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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 ,
- (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:
(
