Understanding Wavelets & Filter Banks: Mallat Pyramid Algorithm & Multiresolution Analysis, Slides of Banking and Finance

Download Understanding Wavelets & Filter Banks: Mallat Pyramid Algorithm & Multiresolution Analysis and more Slides Banking and Finance in PDF only on Docsity! Course 18.327 and 1.130 Wavelets and Filter Banks Mallat pyramid algorithm � φ � φ � Pyramid Algorithm for Computing Wavelet Coefficients Goal: Given the series expansion for a function fj(t) in Vj fj(t) = aj[k] j,k(t) k how do we find the series fj-1(t) = aj-1[k] j-1,k(t) k in Vj-1 and the series gj-1(t) = bj-1[k]wj-1,k(t) k in Wj-1 such that fj(t) = fj-1(t) + gj-1(t) ? 2 1 Docsity.com 3 Example: suppose that φ(t) = box on [0,1]. Then functions in V1 can be written either as a combination of  φ(2t) 1 0 ½ , 1 0 ½ 1 , φ(2t-1)  or as a combination of  φ(t) 1 0 1 , ,  1 0 1 2 φ(t-1) 4 plus a combination of  1 0 1 , , w(t) w(t-1)  1 0 1 2 Easy to see because φ(2t) = ½[φ(t) + w(t)] φ(2t –1) = ½[φ(t) - w(t)] 2 Docsity.com φ � φ � φ � φ � � φ φ � � φ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ Multiresolution reconstruction equation Start with fj(t) = fj-1(t) + gj-1(t) Multiply by j,n(t) and integrate fj(t) j,n(t) dt = fj-1(t) j,n(t)dt + gj-1(t) j,n(t) dt - - - So aj[n] = aj-1[k] j-1,k(t) j,n(t) dt + k - bj-1[k] wj-1,k(t) j,n(t) dt k - 9 10  φj-1,k(t) φj,n(t) dt = √2 � h0[] � φj,2k+(t) φj,n(t) dt = √2 � h0[] δ[2k +  - n] = √2 h0[n – 2k]  -∞ ∞ -∞ ∞  Similarly � wj-1,k(t)φj,n(t) dt = √2 h1[n –2k] Result: aj[n] = √2 � aj-1[k]h0[n - 2k] + √2 � bj-1[k]h1[n – 2k] ∞ -∞ k k 5 Docsity.com 11 ↓2 ↓2 aj[n] aj 1[n] b ] ↑2 ↑2 Analysis u0[n] u1[n] aj[n] Synthesis v1[n] v0[n] Filter Bank Representation √2h0[n] ~ √2h1[n] ~ √2h1[n] √2h0[n] ⊕ time reversalh0[n] = h0[-n] h1[n] = h1[-n] Verify that filter bank implements MRA equations: u0[n] = √2 � h0[n - k]aj[k] = √2 � h0[k – n]aj[k] ~ ~ ~ k k 12 aj-1[n] = u0[2n] downsample by 2 = √2 � h0[k – 2n]aj[k] bj-1[n] = u1[2n] = √2 � h1[k – 2n]aj[k] aj[n] = √2 � h0[n - ]v0[] + √2 � h1[n - ]v1[] v0[] = k k   � � � � � � � � � � � � aj-1[/2] ;  even 0 ; otherwise So aj[n] = √2 � h0[n - ]aj-1[/2] + √2 � h1[n - ]bj-1[/2] = √2 � h0[n –2k]aj-1[k] + √2 � h1[n – 2k]bj-1[k]  even  -1 0  1 2 n v0[n]aj-1[0] aj-1[1] upsample by 2 k k 6 Docsity.com