Chapter 3
FIR Filter Banks and Compactly Supported
Wavelets
Contents
3.1
3.2
3.3
3.4
3.5
3.6
Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : :
Biorthogonal perfect reconstruction lter banks and wavelets
Algebraic structure of FIR solutions : : : : : : : : : : : : : :
Design results : : : : : : : : : : : : : : : : : : : : : : : : : :
Conclusion : : : : : : : : : : : : : : : : : : : : : : : : : : : :
Appendix : : : : : : : : : : : : : : : : : : : : : : : : : : : :
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3.1 Introduction
In this chapter we shall address the question of the design of nite impulse response
(FIR) lter banks. The rst solutions to the perfect reconstruction problem were of
this form 68, 78], and because of their simplicity they remain the most widely used
in practical subband coding systems. Our aim will be to give a complete examination
of the structure of two-channel lter banks where all of the lters involved are FIR.
We shall in particular examine the special cases, where in addition to being FIR,
52
53
the system is either orthogonal or linear phase. It is found that that a biorthogonal
two-channel lter bank leads, under certain constraints, to a biorthogonal wavelet
basis. Using the fact that the condition for perfect reconstruction in a two-channel
lter bank can be written as a polynomial Bezout identity we derive several new
results on the structure of the solutions. In particular, a complete characterization of
linear phase solutions is given these lead to symmetric biorthogonal bases of bounded
support wavelets.
3.1.1 Properties of two-channel lter banks
To begin we give an easy condition that characterizes FIR lter banks, and makes
use of the denition of valid functions introduced in (2.48)
Lemma 3.1 To have an FIR perfect reconstruction lter bank it is necessary and
sucient that P (z) be FIR, and that H0(z ) and G0 (z ) be its FIR factors.
Proof: Clearly, following Lemma 2.2, for H0(z) and G0(z) both to be FIR it is
necessary and sucient that P (z) be FIR. Equally, for H1(z) and G1(z) both to be
FIR it is necessary and sucient that P (;z) be FIR. 2
Thus, in order to design an FIR lter bank, we need merely nd an FIR function P (z)
satisfying (2.48). That is, all but one of the even-indexed samples will be zero, as
indicated in Figure 2-18, but in addition P (z) will have only a nite number of terms.
Given a lter H0(z) we call any lter H1(z) such that P (z) = H0 (z)H1(;z)z2l+1
satises (2.48) a complementary lter to H0(z). There are constraints, however
H0 (z) cannot be chosen arbitrarily if we wish the complementary lter to be FIR also.
Necessary and sucient conditions on H0 (z) for it to possess an FIR complementary
lter will be given below. An equivalent condition to the above lemma can be written
in terms of the matrix Hm (z) dened in (2.7).
54
Corollary 3.2 For perfect reconstruction with FIR synthesis lters after an FIR
analysis, it is necessary and sucient that:
detHm(z)] = 2 z;2l;1
where l 2 Z
and Hm (z) is the matrix de ned in (2.7).
Proof: From (2.8) if the analysis lters are FIR then the synthesis lters G0(z) and
G1 (z) are FIR , m(z) = 2 z2l+1:2
This reects a fact that we will encounter repeatedly: very often the constraints we
wish to impose on a lter bank can be expressed in a number of dierent ways. When
this is the case we will use whichever of the equivalent forms is most convenient for
our purposes. Clearly then for an FIR solution P (z) = H0(z)H1(;z)z2l+1. Generally
we shall be interested in lter banks that have also the properties of orthogonality or
linear phase, in addition to being FIR. We introduce conditions for these properties
to be satised, also in terms of P (z). First note that an autocorrelation is dened as
a function whose z-transform has the form A(z)A(z;1) for some function A(z) 86].
Lemma 3.3 To have an orthogonal lter bank it is necessary and sucient that P (z)
be an autocorrelation, and that H0(z ) and G0 (z ) be its spectral factors.
Proof: Equation (2.15) implies that every second term of the cross-correlation between the lters H0 (z) and H1 (z) is zero
H0 (z)H1(z;1) + H0(;z)H1(;z;1) = 0
(3.1)
so that H0(z)H1(z;1) has the form shown in Figure 2-8 (b). This implies that
H1(z) = z;1(z2)H0(;z;1)
(3.2)
55
for some function (z). Equally, (2.11) and (2.12) together imply that every second
term but one of the autocorrelations of H0(z) and H1(z) are zero
Hi (z)Hi(z;1) + Hi (;z)Hi(;z;1) = 2
i 2 f0 1g
(3.3)
so that Hi (z)Hi(z;1) has the form shown in Figure 2-8 (a). In order for this to be
true for both i = 0 and i = 1 requires that (z)(z;1 ) = 1, that is (z) is an allpass
function.
Now construct P (z)
P (z) = 2 H0(z)H1(;z)=m(z)(z)
;1
;1
2)
= ;z;1(z2;)2H z(z)HH0((zz;)1H)0+(zH )(;(zz)H
;1
0
0
0
0 (;z )]
= H0 (z)H0(z;1)
(3.4)
(3.5)
(3.6)
which is always an autocorrelation. Since G0(z) = H0(z;1) above it is clear that
H0 (z) and G0 (z) are the spectral factors of P (z). 2
Lemma 3.4 To have a linear phase lter bank it is necessary and sucient that
P (z) be linear phase, and that H0(z) and G0(z) be its linear phase factors.
Proof: Clearly for H0(z) and G0(z) both to be linear phase it is necessary that P (z)
be linear phase. If P (z) is linear phase it is sucient to take H0(z) and G0(z) as its
linear phase factors. If these lters are linear phase H1(z) and G1(z) will also be so.
2
Our designs of perfect reconstruction lter banks will thus involve nding a P (z)
with the appropriate properties, and which also satises (2.48), and then factoring
56
it. Often in fact we will impose more than a single constraint for example if we can
make P (z) linear phase and FIR then we can get an FIR linear phase lter bank,
and symmetric compactly supported wavelets. The structure of such designs are
examined in Section 3.2, and examples are given in Section 3.4.
3.2 Biorthogonal perfect reconstruction lter banks and
wavelets
Many of the ideas involved in the construction of linear expansions, wavelets and
multiresolution analysis were introduced using the case where orthogonal bases were
involved. This simplied the analysis considerably however, consideration of the
most general case allows greater freedom, and can be explained using simple extension
of the ideas involved for the orthogonal case.
Observe that, as a consequence of (2.47) G0(z)H1(z), i.e. the cross-correlation
of g1(n) and the time-reversed lter h0(;n), and G1(z)H0(z), the cross-correlation
of g1(n) and h0(;n), have only odd-indexed coecients, just as for the function in
Figure 2-8 (b), that is:
< g0(n) h1(2k ; n) > = 0
(3.7)
< g1(n) h0(2k ; n) > = 0
(3.8)
(note the time reversal in the inner product). If we now have H0 and H1 as dened
in (2.13), and dene
2
3
...
...
...
...
...
...
...
66
77
66 g (0) g (1) g (L ; 1)
77
0
0
0
0
0
.
.
6
T
.
.
. 777 (3.9)
G0 = 66 .
66
0
0 g0(0) g0(L ; 3) g0(L ; 2) g0(L ; 1) 77
4
5
...
...
...
...
...
...
...
57
so G0 has as its k-th column the elements of the sequence g0 (n ; 2k), we nd that
(3.7) gives that all rows of H1 are orthogonal to all columns of G0. Similarly dening
G1 to be the matrix containing the elements of the sequences g1(n ; 2k) in its k-th
column, we nd, from (3.8), that all of its columns are orthogonal to the rows of H0.
So in matrix notation:
H0G1 = 0 = H1G0:
(3.10)
Now P (z) = G0(z)H0(z) = z;1H0(z)H1(;z) and P (;z) = G1(z)H1(z) are both
valid and have the form given in Figure 2-18 (a). Hence the impulse responses of
gi (n) and hi(n) are orthogonal with respect to even shifts
In operator notation:
< gi(n) hi(2l ; n) > = l:
(3.11)
H0G0 = I = H1G1:
(3.12)
Since we have a perfect reconstruction system we get:
G0H0 + G1H1 = I:
(3.13)
Of course the last equation above indicates that no non-zero vector can lie in the
column nullspaces of both H0 and H1. Note that (3.12) implies that G0H0 and
G1H1 are each projections (since GiHiGiHi = Gi Hi). They project onto subspaces
which are not in general orthogonal (since the operators are not self-adjoint). Because
of (3.7), (3.8) and (3.11) the analysis/synthesis system is termed biorthogonal. If we
interleave the rows of H0 and H1, much as was done in the orthogonal case, and form
58
again a block Toeplitz matrix
2
...
...
...
66
66
66 h0(L ; 1) h0(L ; 2)
66 h (L ; 1) h (L ; 2)
1
A = 666 . . . 1
66
0
0
h0(L ; 1)
66
66
0
0
h1(L ; 1)
4
...
...
...
...
...
3
...
...
...
77
77
h0(0) 0
0
77
h1(0) 0
0 . . 777
.77 h0(2) h0(1) h0(0) 77
7
h1(2) h1(1) h1(0) 777
5
...
...
...
we nd that the rows of A form a basis for l2(Z ). If we form B by interleaving the
columns of G0 and G1 we nd
B A = I:
In the special case where we have a paraunitary solution one nds: G0 = HT0
and G1 = HT1 , and (3.10) gives that we have projections onto subspaces which are
mutually orthogonal. The system then simplies to the orthogonal case examined in
Section 2.4.1, that is B = AT .
3.2.1 Linear phase FIR lter banks
One reason to use biorthogonal rather than orthogonal bases is that the additional
freedom allows us to have arbitrary length linear phase lters. It is well known
23, 81, 107] that the only orthogonal (or paraunitary) real FIR lter bank having
linear phase has only two non-zero taps, and is given by H0(z) = z;l + z;l;2n;1 and
H1 (z) = (z;l ; z;l;2n;1 )z;2n . (Note that paraunitary lters can be factorized so as
to approximate linear phase 24, 91]).
To obtain longer real FIR lters, and still have exact linear phase, one has to give
up orthogonality. The classes of linear phase solutions are indicated below:
59
Proposition 3.5 Linear phase perfect reconstruction real FIR lter banks using l-
ters H0(z) and H1 (z) have one of the following forms:
(a) Both lters are symmetric and of odd lengths, di
ering by an odd multiple of 2.
(b) One lter is symmetric and the other is antisymmetric
both lengths are even, and are equal or di
er by an even multiple of 2.
(c) One lter is of odd length, the other is even both have all zeros on the unit circle.
Either both lters are symmetric, or one is symmetric and the other is antisymmetric.
Proof see Appendix 3.6.1.
We shall refer to any symmetric or antisymmetric lter that has a central term as
having whole sample symmetry (WSS) or whole sample antisymmetry (WSA), and
one that does not have a central term as having half sample symmetry or antisymmetry (HSS or HSA). The four types of symmetry are illustrated in Figure 3-1. In the
case of FIR lters WSS and WSA correspond to lters of odd length, and are often
referred to Type I and Type III lters respectively 83] whereas HSS and HSA imply
lters of even length of Type II and Type IV respectively. Clearly in Proposition 3.5
(a) above both lters are WSS, while in (b) one is HSS and the other is HSA. From
(2.47) the synthesis lter symmetries follow from those of the analysis. In class (c)
we nd that P (z) = H0(z)H1(;z)z2l+1 has zeros only at the 2n ; 1 roots of 1 so
the lters are of very little practical interest. We will not consider this trivial solution
further.
Note that while in the FIR case it is necessary to give up orthogonality to get nontrivial linear phase solutions, in the IIR case it is possible to have both. Such solutions
will be detailed in Chapter 4.
60
s
(a) WSS
s
s
s
s
s
(b) WSA
(c) HSS
s
(d) HSA
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
Figure 3-1: The various types of symmetry possible for a discrete sequence (a) Whole
Sample Symmetry. (b) Whole Sample Antisymmetry. (c) Half Sample Symmetry.
(d) Half Sample Antisymmetry.
3.2.2 Biorthogonal wavelets based on lter banks
Having above established the biorthogonality relations in the discrete-time case, we
now show how such bases may be used to generate continuous-time biorthogonal
bases. The analysis again follows readily by extension from the orthogonal case
presented in Section 2.6.1.
We denote by H0(i)(z) and G(0i)(z) the lters which are equivalent to the cascade
of i blocks of ltering and subsampling in the analysis, and ltering and upsampling
in the synthesis sections respectively. Recall from Section 2.2.3 that these can be
written
iY
;1
(i)
H0 (z) = H0(z2l )
i = 1 2 (3.14)
l=0
G(0i)(z) =
iY
;1
l=0
G0(z2l )
i = 1 2 :
(3.15)
Again, dene H0(0)(z) = G(0)
0 (z ) = 1 to initialize the recursions, and normalize
H0 (1) = G0(1) = 1. From these dene functions which are piecewise constant on
61
intervals of length 1=2i :
f (i)(x) = 2i=2 h(0i)(n)
f(i)(x) = 2i=2 g0(i)(;n)
n=2i x < (n + 1)=2i
n=2i x < (n + 1)=2i :
(3.16)
(3.17)
Note that we assume regularity, that is in the limit as i ! 1 both f (i)(x) and f(i)(x)
converge to continuous functions. Using the analysis in Section 2.6.2 we easily show
that f (i)(x) and f(i)(x) lead to limit functions which satisfy the two-scale dierence
equations:
(x) = 2 1=2
(x) = 21=2 LX
;1
n=0
LX
;1
n=0
h0(n) (2x ; n)
(3.18)
g0(;n) (2x ; n):
(3.19)
Dene also the associated bandpass functions:
(x) = 2 1=2
(x) = 21=2 LX
;1
n=0
LX
;1
n=0
h1(n) (2x ; n)
(3.20)
g1(;n) (2x ; n):
(3.21)
That (x) and (x) are orthogonal with respect to integer shifts is again shown inductively. We immediately get orthogonality at the 0-th level since f(0)(x) and f (0)(x)
are each equal to the indicator function on the interval 0 1): < f(0)(x) f (0)(x ; k) >
= kl: Again assume orthogonality at the i-th level: < f(i)(x) f (i)(x ; k) > = kl:
And then get:
< f (i+1)(x) f (i+1)(x ; k) >
= 2<
X
n
g0 (;n) f (i)(2x ; 2k ; n)
X
m
h0(m)f (i)(2x ; m) > m;2k;n =2
62
=
X
n
= k :
g0(;n) h0 (n ; 2k)
In the limit this gives:
Once this is established
< (x) (x ; k) > = k :
(3.22)
< (x) (x ; k) > = k (3.23)
follows immediately from (3.11) and the relations
< (x) (x ; k) > = 0
(3.24)
< (x) (x ; k) > = 0
(3.25)
come from (3.7) and (3.8) respectively.
We have shown that the conditions for perfect reconstruction on the lter coefcients lead to functions that have the biorthogonality properties as shown above.
Orthogonality across scale is also veried following the analysis of Section 2.6.2:
< (2j x) (2ix ; k) > = i;j k :
Thus the set f(2j x) (2ix ; k) i j k Z g is biorthogonal. That it is complete can
be veried as in the orthogonal case 16]. Hence any function from L2(R) can be
written:
XX
f (x) =
< f (x) 2;j=2(2j x ; l) > 2;j=2(2j x ; l):
j
l
Note that (x) and (x) play interchangeable roles.
So, regular biorthogonal FIR lter banks lead to biorthogonal bases of functions of
63
nite length it is easily shown that the converse is also true. Assume that we have
functions (x) (x) (x) (x) satisfying (3.18)-(3.21) and (3.22)-(3.25). Then it is
easy to verify that they can be used to derive biorthogonal lter banks. For example
using (3.22):
X
X
< (x) (x ; k) > = < g0 (;n) (2x ; n) h0(m) (2x ; 2k ; m) >
n
m
XX
=
g0(;n) h0(m) < (2x ; n) (2x ; 2k ; m) >
n m
X
=
g0(;n) h0(n ; 2k) = k :
n
The other lter biorthogonality relations follow from (3.23), (3.24) and (3.25).
3.2.3 Properties of the wavelets
Much emphasis has been placed on the design of lter banks, so that they could be
used to generate wavelets. In Section 2.6.2 it was seen that using an orthogonal lter
set in the iteration gave rise to an orthogonal wavelet basis. In like manner, above we
saw that biorthogonality of the discrete-time basis sequences implied biorthogonality
of the wavelet basis. Other properties of the lter bank are carried over to the wavelet
basis also.
Lemma 3.6 If the lters belong to an FIR lter bank then (x), (x), (x) and
(x) will have support on some nite interval.
Proof: The lter H0(i)(z) and G(0i)(z) dened in (2.34) have respective lengths (2i ;
1)(La ; 1) + 1 and (2i ; 1)(Ls ; 1) + 1 where La and Ls are the lengths of H0(z)
and G0(z). Hence f (i)(x) in (2.35) is supported on the interval 0 La ; 1) and f(i)(x)
on the interval 0 Ls ; 1). This holds 8 i hence in the limit i ! 1 this gives the
support of the scaling functions (x) and (x). That (x) and (x) have bounded
support follow from (3.20) and (3.21). 2
64
Lemma 3.7 If the lters belong to a linear phase lter bank then (x), (x), (x)
and (x) will be symmetric or antisymmetric.
Proof: The lter H0(i)(z) will have linear phase if H0 (z) does. If H0(i)(z) has length
(2i ; 1)(La ; 1) + 1 the point of symmetry is (2i ; 1)(La ; 1)=2 which need not be
an integer. The point of symmetry for f (i)(x) will then be (2i ; 1)(La ; 1) + 1]=2i+1
or (2i ; 1)(La ; 1) + 2]=2i+1 . In either case, by taking the limit i ! 1 we nd that
(x) is symmetric about the point (La ; 1)=2. Similarly for the other cases. 2
3.3 Algebraic structure of FIR solutions
3.3.1 Bezout's identity
From Corollary 3.2 we saw that in the FIR case the condition to achieve perfect
reconstruction (2.48) can be expressed
H0(z)H1(;z)z2l+1 ; H0(;z)H1(z)z2l+1 = 2:
(3.26)
It will prove convenient to introduce a matrix consisting of the polyphase components
of the lters
2
3 2
32
32
3
2
2
64 H00(z ) H01(z ) 75 = 1 64 H0 (z) H0(;z) 75 64 1 1 75 64 1 0 75 (3.27)
2 H1 (z) H1(;z) 1 ;1
H10(z2) H11(z2)
0 z
or:
2
32
3
1 1 76 1 0 7
Hp(z2) = 2;1 Hm(z) 64
54
5
1 ;1
0 z
where Hp(z) is called the polyphase matrix. Now, obviously
detHm(z)] = 2z;1 detHp(z2)]
(3.28)
65
and hence, from Corollary 3.2, in order to have an FIR system it is necessary and
sucient that
detHp(z)] = H00(z)H11(z) ; H01(z)H10(z)
(3.29)
= z;l:
Each of the equations (3.26) and (3.29) are equivalent to nding an FIR solution
to (2.48). These requirements of course considerably constrain the possible solutions
both (3.26) and (3.29) are special forms of what is known as a Bezout identity 57, 93].
This identity arises in the Euclidean algorithm which calculates the greatest common
divisor of two polynomials a0(x) and a1(x):
ar (x) = gcd(a0(x) a1(x))
where ar (x) is the last divisor of the algorithm and is a constant. It is well known
that we can always write:
ar (x) = p0(x)a0(x) + p1(x)a1(x)
for some p0(x) p1(x):
(3.30)
If a0(x) and a1(x) are coprime then ar (x) has zeros at 0 or 1 only and the identication between (3.30), the Bezout identity, and (3.26) and (3.29) becomes clear. By
examining the implications of this, new results emerge also from this angle certain
results that previously had more involved proofs (e.g. 55]) now become simpler. A
review of the necessary properties of the Euclidean algorithm can be found for example in 8]. The importance of the algorithm in the context of biorthogonal systems
has also been noted in 16] and 32]. The following easily proved fact will play an
important role in the rest of this section 8].
66
Fact 3.8 Given a(x) and b(x)
a(x)p(x) + b(x)q(x) = c(x)
has a solution p(x) q (x)] , gcd(a(x) b(x)) divides c(x).
We now make use of these observations to examine the constraints on the lter banks.
Proposition 3.9 Assume that the lters H0(z) and H1(z) are FIR and causal. Then
given one of the pairs H00(z) H01 (z )],H10(z ) H11(z )],H00(z ) H10 (z )] or
H01(z) H11(z)] in order to calculate the other pair necessary to achieve perfect reconstruction it is necessary and sucient that the given pair be coprime (except for
possible zeros at z = 1).
Proof From Fact 3.8 the gcd of each of these pairs must divide the right hand side
of (3.29). Hence the only factors that they can have in common are zeros at z = 1.2
The above Proposition is also proved using a dierent argument in 55].
Proposition 3.10 A lter H0(z) has a complementary lter if and only if it has no
zeros in pairs at z = and z = ;.
Proof H0(z) has a zero pair at ( ;) if and only if H0 (z) has a factor A(z2). This
happens if and only if both H00(z) and H01(z) have a common factor A(z), that
is they are not coprime. Thus the absence of zero pairs of the form ( ;) and
coprimeness are equivalent and the proof is completed by using Proposition (3.9). 2
Proposition 3.11 There is always a complementary lter to the binomial lter:
H0(z) = (1 + z;1)k = H00(z2) + z;1H01(z2):
(3.31)
67
Proof If H00(z) and H01(z) had a common factor it would appear as a pair of zeros
of H0(z) at ( ;) since H0(z) has zeros only at z = ;1 it cannot have such a
factor. 2
In fact a closed form for the coecients of a linear phase lattice structure that generates the complementary lter to the binomial is given in Appendix 3.6.2.
It should also be clear that for H0(z) and H1(z) to form a perfect reconstruction pair
it is necessary that they be coprime. From Fact 3.8 gcdH0(z) H1(z)] must divide
the right hand side of (3.26). This has a very clear signal processing interpretation:
a zero common to H0(z) and H1(z) would imply a transmission null in both channels
of the lter bank at some frequency, making reconstruction impossible.
3.3.2 An analogy with Diophantine equations
The conditions under which a complementary lter to some chosen H0(z) exists were
detailed in Section 3.3.1. It has already been pointed out that such a lter could
be found using Euclid's algorithm another method of nding a complementary lter
based on solving a set of linear equations is given in Section 3.4.2. The complement
is not unique. If given a lter H0(z), we can calculate one of its complements H1(z),
it is natural to wonder how we may nd others. The results of this section will show
that given any complementary lter, a simple mechanism exists for nding all others.
The basic idea stems from the form of the condition for perfect reconstruction
(2.48). Clearly if P (z) is valid, so is P (z) = P (z) + (z2)z for any (z)
0
P (z) + P (;z) = P (z) + P (;z) + z(z2) ; z(z2)]
0
0
= P (z) + P (;z):
Thus if (z2) contains H0(z) as a factor (i.e. (z2) = H0(z)
(z) for some FIR (z))
68
then
P (z) = H0(z)H1(;z) + z;2l
(z)] z2l+1
0
and hence H1(z) + z;2l
(;z) is also a complementary lter to H0(z). In fact, as we
shall show, this form is not merely sucient, but is also necessary. In addition (z)
has the form (z) = E (z2)H0(;z) and can be easily constrained to be linear phase.
For the following proposition we make use of Proposition 3.18 which gives that
the length N ; 2 linear phase complementary lter to an odd length N linear phase
lter is unique if it exists.
Proposition 3.12 Given a linear phase H0(z) of odd length N , and its length N ;
2 linear phase complement H1(z), all higher degree odd length linear phase lters
complementary to H0 (z) are of the form:
H1(z) = H1(z) + E (z)H0(z)
0
where
E (z) =
m
X
i=1
i(z;2(i;m;1) + z;2(m;i)):
Proof First note that H1(z) has length N +4m;2 and that, from Proposition 3.5, the
length of any higher degree complementary lter is indeed N +4m;2. Assume, for the
0
sake of deniteness, that the lters are causal, and that the coecient of z0 is the rst
non-zero one. Hence H0(z) is WSS where the central term is h0(l +1) and H1(z) is also
WSS with central term h1(l), where l = (N ; 3)=2. Thus P (z) = z2l+1H0(z)H1(;z)
is WSS with symmetry about the origin.
Observe that H1(z) above is indeed linear phase since z;1E (z), zl+1H0 (z) and
zlH1(z) are each symmetric about the origin, and then
0
zlH1(z) = zlH1(z) + z;1E (z)H0(z)zl+1
0
(3.32)
69
= zlH1(z) + E (z)H0(z)]
(3.33)
is also symmetric about the origin, while the point of symmetry for H1(z) is l. That
H1 (z) is indeed a complementary lter to H0(z) is shown by
0
0
z2l+1H0 (z)H1(;z) = z2l+1H0(z)H1(;z) + z2l+1H0(z)H0(;z)E (z) (3.34)
0
= P (z) + z(z2)
(3.35)
where (z2) = z2lH0(z)H0(;z).
We now show that all length N + 4m ; 2 solutions are of this form. Suppose
that F1(z) is a complementary lter of this length, and symmetric about the origin
i.e. zl+1H0 (z)F1(;z) satises (2.48). Thus the two functions zl+1 H0 (z )F1(;z ) and
z2l+1H0(z)H0(;z)E (z) are both of length 2N + 4m ; 3 and are symmetric about the
origin since the rst term is valid, and the second is of the form z(z2) we get that
P (z) = zl+1H0 (z)F1(;z) ; zlH0(;z)E (z)]
00
is valid. We can choose the coecients of E (z) to set some of the end terms of P (z)
to zero. That is for some choice of 1 the coecients of z;(N +2m;2) and zN +2m;2
become zero, so that P (z) is reduced in length by 4 (the coecients of z;(N +2m;3)
and zN +2m;3 are already zero). Similarly, for each of the i we can reduce the length
of P (z) by 4. When 1 m have been appropriately xed P (z) has length
2N ; 3, has powers of z in the range z;(N ;2) to zN ;2 and is still valid. Since it
contains H0(z) as a factor it must have the form:
00
00
00
00
P (z) = z2l+1H0(z)H1(;z)
00
since the length N ; 2 solution is unique by Proposition 3.18.2
70
From Proposition 3.5 it follows that in the two non-trivial cases of linear phase
solutions, the length of P (z) = H0(z)H1(;z) is 4n ; 1. In Appendix 3.6.3 it is shown
that the solutions indicated in Proposition 3.5 (b) are special cases of those in (a)
that is they can always be refactored into the form (a). It follows that all higher
degree complementary lters to a xed H0(z) are given by Proposition 3.12, unless
they are trivial, in the sense of belonging to class (c) of Proposition 3.5.
Example Consider H0 (z) = 1 4 6 4 1] and its unique length 3 complementary
lter H1(z) = 1 4 1]=16 we're using the vector h0(0) h0(1) h0(2) ] to list the
coecients of the impulse response. Let m = 2. So we get:
z5 z4 z3 z2
z41H0(z) 1 41 61 41
z22H0(z)
2 42
2H0(z)
z;21H0(z)
H1(z)
z1 z0 z;1 z;2 z;3 z;4 z;5
1
62 42 2
2 42 62 42 2
1 41 61 41 1
1=16 1=4 1=16
Giving for H1(z):
0
H1 (z) = 1 41 61 + 2 41 + 62 1=16 + 1 + 72 82 + 1=4 1=16 + 1 + 72
0
41 + 62 61 + 2 41 1]
which is linear phase, and complementary to H0(z).
A further result allows us to use the Diophantine equations to reach more general
solutions. For this we use the result (from Proposition 3.17) that a length N lter
has N ; 2 length N ; 2 complementary lters.
71
Proposition 3.13 All lters of length N + 2m ; 2 which are complementary to a
length N lter H0 (z) have the form:
H1 (z) = z;2k H1(z) + E (z)H0(z)
0
where E (z) = E (;z) is a polynomial of degree 2(m ; 1), k f0 1 mg and H1(z )
is a length N ; 2 complementary lter.
Proof That this is a solution is easily veried by substitution. If
R(z) =
mX
;1
i=0
iz;2i
then P (z) = H0(z)H1(;z) ; R(z)H0(;z)] is reduced in length by 2 for some choice
of 0 or m;1. Again the argument is repeated the length of P (z) can be reduced
by 2 at each stage by xing one of the i. In this way we reduce the degree of P (z)
by 2m when all of the i's are appropriately chosen. Since H0(z) is still a factor, the
remaining factor must be one of the length N ; 2 solutions of which there are only
N ; 2.2
00
0
00
00
So far we have addressed only the problem of generating higher degree solutions
from lower ones the next proposition shows that we can also go in the opposite
direction. We examine the linear phase case only the extension to the general case
is obvious. For the linear phase case this says that given any complementary lter
we can generate all others using the results of Proposition 3.12.
Proposition 3.14 If H1(1)(z) and H1(2)(z) are length N + 4m1 ; 2, and length N +
4m2 ; 2 linear phase complementary lters to the linear phase odd length N lter
H0 (z), with m2 > m1 then we can generate the lower degree solution from the higher
72
as follows:
H (1)(z) = H1(2)(z) + E3(z)H0(z)]
where E3 (z) = E1(z) ; E2 (z).
Proof Direct substitution.2
The subclass of all valid P (z)'s which can be factored as P (z) = H0(z)H0(z;1)
corresponds to paraunitary FB's, and generates orthonormal bases of wavelets. Since
all such P (z)'s are symmetric and positive they form a subclass of those generated by
the construction of Proposition 3.12 above. The case for which H0(z) is chosen to be
an even power of the binomial is treated in detail by Daubechies. In Appendix 3.6.5
we establish the close relation between her results and Proposition 3.12 above. The
fact that higher degree solutions contain lower degree ones is also given in 32]. The
freedom gained by using the Diophantine approach is examined in Appendix 3.6.6.
3.3.3 Continued fraction expansions
In Section 3.3.2 it was shown that any solution to (3.26) or (3.29) could be written
as the sum of lower degree solutions and trivial higher degree solutions (trivial in
the sense that the right hand side of (2.48) becomes zero). In Section 3.3.2 we dealt
exclusively with the modulation domain but the results there, like those of this
section could be expressed in either modulation or polyphase notation. For the sake
of deniteness consider the polyphase version of the perfect reconstruction condition
(3.29), and assume that H00(z) and H01(z) are given. It can be seen from Proposition
3.13 that this lowest degree solution is in some sense xed, and embedded at the core
of any higher degree solution. The strong connections between Euclid's algorithm,
the Bezout identity and continued fraction expansions (CFE's) is well known 93, 61].
In fact we now show that the canonic CFE of H10=H11 is the same as that of H00=H01
73
except for the last term and further that higher degree solutions are formed by adding
terms to the CFE, the rst terms remaining unchanged.
We dene D;1 = H00, D0 = H01, A;1 = H10 and A0 = H11. For the sake of
simplicity we remove the phase factor in (3.29). In this notation (3.29) becomes:
D;1 (z)A0(z) ; A;1(z)D0 (z) = 1:
(3.36)
Now use Euclid's algorithm starting with the pair D;1(z) D0(z). The rst step gives:
D;1 (z) = b0(z)D0(z) + D1 (z)
deg D0 > deg D1 :
Also do one division of the pair A;1(z) A0(z), denoting the remainder A1(z):
A;1(z) = a0(z)A0(z) + A1(z)
deg A0 > deg A1:
Together these equations give:
1 = D;1 (z)A0(z) ; A;1(z)D0(z)
= (b0(z) ; a0(z))A0(z)D0(z) + D1(z)A0(z) ; A1(z)D0(z):
However since deg A0D0 > deg D1A0 and deg A0D0 > deg D0A1 we must have a0 =
b0, and hence:
D0 (z)A1(z) ; A0(z)D1(z) = ;1:
Since this is of the same form, but of lower degree, than the equation that we started
with (3.36), we can compare the second step of Euclid's algorithm with a division of
A0(z) A1(z) and this gives a1(z) = b1(z). The result is that we get a succession of
74
Bezout identities:
Dj;1 (z)Aj (z) ; Dj (z)Aj;1(z) = (;1)j (3.37)
which are of decreasing degree. We nd in turn that a0(z) = b0(z) a1(z) = b1(z)
aj (z) = bj (z) aN (z) = bN (z). Note that these outputs of Euclid's algorithm
are the partial denominators of the canonical CFE of D;1 (z)=D0(z) 61, 93]:
D;1 (z) = H00(z) = b (z) +
0
D0(z) H01(z)
1
1
b1(z) +
1
b2(z) +
.. . +
1
bN (z )
= b0(z) b1(z) b2(z) bN (z)]
where we have used the standard notation b0 b1 b2 bN ] to denote a terminating
CFE 60]. It follows that the CFE's of D;1 (z)=D0(z) and A;1(z)=A0(z) are identical
for the rst N + 1 terms. The terminal equation for j = N gives:
N (z ) ; (;1)N
AN ;1(z) = DN ;1(z)AD
:
N
(3.38)
Remark that the Dj (z)'s are known, and DN is a scalar (since it is the last divisor
and D;1(z) D0(z) are assumed to be coprime) hence we can choose AN (z) to be any
polynomial and we are assured that Aj (z)'s are polynomial for j = N ; 1 0 ;1.
Therefore we get a valid complementary lter A0(z) A;1(z)] for any appropriate
AN (z). This can be expressed by writing aN +1(z) = (;1)N AN (z)DN . In summary:
H00(z) = b (z) b (z) b (z) b (z)]
0
1
2
N
H01(z)
(3.39)
75
H10(z) = b (z) b (z) b (z) b (z) ;(;1)N A (z)D ]:
0
1
2
N
N
N
H11(z)
(3.40)
It can be shown that choosing AN (z) = 0 gives the same solution as that produced
by Euclid's algorithm. That is:
H10(z) = b (z) b (z) b (z) b (z) 0]
0
1
N ;1
N
H11(z)
= b0(z) b1(z) bN ;1(z)]:
If we divide (3.37) by Dj (z)Dj;1 (z) we get:
Aj (z) ; Aj;1(z) = (;1)j Dj (z) Dj;1 (z) Dj (z)Dj;1 (z)
(3.41)
which is the expression relating successive convergents of a CFE 11]. This is precisely
the continued fraction in (3.39), (3.40). Note that since the successive convergents
satisfy the decreasing degree Bezout identities (3.37), truncating (3.39) and (3.40)
gives a lower degree FIR PRFB.
There is more structure yet. Dividing (3.37) by Dj;1 (z)Aj;1(z) gives:
Aj (z) ; Dj (z) =
(;1)j
Aj;1(z) Dj;1 (z) Aj;1(z)Dj;1 (z) :
Like (3.41) this relates successive convergents of a continued fraction but the CFE
here is that of a dierent grouping of the polyphase pairs. It turns out that we get:
H10(z) = ;(;1)N A (z)D b (z) b (z) b (z)]
N
N N
N ;1
0
H00(z)
H11(z) = ;(;1)N A (z)D b (z) b (z) b (z)]:
N
N N
N ;1
1
H01(z)
Of course these continued fraction relations give another method for generating
76
higher degree solutions to the perfect reconstruction condition. By varying AN (z) in
(3.40) for example we generate dierent complementary lters. Clearly this structure
is complete all possible complementary lters can be generated by appropriate choice
of AN (z).
It warrants reiteration that the above analysis using the polyphase notation, can
be used to give equivalent results in the modulation notation.
3.4 Design results
3.4.1 Filter design
Proposition 2.1 gives a sucient condition to ensure pointwise convergence to a continuous function, which hinges on the number of zeros at . For this reason in the
rst designs of orthogonal compactly supported wavelet bases 23] the emphasis was
placed on using lters that have a maximum number of zeros at z = ;1. In addition,
a zero of order N at z = ;1 in H0(z) implies N vanishing moments for the wavelet
23], see Appendix 3.6.4
Z1
k (x)dx = 0 k = 0 1 N ; 1:
x
;1
(3.42)
It can be shown that having a maximum number of zeros at z = ;1, implies a
maximally at characteristic for the lters involved 23, 94]. This implies that both
the wavelet and the lter spectrum have considerable smoothness, which may be
advantageous in certain contexts.
We shall for these reasons consider maximally at lters. Our procedure to design
linear phase FIR lters amounts then to the following:
(i) Choosing B2N (z) = (1 + z;1)2N for some N,
77
(ii) Finding the least degree FIR K2N (z) such that P (z) = B2N (z)K2N (z) is valid,
and
(iii) Factoring P (z) = H0(z)H1(;z)z2l+1:
Figure 3-2: Scaling function generated by using the lowpass lter 1 ; 1] and
2 3 ;3]. Sixth iteration is shown.
The question of regularity is more involved for the biorthogonal case than it is in
the orthogonal case, since now we have to check the regularity of both the analysis
(H0 (z)) and synthesis (H1(;z)) lowpass lters. Figure 3-2 illustrates the diculty of
making both H0(z) and H1(;z) regular for the length 4 linear phase case. Recall that
the impulse responses of H0 (z) and H1(;z) are given by 1 1] and 1 ; ; 1]
respectively 107]. The gure shows the scaling function generated by 1 1] for
;3 3]. At = 3 we have that H0(z) is very regular, while H1(;z) is very
irregular (as shown also in Figure 2-17). The dierence between the regularity of the
two lters clearly decreases as gets smaller. From the gure it would appear that
78
this never leads to two linear phase perfect reconstruction lters of length 4, a fact
that was proved in 112].
Actually at = 0 there is neither pointwise nor L2 convergence 23], (the gure
which shows the sixth iteration gives therefore an erroneous impression at = 0).
3.4.2 Linear phase lters with a maximum number of zeros at Step (ii) of the procedure above amounts to nding the complementary lter to the
binomial. First observe that if P (z) has even a single zero at z = ;1 it cannot have
any at z = 1 and vice versa. This should be clear since otherwise we could not have
P (z) + P (;z) = 2. We consider only the case where P (z) is of odd length (see
Proposition 3.5) hence it has an even number of zeros since P (z) is linear phase its
zeros must occur in quads, and pairs on the unit circle or real axis. It follows that
P (z) must have an even number of zeros at z = ;1 or none at all.
So we seek a linear phase FIR K2N (z) such that P (z) = B2N (z)K2N (z) is valid.
Note that by Proposition 3.16 we are assured that there exists a complementary lter
to the binomial of any degree. The lowest degree solution will be of length 2N ; 1,
and so P (z) will have length 4N ; 1. Since we know that p(2n) = n there are 2N ; 2
terms to be set to zero, but, because of symmetry, only half of them are independent
so there are N ; 1 zero constraints, which along with the p(2n) = n constraint gives
a total of N . K2N (z) has N independent coecients. Since the number of constraints
equals the number of degrees of freedom, and the system is linear, we have merely to
solve a square system.
Example If we choose N = 3 we must nd the complement to (1 + z;1)6 so we
solve the 3 by 3 system found by imposing the constraints on the coecients of the
odd powers of z;1 of
P (z) = (k0 + k1 z;1 + k2z;2 + k1z;3 + k0z;4)
79
(1 + 6z;1 + 15z;2 + 20z;3 + 15z;4 + 6z;5 + z;6) z5:
So we solve:
0
BB 6 1 0
BB 20 16 6
B@
12 30 20
10
CC BB k0
CC BB k
CA B@ 1
k2
1 0 1
CC BB 0 CC
CC = BB 0 CC CA B@ CA
1
giving k6 = (3=2 ;9 19)=128.
In general therefore we solve the system:
F2N k2N = e2N (3.43)
where F2N is the N N matrix, k2N = (k0 k(k;1) ) and e2N is the length k vector
(0 0 1).
The P (z) = B2N (z)K2N (z) functions for N = 1 2 5 are tabulated in Table 4.1.
Having found the coecients of K2N (z) we factor it into linear phase components
and then regroup these factors of K2N (z), and the 2N zeros at z = ;1 to form two
lters: H0(z) and H1(;z), both of which are to be regular.
It turns out that for small N , ensuring that both H0(z) and H1(;z) meet the
bound of Proposition 2.1 can force one to choose lters of quite unequal length. For
larger N however this problem eases, and it becomes possible to get lters of the
same, or approximately the same, length that generate regular symmetric wavelets.
For example Figure 3-3 shows the analysis and synthesis wavelets and their spectra
for one particular factorization of the N = 9 case, and Table 3.1 lists the coecients
of the lters H0(z) and H1(;z). Each of these lters has a factor (1 + z;1)9. We get
H0 (z) = (1 + z;1)=2]9F0(z) where B = sup!02] jF0(ej! )j ' 115:06 < 28 so H0(z)
satises the bound of Proposition 2.1. The second condition, (2.36) is always satised
1.5
1.2
1
1
0.5
0.8
Magnitude
Amplitude
80
0
0.6
-0.5
0.4
-1
0.2
-1.5
-8
-6
-4
-2
0
Time
2
4
6
0
0
8
5
10
15
1.5
1.2
1
1
0.5
0.8
0
0.4
-1
0.2
-8
-6
-4
-2
0
Time
(c)
35
40
45
30
35
40
45
0.6
-0.5
-1.5
30
(b)
Frequency
Frequency
(a)
20
25
Frequency
2
4
6
8
0
0
5
10
15
20
25
Time
(d)
Figure 3-3: Biorthogonal wavelets generated by lters of length 18 given in Table 3.1.
(a) Analysis wavelet function (x). (b) Spectrum of analysis wavelet. (c) Synthesis
wavelet function (x). (d) Spectrum of synthesis wavelet.
81
H0(z)
H1(z)
h0(0) 0.00122430 h0(17)
h1(0) 0.00122430 ;h1(17)
h0(1) -0.00069860 h0(16)
h1(1) 0.00069979 ;h1(16)
h0(2) -0.01183749 h0(15)
h1(2) -0.01134887 ;h1(15)
h0(3) 0.01168591 h0(14)
h1(3) -0.01141245 ;h1(14)
h0(4) 0.07130977 h0(13)
h1(4) 0.02347331 ;h1(13)
h0(5) -0.03099791 h0(12)
h1(5) 0.00174835 ;h1(12)
h0(6) -0.22632564 h0(11)
h1(6) -0.04441890 ;h1(11)
h0(7) 0.06927336 h0(10)
h1(7) 0.20436993 ;h1(10)
h0(8) 0.73184426 h0(9)
h1(8) 0.64790805 ;h1(9)
Table 3.1: Impulse response coecients of the lters H0(z) and H1(z) for the biorthogonal solution. The innite iteration of these lters give the wavelets shown in Figure
3-3.
1.2
1
1
0.5
Magnitude
Amplitude
0.8
0
0.6
0.4
-0.5
0.2
-8
-6
-4
-2
0
Time
(a)
2
4
6
8
0
0
5
10
15
20
25
Frequency
30
35
40
45
(b)
Figure 3-4: Orthogonal wavelet generated by lter of length 18 designed in 23]. (a)
Wavelet function (x). (b) Spectrum of wavelet.
82
for FIR sequences. Hence the scaling function (x) and wavelet (x) generated by
the innite iterations, following (3.18) and (3.20), converge to continuous functions.
H1 (;z) can be similarly factored, and B = sup!02] jF1(ej!+ )j ' 211:3 < 28 hence
(x) and (x) converge to continuous functions also.
Various methods for estimating the regularity index such that (x) (x) C are available 23, 92]. These are especially useful if a lter fails the test of Proposition
2.1. The method outlined in 92] yields the following estimates for the regularity index
: for (x) we nd 2:46 < , and for (x) we get 3:55 < . For comparison purposes
we show in Figure 3-4 the corresponding plots for the orthonormal wavelet generated
by a lter of the same length, as presented in 23].
3.4.3 More general solutions: Diophantine approach
While the design procedure of the previous section is very simple, the spectra of the
scaling function (!) and the wavelet (!) are not as one might wish from lowpass
and bandpass lters.
Proposition 3.12 showed how to generate any valid linear phase P (z) containing
a given factor. For example to design a P (z) with 2N zeros at , we can calculate
K2N (;z), as in the previous section. This solution, which has degree 2N ; 2, can
then be used to generate all possible solutions of degree 2N ; 2 + 2m.
A second approach is to note that we need not place all of the zeros at z = ;1
to start with. We could for example calculate the complementary lter to a factor
(1 + z;1)2j u1(z)u2(z) ul(z)]2 where ui(z) represents a zero pair on the unit circle.
We are then assured of having a factor (1 + z;1)j u1(z)u2(z) ul(z)] to place in the
stopband of each of the lters.
In Figure 3-5 we show the two symmetric wavelets (t) and (t) and their spectra.
These were generated by linear phase lters of lengths 24 and 20, the coecients of
1.5
1.2
1
1
0.5
0.8
Magnitude
Amplitude
83
0
0.6
-0.5
0.4
-1
0.2
-1.5
-10
-8
-6
-4
-2
Time
0
2
4
0
0
6
5
10
15
1.5
1.2
1
1
0.5
0.8
0
0.4
-1
0.2
-10
-8
-6
-4
Time
(c)
35
40
45
30
35
40
45
0.6
-0.5
-1.5
-12
30
(b)
Magbitude
Amplitude
(a)
20
25
Frequency
-2
0
2
4
0
0
5
10
15
20
25
Frequency
(d)
Figure 3-5: Biorthogonal wavelets generated by lters of lengths 20 and 24 given in
Table 3.2. The lters were designed using the Diophantine approach. (a) Analysis
wavelet function (x). (b) Spectrum of analysis wavelet. (c) Synthesis wavelet
function (x). (d) Spectrum of synthesis wavelet.
84
which are listed in Table 3.2. The lters were designed using a combination of the two
approaches described above. Each of the lters has only a single zero at z = ;1, and
neither of them meets the bound given in Proposition 2.1. However again using the
estimation methods of 92] we nd 0:79 < and 0:95 < . Note that the stopband
performance is much better than in Figures 3-3 and 3-4.
H0(z)
h0(0) 0.00133565 h0(23)
H1(z)
h0(1) -0.00201229 h0(22)
h1(0) 0.00465997 ;h1(19)
h0(2) -0.00577577 h0(21)
h1(1) 0.00702071 ;h1(18)
h0(3) 0.00863853 h0(20)
h1(2) -0.01559987 ;h1(17)
h0(4) 0.01279957 h0(19)
h1(3) -0.02327921 ;h1(16)
h0(5) -0.02361445 h0(18)
h1(4) 0.05635238 ;h1(15)
h0(6) -0.01900852 h0(17)
h1(5) 0.10021543 ;h1(14)
h0(7) 0.04320273 h0(16)
h1(6) -0.06596151 ;h1(13)
h0(8) -0.00931630 h0(15)
h1(7) -0.13387993 ;h1(12)
h0(9) -0.12180846 h0(14)
h1(8) 0.38067810 ;h1(11)
h0(10) 0.05322182 h0(13)
h1(9) 1.10398118 ;h1(10)
h0(11) 0.41589714 h0(12)
Table 3.2: Impulse response coecients of the lters H0(z) and H1(z) for the improved biorthogonal solution. The innite iteration of these lters give the wavelets
shown in Figure 3-5.
Figure 3-6 shows the wavelet and its spectrum for a better orthogonal set of lters.
The associated lter bank is paraunitary so H1(z) = H0(z;1). The coecients are
listed in Table 3.3. The regularity estimate here is 0:97 < .
3.5 Conclusion
We have demonstrated that the ideas used in the construction of orthogonal wavelet
bases, are readily extended to the non-orthogonal case. This allows use of general
perfect reconstruction lter banks to derive wavelet bases. The focus of this chapter
85
H0(z)
h0(0) 0.055739
h0(1) 0.288322
h0(2) 0.614682
h0(3) 0.608634
h0(4) 0.113646
h0(5) -0.290892
h0(6) -0.131805
h0(7) 0.162510
h0(8) 0.085330
h0(9) -0.099666
h0(10) -0.042965
h0(11) 0.060044
h0(12) 0.015233
h0(13) -0.032323
h0(14) -0.001634
h0(15) 0.014199
h0(16) -0.002305
h0(17) -0.004433
h0(18) 0.001808
h0(19) 0.000646
h0(20) -0.000577
h0(21) 0.000111
Table 3.3: Impulse response coecients of the lter H0(z) for the improved orthonormal solution. The innite iteration gives the wavelet shown in Figure 3-6.
86
1
1.2
0.8
1
0.6
0.8
0.2
Magnitude
Amplitude
0.4
0
-0.2
0.6
0.4
-0.4
-0.6
0.2
-0.8
-1
-10
-8
-6
-4
-2
Time
0
2
4
(a)
6
0
0
5
10
15
20
25
Frequency
30
35
40
45
(b)
Figure 3-6: Orthonormal wavelet generated by lter of length 22 whose coecients
are listed in Table 3.3. The lter was designed using the Diophantine approach. (a)
Wavelet function (x). (b) Spectrum of wavelet.
was the FIR case. Observing that when only FIR lters are involved the conditions
for perfect reconstruction reduce to a polynomial Bezout identity, we were able to
illuminate the very strong structure of this class of solutions and give new designs.
3.6 Appendix
3.6.1 Proof of Lemma 3.4
Proof: Since each of the lters is either symmetric or antisymmetric, P (z) is
symmetric or antisymmetric. In order to be valid, and have some symmetry, P (z)
must be symmetric and of odd length, so that the single non-zero coecient of an
odd power of z;1 is at the center. Thus H0(z) and H1(;z) have to be either both
symmetric or both antisymmetric and both must be either of even or of odd length.
Let L0 and L1 be the lengths of the lters H0(z) and H1(z) respectively. Note that
A(;z) has the same symmetry as A(z) if it is an odd length.
(a) L0 and L1 both odd. Now L0 + L1 ; 1 is odd. The center of symmetry, which is
(L0 + L1)=2 ; 1 samples away from the end points, has to be odd. Thus (L0 + L1)=2 =
87
L0 + (L0 ; L1)=2 has to be even. Thus (L0 ; L1)=2 is odd , and the length dierence
L0 ; L1 is an odd multiple of 2. In particular there are no same length solutions.
Suppose that H0(z) and H1(;z) can be both symmetric or both antisymmetric. The
latter possibility is ruled out however, because the two polyphase components are
also antisymmetric when the length is odd, and are therefore not coprime. Perfect
reconstruction is then impossible following Proposition 3.9.
(b)L0 and L1 both even. Again (L0 + L1)=2 ; 1 has to be odd so that the center term
is not in the same set as the end terms. Then L0 + (L0 ; L1)=2 has to be even, and
since L0 is even, this means that the length dierence L0 ; L1 is an even multiple of
2 and for example same length solutions do exist. Also it was assumed that H0(z)
and H1(;z) were both symmetric, but since H1(;z) has opposite symmetry to H1(z)
when the length is even, the even length solution leads to a symmetric/antisymetric
pair.2
3.6.2 Lattice structures generating the binomial lter and its complementary lter
Fact 3.15 Let the polyphase components of H0(z) and H1(z) be generated by the
following lattice structure:
2
3 2
32
3
N 1 0
1
1
Y
1
i 7
75 64
75 64
Hp(z) = 64
5
1 ;1 i=1 0 z
i 1
then the choice:
(1 2 N ) = ((2N + 1)=(2N ; 1) (2N + 1)=(2N ; 3) 2N + 1)
leads to:
H0(z) = (1 + z;1)2N +1:
(3.44)
88
An elegant proof of this fact was pointed out to us by R. Gopinath 37]. It is clear
that the complementary lter has only rational coecients.
3.6.3 Facts concerning FIR lter banks
A valid P (z) will in general be of length 2n ; 1, and in the case where it is symmetric
it will be of length 4n ; 1.
Fact 3.16 There is always a complementary lter to the binomial lter:
H0(z) = (1 + z;1)k = H00(z2) + z;1H01(z2):
(3.45)
Proof If H00(z) and H01(z) had a common factor it would appear as a pair of zeros
of H0(z) at ( ;) since H0(z) has zeros only at z = ;1 it cannot have such a
factor. 2
Fact 3.17 For the general case a length N lter has N ; 2 complementary lters of
length N ; 2.
Proof In this case the length of H (z)C (;z) = 2N ; 3, where C (z) is the complementary lter. So we have N ; 2 coecients of odd powers of z;1. The single non-zero
coecient can be placed in any of the N ; 2 positions. We solve an N ; 2 by N ; 2
linear system where the N ; 2 unknowns are the coecients of C (z).2
Fact 3.18 For the linear phase case, an even length N lter has a unique same length
complementary lter. For N odd there is a unique length N ; 2 complementary lter.
Proof N even gives N=2 equations in N=2 unknowns the non-zero coecient must
be in the centre. Similarly N odd gives (N ; 1)=2 equations in (N ; 1)=2 unknowns.
2
89
Note: If we have a linear phase lter of length N and it's length N complement,
note that this can always be refactored to give two lters of length N + 1 and N ; 1
but the converse is not true.
Fact 3.19 Given a lter H0 (z) and H1(z) which is any of its complementary lters of length N ; 2, then all higher degree complementary lters H1 (z ) must have:
degH1 (z)] = degH1(z)] + 2(m ; 1), for m = 1 2 3 0
0
Proof A valid P (z) is constrained to have degree 2n ; 2:2
Fact 3.20 Given an odd length linear phase lter H0 (z) and its unique complementary lter H1 (z) of length N ; 2, then all linear phase higher degree complementary
lters H1 (z) must have: degH1(z )] = degH1(z )] + 4m, for m = 1 2 3 0
0
Proof For a linear phase solution P (z) is constrained to have degree 4n ; 2. 2
Note: The above facts show that the complementary lters of the appropriate degree
will exist in general provided that the system of equations to be solved is not singular.
The condition for non-singularity was given in Fact 3.9. Also observe that there
possibly exist complementary lters of lower degree than those mentioned this would
imply that there are more equations than unknowns, but that a solution nonetheless
exists.
3.6.4 Zeros at z = ;1 and moments of the wavelet
If a lter H0(z) has a zero of order N at z = ;1, and is from an orthogonal lter
bank, so that, by Lemma 4.1(c):
Hi (ejw )Hi(e;jw ) + Hi(;ejw )Hi(;e;jw ) = 2
(3.46)
90
then together these imply:
dk Hi(ejw ) = 0 k = 1 2 N ; 1
dwk w=0
(3.47)
and H0(1) = H1(0) = 0, come from (2.27), (2.30). For this reason lters which have a
maximum number of zeros at z = ;1 are refered to a maximally at. Indeed Shensa
94] has pointed out that Daubechies' lters are in fact identical to the maximally
at lters derived by Hermann in a dierent context 50].
To see how this property aects the wavelet observe that, from (2.28):
(w) = H1(w=2) (w=2):
(3.48)
This clearly gives:
!
jw !
dH
d
(
w
)
d(w) 1 (e ) =
(
w=
2)
+
H
1 (w=2) dw w=0
dw
dw w=0
w=0
= 0
since H1(0) = H1(0) = 0. This immediately gives:
0
Z1
;1
Similarly:
which gives:
x(x)dx = 0:
dk (w) = 0
dwk w=0
Z1
;1
k = 0 1 N ; 1
xk (x)dx = 0 k = 0 1 N ; 1:
Entirely similar analysis can be followed for the non-orthogonal case.
(3.49)
91
3.6.5 Relation to work of I. Daubechies
We here establish the connection between Lemma 3.12 and a result given in a dierent
form in 23].
Daubechies writes: P (1)(z) = (1 + z;1)2k Q(1)(z) except for a phase factor this
gives:
;jw=2 + ejw=2
)2k Q(w)
P (w) = ( e
2
Since Q(w) is symmetric it can be written as a polynomial in cos(w) = 1=2(1 ;
cos2(w=2)) when restricted to the unit circle. So if we dene y = cos2(w=2) and we
can rewrite the above as:
P (w) = cos2(w=2)]k Q1=2(1 ; cos2(w=2))
P (y) = yk Q1=2(1 ; y)
Using the results of lemma (3.12) we know that higher degree solutions are written:
Q (;z) = z;2m Q1=2(;z) + E (z)(1 ; z;1)2k
0
Since this is symmetric it can be written as a function of cos(w).
Q1=2(1 ; cos2(w=2)) = Q1=2(1 ; cos2(w=2)) + E1=2(cos(w))sin2(w=2)]k
0
Clearly this gives:
Q1=2(y) = Q1=2(y) + yk E1=2(y)
0
where our constraint E (z) = E (;z) translates to E1=2(y) = E1=2(1 ; y), giving that
E1=2(y) is symmetric about the point 1=2. The fact that E (z) is of degree 4m ; 2
gives that E1=2(y) has degree 2m ; 1. If we alter our notation to use zero phase
92
polynomials throughout we would nd E (z) = ;E (;z) or E1=2(y) = ;E1=2(1 ; y),
which is precisely the requirement given in 23].
Further I.Daubechies gives a closed form for Q1=2(y). In our notation this translates to the complementary lter to the binomial.
3.6.6 Root loci of higher degree solutions
While we have shown that higher degree solutions can give better results, it is still
clear from Sections 3.3.2 and 3.3.3 that these solutions are nonetheless very constrained. To give a concrete example we briey examine the case where H0(z) =
(1 + z;1)6 and the unique linear phase degree 4 complementary lter has impulse
response coecients h1(n) = 3 18 38 18 3]=28 . We examine the m = 1 solution
(from Proposition 3.12):
H1(z) = z;2H1(z) + (1 + z;2)H0(z)
0
(3.50)
and plot the trajectory in the z-plane of the roots as is varied.
Figure 3-7 shows the root locus for the region: ;2 2]. While the increased degree solution has more freedom, it is clear that the roots move along very constrained
paths. In other words it would be necessary to look at solutions of considerably
higher degree to get substantial design freedom.
93
Figure 3-7: Locus of the movement of the roots of (3.50) in the z-plane for 2 ;2 2]: