Course 18.327 and 1.130
Wavelets and Filter Banks
Filter Banks: time domain
(Haar example) and frequency domain;
conditions for alias cancellation
and no distortion
Haar Filter Bank
Simplest (non-trivial) example of a two channel FIR
perfect reconstruction filter bank.
v0[n]
r0[n]
t0[n]
å
2
é
2
h0[n]
f0[n]
y0[n]
^
x[n]
x[n]
Synthesis
Analysis
t1[n]
r1[n]
v [n]
å2
f1[n] 1
é2
h1[n]
y1[n]
h0[n] =
0
1
1
µ2
1
µ2
1
µ2
1
µ2
f0[n] =
-1
0
2
1
1
µ2
1
µ2
h1[n] =
0
1
f1[n] =
-1
µ2
-1
-1
µ2
0
Analysis:
1
r0[n] = µ2 (x[n] + x[n œ 1])
lowpass filter
y0[n] = r0[2n]
1
y0[n] =
(x[2n] + x[2n œ 1])
µ2
Similarly
1
y1[n] =
(x[2n] œ x[2n œ 1])
µ2
downsampler
-----------------j
j
------------------k
k
3
Matrix form
4
y0[0]
y0[1]
:
:
y1[0]
y1[1]
4
4
3 1
3 0
=
1
µ2
1
0
0
1
0 3
1 3
0
-1
0 3
1 3
:
x[-1]
x[0]
x[1]
4
3 -1 1
3 0 0
4
yo
y1
x[2]
:
:
L
=
B
x
-------------------l
l
4
2
t0[n] =
&)')(
Synthesis
y0[n/2]
upsampler
0
1
µ2
v0[n] =
&)')(
=
n even
n odd
( t0[n + 1] + t0[n])
1
µ2
y0[n/2]
n even
1
µ2
y0[ n + 1 ]
n odd
2
lowpass filter
5
v1[n] =
&)')(
Similarly
1
µ2
y1[n/2]
n even
1
µ2
y1[ n + 1 ]
n odd
-
2
So, the reconstructed signal is
^
x[n]
= v0[n] + v1[n]
&)')(
=
1
µ2
(y0[n/2] + y1[n/2])
n even
1
µ2
(y0[ n + 1] - y1[ n + 1 ])
n odd
2
2
6
3
i.e.
^
x[2n-1]
=
1
µ2
^
x[2n]
1
µ2
µ
(y0[n] œ y1[n]) = x[2n-1]
from j and k
=
^
So x[n]
=
(y0[n] + y1[n]) = x[2n]
Ω Perfect reconstruction!
x[n]
In general, we will make all filters causal, so we will
have
^
x[n]
= x[n œ n0]
Ω PR with delay
7
Matrix form
4
^
x[-1]
^
x[0]
1
=
^
µ2
x[1]
^
x[2]
4
^
x
4 4
-1 0
1 1
4 4
1 0
1 0
3 3
3
0 1
0 1
4 4
0 -1
-1 1
4 4
3
4
y0[0]
y0[1]
4
4
y1[0]
y1[1]
4
y0
=
LT
----------------m
m
BT
y1
8
4
Perfect reconstruction means that the synthesis
bank is the inverse of the analysis bank.
L
^ = x Ω
x
LT
BT
= I
B
&'(
&'(
W-1
W
≈ Wavelet transform ’
÷
Δ matrix
◊
«
In the Haar example, we have the special case
Wœ1 = WT ç orthogonal matrix
So we have an orthogonal filter bank, where
Synthesis bank = Transpose of Analysis bank
f0[n]
=
h0[- n]
=
h1[- n]
f1[n]
9
Perfect Reconstruction Filter Banks
General two-channel filter bank
H0(z)
r0[n]
é2
y0[n]
3
å2
t0[n]
v0[n]
F0(z)
x[n]
H1(z)
r1[n]
é2
y1[n]
3
å2
t1[n]
F1(z)
^
x[n]
§
v1[n]
z-transform definition:
Ñ
X(z) = ƒ x[n]z-n
n=-Ñ
Ñ
Put z = ei w to get DTFT
10
5
Perfect reconstruction requirement:
^
x[n]
= x[n - ?]
^
X(z) = z- ? X(z)
(?? time delays)
H0(z) and H1(z) are normally lowpass and highpass,
but not ideal
H1(w
w)
1
w)
H1(w
w)
H0(w
p
w
0
-p
p
p
2
2
Ω Downsampling operation in each channel can
produce aliasing
-p
p
11
Let‘s see why:
Lowpass channel has
Y0(z) = ²{R0(z²) + R0(-z²)}
(downsampling)
= ²{H0(z²) X (z²) + H0(-z²)X(-z²)}
In frequency domain:
X(z) ç X(w)
w
X(-z) ç X(w
w + p)
p
w
²
X(z ) ç X( 2 )
Y0(w)
w = ²{H0(
w
)
2
X(
w
)
2
or X(eiww)
+ H0(
w
2
+ p)X(
p
w
2
+ p)}
p
12
6
Suppose X(w
w) = 1 (input has all frequencies)
w) = H0(w
w), so that after downsampling we have
Then R0(w
²R0(
w
2
+ p)
²R0( w )
²
2
²R0(
w
2
+ p)
Y0(w
w) =
-p
p
0
p
w
aliasing
Goal is to design F0(z) and F1(z) so that the overall
system is just a simple delay - with no aliasing term:
V0(z) + V1(z) = z- ? X(z)
13
V0(z) = F0(z) T0(z)
= F0(z) Y0(z2)
(upsampling)
= ²F0(z){ H0(z) X(z) + H0(-z) X(-z)}
V1(z) = ²F1(z){ H1(z) X(z) + H1(-z) X(-z)}
So we want
² {F0(z) H0(z) + F1(z) H1(z) } X(z)
+
= z- ? X(z)
² {F0(z) H0(-z) + F1(z) H1(-z) } X(-z)
14
7
Compare terms in X(z) and X(-z):
1) Condition for no distortion (terms in X (z) amount
to a delay)
F0(z) H0(z) + F1(z) H1(z) = 2z- ?
--------------j
j
2) Condition for alias cancellation (no term in X(-z))
F0(z) H0(-z) + F1(z) H1(-z) = 0
--------------k
k
To satisfy alias cancellation condition, choose
F0(z) = H1(-z)
F1(z) = -H0(-z)
----------------------l
l
15
What happens in the time domain?
F0(z) = H1(-z)
F0(w)
w = H1(w
w + p)
p
= ƒ h1[n] (-z)-n
n
= ƒ (-1)n h1[n] z-n
n
So the filter coefficients are
f0[n] = (-1)n h1[n]
f1[n] = (-1)n+1 h0[n]
Example
h0[n] = { a0, a1, a2}
h1[n] = { b0, b1, b2}
alternating signs
rule
f0[n] = { b0, -b1, b2 }
f1[n] = {-a0, a1, -a2}
16
8
Product Filter
Define
P0(z) = F0(z) H0(z)
------------------------------m
m
Substitute F1(z) = -H0(-z) , H1(z) = F0(-z)
in the zero distortion condition (Equation j)
j
F0(z) H0(z) - F0(-z) H0(-z) = 2z- ?
P0(z) - P0(-z) = 2z-?? ----------------------------n
n
i.e.
Note: ? must be odd since LHS is an odd function.
17
Normalized Product Filter
Define
P(z) = z? P0(z)
----------------------------o
o
P(-z) = -z? P0(-z) since ? is odd
So we can rewrite Equation n as
z- ? P(z) + z- ? P(-z) = 2z- ?
i.e.
P(z) + P(-z) = 2
---------------------------p
p
This is the condition on the normalized product filter
for Perfect Reconstruction.
18
9
Design Process
1. Design P(z) to satisfy Equation p
p. This gives
P0(z). Note: P(z) is designed to be lowpass.
2. Factor P0(z) into F0(z) H0(z). Use Equations l to
find H1(z) and F1(z).
Note: Equation p requires all even powers of z
(except z0) to be zero:
ƒ p[n]z-n + ƒ p[n](-z)-n = 2
n
n
&'(
&'(
Ω p[n] =
1 ;
0 ;
n=0
all even n (n ò 0)
19
For odd n, p[n] and œp[n] cancel.
The odd coefficients, p[n], are free to be designed
according to additional criteria.
Example: Haar filter bank
1
H0(z) = µ2 (1 + z-1)
1
H1(z) = µ2 (1 œ z-1)
1
(1 + z-1)
µ2
1
F1(z) = -H0(-z) = µ2 (1- z-1)
1
P0(z) = F0(z) H0(z) = 2 (1 + z-1)2
F0(z) = H1(-z) =
20
10
So the Perfect Reconstruction requirement is
P0(z) œ P0(-z) =
1
1 + 2z-1 + z-2) 2
Ω
= 2z-1
P(z) = z? P0(z) =
1
1 œ2z-1 + z-2)
2
? = 1
1
-1
2 (1 + z)(1 + z )
lm
z
Zeros of P(z):
1+z = 0
1 + z-1 = 0
1
2nd order
zero at
z = -1
Re
21
11