Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Iteration step of Wavelets - Essay - Physics, Essays (high school) of Physics

The theory, in places, starts to look harder than it is in practice. Hang in there... I will do a lot of examples along the way. The text does this material in a lot of generality (in addition to doing everything with convolutions so as to insure a fast algorithm). This makes the text hard to read and understand at points. I'm going to stick to a special construction called \repeated lters". See p.216 of the text

Typology: Essays (high school)

2011/2012

Uploaded on 03/15/2012

janeka
janeka 🇺🇸

4.1

(15)

260 documents

1 / 7

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
110.312 - Introduction to Wavelets (Metcalfe)
Fall 2001
October 24, 2001
Lecture 16 - The Iteration Step for Wavelets on ZN
The theory, in places, starts to look harder than it is in practice. Hang in there... I will do a lot
of examples along the way. The text does this material in a lot of generality (in addition to doing
everything with convolutions so as to insure a fast algorithm). This makes the text hard to read
and understand at points. I’m going to stick to a special construction called “repeated filters”. See
p.216 of the text. This is the construction used in many (if not most) examples, including almost
every case that I’ve ever seen. It is, also, this case which extends to wavelets on all of R.
The basic idea here is to iterate the first-stage wavelet basis. If Nis divisible by 2n, we will
continue to iterate for nsteps. The corresponding basis at each point will be called the pth stage
wavelet basis.
The main question is: How can we construct a first-stage wavelet basis for Vn1based on that
of Vn?
Lemma 1 (The Folding Lemma) Suppose MN,N= 2M, and u1`2(ZN) = CN. Define
u2`2(ZN
2
) = CN
2by
u2(n) = u1(n) + u1n+N
2
Then,
ˆu2(m) = ˆu1(2m)
Proof:
1
pf3
pf4
pf5

Partial preview of the text

Download Iteration step of Wavelets - Essay - Physics and more Essays (high school) Physics in PDF only on Docsity!

110.312 - Introduction to Wavelets (Metcalfe)

Fall 2001

October 24, 2001

Lecture 16 - The Iteration Step for Wavelets on ZN

The theory, in places, starts to look harder than it is in practice. Hang in there... I will do a lot

of examples along the way. The text does this material in a lot of generality (in addition to doing

everything with convolutions so as to insure a fast algorithm). This makes the text hard to read

and understand at points. I’m going to stick to a special construction called “repeated filters”. See

p.216 of the text. This is the construction used in many (if not most) examples, including almost

every case that I’ve ever seen. It is, also, this case which extends to wavelets on all of R.

The basic idea here is to iterate the first-stage wavelet basis. If N is divisible by 2n, we will

continue to iterate for n steps. The corresponding basis at each point will be called the pth^ stage

wavelet basis.

The main question is: How can we construct a first-stage wavelet basis for V n−^1 based on that

of V n?

Lemma 1 (The Folding Lemma) Suppose M ∈ N, N = 2M , and u 1 ∈ `^2 (ZN ) = CN^. Define

u 2 ∈ `^2 (Z (^) N 2

) = C

N (^2) by

u 2 (n) = u 1 (n) + u 1

n +

N

Then,

ˆu 2 (m) = ˆu 1 (2m)

Proof:

In the second sum below, make the substitution k = n +

N 2 =^ n^ +^ M^. Then,

uˆ 2 (m) =

M∑ − 1

n=

u 2 (n)e

− 2 πimn/M

M∑ − 1

n=

u 1 (n)e−^2 πimn/M^ +

M∑ − 1

n=

u 1

n +

N

e−^2 πimn/M

M∑ − 1

n=

u 1 (n)e−^2 πimn/M^ +

2 M∑ − 1

k=M

u 1 (k)e−^2 πi(k−M^ )m/M

M∑ − 1

n=

u 1 (n)e−^2 πimn/M^ +

N∑ − 1

k=M

u 1 (k)e−^2 πikm/M

N∑ − 1

k=

u 1 (k)e

− 2 πikm/M

N∑ − 1

k=

u 1 (k)e−^2 πik(2m)/N

= ˆu 1 (2m)

You can iterate the above result to get:

Corollary 2

Suppose that N is divisible by 2 . Define u^2 (Z (^) N 2−^1

) by

u`(n) =

2 `∑−^1 − 1

k=

u 1

n +

kN

2 `−^1

Then,

ˆu(m) = ˆu 1 (2−^1 m)

You should try to prove the above corollary as an exercise. Just use the Folding Lemma and

induction.

The Folding Lemma, then, gives us the answer to our question:

Theorem 3 Suppose that N is divisible by 4 and that u 1 , v 1 ∈ `^2 (ZN ) = CN^ are such that

{R 2 ku 1 }

(N/2)− 1 k=0 ∪ {R^2 kv^1 }

(N/2)− 1 k=

is a first-stage wavelet basis for ^2 (ZN ) = CN^. If u 2 , v 2 ∈^2 (Z N 2

) = C

N (^2) are given by:

u 2 (n) = u 1 (n) + u 1

n +

N

v 2 (n) = v 1 (n) + v 1

n +

N

v =

Let’s try to write the signal

z = (3, 8 , 5 , 2 , 4 , 10 , − 3 , −1)

with respect to the second-stage Haar wavelet basis.

Let’s begin by first writing this vector with respect to the first-stage Haar basis:

{u, R 2 u, R 4 u, R 6 u, v, R 2 v, R 4 v, R 6 v}

Recall that since this is an orthonormal basis, the components will be given by the inner

products:

< z, u >, < z, R 2 u >, < z, R 4 u >, < z, R 6 u >, < z, v >, < z, R 2 v >, < z, R 4 v >, < z, R 6 v >

So, we see that:

z →

Now, let’s recurse on the trend signal

a 1 =

Recall, that we will be writing this with respect to the new basis given by the folding lemma.

That is, we define:

u 2 (n) = u(n) + u(n + 4)

v 2 (n) = v(n) + v(n + 4)

Thus,

u 2 =

v 2 =

and the basis that we want to write a 1 in terms of is:

{u 2 , R 2 u 2 , v 2 , R 2 v 2 }

Since this is orthonormal, we can, again, just look at inner products.

a 1 =

Thus,

z →

Thus, z written with respect to the second stage Haar wavelet basis is: (

9 , 5 , 2 , 9 ,

  1. (Shannon basis) Let N = 8. Let’s write

z = (1, 0 , − 1 , 0 , 1 , 0 , − 1 , 0) ∈ C^8 = `^2 (Z 8 )

with respect to the second-stage Shannon wavelet basis.

Recall that the Shannon father wavelet u and the Shannon mother wavelet v are given (re- spectively) by:

uˆ = (

vˆ = (0, 0 ,

We start by writing z with respect to the first-stage wavelet basis. To do so, we’ll need to

compute the inner products (since the basis is orthonormal):

< z, u >, < z, R 2 u >, < z, R 4 u >, < z, R 6 u >, < z, v >, < z, R 2 v >, < z, R 4 v >, < z, R 6 v >

By Parseval’s relation, we could instead calculate:

< z,ˆ u >,ˆ

< ˆz, (R 2 u)ˆ >,

< ˆz, (R 4 u)ˆ >,

< z,ˆ (R 6 u)ˆ >,

< z,ˆ v >,ˆ

< z,ˆ (R 2 v)ˆ >,

< z,ˆ (R 4 v)ˆ >,

< z,ˆ (R 6 v)ˆ >

So, we need to find ˆz. Notice that

zeven = (1, − 1 , 1 , −1)

and

zodd = (0, 0 , 0 , 0)

Here, zeven = 2E 2. Thus, since ˆw(m) =

N < w, Em > and the fact that the Em’s form an

orthonormal set, we have: zˆeven = (0, 0 , 4 , 0)

On the other hand, we have:

ˆzodd = (0, 0 , 0 , 0)

Thus, by the Fast Fourier transform, we have:

zˆ = (0, 0 , 4 , 0 , 0 , 0 , 4 , 0)

Recall that

(R 2 ku)ˆ(m) = e−^2 πim(2k)/N^ ˆu(m)

Thus,

< z, u >=

< z,ˆ u >ˆ =

< z, R 2 u >=

< z,ˆ (R 2 u)ˆ >=

e

2 πi(12)/ 8 = −

< a 1 , R 2 v 2 >=

< ˆa 1 , (R 2 v 2 )ˆ >=

2 e−^2 πi(2)(1)/^4 ) = 1

Thus, z written with respect to the second-stage Shannon basis is:

(