1. EKT430/4 DIGITAL SIGNAL PROCESSING 2007/2008 CHAPTER 7 ANALOG FILTER

2. Analog Low-Pass Filters: a review In this section, we shall take a review of performance of polynomial based analog low pass filters, such as Butterworth filter, Tchebyshev filter, Inverse Tchebyshev filter. Elliptical filters or Cauer filter, Bessel’s filter. & transformability from LPF to lowpass, band-pass, high-pass and band-reject configurations; with frequency scaling.

3. A preview to Low Pass Analog Filters Low pass (LP) analog filters are all-pole filters. The pass band extends from DC to usually -3dB or, half power frequency. A range beyond pass band is [transit + stop band]. Transit band lie between pass and stop- band. The roll-off rate in transit band is @ -nx20 dB/dec. Here n represents number of poles in excess of zeros.

4. A preview to Low Pass Analog Filters HP and LP filter have one pass and one stop band. Band-reject (BR) has 2 pass band and 1 stop band. Band-pass (BP) has 2 stop-bands and 1 pass band. An steeper transit-band is always associated with: (a) abrupt non-linear phase characteristic (b) abrupt peak at the end of pass-band.

5. Properties of Butterworth Filters The pass-band is mathematically maximal-flat. All the poles lie on a left hand semi-circle in s- plane. They are equi-distant. Odd pole always lie on negative real axis. Remaining poles are complex-conjugate. The roll off rate of transit characteristic is @–20n dB/dec, n is the number of poles.

6. Properties of Butterworth Filters Stop band response asymptotically goes to zero. Location of poles being known, they are easy to design. follows the equation of a circle  k 2 +  k 2 = a 2. or  k + j  k = a

7. Coefficients of the Butterworth polynomials for various number of poles at  c =1. Naoa1a2a3a4a5a6a7a8 111 211.41421 31221 412.61313.41422.61311 513.23615.2361 3.23611 613.86377.46419.14167.46413.86371 714.49410.09814.592 10.0984.4941 815.125813.13721.84625.68821.86413.1375.12581 See the symmetry of coefficients

8. Poles, angles, Polynomial and Factor form For N=3: Angles: [ 0, 60] D(s)=1+2s+2s 2 +s 3 = (1+s)(1+s+s 2 ) F or N=4: Angles: [22.5, 67.5] D(s) = 1+2.613s+3.414s 2 +2.613 s 3 +s 4 = (1+0.76536s +s 2 )(1+1.84776s+s 2 ) For N=5: Angles: [0, 36, 72] D(s) =1+3.236s+5.236s 2 + 5.236s 3 + 3.236s 4 + s 5 = (1+s)(1+0.618s+s 2 ) (1+1.618s+s 2 ) For N=6: Angles: [15, 45, 75] D(s)=1+3.864s+7.464s 2 +9.141s 3 + 7.464s 4 +3.864 s 5 + s 6 = (1+0.5176s+s 2 ) (1+1.414s+s 2 ) (1+1.9318s+s 2 )

9. Pole location on ellipse from a circle: Butterworth to Tchebyshev. Major axis Minor axis  = (1/n) sinh -1 (1/  )

10. Pass band ch. of 6 th order Butterworth and Tchebyshev pp

11. Tchebyshev Filters is also an all pole filter. Poles lie on ellipse and follows following equation: Here sinh  and cosh  represent the radius of minor and major axes respectively. Since  is the angle with respect to y axis, -sin  + jcos  are the Butterworth poles, yielding Tchebyshev_1 poles at p k =  k + j  k = -sin  sinh  + j cos  cosh , while  = (1/n) sinh -1 ( 1 /  ) [Ambarder p 415] Or ( 1 /  ) = sinh (  n)

12. Tchebyshev Filter contd… For the same order of filter, the poles are obtainable from Butterworth circle by horizontally shifting them on to the ellipse having minor axis sinh  and mazor axis cosh . In case of Tchebyshev, ripples are in pass-band while stop band is flat. For an ‘n’ pole filter, there are n numbers of maxima and minima of equal heights. Distances between maxima and minima decrease toward “corner” frequencies. It is also called equi-ripple filter.

13. Tchebyshev filter contd…… For the normalized value of maxima =1, denoting r =  (1+  2 ) = 1/(1-  p ),  p is the peak to peak ripple factor the minima has the value 1/r resulting in: A trade off has to be maintained between the pass band ripple and stop band gain: since any decrease in the pass-band ripple, the gain at stop band increases resulting in poor stop-band performance. These filters have superior transit-band characteristic.

14. Calculations of  and  n from ripple… say -1.1dB We know that 20 log 10 (r )= 10 log (r 2 ) = 1.1 thus r 2 = 1.288=1 +  2  2 = 0.288 hence 1/  = 1.8626 .n = sinh -1 (1/  ) = 1.3804 Where n is the number of poles. Thus  = 1.3804/n

15. Calculation of poles…. Let n = 3   = 1.3804/3 = 0.46. Hence: sinh(0.46)= 0.4764 & cosh(0.46)= 1.1077. Butterworth poles for n=3 are: [-1, cos(  /3)  j sin(  /3)] = [ -1 -0.5  j 0.866] Multiply real parts by ‘sinh’ and imaginary, by ‘cosh’ we get: [-0.4764 -0.2382  j 0.9593] These are the locations of poles in Tchebyshev_1filter. Work out the denominator polynomial. Numerator is a constant and can be calculated from H(0)=1. Soln. H lp (s) =0.4656/(s+0.4766)(s 2 +0.4766s+0.977) Try the formulae p k =  k + j  k = -sin  sinh  + cos  cosh .

16. Transformation of analog filter from LP  LP; LP  HP: Low Pass to Low Pass: The transformation of cut-off frequency from 1rad/sec to  p radians per second can be carried out by the transformation: s  (s/  p ) Low Pass to High Pass: A low pass prototype can be converted into a high pass filter of cut-off frequency  c by following transformation. s  (  c /s)

17. Transformation of analog filter..... LP  BP; LP  BR Let B rad/sec be the –3dB bandwidth, and  o rad/sec be the center frequency, the transformation to operate on Low Pass filter to get band-pass polynomial is: s  (s 2 +  o 2 )/ Bs and to get band-stop polynomial is: s  Bs/(s 2 +  o 2 )

18. Inverse Tchebyshev Filter This filter has flat pass band with ripple in stop band. It’s TF can be derived from that of Tchebyshev filter :: Step-1: Employ s  (1/s) transformation to convert the low-pass filter into a high pass filter; H T (1/s). The ripples of H T (1/s) are in pass band of this high pass filter.

19. Inverse Tchebyshev Filter Step-2: Subtract the so obtained high pass filter from H T (0). Thus H IT (s) = [H T (0) – H T (1/s)]. Thus this filter has zeros in the transfer function. All other characteristics of Tchebyshev filter are maintained. Thus: The transit band characteristics are same, Peak to Peak height of the stop-band ripple is decided by the minor to major axis ratio.

20. Elliptical or, Cauer filter.  It is an extension of Tchebyshev filter.  Here pass-band as well as stop-band have ripples.  Transit band characteristics are at best.  This filter has a zero at a frequency close to stop band. [details not to discuss]

21. BESSEL’S FILTER All above filters care only for amplitude characteristics. Their phase characteristics become poor as we increase the order of filter. Philosophy of this filter is based on Bessel’s Polynomial. This filter takes care of both amplitude and phase characteristics. (details not to discuss)

22. Butterworth low-pass polynomial The Butterworth Polynomial for n pole filter is given by: H(s)H(-s) = [1+(s/j) 2n ] -1 The poles lie at s 2N = (-1)(j) 2n  s 2n = (-1)(j) 2n Since j = e j  /2 & hence j 2n = e j  n for integer k. and -1 = e j  (2k+1) for k  0 S 2n = e j  (2k+1) e j  n =e j  (2k+n+1) s k = e j  (2k+n+1)/2n) = cos  (2k+n+1)/2n +j sin  (2k+n+1)/2n for k  0. Poles are separated by angle  /N.

23. Butterworth filter For odd number of poles, one pole has to lie on the real axis while remaining all other are complex conjugate in the polynomial: H(s) = 1/{(s+a 1 )(s+a 2 ) …. (s+a n )}. At most one pole will lie on real axis rest are complex conjugates. The frequency scaling is carried out by substituting s by s/  c, where  c is the -3 dB frequency in rad./sec.

24. Determination of order n of the filter: Given that H(s)H(-s) = [1+(s/j) 2n ] -1, G x = 20 log 10 |H(j  )| = -10log 10 [1 + (  x /  c ) 2n ] Where s =  x,  c = cut-off frequency, n is order of filter. Letting Pass band gain G p dB occur at  p rad/sec and stop band gain G s dB occur at  s rad/sec. We get expressions as: G p = -10log 10 [1 + (  p /  c ) 2n ]  (  p /  c ) 2n = 10 -Gp/10 -1 G s = -10log 10 [1 + (  s /  c ) 2n ]  (  s /  c ) 2n = 10 -Gs/10 –1 Hence: (  s /  p ) 2n = (10 -Gs/10 –1)/ (10 -Gp/10 –1)

25. Conclusion From here we conclude: (a) the order of the filter n  log 10 {(10 -Gs/10 –1)/ (10 -Gp/10 –1)} / 2log 10 (  s /  p ). (b)  c, the cut off frequency =  p /{10 -Gp/10 –1} 1/2n ALTERNATIVELLY AS =  s /{10 -Gs/10 –1} 1/2n

26. Ex.9.01: Designing a maximally flat LP filter specifications: Pass band gain Gp= -2 dB for 0   rad/sec, and stop band gain Gs = -20 dB for   20 rad/sec. Soln: (a) Calculating number of poles: Given  p = 10, G p = -2dB and  s = 20, G s = -20 dB. Substituting the values in the equation, n = log {(10 -Gs/10 –1)/ (10 -Gp/10 –1)} / 2log(  s /  p ). we get: n = 3.7  4. ….. Should be integer

27. design contd..… (b) since n is taken 4 in place of 3.7, there will be two values of  c, the -3dB cut off frequency: =  p /{10 -Gp/10 –1} 1/2n and =  s /{10 -Gs/10 –1} 1/2n pass band consideration: For n=4,  p = 10, we get -3dB Cut-off frequency  c = 10.693 rad/sec. Stop-band consideration: For n=4,  s = 20, we get -3dB Cut-off frequency  c = 11.261 rad/sec.

28. Contd: We can choose any frequency between 10.693 and 11.261 rad/sec but least width of transit ch. obtained when lowest value is chosen. So we choose :  c = 10.693 rad/sec. Use the prototype denominator polynomial: D(s)=S 4 +2.6131S 3 +3.4142S 2 + 2.6163S +1 Or, can calculate factored polynomial from circle at angles 22.5  and 67.5 . It is: D(s) = (s 2 +1.842s+1) (s 2 +0.771s+1)

29. design continued….. (c) The proto-filter function for n =4 is: H(s) = 1/{s 4 + 2.6131s 3 + 3.4142 s 2 +2.6131s +1} Or, H(s) -1 = (s 2 +1.842s+1) (s 2 +0.771s+1) Frequency scaling, replace s by s’=(s/10.693), H(s’) = 13073.7/{s 4 +27.942s 3 +390.4s 2 +3194.88s+13073.7} Cascade realization: H(s) = H 1 (s) H 2 (s) = 13073.7/{(s 2 +8.1844s+114.34) (s 2 +19.578s+114.34)} Parallel realization: H(s) = H 3 (s) + H 4 (s) = 95.5858/( s 2 +19.578s+114.34) + 136.7746/( s 2 +8.1844s+114.34)

30. design contd…….. Each one of these forms can be converted in z-domain format using impulse-invariance or, bilinear transformation. Alternatively, one can be converted in desired format and other realizations can be derived from there.

31. Designing Tchebyshev LP filter: Amplitude of a normalized Tchebyshev filter is given by : The nth order Tchebyshev polynomial Cn(  ) is alternatively given by Use of first one is preferred used when |  | <1. C n can also be represented by polynomial.

32. Designing Tchebyshev LP filter: The following properties of Tchebyshev Polynomial helps calculate higher order polynomials iteratively.. C n (  ) = 2  C n-1 (  ) -C n-2 (  ) for n>2, and C 0 (  ) =1 and C 1 (  ) = .

33. Tchebyshev polynomials Order n Cn()Cn() 001 01  02 2  2 -1 03 4  3 -3  04 8  4 - 8  2 +1 05 16  5 -20  3 +5  06 32  6 -48  4 +18  2 -1 07 64  7 -112  5 +56  3 -7  08 128  8 -256  6 +160  4 -32  2 + 1 09 256  9 - 576  7 + 432  5 - 120  3 + 9  10 512  10 -1280  8 +1120  6 - 400  4 +50  2 -1

34. Tchebyshev filter design algorithm The normalized gain in dB is G = 20log|H(j  ) = -10log{1+  2 C n 2 (  )}. Letting G s as gain at the stop-band frequency  s, G s = -10log 10 {1+  2 C n 2 (  s )}, Or C n 2 (  s ) = {[10 -Gs /10 ] -1}/  2

35. Tchebyshev filter design algorithm Again, letting r =  (1+  2 ) = 1/(1-  p ), taking logarithm on both the sides, 20log(r) = R = 10 log 10 (1+  2 )dB or,  2 = [10 R/10 ] –1.

36. Tchebyshev algorithm contd…  C n 2 (  s ) = [10 -Gs /10 -1]/ [10 R/10 –1] C n (  s ) =  {[10 -Gs /10 -1]/ [10 R/10 –1]} Since, Equating RHS,

37. Tchebyshev…… To normalize with pass-band frequency we substitute  s by,  s /  p to get The number of poles n can now be calculated as:

38. Tchebyshev Algorithm contd…. Location of poles are given by the equation Where  = [sinh-1(1/  )]/n and k= [0,1, 2,…..,n]. Lathi, ’Signal Processing & Linear Systems’, Oxford,1998, pp. 505-524.

39. Ex. 9.02: Designing a Tchebyshev LP filter. Design to meet the parameters: Pass band gain G p = -2 dB for 0   rad/sec; stop band gain G s = -20 dB for   16.5 rad/sec. Soln: (a) Calculations for no. of Poles: (b) Here pass band gain= peak to peak ripple. n = {1/cosh -1 (16.5/10) cosh -1 [{10 2 –1}/{10 0.2 -1}] 1/2 = 2.999, say 3.0 (c) Calculation of  and x. (i) In the equation  2 = 10 R/10 –1. For R = 2 dB,  = 0.7647. (ii) Putting the values of n and  in  = [sinh -1 (1/  )]/n=(1/3)sinh -1 (1/0.7647) = 0.3610. Contd..

40. Design of Tchebyshev LP filter (d) Calculation of location of poles: (Tchebyshev polynomial can also be used) s 1 = -0.3689, s 2,s 3 = -0.1844  j0.9231 (e) Calculation of normalized Transfer function: that is to satisfy; H(0) = 1, H(s) = 0.3269/[s 3 + 0.7378s 2 + 1.0222s + 0.3269] (f) Calculation of transfer function for the filter: Since the pass band frequency  p = 10 rad/sec, replace s  s/10 in the normalized transfer function equation. We get: H(s) = 326.9/[s 3 + 7.378s 2 + 102.22s + 326.9]

41. Elliptical or, Cauer filter.  It is an extension of Tchebyshev filter.  Here pass-band as well as stop-band have ripples.  Transit band characteristics are at best.  This filter has a zero at a frequency close to stop band. [details not to discuss]

42. Magnitude-frequency response

43. Phase-frequency characteristic

44. Transformation of analog filter..... LP  BP; LP  BR Let B rad/sec be the –dB bandwidth at center frequency of  o rad/sec, the polynomial transformation for Low Pass to band-pass is: s  (s 2 +  o 2 )/ Bs and Low Pass to Band stop is: s  Bs/(s 2 +  o 2 )

45. A note of BP and BR filters These filters observes the geometric mean.  o 2 =  l  h where  l,  h are the two frequency points obtained by intersection of any line parallel to  -axis on the filter characteristic. If case the edges given in specifications do not satisfy the above equation, then they need to be modified to satisfy the above requirement.

46. Ex. 9.3: (a) Design a low pass filter with f p = 200 Hz and f s = 500 Hz. (b) Design a high pass filter with  p = 500 ;  s =200 rad/s (c) Design a bandpass filter with band edges [16, 18, 32, 48]. (d) Design a bandstop filter with band edges [16, 18, 32, 48]. Soln: (a) Choose a low pass proto-type filter with {  s /  p } = 2.5 and operate with following transformation: s  (s/  p ) = s/[200x2  ].

47. Soln. 9.3 contd: (b) The ratio of stop band to pass band characteristic of high-pass filter is {  p /  s } = 2.5. The needed characteristic of a low pass filter is {  s /  p } = 2.5. Choose the low-pass characteristic that meet above requirements, operate transformation as s  (2.5x200/s).

48. Soln. 9.3 contd… Given that BP filter has edges [  1,  2,  3,  4 ] = [16, 18, 32, 48]. Since bandwidth chosen is B =  3 -  2 = 32-18 = 14 rad/sec. yielding  o1 2 =  2  3 = 32 x 18 = 576. According to the hypothesis,  o2 2 =  1 x  4. = 768 >  o1 2 Hence we need to modify one or, both of the two edge frequencies. To compute design within transition band,  4 =  o 2 /  1 Or  4 = 36 rad/sec.

49. Soln. 9.3 contd… The modified edges are [16, 18, 32, 36]. Thus  3 -  2 = 14 rad/sec and  4 -  1 = 20 rad/sec choose the low pass filter with characteristics Ratio of stop band width/pass bandwidth = 20/14=1.4286 for the specified G s and G p (G s and G p ignored here). Design a LPF to workout the order of TF and polynomial to be used. use the transformation: S  (s 2 + 576)/14s

50. Soln 9.3… Consider a band-stop filter with band edges: [  1,  2,  3,  4 ] = [16, 18, 32, 48]. The bandwidth B =  4 -  1 = 32 rad/sec It makes  o 2 = 768. But  2  3 = 576. This necessitates to modify  3 =  o 2 /  2 = 42.667. This has been done to limit the design in the given band.

51. Soln 9.3… The modified edges are: [  1,  2,  3,  4 ] = [16, 18, 42.667, 48]. Thus  4 -  1 =32 &  3 -  2 = 24.667 rad/sec. We choose a low pass filter with Stop bandwidth/pass bandwidth = 32/(24.667) = 1.2973 for given (G s and G p, ignored) and transformation: s  32s/(s 2 + 768).

52. Direct determination of normalized LPF characteristcs from Band Pass One The LP  BP transformation is: s  (s 2 +  o 2 )/ Bs For a given value of s=j  x in RHS, we get normalized frequency  n  (  x 2 -  o 2 )/B  x For B= 14,  0 2 =576 :  n  (  x 2 - 576)/14  x To get normalized LPF characteristics,  n = [  1  2  3  4 ], put for  x = [16,18 32 48]; Hence   1 = (16 2 – 576)/14x16 = -1.429

53. Direct determination of LPF ch. from BP  2 = (182 – 576)/14x18 = -1  3 = (322 – 576)/14x32 = 1  4 = (482 – 576)/14x48 = 2.57 We get normalized LP Frequencies [-1.429, -1, 1, 2.57] For least transit band,  s /  p = min[  1 /  2 ;  4 /  3 ] = 1.429 The Low Pass equivalent frequency normalized specifications are [1 1.429] corresponding to [G p G s ]. The values are same as before.

54. Direct determination of normalized LPF characteristics from Band Stop One The LP to BS transformation is: s  Bs/(s 2 +  0 2 ) For a given value of s=j  x in RHS, we get normalized frequency  n  B  x /(  o 2 -  x 2 ) For B= 32,  0 2 =768 :  n  32  x /(768 -  x 2 ) To get normalized LPF freequencies,  n = [  1  2  3  4 ], put for  x = [16,18 32 48]; Hence  1 = 32x 16/(768 – 162) = 1

55. Direct determination of LPF ch. Band Stop  2 = 32x 18/(768 – 182) = 1.297  3 = 32x 32/(768 – 322) = -4.0  4 = 32x 48/(768 – 482) = -1 The normalized frequencies are [1, 1.297, -4, -1] For least transit band  s /  P = min(  2/  1  3/  2) = 1.297 The Low Pass equivalent frequency normalized specifications are [1 1.297] for the given [Gp Gs] Same as derived earlier.

56. Example_9.4 A band stop filter to satisfy i. Butterworth Criteria and ii. Tchebyshev Criteria is required to meet following specifications. to meet the following specifications. 1. Stop band 100 to 600 Hz (assumed at -3dB) 2. At and between 200 and 400 Hz, the magnitude should be at least -20 dB. 3. Maximum Gain (at the zero) =1. 4. Pass band ripple  1.1 dB. 5. Sampling frequency 2000 Hz.

57. Conversion to normalized LPF specs. f n = [f 1 f 2 f 3 f 4 ]; f x = [100 200 400 600] Hz -3 dB bandwidth B= f 4 -f 1 = 600-100=500 Hz. f 0 2 = f 4 f 1 = 6x 10 4 =(center frequency) 2 Controlling LP to BS transformation is: f n = Bf x /(f 0 2 -f x 2 ) f n = 500xf x /(6x10 4 – f x 2 ) f 1 = 500x100/(6x10 4 -1x10 4 ) = 1 f 2 = 500x200/(6x10 4 -4x10 4 ) = 5 f 3 =500x400 /(6x10 4 -16x10 4 ) = -2 f 4 =500x600 /(6x10 4 -36x10 4 ) = -1  (f s /f p ) = min(5, 2) = 2.

58. 1. Butterworth filter. Number of poles of LP filter  = log 10 {(10 -Gs/10 –1)/ (10 -Gp/10 –1)} / 2log 10 (  s /  p ) = log 10 {(10 2 -1)/(10 0.3 -1)/2 log 10 (2) = 3.318  4. (for better transition characteristic) The Denominator of LP normalized TF = (s 2 +0.765s+1)(s 2 +1.8484s+1) After transforming s  1000  s/(s 2 + 2.3687 x 10 6 ); The resulting polynomial would be: H BS (s)= N(s)/D(s) where N(s) = (s 2 +  0 2 ) 4 D(s) = (s 4 +a 1 s 3 +b 1 s 2 +c 1 s+d 1 ) (s 4 +a 2 s 3 +b 2 s 2 +c 2 s+d 2 )

59. Butterworth filter Where  0 2 = 2.3687 x 10 6 a 1 = 2.4045 x 10 3 a 2 = 5.8049 x 10 3 b 1 = 1.4607 x 10 7 b 2 =1.4607 x 10 7 c 1 = 5.6955 x 10 9 c 2 = 1.3750 x 10 10 d 1 = 5.6108 x 10 12 d 2 = 5.6108 x 10 12 and location of poles: -180  620j -1022  3523j -667  512j - 2235  1714j

60. Designing Tchebyshev_1 1/cosh -1 (2) =0.7593 cosh -1 ( )=cosh -1 (18.53) = 3.612 n= 2.74  3 for better transit characteristic The control equations is where Given:G s =-20 dB, R=-1.1 dB and Calculated  s /  p = 2

61. Tchebyshev… Normalized Tchebyshev LP TF for R=1.1 dB H lp (s) =0.4656/(s+0.4766)(s 2 +0.4766s+0.977) Transforming to BS by s  1000  s/(s 2 + 2.3687 x 10 6 ); H BP (s) = N(s)/D(s) where N(s) = (s 2 +  0 2 ) 3 & D(s) = (s 2 +e 1 s+f 1 ) (s 4 +a 3 s 3 +b 3 s 2 +c 3 s+d 3 ) Where  0 2 = 2.3687 x 10 6 e 1 = 6.5923 x 10 3 f 1 =  0 2 a 3 = 1.5322 x 10 3 b 3 = 1.4838 x10 7 c 3 = 3.6294 x 10 9 d 3 = 5.6108 x 10 12 and poles are at : [-381.38, -6210.9, -109.8  620j -656.3  3704.4j]