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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
Copyright 2001, S. K. Mitra2
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 =
Copyright 2001, S. K. Mitra3
Comb FiltersComb Filters
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
Copyright 2001, S. K. Mitra4
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
Copyright 2001, S. K. Mitra5
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
Copyright 2001, S. K. Mitra6
Comb FiltersComb Filters
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
Copyright 2001, S. K. Mitra7
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
Copyright 2001, S. K. Mitra8
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 ......)(
Copyright 2001, S. K. Mitra9
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
Copyright 2001, S. K. Mitra10
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 =
~
Copyright 2001, S. K. Mitra11
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
Copyright 2001, S. K. Mitra12
AllpassAllpass Transfer Function Transfer Function
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
Copyright 2001, S. K. Mitra13
AllpassAllpass Transfer Function Transfer Function
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
Copyright 2001, S. K. Mitra14
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
Copyright 2001, S. K. Mitra15
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
Copyright 2001, S. K. Mitra16
AllpassAllpass Transfer Function Transfer Function
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
Copyright 2001, S. K. Mitra17
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
Copyright 2001, S. K. Mitra48
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
Copyright 2001, S. K. Mitra49
Complementary TransferComplementary TransferFunctionsFunctions
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
Copyright 2001, S. K. Mitra50
Complementary TransferComplementary TransferFunctionsFunctions
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
Copyright 2001, S. K. Mitra51
Complementary TransferComplementary TransferFunctionsFunctions
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
Copyright 2001, S. K. Mitra52
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
Copyright 2001, S. K. Mitra53
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
Copyright 2001, S. K. Mitra54
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
Copyright 2001, S. K. Mitra55
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 ,
Copyright 2001, S. K. Mitra56
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- -
Copyright 2001, S. K. Mitra57
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
==== ,,,
Copyright 2001, S. K. Mitra58
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
Copyright 2001, S. K. Mitra59
Two-Pair InterconnectionTwo-Pair InterconnectionSchemesSchemes
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
Copyright 2001, S. K. Mitra60
Two-Pair InterconnectionTwo-Pair InterconnectionSchemesSchemes
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
Copyright 2001, S. K. Mitra61
Two-Pair InterconnectionTwo-Pair InterconnectionSchemesSchemes
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
Copyright 2001, S. K. Mitra62
Two-Pair InterconnectionTwo-Pair InterconnectionSchemesSchemes
Constrained Two-Pair
It can be shown that1Y1X 2Y
2XG(z)
H(z)
)()()(
1
1zGBAzGDC
XYzH
++
==
)(1)(
22
211211 zGt
zGttt
+=
Copyright 2001, S. K. Mitra63
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
Copyright 2001, S. K. Mitra64
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
Copyright 2001, S. K. Mitra65
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
Copyright 2001, S. K. Mitra66
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
Copyright 2001, S. K. Mitra67
Algebraic Stability TestAlgebraic Stability Test
The region in the ( )-plane where thetwo coefficient condition are satisfied,called the stability triangle, is shown below
21,dd
Stability region
Copyright 2001, S. K. Mitra68
Algebraic Stability TestAlgebraic Stability Test
Example - Consider the two 2nd-orderbandpass transfer functions designedearlier:
21
2
3763817343424011188190
+=
zzzzHBP
..
.)('
21
2
726542530533531011136730
+=
zzzzHBP
..
.)("
Copyright 2001, S. K. Mitra69
Algebraic Stability TestAlgebraic Stability Test
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 +
Copyright 2001, S. K. Mitra70
Algebraic Stability TestAlgebraic Stability Test
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
=
Copyright 2001, S. K. Mitra71
Algebraic Stability TestAlgebraic Stability Test
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(
Copyright 2001, S. K. Mitra72
Algebraic Stability TestAlgebraic Stability Test
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||
Copyright 2001, S. K. Mitra73
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
Copyright 2001, S. K. Mitra74
Algebraic Stability TestAlgebraic Stability Test
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
Copyright 2001, S. K. Mitra75
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
Copyright 2001, S. K. Mitra76
Algebraic Stability TestAlgebraic Stability Test
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
Copyright 2001, S. K. Mitra77
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
Copyright 2001, S. K. Mitra78
Algebraic Stability TestAlgebraic Stability Test
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
Copyright 2001, S. K. Mitra79
Algebraic Stability TestAlgebraic Stability Test
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 >
Copyright 2001, S. K. Mitra80
Algebraic Stability TestAlgebraic Stability Test
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
Copyright 2001, S. K. Mitra81
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
Copyright 2001, S. K. Mitra82
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
Copyright 2001, S. K. Mitra83
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
Copyright 2001, S. K. Mitra84
Algebraic Stability TestAlgebraic Stability Test Note: Following the above procedure, we derive
the next two lower-order allpass functions:
1)(151'
333
Copyright 2001, S. K. Mitra85
Algebraic Stability TestAlgebraic Stability Test
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