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An in-depth analysis of orthogonal filter banks, paraunitary matrices, and their conditions in the time domain, polyphase domain, and modulation domain. It also discusses the unitary matrix and its properties. A useful resource for students studying the course 18.327 and 1.130: wavelets and filter banks.
Typology: Slides
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The constant complex matrix A is said to be unitary if
†
example:
μμ μμ 2
1 -i
i -
μ μμ
μ 2
1 i
-i -
μμμμ 2
-1 i
-i 1
† =
T
μμμμ 2
1 -i
i -
†
òòò
w ww
w ww
w ww
www www
The matrix function H(z) is said to be paraunitary if
it is unitary for all values of the parameter z
T
(z
) H(z) = I for all z ò 0 -----------------(1)
Frequency Domain:
T
(-w) H(w) = I for all w
or H*
T
(w) H(w) = I
Note: we are assuming that h[n] are real.
3
4
Centered form (PR with no delay):
y
0
[n]
x[n]
x[n]
h
0
[n]
h
1
[n]
éé éé 2
éééé 2
åååå 2
å åå
å 2
h
0
[-n]
h
1
[-n]
y
1
[n]
Synthesis bank = transpose of analysis bank
h
0
[n] causal Ω ΩΩ
Ω f
0
[n] ô ôô
ô h
0
[-n] anticausal
7
x[-3]
x[-2]
x[-1]
x[0]
x[1]
x[2]
x[3]
x[4]
x[5]
x[6]
h
0
[3]
h
0
[2]
h
0
[1] h
0
[3]
h
0
[0] h
0
[2]
h
0
[1] h
0
[3]
h
0
[0] h
0
[2]
h
0
[1] h
0
[3]
h
0
[0] h
0
[2]
h
0
[1]
h
0
[0] 4 44
4
h
1
[3]
h
1
[2]
h
1
[1] h
1
[3]
h
1
[0] h
1
[2]
h
1
[1] h
1
[3]
h
1
[0] h
1
[2]
h
1
[1] h
1
[3]
h
1
[0] h
1
[2]
h
1
[1]
h
1
[0]
y
0
[0]
y
0
[1]
y
0
[2]
y
0
[3]
4444
4 44
4
y
1
[0]
y
1
y
1
[2]
y
1
[3]
T
. .
4444
8
Orthogonality condition (Condition O) is
T
T
ΩΩΩΩ W orthogonal matrix
Block Form:
T
T
T
T
T
T
T
= I ΩΩΩΩ ƒƒƒƒ h
0
[n] h
0
[n œ 2k] = dddd[k] --------------(4)
n
T
= 0 ΩΩΩΩ ƒƒƒƒ h
0
[n] h
1
[n œ 2k] = 0 --------------(5))))
n
T
Ω ƒ ƒƒ
ƒ h
1
[n] h
1
[n œ 2k] = d dd
d[k] --------------(6)
n
êêê
ddd ƒƒƒ ƒƒƒ
ƒƒƒ ??? ???
? ??
Good choice for h
1
[n]:
h
1
[n] = (-1)
n
h
0
[N-n] ; N odd --------------(7)
Alternating flip
Example: N = 3
h
1
[0] = h
0
h
1
[1] = -h
0
h
1
[2] = h
0
h
1
[3] = -h
0
With this choice, Equation (5) is automatically satisfied:
k = -1: h
0
[0]h
0
[1] - h
0
[1]h
0
k = 0: h
0
[0]h
0
[3] - h
0
[1]h
0
[2] + h
0
[2]h
0
[1] œ h
0
[3]h
0
k = 1: h
0
[2]h
0
[3] œ h
0
[3]h
0
k = ê2: no overlap
9
Also, Equation (6) reduces to Equation (4)
d[k] = ƒ h
1
[n] h
1
[n-2k] = ƒ (-1)
n
h
0
[N-n] (-1)
n-2k
h
0
[N-n+2k]
n n
= ƒ h
0
[?] h
0
[? + 2k]
?
So, Condition O on the lowpass filter + alternating flip
for highpass filter lead to orthogonality
10
&'(&'(&'( &'(&'(&'(
0,even
(z) H
0,even
(z
0,odd
(z) H
0,odd
(z
Express Condition O as a condition on H
0,even
(z),
0,odd
(z):
Frequency domain:
13
úúúúH
0,even
(wwww)úúúú
2
0,odd
(wwww)úúúú
2
The alternating flip construction for H
1
(z) ensures
that the remaining conditions are satisfied.
0
(z) = H
0,even
(z
2
) + z
0,odd
(z
2
1
(z) = -z
-N
0
(-z
) alternating flip
= -z
-N
0,even
(z
) - z H
0,odd
(z
= -z
-N
0,even
(z
) + z
-N+
0,odd
(z
&'( &'(
z
1,odd
(z
2
1,even
(z
2
So
1,even
(z) = z
(-N+1)/
0,odd
(z
1,odd
(z) = -z
(-N+1)/
0,even
(z
0,even
(z) H
1,even
(z
0,odd
(z) H
1,odd
(z
and H
1,even
(z) H
1,even
(z
1,odd
(z) H
1,odd
(z
14
15
x[n]
éééé 2
éééé 2
0
(z)
y
0
[n]
y
1
[n]
0
(z
åååå 2
å åå
å 2
x[n]
1
(z
1
(z)
PR conditions:
0
(z) H
0
(z
1
(z) H
1
(z
0
(-z) H
0
(z
1
(-z) H
1
(z
0
(z
1
(z
0
(z) H
0
(-z)
1
(z) H
1
(-z)
&'(&'(&'(&'(
m
(z) modulation matrix
No
distortion
Alias
cancellation
16
Replace z with œz in Equations (10) and (11)
0
(-z) H
0
(-z
1
(-z) H
1
(-z
0
(z) H
0
(-z
1
(z) H
1
(-z
0
(z
1
(z
0
(-z
1
(-z
0
(z) H
0
(-z)
1
(z) H
1
(-z)
&'( &'(&'(
&'(
&'(&'(&'(&'( &'( &'(&'(
&'(
m
T
(z
m
(z) 2I
Condition O:
m
T
(z
m
(z) = 2I ΩΩΩΩ H
m
(z) is paraunitary
