wavelets and filter banks 彭思龙 [email protected] 中国科学院自动化研究所...
TRANSCRIPT
• 参考书– Wavelet and filter banks, G. Strang, T. Nguyen,
Wellesley-Cambridge Press, 1997 （据说有翻译版，也有 MIT 的 ppt 中文有翻译，也可参考瑞士联邦工学院M. Vetterlli 的 ppt）
– 多抽样率信号处理，宗孔德，清华大学出版社， 1996。– Multirate systems and filter banks, Vaidyanathan, PP.,
Englewood Cliffs, New Jersey, Prentice Hall Inc. 1993.– A wavelet tour of signal processing, S. Mallat, Academic
Press. NY, 1998– Ten Lectures on Wavelets, Ingrid Daubechies, 1992– Matlab 6.5, Mathworks.com.– 其他部分小波应用的书，《小波域图像处理》， 2009年出版
Background needed
• Mathematics: – Linear algebra– Polynomial – Mathematical analysis– Functional analysis
• Signal processing• Image processing• Matlab programming
Contents
• Signal processing basic• Filter bank• Mathematical basic• MRA (multiresolution analysis)• Wavelet lifting scheme• Two dimensional wavelet• The application of wavelet
– Wavelet domain denoising– Fast object searching
Contents (cont.)
– Wavelet domain image deconvolution– Wavelet domain image super-resolution– Wavelet domain image compression and post-
processing– Wavelet domain image fusion and mosaicing– Filter approximation– Adaptive wavelet (pyramid)
Contents (cont.)
• Advances of wavelet now:– Nonlinear signal transform:
• Empirical Mode Decomposition (Hilbert-Huang transform)
• Local narrow band signal based decomposition
– Geometry wavelet in 2D (optional)– Image decomposition (optional)
Contents (cont.)
• Some ideas in life and research– How to win before forty– …
Lecture 1
• Introduction– Filter banks=a set of filters, filter is widely used in
many fields of engineering and science for a long time.
– Wavelet, an old and new tool to produce filter banks, have been thoroughly studied in past 20 years. Here we use wavelets to indicate many kinds of wavelets with different properties.
– Application: image compression, pattern recognition, image processing, video processing…
Some basic concepts• Signal: x(t) or {x(n)}• Filter: a vector, h={h(n)}, for a given signal
{x(n)}, the process of filtering: y=h*x, where * is the convolution operator:
• FIR=Finite Impulse Response=finite length• IIR=Infinite Impulse Response=infinite length• Example of filtering:x=sin(-
4:0.08:4)+0.1*randn(1,101);h=[1 1 1 1]/4;y=x*h
k
knxkhny )()()(
0 20 40 60 80 100 120-1.5
-1
-0.5
0
0.5
1
1.5
0 20 40 60 80 100 120-1.5
-1
-0.5
0
0.5
1
1.5
• Continuous Fourier Transform
• Some basic properties:– Linearity– Parseval Identity:
22( ) ( ) { | ( ) | }
Fourier Transform:
1ˆ ( ) ( )2
Inverse Fourier Transform:
1 ˆ( ) ( )2
R
i t
R
it
R
f t L R f t dt
f f t e dt
f t f e d
gfgf ˆ,ˆ,
• Z transformGiven a signal or filter ( ), Z transform is defined as:
( ) ( )
Discrete Time Fourier Transform (DTFT)
( ) ( )
-1
Filtering:
* ( ) ( ) ( ) ( ) ( ) ( )
n
jn
s n
S z s n z
S s n e
j
y x h Y z X z H z Y X H
• Lowpass Filter=moving average=passing low frequency– h={h(n)}, if sum of h is not zero, we call it a
lowpass filter, in most time, the sum of h is 1.– H(z), H(1)=1– Example:
• Simplest: H={1 1}/2; (average)• Spline: {1 2 1}/4• General: {h(n)}• Previous figure
• Highpass Filter=moving difference=passing high frequency
– h={h(n)}, is called highpass filter is the sum of h is zero.
– H(z), H(1)=0– Examples:
• Simplest: h={1 –1}/2, difference
• Dual spline: {-1 2 –1}/4;
• General:{h(n)} sum of h is 0.
• Figure,x as before, h={-1 2 –1}/4, y=x*h;
0 20 40 60 80 100 120-1.5
-1
-0.5
0
0.5
1
1.5
0 20 40 60 80 100 120-0.4
-0.2
0
0.2
0.4
-3 -2 -1 0 1 2 30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Frequency response of {1 1}/2
Frequency response of {1 2 1}/4
-3 -2 -1 0 1 2 30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Frequency response of {-1 2 -1}/4
-3 -2 -1 0 1 2 30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
• Phase
– is called the phase of H– If , we say H has linear phase– H has linear phase is equivalent to say H is
symmetric or antisymmetric, – Previous filters are symmetric, have linear
phase
)(|)(|)( ieHH
)(
ba )(
• Invertibility or noninvertibility– Y(z)=H(z)X(z), If we want to reconstruct X, H
can not be 0 at any point z. If H does not equal to 0 at any |z|=1, we say H is invertible, that is to say, we can reconstruct X by:
X(z)=Y(z)/H(z), which is a inverse filtering, the filter is 1/H(z).
But in most cases, H equals to 0 at some points, we can not reconstruct X exactly.
– Example: H(z)=1+0.5z is invertible, but H(z)=(1+z)/2 is not invertible. How to reconstruct a signal? We can use filter banks
• Filter banks=Lowpass+Highpass(inter-complement), – Simplest idea: H0 and H1, where H0 is a
lowpass filter, and H1 is a highpass filter, the lost information in the process of lowpass filtering can be fund in the output of the highpass filter.
– Some problems: • How to reconstruct the signal?
• How to find such filter bank?
• How to reduce the computation and/or storage?
• Any more properties beside reconstruction?
• Inner product– F_1 and F_2 are two functions in L_2, the inner
product of these two functions is defined as:
• Orthogonality– If <F_1, F_2>=0, we say they are orthogonal.
2121, FFFFR
• Biorthogonality– Two sets of function {F_j} and {G_j}, if <F_j,
G_k>=1 if j=k and 0 otherwise. We call the two set of functions are biorthogonal.
• Compact support– For a given function f, supp(f)={x|f(x) is not 0}– If measure(supp(f)) is finite, we say f is
compactly supported or f has compact support.– Corresponding to FIR
Filters• Signal
– Sequence of numbers, {…x(-1), x(0), x(1), …}– Unit impulse: x(n)=1 if n=0 and 0 otherwise– Usually use Dirac delta symbol
-5 -4 -3 -2 -1 0 1 2 3 4 50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
)(n
• is called sampling function:– – – Continuous signal x(t)discrete signal x(n)
• Sampling rules:– For band-limited and energy limited – Nyquist rate– Sharp: Nyquist, – Redundant: fast than Nyquist– Alias: slow than Nyquist
)(n
)0()()( xnnxxT
)0()()()(),( xdtttxttx
• Shannon sampling theorem:– If signal f(t) satisfies: supp( ) is included in the
interval [-T, T], and sampling rate is r, then• If r> , we can not reconstruct signal f(t);
• If r<= , we have• Where
• is called Nyquist sampling rate.
•
• Simple Proof:
T
T
)()()( krxkrfrtf
.1|ˆ /r], /r,[-)ˆsupp(
satisfieshich function wany is
T] [-T,
T
x
x)sin(T :lyPurticular
f̂
ikr
ikr
ik
ˆ [ , ] , , T
r= / ,
therefore
1 ˆf(kr) f( )e d2
1 ˆ f( )e d 2
1 ˆ f( )e d2 r
such that,
R
f T T
r
-ik
-ikr
[ , ]
\[ , ]
-ikr
f̂( ) f(kr)e , for , ,that isr
f̂( ) f(kr)e , for ,
ˆFor any function with | 1 and
ˆ | 0
ˆ ˆf( ) f(kr)e ( )
Inverse transform will be
f(t) ( ) ( )
Z
Z
T T
R
Z
Z
r
r
r
r f kr t kr
• Delay– Sx(n)=x(n-1)
• Advance– S-1x(n)=x(n+1)
• SS-1=S-1S=I, where I represent the unit operator.
• Time-invariant filters: H is a linear filter, if H(Sx)=S(Hx): a shift of the input produces a shift of the output.
• Ideal filters, – Ideal lowpass:
– Ideal Highpass:
– For ideal lowpass fitler: –
– But in practice, we must use finite filter for convolution, the first idea is to use finite part of this filter, it leads to Gibbs phenomenon.(see ex)
Zk
ikekhH
||/2 ,0
/2||0 1, )()(
Zk
ikekhH
||/2 ,1
/2||0 0, )()(
n
n
nh
2
sin)(
• function h=idealh(N)• step=pi/200;• o=-pi;• h=zeros(400,1);• for kk=1:400• h(kk)=1/2;• sn=1;• for ii=1:N• nu=2*ii-1;• h(kk)=h(kk)+2*sn*cos(nu*o)/pi/nu;• sn=-sn;• end• o=o+step;• end• figure;plot(h);
• We can verify that
• We will meet this equation later in constructing filter banks.
• Traditional filter design methods:
• Firstly, we note that we only need to construct lowpass filter, and shift in phase by pi we can get correspondent highpass filters.
1|)(| |)(| 22 HH
• Window method: h(n)=hI(n)w(n)
– Hamming window:
– Hanning window:
– Kaiser window:
1)/2-(N|n| )2
cos()1()( N
nnw
1)/2-(N|n| )2
cos(2/12/1)( N
nnw
1
22
0
0
2
0
!
)5.0(1)(I
where
N/2|n| )(/2
12
1)(
k k
xx
IN
nInw
• Equiripple method:– The filter with the smallest maximum error in
passband and stopband is an equiripple filter. Means the ripples in passband and stopband is equal height.
– Remez exchange algorithm
• Weighted least squares(eigenfilters)– This method is to minimize the function:
response.frequency desired is )D( where
weight)d(|)H(e - )D(| 2j
E
• In discrete case, we can rewrite the above function into E=hTPh, where h is the unknown coefficient vector. To minimize the function, h must satisfies Ph=u h, that is to say, h is an eigenvector of the matrix P, such that the optimal problem reduces to find the eigenvector of P.
• In detail for designing lowpass filter:– Given a unknown symmetric filter h(n)=h(2L-1-
n) of length 2L– –
12
0
1
0
)2/1( )()()()(L
n
L
nn
Ljnj cnheenhH
in stop band, we need the desired response from ws to PI is zero., then the error function is:
In the passband, from 0 to wp, we need the desired response all pass, such that we can use normalized constant hTc(0) to rewrite the error in pass band into :
])2/1cos[(2)(c where n n
hPhhdcchdHE sTTT
rstops
)()()(2
p
hPhhdcccchE pTTT
pass
0))()0())(()0((
Where the matrix P is known since we know c.
We consider the weighted error:
E=aEp+(1-a)Es, find the eigenvectors of
P=Pp+Ps which will be the suitable filters we needed.
• Halfband and Mth band filter design(filter banks).
• Maximally Flat Filter: (Daubechies wavelets).
• Poisson Summation Formula:– We use Dirac function as a sampling function:
– By sampling rules, we need x(kr), k is any integer, consider is called Dirac comb which likes a comb in figure.
– Poisson summation formula:
– Simple Proof: using distribution and Parseval Identity.
)()()( axdtattx
Zk
kt )2(
Zk Zn
itnekt
2
1)2(
• Another equivalent form:
• Heisenberg’s Uncertainty Principle:
• Define two window width:time and frequency:
• Then: if ||f||=1, we have
• If f is the Gaussian function, the minimum value is reached. The inverse is true.
k n
nGkG )(ˆ2
1)2(
dfdttft 222222 |)(ˆ|ˆ ;|)(|
2/1ˆ
• Basis and frames– Basis: unique representation; linear
independence and completeness– Frame: linear independence and completeness
but stable– Riesz basis: stable basis
• The Wigner-Ville Transform—time-frequency analysis:
• Given a signal f(t), the WV transform is:
• Analyze signal in time-frequency plane.
• Some properties:
R
if detftftW )2/()2/(),(
1.|c| multiplierconstant a toup )( determine ),(
);1
(a
1 of transform theis ),(
);( of transform theis ),(
;|)(ˆ|),(2
1 ;|)(|),(
2
1 22
tftW
afa
a
tW
TtfeTtW
fdttWtfdtW
ti
ff
200 250 300 350 400 450
-2
-1
0
1
2
Time
Frequency
200 250 300 350 400 4500
20
40
60
80
100
120
• Shortcomings:– Not positive;– Interactive parts between separated frequency
parts.
• Cohen class:• General theory reference:
– Time-frequency signal analysis Leon Cohen,– 中译本：时频信号分析， L.科恩 .
• Related theory: Matching Pursuit ( S. Mallat,
Refer to “A Wavelet tour of signal processing”)
200 250 300 350 400 450
-2
-1
0
1
2
Time
Frequency
200 250 300 350 400 4500
20
40
60
80
100
120
• Downsampling and Upsamping– DS: {x(n)}{y(n)}, y(n)=x(2n);– US: {x(n)}{y(n)}, y(2n)=x(n), y(2n+1)=0;– Examples of DS and US.– Two functions: ds and us– Recoverable for half-band signal by using
Shannon sampling theorem. – Otherwise, we can not recover the signal
always.– We use D denotes the Downsamping operator,
and U denote the Upsampling operator.– DU=I, that is, D after U does not change the
signal.
• Downsmapling in the frequency domain.
• Simple proof:
• Example:
• Aliasing: (see alias.m)– Extreme aliasing– No aliasing– Typical aliasing
• Upsamping in the frequency domain:
)]2
()2
([2
1)V( then ),2or ( , XXxvDxv
)2()( ,)2or ( XVxvUxv
• Imaging: (see imaging.m)
• ……
• Upsamping after downsampling
•
• Produce both aliasing and imaging
• In the Z-domain:
2/))()(()( XXUDx
)()(
:Upsamping
2/)]()([)(
:ngDowndampli
2
2/12/1
zXzV
zXzXzV
• To remove the aliasing and imaging, we use filtering before downsampling to remove aliasing and after upsampling to remove imaging.
• ……
• M-channel subsampling:• V(n)=x(Mn), and u(Mn)=x(n) and 0
otherwise.
• In the z domain:
)()(
)2)1(
(...)(1
)(
MXU
M
MX
MX
MV
)()(
)(1
)(1
0
/2/1
M
M
k
MikM
zXzU
ezXM
zV
• Fractional sampling rate:
• DL and UM can commute if and only if L and M are relatively prime.
• Fundamental rule: if L and M are relatively prime, then {Mk, k=0, …., L-1} is the same as {0, 1, …, L-1} besides a integer times of L.
• Simple proof:…
• Filters exchanged with samplers.– G(z)DM=DMG(zM)
• Proof:
– UMG(z)=G(zM)UM
• Proof:
Filter Bank
• Lowpass+Highpass
X(n)
Ideal Lowpass
Ideal Highpass
X(n)
Improved Lowpass+Highpass
X(n)
Ideal LP
Ideal HP
X(n)
2
2
2
2
Ideal LP
Ideal HP
Filter bank
• General lowpass and highpass
X(n)
Lowpass
Highpass
X(n)?
Filter banks
• Perfect Reconstruction condition
• We need T=X to recover the original signal.
X(n)
H0
H1
y0(n)
y1(n)
2
2
v0
v1
2
2
u0(n)
u1(n)F1
F0
T(n)
0 0
1 1
2 20 0 0
20 0
0 0
0 0 0
1 1 1
0 0 1 1
( ) ( ) ( )
( ) ( ( ) ( )) / 2
( )
( ) ( ) ( )
( ) ( )( ( ) ( ) ( ) ( )) / 2
( )( ( ) ( ) ( ) ( )) / 2
Therefore
( ) ( )( ( ) ( ) ( ) ( )) / 2
(
Y z H z X z
V z Y z Y z
U V Z
T z U z F z
T z F z H z X z H z X z
F z H z X z H z X z
T z X z F z H z F z H z
X z
0 0 1 1)( ( ) ( ) ( ) ( )) / 2F z H z F z H z
• Theorem 4.1.– A 2-channel filter bank gives perfect
reconstruction when
– F0(z)H0(z)+F1(z)H1(z)=2z-L
– F0(z)H0(-z)+F1(z)H1(-z)=0
– F0(z)H0(-z)=-F1(z)H1(-z)
• Alias Cancellation and the product filter – F0(z)=H1(-z), and F1(z)=-H0(-z)
– Let P0(z)=F0(z)H0(z) and P1(z)=F1(z)H1(z), then
– P1(z)=-P0(-z), and then
– P0(z)-P0(-z)=2z-L
– L must be odd, so let P(z)=zLP0(z)
– Then we have P(z)+P(-z)=2– Because all even terms in P(z) are zero, we can
conclude that P(z) is a half band filter.– Some examples.– Haar Filter
• Ex3:– P(z)=(-z3+9z+16+9-1-z-3)/16
– P=H0F0
– The roots of P is: c=2+31/2, 2-31/2, -1(4)
– H0 or F0 can be:
– The order N can be:• N=0, 1
• N=1, 1+z-1
• N=2, (1+ z-1)2 or (1+z-1)(c-z-1)
• N=3, (1+ z-1) 3 or (1+z-1) 2(c-z-1)
• (1+z-1)3 and (-1+ 3 z-1 +3 z-2 - z-3)
• (1+z-1)2(c- z-1) and (1+z-1)2(1/c- z-1), Daubechies wavelet of length 4
• Modulation matrix:
• Theorem 4.2.– If all filters are symmetric ( or anti-symmetric)
around zero, h(k)=h(-k), then the condition of PR becomes a statement about inverse matrices
– Fm(z)Hm(z)=2I
l
l
z
z
zHzH
zHzH
zFzF
zFzF
)(20
02
)()(
)()(
)()(
)()(
11
00
10
10
• Early choice:– Croisier-Esaban-Galand(1976)– H1(z)=H0(-z)– H0
2(z) - H02(-z) =2z-L
– So called QMF(Quadrature Mirror Filter) just because |H1(z)|=|H0(-z)|, they are symmetric about PI/2---quadrature frequency.
– No FIR filters (except Haar).– Simple proof: use polyphase expansion
• Better choice:– Smith and Barnwell (1984-6), Mintzer(1985)– H1(z)=-zNH0(-z-1), – Orthogonal filter banks.– Simple example: db4
• General choice: F0(z)H0(z) is a half band filter. Biorthogonal
• Theorem 4.3: In a biorthogoanl linear-phase filter bank with two channels, the filter lengths are all odd or all even. The analysis filters can be:– A)both symmetric, of odd length
– B)one symmetric, and one antisymmetric of even lenth.
– Proof:
• Perfect reconstruction with M Channels.• Modula matrix:
– Hm(z)=(Hjk(z))jk
– Where Hjk(z)=Hj(zWk) for j,k=0, …, M-1
• Polyphase matrix– Meaning of polyphase
– Purpose of polyphase, efficient for computing.
2 2 1 2 22 2 1 0 1
1
0
( ) ( ) ( )
( ) ( )
( ) is called a phase of ( )
k k kk k k
MM k
kk
k
x t a t a t a t x t tx t
x t x t t
x t x t
x H 2 v
1/ 2 1/ 2 1/ 2 1/ 2
2 1 2
2 1 2
1/ 2 1/ 2 1/ 2 1/ 2
1
1( ) ( ) ( ) ( ) ( )
2
( ) ( ) ( ),
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
e o
e o
e o e o
e e o o
V z X z H z X z H z
X z X z z X z
H z H z z H z
X z H z X z z X z H z z H z
X z H z z X z H z
1/ 2
1/ 2 1/ 2 1/ 2 1/ 2
1
1/ 2
1/ 2 1/ 2
( ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
- ( ( ) ( ) ( ) ( )
1( ) ( ) (
2
e o o e
e o e o
e e o o
e o o e
z X z H z X z H z
X z H z X z z X z H z z H z
X z H z z X z H z
z X z H z X z H z
X z H z X z
1/ 2 1/ 2 1) ( ) ( ) ( ) ( ) ( )e e o oH z X z H z z X z H z
• (HX)e=HeXe+z-1 HoXo
• Polyphase matrix:
x
2
2
Ho
He
(HX)e
z-1
X(n)
H0
H1
y0(n)
y1(n)
2
2
v0
v1
.
• Efficient Filter bank by using polyphase
00 010 0 01 1
10 111 1 1
00 01
10 11
( ) ( )( ) ( ) ( )( )
( ) ( )( ) ( ) ( )
( ) ( )( )
( ) ( )
p
p
H z H zV z X z X zH z
H z H zV z z X z z X z
H z H zH z
H z H z
2 2 10 1
2 2 10 00 01
2 2 11 10 11
10 0 00 0 01 1
11 1 10 0 11 1
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) 2( )( ) ( ) ( ) ( ) ( )
( ) 2( )( ) ( ) ( ) ( ) ( )
X z X z X z z
H z H z H z z
H z H z H z z
V z H X z H z X z z H z X z
V z H X z H z X z z H z X z
2 0 ( )X z
( )X z
1z 2 11( )z X z
1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 1/ 21
1( ) ( ) ( )
2z X z z X z z z X z
X(n)
z-1
2
2
Hp
v0(n)
v1(n)
X(n)
H0
H1
y0(n)
y1(n)
2
2
v0
v1
• Relations with Modula matrix
11
11
)()(
)()(
2
11
)()(
)()(
11
00
12,1
2,1
2,0
2,0
zHzH
zHzH
zzHzH
zHzH
oddeven
oddeven
• Polyphase for upsampling and reconstruction
22 1 2
2
2 21 1
2 2
( )( ) ( ) ( ) 1 ( )
( )
( ) ( ) ( 2) ( ) ( )( ) 1 1
( ) ( ) ( 2) ( ) ( )
even
odd
even even
odd odd
F zW z F z v z z v z
F z
F z v z F z v zW z z z
F z v z F z v z
v(n) 2u(n)
F w(n)
v(n)
Feven
Fodd
2
2z-1
w(n)
• Synthesis bank: direct and polyphase
v0(n) 2 F0
v1(n) 2 F1
ˆ( )x n
Polyphase matrix of synthesis filter bank:2 2
0 0 1 1
2 2 21 00 10 0
2 2 201 11 1
00 10 01
01 11 1
01 111
00
ˆ ( ) ( ) ( ) ( )
( ) ( ) ( )1
( ) ( ) ( )
( ) ( ) ( )2 01
( ) ( ) ( )0 2
( ) ( )2 0ˆ 1(0 2
X F z V z F z V z
F z F z V zz
F z F z V z
F z F z V zz
F z F z V z
F z F zX z
F z
0
10 1
00 10
01 11
( )
) ( ) ( )
( ) ( )
( ) ( )I
p
V z
F z V z
F z F zF
F z F z
v0(n)
v1(n) 2
2
z-1ˆ( )x n00 10
01 11
F F
F F
v0(n) 2 F0
v1(n) 2 F1
ˆ( )x n
X(n)
H0
H1
y0(n)
y1(n)
2
2
v0
v1
2
2
u0(n)
u1(n)F1
F0
ˆ( )x n
X(n)
z-1
2
2
Hp
v0(n)
v1(n) 2
2
z-1ˆ( )x nI
pF
• Type 2 polyphase:
v0(n)
v1(n)
Fp-type 2
2
2 z-1
ˆ( )x n
00 10
01 11
01 11
00 10
( ) ( )( )
( ) ( )
( ) ( )0 1
( ) ( )1 0
Ip
II Ip P
F z F zF z
F z F z
F z F zF F
F z F z
IIpF
X(n)
H0
H1
y0(n)
y1(n)
2
2
v0
v1
2
2
u0(n)
u1(n)F1
F0
ˆ( )x n
v0(n)
v1(n)
Fp-type 2
2
2 z-1
ˆ( )x n
X(n)
z-1
2
2
Hp
v0(n)
v1(n)
• Polyphase matrix
0, 0,
1, 1,
0, 1,
0, 1,
0, 1,
0, 1,
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
even oddp
even odd
odd oddIIp
even even
even evenIp
odd odd
H z H zH
H z H z
F z F zF
F z F z
F z F zF
F z F z
• Perfect reconstruction– Theorem 4.9:
• If
• Ex: QMF: F0(z)=H1(-z), and F1(z)=-H0(-z)
• banks give perfect reconstruction when Fp and Hp are inverse:
– Fp(z)Hp(z)=I or z-LI
• Hp is of type 1, and Fp is type 2, transposed, for synthesis
2 1ˆ ( ) ( ) LX z X z z
0 021
1 1
0 121
0 1
2 21 1
( ) ( )1 0 1 11( )
( ) ( )0 1 12
( ) ( )0 1 1 11( )
( ) ( )0 1 12
2(L is odd)
2( )
0 1 1 0 1 11( ) ( )
0 0 14
p
IIp
L
M M L
IIp p
H z H zH z
H z H zz
F z F zF z
F z F zz
zF H
z
F z H zz z
2 2 1
1 12
1 1 12( )
1 1 1 1 1
1 1 1 1 12
2 1
2 12
1 1 00( ) ( )
1 01 0
L
L
L
LL
II L Lp p
z
z
z
zz
zF z H z z z I
z
1 22 2
1
/ 2
1
generally,
2
2( )
if and only if
(1 ( 1) ) (1 ( 1) )1( ) ( )
2 (1 ( 1) ) (1 ( 1) )
If is odd,
0( ) ( )
1 0
det( ( ) ( )) ( )
L
M M L
L LII Lp p L L
II Lp p
II Lp p
zF H
z
z zF z H z z
z
L
zF z H z z
F z H z z
• Lattice structure– How to find a solution of above equation? To
find suitable analysis and synthesis bank?– The simplest example:
– To ensure that Fp is also a FIR, det(Hp) must be a monomial.
– Particularly,
1 11 1 2 2
1 13 3 4 4
p
a b z a b zH
a b z a b z
1 1
1 1p
a cz b dzH
a cz b dz
• We can get Fp=inverse(Hp)
• In fact,
• The condition of linear phase to require:– a=d, and b=c
1
1
1
1 1 1( )
1 1
Such that
1 1 11( )
1 12
p
p
a bH z
z c d
a bH z
c d z
• Another case: a=d, and b=-c: Orthogonal filter
• when |z|=1, we can have Fp is a unitary matrix, we call it paraunitary.
• In general, let
-1
1 1 0
cos sin 1 and (z)=
sin cos z
then the filter given by
( ) ( 1) ( ) ... ( )
is a proper orthogonal filter.
p l l
R
H z R z R R z R
11( ) ( )
det( )T
p pp
F z H zH
X(n) 2
2
0cos
0cos
0sin0sin
z-1
z-1
• If H0=0 at z=-1, then the polyphase matrix has:
• Which means in orthogonal cases, the angels of the lattices add to PI/4.
0
1
2 2 2 2
1 11(1)
1 12
Proof:
Let (1) ,
( 1) 0
(1) 0
is orthogonal 1& 1
p
p
p
H
a bH
c d
H a b
H c d
H a b c d
• Synthesis filter banks:
• Theorem 4.7– Every lowpass-highpass orthonormal filter bank
has a polyphase matrix of a lattice form as above.
-1
1 1 10 1 2
cos sin 1 and (z)=
sin cos z
then the synthesis filter bank is given by
( ) ( ) ( ) ... ( 1)T T T Tp l
R
H z R z R z R R
1 1
0
0
0
2 2 2 2
Proof:
( ) ( ) ( ) ( )
Let ( )
therefore
is the coefficient of , so it must be 0.
Let and
In general, assume that 0 and 0
Defin
T Tp p p p
Nk
p kk
T NN
TN
H z H z H z H z I
H z H z
H H z
a b e fH H
c d g h
a b e g
2 2 2 2e where = ,
We can verify that is a unitary matrix and
.
a b
t tR l e g t a be g
l lR
0
11 121 1
21 22
1
11 12
21 22
* * 0 0 and
0 0 * *
that is
( ) ( )( )
( ) ( )
where the matrices are polynomial of with degree 1
( ) ( )Let ( ) , then
( ) ( )
I
N
p
Tnew p new
RH RH
M z M zRH z
M z z M z z
z N
M z M zH z H R H
M z M z
teratively, we prove that ( ) has the lattice decompostionpH z
• Theorem 4.8:– Every linear phase PR filter bank with equal
(even) length filters has a lattice factorization:
– We can collect all a’s together to be one constant to reduce the computational complex.
1 1 0
1
1 1( ) ( ) ( )... ( )
1 1
11 0( ) and
10
p L L
i i ii i
i i i
H z S z S z S z S
where
a b kz S a
b a kz
• Theorem 4.9:– An FIR analysis bank has an FIR synthesis
bank that gives PR if and only if the determinat of Hp is a nonzero monomial.
• The lattice complexity is approximately half of the polyphase complexity.
1 1 0
-1
( ) ( 1) ( ) ... ( )
1 1 and (z)=
1 z
p l l
ll
l
H z R z R R z R
kR
k
X(n) 2
2kk
z-1
z-1
Orthogonal Filter banks
• Paraunitary matrices– Definition 5.1, The matrix H(z) is paraunitary if
it is unitary for all |z|=1:– HT(1/z)H(z)=I, for all |z|=1;– The above formula is true for all z.
• Theorem 5.1,– Det(H(z))=+z-L or -z-L
• Condition O in polyphase form:
– A filter bank is orthogonal when its polyphase matrix is Paraunitary.
• Theorem 5.2:
– For an orthogonal filter bank the lowpass filter H must satisfy Condition O:
2 2even odd
2 2
Polypase form: |H ( ) | |H ( ) | 1
Modula form: |H( )| |H( )| 2
Coefficient form: h(n)h(n-2k)= (k)
j je e
• Proof:
2 1 2
2 2 2 2 2
2 2 1 2 2
2 2 2 2 2 2
2
( ) ( ) ( )
| ( ) | | ( ) | | ( ) |
( ) ( ) ( ) ( )
| ( ) | | ( ) | 2(| ( ) | | ( ) | )
( ) | ( ) |
is halfband,
even odd
even odd
even odd even odd
even odd
kk
k n n k
H z H z z H z
H z H z H z
zH z H z z H z H z
H z H z H z H z
P z H z p z
p h h
P
2 0 0 except 0, 1kp k p
• Theorem: 1 1
1 1
( ) ( ) ( ) ( ) 2
If and only if
( ) ( ) ( ) ( )
T Tm m m m
T Tp p p p
H z H z H z H z I
H z H z H z H z I
2 2
0
0
22
L
0
Lemma: If |H( )| |H( )| 2, and
( )
then is odd
Proof:
If is even, assume that 0,
|H( )| 0( 0)
c 0
0 0
Lk
kk
Lk
k kk L
L L L
H z h z
L
L h
c e c k
c h h h
• Theorem 5.3, A symmetric orthogonal FIR filter can only have two nonzero coefficients.
• Proof: 1
2 1 2
2 2 2 1 2
2 2
( ) is symmetric, then ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
If is odd
( ) ( ) | ( ) | | ( ) |
| ( ) | | ( ) | 1/ 2
L
even odd
L Leven odd even odd
Lodd even odd even
odd even
H z H z z H z
H z H z z H z
H z zH z z H z z H z
L
zH z z H z H z H z
H z H z
• Let P(z)=|H(z)|2, we have P(z)+P(-z)=2, that is to say, P(z) is a halfband filter:– p(2m)=1 if m=0, and 0 otherwise.– P(z) is real, symmetric, nonnegative, halfband
• Spectral Factorization– Given P(z), how to find H?– Can any nonnegative P(z) have such
factorization?– How to factorize?– The root of P(z) (polynomial),
• z, 1/z, con(z), 1/con(z), for z is complex
• z, z for z is real.
2
1
( ) ( ) ( )( 1/ )( )( 1/ ) ( )M
i i i i ji
P x P N x z x z x z x z x z
• For orthogonal factorization,
• If all z’s are in unit circle, we call it minimum phase spectral factor .
• We can choose F and H freely to be biorthogonal filter bank. To ensure that H and F are real, relatively conjugate, linear phase needs symmetric about unit circle.
1/ 2
1
( ) | ( ) | ( )( ) ( )M
i i ji
H z P N x z x z x z
• Which one is better?(length=10)
0 50 100 150 200 250 300 350 4000
0.5
1
1.5
0 50 100 150 200 250 300 350 4000
0.5
1
1.5
Same?
• Maxflat (Daubechies) filters
– Condition O: P=|H|2 is a normalized halfband filter:
• p(0)=1, p(2)=….=p(2p-2)=0
– Condition Ap: H has a zero of order p at π:
•
( 1)
2 1
0
( ) ( ) ... ( ) 0
( 1) ( ) 0, for 0,1,..., 1
1( ) ( )
2
1 cos( ) has a factor
2
p
pn k
n
pi
p
H H H
n h n k p
eH R
P
2 1 2 12 2
1 2 0
( ) ( ) equals to | ( ) | | ( ) |p p
in in
p
P p n e h h n e
• Lemma:
2
2
2
( )
| ( ) | (cos )
Proof:
| ( ) | *
| ( ) | cos( )
cos( ) is a polynomial of cos( )
k
k
kk
k k kk k k
k
H h e
then
H c
H h e h e d e
H d k
k
• Formulas for P: Daubechies method:( ) is a polynomial of cos( )
1 cos( )set , then ( ) (1 ) ( )
2the orthogonal condition is :
(1 ) ( ) (1 ) 2
if the degree of R as a polynomail of y is less than p,
then according the theory o
p
p p
P
y P y R y
y R y y R y
1
0
f polynomial, the solution is
unique.
( ) (1 ) (2 (1 ))
1(1 ) ( )
p p
pp k p
k
R y y y R y
p ky y y Q y
k
• 1
0
1
0
1
0
0
1( ) ( ( ))(2 (1 ))
1 =2 ( )
Since ( ) is polynomial of degree p-1, we can say 0.
1( )=2
1( ) 2(1 )
pk p p
k
pk p
k
pk
k
pp k
k
p kR y y y Q y y R y
k
p ky y Q y
k
R y Q
p kR y y
k
p kP y y y
k
1
– Unit factorization method:2 1 2 1
2 12 1
0
1 2 12 1 2 1
0
11
0
1 11 1 ( )
2 22 1 1 1
( ) ( )2 2
2 1 2 11 1 1 1( ) ( ) ( ) ( )
2 2 2 2
2 1 2 11 1 1( ) ( ) ( ) (
2 2 2
p p
pk p k
k
p pk p k k p k
k k p
pp k p k
k l
y y
p y y
k
p py y y y
k k
p py y y
k l p
1
1
0
1 11 1
0 0
1 11 1
0 0
1 1) ( )
2 2
2 1 2 11 1 1 1 1 1( ) ( ) ( ) ( ) ( ) ( )
2 2 2 2 2 2
2 1 2 11 1 1 1 1( ) ( ) ( ) ( ) ( )
2 12 2 2 2 2
pl p p l
p pp k p k p l p l
k l
p pp k p k p p l
k l
y y
p py y y y y y
k l p
p py y y y y
k p l
1 1
1 1
0 0
1( )
2
2 1 2 11 1 1 1 1 1( ) ( ) ( ) ( ) ( ) ( )
2 2 2 2 2 2
l
p pp k p k p p l l
k l
y
p py y y y y y
k l
• Two formulas:1
1
0
1
0
2 1( ) 2(1 ) (1 )
1( ) 2(1 )
pp p k k
k
pp k
k
pp y y y y
k
p kP y y y
k
1 1 1 1( ) ( ) 1
2 2 2 2
1 1Let , then 1
2 2
(1 ) (1 ) ( ) 1
p p
p p
y y y yQ Q
y yt t
t Q t t Q t
• Meyer’s method:
1 1
Let ( ) be a polynomail of degree 2 -1, satisfies:
has oder p zeros at 1, and oder at 0 except (0) 2.
then ( ) (1 ) , all such polynomials satisfies P(0)=2
must have ( ) (1 ) 2, sin
p p
P y p
y p y P
P y cy y
P y P y
2 1
2 1
0
ce the codition at 0 and 1
decide the polynomial uniqely.
1 cos( )y= will give
2
( ) sin
( ) 2 sin
where c is chosen for ( ) 0.
p
p
P c
P c d
P
• Transition band for maxflat filters– Theorem 5.6, the maxflat filter has center slope
proportional to .The transition rom 0.98 to 0.02 is over an interval of length 4/ , where N is the order of the filter.
N
N