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Copyright © 2001, S. K. Mitra1
Perfect Reconstruction Two-Perfect Reconstruction Two-Channel FIR Filter BanksChannel FIR Filter Banks
• A perfect reconstruction two-channel FIRfilter bank with linear-phase FIR filters canbe designed if the power-complementaryrequirement
between the two analysis filters and is not imposed
1|)(||)(| 21
20 =+ ωω jj eHeH
)(zH0)(zH1
Copyright © 2001, S. K. Mitra2
Perfect Reconstruction Two-Perfect Reconstruction Two-Channel FIR Filter BanksChannel FIR Filter Banks
• To develop the pertinent design equationswe observe that the input-output relation ofthe 2-channel QMF bank
can be expressed in matrix form as
)()}()()()({)( zXzGzHzGzHzY 110021 +=
)()}()()()({ zXzGzHzGzH −−+−+ 110021
[ ]
−
−−=
)()(
)()()()(
)()()( zXzX
zHzHzHzHzGzGzY
1100
1021
Copyright © 2001, S. K. Mitra3
Perfect Reconstruction Two-Perfect Reconstruction Two-Channel FIR Filter BanksChannel FIR Filter Banks
• From the previous equation we obtain
• Combining the two matrix equations we get
[ ]
−
−−−−=−
)()(
)()()()(
)()()( zXzX
zHzHzHzHzGzGzY
1100
1021
−
−−
−−=
− )()(
)()()()(
)()()()(
)()(
zXzX
zHzHzHzH
zGzGzGzG
zYzY
1100
1010
21
−=)(
)()]()[( )()(zX
zXzz Tmm HG21
Copyright © 2001, S. K. Mitra4
Perfect Reconstruction Two-Perfect Reconstruction Two-Channel FIR Filter BanksChannel FIR Filter Banks
where
are called the modulation matrices
−−=)()(
)()()()(
zGzGzGzGzm
1010G
−−=)()(
)()()()(
zHzHzHzHzm
1010H
Copyright © 2001, S. K. Mitra5
Perfect Reconstruction Two-Perfect Reconstruction Two-Channel FIR Filter BanksChannel FIR Filter Banks
• Now for perfect reconstruction we must have and correspondingly
• Substituting these relations in the equation
we observe that the PR condition is satisfiedif
)()( zXzzY l−=)()()( zXzzY −−=− −l
−=
− )()()]()[(
)()( )()(
zXzXzzzY
zY Tmm HG21
−
= −
−
l
l
)()]()[( )()(
zzzz Tmm0
021 HG
Copyright © 2001, S. K. Mitra6
Perfect Reconstruction Two-Perfect Reconstruction Two-Channel FIR Filter BanksChannel FIR Filter Banks
• Thus, knowing the analysis filtersand , the synthesis filters and
are determined from
• After some algebra we arrive at
)(zH0)(zH1 )(zG0
)(zG1
( ) 1
002
−−
−
−
= Tmm zz
zz )]([)(
)( )()( HG l
l
Copyright © 2001, S. K. Mitra7
Perfect Reconstruction Two-Perfect Reconstruction Two-Channel FIR Filter BanksChannel FIR Filter Banks
where
and is an odd positive integer
)()](det[
)()(
zHz
zzG m −⋅=−
102H
l
)()](det[
)()(
zHz
zzG m −⋅−=−
012H
l
)()()()()](det[ )( zHzHzHzHzm1010 −−−=H
l
Copyright © 2001, S. K. Mitra8
Perfect Reconstruction Two-Perfect Reconstruction Two-Channel FIR Filter BanksChannel FIR Filter Banks
• For FIR analysis filters and ,the synthesis filters and willalso be FIR filters if
where c is a real number and k is a positiveinteger
• In this case
)(zH0 )(zH1)(zG0 )(zG1
km czz −=)](det[ )(H
)()( )( zHzzG kc −= −−
12
0l
)()( )( zHzzG kc −−= −−
02
1l
Copyright © 2001, S. K. Mitra9
Orthogonal Filter BanksOrthogonal Filter Banks
• Let be an FIR filter of odd order Nsatisfying the power-symmetric condition
• Choose• Then
)(0 zH
1)()()()( 100
100 =−−+ −− zHzHzHzH
)()( 101
−− −= zHzzH N
)](det[ )( zmH
( ) NN zzHzHzHzHz −−−− −=−−+−= )()()()( 100
100
Copyright © 2001, S. K. Mitra10
Orthogonal Filter BanksOrthogonal Filter Banks• Comparing the last equation with
we observe that and k = N• Using in
with we get
km czz −=)](det[ )(H1−=c
)()( 101
−− −= zHzzH N
)()( )( zHzzG kc −= −−
12
0l
)()( )( zHzzG kc −−= −−
02
1l
Nk ==l)(2)(),(2)( 1
111
00−−−− == zHzzGzHzzG NN
Copyright © 2001, S. K. Mitra11
Orthogonal Filter BanksOrthogonal Filter Banks
• Note: If is a causal FIR filter, theother three filters are also causal FIR filters
• Moreover,• Thus, for a real coefficient transfer function
if is a lowpass filter, then is ahighpass filter
• In addition,
)(0 zH
|)(||)(| 01ωω −= jj eHeH
)(0 zH )(1 zH
2,1,|)(||)(| == ωω ieHeG ji
ji
Copyright © 2001, S. K. Mitra12
Orthogonal Filter BanksOrthogonal Filter Banks• A perfect reconstruction power-symmetric
filter bank is also called an orthogonal filterbank
• The filter design problem reduces to thedesign of a power-symmetric lowpass filter
• To this end, we can design a an even-order whose spectral
factorization yields
)(0 zH
)()()( 100
−= zHzHzF)(0 zH
Copyright © 2001, S. K. Mitra13
Orthogonal Filter BanksOrthogonal Filter Banks• Now, the power-symmetric condition
implies that F(z) be a zero-phase half-bandlowpass filter with a non-negativefrequency response
• Such a half-band filter can be obtained byadding a constant term K to a zero-phasehalf-band filter Q(z) such that
1)()()()( 111
100 =−−+ −− zHzHzHzH
)( ωjeF
ω≥+= ωω allfor0)()( KeQeF jj
Copyright © 2001, S. K. Mitra14
Orthogonal Filter BanksOrthogonal Filter Banks
• Summarizing, the steps for the design of areal-coefficient power-symmetric lowpassfilter are:
• (1) Design a zero-phase real-coefficientFIR half-band lowpass filter Q(z) of order2N with N an odd positive integer:
∑−=
−=N
Nn
nznqzQ ][)(
)(0 zH
Copyright © 2001, S. K. Mitra15
Orthogonal Filter BanksOrthogonal Filter Banks• (2) Let δ denote the peak stopband ripple of
• Define F(z) = Q(z) + δ which guaranteesthat for all ω
• Note: If q[n] denotes the impulse responseof Q(z), then the impulse response f [n] ofF(z) is given by
• (3) Determine the spectral factor ofF(z)
)( ωjeQ
0)( ≥ωjeF
0for0for
≠=+
nnqnnq
],[,][ δ=][nf
)(0 zH
Copyright © 2001, S. K. Mitra16
Orthogonal Filter BanksOrthogonal Filter Banks
• Example - Consider the FIR filter
where R(z) is a polynomial in of degree with N odd
• F(z) can be made a half-band filter bychoosing R(z) appropriately
• This class of half-band filters has beencalled the binomial or maxflat filter
)()()( zRzzzF NN 111 +−+=1−z
1−N
Copyright © 2001, S. K. Mitra17
Orthogonal Filter BanksOrthogonal Filter Banks
• The filter F(z) has a frequency responsethat is maximally flat at ω = 0 and at ω = π
• For N = 3,resulting in
which is seen to be a symmetric polynomialwith 4 zeros located at , a zero at
, and a zero at
)()( 21161 41 −− −+−= zzzR
)()( 313161 9169 −− −+++−= zzzzzF
1−=z32 −=z 32 +=z
Copyright © 2001, S. K. Mitra18
Orthogonal Filter BanksOrthogonal Filter Banks
• The minimum-phase spectral factor istherefore the lowpass analysis filter
• The corresponding highpass filter is givenby
])([)(.)( 1210 321134150 −− −−+−= zzzH
)...(. 321 268046407321134150 −−− −++−= zzz
)...(. 321 7321464102679034150 −−− +−+−= zzz)()( 1
03
1−− −= zHzzH
Copyright © 2001, S. K. Mitra19
Orthogonal Filter BanksOrthogonal Filter Banks• The two synthesis filters are given by
• Magnitude responses of the two analysisfilters are shown on the next slide
)()( 10
30 2 −−= zHzzG
)...(. 321 732146410267906830 −−− +++−−= zzz)()( 1
13
1 2 −−= zHzzG)...(. 321 2679046410732116830 −−− ++−−= zzz
Copyright © 2001, S. K. Mitra20
Orthogonal Filter BanksOrthogonal Filter Banks
• Comments: (1) The order of F(z) is of theform 4K+2, where K is a positive integer
• Order of is N = 2K+1, which isodd as required
)(zH0
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω/π
Mag
nitu
de
H0(z) H
1(z)
Copyright © 2001, S. K. Mitra21
Orthogonal Filter BanksOrthogonal Filter Banks• (2) Zeros of F(z) appear with mirror-image
symmetry in the z-plane with the zeros onthe unit circle being of even multiplicity
• Any appropriate half of these zeros can begrouped to form the spectral factor
• For example, a minimum-phase canbe formed by grouping all the zeros insidethe unit circle along with half of the zeroson the unit circle
)(zH0
)(zH0
Copyright © 2001, S. K. Mitra22
Orthogonal Filter BanksOrthogonal Filter Banks
• Likewise, a maximum-phase can beformed by grouping all the zeros outside theunit circle along with half of the zeros onthe unit circle
• However, it is not possible to form aspectral factor with linear phase
• (3) The stopband edge frequency is thesame for F(z) and
)(zH0
)(zH0
Copyright © 2001, S. K. Mitra23
Orthogonal Filter BanksOrthogonal Filter Banks
• (4) If the desired minimum stopbandattenuation of is dB, then theminimum stopband attenuation of F(z) is
dB• Example - Design a lowpass real-coefficient
power-symmetric filter with thefollowing specifications: , and
dB
sα)(zH0
0262 .+sα
)(zH0πω 630.=s
12=sα
Copyright © 2001, S. K. Mitra24
Orthogonal Filter BanksOrthogonal Filter Banks• The specifications of the corresponding zero-
phase half-band filter F(z) are therefore: and dB
• The desired stopband ripple is thuswhich is also the passband ripple
• The passband edge is at• Using the function remezord we first
estimate the order of F(z) and then using thefunction remez design Q(z)
πω 630.=s 40=sα010.=sδ
πππω 370630 .. =−=p
Copyright © 2001, S. K. Mitra25
Orthogonal Filter BanksOrthogonal Filter Banks• The order of F(z) is found to be 14 implying
that the order of is 7 which is odd asdesired
• To determine the coefficients of F(z) we adderr (the maximum stopband ripple) to thecentral coefficient q[7]
• Next, using the function roots we determinethe roots of F(z) which should thereticallyexhibit mirror-image symmetry with doubleroots on the unit circle
)(zH0
Copyright © 2001, S. K. Mitra26
Orthogonal Filter BanksOrthogonal Filter Banks
• However, the algorithm s numerically quitesensitive and it is found that a slightly largervalue than err should be added to ensuredouble zeros of F(z) on the unit circle
• Choosing the roots inside the unit circlealong with one set of unit circle roots we getthe minimum-phase spectral factor )(zH0
Copyright © 2001, S. K. Mitra27
Orthogonal Filter BanksOrthogonal Filter Banks
• The zero locations of F(z) and areshown below
)(zH0
-1 0 1-1
-0.5
0
0.5
1
Real Part
Imag
inar
y P
art
Zero locations of H0(z)
7
-1 0 1 2
-1
-0.5
0
0.5
1
Real Part
Imag
inar
y P
art
Zero locations of F(z)
2
2
2
214
Copyright © 2001, S. K. Mitra28
Orthogonal Filter BanksOrthogonal Filter Banks
• The gain responses of the two analysisfilters are shown below
0 0.2 0.4 0.6 0.8 1-50
-40
-30
-20
-10
0
10
ω/π
Gai
n, d
B
H0(z) H
1(z)
Copyright © 2001, S. K. Mitra29
Orthogonal Filter BanksOrthogonal Filter Banks
• Separate realizations of the two filtersand would require 2(N+1)multipliers and 2N two-input adders
• However, a computationally efficientrealization requiring N+1 multipliers and 2Ntwo-input adders can be developed byexploiting the relation
)(zH0)(zH1
)()( 101
−− −= zHzzH N
Copyright © 2001, S. K. Mitra30
ParaunitaryParaunitary System System• A p-input, q-output LTI discrete-time
system with a transfer matrix iscalled a paraunitary system if is aparaunitary matrix, i.e.,
• Note: is the paraconjugate ofgiven by the transpose of witheach coefficient replaced by its conjugate
• is an identity matrix, c is a realconstant
)(zpqT)(zpqT
ppqpq czz ITT =)()(~
)(zpqT)( 1−zpqT
)(zpqT~
pI pp×
Copyright © 2001, S. K. Mitra31
ParaunitaryParaunitary Filter Banks Filter Banks
• A causal, stable paraunitary system is also alossless system
• It can be shown that the modulation matrix
of a power-symmetric filter bank is aparaunitary matrix
−−=)()(
)()()()(
zHzHzHzHzm
1010H
Copyright © 2001, S. K. Mitra32
ParaunitaryParaunitary Filter Banks Filter Banks• Hence, a power-symmetric filter bank has
also been referred to as a paraunitary filterbank
• The cascade of two paraunitary systemswith transfer matrices and isalso paraunitary
• The above property can be utilized indesigning a paraunitary filter bank withoutresorting to spectral factorization
)()( zpq1T )()( zqr
2T
Copyright © 2001, S. K. Mitra33
Power-Symmetric FIRPower-Symmetric FIRCascaded Lattice StructureCascaded Lattice Structure
• Consider a real-coefficient FIR transferfunction of order N satisfying thepower-symmetric condition
• We shall show now that can berealized in the form of a cascaded latticestructure as shown on the next slide
)(zHN
NNNNN KzHzHzHzH =−−+ −− )()()()( 11
)(zHN
Copyright © 2001, S. K. Mitra34
Power-Symmetric FIRPower-Symmetric FIRCascaded Lattice StructureCascaded Lattice Structure
• Define
,)()(
)(zXzXzH i
i0
=)(
)()(
zXzYzG i
i0
=
Ni ≤≤1
Copyright © 2001, S. K. Mitra35
Power-Symmetric FIRPower-Symmetric FIRCascaded Lattice StructureCascaded Lattice Structure
• From the figure we observe that
• Therefore,
• It can be easily shown that
)()()( zXzkzXzX 01
101−+=
)()()( zXzzXkzY 01
011−+−=
111
111 1 −− +−=+= zkzGzkzH )(,)(
)()( 11
11
−− −= zHzzG
Copyright © 2001, S. K. Mitra36
Power-Symmetric FIRPower-Symmetric FIRCascaded Lattice StructureCascaded Lattice Structure
• Next from the figure it follows that
• It can easily be shown that
provided
)()()( zGzkzHzH iiii 22
2 −−
− +=)()()( zGzzHkzG iiii 2
22 −
−− +−=
)()( 1−− −= zHzzG ii
i
)()( )( 12
22
−−
−−− −= zHzzG i
ii
Copyright © 2001, S. K. Mitra37
Power-Symmetric FIRPower-Symmetric FIRCascaded Lattice StructureCascaded Lattice Structure
• We have shown thatholds for i = 1
• Hence the above relation holds for all oddinteger values of i
• N must be an odd integer• It is a simple exercise to show that both
and satisfy the power-symmetrycondition
)()( 1−− −= zHzzG ii
i
)(zGi
)(zHi
iiiii KzHzHzHzH =−−+ −− )()()()( 11
Copyright © 2001, S. K. Mitra38
Power-Symmetric FIRPower-Symmetric FIRCascaded Lattice StructureCascaded Lattice Structure
• In addition, and are power-complementary, i.e.,
• To develop the synthesis equation weexpress and in terms of
and :
)(zGi)(zHi
)(zGi)(zHi
)(zGi 2−)(zHi 2−
)()()()( zGzHkzGzk iiiii +=+ −−
2221
)()()()( zGkzHzHk iiiii −=+ −221
)()()()( zGzHkzGzk iiiii +=+ −−
2221
Copyright © 2001, S. K. Mitra39
Power-Symmetric FIRPower-Symmetric FIRCascaded Lattice StructureCascaded Lattice Structure
• Note: At the i-th step, the coefficient ischosen to eliminate the coefficient of ,the highest power of in
• For this choice of the coefficient ofalso vanishes making a polynomialof degree
• The synthesis process begins with i = N andcompute using
ik
ikiz−
1−z )()( zGkzH iii −
)(zHi 2−2−i
)(zGN )()( 1−− −= zHzzG NN
N
Copyright © 2001, S. K. Mitra40
Power-Symmetric FIRPower-Symmetric FIRCascaded Lattice StructureCascaded Lattice Structure
• Next, the transfer functions and are determined using the synthesis
equations
• This process is repeated until all coefficientsof the lattice have been determined
)()()()( zGkzHzHk iiiii −=+ −221
)()()()( zGzHkzGzk iiiii +=+ −−
2221
)(zH N 2−)(zGN 2−
Copyright © 2001, S. K. Mitra41
Power-Symmetric FIRPower-Symmetric FIRCascaded Lattice StructureCascaded Lattice Structure
• Example - Consider
• It can be easily verified that satisfiesthe power-symmetric condition
• Next we form
54 20060 −− +− zz ..
3215 376020301 −−− −++= zzzzH ...)(
)(zH5
115
55 06020 −−− −−=−= zzHzzG ..)()(
5432 30203760 −−−− +−++ zzzz ...
Copyright © 2001, S. K. Mitra42
Power-Symmetric FIRPower-Symmetric FIRCascaded Lattice StructureCascaded Lattice Structure
• To determine we first form
• To cancel the coefficient of in the abovewe choose
)(zH51
55555 06030201 −+++=− zkkzGkzH )..().()()(
35
25 203760376020 −− −−+−+ zkzk )..()..(
55
45 2030060 −− −++−+ zkzk ).()..(
5−z
205 .=k
Copyright © 2001, S. K. Mitra43
Power-Symmetric FIRPower-Symmetric FIRCascaded Lattice StructureCascaded Lattice Structure
• Then
• We next form
• Continuing the above process we get
[ ])()()( zGkzHzHk 55511
3 25
−=−
)....(.
3210411 4160124803120041 −−− −++= zzz
32113
33 3012040 −−−−− +−+=−= zzzzHzzG ...)()(
3040 13 .,. =−= kk
Copyright © 2001, S. K. Mitra44
Power-Symmetric FIR BanksPower-Symmetric FIR Banks
• Using the method outlined for therealization of a power-symmetric transferfunction, we can develop a cascaded latticerealization of the 2-channel paraunitaryQMF bank
• Three important properties of the QMFlattice structure are structurally induced
Copyright © 2001, S. K. Mitra45
Power-Symmetric FIR BanksPower-Symmetric FIR Banks• (1) The QMF lattice guarantees perfect
reconstruction independent of the latticeparameters
• (2) It exhibits very small coefficientsensitivity to lattice parameters as eachstage remains lossless under coefficientquantization
• (3) Computational complexity is about one-half that of any other realization as itrequires (N+1)/2 total number of multipliersfor an order-N filter
Copyright © 2001, S. K. Mitra46
Power-Symmetric FIR BanksPower-Symmetric FIR Banks• Example - Consider the analysis filter of the
previous example:
• We place a multiplier h[0] = 0.3231 at theinput and synthesize a cascade latticestructure for the normalized transferfunction
217 30134051935032310 −− ++= zzzH ...)(
543 321013767007810 −−− +−− zzz ...76 04900790 −− −+ zz ..
][/)( 07 hzH
Copyright © 2001, S. K. Mitra47
Power-Symmetric FIR BanksPower-Symmetric FIR Banks
• The lattice coefficients obtained for thenormalized analysis transfer function are:
• Note: Because of the numerical problem,the coefficients of the spectral factorobtained in the previous example are notvery accurate
,.,. 23540151650 57 =−= kk611483930 13 .,. =−= kk
Copyright © 2001, S. K. Mitra48
Power-Symmetric FIR BanksPower-Symmetric FIR Banks
• As a result, the coefficients of of thetransfer function generated fromthe transfer function using therelation
is not exactly zero, and has been set to zeroat each iteration
)(zHi
)(2 zHi−
)1( −− iz
)]()([)( 211
2 zGkzHzH iiikii
−=+−
Copyright © 2001, S. K. Mitra49
Power-Symmetric FIR BanksPower-Symmetric FIR Banks
• Two interesting properties of the cascadedlattice QMF bank can be seen by examiningits multiplier coefficient values
• (1) Signs of coefficients alternate betweenstages
• (2) The values of the coefficientsdecrease with increasing i
}{ ik
Copyright © 2001, S. K. Mitra50
Power-Symmetric FIR BanksPower-Symmetric FIR Banks
• The QMF lattice structure can be useddirectly to design the power-symmetricanalysis filter using an iterativecomputer-aided optimization technique
• Goal: Determine the lattice parametersby minimizing the energy in the stopband of
)(0 zH
)(0 zH
ik
Copyright © 2001, S. K. Mitra51
Power-Symmetric FIR BanksPower-Symmetric FIR Banks
• The pertinent objective function is given by
• Note: The power-symmetric propertyensures good passband response
∫π
ω
ω ω=φs
deH j 20 )(
Copyright © 2001, S. K. Mitra52
BiorthogonalBiorthogonal FIR Banks FIR Banks
• In the design of an orthogonal 2-channelfilter bank, the analysis filter ischosen as a spectral factor of the zero-phaseeven-order half-band filter
• Note: The two spectral factors and of F(z) have the same magnitude
response
)()()( 100
−= zHzHzF
)(0 zH
)(0 zH)( 1
0−zH
Copyright © 2001, S. K. Mitra53
BiorthogonalBiorthogonal FIR Banks FIR Banks
• As a result, it is not possible to designperfect reconstruction filter banks withlinear-phase analysis and synthesis filters
• However, it is possible to maintain theperfect reconstruction condition with linear-phase filters by choosing a differentfactorization scheme
Copyright © 2001, S. K. Mitra54
BiorthogonalBiorthogonal FIR Banks FIR Banks
• To this end, the causal half-band filterof order 2N is factorized in the form
where and are linear-phasefilters
• The determinant of the modulation matrix is now given by
)(zFz N−
)()()( 10 zHzHzFz N −=−
)(0 zH )(1 zH
)()( zmHNm zzHzHzHzHz −=−−−= )()()()()]([ 1010
)(Hdet
Copyright © 2001, S. K. Mitra55
BiorthogonalBiorthogonal FIR Banks FIR Banks
• Note: The determinant of the modulationmatrix satisfies the perfect reconstructioncondition
• The filter bank designed using thefactorization schemeis called a biorthogonal filter bank
• The two synthesis filters are given by
)()()( 10 zHzHzFz N −=−
)()(),()( 0110 zHzGzHzG −−=−=
Copyright © 2001, S. K. Mitra56
BiorthogonalBiorthogonal FIR Banks FIR Banks
• Example - The half-band filter
• can be factored several different ways toyield linear-phase analysis filters and
• For example, one choice yields
)41()1()( 21413161 −−− −+−+= zzzzzF
)(0 zH)(1 zH
)2621()( 432181
0−−−− −+++−= zzzzzH
)21()( 2121
1−− +−= zzzH
Copyright © 2001, S. K. Mitra57
BiorthogonalBiorthogonal FIR Banks FIR Banks• Since the length of is 5 and the
length of is 3, the above set ofanalysis filters is known as the 5/3 filter-pair of Daubechies
• A plot of the gain responses of the 5/3 filter-pair is shown below
)(0 zH)(1 zH
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω/π
Mag
nitu
de
H0(z)
H1(z)
Copyright © 2001, S. K. Mitra58
BiorthogonalBiorthogonal FIR Banks FIR Banks• Another choice yields the 4/4 filter-pair of
Daubechies
• A plot of the gain responses of the 4/4 filter-pair is shown below
)331()( 32181
0−−− +++= zzzzH
)331()( 32121
1−−− ++−−= zzzzH
0 0.2 0.4 0.6 0.8 10
0.5
1
1.5
ω/π
Mag
nitu
de
H0(z)
H1(z)