6. Design Of IIR Filters - Simon Fraser University

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6. Design of IIR Filters Reference: Sections 7.1 of Textjω A digital filter, H ( e ) , with infinite impulse response (IIR), can bedesigned by first transforming it into a prototype analog filter Hc ( jΩ ) andthen design this analog filter using a standard procedure. Once the analogfilter is properly designed, it is then mapped back to the discrete-timedomain to obtain a digital filter that meets the specifications.The commonly used analog filters are1. Butterworth filters – no ripples at all,2. Chebychev filters - ripples in the passband OR in the stopband, and3. Elliptical filters - ripples in BOTH the pass and stop bands.The design of these filters are well documented in the literature.A disadvantage of IIR filters is that they usually have nonlinear phase.Some minor signal distortion is a result. There are two main techniques used to design IIR filters:1. The Impulse Invariant method, and2. The Bilinear transformation method.6-1

6.1 The Impulse Invariant Method In the impulse invariant method, the impulse response of the digital filter,h[n ] , is made (approximately) equal to the impulse response of an analogfilter, hc (t ) , evaluated at t nTd , where Td is an (abitrary) sampling period.Specificallyh[ n] Td hc ( nTd ) From our discussion in Chapter 2,H (ejω) ω2π k Hj j c T Td k d and aliasing would occur if H c ( jΩ ) is not bandlimited to π / Td (in rad/s).If H c ( jΩ ) is bandlimited to π / Td , thenH ( e jω ) H c ( jω / Td ) .In this case, it is straight forward to specify the prototype analog filter.However, all the commonly used prototype analog filters used in theimpulse invariant design method are indeed non-bandlimited. So there isaliasing. However, the aliasing can be minimized if we over-design theanalog filter (especially in the stop band). The picture below illustrates the design procedure. We first specify thedigital filter as shown in the first diagram. Then we map the digital6-2

frequency ω onto the analog frequency Ω ω / Td and makeHc ( jΩ ) H ( e j ΩT ) . Notice from the diagram that we can only controlthe magnitude of the responses because of the nature of the analog filtersused.dH ( e jω )11 δ1δ2ωωpωsπDigital filter specificationsH c ( jΩ )11 δ1δ2ΩωpTdωsTdπTdPrototype analog filter6-3

Let H c ( s) be the Laplace transform of hc (t ) , where the complex numbers σ jΩis the Laplace domain variable. Assuming H c ( s) has only single-orderpoles sk , thenNAkk 1 s s kH c (s ) This implies N Ak e s thc (t ) k 1 0t 0kt 0,h[ n] Td hc (nTd )N Td Ak e sk nTd u [n ]k 1, AkTd ( e sk Td ) u[n ]Nnk 1andH ( z) h[ n] z nn AkTd ( esk Td ) z nNnn 0 k 1N k 1n 0 AkTd ( z 1esk Td )nNAkTd 1 sk Tdk 1 1 z e 6-4

It is observed that the pole sk σ k jΩ k in the s-plane is mapped into thepolepk e sk Td eσ k Td e jΩk Tdin the z-plane. If all the sk ’sare on the left half of the s-plane, i.e.σ k 0, k 1,2,., N , then all the pk ’s are within the unit-circle in the zplane (i.e. pk 1 ). This means that a stable analog filter always yields astable digital filter with the impulse invariant method. It should be emphasized thatz e sTdin general. Example: Design a digital low pass IIR filter with the followingspecificiations:0.89125 H ( e jω ) 1,0 ω 0.2πH ( e jω ) 0.17783,0.3π ω πUse the impulse invariant technique and an analog Butterworth filter.Assume Td 1 and that there is minimal aliasing.6-5

Solution- with Td 1 , this means Ω ω , and H c ( jΩ) H ( e jΩ ) . Consequently, thespecifications of the analog prototype filter are0.89125 H c ( jΩ ) 1,0 Ω 0.2πHc ( jΩ ) 0.17783,0.3π ΩNotice that we did not specify an upperlimit for the stopband frequency.- The magnitude square response of a Butterword filter of order N isH c ( j Ω) 121 ( )ΩΩc2Nwhere Ωc is the 3-dB frequency of the filter. It is observed that themagnitude square response decreases monotonically with frequency.The larger N is, the closer the Butterworth filter is to an ideal low passfilter.The term H a ( jΩ) in this figure is the same as our H c ( jΩ )6-6

- The 3-dB frequency and the filter order N are solutions of the twosimultaneous equations: 0.2π 1 Ωc 2N 1 .89125 0.3π 1 Ωc 2N 1 .17783 22The exact results are N 5.8858 and Ωc .70474 . After rounding, wehave N 6 and Ωc .7032 . This latter set of results means the passbandrequirement is met exactly at ω p 0.2π and met with margin atω s 0.3π . Specifically, H c ( j0.3π ) 0.1700 0.17783 .- The next step is to find the poles of the Butterworth filter. Note thatH c ( s) H c( s) (11 jΩs c 2N)has 2N poles whose locations in the s-plane are depicted in the diagrambelow.6-7

Mathematically, these poles are π π2πrk 0.7032exp j ( k 1)12 12 2 ; k 1,2,.,2NHalf of these poles are the poles of the Butterworth filter. Specificallywe choose those rk ’s on the left-half s-plane to be the poles of H c ( s) .Consequently the poles of H c ( s) are 0.182 0.679 j 0.497 0.497 j 0.679 0.182 jThe poles of the corresponding digital filter are0.649 0.524 j0.535 0.290 j0.499 0.092 j- The transfer function of the Butterworth filter is6H c ( s) k 1 1 s 1 s k Ω6c6 ( s s )kk 10.12093( s 2 0.3640s 0.4945 )( s 2 0.9945s 0.4945 )( s 2 1.3585 s 0.4945 )6Akk 1 s sk where the Ak s are the partial fraction expansion coefficients.6-8

- The transfer function of the digital filter H ( z ) is (rememberTd 1 )NAk 1 skk 1 1 z eH ( z) 0.2871 0.4466 z 1 2.1428 1.1455 z 1 1 1.2971z 1 0.6949z 2 1 1.0691z 1 0.3699 z 21.8557 0.6303 z 1 1 0.9972 z 1 0.2570 z 26-9

- If Matlab is used to design the above prototype analog filter, first setWp 0.2π 0.62832Ws 0.3π 0.94248Rp 20log10 (0.89125) 1Rs 20log10(0.17783) 15Then issue the Matlab command[N,Wn] buttord(Wp,Ws,Rp,Rs,'s')Matlab with return the filter order in N and its 3db frequency Ωc in thevariable Wn. The results I got is N 6, Wp 0.70866. The latter result isslightly different from Ω c 0.7032 .The poles of the corresponding Butterworth filter can be obtained byissuing the Matlab command[Z,P,K] butter(N,Wn,'s')where the array Z contains the zeros, the array P contains all the poles,and the variable K is the gain. The poles I obtained for this example are-6.8451e-001 1.8341e-001i-6.8451e-001 -1.8341e-001i-5.0109e-001 5.0109e-001i-5.0109e-001 -5.0109e-001i-1.8341e-001 6.8451e-001i-1.8341e-001 -6.8451e-001iwhich are slightly different from those in the text.Finally use the matlab command ‘impinvar” to convert the analogprototype filter into a corresponding digital filter.6-10

Example: Repeat the last example using a Chebyshev filter.- The magnitude square response of a N-th order Chebyshev filter with aripple parameter of ε isH c ( jΩ) 211 ε 2VN2( ),ΩΩcwhere VN ( x ) is the N-th order Chebysheve polynomial, defined as(VN ( x ) cos N cos 1 x)For example, V0 ( x) 1 , V1 ( x) x , V2 ( x) 2 x 2 1 . In general,VN 1 ( x) 2xVN ( x ) VN 1 ( x)- It is observed that when x is between 0 and unity, VN ( x) varies between0 and unity. However, when x is greater than 1, cos 1 x is imaginaryand so VN ( x) behaves like a hyperbolic cosine and consequentlyincreases monotonically for x greater than unity. Consequently, themagnitude response of a Chebyshev filter looks like the following(correction: replace 1 ε by 1/ 1 ε 2 )6-11

- We will design a Chebyshev filter such that the passband requirement ismet exactly at Ω 0.2π . Consequently, Ω c 0.2π and0.891252 1, or1 ε 2ε 0.50885 .- To determine the filter order N, we calculate H c (0.3π ) for differentvalues of N and pick the smallest N that exceeds the specification. Note2that the requirement is H c (0.3π ) 0.17783 or H c (0.3π ) 0.031624 . It canbe shown that2N 3, H c(0.3π ) 0.045512N 4, H c (0.3π ) 0.006942Consequently the filter order is N 4.- The poles of the Chebyshev filter lie on an ellipse in the s-plane withminor axis aΩ c and major axis bΩ c , wherea 1 1/ Nα α 1/ N ) 0.3646 ,(2b 1 1/ Nα α 1/ N ) 1.0644 ,(2andα ε 1 1 ε 2 4.1702 .The equation of this ellipse is given byσ2Ω2 2 2 12 2a Ωc b Ωc6-12

where s σ j Ω is the complex s variable. In addition to the aboveellipse, we defined the major circle as the circle centered at s 0 andwith a radius bΩ c . Similarly, the minor circle is the circle centered ats 0 and with a radius aΩ cTo locate the poles, we first identify the points on the major and minorcircles that are equally spaced in angle with a spacing of π / N andarranged in such a way that the points are symmetrically located withrespect to the imaginary axis (but never fall on the imaginary axis) and apoint occurs on the real axis for N odd but not for N even.The poles of a Chebyshev filter fall on the above ellipse with theordinate specified by the points identified on the major circle and theabscissa specified by the points identified on the minor circle.The diagram below illustrates how the poles can be located for the caseN 3.An example for finding the poles of a Chebyshev filter with N 3.It should be emphasized that only poles on the left-half plane are used inthe transfer function of the filter.6-13

- Once all the 4 poles, s1 , s2 , s3 , s4 , are found, the transfer function of theChebyshev filter can be written as (please verify)4H c (s) sk11 ε 2k 14 (s s )k (s0.0382862)( 0.4233s 0.1103 s 2 0.1753s 0.3894k 1The magnitude and phase of this transfer function are shown below6-14)

- If you use Matlab to determine this filter, first setWp 0.2π 0.62832Ws 0.3π 0.94248Rp 20log10 (0.89125) 1Rs 20log10(0.17783) 15and then issue the Matlab command[N,Wn] cheb1ord(Wp,Ws,Rp,Rs,'s')Finally complete the filter design by issuing the command[Z,P,K] cheby1(N,-Rp,Wn,’s’)The zeros of the filter will be returned in the array Z, the poles stored inP, and the gain stored in K. What I found for the poles were-2.1166e-001 2.5593e-001i-2.1166e-001 -2.5593e-001i-8.7673e-002 6.1788e-001i-8.7673e-002 -6.1788e-001Iand the gain was K 0.038286. There are no zeros. These results agreewith those obtained through analysis.- The transfer function of the corresponding digital IIR filter is0.08327 0.0239 z 10.08327 0.0246 z 1H ( z) 1 1.5658 z 1 0.6549 z 2 1 1.4934 z 1 0.8392 z 2- It is interesting to point out the Chebyshev filter has a lower order thanthe Butterworth filter in the last example.6-15

6.2 The Bilinear Transformation Method In the impulse invariant method, aliasing occurs when the prototype analogfilter is transformed back into the digital filter. To reduce the distortionintroduced by aliasing, we start off by tightening the specifications on thedigital filter. This is somewhat cumbersome and may lead to severaliterations before the “optimal” filter is found. Aliasing occurs because points in the Ω axis separated by 2π / Td aremapped into the same digital frequency ω . In the Bilinear transformationmethod, there is a one-to-one correspondence between Ω and ω . Soaliasing is avoided in transforming the prototype analog filter back into thedigital filter.Since ω is limited to [ π , π ] but Ω varies from to , it becomes clearthat Ω must be compressed when it is mapped to ω . In other word,Bilinear transformation is non-linear in nature. Let H c (s ) be the transfer function of the protype analog filter. The transferfunction H ( z ) of the digital filter is obtained by substituting2 1 z 1s Td 1 z 1into the expression of H c (s ) . In other word 2 1 z 1 H ( z) H c 1 T1 z d 6-16

The Bilinear transformation can be written alterantively as:(1 z ) sT 1d 2 (1 z 1 ) , orsTd z 1 sTd 2 2 z 1 , or( 2 sTd ) z 1 2 sTd ,z or2 sTd, or2 sTd1 sT2d 1 sT2d If s σ jΩ , where Ω is the analog frequency, then1 ) j(z (1 ) j(1 ) ( ) exp { j arctan ( ) (1 ) } (1 ) ( ) exp{ j arctan ( ) (1 ) }σ Td2ΩTd2σ Td2ΩTd22σ Td2σ Td222ΩTd2ΩTd2ΩTd22ΩTd2σ Td2σ Td2 z exp { jθ }wherez () ( )(1 ) ( )1 σT2d2σ Td22 ΩTd22ΩTd22and6-17

( ) ( 1 ) arctan ( ) (1 ) θ arctan ΩTd2σ Td2ΩTd2σ Td2are respectively the magnitude and phase of z .Observations:1. When σ 0 , z 1 . So left-plane poles of H c (s ) will be mapped intopoles within the unit circle in the z-plane. In other word, a stableprototype analog filter will lead to a stable digital filter.2. When σ 0 , z 1 . In other word, z lies on the unit circle and can bewritten asz e jω (when σ 0)where ΩT ω θ 2arctan d 2 represents the digital frequency. Alternatively, we can express theanalaog frequency Ω in terms of the digital frequency ω asΩ 2 ω tan Td 2 The figure below illustrates the relationship between Ω and ω . It isclear that compression occurs in the mapping process.These results indicate that the entire left-half s-plane is mapped into adisc of radius 1 in the z-plane.6-18

Example: Design a digital low pass filter with the following specifications0.89125 H ( e jω ) 1,0 ω 0.2πH ( e jω ) 0.17783,0.3π ω πusing the Bilinear transformation method and a Butterworth prototypefilter. Compare the results with those obtained through the impulseinvariant method.Solution:For simplicity, set Td 1 . So Ω 2tan (ω / 2) . This means the digitalpassband and stopband frequencies,6-19

ω p 0.2π ,ω s 0.3πare mapped into the analog passband and stopband frequenciesΩ p 2tan ( ω p / 2 ) 0.64984,Ωs 2tan (ω s / 2) 1.0191The specifications of the prototype analog filter become0.89125 H c ( jΩ ) 1,Hc ( jΩ ) 0.17783,0 Ω 0.649841.0191 Ω Next we define the following Matlab variablesRp 20log10(0.89125) -1Rs 20log10(0.17783) -15Wp 0.64984Ws 1.0191and issue the command [N,Wn] buttord(Wp,Ws,Rp,Rs,’s’). Matlab willreturn the order of the Butterworth filter in the variable N and the 3-dbfrequency in the variable Wn. It was found thatN 6Wn 0.76627 Ω cTo complete the design of the protoype Butterworth filter, we next issue theMatlab command [Z,P,K] butter(N,Wn,’s’). The zeros of H c ( s ) will thenbe returned in the array Z, the poles in the array P, and the filter gain (sameNas Ωc ) in the variable K. It was found that there are no zeros, the poles are6-20

-7.4016e-001 1.9832e-001i-7.4016e-001 -1.9832e-001i-5.4183e-001 5.4183e-001i-5.4183e-001 -5.4183e-001i-1.9832e-001 7.4016e-001i-1.9832e-001 -7.4016e-001Iand the gain is 0.20243. Denoting the poles as s1 , s2 ,., s6 , the transferfunction of the Butterworth filter isH c (s ) K 6k 1(s sk )0.20243( s 0.39665s 0.58716 )( s 1.0837s 0.58716 )( s 2 1.4803s 0.58716)22This result is very close to that in Example 7.3 of the Text. Compared tothe result obtained under the impulse invariant method, we observe anoticeable difference in H c ( s) .Finally, the last step in the design exercise is to map the above transferfunction into a digital filter using Bilinear transformation, i.e. 1 z 1 H (z ) H c 2 1 1 z This can be done by using the Matlab command[Zd,Pd,Kd] bilinear(Z,P,K,1)where Z, P, and K are same as before, 1 corresponds to the samplingfrequency 1/ Td , Zd is an array containing the zeros of H ( z ) , Pd is the array6-21

storing the poles of H ( z ) , and Kd is the gain of H ( z ) . The results Iobtained are:Kd 0.00073798Zd -1-1-1-1-1-1Pd 4.5216e-001 1.0510e-001i4.5216e-001 -1.0510e-001i5.0527e-001 3.2087e-001i5.0527e-001 -3.2087e-001i6.3430e-001 5.5026e-001i6.3430e-001 -5.5026e-001ICalling the 6 zeros as z1 , z2 ,., z 6 and the 6 poles as p1 , p2 ,., p6 . Then thetransfer function of the digital filter H ( z ) can be written as1 z z )( H ( z ) Kd (1 p z )6 1k 16kk 1k(1 1.2686 z 1(0.00073798 1 z 1 1)()6 0.70512 z 2 1 1.0105 z 1 0.35826z 2(1 0.90433 z1 1 0.21550z 2) )The result is once again close to that in Example 7-3 of the text.6-22

6.3 Frequency Transformation of Lowpass IIR Filters So far, we focus our discussion on the design of lowpass IIR filters. Howabout highpass, bandpass, and bandstop filters?Frequency responses of lowpass, highpass, bandpass, and bandstop filters. The impulse invariant method will not be suitable for the highpass andbandstop filters because of heavy aliasing.6-23

It is possible to get around the problem by first designing a digitalportotype lowpass filter and then perform an algebraic transformation onthe digital lowpass filter to obtain the desired frequency selective filter. The tranfer function of the digital prototype low pass filter is denoted byH lp ( p ) where p plays the same role as z , the conventional z -transformvariable.The transfer function of the desired frequency selective filter is denoted byH (z) .We want to find a relationship between p and z, denoted by( )p 1 G z 1such thatH ( z ) H lp ( p )( )p 1 G z 1It should be emphasized that the transformation relates p 1 to z 1 , not p toz. This is due to the fact that H lp ( p ) is usually expressed as a function inp 1 instead of p. We want a stable, rational H lp ( p ) be mapped into a stable rational H ( z ) .Consequently, the requirements on the mapping function p 1 G ( z 1 ) are:1. G ( z 1 ) must be a rational function of z 1 .2. The inside of the unit circle in the p-plane is mapped into the inside ofthe unit circle in the z-plane.6-24

3. The unit circle in the p-plane is mapped into the unit circle in the zplane. So if θ and ω are the frequency variables in the p-plane and thez-plane, thene jθ G e jω(Consequently)G ( e jω ) 1and()θ R G e jω . The most general form of the function G ( z 1 ) that satisfies all the aboverequirements is:z 1 α kp G ( z ) ; 1k 1 1 α k z 1N 1αk 1By choosing appropiate values for N and the α k ’s, a variety of mappingscan be obtained. Low-pass to Low-pass transformation:z 1 αp 1 α z 1 1This means the relationship between θ and ω ise jθe jω α 1 α e jω6-25

ore jωe jθ α 1 α e jθe jθ α )(1 α e jθ )( (1 α e jθ )(1 α e jθ ) or2α (1 α 2 ) cos(θ ) j (α 2 1) sin(θ )1 α 2 2α cos(θ ) (1 α 2 ) sin θ ω arctan 2 2α (1 α ) cosθ For a given pair of θ p and ω p , the parameter α isα sin (θ p ω p ) / 2 sin (θ p ω p ) / 2 It can be shown that the absolute value of α is less than 1.6-26

Transformations from LP to LP, HP, BP, and BS filters are shown below:The corresponding definitions of filter cutoff frequencies are shown in thenext page.6-27

(a) low pass, (b) high pass, (c) bandpass, (d) bandstop.6-28

Example: The prototype low pass filter0.001836(1 p 1 )4H lp ( p) (1 1.5548 p 1 0.6493 p 2 )(1 1.4996 p 1 0.8482 p 2 )satisfies0.89125 H ( e jθ ) 1,0 θ 0.2πH ( e jθ ) 0.17783,0.3π θ πDetermine the corresponding high pass filter that satisfies the requirements0.89125 H ( e jω ) 1;0.6 π ω πSolution:The passband frequency of the high pass filter is ω p 0.6π and thepassband frequency of the digital prototype filter is θ p 0.2π . So accordingto the design formula,(α cos (cosθ p ω p2θ p ω p2) 0.38197)This meansz 1 α z 1 0.38197p 1 α z 1 1 0.38197 z 1 1Consequently,6-29

H ( z ) H lp ( p ) ()(p 1 z 1 0.38197 / 1 0.38197 z 1(1 1.0416z)0.02426 (1 z 1 ) 14 0.4019z 2 )(1 0.5661z 1 0.7647 z 2 )It should be pointed out that in actual design, the stopband frequency of thedigital prototype filter will be determined by the stopband frequency of thehigh pass (or desired) filter. We assume in this example that the stop bandfrequency of the high pass filter is mapped into θ s 0.3π based onα 0.38197.6.4 Implementation Structures for IIR Filters Reference: Section 6.3 of Text The transfer function of an IIR filter can always be expressed as a ratio oftwo polynomials, i.e.H ( z) B (z ),A (z )where A( z) and B( z) are respectively polynomials of orders N and Min z 1 .It is the denominator polynomial that makes the impulse response of thefilter infinitely long.6-30

Different expressions for A( z ) , B ( z ) , and H ( z ) lead to differentimplementation structures. For example, if these two polynomials arewritten asNA( z ) 1 ak z kk 1andMB( z ) bk z k ,k 0thenMH ( z) b zk 0N kk1 ak z k.k 1This means in the time domain, the input x [ n] and the output y[ n] of thefilter satisfiesNy[n ] ak y [n k ] k 1M b x[ n k ] ,k 0kwhich is simply a linear constant coefficient different equation.A a possible computational structure for y[n ] is shown in the signal flowgraph in the next page. There, a branch with a transmittance of z 1 isequivalent to a delay of 1 sample, and a branch with a transmittance ai (orb j ) implies a scaling of the signal at the originating node of that branch bythe constant ai (or b j ). As in any signal flow graph, the signal at any node,i.e. the node value, is the sum of products of the signal at an originatingnode and the corresponding branch transmittance.6-31

The above computational structure is called the Direct Form Iimplementation of an IIR filter. Note that for simplicity, we assume thatthe order of the numerator and denominator polynomials are identical, i.e.N M .It is observed that the Direct Form I implementation structure requires2 N 1 multiplications, 2N additions, and 2N delay elements. The number of delay elements can be reduced to N if we interchange thetwo sections in the Direct Form I structure. This leads us to the DirectForm II structure shown below.6-32

It is also possible to express the transfer function of an IIR filter in productform as (1 f z ) (1 gM1H (z ) C 1M2kk 1N1k 1N2k 1k 1kz 1 )(1 g*k z 1 ) (1 ck z 1 ) (1 d k z 1 )(1 d*k z 1 ),where C is a constant, the f k ‘s are the real zeros of the transfer function,the g k ‘s are the complex zeros (which always exist in conjugate pairs), theck ’s are the real poles, and the d k ’s are the complex poles (which alwaysexist in conjugate pairs).Note that the products(1 g z )(1 g 1k*kz 1 ) and(1 d z )(1 d z ) 1k* 1kare 2ndorder polynomials in z 1 with real coefficients.6-33

For simplicity, we assume that N1 M 1 2 K (K an integer) and N 2 M 2 .This means we can combine pairs of real poles and combine pairs of realzeros and rewrite H ( z ) asb0 k b1k z 1 b2 k z 2H (z) 1 2k 1 1 a1k z a2 k zNsNs H k ( z),k 1whereb0 k b1k z 1 b2 k z 2H k ( z) 1 a1k z 1 a 2k z 2andN s K N2 .This expression for the transfer function enables us to visualize the IIRfilter as the serial concatentation of N s subsystems. Subsequently it leadsto the Cascade implementation structure; see for example the case ofN s 3 below. Exercise: Draw the Cascade form of the high pass filter in Section 6-3.6-34

With the Cascade form implementation, the locations of the poles and zeroswill not change dramatically when the aij ’s and the bij ’s are quantized (asin fixed point implementation). This is in contrast to the Direct formimplementation where the poles and zeros can change substantially becauseof quantization. This issue will be discussed in details in Chapter 7 of thelecture notes. If we express the transfer function as a partial fraction expansion, thene0k e1k z 1H ( z) 1 a2k z 2k 1 1 a1k zNsNs Gk ( s),k 1wheree0 k e1k z 1Gk (s ) 1 a1k z 1 a2 k z 2and all the polynomial coefficients are real.The transfer function suggests that the IIR filter can be viewed as a “large”system comprising of N s parallel subsystems. The input to all thesubsystems is x[n ] and the output of the IIR filter, y[ n] , is the sum of theyk [ n] ’s , the individual outputs of the subsystems.The figure below shows this parallel form of the IIR filter.6-35

The parallel form offers robustness against quantization error for the poles.The zeros, however, will be affected by the shifting of the poles in theindividual subsystems. Exercise: While it is straight forward to incorporate any given initialcondition of an IIR filter in the Direct Form I structure, it is not clear howthis can be done in the other structures. Is this a matter of great concern? Ifso, suggest how a given initial condition can be incorporated into the DirectForm II, the Cascade form, and the Parallel form.6-36

domain to obtain a digital filter that meets the specifications. The commonly used analog filters are 1. Butterworth filters - no ripples at all, 2. Chebychev filters - ripples in the passband OR in the stopband, and 3. Elliptical filters - ripples in BOTH the pass and stop bands. The design of these filters are well documented in the literature.

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