Lecture 7 Multiresolution Analysis

Transcription

1 David Walnut Department of Mathematical Sciences George Mason University Fairfax, VA USA Chapman Lectures, Chapman University, Orange, CA

2 Outline Definition of MRA in one dimension Finding the wavelet from the scaling function The Daubechies wavelets MRA in higher dimensions

3 Multiresolution Analysis Definition A multiresolution analysis on R is a sequence of subspaces {V j } jz L (R) satisfying: (a) For all j Z, V j V j+1. (b) span{v j } jz = L (R). That is, the set [ jz V j is dense in L (R). (c) \ jz V j = {0}. (d) A function f (x) V 0 if and only if D j f (x) V j. (e) There exists a function '(x), L on R, called the scaling function such that the collection {T n '(x)} is an orthonormal basis for V 0.

4 An MRA is completely determined by the scaling function '(x). Given ' with the property that {T n '(x)} is an orthonormal system, define the subspace V 0 by V 0 = span{t n '(x)}, and the subspaces V j by V j = D j V 0, that is, f V j if and only if D j f V 0. Then verify that (a) (e) hold for this sequence of subspaces. The following lemma holds. Lemma Given ' L (R), the system {T n '(x)} is an orthonormal system if and only if X b'( + n) 1. nz

5 The Haar MRA If we let '(x) =1 [0,1] (x), then the MRA so generated is called the Haar MRA and leads to the construction of the Haar wavelet. In this case, V 0 is the space of scale-0 dyadic step functions, and clearly {'(x n): n Z} is an orthonormal basis for V 0. V j is the space of scale-j dyadic step functions.

6 The Shannon MRA If we let '(x) be defined by b'( )=1 [ 1/,1/] ( ), then the MRA so generated is called the bandlimited MRA and leads to the construction of the Bandlimited wavelet. By the Shannon Sampling Theorem, sin (x n) {'(x n): n Z} = : n Z (x n) is an orthonormal basis for V 0. The space V j consists of those functions bandlimited to the interval [ j 1, j 1 ].

7 The Meyer MRA 8 0 if /3 >< 1 if apple1/3 b'( )= c( 1/) if x (1/3, /3) >: s( + 1/) if x ( /3, 1/3)

8 Because X b'( + n) 1, {'(x n): n Z} is an nz orthornomal system. Define V 0 = span{'(x n): n Z}. We can describe V 0 as follows. ( V 0 = f L (R): b f ( )= X ) c n e in b'( ): (c n ) `(Z) nz

9 Wavelets from MRA Our goal will be to prove the following theorem. Theorem If {V j } is an MRA, then there exists a function L (R) such that { j,k } is an orthonormal wavelet basis for L (R).

10 Outline of proof. For each j we define W j to be the orthogonal complement of V j in V j+1, i.e. V j+1 = V j W j. Find a function (x) with the property that {T k } kz is an orthonormal basis for the space W 0. Then {D j T k } kz is an orthonormal basis for W j. Finally we observe that L (R) = M jz W j so that {D j T k } j,kz is an orthonormal basis for L (R).

11 The dilation equation There exists {h(k)} ` such that '(x) = X k h(k) 1/ '(x k). This equation is referred to as the two-scale dilation equation and the sequence {h(k)} is referred to as the scaling sequence or scaling filter. That ' satisfies such an equation is a simple consequence of the fact that ' V 0 V 1 and that { 1/ '(x k): k Z} is an orthonormal basis for V 1.

12 We may write b'( )=m 0 ( /) b'( /), where m 0 ( )= p 1 X h(k) e ik is called the auxiliary function. k b'( ) = X k h(k)(d T k ')^( ) = X h(k)(d 1/ M k b')( ) k X = h(k) 1/ e in( /) b'( /) k = m 0 ( /) b'( /).

13 Haar case. With '(x) =1 [0,1] (x), '(x) ='(x)+'(x 1) = 1 p ( 1/ '(x)+ 1/ '(x 1)). Therefore, h(k) = ( p 1 if k = 0, 1 0 otherwise

14 m 0 ( )= 1 (1 + e i )=e i cos( )

15 Shannon case solving yields b'( )=1 [ 1/,1/] ( ), b'( )=m 0 ( /) b'( /)

16 m 0 ( )=1 [ 1/4,1/4] ( ) on [ 1/, 1/]

17 Meyer case

18 The auxiliary function Lemma If {T n '(x)} is an orthonormal system and if '(x) satisfies the two-scale dilation equation with scaling filter {h(k)}. Then the auxiliary function m 0 ( ) satisfies m 0 ( ) + m 0 ( + 1/) 1. Proof: 1 = X n = X k b'( + n) = X n + k m 0 b' + k m 0 m 0 + n b' + k + k + 1 b' + n

19 = X k m 0 ( / + k) b'( / + k) + X k m 0 ( / + 1/ + k) b'( / + 1/ + k) = m 0 ( /) X k b'( / + k) + m 0 ( / + 1/) X k b'( / + 1/ + k) = m 0 ( /) + m 0 ( / + 1/).

20 The wavelet recipe. We seek a function such that { (x orthonormal basis for W 0. Since W 0 V 1, k): k Z} is an (x) = X k g(k) 1/ '(x k) or equivalently b ( )=m1 ( /) b'( /) where m 1 ( )= p 1 X g(k) e ik. How does the function m 1 relate to m 0? k

21 Given a function f V 1, we write where f 0 V 0 and g 0 W 0. By our assumptions f = f 0 + g 0 f (x) = X k a(k) 1/ '(x k), f 0 (x) = X k b(k) '(x k), g 0 (x) = X k c(k) (x k).

22 Taking Fourier transforms gives b f ( ) = A b', b f0 ( ) = B( ) b'( ), bg 0 ( ) = C( ) b ( ) where A( ), B( ), and C( ) all have period 1. A b' 4 = B( ) b'( )+C( ) b ( ) = B( ) m 0 b' + C( ) m 1 m 0 m m 0 m B( ) = 4 C( ) A A b'

23 Lemma If m 1 ( )=e i( +1/) m 0 ( 4 is unitary. Moreover, if + 1/) then the matrix m 0 m m 0 m 1 m 0 ( )= p 1 X h(k) e ik and m 1 ( )= 1 X p g(k) e ik k 3 5 k then g(k) =( 1) k h(1 k).

24 Theorem Let {V j } be an MRA with scaling function '(x) and scaling filter h(k). Define the wavelet (x) by (x) = X k ( 1) k h(1 k) 1/ '(x k). Then { j,k (x): j, k Z} = { j/ ( j x k): j, k Z} is a wavelet orthonormal basis on R.

25 Haar wavelet h(k) = Therefore, ( p 1 if k = 0, 1 0 otherwise 8 >< g(k) = >: p1 if k = 0 p1 if k = 1 0 otherwise (x) = 1 p ( 1/ '(x) 1/ '(x 1)) = 1 [0,1/] (x) 1 [1/,1] (x).

26 Shannon wavelet

27 Meyer wavelet

28 Smooth Compactly supported wavelets Theorem Let (x) be such that for some N N, both x N (x) and N+1 b ( ) are in L 1 (R). If { j,k (x)} j,kz is an orthogonal system on R, then R R x m (x) dx = 0 for 0 apple m apple N. Theorem says that smooth wavelets have vanishing moments. We also want wavelets with compact support, which means that m 0 ( ) is a polynomial.

29 Daubechies wavelets Theorem Let '(x) be a compactly supported scaling function associated with an MRA with finite scaling filter h(n). Let (x) be the corresponding wavelet. Then for each N N, Z x k (x) dx = 0 for 0 apple k apple N 1 R if and only if m 0 ( ) can be factored as 1 + e i m 0 ( )= N L( ) for some period 1 trigonometric polynomial L( ).

30 Daubechies s Strategy. We seek a trig polynomial m 0 ( )= p 1 X h(k) e ik satisfying 1 + e i m 0 ( )= and the QMF conditions. We have k N L( ). m 0 ( ) = 1 + e i N L( ) = cos N ( ) L( ). Since L( ) is a real-valued trig polynomial with real coefficients, we arrive at L( )=P(sin ( )) for some polynomial P.

31 This polynomial P must satisfy with P(y) For example, 1 =(1 y) N P(y)+y N P(1 y) 0 for all 0 apple y apple 1. and we arrive at P N 1 (y) = NX 1 k=0 P 0 (y) = 1, N 1 k P 1 (y) = 1 + y, P (y) = 1 + 3y + 6y, y k (1 y) N 1 k. P 3 (y) = 1 + 4y + 10y + 0y 3.

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