Course 18.327 and 1.130
Wavelets and Filter Banks
Signal and Image Processing: finite
length signals; boundary filters and
boundary wavelets; wavelet
compression algorithms.
Finite-Length Signals
y-2
y-1
y0
y1
yN-1
=
y[n]
h0
h1 h0 h-1
h2 h1 h0 h-1
h1 h0 h-1
x-2
x-1
x0
x1
xN-1
unknown
H(ω
ω)
x[n]
length N
(finite length)
unknown
2
1
1) zero-padding
x[n]
-2 -1 0 1 2 3 n
filtered by [1, 1]
y[n]
n
-2 -1 0 1 2 3
filtered by [1, -1]
y[n]
-2 -1 0 1 2 3
artificial edge resulting
from zero-padding
2) Periodic Extension
…
n
3
x[N] = x[0]
x[n]
…
…
n
- 1 0 1 2 … N-1 N
wrap-around
y0
y1
=
yN-1
N-output
h1 h0 h-1 h-2 h2 h1
h1 h0 h-1 h3 h2
h-1 h-2
h2 h1 h0
xN-2
xN-1
x0
x1
x2
xN-1
x0
circulant matrix = H
4
2
What is the eigenvector for the circulant matrix?
[ 1 eiωω ei2ωω ei(N-1)ωω ] T
We need
eiNωω = 1 = ei0ωω
∴ Nω
ω = 2π
πk
,
ω =
2π
πk
N
discrete set of ω’s
For the
0th
row,
N-1
2π
πk
H[k] = ¦ h[n] e-i N n
n=0
5
1 1 1 1 w w2
[H]
1 w2 w4
1 wN-1
1
H[0]
N-1
H[1]
w
w2(N-1) = [F]
H[N-1]
k=0k=1
k=N-1
2π
π
w = ei N
F
HF = FΛ
Λ
Λ contains the Fourier coefficients
H[k] = ¦ h[n]e-i
2π
πk
Nn
n
¦ ¦ h[n - ]x[]e-i
n If x[] = ei
2π
π k0
N 2π
πk
Nn
= H[k]X[k]
Ÿ X[k] = δ[k – k0]
Ÿ H[k]X[k] = H[k0]X[k]
6
3
3) Symmetric Extension
1) Whole point symmetry – when filter is whole
point symmetric.
2) Half point symmetry – when filter is half point
symmetric.
e.g. Whole point symmetry: filter and signal
h0 h1
h1x2 + h0x1 + h1x0
h1x1 + h0x0 + h1x1
h1 h0 h1
=
h1x0 + h0x1 + h1x2
h1 h0 h1
x2
x1
x0
x1
x2
7
e.g. whole point symmetry – filter,
half-point symmetry - signal
h1x2 +
h1x1 +
h1x0 +
h1x0 +
h0x1 + h1x0
h0x0 + h1x0
h0x0 + h1x1
h0x1 + h1x2
Half point symmetry
h1 h0 h1
h1 h0 h1
=
h1 h0 h1
x2
x1
x0
x0
x1
x2
Whole point symmetry
8
4
Downsampling a whole-point symmetric signal with
even length N
at the left boundary:
x
x
-2 -1 0
1
Ÿ still whole-point symmetric after ↓2.
2
at the right boundary:
x
x
Ÿ half-point symmetric after ↓2.
x
N-1
odd
E.g. 9/7 filter: whole-point symmetric
use the above extension for signal Ÿ N
N/2
exactly
N/2
9
Downsample a half-point symmetric signal
x
x
-3 -2 -1 0 1
x
Ÿ nothing guaranteed
2
Linear-phase filters
ωα
H(ω
ω) = A(ω
ω)e-iωα
1) half-point symmetric,
α = fraction
2) whole-point symmetric,
α = integer
Symmetric extension of finite-length signal
ωβ
X(ω
ω) = B(ω
ω)e-iωβ
10
5
The output:
Y(ω
ω) = H(ω
ω)X(ω
ω)
W
W
H
H
W
H
H
W
W
H
W
H
W = whole-point symmetry
H = half-point symmetry
The above extensions ensure the continuity of function
values at boundaries, but not the continuity of
derivatives at boundaries.
11
4) Polynomial Extrapolation (not useful in image
processing)
• Useful for PDE with boundary conditions.
x0
x1
0 1
x2
x3
4 coefficients Ÿ fits up to 3rd order
polynomials.
2 3
a + bn + cn2 + dn3 = x(n)
1 0 0 0
1 1 1 1
1 2 22 23
1 3 32 33
A
a
b
c
d
=
x0
x1
x2
x3
12
6
Then,
x–1 = [1 -1 1 -1]
a
b = [1 -1 1 -1] A–1
c
d
x0
x1
x2
x3
PDE
f(x) = ¦ ckφ(x – k)
k
Assume f(x) has polynomial behavior near boundaries
p-1
¦ αixi = f(x) = ¦ ckφ(x – k)
k
i=0
{φ
φ (•• - k)} orthonormal
p-1
Ÿ ¦ αi ³ φ(x – k)xidx = ck
i=0
µik
13
µ00
µ10 µ p-1
0
α0
µ01
µ11 µ21
α1
c0
=
αp-1
c1
cp-1
Using the computed αi’s, we can extrapolate,
α0
0
1
p-1
µ–1 µ –1 µ –1 ] e.g. c–1 = [µ
αp-1
DCT idea of symmetric extension
cf. DFT X[k] = ¦ x[n]e-i
n
complex-valued
2π
πk
n
N
Want real-valued results.
14
7
1 0
N-1
N
2N-1
2N
DFT of this extended signal:
N-1
2π
πk
2N-1
2π
πk
¦ x[n]e–i 2N n + ¦ x[2N-1-n]e-i 2N n
n=N
n=0
N-1
2π
πk
¦ x[m]e-i 2N
(2N-1-m)
m=0
N-1
2π
πk
2π
πk
= ¦ x[n] {e-i 2N n + e-i 2N (2N-1-n)}
N-1
ck ¦ ¥N2 x[n]cos
n=0
πk
(n+½)
N
n=0
X(k)
ck =
1
DCT – II used in JPEG
¥2
k=0
1
k = 1, 2, …, N - 1
15
8