Perfect Reconstruction Two-Channel FIR Filter...

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



• 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



• 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



where

are called the modulation matrices

−−=)()(

)()()()(

zGzGzGzGzm

1010G

−−=)()(

)()()()(

zHzHzHzHzm

1010H



• 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



• 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



where

and is an odd positive integer

)()](det[

)()(

zHz

zzG m −⋅=−

102H

l

)()](det[

)()(

zHz

zzG m −⋅−=−

012H

l

)()()()()](det[ )( zHzHzHzHzm1010 −−−=H

l



• 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


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


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



• 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


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


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



• 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


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



• 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



• 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



• 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


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



• 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)


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



• 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



• (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


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


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



• 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



• 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



• 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)



• 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


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×


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


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


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



• Define

,)()(

)(zXzXzH i

i0

=)(

)()(

zXzYzG i

i0

=

Ni ≤≤1



• 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



• 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



• 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



• 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


2221



• 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



• 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


2221

)(zH N 2−)(zGN 2−



• 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 ...



• 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



• Then

• We next form

• Continuing the above process we get

[ ])()()( zGkzHzHk 55511

3 25

−=−

)....(.

3210411 4160124803120041 −−− −++= zzz

32113

33 3012040 −−−−− +−+=−= zzzzHzzG ...)()(

3040 13 .,. =−= kk


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


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


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



• 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



• 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

−=+−



• 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



• 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



• The pertinent objective function is given by

• Note: The power-symmetric propertyensures good passband response

∫π

ω

ω ω=φs

deH j 20 )(


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



• 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



• 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



• 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 −−=−=



• 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


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)


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)

Perfect Reconstruction Two-Channel FIR Filter...

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