小波与滤波器组：Wavelets and Filter Banks(PPT-11)

2 ^ /2 H /2 ) 0( + ( + + 2k)
But each of the two infinite sums is equal to 1
So the discrete time filter H0() must be orthogonal:
1
= H0() + H0 ( + )
k
aj[k] = f(t), j,k(t)
f(t) fj(t)
Haar example
2j/2aj[0]
2j/2aj[1]
0
1/2j
2/2j
3/2j
4/2j
t
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Projection of f(t) onto Wj: gj(t) = bj[k]wj,k(t)
k
with
f(t), wj,k(t) bj[k] =
k=- -
Use Parseval's theorem 1 ^ a[n] = 2 () () e-in d = ^
-
d
=
1 2
- k=-
2 in ^ ( + 2k) e d 14
Take the Discrete Time Fourier Transform of both sides ^ in = ( + 2k)2 A() a[n]e If n) are orthonormal, then a[n] = (t [n] A() = 1 So the scaling function and it shifts are orthogonal if ^ 2 ( + 2k) = 1
[n] = 2 h0[k] h0[k 2n]
This is the —double shift" orthogonality condition that we previously encountered for orthogonal filters.
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So orthogonal filters are necessary for orthogonal wavelets. Are they sufficient? Consider the cascade algorithm
6
Biorthogonal Wavelet Bases Two scaling functions and two wavelets:
Synthesis: (t) = 2 0[k] k) f (2t
k k
w(t) = 2 1[k] k) f (2t
Analysis: ~ ~ (2t (t) = 2 h0[-k] k) k ~ ~ w(t) = 2 h1[-k] k) (2t
2
v , i = [vx vy]
0 1
= vx
Inner Product
Orthogonal multiresolution spaces: Vj has an orthonormal basis {2 j/2 (2j t k): - k }
Wj has an orthonormal basis {2j/2 w(2j t k): - k }
-
For orthogonal multiresolution spaces, we have Vj
So j,k(t) , wj,l(t) = 0 Wj
4
Projection of an L2 function, f(t), onto Vj: with fj(t) = aj[k] j,k(t)
16
Trick: split sum into even and odd /2 1 = H0( + 2k)2 (/2 + 2k)2 + ^
k=-
^ /2 = 0( ) ( + 2k)2 + H /2
2 k=- 2 k=-
k=-
^ /2 H0( + (2k + 1))2( + (2k + 1)2 /2
-
[n] (i + 1)(t) (i + 1)(t n)dt =
we have n)dt = [n] as the limit, provided that (t) (t
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the cascade algorithm converges.
Note: with the alternating flip requirement, which was necessary for the highpass filter in the case of orthogonal filters, we can show that
-
w(t) w(t n)dt = [n]
and
-
(t) w(t n)dt = 0
13
Orthogonality in the Frequency Domain Let a[n] = (t) (t n)dt
-
1 ^ 2 () ein 2 -
Trick: split integral over entire axis into a sum of integrals over 2 intervals ^ 2 i( + 2k)n 1 a[n] = 2 ( + 2k) e d
- k l
And then change scale using the refinement equation: = 4 h0[k] h0[l] + k 2n - l)d/2 () ( = 2 h0[k] h0[l] + 2n + l] [-k
k l k l k -
wavelet bases.
Orthogonal Wavelets
2D Vector Space: Basis vectors are i, j y orthonormal basis
vy
v vx x
vx and vy are the projections of v onto the x and y axes: vx = v , i i vy = v , j j
j,k(t)
wj,k (t)
3
Orthonormal means
(2 (2 [k j,k(t) , j,l(t) = 2j/2 jt k) 2j/2 jt - l)dt = - l]
-
j/2 w [k j,k(t) , wj,l(t) = 2 w(2jt k) 2j/2w(2jt - l)dt = - l]
n k=-
k=-
^(0) = 1, we see that and since ^ (2k) = 0 for k 0
k
Note: if we set = 0, then ^ (2k)2 = 1
15
Connection between orthogonal wavelets and orthogonal filters (in frequency domain): Start with an orthogonal scaling function: ^ 1 = ( + 2k)2
Representation of functions in a biorthogonal basis: f(t) = ck k) + dj,k 2j/2 w(2jt k) (t
k j=0 k
ck dj,k
~ = f(t) k) dt (t ~ = 2j/2 f(t) w(2jt k) dt
8
Bases are orthogonal w.r.t. each other i.e. ~ ~ (t) (t k) dt = [k] (t) w(t k)dt = 0 ~ ~ w(t) (t k) dt = 0 w(t) w(t k)dt = [k] Equivalent to perfect reconstruction conditions on filters
V0 has a basis {I(t k) :
- f k f} I
~
~
~
~
~
~
Vj
h
Wj
~
~
Vj
h
Wj
~
V0 has a basis {I(t k) : - f k f}
f I
~
W0 has a basis {w(t k): - f k f} f
~ W0 has a basis {w(t k): - f k f} f ~
k=-
and then change scale using the refinement equation in the frequency domain: ^ ^ () = H0(/2 ) ( /2) 2 ^ 2 1 = H0( + k) ( /2 + k) /2
k=-
k
7
Two sets of multiresolution spaces: {0} … V0 V1 … Vj … L2() {0} … V0 V1 … Vj … L2() Vj + Wj = Vj+1 Vj + Wj = Vj+1 Spaces are orthogonal w.r.t. each other i.e.
Course 18.327 and 1.130
Wavelets and Filter Banks
Orthogonal wavelet bases: connection
to orthogonal filters; orthogonality in
the frequency domain. Biorthogonal
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Similarly, we can represent f(t) in the dual basis ~ ~ ~ ~ k) + d 2j/2 w(2j t k) f(t) = c (t
k k j=0 k j,k
~
(t ck = f(t) k) dt
~Baidu Nhomakorabea
dj,k = 2j/2 f(t) w(2jt k) dt
Note: When f0[k] = h0[-k] and f1[k] = h1[-k], we have ~ ~ (t) = (t) Vj = Vj ~ ~ w(t) = w(t) Wj = Wj i.e. we have orthogonal wavelets!
10
Connection between orthogonal wavelets and orthogonal filters Start with the orthonormality requirement on scaling functions [n] = - n) dt (t) (t [n] = 2 0[k](2t k) 2 0[l](2(t n)-l) dt h h
N k=0
(i + 1) (t) = 2 h0[k](i) (2t k) (i) (t) (i)(t n)dt = [n]
If the filters are orthogonal and
-
then
-
Our initial guess for the iteration was (0)(t) = box on [0,1] which is orthonormal w.r.t. its shifts. By induction,
2 2
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