Slides5 - Course 18.327 and 1.130 Course Wavelets and Filter Banks Modulation and Polyphase Modulation Representations Noble Identities Block

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Unformatted text preview: Course 18.327 and 1.130 Course Wavelets and Filter Banks Modulation and Polyphase Modulation Representations: Noble Identities; Block Toeplitz Matrices and Block z-transforms; and transforms; Polyphase Examples Polyphase Modulation Matrix Matrix form of PR conditions: [F0 (z) F1 (z)] H0(z) H0(-z) = [ 2z –l 0 ] (z)] z) H1(z) H1(-z) 123 Modulation matrix, Hm(z) So So [ F0(z) F1(z)] = [2z –l 0] Hm –1(z) (z)] 0] Hm–1(z) = 1 (z) H1(-z) -H0(-z) z) z) ? -H1(z) H0(z) ? = H0(z) H1(-z) - H0 (-z) H1 (z) (must be non-zero) z) 2 1 2z-l H0(-z) F1(z) = (z) ? 678 678 1 2z-l H1(-z) ⇒ F0(z) = (z) ? Require these Require to be FIR to Suppose we choose ? = 2z - l Suppose 2z Then F1(z) = -H0(-z) (z) 678 678 F0(z) = H1(-z) 3 Synthesis modulation matrix: Complete the second row of matrix PR conditions by replacing z with –z: by F0(z) F1(z) H0(z) z -l H0(-z) 0 =2 123 F0(-z) F1(-z) H1(z) H 1 (-z ) z) 0 (-z) - l z) Synthesis Synthesis modulation modulation matrix, Fm(z) Note the transpose convention in Fm(z). 4 Noble Identities 1. Consider x [n] H(z2) u[n] U(z) = H(z2)X(z) Y(z) = ½ {U(z ½ ) + U( -z ½ )} Y(z) U( )} ↓2 y[n] (downsampling) = ½ { H (z) X (z ½ ) + H(z) X (-z½ )} (z) H(z) )} = H(z) • ½ {X (z ½) + X (-z ½ )} ⇒ can downsample H(z) (z )} can first First Noble identity: First x [n] y[n] x[n] y[n] 2) ≡ H(z) ↓2 H(z ↓2 5 2. Consider 2. Consider x[n] [n] H(z) u[n] U(z) = H(z) X(z) U(z) Y(z) = U(z2) = H(z2) X(z2) ↑2 y[n] (upsampling) ⇒ can upsample first can first Second Noble Identity: x[n] y[n] x[n] [n] ≡ H(z2) ↑2 H(z) ↑2 y[n] 6 Derivation of Polyphase Form 1. Filtering and downsampling: Filtering downsampling x[n] y[n] H(z) ↓2 H(z) = Heven(z2) + z -1 Hodd(z2); heven[n] = h[2n] H(z) ); hodd[n] = h[2n+1] Heven(z2) x[n] + z-1 ↓2 y[n] Hodd(z2) 7 Heven(z2) ↓2 x[n] y[n] + z-1 Hodd(z2) ↓2 x[n] xeven[n] ↓2 Heven(z) + z-1 ↓2 xodd[n-1] Hodd(z) y[n] Polyphase Form 8 2. Upsampling and filtering Upsampling and x[n] ↑2 F(z) y[n] F(z) = Feven(z2) + z-1 Fodd (z2) F(z) x[n] Feven(z2) + ↑2 Fodd(z2) y[n] z-1 9 x[n] ↑2 Feven(z2) + ↑2 x[n] Feven(z) Fodd(z2) yeven[n] z-1 ↑2 + Fodd(z) yodd[n] ↑2 y[n] y[n] z-1 Polyphase Form 10 10 Polyphase Matrix Consider the matrix corresponding to the analysis Consider filter bank in interleaved form. This is a block interleaved form. Toeplitz matrix: matrix: M L L 0 0 0 0 h0[3] h0[2] h1[3] h1[2] 0 0 0 0 h0[1] h0[0] h1[1] h1[0] L L L L L Hb= [0] L h0[3] h0[2] h0[1] h0[0] L h1[3] h1[2] h1[1] h1[0] 4-tap Example 11 11 Taking block z-transform we get: Hp(z) = = = h0[0] h0[1] h1[0] h1[1] [1] + h0[0] + z-1 h0[2] [2] h1[0] + z-1 h1[2] H0,even (z) H1,even (z) z-1 h0[2] h0[3] h1[2] h1[3] h0[1] + z-1 h0[3] h1[1] + z-1 h1[3] H0,odd (z) H1,odd (z) 1,odd (z) This is the polyphase matrix for a 2-channel filter bank. 12 12 Similarly, for the synthesis filter bank: M Fb = M M f0[0] f1[0] f0[1] f1[1] M 0 0 0 0 L f0[2] f1[2] f0[0] f1[0] f0[3] f1[3] f0[1] f1[1] 0 0 0 f0[2] f1[2] 0 f0[3] f1[3] M M L M M 13 13 Fp(z) = = f0[0] f1[0] f0[1] f1[1] + z-1 F0,even [z] F1,even [z] 0,even [z] 1,even [z] F0, odd [z] F1, odd [z] [z] 1, [z] 0, f0[2] f1[2] f0[3] f1[3] Note transpose Note convention for synthesis synthesis polyphase matrix • Perfect reconstruction condition in polyphase domain: Fp(z) Hp(z) = I (centered form) This means that Hp(z) must be invertible for all z on the (z) unit circle, i.e. unit det Hp(eiω) ≠ 0 for all frequencies ω. for 14 14 • Given that the analysis filters are FIR, the Given requirement for the synthesis filters to be also FIR is: FIR det Hp(z) = z-l (simple delay) det because Hp-1(z) must be a polynomial. • Condition for orthogonality: Fp(z) is the transpose Condition (z) of Hp(z), i.e. of HpT(z-1) Hp(z) = I i.e. Hp(z) should be paraunitary. (z) paraunitary 15 15 14243 14243 Relationship between Modulation Relationship and Polyphase Matrices and h0,even[n] = h0[2n] H0(z) = H0,even(z2) + z-1 H0,odd(z2) ; h0,odd[n] = h0[2n+1] H1(z) = H1,even(z2) + z-1 H1,odd(z2) Two more equations by replacing z with -z. Two So in matrix form: H0(z) H0(-z) H 0,even(z2) H0,odd(z2) 1 = H1(z) H1(-z) H 1,even(z2) H1,odd(z2) z-1 123 123 1 -z-1 Hm(z) Hp(z2) Modulation matrix Polyphase matrix 16 16 But 1 1 z-1 -z-1 = 1 z-1 123 11 1 -1 123 D2(z) F2 Delay Matrix 2-point DFT Matrix FN = -1 FN1 = 1 1 1 . . . 1 1 N 1 1…1 w w2 … w N-1 w2 w4 … w 2(N-1) 2π iN ; w=e . . . . . . N-1 w 2(N-1) w (N-1)2 w N-point DFT Matrix FN Complex conjugate: replace w with w = Complex with -i 2π eN 17 17 So, in general 1 Hm(z) F-N = H p(zN) DN(z) N = # of channels in filterbank of filterbank (N = 2 in our example) 18 18 Polyphase Matrix Example: Daubechies 4-tap filter tap 1+√3 3 + √3 1+ h0[0] = h0[1] = 4 √2 4 √2 3 -√ 3 1 - √3 h0[3] = h0[2] = 4 √2 4 √2 1 {(1 + √3 ) + (3 + √3 ) z-1 + (3 - √ 3) z-2 + (1 - √3) z-3} {(1 (3 H0(z) = 4√2 1 {(1 - √3) – (3 - √3) z-1 + (3 + √3)z-2 – (1 + √3)z-3} {(1 3) (3 (3 (1 H1(z) = (z) 4√2 19 19 Time domain: h0[0]2 + h0[1]2 + h0[2]2 + h0[3]2 = h0[0] h0[2] + h0[1] h0[3] = [3] 1 32 + 2√3) + (12 + 6 √3) + 3) (12 – 6 √3) + (4 – 2 √3)} (12 3) =1 {(2√3) + (-2√3)} 1 {(4 {(4 32 =0 i.e. filter is orthogonal to its double shifts 20 20 Polyphase Domain: 1 {(1 + √3) + (3 - √3) z-1} H0,even(z) = {(1 3) 4 √2 = 1 {(3 + √3) + (1 - √3) z-1} {(3 3) 4 √2 H1,even(z) = 1 {(1 - √3) + (3 + √3) z-1} {(1 3) 4 √2 H1,odd(z) 1 { - (3 - √3) – (1 + √3) z-1} (3 3) (1 4 √2 H0,odd(z) 1 Hp(z) = 4√2 = 1 + √3 1 - √3 3 + √3 -(3 - √3) (3 3) 123 A 1 + 4√2 3 - √3 1 - √3 z-1 3 + √3 -(1 + √3) (1 3) 123 B 21 21 Hp(z) = A + B z-1 HpT(z-1) Hp(z) = (AT + BT z)(A + Bz-1) z)(A Bz = (AT A + BTB) + ATBz-1 + BTAz B) AT A 1 1 + √3 1 - √3 1 = (3 4√2 3 + √3 - (3 - √3) 4√2 1 = 32 = 1 + √3 1 - √3 3 + √3 -(3-√3) (4 + 2√3) + (4 - 2√3) (6 + 4√3) - (6 - 4√3) 3) 3) (6 (6 + 4√3) - (6 - 4√3) (12 + 6√3) + (12 - 6√3) 3) (6 3) ¼ √3/4 √3/4 3/4 ¾ 22 22 BT B 1 3 - √3 3 + √3 1 3 - √3 1 - √3 = (1 (1 4 √2 1 - √3 -(1 + √3) 4 √2 3 + √3 - (1 + √3) = = (12 – 6√3) + (12 + 6√3) (6 - 4√3) - (6 + 4√3) (12 3) 3) (4 - 2√3) + (4 + 2√3) (6 3) 3) 32 (6 - 4√3) – (6 + 4√3) 1 ¾ - √3/4 3/4 - √3/4 ¼ ⇒ AT A + BTB = I 23 23 =1 4 √2 1 + √ 3 1 -√ 3 1 3 - √ 3 1 -√ 3 3 + √3 -(3-√3) 4 √2 3 + √3 -(1+√3) 1 = 32 ATB (2 √3) + (-2√3) (-2) – (-2) (2 2) (6) – (6) ( -2 √3) + (2 √3) (6) 3) = BT A = 0 (ATB)T = 0 So HpT(z-1) Hp(z) = I (z) i.e. Hp(z) is a Paraunitary Matrix i.e. (z) Paraunitary 24 24 Modulation domain: 1 H0(z) = P(z) = (-z3 + 9z + 16 + 9z-1 – z-3) 16 1 H0(-z) H0(-z-1) = P(-z) = (z3 – 9z + 16 – 9z-1 + z-3) 9z 16 H0(z-1) So H0(z) H0(z-1) + H 0(-z) H0(-z-1) = 2 i.e. |H0(ω)|^2 + |H0(ω + π)|^2 = 2 25 25 Magnitude Response of Daubechies 4-tap filter. Magnitude response of Daubechies 4-tap filter. 2.5 Frequency response phase 2 1.5 1 0.5 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 Angular frequency (normalized by π) 0.6 0.8 1 26 26 Phase response of Daubechies 4-tap filter. Phase response of Daubechies 4-tap filter. 4 3 Frequency response phase 2 1 0 -1 -2 -3 -4 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 Angular frequency (normalized by π) 0.6 0.8 1 27 27 ...
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This note was uploaded on 12/04/2011 for the course ESD 18.327 taught by Professor Gilbertstrang during the Spring '03 term at MIT.

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