6. Design Of IIR Filters - Simon Fraser University

( ) ( 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.