Course and Wavelets and Filter Banks. Filter Banks (contd.): perfect reconstruction; halfband filters and possible factorizations.

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Wilfred Ferguson
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1 Course and Wavelets and Filter Banks Filter Banks (contd.): perfect reconstruction; halfband filters and possible factorizations.

2 Product Filter Example: Product filter of degree 6 P 0 (z) = (-1( 1 + 9z z z -4 - z -6 ) P 0 (z) P 0 (- z) = 2z -3 Expect perfect reconstruction with a 3 sample delay Centered form: 1 16 P(z) = z 3 P 0 (z) = (-( z 3 + 9z z -1 z -3 ) P(z) + P(- z) = 2 i.e. even part of P(z) = const In the frequency domain: P(ω) ) + P(ω + π) ) = 2 Halfband Condition 2

3 2 P(ω) Note antisymmetry about ω = π/2 0 π 2 π ω P(ω) ) is said to be a halfband filter. How do we factor P 0 (z) into H 0 (z) F 0 (z)? P 0 (z) = 1/16(11 16(1 + z -1 ) 4 ( z -1 - z -2 ) = -1/16(1 + z -1 ) 4 (2 + 3 z -1 )(2-3 z -1 ) 3

4 So P 0 (z) has zeros at z = -11 (4 th order) z = 2 ± 3 3 Note: = 4 th order zero at z = -1 Im P 0 (z) Re 4

5 Some possible factorizations H 0 (z) (or F 0 (z) ) F 0 (z) (or H 0 (z) ) (a) 1-1/16(1 + z - 1 ) 4 ( z - 1 )(2-3 - z - 1 ) (b) ½(1 + z - 1 ) -1/8(1 + z - 1 ) 3 ( z - 1 )(2-3 - z - 1 ) (c) ¼(1 + z - 1 ) 2-1/4(1 + z - 1 ) 2 ( z - 1 )(2-3 - z - 1 ) (d) ½(1 + z - 1 )( z - 1 ) -1/8(1 + z - 1 ) 3 (2-3 - z - 1 ) (e) 1/8(1 + z - 1 ) 3-1/2(1 + z - 1 )( z - 1 )(2-3 - z - 1 ) (f) (g) ( 3 1) 4 2 (1 + z - 1 ) 2 ( z - 1 ) (1 + z - 1 ) 2 (2-3 - z - 1 ) 1/16(1 + z - 1 ) 4 -( z - 1 )(2-3 - z - 1 ) ( 3( 1) 5

6 Case (b) -- Symmetric filters (linear phase) 3rd order filter length = 2 filter length = 6 ½{ { 1, 1 } ¹/8 {-1, 1, 8, 8, 1, -1} 6

7 Case (c) -- Symmetric filters (linear phase) 2 nd order -1 2 nd order filter length = 3 filter length = 5 ¼ { 1, 2, 1 } ¼ { -1, 2, 6, 2, -1} 7

8 Case (f) -- Orthogonal filters (minimum phase/maximum phase) 2 nd order nd order filter length = 4 filter length = , 3+ 3, 3, 3-3, 3 3, , 3-3, 3 3, 3+ 3, 3, Note that, in this case, one filter is the flip (transpose) of the other: f 0 [n] = h 0 [3 - n] F 0 (z) = z -3 H 0 (z -1 )

9 General form of product filter (to be derived later): 1 + z z P(z) = 2( ) p ( 1 + z ) p ( p + k z )( ) k ( 1 z -1 ) k P 0 (z) = z -(2p 1) P(z) 2-1 p - 1 p - 1 k = 0 = (1 + z -1 ) 1 ( )(-1) z -(p - 1) + k 1 z ( ) 2 k k = Binomial Q(z) (spline) Cancels all odd powers filter except z (2p (2p-1) k k P 0 (z) has 2p zeros at π (important for stability of iterated filter bank.) Q(z) factor is needed to ensure perfect reconstruction. 9

10 p = 1 P 0 (z) has degree 2 leads to Haar filter bank. 1, 1, 1, z , z z z , 0 F 0 (z) = 1 + z - 1, H 0 (z) = Synthesis lowpass filter has 1 zero at π Leads to cancellation of constant signals in analysis highpass channel. Additional zeros at π would lead to cancellation of higher order polynomials. 1 + z 1 + z

11 p = 2 P 0 (z) has degree 4p 2 = 6 P 0 (z) = (1 + z -1 ) { ( ) z -1 2 ( )( ) 2 } = (1 + z -1 ) 4 ( z -1 - z -2 ) = {-{ 1 + 9z z z -4 z -6 } 1 1 z th -1 th order Possible factorizations 1/8 trivial 2/6 linear phase 3/5 4/4 orthogonal (Daubechies-4)

12 p = 4 P 0 (z) has degree 4p 2 = 14 8 th th order -1 12

13 Common factorizations (p = 4): (a) 9/7 Known in Matlab as bior4.4 4 th order -1 4 th order -1 13

14 (b) 8/8 (Daubechies( 8) -- Known in Matlab as db4 4 th order -1 4 th order -1 14

15 Why choose a particular factorization? Consider the example with p = 2: i. One of the factors is halfband The trivial 1/8 factorization is generally not desirable, since each factor should have at least one zero at π. However, the fact that F 0 (z) is halfband is interesting in itself. V(z) X(z) Y(z) 2 F 0 (z) Let F 0 (z) be centered, for convenience. Then F 0 (z) = 1 + odd powers of z Now X(z) = V(z 2 ) = even powers of z only 15

16 So Y(z) = F 0 (z) X(z) = X(z) + odd powers y[n] = x[n] ; n even 678 f 0 [k]x[n k] ; n odd k odd f 0 [n] is an interpolating filter Another example: f 0 [n] = (ideal bandlimited interpolating filter) sin π ( ) πn 2 n x[n] y[n] n n 16

17 ii. Linear phase factorization e.g. 2/6, 5/3 Symmetric (or antisymmetric) ) filters are desirable for 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( i(ω α + θ) real delays all 0 if symmetric frequencies π if by α samples 2 antisymmetric Linear phase may not necessarily be the best choice for audio applications due to preringing effects. 17

18 iii. Orthogonal factorization This leads to a minimum phase filter and a maximum 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: H 0 (z) = (1 + z -1 ) 2 [(2 + 3) z -1 ] 1 = {(1 + 3) + (3 + 3)z -1 + (3-3)z -2 + (1-3)z -3 } F 0 (z) = (1 + z -1 ) 2 [(2-3) z -1 ] 4( 3-1) 3-1 = z -3 (1 + z 2 )[(2 + 3) - z] 4 2 = z - 3 H 0 (z - 1 ) 18

19 P(z) = z l P 0 (z) = H 0 (z) H 0 (z -1 ) From alias cancellation condition: H 1 (z) = F 0 (-z) = -z -3 H 0 (-z -1 ) F 1 (z) = -H 0 (-z) = z -3 H 1 (z -1 ) 19

20 Special Case: Orthogonal Filter Banks Choose H 1 (z) so that H 1 (z) = - z -N H 0 (- z -1 ) N odd Time domain h 1 [n] = (-( 1) n h 0 [N n] F 0 (z) = H 1 (- z) = z -N H 0 (z 1 ) f 0 [n] = h 0 [N n] F 1 (z) = - H 0 (- z) = z -N H 1 (z -1 ) f 1 [n] = h 1 [N n] So the synthesis filters, f k [n], are just the time-reversed versions of the analysis filters, h k [n], with a delay. 20

21 Why is the Daubechies factorization orthogonal? Consider the centered form of the filter bank: x[n] H 0 [z] H 1 [z] 2 2 Analysis bank causal only negative powers of z y 0 [n] y 1 [n] 2 H 0 (z -1 ) 2 H 0 (z -1 ) Synthesis bank anticausal only positive powers of z x[n] no delay in centered form 21

22 In matrix form: Analysis Synthesis y o y 1 = L B x x = L T B T W 123 W T W y 0 y 1 So x = W T W x for any x W T W = I = WW T An important fact: symmetry prevents orthogonality 22

23 Matlab Example 2 1. Product filter examples 23

24 Degree-2 2 (p=1): pole-zero plot 24

25 Degree-2 2 (p=1): Freq. response 25

26 Degree-6 6 (p=2): pole-zero plot 26

27 Degree-6 6 (p=2): Freq. response 27

28 Degree-10 (p=3): pole-zero plot 28

29 Degree-10 (p=3): Freq. response 29

30 Degree-14 (p=4): pole-zero plot 30

31 Degree-14 (p=4): Freq. response 31

MULTIRATE DIGITAL SIGNAL PROCESSING

MULTIRATE DIGITAL SIGNAL PROCESSING Signal processing can be enhanced by changing sampling rate: Up-sampling before D/A conversion in order to relax requirements of analog antialiasing filter. Cf. audio