22{. . . , 37

8,−5

8, 1,−5

8, 37

8, 263

4, 263

4, 37

8, . . .},

and Tu1v is 7-periodic and is symmetric about −1

2, 3:

w1 =√

22 Tu1

v =√

22 {. . . ,−3

4, 0, 0,−34,−33

2 , 1,−332 ,−3

4, . . .}.


Example: Tight Framelet Filter Bank

Since Su0 = Su1 = Su2 = 1, extend v b with bothendpoints non-repeated (EN). Then all Tu0

v , Tu1v , Tu2

vare 7-periodic and symmetric about 0 and 7/2:

w0 =√

22Tu0

v =√

22{. . . , 58, 51

4,−3

4, 1

2,−3

4, 51

4, 58, 58, 51

4,−3

4, . . .},

w1 =√

22Tu1

v =√

22{. . . , 0, 67

4, 1

4,−1

2, 1

4, 67

4, 0, 0, 67

4, 1

4, . . .}.

w2 =√

22Tu1

v = {. . . ,−29,−594, 1

4, 0, 1

4,−59

4,−29,−29,−59

4, 1

4, 0, . . .}.


Algorithms of Wavelets and Framelets

Bin Han

Department of Mathematical and Statistical SciencesUniversity of Alberta, Canada

Present at Summer School onApplied and Computational Harmonic Analysis

Edmonton, Canada

July 29–31, 2011


Outline of Tutorial: Part II

Design of interpolatory filters

Design of filters with linear-phase moments

Real-valued orthogonal wavelet filter banks

Symmetric complex orthogonal wavelet filter banks

Biorthogonal wavelet filter banks

Tight wavelet framelet filter banks


Biorthogonal Wavelet Filter Banks

A biorthogonal wavelet filter bank ({a; b}, {a; b}):[

â(ξ) ˆb(ξ)

â(ξ + π) ˆb(ξ + π)

] [a(ξ) b(ξ)

a(ξ + π) b(ξ + π)

]⋆= I2.

We must have: for some c ∈ C\{0}, n ∈ Z,

b(ξ) = ce i(2n−1)ξâ(ξ + π), ˆb(ξ) = 1ce i(2n−1)ξa(ξ + π).

PR becomes a simple condition:

a(ξ)â(ξ) + a(ξ + π)â(ξ + π) = 1

with standard high-pass filters:

b(ξ) = e−i(ξ+π)â(ξ + π), ˆb(ξ) = e−i(ξ+π)a(ξ + π).Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 3 / 72

Orthogonal Filters and Refinable Functions

An orthogonal low-pass filter is defined by

|a(ξ)|2 + |a(ξ + π)|2 = 1

with standard high-pass filter:

b(ξ) = e−i(ξ+π)a(ξ + π).

Refinable function φa satisfies refinement equation:

φa = 2∑

k∈Z

a(k)φa(2 · −k).

For b = {b(k)}k∈Z, a wavelet function ψa,b by

ψa,b := 2∑

k∈Z

b(k)φa(2 · −k).


Basic Quantities

vm(a) := m, the highest order of vanishingmoments satisfied by u: u(ξ) = O(|ξ|m), ξ → 0.sr(a) := m, the highest order of sum rules satisfiedby u: u(ξ + π) = O(|ξ|m) as ξ → 0.lpm(a) := m, highest order of linear-phase momentssatisfied by u: u(ξ) = e−icξ + O(|ξ|m), ξ → 0.If u(m)u(n) 6= 0 and u(k) = 0 for all k ∈ Z\[m, n],we define filter support fsupp(u) := [m, n].The smoothness exponent of filter u is defined by

sm(u) := −1/2 − log2

√ρ(u),

where ρ(u) is spectral radius of (w(2j − k))−K6j ,k6K

with∑K

k=−K w(k)e−ikξ := |v(ξ)|2.Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 5 / 72

Other Quantities

Expectation of a filter u:

E(u) :=

∑k∈Z

|u(k)|2k‖u‖2

l2(Z)

, ‖u‖2l2(Z) :=

∑

k∈Z

|u(k)|2.

Variance Var(u):

Var(u) :=

∑k∈Z

|u(k)|2(k − E(u))2

‖u‖2l2(Z)

.

frequency separation indicator Fsi(u, v):

Fsi(u, v) :=

∫π

−π|u(ξ)|2|v(ξ)|2dξ

√∫π

−π|u(ξ)|4dξ

√∫π

−π|v(ξ)|4dξ

.

In particular,

Fsi(u) := Fsi(u, v) with v(ξ) := e−i(ξ+π)u(ξ + π).


Coset Sequences

coset sequence of u at γ is u[γ] := {u(γ + 2k)}k∈Z,

u[γ](ξ) =∑

k∈Z

u(γ + 2k)e−ikξ.

(a, a) is a pair of biorthogonal filters ⇐⇒

a[0](ξ)a[0](ξ) + a[1](ξ)a[1](ξ) = 1/2.


Interpolatory filters

(a, a) is a pair of biorthogonal filters ⇐⇒ theircorrelation filter u := a⋆ ∗ a is interpolatory:

u(ξ) + u(ξ + π) = 1.

Equivalently, u is interpolatory if u[0](ξ) = 1/2 or

u(0) = 1/2 and u(2k) = 0 ∀ k ∈ Z\{0}.

u is an interpolatory filter ⇐⇒ (u, δ) is a pair ofbiorthogonal filters, where δ(0) = 1 and δ(k) = 0for all k ∈ Z\{0}.


Interpolation

PropositionA filter u ∈ l0(Z) is an interpolatory filter if and only if

(Suv)↓2 = v , that is, [Suv ](2k) = v(k), ∀ k ∈ Z.


Polynomials

For any real number m and any nonnegative integer n,

Pm,n(x) = (1 − x)−m + O(xn), x → 0.

More explicitly,

Pm,n(x) :=

n−1∑

j=0

(m + j − 1

j

)x j ,

where(c

0

):= 1,

(c

j

):=

c(c − 1) · · · (c − j + 1)

j!, j ∈ N, c ∈ R

and j ! = 1 · 2 · · · (j − 1)j .Basic identity:

(1− x)mPm,m(x)+ xmPm,m(1− x) = 1 ∀ x ∈ R,m ∈ N.


Construction of Interpolatory Filters

TheoremFor every positive integer m, there exists a uniqueinterpolatory filter aI

2m such that aI2m has 2m sum rules

and vanishes outside [1 − 2m, 2m − 1]. Moreover, aI2m is

a real-valued filter, symmetric about the origin, and hasthe following explicit expression:

aI2m(ξ) = cos2m(ξ/2)Pm,m(sin2(ξ/2)).


Examples of Interpolatory Filters

aI2 = {1

4,12, 1

4}[−1,1],

aI4 = {− 1

32, 0,932,

12, 9

32, 0,− 132}[−3,3],

aI6 = { 3

512, 0,− 25512, 0,

75256,

12, 75

256, 0,− 25512, 0,

3512}[−5,5],

aI8 = {− 5

4096, 0,49

4096, 0,− 2454096, 0,

12254096,

12, 1225

4096, 0,− 2454096,

0, 494096, 0,− 5

4096}[−7,7],

aI10 = { 35

131072, 0,− 405

131072, 0, 567

32768, 0,− 2205

32768, 0, 19845

65536, 1

2,

1984565536, 0,− 2205

32768, 0,567

32768, 0,− 405131072, 0,

35131072}[−9,9].


Table

m 1 2 3 4 5 6 7 8

sm(aI2m) 1.5 2.440765 3.175132 3.793134 4.344084 4.8620120 5.362830 5.852926

‖aI2m‖2

l2(Z) 0.375 0.410156 0.426498 0.436333 0.443063 0.448035 0.451899 0.455014

Var(aI2m) 0.333333 0.428571 0.507137 0.574308 0.633798 0.687718 0.737374 0.783634

Fsi(aI2m) 0.085714 0.048316 0.035932 0.029520 0.025506 0.022714 0.020637 0.019019

The smoothness exponents sm(aI2m), variances Var(aI

2m),and frequency separation indicators Fsi(aI

2m).


Graphs of Refinable Functions

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.2

0

0.2

0.4

0.6

0.8

1

(a) φaI1

−3 −2 −1 0 1 2 3−0.2

0

0.2

0.4

0.6

0.8

1

1.2

(b) φaI2

−5 −4 −3 −2 −1 0 1 2 3 4 5−0.2

0

0.2

0.4

0.6

0.8

1

1.2

(c) φaI3

−6 −4 −2 0 2 4 6−0.2

0

0.2

0.4

0.6

0.8

1

(d) φaI4

−8 −6 −4 −2 0 2 4 6 8−0.2

0

0.2

0.4

0.6

0.8

1

(e) φaI5

−10 −8 −6 −4 −2 0 2 4 6 8 10−0.2

0

0.2

0.4

0.6

0.8

1

(f) φaI6


Convolution Method

TheoremLet a be an interpolatory filter having n sum rules. Letm be a positive integer and P be a polynomial satisfying

(1 − x)mP(x) + xmP(1 − x) = 1

(For example, P = Pm,m.) Define a filter u by

u(ξ) := (a(ξ))mP(a(ξ + π)).

Then u is an interpolatory filter and has mn sum rules.

If plug aI2(ξ) = cos2(ξ/2) as a, then we have aI

2m.


Linear-phase of Filters

PropositionLet u ∈ l0(Z) and ξ0 ∈ (−π, π) such that u(ξ0) 6= 0.Write u(ξ0) = M0e

−iθ0 for some M0, θ0 ∈ R. Then thereare unique real-valued continuous functionsM , θ : (−π, π) → R such that

u(ξ) = M(ξ)e−iθ(ξ) ∀ ξ ∈ (−π, π)

withM(ξ0) = M0, θ(ξ0) = θ0.

Moreover, for a real-valued filter u, the filter u hassymmetry if and only if θ is a linear function.


Linear-phase Moments

Propositionsr(u) = lpm(u) for any interpolatory filter u.

PropositionLet M ,N ∈ N and c ∈ R. For any subset Λ of Z suchthat #Λ = N, there exists a unique solution {ck}k∈Λ to

u(ξ) = e−icξ + O(|ξ|N), ξ → 0,

where u(ξ) := (1 + e−iξ)M∑

k∈Λ cke−ikξ.


Filters Having Linear-phase Moments

TheoremFor any positive integers m and n, set c = 0, M = 2m,N = 2n − 1, and Λ = {1− n −m, . . . , n −m − 1}. Thenthe unique filter in Proposition, denoted by a2m,2n, musttake the form

a2m,2n(ξ) = cos2m(ξ/2)Pm,n(sin2(ξ/2)),

The filter a2m,2n has 2m sum rules, 2n linear-phasemoments with phase 0, is symmetric about the origin,and has filter support [1 − m − n,m + n − 1]. Moreover,when n = m, the above filter a2m,2m is obviously theinterpolatory filter aI

2m.Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 18 / 72

Examples of a2m,2n

a4,2 = { 116,

14,

38, 1

4,116}[−2,2],

a6,2 = { 164,

332,

1564,

516, 15

64,332,

164}[−3,3],

a6,4 = {− 3256,− 1

32,364,

932,

55128, 9

32,364,− 1

32,− 3256}[−4,4],

a8,4 = {− 1256,− 5

256,

− 5256,

564,

35128,

49128, 35

128,564,− 5

256,− 5256,− 1

256}[−5,5],

a8,6 = { 52048,

3512,− 15

1024,− 25512,

752048,

75256,

231512, 75

256,75

2048,

− 25512,− 15

1024,3

512,5

2048}[−6,6].


Table of a2m,2n

(m, n) (2,1) (3,1) (3,2) (4,2) (4,3) (5,2) (5,3) (5,4)sm(a2m,2n) 3.5 5.5 4.098191 5.820374 4.659613 7.586972 6.219362 5.173103

‖a2m,2n‖2l2(Z) 0.273438 0.225586 0.349457 0.309845 0.383178 0.281315 0.351056 0.402620

Var(a2m,2n) 0.571429 0.818182 0.565889 0.704225 0.611512 0.842854 0.716001 0.661298Fsi(a2m,2n) 0.005439 0.000342 0.007165 0.000916 0.007757 0.000105 0.001436 0.007946

The smoothness exponents sm(a2m,2n), variancesVar(a2m,2n), and frequency separation indicatorsFsi(a2m,2n), where the filters a2m,2n. Note thatE(a2m,2n) = 0.


Graph of Refinable Functions

−2 −1.5 −1 −0.5 0 0.5 1 1.5 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

(a) φa4,2

−3 −2 −1 0 1 2 30

0.1

0.2

0.3

0.4

0.5

(b) φa6,2

−4 −3 −2 −1 0 1 2 3 4−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

(c) φa6,4

−5 −4 −3 −2 −1 0 1 2 3 4 5−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

(d) φa8,4

−6 −4 −2 0 2 4 6−0.2

0

0.2

0.4

0.6

0.8

(e) φa8,6

−6 −4 −2 0 2 4 6−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

(f) φa10,4


Filters a2m−1,2n

TheoremFor any positive integers m, n, set c = 1/2,M = 2m − 1, N = 2n − 1, andΛ = {2 − n − m, . . . , n − m}. Then the unique filter inTheorem, denoted by a2m−1,2n, must take the form

a2m−1,2n(ξ) = 2−1(1 + e−iξ) cos2m−2(ξ/2)Pm−1/2,n(sin2(ξ/2)),

The filter a2m−1,2n has 2m− 1 sum rules, 2n linear-phasemoments with phase 1/2, is symmetric about the point1/2, and has filter support [2 − m − n,m + n − 1].


Examples of a2m−1,2n

a3,2 = {18,

38, 3

8 ,18}[−1,2],

a5,2 = { 132,

532,

516, 5

16,532,

132}[−2,3],

a5,4 = {− 5256,− 7

256,35256,

105256, 105

256,35256,− 7

256,− 5256}[−3,4],

a7,4 = {− 71024,− 27

1024, 0,21128,

189512, 189

512,21128, 0,

− 271024,− 7

1024}[−4,5],

a7,6 = { 6316384,

7716384,− 495

16384,− 69316384,

11558192,

34658192

, 34658192,

11558192,− 693

16384,− 49516384,

7716384,

6316384}[−5,6].


Table of a2m−1,2n

(m, n) (2,1) (3,1) (3,2) (4,2) (4,3) (5,2) (5,3) (5,4)sm(a2m−1,2n) 2.5 4.5 3.259609 4.952718 3.906495 6.699013 5.431405 4.472525

‖a2m−1,2n‖2l2(Z) 0.3125 0.246094 0.376099 0.327847 0.403048 0.294540 0.366042 0.418425

Var(a2m−1,2n) 0.45 0.694444 0.496998 0.634999 0.559294 0.773519 0.663753 0.617812

Fsi(a2m−1,2n) 0.021645 0.001364 0.019019 0.002602 0.017058 0.000314 0.003395 0.015613

The smoothness exponents sm(a2m−1,2n), variancesVar(a2m−1,2n), and frequency separation indicatorsFsi(a2m−1,2n), where the filters a2m−1,2n.


Graphs of a2m−1,2n

−1 −0.5 0 0.5 1 1.5 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

(a) φa3,2

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

(b) φa5,2

−3 −2 −1 0 1 2 3 4−0.2

0

0.2

0.4

0.6

0.8

(c) φa5,4

−4 −3 −2 −1 0 1 2 3 4 5−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

(d) φa7,4

−5 −4 −3 −2 −1 0 1 2 3 4 5 6−0.2

0

0.2

0.4

0.6

0.8

(e) φa7,6

−5 −4 −3 −2 −1 0 1 2 3 4 5 6−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

(f) φa9,4


Real-valued Orthogonal Filters

TheoremIf a ∈ l0(Z) is a real-valued orthogonal filter with m sumrules,

|a(ξ)|2 = cos2m(ξ/2)P(sin2(ξ/2))

for some polynomial P with real coefficients such thatbasic identity xmP(1 − x) + (1 − x)mP(x) = 1 and

P(x) > 0, ∀ x ∈ [0, 1].

Conversely, if both identities are satisfied for apolynomial P with real coefficients, then there exists areal-valued orthogonal filter a with a(0) = 1 such that ahas m sum rules.Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 26 / 72

Daubechies Orthogonal Filters

Take P = Pm,m. Then P(x) > 0, x ∈ [0, 1]. There is areal-valued sequence QD

m ∈ l0(Z) such that

|QDm(ξ)|2 = Pm,m(sin2(ξ/2)), QD

m(0) = 1, fsupp(QDm) = [0,m−1].

The Daubechies orthogonal filter of order m is

aDm(ξ) := 2−me i(m−1)ξ(1 + e−iξ)mQD

m (ξ), m ∈ N.

fsupp(aDm) = [1 − m,m], sr(aD

m) = m, and aDm satisfies

|aDm(ξ)|2 = aI

2m(ξ) = cos2m(ξ/2)Pm,m(sin2(ξ/2)).


Examples of Daubechies Filters

aD1 = {1

2, 1

2}[0,1],

aD2 = {1+

√3

8 , 3+√

38, 3−

√3

8 , 1−√

38 }[−1,2]

aD3 = {0.235233603893, 0.570558457918,

0.325182500266,−0.095467207782,

− 0.060416104155, 0.024908749869}[−2,3].


More Examples

QD4 = {−0.857191211347, 3.093477124385,

− 1.600848680204, 0.364562767168}[0,3],

QD5 = {0.618476735277,−2.424433845637,

5.051894897560,−2.688052234523,

0.442114447325}[0,4]

QD6 = {0.697110410451,−4.024690866806,

8.351866150884,−5.869382002997,

2.198116068769,−0.35301976030}[0,5]


Table for Daubechies Filters

m 1 2 3 4 5 6 7 8

sm(aDm) 0.5 1.0 1.415037 1.775565 2.096787 2.388374 2.658660 2.914722

E(aDm) 0.5 -0.149518 -0.835864 -0.153565 0.449967 -0.154343 -0.869398 -0.154614

Var(aDm) 0.25 0.328124 0.453684 0.425360 0.559572 0.531640 0.569226 0.631786

Fsi(aDm) 0.333333 0.219047 0.172336 0.145914 0.128509 0.115986 0.106442 0.098868

The smoothness exponents sm(aDm), means E(aD

m),variances Var(aD

m), and frequency separation indicatorsFsi(aD

m), where aDm is a Daubechies orthogonal filter of

order m (with the smallest variance). Note that‖aD

m‖2l2(Z) = 1/2.


Graphs of Daubechies Wavelets

−1 −0.5 0 0.5 1 1.5 2

0

0.1

0.2

0.3

0.4

0.5

(a) Filter aD1

−3 −2 −1 0 1 2 3

−1.5

−1

−0.5

0

0.5

1

1.5

(b) aD1

−1 −0.5 0 0.5 1 1.5 2−0.2

0

0.2

0.4

0.6

0.8

1

(c) φaD1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

(d) ψaD1

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

(e) Filter aD2

−3 −2 −1 0 1 2 3

−0.2

0

0.2

0.4

0.6

0.8

1

(f) aD2

−1 −0.5 0 0.5 1 1.5 2

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

(g) φaD2

−1 −0.5 0 0.5 1 1.5 2

−1.5

−1

−0.5

0

0.5

1

(h) ψaD2


Graphs

−5 −4 −3 −2 −1 0 1 2 3 4 5 6

−0.1

0

0.1

0.2

0.3

0.4

0.5

(a) Filter aD5

−3 −2 −1 0 1 2 3

−1.5

−1

−0.5

0

0.5

1

1.5

(b) aD5

−1 −0.5 0 0.5 1 1.5 2−0.2

0

0.2

0.4

0.6

0.8

1

(c) φaD5

−4 −3 −2 −1 0 1 2 3 4 5

−1

−0.5

0

0.5

1

(d) ψaD5

−6 −4 −2 0 2 4 6

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

(e) Filter aD6

−3 −2 −1 0 1 2 3

−0.2

0

0.2

0.4

0.6

0.8

1

(f) aD6

−5 −4 −3 −2 −1 0 1 2 3 4 5 6

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

(g) φaD6

−5 −4 −3 −2 −1 0 1 2 3 4 5 6−1.5

−1

−0.5

0

0.5

1

(h) ψaD6


Shortcoming of aDm: No Symmetry

PropositionLet a ∈ l0(Z) be an orthogonal filter. Suppose thata(c − k) = ǫa(k) with ǫ ∈ {−1, 1} and c ∈ Z. Then

1 if c is an even integer, then a must take the form

a(ξ) =√

22 e iαe−inξ;

2 if c is an odd integer and a is real-valued, then

a(ξ) = e iα

2 (e i2mξ + ǫe−i(2m+2n+1)ξ),

for some real number α and some integers m, n.


Orthogonal Filter with LPM

PropositionLet a ∈ l0(Z) be a real-valued orthogonal filter such thata(0) = 1. Then one of the following two cases musthold:

(i) lpm(a) = 2 sr(a) and lpm(a) is an even integer;

(ii) lpm(a) < 2 sr(a) and lpm(a) must be an oddinteger.

CorollaryIf a ∈ l0(Z) is a real-valued orthogonal filter witha(0) = 1 and sr(a) > 2, then lpm(a) > 3.


Shortcoming: Lack Linear-phase Moments

Every real-valued orthogonal filter has at least 3linear-phase moments;

Quite often aDm has no more than 3 linear-phase

moments.

It remains unsolved whether there is a family ofreal-valued orthogonal filters having increasing orderof linear-phase moments.


Algorithm for Orthogonal Filter with LPM

a is a real-valued orthogonal filter having m sum rulesand n linear-phase moments with phase c ∈ R ⇐⇒

a(ξ) = e−iβξ2−m(1 + e−iξ)m(θm,n(ξ) + (1 − e−iξ)nθ(ξ)),

where β ∈ Z is the left endpoint of filter support of a,1 θm,n(ξ) =

∑n−1j=0 λje

−ijξ with λ0, . . . , λn−1 uniquelydetermined by

θm,n(ξ) = e−i(c−β)ξ2m(1+e−iξ)−m+O(|ξ|n), ξ → 0,

2 θ(ξ) =∑ℓ−1

j=0 tje−ijξ for some ℓ ∈ N, where unknown

t0, . . . , tℓ−1 are to be determined by theorthogonality condition.


Example

Let m = 2, n = 3 in Algorithm. Then

θ2,3(ξ) = (12c2− 5

2c + 11

4)+(−c2 +4c− 5

2)e−iξ +(1

2c2− 3

2c + 3

4)e−i2ξ.

For ℓ = 1, set c = 1. Then t0 = 3−√

158 . For c = 0 and

β = −1 satisfying c − β = c = 1, the real-valuedorthogonal filter aOL

2,3 is supported inside [−1, 4] with

sr(aOL2,3) = 2 and lpm(aOL

2,3) = 3 with phase c = 0.

By calculation, E (aOL2,3) ≈ 0.38393, Var(aOL

2,3) ≈ 0.465706,

Fsi(aOL2,3) ≈ 0.184371, and sm(aOL

2,3) ≈ 1.232138.


Graphs

−2 −1 0 1 2 3 4 5

−0.1

0

0.1

0.2

0.3

0.4

0.5

(a) Filter aOL2,3, c =

0

−3 −2 −1 0 1 2 3

−3

−2

−1

0

1

2

3

(b) aOL2,3, c = 0

−1 −0.5 0 0.5 1 1.5 2 2.5 3 3.5 4

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

(c) φaOL2,3 , c = 0

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3

−1.5

−1

−0.5

0

0.5

1

(d) ψaOL2,3 , c = 0

−2 −1 0 1 2 3 4 5

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

(e) Filter aOL2,3, c =

− 15

−3 −2 −1 0 1 2 3

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

(f) aOL2,3, c = − 1

5

−1 −0.5 0 0.5 1 1.5 2 2.5 3 3.5 4

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

(g) φaOL2,3 , c = − 1

5

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3

−1.5

−1

−0.5

0

0.5

1

(h) ψaOL2,3 , c = − 1

5


Symmetric Complex Orthogonal Filters

LemmaLet m be a positive odd integer and a ∈ l0(Z) havecomplex coefficients. Then a is a filter having symmetrya(1 − k) = a(k) and m sum rules ⇐⇒ there is apolynomial Q with complex coefficients such that

a(ξ) = 2−me iξ(m−1)/2(1 + e−iξ)mQ(sin2(ξ/2)).

Moreover, the filter a above is an orthogonal filter ⇐⇒for all x ∈ R,

(1 − x)m|Q(x)|2 + xm|Q(1 − x)|2 = 1, x ∈ R.



LemmaLet P be a polynomial with real coefficients andP(0) 6= 0. Then P(x) = |Q(x)|2, x ∈ R for somepolynomial Q with complex coefficients andQ(0) =

√P(0) ⇐⇒ the polynomial P is nonnegative

on the real line:

P(x) > 0, ∀ x ∈ R.



LemmaFor any positive odd integer m, the polynomial Pm,m

satisfies Pm,m(x) > 0 for all x ∈ R.

There is a subset Y Sm (S=Symmetry) of C such that

∣∣∣∏

y∈Y Sm

(1 + yx)∣∣∣2

= Pm,m(x).

aSm(ξ) := 2−me iξ(m−1)/2(1 + e−iξ)m

∏

y∈Y Sm

(1 + y sin2(ξ/2)).

1 aSm is orthogonal and aS

m(1 − k) = aSm(k);

2 |aSm(ξ)|2 = |aD

m(ξ)|2 = aI2m(ξ) and aS

m(0) = 1;3 sr(aS

m) = m, fsupp(aSm) = [−(m − 1)/2, (m + 1)/2].


Examples

Y S3 = {3−

√15i

2 }Y S

5 = {2.67984516848 + 1.60066496071i ,

− 0.179845168483− 2.67428235144i}Y S

7 = {3.22858920491 + 1.30036579467i ,

1.30198701500 + 2.95813693603i ,

− 1.03057621991− 2.49790321151i}


Table

m 3 5 7 9 11 13 15 17

sm(aSm) 1.415037 2.096787 2.658660 3.161667 3.639798 4.106047 4.565135 5.019141

Var(aSm) 0.468752 0.652384 0.618386 0.701961 0.767464 0.839722 0.893810 0.937622

Fsi(aSm) 0.172338 0.128507 0.106441 0.0926729 0.0830663 0.0758893 0.0702697 0.0657165

The smoothness exponents, variances, and frequencyseparation indicators of aCS

m . Note that‖aD

m‖2l2(Z) = E(aS

m) = 1/2, and sm(aSm) = sm(aD

m).


Graphs

−3 −2 −1 0 1 2 3 4

0

0.1

0.2

0.3

0.4

0.5

(a) Filter |aS3 |

−3 −2 −1 0 1 2 3

−0.5

0

0.5

1

1.5

2

(b) aS3

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3−0.5

0

0.5

1

(c) φaS3

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

(d) ψaS3

−5 −4 −3 −2 −1 0 1 2 3 4 5 6

0

0.1

0.2

0.3

0.4

0.5

(e) Filter |aS5 |

−3 −2 −1 0 1 2 3

−0.5

0

0.5

1

1.5

2

2.5

(f) aS5

−4 −3 −2 −1 0 1 2 3 4 5−0.5

0

0.5

1

(g) φaS5

−4 −3 −2 −1 0 1 2 3 4 5

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

(h) ψaS5


Graph

−10 −8 −6 −4 −2 0 2 4 6 8 10 12

0

0.1

0.2

0.3

0.4

0.5

(a) Filter |aS11|

−3 −2 −1 0 1 2 3

−1

−0.5

0

0.5

1

1.5

(b) aS11

−10 −8 −6 −4 −2 0 2 4 6 8 10

−0.2

0

0.2

0.4

0.6

0.8

1

(c) φaS11

−10 −8 −6 −4 −2 0 2 4 6 8 10

−1

−0.5

0

0.5

1

(d) ψaS11

−10 −5 0 5 10

0

0.1

0.2

0.3

0.4

0.5

(e) Filter |aS13|

−3 −2 −1 0 1 2 3

−1

−0.5

0

0.5

1

1.5

2

(f) aS13

−10 −5 0 5 10

−0.2

0

0.2

0.4

0.6

0.8

1

(g) φaS13

−10 −5 0 5 10

−1

−0.5

0

0.5

1

(h) ψaS13


Shortcoming: Lack Linear-phase Moments

However, the filter aSm in general has no more than two

linear-phase moments, and often has only onelinear-phase moment .


Symmetry and Linear-phase Moments

TheoremLet a ∈ l0(Z) be a complex-valued orthogonal filter witha(0) = 1 such that a has symmetry a(1 − k) = a(k).Then

lpm(ar) = min(2 vm(ai), 2 sr(a)),

lpm(ar) = min(2 lpm(a), 2 sr(a)),

lpm(a) = min(vm(ai), 2 sr(a)).

For an integer n, we define

odd(n) :=

{1, if n is odd,

0, if n is even.


Symmetry and Linear-phase Moments

TheoremLet m and n be positive integers such that n 6 m. Then

Hm,n(x) := 1−x2m−1[Pm−1/2,n(1−x)]2−(1−x)2m−1[Pm−1/2,n(x)]2 > 0 ∀

and Hm,n(x) = xn(1 − x)nRm,n(4x(1 − x)) withRm,n(x) > 0, x ∈ [0, 1].for a polynomial Rm,n. Define am;n by

am;n(ξ) := arm;n(ξ) + i ai

m;n(ξ),

arm;n(ξ) := 21−2me i(m−1)ξ(1 + e−iξ)2m−1Pm−1/2,n(sin

2(ξ/2)),

aim;n(ξ) := 2−2n−1(e iξ − e−iξ)n[Qm,n(2ξ) + (−1)ne−iξQm,n(−2ξ)],

where |Qm,n(ξ)|2 = Rm,n(sin2(ξ/2)).


Continued...

TheoremThen

am;n is a complex-valued orthogonal filter witham;n(0) = 1;

am;n(1 − k) = a(k) for all k ∈ Z;

has at least n + odd(n) linear-phase moments (withphase 1/2);

at least n + 1 − odd(n) sum rules.


Filters aHm

aHm(ξ) :=21−2me i(m−1)ξ(1 + e−iξ)2m−1Pm−1/2,m(sin2(ξ/2))

+ i2−2m−1(e iξ − e−iξ)m[QHm(2ξ)

+ (−1)me−iξQHm(−2ξ)],

where QHm := Qm,m is a finitely supported sequence which

satisfies |QHm(ξ)|2 = Rm,m(sin2(ξ/2)).


Examples

QH2 = {

√152 }[0,0],

QH3 = {0.383876505437,−4.00672069200}[0,1],

QH4 = {−7.93455097214, 1.20587188834

− 0.121275295606}[−1.1],

QH5 = {0.370734915266,−1.12662000411,

15.7161958860,−1.88630270865}[−1,2]

QH6 = {−0.537573675927, 7.07825179788,

− 30.4983020577,−0.150565196702,

− 1.02128030839}[−2,2]


Table

m 2 3 4 5 6 7 8 9

sm(aHm) 1.415037 1.72798 1.95884 2.61918 3.43832 3.71051 3.89248 4.23551

Var(aHm) 0.468752 0.753788 1.13706 0.992888 0.944372 1.03059 2.74195 1.16315

Fsi(aHm) 0.172338 0.112118 0.0932405 0.0817623 0.0869154 0.0801046 0.101746 0.0731063

The smoothness exponents, variances, and frequencyseparation indicators of aH

m. Note that‖aH

m‖2l2(Z) = E(aH

m) = 1/2.


Graphs

−3 −2 −1 0 1 2 3 4

0

0.1

0.2

0.3

0.4

0.5

(a) Filter |aH2 |

−3 −2 −1 0 1 2 3

−0.5

0

0.5

1

1.5

2

(b) aH2

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3−0.5

0

0.5

1

(c) φaH2

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

(d) ψaH2

−5 −4 −3 −2 −1 0 1 2 3 4 5 6

0

0.1

0.2

0.3

0.4

0.5

(e) Filter |aH3 |

−3 −2 −1 0 1 2 3−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

3

(f) aH3

−4 −3 −2 −1 0 1 2 3 4 5

−0.2

0

0.2

0.4

0.6

0.8

1

(g) φaH3

−4 −3 −2 −1 0 1 2 3 4 5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

(h) ψaH3


Graphs

−10 −8 −6 −4 −2 0 2 4 6 8 10 12

0

0.1

0.2

0.3

0.4

0.5

(a) Filter |aH6 |

−3 −2 −1 0 1 2 3

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

3

(b) aH6

−10 −8 −6 −4 −2 0 2 4 6 8 10

−0.2

0

0.2

0.4

0.6

0.8

1

(c) φaH6

−10 −8 −6 −4 −2 0 2 4 6 8 10

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

(d) ψaH6

−10 −5 0 5 10

0

0.1

0.2

0.3

0.4

0.5

(e) Filter |aH7 |

−3 −2 −1 0 1 2 3

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

3

(f) aH7

−10 −5 0 5 10

−0.2

0

0.2

0.4

0.6

0.8

1

(g) φaH7

−10 −5 0 5 10

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

(h) ψaH7


Biorthogonal Filters: Splitting

Interpolatory aI2m(ξ) = cos2m(ξ/2)Pm,m(sin2(ξ/2)).

Factorize Pm,m as Pm,m = PP, where P, P arepolynomials with real coefficients andP(0) = P(0) = 1.Define a pair (aIS

m , aISm ) (IS=Interpolatory Splitting)

of biorthogonal filters by splitting aI2m as follows:

aISm (ξ) = 2−me iξ⌊m/2⌋(1 + e−iξ)mP(sin2(ξ/2)),

aISm (ξ) = 2−me iξ⌊m/2⌋(1 + e−iξ)mP(sin2(ξ/2)).

Primal high-pass filter bISm , dual high-pass filter bIS

m :

bISm (ξ) := e−i(ξ+π)aIS

m (ξ + π),bIS

m (ξ) := e−i(ξ+π)aISm (ξ + π).


Examples: Le-Gall Filter

For m = 2, we have P2,2(x) = 1 + 2x . Choose

P(x) = 1+2x and P(x) = 1 so that P(x)P(x) = P2,2(x).

aIS2 = u0 = {−1

8 ,14 ,

34, 1

4,−18}[−2,2],

aIS2 = u0 = {1

4 ,12, 1

4}[−1,1].


Example

For m = 3, we have P3,3(x) = 1 + 3x + 6x . Choose

P(x) = 1 + 3x + 6x2 and P(x) = 1 so thatP(x)P(x) = P3,3(x).

aIS3 = { 3

64,− 964,− 7

64,4564, 45

64,− 764,− 9

64,364}[−3,4],

aIS3 = {1

8,38, 3

8 ,18}[−1,2].


Example: 9-7 biorthogonal filterFor m = 4, we have P4,4(x) = 1 + 4x + 10x2 + 20x3.Choose P(x) = 1 + (4 − t)x + (10 − 4t + t2)x2 andP(x) = 1 + tx so that P(x)P(x) = P4,4(x), where

t := 30t0t20+5t0−35

and t0 := (350 + 105√

15)1/3.

aIS4 = { t2−4t+10

256,

t−464

,−t2+6t−14

64,

20−t64

,3t2−20t+110

128,

20−t64

,−t2+6t−14

64,

t−464

,t2−4t+10

256}[−4,4]

aIS4 = {− t

64,2−t32 ,

16+t64 ,

6+t16, 16+t

64 ,2−t32 ,− t

64}[−3,3].

aIS4 |[0,4] = {0.602949018236, 0.266864118443,−0.0782232665290,−0.0168641184429

0.0267487574108}[0,4]

andaIS4 |[0,3] = {0.557543526229, 0.295635881557,−0.0287717631143,−0.0456358815571


Table

m 2 3 4 5 6 7 8

sm(aISm ) 0.440765 0.175132 1.409968 1.530850 2.127349 2.018217 2.7293236 2.648124

sm(aISm ) 1.5 2.5 2.122644 2.662135 2.638762 3.268119 3.096332 3.675177

‖aISm ‖2

l2(Z) 0.718750 1.05664 0.520217 0.871940 0.449413 0.421335 0.413449 0.788519

‖aISm ‖2

l2(Z) 0.375 0.3125 0.491478 0.315518 0.564468 0.633216 0.617861 0.352030

Var(aISm ) 0.347826 0.569778 0.421747 0.687719 0.477656 0.611810 0.520229 0.985709

Var(aISm ) 0.333333 0.45 0.445416 0.441148 0.588186 0.735834 0.733994 0.499588

Fsi(aISm , bIS

m ) 0.215436 0.159414 0.152107 0.107347 0.120248 0.121768 0.100035 0.0767112

Ofi(aISm ) 0.318359 2.36290 0.0192038 0.805697 0.0170676 0.0566292 0.0412137 0.590129

Ofi(aISm ) 0.09375 0.210938 0.0122206 0.177758 0.0279181 0.134896 0.0814951 0.136545

Ofi(aISm , bIS

m ) 0.0136919 0.0308002 0.00796381 0.00003864 0.00583666 0.0207335 0.00466501 0.000048884

The smoothness exponents, variances, frequencyseparation indicators, and orthogonal filter indicators ofaIS

m and aISm . Note that E(aIS

m ) = E(aISm ) = odd(m)/2,

Fsi(aISm , b

ISm ) = Fsi(aIS

m , bISm ), and

Ofi(aISm , b

ISm ) = Ofi(aIS

m , bISm ).


Graphs

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2

−1

0

1

2

3

4

5

(a) φaIS2

−1 −0.5 0 0.5 1 1.5 2

−6

−5

−4

−3

−2

−1

0

1

2

3

4

(b) ψaIS2 ,aIS

2

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

(c) φaIS2

−1 −0.5 0 0.5 1 1.5 2

−1.5

−1

−0.5

0

0.5

(d) ψaIS2 ,bIS

2

−3 −2 −1 0 1 2 3 4

−2

−1

0

1

2

3

(e) φaIS3

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3

−4

−3

−2

−1

0

1

2

3

4

(f) ψaIS3 ,bIS

3

−1 −0.5 0 0.5 1 1.5 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

(g) φaIS3

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

(h) ψaIS3 ,bIS

3


Graphs

−6 −4 −2 0 2 4 6−0.2

0

0.2

0.4

0.6

0.8

1

(a) φaIS6

−5 −4 −3 −2 −1 0 1 2 3 4 5 6

−1

−0.5

0

0.5

(b) ψaIS6 ,bIS

6

−5 −4 −3 −2 −1 0 1 2 3 4 5

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

(c) φaIS6

−5 −4 −3 −2 −1 0 1 2 3 4 5 6

−1.5

−1

−0.5

0

0.5

1

(d) ψaIS6 ,bIS

6

−6 −4 −2 0 2 4 6 8

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

(e) φaIS7

−6 −4 −2 0 2 4 6−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

(f) ψaIS7 ,bIS

7

−5 −4 −3 −2 −1 0 1 2 3 4 5 6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

(g) φaIS7

−6 −4 −2 0 2 4 6

−1.5

−1

−0.5

0

0.5

1

1.5

(h) ψaIS7 ,bIS

7


CBC Algorithm

Let a ∈ l0(Z) with a(0) 6= 0 and a has a dual filter a.(S1) choose Λ ⊂ Z with #Λ > m. Then there exists a

solution {cn}n∈Λ with (#Λ − m) free parameters to

∑

n∈Λ

cne−inξ =

e iξ/2â(ξ/2 + π)

a(ξ/2), ξ → 0.

In particular, if #Λ = m, a unique solution {cn}n∈Λ;(S2) construct a filter a coset by coset as follows:

a(2k) = a(2k) −∑

n∈Λ

cn a(1 + 2n − 2k), k ∈ Z,

a(1 + 2k) = a(1 + 2k) +∑

n∈Λ

cn a(2n − 2k), k ∈ Z.

Then a is a dual filter of a with m sum rules.Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 62 / 72

Symmetry

LemmaLet (a, a) be a pair of biorthogonal filters such that a hassymmetry: Sa(ξ) = ǫe−icξ for some ǫ ∈ {−1, 1} andc ∈ Z. Define

aS(ξ) := (â(ξ) + ǫe−icξâ(−ξ))/2.

Then aS is also a dual filter of a and aS has the samesymmetry type as a: S aS = Sa.


ExampleLet a = aI

4 = {− 132, 0,

932,

12 ,

932, 0,− 1

32}[−3,3]. Then a = δ

is a dual filter of a such that Sa(ξ)Sâ(ξ) = 1.

ha,a(0) =1

2, ha,a(1) = 0, ha,a(2) =

9

64, ha,a(3) =

45

128, ha,a(4) =

For m = 1 and Λm = {0}, then t0 = 12 is a solution and

a = { 1

64, 0,−1

8,1

4,23

32,1

4,−1

8, 0,

1

64}[−4,4].

For m = 2 and Λm = {0, 1}, then {t0 = 916, t1 = − 1

16}

a = {− 1

512, 0,

9

256,− 1

32,− 63

512,

9

32,

87

128,

9

32,− 63

512,− 1

32,

9

256, 0,− 1

512}[−6,6].


Example

a = a6,4 = {− 3

256,− 1

32,

3

64,

9

32,

55

128,

9

32,

3

64,− 1

32,− 3

256}[−4,4].

Then a = { 980,− 3

10, 3

16, 1, 3

16,− 3

10, 9

80}[−3,3] is a dual filter

of a.

ha,a(0) = −1

10, ha,a(1) = −

3

5, ha,a(2) = −

15

16, ha,a(3) = −

4389

640, ha,a(4) = −

78117

1024.

For m = 2 and Λm = {0, 1}, then {t0 = 316, t1 = −23

80} is

a solution.

a = {69

20480,−

23

2560, −

321

20480,

111

1280, −

91

20480,−

681

2560,

5463

20480,

561

640,

5463

20480,

−681

2560,−

91

20480,

111

1280,−

321

20480, −

23

2560,

69

20480}[−7,7].


Table

CBC Algorithm sr ‖ · ‖l2(Z) Var sm Ofi Ofi(a, b) Fsi(a, b)

a = aI4 4 0.410156 0.428571 2.440765 0.0650482

a with m = 1 2 0.673340 0.382886 0.593223 0.249223 0.0339170 0.188788a with m = 2 4 0.654892 0.514178 1.179370 0.218591 0.0101275 0.141911

a = a6,4 6 0.349457 0.565889 4.098191 0.176972

a with m = 2 4 1.06795 0.906996 0.349587 3.29133 0.0463589 0.135131a with m = 3 6 1.03806 1.01883 0.649332 3.08117 0.0320641 0.112711

a = a5,4 5 0.376099 0.496998 3.259609 0.122794

a with m = 2 3 0.846221 0.543673 0.346291 1.07161 0.0566240 0.183283a with m = 3 5 0.801151 0.740546 1.042980 0.904897 0.0193244 0.121141

The smoothness exponents, variances, frequencyseparation indicators, and orthogonal filter indicators of aand a in Examples. Note that Ofi(a, b) = Ofi(a, b) andFsi(a, b) = Fsi(a, b).


Graphs

−3 −2 −1 0 1 2 3

0

0.2

0.4

0.6

0.8

1

(a) φa with a = aI4

−3 −2 −1 0 1 2 3 4

−1.4

−1.2

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

(b) ψa,b with m =1

−4 −3 −2 −1 0 1 2 3 4

−1

−0.5

0

0.5

1

1.5

2

2.5

3

(c) φa with m = 1

−3 −2 −1 0 1 2 3 4

−4

−3

−2

−1

0

1

2

3

(d) ψa,b, m = 1

−3 −2 −1 0 1 2 3

0

0.2

0.4

0.6

0.8

1

(e) φa with a = aI4

−4 −3 −2 −1 0 1 2 3 4 5

−1.4

−1.2

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

(f) ψa,b with m =2

−6 −4 −2 0 2 4 6

−0.5

0

0.5

1

1.5

(g) φa with m = 2

−4 −3 −2 −1 0 1 2 3 4 5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

(h) ψa,b, m = 2


Polyphase and Laurent Polynomials

For a sequence u = {u(k)}k∈Z,

u(z) :=∑

k∈Z

u(k)zk, z ∈ C\{0}. (1)

u⋆(k) := u(−k)T, k ∈ Z.

u(k) = ǫu(2c − k) for all k ∈ Z ⇐⇒Su(z) := u(z)

u(1/z) = ǫz2c .

u has m linear-phase moments with phase c ⇐⇒u(z) = z c + O(|z − 1|m), z → 1.

biorthogonality condition becomes

a⋆(z)a(z) + a⋆(−z)a(−z) = 1, z ∈ C\{0}.Bin Han (University of Alberta) Algorithms of Wavelets and Framelets Summer School, Edmonton 68 / 72

Biorthogonal Filters

PropositionFor a ∈ l0(Z), a has at least one finitely supported dualfilter, if and only if, a(z) and a(−z) have no commonzeros in C\{0}.


Chain Structure

TheoremLet a ∈ l0(Z) be a filter such that a has at least onefinitely supported dual filter a. Assume that the filtersupport of a is [m, n] for some integers m and n. Iflen(a) = n −m > 0, then there exists a unique dual filtera of a such that fsupp(a) ⊆ [m, n − 1].

(i) if len(a) is a positive even integer, then there existsa unique dual filter a of a such thatfsupp(a) ⊆ [m + 1, n − 1];

(ii) if len(a) is odd, then there exists a unique dual filtera of a such that fsupp(a) ⊆ [m, n − 2];

(iii) if len(a) = 1, a = 1

2a(m)δ(· − m), a = 1

2a(n)δ(· − n).


Dual Framelet Filter Banks

({a; b1, . . . , bs}, a; b1, . . . , bs})Θ is an OEP-baseddual framelet filter bank if[

b1(z) · · · bs(z)

b1(−z) · · · bs(−z)

] [b1(z) · · · bs(z)

b1(−z) · · · bs(−z)

]⋆

= MΘ,a,a(z),

where MΘ,a,a(z) :=[Θ(z) − Θ(z2)a(z)a⋆(z) −Θ(z2)a(z)a⋆(−z)−Θ(z2)a(−z)a⋆(z) Θ(−z) − Θ(z2)a(−z)a⋆(−z)

].

Equivalently,[b

[0]1 (z) · · · b

[0]s (z)

b[1]1 (z) · · · b

[1]s (z)

][b

[0]1 (z) · · · b

[0]s (z)

b[1]1 (z) · · · b

[1]s (z)

]⋆

= NΘ,a,a(z),


Tight Framelet Filter Banks

{a; b1, . . . , bs}θ is an OEP-based tight framelet filterbank fif

[b

[0]1 (z) · · · b

[0]s (z)

b[1]1 (z) · · · b

[1]s (z)

][b

[0]1 (z) · · · b

[0]s (z)

b[1]1 (z) · · · b

[1]s (z)

]⋆

= NΘ,a(z),

where NΘ,a(z) :=[

12Θ[0](z) − Θ(z)a[0](z)(a[0](z))⋆ 1

2zΘ[1](z) − Θ(z)a[0](z)(a[1](z))⋆

12Θ[1](z) − Θ(z)a[1](z)(a[0](z))⋆ 1

2Θ[0](z) − Θ(z)a[1](z)(a[1](z))⋆

].


Documents

Algorithms of Wavelets and Framelets - University of Albertabhan/notes.pdf · 2011-07-22 · Algorithms of Wavelets and Framelets Bin Han Department of Mathematical and Statistical