Filter Banks: time domain
Filter Banks: time domain
(
(Haar
Haar example) and frequency domain;
example) and frequency domain;
conditions for alias cancellation
conditions for alias cancellation
and no distortion
and no distortion
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Perfect Reconstruction Filter Banks: Haar Example and Conditions for Alias Cancellation, Slides of Banking and Finance
The concept of perfect reconstruction filter banks using the haar example. It explains the conditions for alias cancellation and no distortion in the context of filter banks. The document also provides the simplest example of a two-channel fir perfect reconstruction filter bank.
Typology: Slides
2012/2013
1 / 21
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Filter Banks: time domain
Filter Banks: time domain
(HaarHaar example) and frequency domain;example) and frequency domain;
conditions for alias cancellation
conditions for alias cancellation
and no distortion
and no distortion
22
Simplest (non
Simplest (non--trivial) example of a two channel FIRtrivial) example of a two channel FIR
perfect reconstruction filter bank.
perfect reconstruction filter bank.
h
h
0
0
[n]
[n]
h
h
1
1
[n]
[n]
fl
fl 2
fl
fl 2
x[n]
x[n]
y
y
0
0
[n]
[n]
y
y
1
1
[n]
[n]
Analysis
Analysis
r
r
0
0
[n]
[n]
r
r
1
1
[n]
[n]
f
f
0
0
[n]
[n]
f
f
1
1
[n]
[n]
x[n]
x[n]
^
^
Synthesis
Synthesis
v
v
0
0
[n]
[n]
v
v
1
1
[n]
[n]
t
t
1
1
[n]
[n]
t
t
00
[n]
[n]
h
h
0
0
[n] =
[n] =
f
f
0
0
[n] =
[n] =
44
Matrix form
Matrix form
M
M
M
M
M
M
y
y
0
0
[0]
[0]
y
y
00
[1]
[1]
M
y
y
1
1
[0]
[0]
y
y
1
1
[1]
[1]
M
M
L
L
L
L
L
L
L
L
L
L
L
L
L
L
L
L
M
M
x[
x[--1]1]
x[0]
x[0]
x[1]
x[1]
x[2]
x[2]
l
l
y
y
o
o
y
y
1
1
L
L
B
B
x
x
55
Synthesis
Synthesis
y
y
0
0
[n/2] n even
[n/2] n even
t
t
0
0
[n] =
[n] = upsampler
upsampler
0 n odd
0 n odd
v
v
0
0
[n] =
[n] = ( t
( t
0
0
[n + 1] + t
[n + 1] + t
0
0
[n])
[n]) lowpass
lowpass filterfilter
y
y
0
0
[n/2] n even
[n/2] n even
y
y
00
[ ] n odd
[ ] n odd
11
�
� 2
2
1
1
�
� 2
2
1
1
�
� 2
2
n + 1n + 1
2
2
77
i.e.
i.e.
x[2n
x[2n--1] = (y
1] = (y
00
[n]
[n] –– yy
11
[n]) = x[2n
[n]) = x[2n--1]1]
x[2n] = (y
x[2n] = (y
00
[n] + y
[n] + y
11
[n]) = x[2n]
[n]) = x[2n]
So x[n] = x[n]
So x[n] = x[n] �
Perfect reconstruction!
Perfect reconstruction!
In general, we will make all filters causal, so we will
In general, we will make all filters causal, so we will
have
have
x[n] = x[n
x[n] = x[n –– nn
00
]
]
PR with delay
PR with delay
1
1
�
� 2
2
^
^
1
1
�
� 2
2
from
from j
j and
and k
k
^
^
^
^
^
^
88
Matrix form
Matrix form
M
M
M
= M
x[
x[--1]1]
x[0]
x[0]
x[1]
x[1]
x[2]
x[2]
M
M
^
^
^
^
^
^
^
^
y
y
0
0
[0]
[0]
y
y
0
0
[1]
[1]
M
M
M
M
y
y
11
[0]
[0]
y
y
1
1
[1]
[1]
M
M
M
M
M
M M
M
L
L
L
L
M
M
M
M
L
L LL
M
M M
M
M
M
M
M
M
M
^
^
x = L
x = L
T
T
B
B
T
T
y
y
0
0
y
y
1
1
m
m
1010
Perfect Reconstruction Filter Banks
Perfect Reconstruction Filter Banks
General two
General two--channel filter bankchannel filter bank
H
H
0
0
(z)
(z)
H
H
1
1
(z)
(z)
fl
fl 2
fl
fl 2
x[n]
x[n]
y
y
0
0
[n]
[n]
y
y
11
[n]
[n]
r
r
0
0
[n]
[n]
r
r
1
1
[n]
[n]
F
F
00
(z)
(z)
F
F
1
1
(z)
(z)
x[n]
x[n]
^
^
v
v
0
0
[n]
[n]
v
v
1
1
[n]
[n]
t
t
1
1
[n]
[n]
t
t
0
0
[n]
[n]
L
L
L
L
z
z--transform definition:transform definition:
X(z) =
X(z) = �
x[n]z
x[n]z
- -nn
Put z =
Put z = ee
i
i w
w
to get DTFT
to get DTFT
¥
¥
n=n=--¥¥
1111
Perfect reconstruction requirement:
Perfect reconstruction requirement:
x[n] = x[n
x[n] = x[n -- l
l ] (
] (
l
l time delays)
time delays)
X(z) = z
X(z) = z
l
l
X(z)
X(z)
H
H
0
0
(z) and H
(z) and H
1
1
(z) are normally
(z) are normally lowpasslowpass andand highpasshighpass,,
but not ideal
but not ideal
^
^
^
^
Downsampling
Downsampling operation in each channel canoperation in each channel can
produce
produce aliasingaliasing
p
p
p
p 0
0 p
p
p
p
w
w
H
H
11
w
w )
H
H
00
w
w )
H
H
1
1
w
w )
1313
Suppose X(
Suppose X( w
w ) = 1 (input has all frequencies)
) = 1 (input has all frequencies)
Then R
Then R
00
w
w ) = H
) = H
00
w
w ), so that after
), so that after downsamplingdownsampling we havewe have
Y
Y
00
w
w ) =
p
p
p
p
p
p
w
w
½RR
00
½RR
0
0
½RR
0
0
w
w
2
2
w
w
22
w
w
2
2
p
p
aliasing
aliasing
Goal is to design F
Goal is to design F
00
(z) and F
(z) and F
11
(z) so that the overall
(z) so that the overall
system is just a simple delay
system is just a simple delay -- with nowith no aliasingaliasing term:term:
V
V
0
0
(z) + V
(z) + V
1
1
(z) = z
(z) = z
- ll
X(z)
X(z)
1414
V
V
0
0
(z) = F
(z) = F
0
0
(z) T
(z) T
0
0
(z)
(z)
= F
= F
0
0
(z) Y
(z) Y
0
0
(z
(z
2
2
(upsamplingupsampling))
= ½½FF
0
0
(z){ H
(z){ H
0
0
(z) X(z) + H
(z) X(z) + H
0
0
(--z) X(z) X(--z)}z)}
V
V
11
(z) =
(z) = ½½FF
11
(z){ H
(z){ H
11
(z) X(z) + H
(z) X(z) + H
11
(--z) X(z) X(--z)}z)}
So we want
So we want
½ {F{F
0
0
(z) H
(z) H
0
0
(z) + F
(z) + F
1
1
(z) H
(z) H
1
1
(z) } X(z)
(z) } X(z)
- = zz
l
l
X(z)
X(z)
½ {F{F
0
0
(z) H
(z) H
0
0
(--z) + Fz) + F
1
1
(z) H
(z) H
1
1
(--z) } X(z) } X(--z)z)
1616
What happens in the time domain?
What happens in the time domain?
F
F
00
(z) = H
(z) = H
11
(--z) F
z) F
00
w
w ) = H
) = H
11
w
w
p
p )
h
h
1
1
[n] (
[n] (--z)z)
-nn
n
n
h
h
11
[n] z
[n] z
-nn
So the filter coefficients are
So the filter coefficients are
f
f
0
0
[n] = (
[n] = (--1)1)
nn
h
h
1
1
[n] alternating signs
[n] alternating signs
f
f
11
[n] = (
[n] = (--1)1)
n+
n+
h
h
00
[n] rule
[n] rule
Example
Example
h
h
0
0
[n] = { a
[n] = { a
0
0
, a
, a
1
1
, a
, a
2
2
} f
} f
0
0
[n] = { b
[n] = { b
0
0
, --bb
1
1
, b
, b
2
2
h
h
1
1
[n] = { b
[n] = { b
0
0
, b
, b
1
1
, b
, b
2
2
} f
} f
1
1
[n] = {
[n] = {--aa
0
0
, a
, a
1
1
, --aa
2
2
nn
n
n
1717
Product Filter
Product Filter
Define
Define
P
P
0
0
(z) = F
(z) = F
0
0
(z) H
(z) H
0
0
(z)
(z)
Substitute F
Substitute F
11
(z) =
(z) = --HH
00
(--z) , Hz) , H
11
(z) = F
(z) = F
00
(--z)z)
in the zero distortion condition (Equation
in the zero distortion condition (Equation j
j )
F
F
0
0
(z) H
(z) H
0
0
(z)
(z) -- FF
0
0
(--z) Hz) H
0
0
(--z) = 2zz) = 2z
- ll
i.e. P
i.e. P
0
0
(z)
(z) -- PP
0
0
(--z) = 2zz) = 2z
- l
l
Note:
Note: l
l must be odd since LHS is an odd function.
must be odd since LHS is an odd function.
m
m
n
n
1919
Design Process
Design Process
Design P(z) to satisfy Equation
Design P(z) to satisfy Equation p
p
. This gives . This gives
P
P
0
0
(z). Note: P(z) is designed to be
(z). Note: P(z) is designed to be lowpasslowpass..
Factor P
Factor P
0
0
(z) into F
(z) into F
0
0
(z) H
(z) H
0
0
(z). Use Equations
(z). Use Equations l
l to
to
find H
find H
11
(z) and F
(z) and F
11
(z).
(z).
Note: Equation
Note: Equation p
p requires all even powers of z
requires all even powers of z
(except z
(except z
00
) to be zero:
) to be zero:
p[n]z
p[n]z
-nn
p[n](
pnz)
-nn
p[n] =
p[n] =
nn nn
1 ; n = 0
1 ; n = 0
0 ; all even n (n
0 ; all even n (n „
2020
For odd n, p[n] and
For odd n, p[n] and ––p[n] cancel.p[n] cancel.
The odd coefficients, p[n], are free to be designed
The odd coefficients, p[n], are free to be designed
according to additional criteria.
according to additional criteria.
Example:
Example: HaarHaar filter bankfilter bank
H
H
0
0
(z) =
(z) = (1 + z
(1 + z
-1 1
) H
) H
1
1
(z) = (
(z) = (1 –– zz
-1 1
F
F
00
(z) = H
(z) = H
11
(--z) =z) = (1 + z
(1 + z
-1 1
F
F
1
1
(z) =
(z) = --HH
0
0
(--z) = (
z) = (1-- zz
-1 1
P
P
0
0
(z) = F
(z) = F
0
0
(z) H
(z) H
0
0
(z) = (1 + z
(z) = (1 + z
-1- 1
22
1
1
�
� 2
2
1
1
�
� 2
2
11
�
� 2
2
-1- 1
�
� 2
2
1
1
2
2