Tài liệu Digital Signal Processing Handbook P35 ppt

Cormac Herley. “Wavelets and Filter Banks.”
2000 CRC Press LLC. <>.
WaveletsandFilterBanks
CormacHerley
HewlettPackardLaboratories
35.1FilterBanksandWavelets
DerivingContinuous-TimeBasesFromDiscrete-TimeOnes
•
Two-ChannelFilterBanksandWavelets
•
StructureofTwo-
ChannelFilterBanks
•
PuttingthePiecesTogether
References
35.1 FilterBanksandWavelets
Themethodsofdesigningbasesthatwewillemploydrawonideasfirstusedintheconstructionof
multiratefilterbanks.Theideaofsuchsystemsistotakeaninputsystemandsplititintosubsequences
usingbanksoffilters.Thissimplestcaseinvolvessplittingintojusttwopartsusingastructuresuch
asthatshowninFig.35.1.Thistechniquehasalonghistoryofuseintheareaofsubbandcoding:
firstofspeech[1,2]andmorerecentlyofimages[3,4].Infact,themostsuccessfulimagecoding
schemesarebasedonfilterbankexpansions[5,6,7].Recenttextsonthesubjectare[8,9,10].We
willconsideronlythetwo-channelcaseinthissection.If
ˆ
X(z)=X(z),thenthefilterbankhasthe
perfectreconstructionproperty.
FIGURE35.1:Maximallydecimatedtwo-channelmultiratefilterbank.
Itiseasilyshownthattheoutput
ˆ
X(z)oftheoverallanalysis/synthesissystemisgivenby:
ˆ

X(z)=
1
2
[G
0
(z)G
1
(z)]

H
0
(z)ψ H
0
(−z)
H
1
(z)ψ H
1
(−z)

X(z)
X(−z)

(35.1)
=
1
2
[H
0
(z)G

0
(z)+H
1
(z)G
1
(z)]·X(z)
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+
1
2
[H
0
(−z)G
0
(z) + H
1
(−z)G
1
(z)]·X(−z).
Call the above 2 × 2 matrix H
m
(z). This gives that the unique choice for the synthesis filters is

G
0
(z)
G
1

(z)

=

H
0
(z) H
0
(−z)
H
1
(z) H
1
(−z)

−1
·

2
0

=
2

m
(z)

H
1
(−z)

−H
0
(−z)

,
(35.2)
where 
m
(z) = detH
m
(z).
Ifweobservethat
m
(z) =−
m
(−z) anddefineP(z) = 2·H
0
(z)H
1
(−z)/
m
(z) = H
0
(z)G
0
(z),
it follows from (35.2) that G
1
(z)H
1

(z) = 2 ·H
1
(z)H
0
(−z)/
m
(−z) = P(−z). We can then write
that the necessary and sufficient condition for perfect reconstruction ( 35.1) is:
P(z)+ P(−z) = 2.
(35.3)
Sincethisconditionplaysan important roleinwhat follows, wewillrefertoanyfunction having this
propertyas valid. The implicationof thispropert y isthat allbut one ofthe even-indexedcoefficients
of P(z)are zero. That is
P(z)+ P(−z) =

n
(p(n)z
−n
+ p(n)(−z)
−n
)
=

n
2 ·p(2n)z
−(2n+1)
.
For this to satisfy (35.3) requires p(2n) = δ
n
; thus, one of the polyphase components of P(z)must

be the unit sample. By polyphasecomponents we mean the set of even-indexed samples,and the set
of the odd-indexed samples. Such a function is illustrated in Fig. 35.2(a).
FIGURE 35.2: Zeros of the correlation functions. (a) Autocorrelation H
0
(−z)H
0
(z
−1
). (b) Cross-
correlation H
0
(−z)H
1
(z
−1
).
Constructingsuchafunctionisnotdifficult. Ingeneral,however,wewillwishtoimposeadditional
constraints on the filter banks. So, P(z)w ill have to satisfy other constraints in addition to (35.3).
Observe that as a consequence of (35.2) G
0
(z)H
1
(z), i.e., the cross-correlation of g
1
(n) and the
time-reversed filter h
0
(−n), and G
1
(z)H

0
(z), the cross-correlation of g
1
(n) and h
0
(−n), have only
odd-indexed coefficients, just as for the function in Fig. 35.2(b), that is:
<g
0
(n), h
1
(2k −n) > = 0, (35.4)
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<g
1
(n), h
0
(2k −n) > = 0, (35.5)
(note the time reversal in the inner product). Define now the matrix H
0
as
H
0
=







.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
h
0
(L − 1)h
0

(L − 2) ··· ··· h
0
(0) 00
00h
0
(L − 1) ··· h
0
(2)h
0
(1)h
0
(0)
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.






.
(35.6)
whichhasasitskthrowtheelementsof the sequenceh
0
(2k −n). Pre-multiplyingbyH
0
corresponds
to filtering by H
0
(z) followed by subsampling by a factor of 2. Also define
G
T
0
=







.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
g
0
(0)g
0

(1) ··· ··· g
0
(L − 1) 00
00g
0
(0) ··· g
0
(L − 3)g
0
(L − 2)g
0
(L − 1)
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.






,
(35.7)
so G
0
has as its kth column the elements of the sequence g
0
(n − 2k). Define H
1
by replacing the
coefficients ofh
0
(n) with those of h
1
(n) in (35.6) and G
1
by replacing the coefficients of g
0
(n) with

those of g
1
(n) in (35.7).
Wefindthat (35.4)givesthatallrowsofH
1
areorthogonaltoallcolumnsofG
0
. Similarlywefind,
from (35.5), that all of the columns of G
1
are orthogonal to the rows of H
0
. So, in matrix notation:
H
0
G
1
= 0 = H
1
G
0
. (35.8)
Now P(z) = G
0
(z)H
0
(z) = z
−1
H
0

(z)H
1
(−z) and P(−z) = G
1
(z)H
1
(z) are b oth valid and
have theform given inFig. 35.2 (a). Hence, theimpulse responses ofg
i
(n) andh
i
(n) are orthogonal
with respect to even shifts
<g
i
(n), h
i
(2l −n) > = δ
l
. (35.9)
In operator notation:
H
0
G
0
= I = H
1
G
1
. (35.10)

Sincewehaveaperfectreconstructionsystemweget:
G
0
H
0
+ G
1
H
1
= I. (35.11)
Of course (35.11) indicates that no nonzero vector can lie in the column nullspaces of both G
0
and
G
1
. Notethat (35.10) implies thatG
0
H
0
and G
1
H
1
areeachprojections(sinceG
i
H
i
G
i
H

i
= G
i
H
i
).
They project onto subspaces which are not, in general, orthogonal (since the operators are not self-
adjoint). Because of (35.4), (35.5),and (35.9)the analysis/synthesissystem is termed biorthogonal.
If we interleave the rows of H
0
and H
1
, much as was done in the orthogonal case, and form again a
block Toeplitz matrix
A =










.
.
.
.
.

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
h
0
(L − 1)h
0
(L − 2) ··· ··· h
0
(0) 00
h
1
(L − 1)h
1

(L − 2) ··· ··· h
1
(0) 00
00h
0
(L − 1) ··· h
0
(2)h
0
(1)h
0
(0)
00h
1
(L − 1) ··· h
1
(2)h
1
(1)h
1
(0)
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.










,
(35.12)
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wefind thatthe rowsof A form a basisfor l
2
(Z).IfweformBby interleaving the columnsof G
0
and
G
1
,wefind
B · A = I.
In the special case where we have a unitary solution, one finds: G
0
= H
T
0
and G
1
= H
T
1
, and
(35.8) gives that we have projections onto subspaces which are mutually orthogonal. The system
then simplifies to the orthogonal case, where B = A
−1
= A
T
.
A point that we wish to emphasize is that in theconditions for perfect reconstruction, (35.2) and
(35.3), the filters H
0
(z) and G

0
(z) are related via their product P(z). It is the choice of the function
P(z)and the factorization taken that determines the properties of the filter bank. We conclude the
introduction with a proposition that sums up the foregoing.
PROPOSITION35.1 Todesignatwo-channelperfectreconstruction filter bank, itisnecessaryand
sufficient to finda P(z)satisfying (35.3),factorit P(z) = G
0
(z)H
0
(z) andassign thefilters as given
in (35.2).
35.1.1 Deriving Continuous-Time Bases From Discrete-Time Ones
We have seen that the construction of bases from discrete-time signals can be accomplished easily
by using a perfect reconstruction filter bank as the basic building block. This gives us bases that
have a certain structure, and for which the analysis and synthesis can be efficiently performed. The
design of bases for continuous-time signals appears more difficult. However, it works out that we
can mimic many of the ideas used in the discrete-time case, when we go about the construction of
continuous-time bases.
In fact, there is a very close correspondence between the discrete-time bases generated by two-
channel filter banks, and dyadic wavelet bases. These are continuous-time bases formed by the
stretches and translates of a single function, where the stretches are integer powers of two:
{ψ
jk
(x) = 2
−j/2
ψ(2
−j
x −k), j, k, ∈ Z} (35.13)
This relation has been thoroughly explored in [11, 12].
To be precise, a basis of the form in (35.13) necessarily implies the existence of an underly ing

two-channel filter bank. Conversely, a two-channel filter bank can be used to generate a basis as in
(35.13) provided that the lowpass filter H
0
(z) is regular. It is not our intention to go into the details
of this connection, but the generation of wavelets from filter banks goes briefly as follows:
Considering the logarithmic tree of discrete-time filters in Fig. 35.3, one notices that the lower
branch is a cascade of filters H
0
(z) followed by subsampling by 2. It is easily shown [12], that the
cascade of i blocks of filtering operations, followed by subsampling by 2, is equivalent to a filter
H
(i)
0
(z) with z-transform:
H
(i)
0
(z) =
i−1

l=0
H
0
(z
2
l
), i = 1, 2 ···, (35.14)
followed by subsampling by 2
i
. We define H

(0)
0
(z) = 1 to initialize the recursion. Now, in addition
to the discrete-time filter, consider the function f
(i)
(x) which is piecewise constant on intervals of
length 1/2
i
, and equal to:
f
(i)
(x) = 2
i/2
· h
(i)
0
(n), n/2
i
≤ x<(n+ 1)/2
i
. (35.15)
Notethat the normalizationby2
i/2
ensuresthatif

(h
(i)
0
(n))
2

= 1 then

(f
(i)
(x))
2
dx = 1 as well.
Also, it can be checked that h
(i)
0

2
= 1 when h
(i−1)
0

2
= 1. The relation between the sequence
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H
(i)
0
(z) and the function f
(i)
(x) is clarified in Fig. 35.3, where the first three iterations of each is
shown for the simple case of a filter of length 4.
FIGURE35.3: Iterationsofthediscrete-timefilter(35.14)andthecontinuous-timefunction(35.15)
for the case of a length-4 filterH

0
(z). The length ofthe filterH
(i)
0
(z) increases withoutbound, while
the function f
(i)
(x) actually has bounded support.
We are going touse thesequenceof functionsf
(i)
(x) toconvergetothe scalingfunction φ(x)of a
waveletbasis. Hence,a fundamentalquestionis tofind out whetherand towhat the functionf
(i)
(x)
converge s as i →∞. First assume that the filter H
0
(z) has a zero at the half sampling frequency,
or H
0
(e
jπ
) = 0. This together with the fact that the filter impulse response is orthogonal to its
even translates is equivalent to

h
0
(n) = H
0
(1) =
√

2. Define M
0
(z) = 1/
√
2 · H
0
(z), that is
M
0
(1) = 1.NowfactorM
0
(z) intoitsrootsatπ (thereisatleastonebyassumption)andaremainder
polynomial K(z), in the following way:
M
0
(z) =[(1 + z
−1
)/2]
N
K(z).
Note that K(1) = 1 from the definitions. Now call B the supremum of |K(z)| on the unit circle:
B = sup
ω∈[0,2π ]
|K(e
jω
)|.
Then the following result from [11] holds:
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PROPOSITION35.2 [Daubechies 1988] If B<2
N−1
, and
∞

n=−∞
|k(n)|
2
|n|

< ∞, for some >0, (35.16)
thenthe piecewise constantfunction f
(i)
(x) definedin (35.15) convergespointwise toa continuous
function f
(∞)
(x).
This is a sufficient condition to ensure pointw ise convergence to a continuous function, and can
beusedasa simple test. We shallrefertoanyfilterforwhichtheinfiniteproductconvergesasregular.
If we indeed have convergence, then we define
f
(∞)
(x) = φ(x)
as the analysis scaling function, and
ψ(x) = 2
−1/2

h
1
(n)φ(2x − n), (35.17)

as the analysis wavelet. It can be shown that if the filters h
0
(n) and h
1
(n) arefromaperfectrecon-
struction filter bank, then (35.13) indeed forms a continuous-time basis.
In a similar way we examine the cascade of i blocks of the synthesis filter g
0
(n)
G
(i)
0
(z) =
i−1

l=0
G
0
(z
2
l
), i = 1, 2 ···. (35.18)
Again, define G
(0)
0
(z) = 1 to initialize the recursion, and normalize G
0
(1) = 1. From this define a
function which is piecewise constant on intervals of length 1/2
i

:
ˇ
f
(i)
(x) = 2
i/2
· g
(i)
0
(−n), n/2
i
≤ x<(n+ 1)/2
i
. (35.19)
We call the limit
ˇ
f
(∞)
(x), if it exists,
ˇ
φ(x) the synthesis scaling function, and we find
ˇ
φ(x) = 2
1/2
·
L−1

n=0
g
0

(−n) ·
ˇ
φ(2x −n) (35.20)
ˇ
ψ(x) = 2
1/2
·
L−1

n=0
g
1
(−n) ·
ˇ
φ(2x −n). (35.21)
The biorthogonality properties of the analysis and synthesis continuous-time functions follow from
the corresponding properties of the discrete-time ones. That is, (35.9) leads to
<
ˇ
φ(x),φ(x − k) > = δ
k
. (35.22)
and
<
ˇ
ψ(x), ψ(x −k) > = δ
k
.
(35.23)
Similarly

<
ˇ
φ(x),ψ(x − k) > = 0
(35.24)
<
ˇ
ψ(x), φ(x −k) > = 0, (35.25)
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come from (35.4) and (35.5), respectively.
We have shown that the conditions for perfect reconstruction on the filter coefficients lead to
functions that have the biorthogonality properties as shown above. Orthogonality across scales is
also easily verified:
<
ˇ
ψ(2
j
x),ψ(2
i
x −k) > = δ
i−j
δ
k
.
Thus, theset {ψ(2
j
x),
ˇ
ψ(2

i
x −k),i,j,k ∈ Z}is biorthogonal. Thatit is completecan be verified
as in the orthogonal case [13]. Hence, any function from L
2
(R) can be written:
f(x) =

j

l
< f (x), 2
−j/2
ψ(2
j
x −l) > 2
−j/2
ˇ
ψ(2
j
x −l).
Note that ψ(x) and
ˇ
ψ(x) play interchangeable roles.
35.1.2 Two-Channel Filter Banks and Wavelets
We have seen that the design of discrete-time bases is not difficult: using two-channel filter banks as
thebasicbuilding block theycan be easilyderived. We alsoknowthat,using (35.15)and (35.19),we
cangenerate continuous-timebases quite easilyas well. Ifwe were just interestedin the construction
of bases, with no further requirements, we could stop here. However, for applications such as com-
pression, we will often be interested in other propertiesof the basis functions, for example, whether
or not they have any symmetry or finite support, and whether or not the basis is an orthonormal

one. We examinethesethreestructuralproperties fortheremainderofthissection. Chapter36deals
with the design of the filters. Chapter 37 deals with time-varying filter banks, where the filters used,
or the tree structure employing them, varies over time. Chapter 38 deals with the case of Lapped
Transforms, a veryimportantclass of multirate filter banks that have achieved considerable success.
From the filter bank point of view, the properties we are most interested in are the following:
• Orthogonality:
<h
0
(n), h
0
(n + 2k) > = δ
k
= <h
1
(n), h
1
(n + 2k) >, (35.26)
<h
0
(n), h
1
(n + 2k) > = 0. (35.27)
• Linearphase: H
0
(z), H
1
(z), G
0
(z), and G
1

(z) are all linear phase filters.
• Finite support: H
0
(z), H
1
(z), G
0
(z), and G
1
(z) are all FIR filters.
The reason for our interest is twofold. First, these properties are possibly of value in perfect
reconstruction filter banks used in subband coding schemes. For example, orthogonality implies
that the quantization noise in the two channels will be independent; linear phase is possibly of
interestinverylowbit-ratecodingof images, andFIRfiltershavetheadvantageofhavingvery simple
low-complexity implementations. Second, these properties are carried over to the wavelets that are
generated. So, if we design a filter bank with a certain set of properties, then the continuous-time
basis that it generates will also have these properties.
PROPOSITION35.3 If the filters belong to an orthogonal filter bank, we shall have
< φ(x), φ(x +k) > = δ
k
=< ψ(x), ψ(x +k) >,
< φ(x), ψ(x +k) > = 0.
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PROOF35.1 Fromthedefinition (35.15)f
(0)
(x) isjusttheindicatorfunctionontheinterval[0, 1);
so we immediately get orthogonality at the 0th level, that is: <f
(0)

(x − l), f
(0)
(x − k) > = δ
kl
.
Now we assume orthogonality at the ith level:
<f
(i)
(x −l), f
(i)
(x −k) > = δ
kl
, (35.28)
and prove that this implies orthogonality at the (i +1)st level:
<f
(i+1)
(x −l), f
(i+1)
(x −k) > = 2

n

m
h
0
(n)h
0
(m)
<f
(i)

(2x −2l − n), f
(i)
(2x −2k − m) >
δ
n+2l−2k−m
2
=

n
h
0
(n)h
0
(n + 2l −2k)
= δ
kl
.
Hence, by induction (35.28) holds for all i. So in the limit i →∞:
<φ(x−l),φ(x − k) > = δ
kl
. (35.29)
The orthogonal casegives considerable simplification, both in thediscrete-time and continuous-
time cases.
PROPOSITION 35.4 If the filters belong to an FIR filter bank, then φ(x), ψ(x),
ˇ
φ(x), and
ˇ
ψ(x)
will have support on some finite interval.
PROOF 35.2 ThefiltersH

(i)
0
(z) andG
(i)
0
(z) definedin (35.14)haverespectivelengths(2
i
−1)(L
a
−
1) +1 and (2
i
−1)(L
s
−1) +1 whereL
a
andL
s
arethelengths of H
0
(z) and G
0
(z).Hence,f
(i)
(x)
in (35.15) is supported on the interval [0,L
a
− 1) and
ˇ
f

(i)
(x) on the interval [0,L
s
− 1). This
holds ∀ i; hence, in the limit i →∞this gives the support of the scaling functions φ(x) and
ˇ
φ(x).
That ψ(x) and
ˇ
ψ(x) have bounded support follow from (35.20) and (35.21).
PROPOSITION35.5 If the filters belong to a linear phase filter bank, then φ(x), ψ(x),
ˇ
φ(x), and
ˇ
ψ(x) will be symmetric or antisymmetric.
PROOF 35.3 The filter H
(i)
0
(z) will have linear phase if H
0
(z) does. If H
(i)
0
(z) has length (2
i
−
1)(L
a
−1)+1,thepointof symmetry is(2
i

−1)(L
a
−1)/2 whichneed not beaninteger. Thepoint
of symmetry for f
(i)
(x) will then be [(2
i
− 1)(L
a
− 1) + 1]/2
i+1
or [(2
i
− 1)(L
a
− 1) + 2]/2
i+1
.
Ineither case,by takingthe limiti →∞we findthat φ(x)is symmetric aboutthe point(L
a
−1)/2
and similarly for the other cases.
Thushavingestablishedtherelationbetweenwaveletsandfilterbankswecanexaminethestructure
of filter banks in detail, and afterward usethem to generate waveletsas described above. It shouldbe
emphasized thatwe are speaking of the two-channel, one-dimensional case. Multidimensional filter
banks are a large subject in their own right [8, 10].
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35.1.3 Structure of Two-Channel Filter Banks

We saw already that it is the choice of the function P(z)and the factorization taken that determines
the propertiesof thefilterbank. Interms ofP(z),we givenecessaryand sufficientconditionsfor the
three properties mentioned above:
• Orthogonality: P(z)isan autocorrelation, and H
0
(z) and G
0
(z) areits spectral factors.
• Linearphase: P(z)is linear phase, and H
0
(z) and G
0
(z) are its linear phase factors.
• Finite support: P(z)is FIR, and H
0
(z) and G
0
(z) are its FIR factors.
Obviouslythefactorizationisnotuniqueinanyofthecasesabove. TheFIRcasehasbeenexamined
in detail in [11, 12, 14, 15, 16] and the linear phase case in [12, 15, 17]. In the rest of this paper we
will present new results on the orthogonal case, but we shall also review the solutionsthat explicitly
satisfy simultaneous constraints.
PROPOSITION 35.6 To have an orthogonal filter bank it is necessary and sufficient that P(z)be
an autocorrelation, and that H
0
(z) and G
0
(z) be its spectral factors.
PROPOSITION35.7 To have a linear phase filter bank it is necessary and sufficient that P(z)be a
linear phase, and that H

0
(z) and G
0
(z) be its linear phase factors.
PROPOSITION35.8 To have an FIR filter bank itis necessary and sufficient thatP(z)be FIR, and
that H
0
(z) and G
0
(z) be its FIR factors.
Proofs can be found in [18]. Having seen that the design problem can be considered in terms of
P(z)and its factorizations, we consider the three conditions of interest from this point of view.
Orthogonality
In the case where the filter bank is to be orthogonal, we can obtain a complete constructive
characterization of the solutions, as given by the following theorem, taken from [18].
THEOREM 35.1 All orthogonal rational two channel filter banks can be for med as follows:
1. Choosing an arbitrary polynomial R(z), form:
P(z) =
2 ·R(z)R(z
−1
)
R(z)R(z
−1
) + R(−z)R(−z
−1
)
,
2. factor as P(z) = H(z)H(z
−1
),

3. form the filter H
0
(z) = A
0
(z)H (z),whereA
0
(z) is an arbitrary allpass,
4. choose H
1
(z) = z
2k−1
H
0
(−z
−1
)A
1
(z
2
),whereA
1
(z) is again an arbitrary allpass,
5. choose G
0
(z) = H
0
(z
−1
), and G
1

(z) =−H
1
(z
−1
).
Foraproof,see[18, 19].
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EXAMPLE 35.1:
Take R(z) = (1 + z
−1
)
N
as above and N = 7. It works out that in this case there is a closed form
factorization for the filters.
P(z) =
(1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1) · z
7
14z
6
+ 364z
4
+ 2002z
2
+ 3432 +2002z
−2
+ 364z
−4
+ 14z

−6
=
E(z)E(z
−1
)
K(z)K(z
−1
)
,
where
E(z)
K(z)
=
(1 + 7z
−1
+ 21z
−2
+ 35z
−3
+ 35z
−4
+ 21z
−5
+ 7z
−6
+ z
−7
)
√
2 ·(1 + 21z

−2
+ 35z
−4
+ 7z
−6
)
.
Note that we have used the following shorthand notation to list the coefficients of a causal FIR
sequence:
N−1

n=0
a
n
z
−n
= (a
0
,a
1
,a
2
, ···a
N−1
).
So,using thedescription ofthe filters inTheorem 35.1,with thesimplest caseA
0
(z) = A
1
(z) = 1

and k = 0 we find:
H
0
(z) =
(1 + 7z
−1
+ 21z
−2
+ 35z
−3
+ 35z
−4
+ 21z
−5
+ 7z
−6
+ z
−7
)
√
2 ·(1 + 21z
−2
+ 35z
−4
+ 7z
−6
)
H
1
(z) = z

−1
(1 − 7z
1
+ 21z
2
− 35z
3
+ 35z
4
− 21z
5
+ 7z
6
− z
7
)
√
2 ·(1 + 21z
2
+ 35z
4
+ 7z
6
)
G
0
(z) = H
0
(z
−1

)G
1
(z) = H
1
(z
−1
).
In the notation of Proposition35.2, B = 8 < 2
6
so thatfor this choice ofH
0
(z) theleft-hand side
of (35.15) converges to a continuous function. The wavelet, scaling function, and their spectra are
shown in Fig. 35.4.
Finite Impulse Response and Symmetric Solutions
Inthe casewherethe filtersareto beFIR, we merely requirethat P(z)beFIR; itis trivially easy
to designone. Similarly to havesymmetric filters,we merelyforce P(z)to be symmetric. Obviously
any symmetric P(z)which is FIR and satisfies (35.3) can be used to give symmetric FIR filters. We
wouldlike,inaddition, thatthelowpassfiltersareregular,sothatwegetsymmetricboundedsupport
continuous-time basis functions.
One strategy would be to design a P(z) with the desired properties and then factor to find the
filters. Alternatively,we canchoose one of the factors, and then find the other necessary to make the
product P(z) satisfy (35.3). We will use this approach and, to ensure regularity, choose one factor
to be (1 +z
−1
)
2N
. This can be done by solving a linear system of equations [12].
EXAMPLE 35.2:
If we choose N = 3 we must find the complement to (1 + z

−1
)
6
; so we solve the 3 by 3 system
found by imposing the constraints on the coefficients of the odd powers of z
−1
of
P(z) = (k
0
+ k
1
z
−1
+ k
2
z
−2
+ k
1
z
−3
+ k
0
z
−4
)
·(1 + 6z
−1
+ 15z
−2

+ 20z
−3
+ 15z
−4
+ 6z
−5
+ z
−6
) · z
5
.
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FIGURE 35.4: Example of Butterworth orthogonal wavelet; here N = 7, and the closed form
factorization has been used. (a) The wavelet. (b) Spectrum of the wavelet. (c) Scaling function.
(d) Spectrum of the scaling function.
So we solve:


61 0
20 16 6
12 30 20




k
0
k

1
k
2


=


0
0
1


,
giving k
6
= (3/2, −9, 19)/128.
In general, therefore, we solve the system:
F
2N
· k
2N
= e
2N
, (35.30)
whereF
2N
isthe N ×N matrix, k
2N
= (k

0
, ···,k
(k−1)
), and e
2N
isthe length k vecto r(0, 0, ···, 1).
Having found the coefficients of K
2N
(z), we factor it into linear phase components and then
regroupthese factorsof K
2N
(z) andthe 2N zeros at z =−1 to form two filters: H
0
(z) andH
1
(−z),
both of which are to be regular.
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FIGURE 35.5: Biorthogonal wavelets generated by filters of length 18 given in [12]. (a) Analysis
wavelet function ψ(x). (b) Spectrum of analysis wavelet. (c) Synthesis wavelet function
ˇ
ψ(x).
(d) Spectrum of synthesis wavelet.
35.1.4 Putting the Pieces Together
An importantconsideration that isoften encounteredin the design of wavelets, orof the filter banks
thatgeneratethem,isthenecessityofsatisfyingcompetingdesignconstraints. Thismakesitnecessary
to clearly understand whether desired properties are mutually exclusive.
Perfect reconstruction solutions, with the constraint that P(z)be rational with real coefficients,

must satisfy (35.3). Such general solutions, which do not necessarily have additional properties,
weregivenin[14].
ThesolutionsofsetA,whereallofthefiltersinvolvedareFIR,werestudiedin[14,15]. SetBcontains
all orthogonalsolutions, and has been the main focus of this paper. A complete characterization of
this set was given in Theorem 35.1. A very different characterization, based on lattice structures, is
givenin[20]. Particular cases of orthogonal solutions were also given in [21]. Set C contains the
solutions where all filters are linear phase, first examined in [15].
The earliest examples of perfect reconstruction solutions [22, 23] were orthogonal and FIR; i.e.,
theywereinA ∩ B. A constructive parametrization of A ∩B wasgivenin[24]. The construction
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and characterization of examples which converge to wavelets was first done in [11]. Filter banks
with FIR linear phase filters (i.e.,A ∩ C)werefirstgivenin[15], and also studiedin terms oflattices
in [17, 25]. The construction of wavelet examples is given in [13] and [12]. Filter banks, which are
linear phase and orthogonal, were constructed in Chapter 36 and were presented in [18].
That there exist only trivial solutions which are linear phase, orthogonal and FIR is indicated by
the intersection A ∩B ∩C; the only solutions are two tap filters [11, 12, 26].
It warrants emphasis that Fig. 35.6 illustrates the filter bank solutions; if the filters are regular,
then they will lead to wavelets. Of the dyadic wavelet bases known to the authors, the only ones
based onfilters whereP(z)isnot rational arethose of Meyer [27], and theonly ones where thefilter
coefficients are complexare those of Lawton [28]. For the case of the Battle-Lemari
´
e wavelets, while
the filters themselves are not rational, the P(z)function is; hence, the filters would belong to B ∩C
in the figure.
FIGURE 35.6: Two channel perfect reconstruction filter banks. The Venn diagram illustrates which
competingconstraintscanbesimultaneouslysatisfied. ThesetsA, B,C containFIR,orthogonal,and
linearphasesolutions,respectively. SolutionsintheintersectionA∩B areexaminedin[11,14,23,24];
those in theintersection A ∩C aredetailedin [12,13, 15,17, 25];solutions inB ∩C areconstructed

in [18]. The intersection A ∩B ∩ C contains only trivialsolutions.
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