Biorthogonal Wavelets and Filter Banks via PFFS. Multiresolution Analysis (MRA) subspaces V j, and wavelet subspaces W j. f X n f, τ n φ τ n φ.

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Ἀντιόπη Μανωλάς
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1 Chapter 3. Biorthogoal Wavelets ad Filter Baks via PFFS 3.0 PFFS applied to shift-ivariat subspaces Defiitio: X is a shift-ivariat subspace if h X h( ) τ h X. Ex: Multiresolutio Aalysis (MRA) subspaces V j, ad wavelet subspaces W j. f X f, τ φ τ φ. Theorem: Let X be a shift-ivariat subspace. Let {τ φ } be a Bessel sequece i H such that sp{τ φ } X. The there exists PFFS-duals {φ } H (i the sese that f = f, φ τ φ for all f X) such that, φ = τ φ; φ = φ 0 + z, where φ 0 X, & z X. Remark: The importace of this result is that we ow kow that PFFS duals φ ca be geerated by the traslatios of a fuctio φ. We will be usig this result to study the costructio of Bi-wavelets via PFFS. 3. Costructio of Bi-wavelets (Biorthogoal) wavelets ad associated filter baks via PFFS. ) Assume we have a biorthogoal MRA i which {φ(t )} is a exact frame (Riesz basis) of V 0 sp{φ(t )}. The there exists a uique bi-dual {φ(t )} V 0 s.t. f V 0, f = f, φ 0 (t ) φ(t ) PFFS, other PFFS-biorthogoal duals of {φ(t )} ca be writte as φ = φ 0 + z, z V 0 Assume that the followig scalig equatios hold: φ = h φ(t ) = h φ

2 ad φ 0 = h 0 φ 0 (t ) = h 0 φ 0 Usig the biorthogoality of φ ad φ h m = φ, φ m = h φ, φ m, φ, φ m = δ m = φ 0 + z, φ m = φ 0, φ m + z, φ m = h m 0 + h m.to Note: h m = z, φ m = z 0 +z +z +, φ m, where z i W i, (V i Wi = V i+, φ m V, ad h m is a sequece). Therefore, h m = z 0, φ m + i= z i, φ m = z 0, φ m Note : i= z i, φ m = 0 ( high pass sub- The secodary compoet takes iformatio from W 0 space) back ito the low-pass bad. ) Biorthogoal Priciple: Claim: If {h } is the filter sequece geeratig the same φ the {h } satisfy the followig biorthgoal priciple. h h k = 0, To uderstad this, let s recall the usual biorthogoal relatioship (as illustrated below): k

3 Recall: I time domai, the biorthogoal filter bak systems with perfect recostructio requiremet is give by (h k hl + g k g l ) = δ kl (3) (Cohe, Daubechies, ad Feauveau 9) Workig out the Fourier Trasform of the idetity (3), we will have the equivalet relatioship i the frequecy domai Furthermore, H(γ) H(γ) + G(γ) G(γ) = H(γ) H(γ + ) + G(γ) G(γ + ) = 0 (4) G(γ) = e πiγ H(γ + ) = e πiγ (H 0 (γ + ) + H(γ + )) (5) Note: G 0 = e πiγ H 0 (γ + ) G(γ) = e πiγ H(γ + ) (6) Recall also that for biorthogoal multiresolutio aalysis: 0 < A H(γ) + H(γ + ) B. Note that (5) ad (6) are derived from the PRFB relatioships (4) ad (7). Because of (5) ad (6), the biorthogoal relatioship of (3) ca be writte as h h k = δ k0 (9) h (h 0 ( k) + h k ) = δ k0 h h 0 ( k) + h h k = δ k0 }{{}}{{} δ k0 0, k 3

4 therefore, h h k = 0, k (0) Take the Fourier Trasform of (0): ( h h k )e πikγ = 0, we k have H(γ) H(γ) + H(γ + ) H(γ + ) = 0 () where H(γ) = h e πiγ ad H(γ) = h e πiγ. 3.. Solutio to H Assume we are iterested i a fiite legth sequeces {h }, { h } etc., H(γ), H(γ) etc. ca be writte as a product of Trig. polyomials with a power of e πiγ. From (), we see that H(γ) must be zero wheever H(γ + ) is zero, ad because H(γ) is a product of a trig polyomial with a power of e πiγ H(γ) = H(γ + )ˆq(γ), () where ˆq(γ) is a trig polyomial. Take () ito (), we have H(γ)H(γ + )ˆq(γ) + H(γ + )H(γ)ˆq(γ + ) = 0 H(γ)H(γ + )[ˆq(γ) + ˆq(γ + )] = 0 Assumig that H(γ) is such that H 0 a.e. expect for at γ =, we have therefore ˆq(γ) = ˆp(γ) cos πnγ where N = odd. This is because ˆq(γ) + ˆq(γ + ) = ˆp(γ) cos πnγ + ˆp(γ + ) cos(πnγ + Nπ) = ˆp(γ) cos πnγ ˆp(γ) cos(πnγ) = 0 4

5 Note: ˆp(γ) is a arbitrary trig polyomial. I geeral, ˆq(γ) = ˆp(γ)e πinγ, N = odd. Typically, for N = ˆq(γ) = ˆp(γ) cos πγ or ˆq(γ) = ˆp(γ)e πiγ 3)The chages i the filterig System: Remark: The correspodig ϕ, ψ ad ψ are chaged (eve though G did t chage). 3.3 Costructio of liear phase FIR H through H ) Liear phase: H(γ) = e πiλγ H(γ), λ Z reduces to symmetric H () H( γ) = H(γ) { h }is real ϕ( t) = ϕ(t) () ϕ( t) = ϕ(t) ˆϕ( γ) = e πiγ ˆϕ(γ) or H( γ) = πiγ H(γ) ) How to icrease the vaishig momet of ϕ. Thm: Let ψ(t) be a bi-wavelet fuctio geerated by a wavelet equatio (ψ(t) = g ϕ (t)). Assume ˆϕ(γ) is N time cotiuously differetiable at γ = 0. The the followig are equivalet: () ψ has N vaishig momets ( t ψ(t)dt = 0, =0,,,...,N ) 5 R

6 () ˆψ() (0) = 0, = 0,,,..., N (3) H() ( ) = 0, = 0,,,..., N pf:(a) ˆψ () (γ) = d ψ(t)e πiγt dt = ( πit) ψ(t)e πiγt dt dγ R R Therefore ˆψ () (γ) = ( πi) Hece () (). (b) Sice ˆψ(γ) = G( γ ) ˆφ( γ ), This implies R ˆψ(γ) = G(γ) ˆφ(γ) t ψ(t)e πiγt dt = e πiγ H(γ + ) ˆφ(γ) (3) ˆψ(0) = H( ) ˆφ(0). Therefore, H( ) = 0, because ˆφ(0) = φ(t)dt 0. Now take the derivative of (3): ˆψ(γ) = ( H (γ + ) ˆφ(γ)e πiγ + H(γ + )( ˆφe πiγ ) ) The, 0 = ˆψ (0) = ( H ( ) ˆφ(0) + 0) H ( ) = 0 By iductio, ˆψ () (0) = 0 H () ( ) = 0. i.e. () (3). Theorem ( H() ( ) = 0, = 0,,,..., N ) H(γ) = (+e πiγ ) N F (γ) where F (γ) is a trig polyomial such that F ( ) 0. 6

7 A outlie of the proof: f (a)=0 f (x) = a (x) = (x a)a (x) 0=f (a)=a (a) + 0 a (a) = 0 Therefore a (x) = (x a)a (x) f(x)=(x-a) g (x) etc. f(a) = 0 f(x) = (x a)a (x) 3 ) Geeral priciple of icreasig the umber of vaishig momets Assume H 0 (γ) has l zeros at γ =, the if H(γ) = ˆp(γ)H(γ + )e πiγ has the same l zeros at γ = the H(γ) ca have at least (l + ) zeros at γ =. This is because H(γ) = H 0 (γ) + H(γ) = (cos πγ) l F (γ) + (cos πγ) l F (γ)h(γ + )e πiγ = (cos πγ) l[ F (γ) + e πiγ F (γ)h(γ + )] If F ( ) = F ( ) H(0),. the [ F (γ)+e πiγ F (γ)h(γ + )] will have a additioal zero at γ = Example: Haar Wavelets H(γ) = + e πiγ = cos πγe πiγ = e πiγ cos πγ ˆp(γ) = si (πγ) ˆp (γ) ˆp(γ) = si (4πγ) ˆp (γ) = si (πγ) cos (πγ)ˆp (γ) = 4 si (πγ) cos (πγ) cos (πγ)ˆp (γ) H(γ) = H(γ) + H(γ) = e πiγ cos πγ + ˆp(γ)H(γ + )e πiγ = e πiγ cos πγ + 4 si π cos πγ cos πγ ie πiγ si πγ ˆp (γ)e πiγ 7

8 = e πiγ cos πγ [ + 4i si πγ cos πγ ˆp (γ)e πiγ] (ote that ˆp (γ)e πiγ = i 4 ) = e πiγ cos πγ [ + si πγ cos πγ ] ˆp(γ) = si πγ ˆp (γ) = si πγ( i 4 )eπiγ = (p )e πiγ Note: If H 0 (γ) is to be real ad H 0 (γ) has l zeros at γ =, the, H 0 (γ) = ( + cos πγ) l F (γ) = l (cos πγ) l F (γ) ˆp(γ) = si πγ ˆp = si πγ ( i ) e πiγ 4 wat: = (p )e πiγ solutio: H(γ) = ˆqH(γ + ) = ˆp(γ)H(γ + )e πiγ h() = (p l ()) ( ( ) h( ) ) h() = h() + h() 8

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