Comb Filter

Copyright 2001, S. K. Mitra1

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 with a period 2/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

|)(| jeH

|)(| jeH|)(| jeG

|)(| jeGp

o

Lkp /

Lko /

10 Lk

10 Lk


Comb FiltersComb Filters For example, the comb filter generated from

the prototype lowpass FIR filter has a transfer function

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

)( 121 1 + z

=)(zH0

)()()( LL zzHzG +== 121

00

10 Lk , in thefrequency range


Comb FiltersComb Filters For example, the comb filter generated from

the prototype highpass FIR filter has a transfer function

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

|)(| 1jeG

)( 121 1 z

=)(zH1

)()()( LL zzHzG == 121

11

10 Lk , in thefrequency range



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)()( 11

1

=

zMz MzH


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

and notches at , The corresponding comb filter has a transfer

function

whose magnitude has L peaks at , and notches at ,

1M M/2 l= 11 Ml

)()( LML

zMzzG

=1

1

Lk/2=10 Lk )( 1ML

LMk/2= )( 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

= allfor,1|)(| 2jeA

MM

MM

MMMM

M zdzdzdzzdzddzA +

+

++++++++= 1

11

1

11

11

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

MM zdzdzdzD+

++++= 111

11 ...)()(zAM

)()()(

zDzDz

MM

MM

zA1

== jrez

= 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~

)(zDM)()( zDzzD MMM =

~


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

)()()(

zDzDz

MM

MM

zA1

=

321

3213 2.018.04.01

4.018.02.0)(

+++++=

zzzzzzzA

-1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

Real Part

I

m

a

g

i

n

a

r

y

P

a

r

t



To show that we observe that

Therefore

Hence

)()(1

1)( = zDzDz

MM

MM

zA

)()(

)()(1

1

1)()(

=

zDzDz

zDzDz

MMM

MM

M

MM

zAzA

1|)(| =jM eA

1)()(|)(| 12 == =

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 2in the phase ()

321

3213 2.018.04.01

4.018.02.0)(

+++++=

zzzzzzzA

0 0.2 0.4 0.6 0.8 1-4

-2

0

2

4

/

P

h

a

s

e

,

d

e

g

r

e

e

s

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

)(c

0 0.2 0.4 0.6 0.8 1-10

-8

-6

-4

-2

0

/

P

h

a

s

e

,

d

e

g

r

e

e

s

Unwrapped phase



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

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 () denote the group delay functionof an allpass filter A(z), i.e.,

==> 0 is a constantKzHzHzHzH =+ )()()()( 11


Complementary TransferComplementary TransferFunctionsFunctions

It can be shown that the gain function G()of a power-symmetric transfer function at = 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

1Y

2Y1X

2X1Y1X 2Y

2X


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

pair is given by

In the above relation the matrix given by

is called the transfer matrix of the two-pair

=

21

22211211

21

XX

tttt

YY

=

22211211

tttt


Digital Two-PairsDigital Two-Pairs

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

02

112

01

111

12 ====

XX XYt

XYt ,

02

222

01

221

12 ====

XX XYt

XYt ,


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

is in terms of its chain parameters as

where the matrix given by

is called the chain matrix of the two-pair

=

22

11

XY

DCBA

YX

= DC

BA- -


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

parameters and the chain parameters aregiven by

ACt

At

ABCADt

ACt ==== 22211211

1,,,

2122112112

2111

2122

211

tttttDt

tCttBtA

==== ,,,


Two-Pair InterconnectionTwo-Pair InterconnectionSchemesSchemes

Cascade Connection - -cascade

Here

'1Y

'1X

"1Y

'2Y "

2Y"1X

"2X'

2X

''''

DCBA

- -

""""

DCBA

--

=

'2

'2

''''

'1

'1

XY

DCBA

YX

=

"2

"2

""""

"1

"1

XY

DCBA

YX



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

equation on the previous slide andcombining the two equations we get

Hence,

'2

"1 YX = '2"1 XY =

=

"2

"2

""""

''''

'1

'1

XY

DCBA

DCBA

YX

=

""""

''''

DCBA

DCBA

DCBA



Cascade Connection - -cascade

Here

'1Y'

1X "1Y

'2Y

"2Y

"1X

"2X

'2X

'22'21

'12'11tttt

- - --

"22"21

"12"11tttt

=

'2

'1'22'21

'12'11'2

'1

XX

tttt

YY

=

"2

"1

"22"21

"12"11"2

"1

XX

tttt

YY



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

equation on the previous slide andcombining the two equations we get

Hence,

'1

"1 YX = '2"2 YX =

=

'2

'1

'22'21

'12'11"22"21

"12"11"2

"1

XX

tttt

tttt

YY

=

'22'21

'12'11"22"21

"12"1122211211

tttt

tttt

tttt



Constrained Two-Pair

It can be shown that1Y1X 2Y

2XG(z)

H(z)

)()()(

1

1zGBAzGDC

XYzH

++

==

)(1)(

22

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

22

111)( ++= zdzdzD

221

121

12

11 )(1)1)(1()( ++== zzzzzD

212211 ),( =+= 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



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

21,dd

Stability region



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

21

2

3763817343424011188190

+=

zzzzHBP

..

.)('

21

2

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 +



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:

)(zDM

= =Mi

iiM zdzD 0)(

10 =d

)()( 1)( zD

zMDzM M

MzA

=



Or, equivalently

If we express

then it follows that

MM

MM

MMMMM

zdzdzdzdzzdzdzdd

zMA +

+

++++++++++

= 11

22

11

11

22

11

....1....

)(

= =Mi

iiM zzD 1 )1()(

= =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||


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

)(zAM

=

= )(1)(

)(1)()(1 zAd

dzAzzAk

kzAzzAMM

MM

MM

MMM

)1('1

)2('2

1'1

)1()2('1

1'2

'1

....1....

)(1

++++++++

= MM

MM

MMMM

zdzdzdzzdzdd

zMA



where

Hence, is an allpass function oforder

Now the poles of are given bythe roots of the equation

o )(1 zAM

11,1 2

'

= Mi

dddddM

iMMii

MkoMA 1)( =

)(1 zAM 1M


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 oMA

)(zAM

==>oMA

1



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

)(1 zAM )(zAM

12


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)()(

11

11

zAzkzAzkzA

MM

MMM

++=

)(1 zAM )(zAM

)(zAM

MkoMo A 111 )( =

12



Therefore, i.e.,

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

condition But the condition reduces

to if

1|)(| 11 > oMo A

)(1 zAM 1|)(| 1 zAM 1|| z

1|| o

|||)(| 1 ooMA >

1|)(| 1 oMA|||)(| 1 ooMA >

1|)(| 1 > oMA 1|| o



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

we have The above condition does not violate the

condition

1|| oMA

|||)(| 1 ooMA >



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


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


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


Algebraic Stability TestAlgebraic Stability Test Using

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

31,1 24

44'

= i

ddddd iii

}{ 'id

}{ id )(4 zA)(3 zA

151

522

15113

15112

523

151

'3

2'2

3'1

'1

2'2

3'3

31

11

)(+++

+++=

++++++

=zzz

zzz

zdzdzdzdzdzd

zA


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

the next two lower-order allpass functions:

1)(151'

333



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

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Comb Filter