Course and Wavelets and Filter Banks

Transcription

1 Course and Wavelets and Filter Banks Refinement Equation: Iterative and Recursive Solution Techniques; Infinite Product Formula; Filter Bank Approach for Computing Scaling Functions and Wavelets

2 Solution of the Refinement Equation N φ(t) = 2 2 h 0 [k] φ(2t-k) k = 0 First, note that the solution to this equation may not always exist! The existence of the solution will depend on the discrete-time time filter h 0 [k]. If the solution does exist, it is unlikely that φ(t) will have a closed form solution. The solution is also unlikely to be smooth. We will see, however, that if h 0 [n] is FIR with h 0 [n] = 0 outside 0 n N then φ(t) has compact support: φ(t) = 0 outside 0 < t < N 2

3 Approach 1 Iterate the box function φ(t) φ (0) (t) = box function on [0, 1] φ (i + I) N (i + I) (t) = 2 2 h 0 [k] φ (i) (2t k) k = t If the iteration converges, the solution will be given by lim φ (i) (t) i (i) (t) This is known as the cascade algorithm. 3

4 Example: suppose h 0 [k] = {¼,{ ½, ¼} φ (i+1) (t) = ½ φ (i) (2t) + φ (i) (2t 1) + ½ φ (i) (2t 2) Then φ (0) φ (2) (t) (0) (t) φ (0) t (t) (2) (t) (t) (0) (t) φ (3) 1 1 ½ t (t) (3) (t) t t Converges to the hat function on [0, 2] 4

5 Approach 2 Use recursion First solve for the values of φ(t) at integer values of t. Then solve for φ(t) at half integer values, then at quarter integer values and so on. This gives us a set of discrete values of the scaling function at all dyadic points t = n/2 i. At integer points: N φ(n) = 2 h 0 [k] φ (2n k) k = 0 5

6 Suppose N = 3 3 φ(0) = 2 h 0 [k] φ(-k) k = 0 3 φ(1) = 2 h 0 [k] φ(2-k) k = 0 3 φ(2) = 2 h 0 [k] φ(4-k) k = 0 3 φ(3) = 2 h 0 [k] φ(6-k) k = 0 Using the fact that φ(n) = 0 for n < 0 and n > N, we can write this in matrix form as φ(0) h 0 [0] φ(0) φ(1) h 0 [2] h 0 [1] h 0 [0] φ(1) = φ(2) 2 h 0 [3] h 0 [2] h 0 [1] φ(2) φ(3) h 0 [3] φ(3) 6

7 Notice that this is an eigenvalue problem λφ = AΦA where the eigenvector is the vector of scaling function values at integer points and the eigenvalue is λ = 1. Note about normalization: Since (A - λi) Φ = 0 has a non-unique solution, we must choose an appropriate normalization for Φ The correct normalization is φ(n) = 1 n This comes from the fact that we need to satisfy the partition of unity condition, φ(x-n) = 1. n 7

8 At half integer points: φ (n/2) = 2 h 0 [k] φ (n-k) So, for N = 3, we have N k = 0 φ(1/2) h 0 [1] h 0 [0] φ(0) φ(3/2) = 2 h 0 [3] h 0 [2] h 0 [1] h 0 [0] φ(1) φ(5/2) φ(2) h 0 [3] h 0 [2] φ(3) 8

9 Scaling Relation and Wavelet Equation in Frequency Domain φ(t) = 2 h 0 [k] φ(2t k) k φ(t)e -iω t dt = 2 h 0 [k] φ(2t k) e -iω t dt k = 2 h 0 [k] ½ k = h 0 [k]e k [k]e -iωk/2 φ(τ)e iω(τ + k)/2 dτ k/2 φ(τ) ) e - iωτ Ωτ/2 dτ 9

10 LMN LMN ^ i.e. φ(ω) ) = H ^ 0 ( Ω Ω ). φ( 2 ) 2 = H 0 ( Ω ). H 0 ( Ω ). ^ φ ( Ω ) o Ω = Π H 0 ( ) φ^ (0) j = 1 2 j QU` ^ φ(0) = φ(t) dt = 1 (Area is normalized to 1) 10

11 So ^ φ(ω) ) = Π H 0 j = 1 ( ) Infinite Product Formula Ω ( ) 2 j Similarly w(t) = 2 h 1 [k] φ(2t k) leads to k Ω w(ω) ) = H ^ 1 φ ^ ( ) Ω 2 ( ) Desirable properties for H 0 (ω): 2 H(0) = 1, so that ^ φ(0) = 1 H(ω) ) should decay to zero as ω π, so that φ(ω) ^ 2 d Ω < 11

12 Computation of the Scaling Function and Wavelet Filter Bank Approach y 0 [n] y 2 H 0 (ω) 1 [n] y 2 H 0 (ω) 2 [n] Y 1 (ω) Y 2 (ω) 2 H 0 (ω) y 3 [n] Y 3 (ω) x 0 [n] 2 H 1 (ω) x 1 [n] 2 H 1 (ω) x 2 [n] Normalize so that h 0 [n] = 1. n 2 H 1 (ω) 12

13 i. Suppose y 0 [n] = δ[n] and x k [n] = 0. Y 0 (ω)) = 1 Y 1 (ω)) = Y 0 (2ω) ) H 0 (ω)) = H 0 (ω) Y 2 (ω)) = Y 1 (2ω) ) H 0 (ω)) = H 0 (2ω)H 0 (ω) Y 3 (ω)) = Y 2 (2ω) ) H 0 (ω)) = H 0 (4ω) ) H 0 (2ω) ) H 0 (ω) After K iterations: K-1 Y K (ω)) = Π H 0 (2 k ω) k=0 What happens to the sampling period? Sampling period at input = T 0 = 1 (say) Sampling period at output = T K = ½ K 13

14 Treat the output as samples of a continuous time signal, y c (t), with sampling period ½ K : y K [n] = K 1 2 K K y c (n/2 K ) ^ Y K (ω)) = Y c (2 K ω) ) ; -π ω π K (yk(t) c is chosen to be bandlimited) Replace 2 K ω with Ω: ^ K-1 K k=0 Y c (Ω)) = Y K ( Ω/2 k ) = Π H 0 ( Ω/2 K-k ) = Π H 0 ( Ω/2 j ) ; K j=1-2 K π Ω 2 K π So ^ lim Y c (Ω)) = Π H 0 ( Ω/2 j ^ ) = φ(ω) k K j=1 14

15 ii. So 2 K y K [n] converges to the samples of the scaling function, φ(t), taken at t = n/2 K. Suppose y 0 [n] = 0, x 0 [n] = δ[n] and all other x k [n] = 0 Then K-2 Y K (ω)) = H 1 (2 K- 1 ω) Π H 0 (2 k ω) k=0 ^ K-2 Y c (Ω)) = Y K (Ω/2 K Ω ) = H 1 ( ) Π H 0 (Ω/2 K- k ) K Ω 1 = H 1 ( ) Π H 0 (. Ω ) ^ lim Y c (Ω)) = H 1 (Ω/2) φ ^ (Ω/2) = w(ω) ^ K K 2 K y K [n] converges to the samples of the wavelet, 2 2 k=0 K-1 j=1 2 2 j w(t), taken at t = n/2 K. 15

16 Support of the Scaling Function y k-1 [n] 2 v[n] h 0 [n] y k [n] length {v[n]} = 2 length {y{ k-1 [n]} - 1 Suppose that h 0 [n] = 0 for n < 0 and n > N length {y{ k [n]} = length {v[n]} + length {h 0 [n]} 1 = 2 length {y{ K-1 [n]} + N 1 Solve the recursion with length {y 0 [n]} = 1 So length {y{ k [n]} = (2 K 1)N

17 i.e. length {y{ c (t)} = T K. length {y{ K [n]} K = (2 K 1) N + 1 = N - 2 K N-1 2 K φ(t) lim K length {φ(t)}{ = N 0 N t So the scaling function is supported on the interval [0, N] 17

18 18

19 Matlab Example 6 Generation of orthogonal scaling functions and wavelets MATLAB M-file MATLAB M-file

20 By Inverse DWT 20

21 By Recursion 21

22 Comparison 22

23 Matlab Example 7 Generation of biorthogonal scaling functions and wavelets. MATLAB M-file

24 Primary Daub 9/7 Pair 24

25 Dual Daub 9/7 Pair 25