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by: Evert Christiansen


Marketplace > Texas A&M University > Mathematics (M) > MATH 414 > FOURIER SERIES WAVELET
Evert Christiansen
Texas A&M
GPA 3.92

Francis Narcowich

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Francis Narcowich
Class Notes
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This 5 page Class Notes was uploaded by Evert Christiansen on Wednesday October 21, 2015. The Class Notes belongs to MATH 414 at Texas A&M University taught by Francis Narcowich in Fall. Since its upload, it has received 15 views. For similar materials see /class/226026/math-414-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15
Math 414 501 Spring 2003 1 Notes for April 8 2003 Last time We discussed Mallat7s multiresolution analysis MRA gave some examples and covered Theorem 59 We concluded with a formula for the wavelet 1 Before reading these notes you should review what we covered The wavelet The formula for the wavelet is constructed from the two scale relation be Zpk gtlt2z e k 1 k Here to keep things simple we will assume that the pk7s are all real and that only 190191132133 are different from 0 That is we will assume that W Zpk lt2z k P0 lt2 mm 7 1 p2 lt2z 2 p3 lt2z 3 k0 If we translate Ms by Z units right ifZ gt 0 or left ifZ lt 0 then translated function is z 7 Z These are important because the set 7 Z67 is an orthonormalL basis for the space V5 The pk7s are not just random numbers Theorem 59 implies that they satisfy at least four conditions which we will explicitly write out for our case 1 100721100 10172491 10272492 10372 103 5am any integer Z 2 pfp p p 2 3 P0101 P2103 2 4 P0102 1and101103 1 These conditions are not independent For example setting Z 0 in the rst implies the second Moreover since we are assuming pk 0 for k lt 0 and k gt 3 only Z i1 results in an equation which isnt identically 0 In fact 1To even use the word orthonormal requires an inner product The one that we are using here is that of L2lR for realvalued functions namely lt97hgt1jo 9Ihzdzi these two values of Z result in the same equation7 100102 p1p3 0 Finally7 the fourth condition obviously implies the third For the present7 the conditions that concern us are p p p p 2 and pop2p1ps 0 Once we know x we know an orthonormal basis for all of the spaces Vj ln particular7 we know that 2x 7 is an orthonormal basis for V1 The wavelet space W0 is de ned to be all functions in V1 that are orthogonal to the entire space Vb Syrnbolically7 W0 w 6 V1 ltw7fgt 0 for all f E Vb Our aim is to construct a function 714 such that the set ww 7 is an orthonormal basis for the space W0 To do this7 we use our basis for V1 to expand 1 We qu lt2z e k and then we take the inner product of 7 with 3 21 W g Zpkgsm i k 7 2e Z pk721 295 7 k k0 k2 and set the result to 07 obtaining WWL Z 210mm 2P1Q2z1 2P2Q2z2 2P3Q2z3 0 2 Let7s look at Z 0 In that case7 we have 10090 10191 10212 103 0 There is a basic trick to nding qk7s that satisfy the equation Write the pk7s backward and alternate signs That is7 19 s 190 191 p2 pg 173 pg 7192 pl 100 Taking the inner product of the two vectors amounts to multiplying the vertical entries and adding the result Notice there is a cancellation that occurs among the outer pairs and inner pairs7 giving 0 overall This suggests that we use qo 1037 Q1 7102 qz pl7 qs 7100 and qk 0 otherwise 2 Let7s check the Z 1 case From 2 and our choice of q7s7 we have WM 4596 7 1gt 2190191 2p17po 2020 21930 0 The same reasoning for Z 71 gives us WWL ZW 1gt 21000 21010 2102103 2103i 02 0 The other Z all follow because the qk7s involved are all 0 Thus7 if 7 has the form M95 103M295 102M295 1 0145296 i 2 100M296 3 3 then we know that it is in W0 It is easy to show that if we translate 714 to wz 7 m7 where m E Z then wz 7 m is also in W0 We can say even more Using 3 and the orthonorrnality of 2 7 with a little work one can show that ww 7 is an orthonormal set for W0 We leave doing this as an exercise Showing that it is also a basis for M is done in an appendix in the book We now return to the general case discussed in the text Just as for the Haar wavelet7 gkw 2727272 7 is an orthonormal basis for the jth level wavelet space We again have the decomposition V1 V771 69 Wj717 V771 i W171 There is one other very important fact that we want to mention The collection of all of functions g x 2j2w2jx7kjkeg is an orthonormal basis for the whole space of signals7 BUR We will discuss decomposition and reconstruction in class Math 414 501 Spring 2003 1 Notes for April 15 2003 The Daubechies7 Wavelets We want to nd the pk7s scaling coef cients in the Daubechies7 N 2 case This is equivalent to nding the function 1 132 i 1 00 1012 10222 10323 1 Let us make some comments about Pz in the the general case There are three conditions that Pz must satisfy 1 113w2 1P7212 E1 121 1 2 P1 1 3 1Pe 1 gt 0for1t1 7T2 Note that 1 with z 1 gives 1P112 1P7112 1 By 2 P1 1 and so 12 1P7112 1 from which it follows that Pi1 0 When there are only a nite number of non zero pk7s P is a polynomial Since 2 71 is a root of P we see that Pz has 2 1N for some N as a factor that is 132 2 1NP27 H71 7e 07 where is the product of the remaining factors of P after dividing out 2 1 an appropriate number of times For the case in equation 1 P is a cubic The values N can have as thus 1 2 or 3 It turns out that N 1 gives the Haar case p0 p1 1 p2 pg 0 and N 3 doesnt work We thus assume that 132 2 12Cv 62 where a and B are also assumed to be real From 2 1 1 12a 6 so a B 14 Hence we see that P has the form 132 2 1214 i 6 62 The question remaining is does P satisfy 1 and 3 To begin we will try to nd a B for which 1 is satis ed We do this simply by nding a value that works for z z39 17 and check to see if it works for all z with 121 1 We have 132 7 1 Z5214 7 6 6239 7 22114 7 B 6239 7 725 12 7 26h Sirnilarly7 P7z39 72B 7 12 7 2 z39 Consequently7 1132 1134 2926 212 7 252 1662 7 46 12 Since the left side is 1 by 17 we end up with 1662 7 46 12 1 or 1662 7 46 712 0 The roots of this equation are Bi lisi It turns out that both values of 6 provide appropriate pk7s ln fact7 the scaling functions they lead to are related to one another by a simple re ection of the z axis about the line z 32 If we choose the 7 then 132 31 22lt1 1 7 V32 1 1 3 3 3 37 3 17 3 7lt f fz422423gtt 2 4 4 4 4 HF W HF HF PO 171 P2 P3 These are the pk7s given in the text Showing that 132 satis es 1 in our list requires some algebra7 but is not really very hard Verifying 3 is even easier The only points at which 0 are precisely the roots of P narnely7 z 71 a double root and z 1ng w 37 The root at z 71 6 has angle t 7139 gt 7T27 so 3 holds in that case The root at z w 37 has gt 17 so 3 holds there as well Thus7 for all ltl S 7T27 we have that lPe itl gt 0


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