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Unformatted text preview: Course 18.327 and 1.130 Course Wavelets and Filter Banks
Filter Banks (contd.): perfect Filter reconstruction; halfband filters and possible factorizations. possible Product Filter
Example: Product filter of degree 6 P0(z) = P0(z) (z)
1 16 (1 + 9z2 + 16z3 + 9z4  z6) 16z P0( z) = 2z3 ⇒ Expect perfect reconstruction with a 3 sample delay Centered form: P(z) = z3P0(z) = P(z) + P( z) = 2
1 16 ( z3 + 9z + 16 + 9z1 – z3) i.e. even part of P(z) = const In the frequency domain: P(ω) + P(ω + π) = 2 Halfband Condition Condition
2 2 P(ω) Note antisymmetry Note antisymmetry about ω = π/2 about 0 π 2 π ω P(ω) is said to be a halfband filter. is filter. How do we factor P0(z) into H0(z) F0(z)? P0(z) = 1/16(1 + z1)4(1 + 4z1  z2) = 1/16(1 + z1)4(2 + √ 3 – z1)(2  √3 – z1) (2 )(2
3 So P0(z) has zeros at z = 1 (4th order) z = 2 ± √3 Im Note: 2 + √3 = Note: 1 2 √3 P0(z) 4th order zero at z = 1 1 2√3 1 2 + √3 Re 4 Some possible factorizations
H0(z) (a) (b) (c) (d) (e) (f) (g) (or F0(z) ) F0(z) (or H0(z) ) 1 ½(1 + z1) ¼(1 + z1)2 ½(1 + z1)(2 + √3  z1) )(2 1/8(1 + z1)3
(√3 – 1) 4 √2 1/16(1 + z1)4(2 + √3  z1)(2  √3  z1) (2 )(2 1/8(1 + z1)3(2 + √3  z1)(2  √3  z1) (2 )(2 1/4(1 + z1)2(2 + √3  z1)(2  √3  z1) (2 )(2 1/8(1 + z1)3(2  √3  z1) (2 1/2(1 + z1)(2 + √3  z1)(2  √3  z1) )(2 )(2
√2 4 (√3 – 1) (1 + z1)2(2 + √3  z1) (2 (1 + z1)2(2  √3  z1) (1 (2 1/16(1 + z1)4 (2 + √3  z1)(2  √3  z1) (2 )(2 5 Case (b)  Symmetric filters (linear phase) Case  3rd 3rd order order 1 1 2√3 2 + √3 filter length = 2 filter ½{ 1, 1 } 1, filter length = 6 ¹/8 {1, 1, 8, 8, 1, 1} 1, 6 Case (c)  Symmetric filters (linear phase) Case  2nd order 1 2nd order 1 2√3 2 + √3 filter length = 3 ¼ { 1, 2, 1 } 1, filter length = 5 ¼ { 1, 2, 6, 2, 1} 1, 7 Case (f)  Orthogonal filters Case  Orthogonal (minimum phase/maximum phase) (minimum 2nd order 1 2√3 2nd order 1 2 + √3 filter length = 4 filter length = 4 1 1 1+√3, 3+√3, 3√3, 1√3 4√ 2 1√3, 3√3, 3+√3, 1+√3 4√ 2 Note that, in this case, one filter is the flip (transpose) Note of the other: f0[n] = h0[3  n] of [3 F0(z) = z3 H0(z1) 678 678 678 678 678 678
8 General form of product filter (to be derived later): P(z) = 2(
p1 1 + z p 1 + z 1 p p + k  1)( 1  z )k( 1 – z1)k )( 2 ) ∑ ( 2 k 2 2 k=0 (2p 1) P(z) P0(z) = z(2p –1) P(z) p + k  1 )(1)k z (p  1) + k( 1 – z1 )2k (p ∑( k 2p 2 2p1k = 0 2 1 p1 = (1 + 123 1444444244444443 Q(z) Binomial Cancels all odd powers (spline) except z–(2p1) filter 2p z1)2p P0(z) has 2p zeros at π (important for stability of iterated (z) (important filter bank.) filter Q(z) factor is needed to ensure perfect reconstruction. 9 p=1 P0(z) has degree 2 → leads to Haar filter bank. (z) leads filter 1, 1, 1, 1 Ò
1  z1 1 + z 1 2 Ò ↓2 Ò ↓2 Ò 1, 1 1 + z 1 2 1  z1 Ò Ò ↓2 Ò ↓2 Ò 1 0 Ò 0, 0
1 + z 1 2 F0(z) = 1 + z1 , H0(z) = Synthesis lowpass filter has 1 zero at π Synthesis → Leads to cancellation of constant signals in analysis highpass channel. Additional zeros at π would lead to cancellation of Additional would higher order polynomials. higher 10 10 p=2 P0(z) has degree 4p – 2 = 6 (z) P0(z) = (1 + = =
1 z1)4 8 { ( 1) 0 z1 – ( 1 – z1 2 2 )( )} 1 2 1 16 (1 1 16 {  + z1)4(  1 + 4z 1  z2) 1 + 9z2 + 16z3 + 9z4 – z6} Possible factorizations 1/8 trivial 2/6 linear phase 3/5 4/4 orthogonal 2 + √3 (Daubechies4)
678 678 11 11 1 4th order 2√3 1 p=4 P0(z) has degree 4p – 2 = 14 (z) 8th order 1 12 12 Common factorizations (p = 4): (a) 9/7 Kn own in Matlab own Matlab as bior4.4 4th order 1 4th order 1 13 13 (b) 8/8 (Daubechies 8)  Known in Matlab as db4 Daubechies  Known as 4th order 1 4th order 1 14 14 Why choose a particular factorization? Consider the example with p = 2: i. One of the factors is halfband One halfband The trivial 1/8 factorization is generally not desirable, since each factor should have at least one zero at π. since However, the fact that F0(z) is halfband is interesting (z) in itself. in V(z) V(z) ↑2 X(z) F0(z) Y(z) Y(z) Let F0(z) be centered, for convenience. Then F0(z) = 1 + odd powers of z Now X(z) = V(z2) = even powers of z only
15 15 So Y(z) = F0(z) X(z) (z) = X(z) + odd powers y[n] = x[n] ; n even even 678 ∑ f0[k]x[n – k] ; n odd [k]x[n k] odd
k odd ⇒ f0[n] is an interpolating filter π sin ( 2 ) n sin Another example: f0[n] = [n] πn (ideal bandlimited interpolating filter) • • 2 • • 2 • • • •• • • 4 n • 0 2 y[n] •• • • • • • • • • • 4• n 0 2
16 16 • x[n] ii. Linear phase factorization e.g. 2/6, 5/3 Symmetric (or antisymmetric) filters are desirable for Symmetric antisymmetric many applications, such as image processing. All frequencies in the signal are delayed by the same amount i.e. there is no phase distortion. h[n] linear phase ⇒ A( ω)e–i(ω α + θ) h[n]
real delays all frequencies by α samples by 0 if symmetric π if antisymmetric if antisymmetric 2 Linear phase may not necessarily be the best choice for Linear audio applications due to preringing effects. preringing effects.
17 17 iii. Orthogonal factorization This leads to a minimum phase filter and a maximum This phase filter, which may be a better choice for applications such as audio. The orthogonal factorization leads to the Daubechies family of wavelets – a particularly neat and interesting case. 4/4 factorization: H0(z) =
√3  1 4√2 (1 + z1)2[(2 + √3) – z1] [(2 3) 1 = 4√2 {(1 + √3) + (3 + √3)z1 + (3 √3)z2 + (1 √3)z3} {(1 3) (3 (1 + z1)2[(2  √3) – z1] [(2 3) F0(z) = =  √2 4(√31) √31 z 3 (1 + z2)[(2 + √3)  z] )[(2 3) 4√2
18 18 = z3 H0 (z1) P(z) = zlP0(z) = H0(z) H0(z1) From alias cancellation condition: H1(z) = F0(z) = z3 H0(z1) z) F1(z) = H0(z) = z3 H1(z1) (z) 19 19 Special Case: Orthogonal Filter Banks Choose H1(z) so that H1(z) =  zN H0( z1) (z) Time domain h1[n] = ( 1)n h0[N – n] [N F0(z) = H1 ( z) = zN H0(z–1) ⇒ f0[n] = h0[N – n] [N F1(z) =  H0( z) = zN H1(z1) (z) ⇒f1[n] = h1[N – n] [N So the synthesis filters, fk[n], are just the timereversed So reversed versions of the analysis filters, hk[n], with a delay.
20 20 N odd Why is the Daubechies factorization orthogonal? Why factorization Consider the centered form of the filter bank: H0[z] ↓2 ↓2 y0[n] ↑2 H0(z1) x[n] H1[z] y1[n] Analysis bank causal – only causal negative powers negative of z of ⊕ no delay H0(z1) in centered ↑2 form Synthesis bank anticausal – only anticausal positive powers of z x[n] 21 21 In matrix form: Analysis yo y1 L =
123 Synthesis y0 x x = LT BT y1 B 123
WT W So x = WTW x for any x WTW = I = WWT An important fact: symmetry prevents orthogonality An orthogonality
22 22 Matlab Example 2
1. Product filter examples 23 23 Degree2 (p=1): polezero plot 24 24 Degree2 (p=1): Freq. response 25 25 Degree6 (p=2): polezero plot 26 26 Degree6 (p=2): Freq. response 27 27 Degree10 (p=3): polezero plot 28 28 Degree10 (p=3): Freq. response 29 29 Degree14 (p=4): polezero plot 30 30 Degree14 (p=4): Freq. response 31 31 ...
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 Spring '03
 GilbertStrang

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