Download Filter Banks - Banking - Lecture Slides and more Slides Banking and Finance in PDF only on Docsity!
Course 18.327 and 1.
Wavelets and Filter Banks
Filter Banks: time domain
(Haar example) and frequency domain;
conditions for alias cancellation
and no distortion
2
Simplest (non-trivial) example of a two channel FIR
perfect reconstruction filter bank.
Haar Filter Bank
h
0
[n]
h
1
[n]
éé éé 2
éé éé 2
x[n]
y
0
[n]
y
1
[n]
åååå 2
åååå 2
Analysis
r
0
[n]
r
1
[n]
f
0
[n]
f
1
[n]
x[n]
^
Synthesis
v
0
[n]
v
1
[n]
t
1
[n]
t
0
[n]
h
0
[n] =
f
0
[n] =
μμμμ 2
μμ μμ 2
μμμμ 2
μμμμ 2
3
h
1
[n] =
f
1
[n] =
μμ μμ 2
μμμμ 2
μμμμ 2
μμμμ 2
Analysis:
r
0
[n] = (x[n] + x[n œ 1]) lowpass filter
y
0
[n] = r
0
[2n] downsampler
y
0
[n] = (x[2n] + x[2n œ 1]) -----------------jjjj
Similarly
y
1
[n] = (x[2n] œ x[2n œ 1]) ------------------k kk
k
μμμμ 2
μ μμ
μ 2
μμμμ 2
4
Matrix form
y
0
[0]
y
0
[1]
y
1
[0]
y
1
[1]
μ μμ
μ 2
x[-1]
x[0]
x[1]
x[2]
-------------------llll
y
o
y
1
L
B
x
μ μμ
jjj kkk
i.e.
^
x[2n-1] =
1
μ μμ
μ 2
(y
0
[n] œ y
1
[n]) = x[2n-1]
from j and k
^
1
x[2n] =
μ 2
(y
0
[n] + y
1
[n]) = x[2n]
^
So x[n] = x[n] Ω Perfect reconstruction!
In general, we will make all filters causal, so we will
have
^
x[n] = x[n œ n
0
] Ω PR with delay
7
8
Matrix form
x[-1]
x[0]
x[1]
x[2]
μμμμ 2
^
^
^
^
y
0
[0]
y
0
[1]
y
1
[0]
y
1
[1]
^
x = L
T
B
T
y
0
y
1
----------------mmmm
9
Perfect reconstruction means that the synthesis
bank is the inverse of the analysis bank.
x = x ΩΩΩΩ L
T
B
T
= I
L
B
^
&'(&'( &'(&'(
&'(&'(&'(&'(
W
W
Wavelet transform
matrix
÷÷÷÷
In the Haar example, we have the special case
W
œ
= W
T
ç çç
ç orthogonal matrix
So we have an orthogonal filter bank, where
Synthesis bank = Transpose of Analysis bank
f
0
[n] = h
0
[- n]
f
1
[n] = h
1
[- n]
10
Perfect Reconstruction Filter Banks
General two-channel filter bank
H
0
(z)
H
1
(z)
é éé
é 2
éé éé 2
x[n]
y
0
[n]
y
1
[n]
åååå 2
å åå
å 2
r
0
[n]
r
1
[n]
F
0
(z)
F
1
(z)
x[n]
^
v
0
[n]
v
1
[n]
t
1
[n]
t
0
[n]
3333
3333
z-transform definition:
X(z) = ƒƒƒƒ x[n]z
-n
Put z = e
i w ww
w
to get DTFT
Ñ ÑÑ
Ñ
n=-ÑÑÑÑ
???
13
Suppose X(w ww
w) = 1 (input has all frequencies)
Then R
0
(wwww) = H
0
(wwww), so that after downsampling we have
Y
0
(wwww) =
pp pp
p pp
p wwww
²R
0
²R
0
²R
0
wwww
2
w ww
w
2
wwww
2
pppp
aliasing
Goal is to design F
0
(z) and F
1
(z) so that the overall
system is just a simple delay - with no aliasing term:
V
0
(z) + V
1
(z) = z
X(z)
V
0
(z) = F
0
(z) T
0
(z)
= F
0
(z) Y
0
(z
2
) (upsampling)
= ²F
0
(z){ H
0
(z) X(z) + H
0
(-z) X(-z)}
V
1
(z) = ²F
1
(z){ H
1
(z) X(z) + H
1
(-z) X(-z)}
So we want
² {F
0
(z) H
0
(z) + F
1
(z) H
1
(z) } X(z)
-?
X(z)
² {F
0
(z) H
0
(-z) + F
1
(z) H
1
(-z) } X(-z)
14
www www ppp
ƒƒƒ
ƒƒƒ
15
Compare terms in X(z) and X(-z):
- Condition for no distortion (terms in X (z) amount
to a delay)
F
0
(z) H
0
(z) + F
1
(z) H
1
(z) = 2z
-? ??
?
- Condition for alias cancellation (no term in X(-z))
F
0
(z) H
0
(-z) + F
1
(z) H
1
(-z) = 0
To satisfy alias cancellation condition, choose
F
0
(z) = H
1
(-z)
F
1
(z) = -H
0
(-z)
--------------jjjj
--------------kkkk
----------------------l ll
l
What happens in the time domain?
F
0
(z) = H
1
(-z) F
0
(w) = H
1
(w + p)
= ƒ h
1
[n] (-z)
-n
n
= ƒ (-1)
n
h
1
[n] z
-n
n
So the filter coefficients are
f
0
[n] = (-1)
n
h
1
[n] alternating signs
f
1
[n] = (-1)
n+
h
0
[n] rule
Example
h
0
[n] = {
{ b
0
, b
1
, b
2
a
0
, a
1
, a
2
} f
0
[n] = { b
0
, -b
1
, b
2
h
1
[n] = f
1
[n] = {-a
0
, a
1
, -a
2
16
ppp
lll
ppp
ƒƒƒ ƒƒƒ
òòò
Design Process
- Design P(z) to satisfy Equation p. This gives
P
0
(z). Note: P(z) is designed to be lowpass.
- Factor P
0
(z) into F
0
(z) H
0
(z). Use Equations l to
find H
1
(z) and F
1
(z).
Note: Equation p requires all even powers of z
(except z
0
) to be zero:
ƒ p[n]z
-n
-n
n n
1 ; n = 0
Ω p[n] =
0 ; all even n (n ò 0)
19
20
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
H
0
(z) = (1 + z
) H
1
(z) = (1 œ z
F
0
(z) = H
1
(-z) = (1 + z
F
1
(z) = -H
0
(-z) = (1- z
P
0
(z) = F
0
(z) H
0
(z) = (1 + z
2
1
μ μμ
μ 2
1
μ μμ
μ 2
1
μμμμ 2
1
μ μμ
μ 2
1
2
21
So the Perfect Reconstruction requirement is
P
0
(z) œ P
0
(-z) = 1 + 2z
) - 1 œ2z
= 2z
P(z) = z
????
P
0
(z) = (1 + z)(1 + z
1
2
1
2
1
2
nd
order
zero at
z = -
lm
Re
z
Zeros of P(z):
1 + z = 0
1 + z
