Comb Filters

Comb FiltersComb Filters

• The simple filters discussed so far arecharacterized either by a single passbandand/or a single stopband

• There are applications where filters withmultiple passbands and stopbands arerequired

• The comb filter is an example of suchfilters

Comb FiltersComb Filters• In its most general form, a comb filter has a

frequency response that is a periodicfunction of ω with a period 2π/L, where L isa positive integer

• If H(z) is a filter with a single passbandand/or a single stopband, a comb filter canbe easily generated from it by replacingeach delay in its realization with L delaysresulting in a structure with a transferfunction given by )()( LzHzG =

• If exhibits a peak at , thenwill exhibit L peaks at ,in the frequency range

• Likewise, if has a notch at ,then will have L notches at ,

in the frequency range• A comb filter can be generated from either

an FIR or an IIR prototype filter

|)(| ωjeH

|)(| ωjeH|)(| ωjeG

|)(| ωjeGpω

Lkp /ω

Lko /ω

10 −≤≤ Lk

π<ω≤ 20

Comb FiltersComb Filters• For example, the comb filter generated from

the prototype lowpass FIR filter has a transfer function

• has L notchesat ω = (2k+1)π/L and Lpeaks at ω = 2π k/L,

)( 121 1 −+ z

=)(zH0

)()()( LL zzHzG −+== 121

10 −≤≤ Lk , in thefrequency range

π<ω≤ 20

|)(| 0ωjeG

0 0.5 1 1.5 20

Comb filter from lowpass prototype

Comb FiltersComb Filters• For example, the comb filter generated from

the prototype highpass FIR filter has a transfer function

• has L peaksat ω = (2k+1)π/L and Lnotches at ω = 2π k/L,

|)(| 1ωjeG

)( 121 1 −− z

=)(zH1

)()()( LL zzHzG −−== 121

10 −≤≤ Lk , in thefrequency range

π<ω≤ 200 0.5 1 1.5 2

Comb filter from highpass prototype

• Depending on applications, comb filterswith other types of periodic magnituderesponses can be easily generated byappropriately choosing the prototype filter

• For example, the M-point moving averagefilter

has been used as a prototype)(

)( 111

−−=

Comb FiltersComb Filters• This filter has a peak magnitude at ω = 0,

and notches at ,• The corresponding comb filter has a transfer

function

whose magnitude has L peaks at , and notches at ,

1−M M/2 lπ=ω 11 −≤≤ Ml

)()( L

zMzzG −

−−=1

Lk/2π=ω10 −≤≤ Lk )( 1−ML

LMk/2π=ω )( 11 −≤≤ MLk

Allpass Allpass Transfer FunctionTransfer FunctionDefinition• An IIR transfer function A(z) with unity

magnitude response for all frequencies, i.e.,

is called an allpass transfer function• An M-th order causal real-coefficient

allpass transfer function is of the form

ω=ω allfor,1|)(| 2jeA

M zdzdzdzzdzddzA −+−

−−

−+−−−

++++++++±= 1

AllpassAllpass Transfer Function Transfer Function• If we denote the denominator polynomials

of as :

then it follows that can be written as:

• Note from the above that if is apole of a real coefficient allpass transferfunction, then it has a zero at

)(zDM)(zAMM

MM zdzdzdzD −+−−

− ++++= 11

111 ...)(

)()()(

zA1−−

±=φ= jrez

φ−= jr ez 1

AllpassAllpass Transfer Function Transfer Function

• The numerator of a real-coefficient allpasstransfer function is said to be the mirror-image polynomial of the denominator, andvice versa

• We shall use the notation to denotethe mirror-image polynomial of a degree-Mpolynomial , i.e.,

)(zDM~

)()( zDzzD MM

AllpassAllpass Transfer Function Transfer Function• The expression

implies that the poles and zeros of a real-coefficient allpass function exhibit mirror-image symmetry in the z-plane

)()()(

zA1−−

3213 2.018.04.01

4.018.02.0)( −−−

−−−

−+++++−=

zzzzzzzA

-1 0 1 2 3

Real Part

• To show that we observe that

• Therefore

• Hence

1)( −±=−zD

1)()( −

−−=−

1|)(| =ωjM eA

1)()(|)(| 12 == ω=−ω

jezMMj

M zAzAeA

• Now, the poles of a causal stable transferfunction must lie inside the unit circle in thez-plane

• Hence, all zeros of a causal stable allpasstransfer function must lie outside the unitcircle in a mirror-image symmetry with itspoles situated inside the unit circle

AllpassAllpass Transfer Function Transfer Function• Figure below shows the principal value of

the phase of the 3rd-order allpass function

• Note the discontinuity by the amount of 2πin the phase θ(ω)

3213 2.018.04.01

4.018.02.0)( −−−

−−−

−+++++−=

zzzzzzzA

0 0.2 0.4 0.6 0.8 1-4

Principal value of phase

AllpassAllpass Transfer Function Transfer Function• If we unwrap the phase by removing the

discontinuity, we arrive at the unwrappedphase function indicated below

• Note: The unwrapped phase function is acontinuous function of ω

)(ωθc

0 0.2 0.4 0.6 0.8 1-10

Unwrapped phase

• The unwrapped phase function of anyarbitrary causal stable allpass function is acontinuous function of ω

Properties• (1) A causal stable real-coefficient allpass

transfer function is a lossless bounded real(LBR) function or, equivalently, a causalstable allpass filter is a lossless structure

AllpassAllpass Transfer Function Transfer Function• (2) The magnitude function of a stable

allpass function A(z) satisfies:

• (3) Let τ(ω) denote the group delay functionof an allpass filter A(z), i.e.,

<>==><

1for11for11for1

)]([)( ωθ−=ωτω cdd

• The unwrapped phase function of astable allpass function is a monotonicallydecreasing function of ω so that τ(ω) iseverywhere positive in the range 0 < ω < π

• The group delay of an M-th order stablereal-coefficient allpass transfer functionsatisfies:

)(ωθc

π=ω∫ ωτπ

AllpassAllpass Transfer Function Transfer FunctionA Simple Application• A simple but often used application of an

allpass filter is as a delay equalizer• Let G(z) be the transfer function of a digital

filter designed to meet a prescribedmagnitude response

• The nonlinear phase response of G(z) can becorrected by cascading it with an allpassfilter A(z) so that the overall cascade has aconstant group delay in the band of interest

• Since , we have

• Overall group delay is the given by the sumof the group delays of G(z) and A(z)

1|)(| =ωjeA|)(||)()(| ωωω = jjj eGeAeG

G(z) A(z)

Minimum-Phase and Maximum-Minimum-Phase and Maximum-Phase Transfer FunctionsPhase Transfer Functions

• Consider the two 1st-order transfer functions:

• Both transfer functions have a pole inside theunit circle at the same location and arestable

• But the zero of is inside the unit circleat , whereas, the zero of is at

situated in a mirror-image symmetry

11121 <<== +

+++ bazHzH az

bzazbz ,,)(,)(

az −=

bz −=

bz 1−=

)(zH1)(zH2

• Figure below shows the pole-zero plots ofthe two transfer functions

)(1 zH )(2 zH

• However, both transfer functions have anidentical magnitude function as

• The corresponding phase functions are11

11 == −− )()()()( zHzHzHzH

ω+ω−

ω+ω−ω −= cos

sin1cos

sin11 tantan)](arg[ ab

ω+ω−

ω+ω−ω −= cos

sin1cos1

sin12 tantan)](arg[ ab

• Figure below shows the unwrapped phaseresponses of the two transfer functions fora = 0.8 and b = 5.0−

0 0.2 0.4 0.6 0.8 1-4

• From this figure it follows that hasan excess phase lag with respect to

• Generalizing the above result, we can showthat a causal stable transfer function with allzeros outside the unit circle has an excessphase compared to a causal transferfunction with identical magnitude buthaving all zeros inside the unit circle

)(2 zH)(1 zH

• A causal stable transfer function with allzeros inside the unit circle is called aminimum-phase transfer function

• A causal stable transfer function with allzeros outside the unit circle is called amaximum-phase transfer function

• Any nonminimum-phase transfer functioncan be expressed as the product of aminimum-phase transfer function and astable allpass transfer function

Complementary TransferComplementary TransferFunctionsFunctions

• A set of digital transfer functions withcomplementary characteristics often findsuseful applications in practice

• Four useful complementary relations aredescribed next along with some applications

Delay-Complementary Transfer Functions• A set of L transfer functions, ,

, is defined to be delay-complementary of each other if the sum oftheir transfer functions is equal to someinteger multiple of unit delays, i.e.,

where is a nonnegative integer

)}({ zHi10 −≤≤ Li

0≠ββ= −

=∑ onL

ii zzH

• A delay-complementary paircan be readily designed if one of the pairs isa known Type 1 FIR transfer function ofodd length

• Let be a Type 1 FIR transfer functionof length M = 2K+1

• Then its delay-complementary transferfunction is given by

)}(),({ 10 zHzH

)()( 01 zHzzH K −= −

)(0 zH

• Let the magnitude response of beequal to in the passband and less thanor equal to in the stopband where and

are very small numbers• Now the frequency response of can be

expressed as

where is the amplitude response

)(0 zHpδ±1sδ pδ

sδ)(0 zH

)()( 00 ω= ω−ω HeeH jKj ~

)(0 ωH~

• Its delay-complementary transfer function has a frequency response given by

• Now, in the passband,and in the stopband,

• It follows from the above equation that inthe stopband, and in thepassband,

)(1 zH)](1[)()( 011 ω−=ω= ω−ω−ω HeHeeH jKjKj ~ ~

,1)(1 0 pp H δ+≤ω≤δ− ~

ss H δ≤ω≤δ− )(0~

pp H δ≤ω≤δ− )(1~

ss H δ+≤ω≤δ− 1)(1 1~

• As a result, has a complementarymagnitude response characteristic to that of

with a stopband exactly identical tothe passband of , and a passband thatis exactly identical to the stopband of

• Thus, if is a lowpass filter, willbe a highpass filter, and vice versa

)(1 zH

)(0 zH

)(1 zH)(0 zH

)(0 zH

• The frequency at which

the gain responses of both filters are 6 dBbelow their maximum values

• The frequency is thus called the 6-dBcrossover frequency

oω5.0)()( 10 =ω=ω oo HH ~~

• Example - Consider the Type 1 bandstoptransfer function

• Its delay-complementary Type 1 bandpasstransfer function is given by

)45541()1()( 121084242641 −−−−−− +−++−+= zzzzzzzHBS

)45541()1( 121084242641 −−−−−− +++++−= zzzzzz

)()( 10 zHzzH BSBP −= −

• Plots of the magnitude responses ofand are shown below

)(zHBS)(zHBP

0 0.2 0.4 0.6 0.8 10

(z) HBP

Allpass Complementary Filters• A set of M digital transfer functions, ,

, is defined to be allpass-complementary of each other, if the sum oftheir transfer functions is equal to an allpassfunction, i.e.,

)}({ zHi10 −≤≤ Mi

ii =∑

Power-Complementary Transfer Functions• A set of M digital transfer functions, ,

, is defined to be power-complementary of each other, if the sum oftheir square-magnitude responses is equal toa constant K for all values of ω, i.e.,

)}({ zHi10 −≤≤ Mi

ω=∑−

ω allfor,)(1

• By analytic continuation, the aboveproperty is equal to

for real coefficient• Usually, by scaling the transfer functions,

the power-complementary property isdefined for K = 1

ω=∑−

− allfor,)()(1

1 KzHzHM

• For a pair of power-complementary transferfunctions, and , the frequencywhere , iscalled the cross-over frequency

• At this frequency the gain responses of bothfilters are 3-dB below their maximumvalues

• As a result, is called the 3-dB cross-over frequency

)(0 zH )(1 zH5.0|)(||)(| 2

0 == ωω oo jj eHeH

• Example - Consider the two transfer functions and given by

where and are stable allpasstransfer functions

• Note that• Hence, and are allpass

complementary

)(0 zH )(1 zH)]()([)( 102

10 zAzAzH +=

)(0 zA )(1 zA)]()([)( 102

11 zAzAzH −=

)()()( 010 zAzHzH =+)(0 zH )(1 zH

• It can be shown that and arealso power-complementary

• Moreover, and are bounded-real transfer functions

)(0 zH )(1 zH

Doubly-Complementary Transfer Functions• A set of M transfer functions satisfying both

the allpass complementary and the power-complementary properties is known as adoubly-complementary set

• A pair of doubly-complementary IIRtransfer functions, and , with asum of allpass decomposition can be simplyrealized as indicated below

)(0 zH )(1 zH

+)(1 zA

)(0 zA)(zX

)(0 zY

)(1 zY1−

0 )( zXzYzH = )(

)(11 )( zX

zYzH =

• Example - The first-order lowpass transferfunction

can be expressed as

α−+α−

)]()([)( 1021

1zAzAzH

α−+α−+ −

1 1)( −

α−+α−=

zzzA,)( 10 =zA

• Its power-complementary highpass transferfunction is thus given by

• The above expression is precisely the first-order highpass transfer function describedearlier

=− −

α−+α−−= 1

1021 )]()([)(

HP zAzAzH

α−−α+

• Figure below demonstrates the allpasscomplementary property and the powercomplementary property of and)(zHLP

)(zHHP

0 0.2 0.4 0.6 0.8 10

(ejω)|

(ejω) + HHP

(ejω)|

0 0.2 0.4 0.6 0.8 10

de |HHP

(ejω)|2

(ejω)|2 + |HHP

(ejω)|2

Power-Symmetric Filters• A real-coefficient causal digital filter with a

transfer function H(z) is said to be a power-symmetric filter if it satisfies the condition

where K > 0 is a constantKzHzHzHzH =−−+ −− )()()()( 11

• It can be shown that the gain function G(ω)of a power-symmetric transfer function at ω= π is given by

• If we define , then it followsfrom the definition of the power-symmetricfilter that H(z) and G(z) are power-complementary as

constanta)()()()( 11 =+ −− zGzGzHzH

)()( zHzG −=dBK 3log10 10 −

Conjugate Quadratic Filter• If a power-symmetric filter has an FIR

transfer function H(z) of order N, then theFIR digital filter with a transfer function

is called a conjugate quadratic filter ofH(z) and vice-versa

)()( 11 −−= zHzzG

• It follows from the definition that G(z) isalso a power-symmetric causal filter

• It also can be seen that a pair of conjugatequadratic filters H(z) and G(z) are alsopower-complementary

• Example - Let• We form

• H(z) is a power-symmetric transferfunction

)3621)(3621( 32321 zzzzzz ++−++−= −−−)()()()( 11 −− −−+ zHzHzHzH

321 3621)( −−− ++−= zzzzH

)3621)(3621( 32321 zzzzzz −++−+++ −−−

)345043( 313 −− ++++= zzzz100)345043( 313 =−−+−−+ −− zzzz

Digital Two-PairsDigital Two-Pairs

• The LTI discrete-time systems consideredso far are single-input, single-outputstructures characterized by a transferfunction

• Often, such a system can be efficientlyrealized by interconnecting two-input, two-output structures, more commonly calledtwo-pairs

Digital Two-PairsDigital Two-Pairs• Figures below show two commonly used

block diagram representations of a two-pair

• Here and denote the two outputs, andand denote the two inputs, where the

dependencies on the variable z has beenomitted for simplicity

1Y 2Y1X 2X

2X1Y1X 2Y

Digital Two-PairsDigital Two-Pairs• The input-output relation of a digital two-

pair is given by

• In the above relation the matrix τ given by

is called the transfer matrix of the two-pair

22211211

ttttτ

Digital Two-PairsDigital Two-Pairs

• It follows from the input-output relation thatthe transfer parameters can be found asfollows:

12 ====

XX XYt

12 ====

XX XYt

Digital Two-PairsDigital Two-Pairs• An alternate characterization of the two-pair

is in terms of its chain parameters as

where the matrix Γ given by

is called the chain matrix of the two-pair

BAΓ- -

Digital Two-PairsDigital Two-Pairs• The relation between the transfer

parameters and the chain parameters aregiven by

ABCADt

ACt −==−== 22211211

2122112112

tttttDt

tCttBtA −

==−== ,,,

Two-Pair InterconnectionTwo-Pair InterconnectionSchemesSchemes

Cascade Connection - Γ-cascade

• Here

• But from figure, and• Substituting the above relations in the first

equation on the previous slide andcombining the two equations we get

• Hence,

"1 YX = '

1 XY =

Cascade Connection - τ-cascade

• Here

1X "1Y

'22'21

'12'11tttt

- - --

"22"21

"12"11tttt

'1'22'21

'12'11'2

"22"21

"12"11"2

• But from figure, and• Substituting the above relations in the first

equation on the previous slide andcombining the two equations we get

• Hence,

"1 YX = '

2"2 YX =

'22'21

'12'11"22"21

"12"11"2

'22'21

'12'11"22"21

"12"1122211211

Constrained Two-Pair

• It can be shown that1Y1X 2Y

2XG(z)

)()()(

1zGBAzGDC

⋅+⋅+

211211 zGt

zGttt−

Algebraic Stability TestAlgebraic Stability Test• We have shown that the BIBO stability of a

causal rational transfer function requiresthat all its poles be inside the unit circle

• For very high-order transfer functions, it isvery difficult to determine the polelocations analytically

• Root locations can of course be determinedon a computer by some type of root findingalgorithms

Algebraic Stability TestAlgebraic Stability Test

• We now outline a simple algebraic test thatdoes not require the determination of polelocations

The Stability Triangle• For a 2nd-order transfer function the

stability can be easily checked byexamining its denominator coefficients

Algebraic Stability TestAlgebraic Stability Test• Let

denote the denominator of the transferfunction

• In terms of its poles, D(z) can be expressedas

• Comparing the last two equations we get

111)( −− ++= zdzdzD

11 )(1)1)(1()( −−−− λλ+λ+λ−=λ−λ−= zzzzzD

212211 ),( λλ=λ+λ−= dd

Algebraic Stability TestAlgebraic Stability Test• The poles are inside the unit circle if

• Now the coefficient is given by theproduct of the poles

• Hence we must have

• It can be shown that the second coefficientcondition is given by

1||,1|| 21 <λ<λ

1|| 2 <d

21 1|| dd +<

• The region in the ( )-plane where thetwo coefficient condition are satisfied,called the stability triangle, is shown below

Stability region

• Example - Consider the two 2nd-orderbandpass transfer functions designedearlier:

3763817343424011188190 −−

+−−−=

zzzzHBP

...)('

726542530533531011136730 −−

+−−=

zzzzHBP

...)("

• In the case of , we observe that

• Since here , is unstable• On the other hand, in the case of ,

we observe that

• Here, and , and hence is BIBO stable

)(' zHBP3763819.1,7343424.0 21 =−= dd

726542528.0,53353098.0 21 =−= dd

)(' zHBP1|| 2 >d)(" zHBP

21 1|| dd +<1|| 2 <d)(" zHBP

A General Stability Test Procedure• Let denote the denominator of an

M-th order causal IIR transfer function H(z):

where we assume for simplicity• Define an M-th order allpass transfer

function:

∑ =−= M

iM zdzD 0)(

)()( 1

)( zDzMDz

−−=

• Or, equivalently

• If we express

then it follows that

zdzdzdzdzzdzdzdd

zMA −+−−

−−

−+−−−

−−

++++++++++

....)(

∏ =−λ−= M

iM zzD 1 )1()(

∏ = λ−= Mi i

MMd 1)1(

• Now for stability we must have ,which implies the condition

• Define

• Then a necessary condition for stability of , and hence, the transfer function

H(z) is given by

1|| <λi1|| <Md

MMM dAk =∞= )(

12 <Mk

Algebraic Stability TestAlgebraic Stability Test• Assume the above condition holds• We now form a new function

• Substituting the rational form of inthe above equation we get

−=− )(1)(

)(1)()(1 zAd

dzAzzAk

kzAzzAMM

)1()2('1

....)(1 −−

−−−

−−

−−−−−−−

++++++++

=− MM

zdzdzdzzdzdd

• Hence, is an allpass function oforder

• Now the poles of are given bythe roots of the equation

oλ )(1 zAM −

11,1 2

' −≤≤−

−= − Mi

dddddM

MkoMA 1)( =λ

)(1 zAM −1−M

Algebraic Stability TestAlgebraic Stability Test• By assumption• Hence• If is a stable allpass function, then

• Thus, if is a stable allpass function,then the condition holds only if

12 <Mk1|)(| >λoMA

<>==><

1for,11for,11for,1

)(zAM1|)(| >λoMA

• Or, in other words is a stableallpass function

• Thus, if is a stable allpass functionand , then is also a stableallpass function of one order lower

• We now prove the converse, i.e., ifis a stable allpass function and , then

is also a stable allpass function

)(1 zAM −

)(zAM12 <Mk

)(1 zAM −

12 <Mk

Algebraic Stability TestAlgebraic Stability Test• To this end, we express in terms of

arriving at

• If is a pole of , then

• By assumption holds

)(1)()(

zAzkzAzkzA

−−

)(1 zAM −

MkoMo A 11

1 )( −=ζζ −−

12 <Mk

• Therefore, i.e.,

• Assume is a stable allpass function• Then for• Now, if , then because of the above

condition• But the condition reduces

1|)(| 11 >ζζ −

−oMo A

)(1 zAM −1|)(| 1 ≤− zAM 1|| ≥z

1|| ≥ζo

|||)(| 1 ooMA ζ>ζ−

1|)(| 1 ≤ζ− oMA|||)(| 1 ooMA ζ>ζ−

1|)(| 1 >ζ− oMA 1|| ≥ζo

• Thus there is a contradiction• On the other hand, if then from

we have• The above condition does not violate the

condition

1|| <ζo1||for1|)(| 1 <>− zzAM

1|)(| 1 >ζ− oMA

|||)(| 1 ooMA ζ>ζ−

• Thus, if and if is a stableallpass function, then is also a stableallpass function

• Summarizing, a necessary and sufficient setof conditions for the causal allpass function

to be stable is therefore:(1) , and(2) The allpass function is stable

)(1 zAM −)(zAM

12 <Mk

)(1 zAM −

12 <Mk

Algebraic Stability TestAlgebraic Stability Test• Thus, once we have checked the condition

, we test next for the stability of thelower-order allpass function

• The process is then repeated, generating aset of coefficients:

and a set of allpass functions of decreasingorder:

12 <Mk)(1 zAM −

121 ,,...,, kkkk MM −

1)(),(),(.,..),(),( 0121 =− zAzAzAzAzA MM

Algebraic Stability TestAlgebraic Stability Test• The allpass function is stable if and

only if for i• Example - Test the stability of

• From H(z) we generate a 4-th order allpassfunction

• Note:

12 <ik)(zAM

)(dzdzdzdzzdzdzdzd

zzzzzA

++++++++

12341)( 234 ++++

444 <==∞= dAk

Algebraic Stability TestAlgebraic Stability Test• Using

we determine the coefficients of thethird-order allpass function from thecoefficients of :

31,1 2

44' ≤≤−

−= − i

ddddd ii

}{ 'id

}{ id )(4 zA)(3 zA

++++++

zdzdzdzdzdzd

Algebraic Stability TestAlgebraic Stability Test• Note:• Following the above procedure, we derive

the next two lower-order allpass functions:

1)(151'

333 <==∞= dAk

2241592

22479 1

)(2 ++

10153 1

• Note:

• Since all of the stability conditions aresatisfied, and hence H(z) are stable

• Note: It is not necessary to derivesince can be tested for stability usingthe coefficient conditions

1)(22479

22 <=∞= Ak

1)(10153

11 <=∞= Ak

)(4 zA)(3 zA

)(2 zA

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