Unformatted text preview: Course 18.327 and 1.130
Course
Wavelets and Filter Banks
Signal and Image Processing: finite
length signals; boundary filters and
boundary wavelets; wavelet
compression algorithms. FiniteLength Signals y2
y1
y0
y1
o
yN1
o = h0
h1
h2 h0
h1 y[n]
h1
h0 h1
h1 h0 h1
rr r x2
x1
x0
x1
o
xN1
o unknown LMN H(ω) x[n] length N
(finite length) unknown 2 1) zeropadding x[n] m 2 1 0 1 2 3 m n filtered by [1, 1]
y[n]
m n 2 1 0 1 2 3
filtered by [1, 1]
y[n]
m
2 1 0 1 2 3 n artificial edge resulting
from zeropadding 3 2) Periodic Extension
… x[N] = x[0] x[n]
… … n  1 0 1 2 … N1 N wraparound o y0
y1
o = yN1 Noutput h1 h0 h1 h2 m h2 h1
h1 h0 h1 m h3 h2
h1 h2 m h2 h1 h0 xN2
xN1
x0
x1
x2
o
xN1
x0
o circulant matrix = H
4 What is the eigenvector for the circulant matrix?
[ 1 eiω ei2ω m ei(N1)ω ] T
We need
eiNω = 1 = ei0ω
∴ Nω = 2πk , ω= 2πk
N discrete set of ω’s
For the 0th row,
N1 H[k] = ∑ h[n] ei 2πk
Nn n=0 5 111m
1 w w2
[H]
1 w2 w4
oo
o
1 wN1 1
H[0]
H[1]
wN1
r
w2(N1) = [F]
o
H[N1] k=0k=1
k=N1
LOOOOMOOOON
2π
iN
w=e
F
HF = FΛ Λ contains the Fourier coefficients
2πk
Nn H[k] = ∑
n h[n]ei ∑ ∑ h[n  x[ei If x[ = ei n n 2πk0
N 2πk
Nn = H[k]X[k]
H[k]X[k] ⇒ X[k] = δ[k – k0]
⇒ H[k]X[k] = H[k0]X[k] 6 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
h1x2 + h0x1 + h1x0
h0 h1
h1x1 + h0x0 + h1x1
h1 h0 h1
=
h1x0 + h0x1 + h1x2
h1 h0 h1
r rr x2
x1
x0
x1
x2 7 e.g. whole point symmetry – filter,
halfpoint 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
r rr x2
x1
x0
x0
x1
x2 Whole point symmetry 8 Downsampling a wholepoint symmetric signal with
even length N
at the left boundary:
x x 2 1 0 1 ⇒ still wholepoint symmetric after ↓2.
2 at the right boundary:
x
x x ⇒ halfpoint symmetric after ↓2. N1
odd
E.g. 9/7 filter: wholepoint symmetric
use the above extension for signal ⇒ N N/2
exactly
N/2
9 Downsample a halfpoint symmetric signal
x ⇒ nothing guaranteed x x
3 2 1 0 1 2 Linearphase filters
H(ω) = A(ω)eiωα
1) halfpoint symmetric, α = fraction 2) wholepoint symmetric, α = integer Symmetric extension of finitelength signal
X(ω) = B(ω)eiωβ
10
10 The output:
Y(ω) = H(ω)X(ω)
W
W
H
H W
H
H
W W
H
W
H W = wholepoint symmetry
H = halfpoint symmetry The above extensions ensure the continuity of function
values at boundaries, but not the continuity of
derivatives at boundaries. 11
11 4) Polynomial Extrapolation (not useful in image
processing)
• Useful for PDE with boundary conditions.
x0 x1 01 x2 x3 4 coefficients ⇒ fits up to 3rd order
polynomials.
23
a + bn + cn2 + dn3 = x(n)
1000
1111
1 2 22 23
1 3 32 33
LOMON
A a
x0
b = x1
c
x2
d
x3 12
12 Then,
x–1 = [1 1 1 1] PDE a
b = [1 1 1 1] A–1
c
d x0
x1
x2
x3 f(x) = ∑ ckφ(x – k)
k Assume f(x) has polynomial behavior near boundaries
p1 ∑ αixi = f(x) = ∑ ckφ(x – k)
k i=0 {φ (•  k)} orthonormal
p1 ⇒ ∑ αi ∫ φ(x – k)xidx = ck
i=0 LOMON
µik 13
13 µ0 µ1 m µ p1
0
0
0
µ0
1 µ1 µ2
1
1 m o α0 c0 α1 c1 = o
αp1 o
cp1 Using the computed αi’s, we can extrapolate,
α0
e.g. c–1 = [µ01 µ1 1 m µp1] o
–
–
–1
αp1
DCT idea of symmetric extension
cf. DFT X[k] = ∑
n complexvalued 2πk x[n]ei N n
Want realvalued results.
Want 14
14 1m 0 N1 N m 2N1 2N DFT of this extended signal:
N1 2πk
–i 2N n ∑ x[n]e 2N1 +∑ LOOMOON
n=N n=0 N1 N1 2πk π
i 2N (2N1m)
∑ x[m]e
m=0 = ∑ x[n]
n=0 2πk x[2N1n]ei 2N n 2πk i 2πk n
{e 2N + 2πk ei 2N (2N1n)} N1 X(k) ck ∑ √2 x[n]cos πk (n+½) m DCT – II used in JPEG
N
N
n=0 1 ck = √2 k=0 1 k = 1, 2, …, N  1 15
15 ...
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Full Document
 Spring '03
 GilbertStrang
 Trigraph, Symmetry group, Filter Banks, point symmetry

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