Suppose
hn=0
h
n
0
for
n<0
n
0
and
n>N−1
n
N
1
. Then the support of
φt
φ
t
is
0
N−1
0
N
1
as we saw before . How can we find
φt
φ
t
from
hn
h
n
?
φt=2∑n=0N−1hnφ2t−n
φ
t
2
n
0
N
1
h
n
φ
2
t
n
It turns out that we can find samples of
φt
φ
t
using the following procedure.

For example, if
hn
h
n
is of length
N=4
N
4
, then from the dilation equation we have
φ0=2(h0φ0+h1φ-1+h2φ-2+h3φ-3)
φ
0
2
h
0
φ
0
h
1
φ
-1
h
2
φ
-2
h
3
φ
-3
φ1=2(h0φ2+h1φ1+h2φ0+h3φ-1)
φ
1
2
h
0
φ
2
h
1
φ
1
h
2
φ
0
h
3
φ
-1
φ2=2(h0φ4+h1φ3+h2φ2+h3φ1)
φ
2
2
h
0
φ
4
h
1
φ
3
h
2
φ
2
h
3
φ
1
φ3=2(h0φ6+h1φ5+h2φ4+h3φ3)
φ
3
2
h
0
φ
6
h
1
φ
5
h
2
φ
4
h
3
φ
3
As we know that
φt
φ
t
is supported on
0
N−1
0
N
1
(
0
3
0
3
for this example), we can see that
φ-1
φ
-1
,
φ4
φ
4
, etc fall outside the support. Therefore these
equations can be simplified.
φ0=2h0φ0
φ
0
2
h
0
φ
0
φ1=2(h0φ2+h1φ1+h2φ0)
φ
1
2
h
0
φ
2
h
1
φ
1
h
2
φ
0
φ2=2(h1φ3+h2φ2+h3φ1)
φ
2
2
h
1
φ
3
h
2
φ
2
h
3
φ
1
φ3=2h3φ3
φ
3
2
h
3
φ
3
or in matrix form,
φ0φ1φ2φ3=2(
h0000
h2h1h00
0h3h2h1
000h3
)φ0φ1φ2φ3
φ
0
φ
1
φ
2
φ
3
2
h
0
0
0
0
h
2
h
1
h
0
0
0
h
3
h
2
h
1
0
0
0
h
3
φ
0
φ
1
φ
2
φ
3
Remember that
φt
φ
t
is zero outside the range
0
3
0
3
.

This matrix equation reveals that the vector of values of
φt
φ
t
on the integers
0…N−1
0
…
N
1
is an eigenvector of the matrix
P=2(
h0000
h2h1h00
0h3h2h1
000h3
)
P
2
h
0
0
0
0
h
2
h
1
h
0
0
0
h
3
h
2
h
1
0
0
0
h
3
Specifically, it is an eigenvector corresponding to the eigenvalue
11. However, not just any set of
coefficients will give a matrix
PP with
11 as an eigenvalue.

It turns out that PP will always
have 11 as an eigenvalue whenever
the coefficients
hn
h
n
satisfy the condition

∑nh2n=∑nh2n−1=12
n
h
2
n
n
h
2
n
1
1
2

(1)
In other words, whenever the even terms sum up to

12
1
2
, and likewise for the odd terms, we are sure that

PP has

11 as an eigenvalue. This can be
shown as follows. Note that if

∑nh2n=∑nh2n−1=12
n
h
2
n
n
h
2
n
1
1
2
then

11112(
h0000
h2h1h00
0h3h2h1
000h3
)=1111
1
1
1
1
2
h
0
0
0
0
h
2
h
1
h
0
0
0
h
3
h
2
h
1
0
0
0
h
3
1
1
1
1
That is because the columns of

PP
consist of the even and odd terms of

hn
h
n
. Therefore, the vector of ones is an eigenvector of

PT
P
with eigenvalue

11.
Because the transpose

PT
P
has the same eigenvalues as

PP,

λP=λPT
λ
P
λ
P
, the matrix

PP also has

11 as an eigenvalue.

Even though PP and
PT
P
have the same eigenvalues, they do not have the same
eigenvectors.

As a result, if the condition is satisfied, we can calculate the
integer samples of
φt
φ
t
by calculating the eigenvector of
PP corresponding to the eigenvalue
11.

We can also write the equation as follows:
2(
h0000
h2h1h00
0h3h2h1
000h3
)φ0φ1φ2φ3−φ0φ1φ2φ3=0000
2
h
0
0
0
0
h
2
h
1
h
0
0
0
h
3
h
2
h
1
0
0
0
h
3
φ
0
φ
1
φ
2
φ
3
φ
0
φ
1
φ
2
φ
3
0
0
0
0
or
(2(
h0000
h2h1h00
0h3h2h1
000h3
)−(
1000
0100
0010
0001
))φ0φ1φ2φ3=0000
2
h
0
0
0
0
h
2
h
1
h
0
0
0
h
3
h
2
h
1
0
0
0
h
3
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
φ
0
φ
1
φ
2
φ
3
0
0
0
0
(P−I)φ=0
P
I
φ
0

If vv is an eigenvector of
PP, then so is
αv
α
v
.

The solution is only unique up to a scaling. Likewise, the
dilation equation itself does not impose a normalization on

φt
φ
t
.

Recall from the sum-integral equality that
∫φtdt=∑kφk
t
φ
t
k
φ
k
.
Therefore, if we want the normalization
∫φtdt=A
t
φ
t
A
, say, we can get it by the normalization
∑kφk=A
k
φ
k
A
Therefore, we can add this equation to the matrix equation we
already have.

(
P−I
(
1111
)
)φ=0A
P
I
1
1
1
1
φ
0
A

(2)
Now we have one more equation than unknowns, however, the new
equation is consistent with the existing equations. It only
normalizes the solution. Solving this linear system of
equations gives the integer samples of

φt
φ
t
.

Notice that the matrix PP is a
convolution matrix, but with every other row removed. It can be
constructed by using the `convmtx`

command
and then removing every other row:

```
H = convmtx(h(:),N);
P = sqrt(2)*H(1:2:2*N-1,:);
```

We can get more samples of
φt
φ
t
by using the dilation equation.
φt=2∑khkφ2t−k
φ
t
2
k
h
k
φ
2
t
k
To find the value of
φt
φ
t
on half integers
n2
n
2
, for
n∈Z
n
, let
t=n2
t
n
2
in the dilation equation.

φn2=2∑khkφn−k=2hn*φn
φ
n
2
2
k
h
k
φ
n
k
2
h
n
φ
n

(3)
According to the dilation equation, the value of

φt
φ
t
on the half-integers can be obtained by convolving
the scaling filter

hn
h
n
with the integer-samples of

φt
φ
t
, as illustrated in

Figure 1.

Similarly, we can get samples of
φt
φ
t
on the quarter-integers by setting
t=n4
t
n
4
in the dilation equation.

φn4=2∑khkφn2−k=2∑khkgn−2k=2[↑2]hn*gn=2[↑2]hn*φn2
φ
n
4
2
k
h
k
φ
n
2
k
2
k
h
k
g
n
2
k
2
[↑2]
h
n
g
n
2
[↑2]
h
n
φ
n
2

(4)
where

gn≔φn2
≔
g
n
φ
n
2
.
The notation

[↑2]xn
[↑2]
x
n
represents an

upsampled version of

xn
x
n
. Recalling that the transfer function of

[↑2]hn
[↑2]
h
n
is

Hz2
H
z
2
we get the following diagram (

Figure 2).

The value of
φt
φ
t
for
t=n8
t
n
8
is found as

φn8=2∑khkφn4−k=2∑khkgn−4k=2[↑4]hn*gn=2[↑4]hn*φn4
φ
n
8
2
k
h
k
φ
n
4
k
2
k
h
k
g
n
4
k
2
[↑4]
h
n
g
n
2
[↑4]
h
n
φ
n
4

(5)
where

gn≔φn4
≔
g
n
φ
n
4
, with corresponding diagram (

Figure 3),

Similarly,
φn16=2[↑8]hn*φn8
φ
n
16
2
[↑8]
h
n
φ
n
8
Given the value of
φt
φ
t
on the integers, a simple method emerges for finding
φt
φ
t
on the dyadic rationals. In general,
we have Equation 6 and Figure 4.

φn2j+1=2
[
↑
2
j
]
hn*φn2j
φ
n
2
j
1
2
[
↑
2
j
]
h
n
φ
n
2
j

(6)
Let hh be a vector of the scaling
filter coefficients. Let pp be a
vector of samples of
φt
φ
t
on the integers. Then the samples of
φt
φ
t
on the dyadic rationals are obtained by the
following simple loop.

```
for k = 1:5
p = sqrt(2) * conv(h,p);
h = up(h,2);
end
```

We also have the diagram in Figure 5,

Using the Z-transform, we can write

𝒵φn2j=2j2HzHz2⋯Hzj−1𝒵φn
𝒵
φ
n
2
j
2
j
2
H
z
H
z
2
⋯
H
z
j
1
𝒵
φ
n

(7)
Note that
hn
h
n
satisfies the sum rule

∑nh2n=∑nh2n−1
n
h
2
n
n
h
2
n
1

(8)
if and only if

Hz
H
z
has

z-1+1
z
1
as a factor,

Hz=Qz(z-1+1)
H
z
Q
z
z
1
To see this, note that if

Hz=Qz(z-1+1)
H
z
Q
z
z
1
, then

H−1=0
H
1
0
. Writing this out, we get

Hz=∑nhnz−n
H
z
n
h
n
z
n

(9)
H-1=∑nhn−1n
H
-1
n
h
n
1
n

(10)
H-1=∑nh2n−∑nh2n−1
H
-1
n
h
2
n
n
h
2
n
1

(11)
So the condition in

Equation 8 can be
written as

H-1=0
H
-1
0
, or equivalently as

Hz=Qz(z-1+1)
H
z
Q
z
z
1
.

To summarize, the sum rule (Equation 8) can be written in various forms.
∑nh2n=∑nh2n−1⇔∑n−1nhn=0
⇔
n
h
2
n
n
h
2
n
1
n
1
n
h
n
0
⇔H-1=0
⇔
H
-1
0
⇔Hz=Qz(z-1+1)
⇔
H
z
Q
z
z
1
⇔(z-1+1) divides Hz
⇔
z
1
divides
H
z
If we note the following,
∑nh2n=∑nh2n−1=12⇔(∑nhn=2)∧(∑n−1nhn=0)
⇔
n
h
2
n
n
h
2
n
1
1
2
n
h
n
2
n
1
n
h
n
0
then we can state the previous result in the following form.
(∑nhn=2)∧(∑n−1nhn=0)⇒1 is an eigenvalue of P
n
h
n
2
n
1
n
h
n
0
1
is an eigenvalue of
P
Or in terms of
Hz
H
z
:
(H1=2)∧(H-1=0)⇒1 is an eigenvalue of P
H
1
2
H
-1
0
1
is an eigenvalue of
P