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Module-21:Realization structure of IIR Filters
Realization of IIR filters using Parallel and Cascade forms Objectives: β’ To interpret the limitations of Direct Form of realizations β’ To appreciate the various structures of Cascade and parallel realizations
Introduction: β Direct form realization is easy to obtain as the coefficients that appear in
the realization are same as those in the difference equation. β No additional computations are required in going from the difference
equation to the block diagram. β This convenient property is overshadowed by the poor accuracy that results
when the coefficients and signals are discretized and represented by finite length binary numbers
β This computational inaccuracy problem is generally referred to as the parameter quantization effect, and is caused by quantizing or discretizing the coefficients using representations with a finite number of binary positions
β This difficulty arises when poles( or zeros) are close together and becomes a more serious problem when the order of the system becomes large
β This problem does not arise generally for second order systems, which are almost always realized using the direct form realization
β The direct form II has the advantage over Direct Form I as it requires less memory storage for the data samples
β The advantage of the reduced storage is offset by the larger adders. β This structure tends to scale the input to reduce the gain. This could result
in a worse signal to noise ratio for larger order filters. β For structures greater than 2nd order, these structures are not preferred.
Cascade realization βͺ The Cascade realization of an IIR system is obtained by first
decomposing the system function H(z) into the product of several
simple transfer function
π»(π§) = π»1(π§). π»2(π§) β¦ β¦ . π»3(π§)
π»(π§) = β π»π
πΎ
π=1
(π§)
βͺ By expressing the Numerator and Denominator polynomials of the
Transfer function H(z) as a product of polynomials of lower degree, a
digital filter is realized as a cascade of low order Filter sections
βͺ π»(π§) =π(π§)
π·(π§)=
π1(π§).π2(π§)β¦β¦π3(π§).
π·1(π§)π·2(π§)β¦..π·3(π§)
βͺ To avoid Multiplication by complex numbers, pairs of complex
conjugate poles can be combined
βͺ In the Transfer function π»(π§) =β ππ
ππ=0 π§βπ
1+β πππ§βπππ=1
, it is assumed that π β€
π.
βͺ The individual transfer functions in the cascade are generally chosen
to be either first order sections with π»1(π§) =ππ0+ππ1π§β1
1+ππ1π§β1 and second
order sections π»1(π§) =ππ0+ππ1π§β1+ππ2π§β2
1+ππ1π§β1+ππ2π§β2, where the coefficients are
assumed to be real.
βͺ The second order sections are used whenever possible. When N is
odd, one first order is needed.
βͺ The second order denominators are formed by pairing two real poles
or by pairing complex conjugate poles, so that real coefficients result.
βͺ Various different cascade realizations of H(z) can be obtained by
different pole-zero polynomial mappings.
βͺ Additional cascade realizations are obtained by simply changing the
ordering of the sections.
Parallel Realization
βͺ In this structure, the input signal is processed separately by a
different subsystems.
βͺ The system output is then a weighted sum of the outputs of the
subsystems, which together constitute the overall system.
βͺ An IIR Transfer function can be realized in a parallel form by making
use of the partial fraction of expansion of the Transfer function.
βͺ π»(π§) =β ππ
ππ=0 π§βπ
1+β πππ§βπππ=1
=β (1βπ½ππ§β1)π
π=1
β (1βπΌππ§β1)ππ=1
βͺ If π > π and πΌπ β πΌπ i.e. the roots of the denominator polynomial
are distinct, H(z) may be expanded as a sum of N no. of first order
factors as π»(π§) = βπ΄π
(1βπΌππ§β1)ππ=1 , where the coefficients π΄π and πΌπ
are in general complex.
βͺ This expansion corresponds to a sum of N no. Of first order system
functions and may be realized by connecting these systems in
parallel.
βͺ If h(n) is real, the poles of H(z) will occur in complex conjugate pairs,
and these complex roots in the partial expansion may be combined
to form second order systems with real coefficients
Worked out Examples:
1.Obtain Cascade and parallel structures for the following system:
π¦(π) =3
4π¦(π β 1) β
1
8π¦(π β 2) + π₯(π) +
1
3π₯(π β 1)
Taking Z-Transform on both sides results in
π(π§) β3
4π§β1π(π§) +
1
8π§β2π(π§) = π(π§) +
1
3π§β1π(π§)
The corresponding Transfer Function is
π»(π§) =1 +
13
π§β1
1 β34
π§β1 +18
π§β2
β’ Cascasde Form:
π»(π§) =1 +
13
π§β1
1 β34
π§β1 +18
π§β2=
1 +13
π§β1
(1 β12
π§β1).
1
(1 β14
π§β1)
= π»1(π§). π»2(π§)
β’ Parallel Form:
By using partial fraction expansion
π»(π§) =1 +
13
π§β1
1 β34
π§β1 +18
π§β2=
π΄
(1 β12
π§β1)+
π΅
(1 β14
π§β1)
Solving for A and B,
π»(π§) =10
3
1
(1β1
2π§β1)
β7
3
1
(1β1
4π§β1)
2.Determine π1, π2, π1ππππ0, ππ π‘ππππ ππ π1 πππ π2 so that he two systems in
the following figure are equivalent.
Fig.1
x(n) y1(n) y2(n) y(n)
co
a1
a2 c1
+
π§β1
+ +
π§β1
Fig.2
Fig.1: It is a cascade structure
π¦1(π) = π₯(π) + π1π¦1(π β 1)πππ π»1(π§) =π1(π§)
π(π§)=
1
(1 β π1π§β1)
π¦2(π) = π¦1(π) + π2π¦2(π β 1)πππ π2(π§)
π1(π§)=
1
(1βπ2π§β1)
π¦(π) = πππ¦2(π) + π1π¦2(π β 1) πππ π(π§)
π2(π§)=
π2(π§)
π1(π§)= (ππ + π1π§β1)
Hence, π»2(π§) =π(π§)
π1(π§)=
(ππ+π1π§β1)
(1βπ2π§β1)
The overall Transfer function of the Fig.1 is
π»(π§) = π»1(π§). π»2(π§) = (ππ+π1π§β1)
(1βπ2π§β1)(1βπ1π§β1)
Fig.2: It is a parallel Structure.
π»(π§) = π»1(π§) + π»2(π§) =1
(1 β π1π§β1)+
1
(1 β π2π§β1)
=2β(π1+π2)π§β1
(1βπ1π§β1)(1βπ2π§β1)
Comparing H(z) in both the cases,
ππ = 2; π1 = β(π1 + π2); π1 = π1; π2 = π2
3.Obtain the cascade realization of the IIR system described by
π»(π§) =(π§+1)(π§β5)
[π§+(1+π)][π§+(1βπ)][π§β(1
2+π)][π§β(
1
2βπ)]
Soln.: In the above H(z), complex poles can be paired to result in real coefficients.
Thus, π»(π§) =(π§+1)(π§β5)
[(π§+1)+π)][(π§+1)βπ][(π§β1
2)+π)][(π§β
1
2)βπ)]
π»(π§) =(π§ + 1)(π§ β 5)
[(π§ + 1)2 β (π)2)] [(π§ β12
)2 β (π)2)]
π»(π§) =(π§ + 1)(π§ β 5)
[(π§ + 1)2 β (π)2)] [(π§ β12
)2 β (π)2)]
π»(π§) =(π§ + 1)(π§ β 5)
[π§2 + 2π§ + 2] [π§2 β π§ +54
]= π»1(π§)π»π§(π§)
The individual Transfer functions can be realized in Direct Form-II and can be
cascaded.
Exercise Problems:
1.Realize the system π»(π§) =8π§3β4π§2+11π§β2
(π§β1
4)(π§2βπ§+
1
2)
using (i) Cascade and (ii) Parallel
realizations.
2.A system is represented by the its Transfer Function π»(π§) = 3 +4π§
π§β1
2
β2
π§β1
4
.
Justify whether the above system represents an FIR or IIR system?
3. Find the recursive difference equation for the FIR system with unit sample
response β(π) = πΏ(π + 1) β 2πΏ(π) + πΏ(π β 1)
4.Find the Transfer function and the difference equation of the following digital
filter. y(n)
x(n) 3 2
-
5.Show that the following two systems are equivalent.
System1
x(n) y(n)
πππππππ
βππ
π§β1
+
4π§β1
+
π§β1
π§β1
+
System-2
x(n) π. πππππ y(n)
π. πππππ
π. πππππ
βππππππ
Simulation:
Cascade Form:
β H(z) is written as a product of second order sections with real coefficients.
β This is done by factoring the numerator and denominator polynomials into
their respective roots and then combining either a complex conjugate pair
or any two real roots into second order polynomials.
β π»(π§) =π0+π1π§β1+β―+πππ§βπ
1+π1π§β1+β―+πππ§βπ= π0
1+π1π0
π§β1+β―+πππ0
π§βπ
1+π1π§β1+β―+πππ§βπ=
β πΉππ 2 ππ πππππ π πππ‘ππππ ,
=π0 β1+π΅π,1π§β1+π΅π,2π§β2
1+π΄π,1π§β1+π΄π,2π§β2πΎπ=1 , π€βπππ πΎ =
π
2 πππ π΄π,1, π΄π,2, π΅π,1 πππ π΅π,2 are
Real numbers representing the coefficients of second order sections.
Given the coefficients {ππ} and {ππ} of the direct form filter, the coefficients
π0, {π΅π,π}, πππ {π΄π,π}
%This Program Converts Direct Form to Cascade Form and implements
%This Program Converts Direct Form to Cascade Form and implements
+ + π§β1 +
π§β1
% b0=Gain Coefficient %B=kx3 Matrix of real coefficients containing bks %A=kx3 Matrix of real coefficients containing aks %b=Numerator Polynomial Coefficients of Direct Form %a=Denominator polynomial Coefficients of Direct Form %Computation of Gain Coefficient b0 clear all %Enter the coefficients such that βaβ and βbβ are of same length b=input('input enter the numerator coefficients') a=input('input enter the denominator coefficnets') b0=b(1); b=b/b0; a0=a(1);a=a/a0; b0=b0/a0; N=length(a); K=floor(N/2); A=zeros(K,3); B=zeros(K,3); if K*2==N b=[b 0]; a=[a 0]; end broots=cplxpair(roots(b)); aroots=cplxpair(roots(a)); i=1; while i<=2*K||i<=K Brow=broots(i:i+1,:); Brow=real(poly(Brow)); B(fix((i+1)/2),:)=Brow; Arow=aroots(i:i+1,:); Arow=real(poly(Arow)); A(fix((i+1)/2),:)=Arow; i=i+2; end disp(βThe value of b0 isβ) disp(b0)
Ex.: 1)The following system is to be realized by using Cascade Connection.
16π¦(π) + 12π¦(π β 1) + 2π¦(π β 2) β 4π¦(π β 3) β π¦(π β 4)
= π₯(π) β 3π₯(π β 1) + 11π₯(π β 2) β 27π₯(π β 3) + 18π₯(π β 4)
b =[1 -3 11 -27 18]; a=[16 12 2 -4 -1];
B=[1.000 0.000 9.0001.000 β3.000 2.000
]; A=[1.000 1.000 0.5001.000 β0.250 β0.1250
]
π0 = 0.625
This implies, section 1 in the cascade can be π»1(π§) =1+9π§β1
1+π§β1+0.5π§β2 and the second
section can be π»2(π§) =1β3π§β1+2π§β2
1β0.25π§β1β0.125π§β2 . The individual stages can be realized
using Direct form II and the output of the first stage will be scaled by a factor of
π0 = 0.625 and will be applied as the input for the second stage. It is a cascade
of two second order sections.
2) π»(π§) =1+0.875π§β1
(1+0.2π§β1+0.9π§β2)(1β0.7π§β1) =
1+0.875π§β1
(1β0.5π§β1+076π§β2β0.63π§β3)
b =[1 0.875 0 0 ]; a=[1 -0.5 0.76 -0.63];
B=[1.000 0.8750 01.000 0 0
]; A=[1.000 0.200 0.90001.000 β0.70 0
] ; π0 = 1
This implies, section 1 in the cascade can be π»1(π§) =1+0.875π§β1
1+0.2π§β1+0.9π§β2 and the
second section can be π»2(π§) =1
1β0.7π§β1 . The individual stages can be
realized using Direct form II and the output of the first stage will be scaled by a
factor of π0 = 1 and will be applied as the input for the second stage. It is a
cascade of one First order section and one second order section.