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This document delves into the concepts of wavelets and filter banks, focusing on modulation and polyphase matrices. It covers topics such as noble identities, block toeplitz matrices, and the perfect reconstruction condition in the polyphase domain.
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2
Matrix form of PR conditions:
Matrix form of PR conditions:
0
0
(z) F
(z) F
1
1
(z)] H
(z)] H
0
0
(z) H
(z) H
0
0
(--z) = [ 2z
z) = [ 2z
l
l
11
(z) H
(z) H
11
(--z)z)
So
So
0
0
(z) F
(z) F
1
1
(z)] = [2z
(z)] = [2z
m
m
(z)
(z)
mm
–1– 1
(z) =
(z) =
11
123
123
Modulation matrix,
Modulation matrix, HH
m
m
(z)
(z)
00
(z) H
(z) H
11
(--z)z) -- H
00
(--z) Hz) H
11
(z) (must be non
(z) (must be non--zero)zero)
1
1
(--z)z) --HH
0
0
(--z)z)
11
(z) H
(z) H
00
(z)
(z)
?
?
4
4
Complete the second row of matrix PR conditions
Complete the second row of matrix PR conditions
by replacing z with
by replacing z with ––z:z:
0
0
(z) F
(z) F
1
1
(z) H
(z) H
0
0
(z) H
(z) H
0
0
(--z) z
z) z
0
0
(--z) Fz) F
1
1
(--z) H
z) H
1
1
(z) H
(z) H
1
1
(--z) 0
z) 0 (
(--z)z)
l
l
Synthesis
Synthesis
modulation
modulation
matrix, F
matrix, F
m
m
(z)
(z)
123
123
Note the transpose convention in F
Note the transpose convention in F
mm
(z).
(z).
5
5
Consider
Consider
U(z) = H(z
U(z) = H(z
2
2
)X(z)
)X(z)
Y(z) = ½ {U(z
Y(z) = ½ {U(z
½½
) + U(--zz
½½
)} (downsamplingdownsampling))
= ½ { H (z) X (z
= ½ { H (z) X (z
½
½
) + H(z) X
) + H(z) X
z
z
½
½
= H(z)
= H(z) •• ½ {½ {XX (z(z
½½
) + X (--zz
½½
can
can downsampledownsample
first
first
First Noble identity:
First Noble identity:
x [n]
x [n]
x [n]
x [n]
H(z
H(z
22
y[n]
y[n] u[n]
u[n]
2 H(z)
H(z)
y[n]
y[n]
y[n]
x y[n]
x[n][n]
H(z
H(z
22
7
7
Filtering and
Filtering and downsamplingdownsampling::
x
x[n] y[n]
[n] y[n]
H(z)
H(z)
H(z) =
H(z) = HH
eveneven
(z
(z
22
) + z
) + z
-1- 1
oddodd
(z
(z
22
); hh
eveneven
[n] = h[2n]
[n] = h[2n]
h
h
oddodd
[n] = h[2n+1]
[n] = h[2n+1]
x
x[n][n]
++
eveneven
(z
(z
2
2
oddodd
(z
(z
2
2
y[n]y[n]
z
z
8
8
x
x[n][n]
z
z
-1- 1
eveneven
(z
(z
22
oddodd
(z(z
2
2
y[n]
y[n]
z
z
-1 1
eveneven
(z)(z)
odd
odd
(z)
(z)
y[n]
y[n]
x
x[n][n]
Polyphase
Polyphase
Form
Form
x
x
eveneven
[n]
[n]
x
x
odd
odd
[n
[n--1]1]
10
10
even
even
(z
(z
2
2
oddodd
(z
(z
22
y[n]
y[n]
x
x[n][n]
eveneven
(z)
(z)
oddodd
(z)
(z)
y[n]
y[n]
x
x[n][n]
y
y
eveneven
[n]
[n]
y
y
oddodd
[n]
[n]
z
z
Polyphase
Polyphase
Form
Form
z
z
11
11
Consider the matrix corresponding to the analysis
Consider the matrix corresponding to the analysis
filter bank in
filter bank in interleavedinterleaved form. This is a blockform. This is a block
Toeplitz
Toeplitz matrix:matrix:
bb
h
h
00
[3] h
[3] h
00
h
h
1
1
[3] h
[3] h
1
1
h
h
0
0
[1] h
[1] h
0
0
h
h
1
1
[1] h
[1] h
1
1
0 0 h
0 0 h
0
0
[3] h
[3] h
0
0
[2] h
[2] h
0
0
[1] h
[1] h
0
0
0 0 h
0 0 h
11
[3] h
[3] h
11
[2] h
[2] h
11
[1] h
[1] h
11
4--tap Exampletap Example
13
13
Similarly, for the synthesis filter bank:
Similarly, for the synthesis filter bank:
bb
f
f
0
0
[0] f
[0] f
1
1
f
f
0
0
[1] f
[1] f
1
1
f
f
00
[2] f
[2] f
11
f
f
0
0
[3] f
[3] f
1
1
f
f
0
0
[0] f
[0] f
1
1
f
f
0
0
[1] f
[1] f
1
1
f
f
00
[2] f
[2] f
11
f
f
0
0
[3] f
[3] f
1
1
14
14
Note transpose
Note transpose
convention for
convention for
synthesis
synthesis
polyphase matrix
polyphase matrix
f
f
00
[0] f
[0] f
11
f
f
00
[1] f
[1] f
11
f
f
00
[2] f
[2] f
11
f
f
0
0
[3] f
[3] f
1
1
z
z
-1- 1
pp
(z) =
(z) =
0,even0,even
[z] F
[z] F
1,even1,even
[z]
[z]
0, odd0, odd
[z] F[z] F
1, odd1, odd
[z][z]
Perfect reconstruction condition in polyphase domain:
Perfect reconstruction condition in polyphase domain:
p
p
(z) H
(z) H
p
p
(z) = I (centered form
(z) = I (centered form))
This means that H
This means that H
pp
(z) must be invertible for all z on the
(z) must be invertible for all z on the
unit circle, i.e.
unit circle, i.e.
det
det HH
pp
(ee
i
i ω
ω
0 for all frequencies
0 for all frequencies ω
ω .
16
16
h
h
0,even0,even
[n] = h
[n] = h
00
[2n]
[2n]
00
(z) = H(z) = H
0,even0,even
(z(z
2
2
) + z) + z
-1 1
0,odd0,odd
(z(z
2
2
h
h
0,odd
0,odd
[n] = h
[n] = h
0
0
[2n+1]
[2n+1]
1
1
(z) = H
(z) = H
1,even
1,even
(z
(z
22
) + z
) + z
1,odd
1,odd
(z
(z
22
Two more equations by replacing z with
Two more equations by replacing z with --z.z.
So in matrix form:
So in matrix form:
0
0
(z) H
(z) H
0
0
(--z) H
z) H
0,even
0,even
(z
(z
22
0,odd
0,odd
(z
(z
22
11
(z) H
(z) H
11
(--z) H
z) H
1,even1,even
(z
(z
22
1,odd1,odd
(z
(z
22
) z
) z
-zz
mm
(z)
(z)
Modulation matrix
Modulation matrix
p
p
(z
(z
22
Polyphase matrix
Polyphase matrix
14243
14243
17
17
But
But
z
z
-1 1
-zz
-1 1
z
z
-1 1
123
123
22
Delay Matrix 2
Delay Matrix 2--point DFT Matrixpoint DFT Matrix
22
(z)
(z)
N
N
w
w w
w
22
w
w
N-N-1 1
w
w
2
2
w
w
4
4
w
w
2(N
2(N--1)1)
w
w
N
N--1 1
w
w
2(N
2(N--1)1)
w
w
(N
(N--1)1)
22
; w =
; w = ee
ii
N--point DFTpoint DFT
Matrix
Matrix
NN
N
N
1
1
N
N
Complex conjugate: replace
Complex conjugate: replace ww withwith ww = e= e
NN
2
2 π
π
N
N
2
2 π
π
19
19
Example: Daubechies 4
Example: Daubechies 4--tap filtertap filter
3 ) z
3 ) z
z
z
z
z
-3- 3
h
h
0
0
00
(z) =
(z) =
11
(z) =
(z) =
z
z
-1 1
3)z
3)z
-2 2
3)z
3)z
-3 3
h
h
0
0
h
h
0
0
[2] = h
h
00
20
20
Time domain:
Time domain:
h
h
0
0
2
2
h
h
0
0
2
2
h
h
0
0
2
2
h
h
0
0
2
2
1
1
h
h
0
0
[0] h
[0] h
0
0
[2] + h
[2] + h
0
0
[1] h
[1] h
0
0
1
1
i.e. filter is orthogonal to its double shifts
i.e. filter is orthogonal to its double shifts
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