1. Nov '04CS3291: Section 61 UNIVERSITY of MANCHESTER Department of Computer Science CS3291: Digital Signal Processing Section 6 IIR discrete time filter design

2. Nov '04CS3291: Section 62 6.1. Introduction: Many design techniques for IIR discrete time filters have adopted ideas of analogue filters Transform H a (s) for analogue ‘prototype’ filter into H(z) for discrete time filter. Begin with a reminder about analogue filters. 6.2. Analogue filters

3. Nov '04CS3291: Section 63 Analogue filters have transfer functions: a 0 + a 1 s + a 2 s 2 +... + a N s N H a (s) =  b 0 + b 1 s + b 2 s 2 +... + b M s M Replace s by j  for frequency-response. For RE(s)  0, H a (s ) is Laplace Transform of h a (t). In terms of poles and zeros: (s - z 1 ) ( z - z 2 )... ( s - z N ) H a (s) = K  (s - p 1 ) ( z - p 2 )... ( s - p M ) Many known techniques for deriving H a (s); e.g. Butterworth low-pass approximation Analogue filters have infinite impulse-responses.

4. Nov '04CS3291: Section 64 Analogue Butterworth low-pass filter of order n [n/2] is integer part of n/2 & P = 0 /1 if n is even / odd

5. Nov '04CS3291: Section 65

6. Nov '04CS3291: Section 66

7. Nov '04CS3291: Section 67

8. Nov '04CS3291: Section 68 y(t) + RC dy(t)/dt = x(t) Take Laplace transforms: Y(s) + RC sY(s) = X(s) Y(s) 1 H a (s) =  =  X(s) 1 + RC s 1 1 G a (  ) =  H a (j  )  =  =   1 + RC j   [ 1 + (  /  C ) 2 ]  C = 1/(RC) Shown graphically when  C = 1 y(t)x(t) 6.3 First order RC filter R C

9. Nov '04CS3291: Section 69 Gain-response of RC filter

10. Nov '04CS3291: Section 610 Impulse response:-  0 : t < 0 h a (t) =   (1/[RC]) e - t / (R C) : t  0 How to transform this filter to a discrete time filter?

11. Nov '04CS3291: Section 611 Example: A filter has system function H(s) = (1+s) / (1 + 2 s + 3 s 2 ). What is its differential equation. Solution: y(t) + 2dy(t)/dt + 3 d 2 y(t)/dt 2 = x(t) + dx(t)/dt It’s easy!

12. Nov '04CS3291: Section 612 Transform an analogue filter with transfer functn H(s) into a digital filter. Many ways exist. We consider four.

13. Nov '04CS3291: Section 613 6.4. Firstly, dispose of a method that will not work. Replacing s by z (or z -1 ) in H a (s) to obtain H(z). Taking previous example with RC = 1: 1 1 H a (s) =  H(z) =  1 + s 1 + z H(z) has pole at z = -1. Even if we moved it to z=0.99, still no good:-

14. Nov '04CS3291: Section 614 With pole at z=-0.99, what does gain-response look like? High-pass instead of low pass. Forget this one.  G(  ) 

15. Nov '04CS3291: Section 615 Substituting : y[n] + RC ( y[n] - y[n-1] ) / T = x[n] (1 + RC/T) y[n] = x[n] + (RC/T) y[n-1] y[n] = a 0 x[n] - b 1 y[n-1] a 0 = 1/(1+RC/T), b 1 = -RC / (T + RC) Recursive diffnce eqn. 6.5 “Derivative approximation” technique Differential equn of RC filter is: y(t) + RC dy(t)/dt = x(t) Sample x(t), y(t) at T seconds to obtain x[n] & y[n]. Assuming T small:

16. Nov '04CS3291: Section 616 Signal flow graph for y[n] = a 0 x[n] - b 1 y[n-1]

17. Nov '04CS3291: Section 617 For analogue filter :  C = 1/(RC) rad/s. To make cut-off 500 Hz: RC = 1 / (2  x 500) = 0.0003183. Let T = 0.0001 s, (f S =10 kHz.) a 0 = 1/(1 + RC/T) = 0.239057 b 1 = - RC/(T + RC) = - 0.760943 Difference equation is: y[n] = 0.24 x[n] + 0.76 y[n-1] 0.24 1 H(z) =  H a (s) =  (1 - 0.76 z - 1 ) 1 + s/(500  2  ) 0.0576 1  H(e j  )  2 =   H(j  )  2 =  1.58 - 1.52cos(  ) 1 + (  / 3142) 2

18. Nov '04CS3291: Section 618 Compare gain-responses:- Shapes similar, but not exactly the same. Replaces s by (1 - z - 1 )/T Not commonly used

19. Nov '04CS3291: Section 619 6.6. “Impulse-response invariant” technique Transforms analog prototype to IIR discrete time filter. See text-books Not on exam

20. Nov '04CS3291: Section 620 Intro to Bilinear Transformation method Most common method for transforming H a (s) to H(z) for IIR discrete time filter. Consider derivative approximation technique: D[n] = dy(t) /dt at t=nT  ( y[n] - y[n-1]) / T. dx(t) /dt at t=nT  (x[n] - x[n-1]) / T. d 2 y(t)/dt 2 at t=nT  (y[n] - 2y[n-1]+y[n-2])/T 2 d 3 y(t)/dt 3 at t=nT  (y[n]-3y[n-1]+3y[n-2]-y[n-3])/T 3 “Backward difference” approximation introduces delay which becomes greater for higher orders. Try "forward differences" : D[n]  [y[n+1] - y[n]] / T, etc. But this does not make matters any better.

21. Nov '04CS3291: Section 621 Bilinear approximation: 0.5( D[n] + D[n-1])  (y[n] - y[n-1]) / T & similarly for dx(t)/dt at t=nT. Similar formulae may be derived for d 2 y(t)/dt 2, and so on. If D(z) is z-transform of D[n] : 0.5( D(z) + z -1 D(z) ) = ( Y(z) - z -1 Y(z) ) / T

22. Nov '04CS3291: Section 622  replacing s by [ (2/T) (z-1)/(z+1)] is bilinear approximation. If LT of y(t) is Y(s) LT of dy(t)/dt is sY(s). s y(t) dy(t)/dt (2/T) (z-1)/(z+1) {y[n]}  {dy(t)/dt / t=nT }

23. Nov '04CS3291: Section 623 6.7. Bilinear transformation: Most common transform from H a (s) to H(z). Replace s by (2 / T) (z-1) / (z+1) to obtain H(z). For convenience take T=1. Example 1 1 H a (s) =  then H(z) =  1 + RC s 1 + RC z + 1 =  (1 + 2RC)z + (1 - 2RC)

24. Nov '04CS3291: Section 624 Re-express as: 1 + z - 1 H(z) = K  1 + b 1 z - 1 where K = 1 / (1 + 2RC) & b 1 = (1 - 2RC) / (1 + 2RC) Properties: (z - 1) (i) H(z) = H a (s) where s = 2  (z + 1) (ii) Order of H(z) = order of H a (s) (iii) If H a (s) is causal & stable, so is H(z). (iv) H(e j  ) = H a (j  ) where  = 2 tan(  /2)

25. Nov '04CS3291: Section 625 Proof of (iii): Let z p be a pole of H(z). Then s p must be a pole of H a (s) where s p = 2 (z p - 1) / (z p + 1). If s p = a + jb then (z p + 1)(a + jb) = 2 (z p - 1), a + 2 + jb (a + 2) 2 + b 2  z p =  and  z p  2 =  -a + 2 - jb (2 - a) 2 + b 2 Now a < 0 as H a (s) is causal & stable,   z p  must be < 1.  if all poles of H a (s) have real parts < 0, all poles of H(z) must lie inside unit circle.

26. Nov '04CS3291: Section 626 Proof of (iv): When z = e j , then e j  - 1 2(e j  / 2 - e - j  / 2 ) s = 2  =  e j  + 1 e j  / 2 + e -j  / 2 2( 2 j sin (  / 2) =  2 cos (  / 2) = 2 j tan(  / 2)

27. Nov '04CS3291: Section 627 Frequency warping: By (iv), H(e j  ) = H a (j  ) with  = 2 tan(  /2).  from -  to  mapped to  in the range -  to .

28. Nov '04CS3291: Section 628 Mapping approx linear for  in the range -2 to 2. As  increases above 2, a given increase in  produces smaller and smaller increases in .

29. Nov '04CS3291: Section 629 Comparing (a) with (b) below, (b) becomes more and more compressed as    . Frequency warping must be taken into account with this method

30. Nov '04CS3291: Section 630 6.8. Design of IIR low-pass filter by bilinear transfm Given cut-off frequency  C in radians/sample:- (i) Calculate  C = 2 tan(  C /2) radians/sec. (  C is "pre-warped" cut-off frequency) (ii) Find H a (s) for analogue low-pass filter with 1 radian/s cut-off. (iii) Scale cut-off frequency of H a (s) to  C (iv) Replace s by 2(z - 1) / (z+1) to obtain H(z). (v) Rearrange the expression for H(z) (vi) Realise by biquadratic sections.

31. Nov '04CS3291: Section 631 Example Design 2 nd order Butterworth-type IIR low-pass filter with  C =  / 4. Solution: Prewarped frequency  C = 2 tan(  / 8) = 0.828 Analogue Butterworth low-pass filter with c/o 1 radian/second: 1 H a (s) =  1 +  2 s + s 2 Scale c/o to 0.828, 1 H a (s) =  1 +  2 s/0.828 + (s/0.828) 2 then replace s by 2 (z+1) / (z-1) to obtain: z 2 + 2z + 1 H(z) =  z 2 - 9.7 z + 3.4

32. Nov '04CS3291: Section 632 which may be realised by the signal flow graph:-

33. Nov '04CS3291: Section 633 6.9 Higher order IIR digital filters: Normally cascaded biquad sections. Example 6.3: Design 4th order Butterwth-type IIR low-pass digital filter with 3 dB c/o at f S / 16.. Solution: (a) Relative cut-off frequency is  /8. Prewarped cut-off :  C = 2 tan((  /8)/2)  0.4 radians/s. Formula for 1 radian/s cut-off is: Replace s by s/0.4 then replace s by 2 (z-1) / (z+1) to obtain:

34. Nov '04CS3291: Section 634 H(z) may be realised as:

35. Nov '04CS3291: Section 635

36. Nov '04CS3291: Section 636 Compare gain-response of 4th order Butt low-pass transfer function used as a prototype, with that of derived digital filter. Both are 1 at zero frequency. Both are 0.707 at the cut-off frequency. Analogue gain approaches 0 as    whereas digital filter gain becomes exactly zero at  = . Shape of Butt gain response is "warped" by bilinear transfn. For digital filter, cut-off rate becomes sharper as    because of the compression as   .

37. Nov '04CS3291: Section 637 6.10. High-pass band-pass and band-stop IIR filters Apply bilinear transformation to H a (s) obtained by frequency band transformations. Cut-off frequencies pre-warped to find analog c/o frequencies. Example: 4th order band-pass filter with  L =  /4,  u =  /2. Solution: Pre-warp both cutoff frequencies:  L = 2 tan ((  /4)/2) = 2 tan(  /8) = 0.828,  u = 2 tan((  /2)/2)) = 2 tan(  /4) = 2 Need 4 th order H a (s) with pass-band  L to  U. Start from 2nd order Butt 1 radian/sec: H a (s) = 1 / (s 2 +  2 s + 1). Replace s by (s 2 + 1.66 ) / 1.17 s to obtain: 1.37 s 2 H a (s) =  s 4 + 1.65 s 3 + 4.69 s 2 + 2.75 s + 2.76

38. Nov '04CS3291: Section 638 We have a problem as the denominatr is not product of 2nd order expressions in s. We have to re-express it in this form. This cannot be done without a computer! One way is to find the roots of the denom using MATLAB: roots([1 1.64 4.69 2.75 2.76]) -0.54 + 1.7 j -0.54 - 1.7 j -0.28 + 0.88 j -0.28 - 0.88 j (s - ( -0.54 +1.7j ) ) (s - (-.54-1.7j ) ) = ( s 2 + 1.08s + 3.18) (s - (-0.28 + 0.88 j) ) ( s - (-0.28 - 0.88) ) = (s 2 + 0.56s + 0.85)

39. Nov '04CS3291: Section 639 After factorising (using MATLAB): 1.37 s 2 H a (s) =  (s 2 + 1.08 s +3.18)(s 2 +0.56 s + 0.85) Replace s by 2(z - 1)/(z + 1):- 5.48 (z - 1) 2 (z + 1) 2 H(z) =  (9.4 z 2 - 1.57 z + 5 ) ( 6 z 2 - 6.3 z + 3.7)

40. Nov '04CS3291: Section 640 Rearrange into 2 biquad sections:  1 - 2 z -1 + z - 2  1 + 2 z - 1 + z - 2 H(z) = 0.1      1 - 0.17 z -1 + 0.54 z -2  1 - 1.05 z -1 + 0.62 z -2  whose gain response is:

41. Nov '04CS3291: Section 641 There are alternative ways of converting H a (s) into second order sections (SOS) in the MATLAB SP tool-box. Can replace s by 2(z - 1)/(z + 1) in the ‘4th order sectn’ expression for H a (s) then use [sos G] = tf2sos ([a 4 a 3 a 2 a 1 a 0 ], [b 4 … b 0 ]) tf2sos does not like functions of s. So we have to convert to a digital filter before using it. ‘help tf2sos’ to find out abt this function I will try to find a more elegant way of doing this one day!

42. Nov '04CS3291: Section 642 Option of cascading high pass & low-pass digital filters to give band-pass or band-stop filters must be used with care. It is much simpler & avoids the factorisation problem. Calculate  U &  L by prewarping  U &  L Then make sure that analogue prototype is wide-band i.e.  U >> 2  L before applying bilinear transformatn. If it’s not wide-band, use the factorisation method. High-pass filters are easy: Apply transformation s   C /s to H a (s) for a 1 radian/s low-pass filter where  C is prewarped c/o freq. Then apply the bilinear transformation as usual.

43. Nov '04CS3291: Section 643 Final note: All these design techniques, and many more, are available in the MATLAB SP toolbox. End of ‘bilinear transformation’ method

44. Nov '04CS3291: Section 644 6.11. Comparison of IIR and FIR digital filters: Advantage of IIR type digital filters: Economical in use of delays, multipliers and adders. Disadvantages: (1) Sensitive to coefficient round-off inaccuracies & effects of overflow in fixed point arith. These effects can lead to instability or serious distortion. (2) An IIR filter cannot be exactly linear phase.

45. Nov '04CS3291: Section 645 Advantages of FIR filters: (1) may be realised by non-recursive structures which are simpler and more convenient for programming especially on devices specifically designed for DSP. (2) FIR structures are always stable. (3) Because there is no recursion, round-off and overflow errors are easily controlled. (4) An FIR filter can be exactly linear phase. Disadvantage of FIR filters: Large orders can be required to perform fairly simple filtering tasks.

46. Nov '04CS3291: Section 646 Problems: 1 Find H(s) for 3rd order Butterwth low-pass filter with  C = 1. 2. Find H(s) for 3 nd order Butterworth low-pass analog filter with cut-off  C Give its differential equation. Apply derivative approx technique to derive 3rd order IIR Butterwth-type digital filter with cut-off 500 Hz where f S = 10 kHz. 3. A 3rd order low-pass IIR discrete time filter is required with 3 dB cut-off at f S /4. Apply bilinear transfn to Butterwth low- pass transfer function to design it & give its signal flow graph as 2nd & 1st order sections in serial cascade. 4. Give program to implement 3rd order IIR filter above in floating point arithmetic. Then do it in fixed point arithmetic. 5. Low-pass IIR digital filter required with cut-off at f s / 4 & stop- band attenuation at least 20 dB for all frequencies above 3f s /8 & below f s /2. Design by bilinear transfn applied to H(s) for Butterwth low-pass, showing that minimum order required is 3.

47. Nov '04CS3291: Section 647 6 Butterworth-type IIR low-pass digital filter needed with 3 dB c/o at f S / 16. Attenuation must be at least 24 dB above f S / 8. What order is needed? Solution: (a) Relative cut-off frequency is  /8. Prewarped cut-off :  C = 2 tan((  /8)/2)  0.40 radians/s. For n t h order Butt low-pass filter with cutoff  C, gain is: 1  H a (j  )  =   [1 + (  /0.4) 2 n ] Gain of IIR filter must be < -24dB at  =  /4.   H a (j  )  must be < -24 dB at  = 2 tan((  /4)/2)  0.83  20 log 1 0 (1/  [1+(.83/.4) 2 n ]) 10 1.2 Hence 1 + (2.1) 2 n > 10 2. 4 = 252. 2.1 2n > 251 n = 4 is smallest possible

48. Nov '04CS3291: Section 648 7. Design a 4th order band-pass IIR digital filter with lower & upper cut-off frequencies at 300 Hz & 3400 Hz when f S = 8 kHz. 8. Design a 4th order band-pass IIR digital filter with lower & upper cut-off frequencies at 2000 Hz & 3000 Hz when f S = 8 kHz 9. (This is really for Sectn 5). What limits how good a notch filter we can implement on a fixed point DSP processor? In theory we can make notch sharper & sharper by moving poles closer & closer to zeros. What limits us in practice. I wonder how sharp a notch we could get in 16-bit fixed pt arithmetic?