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University of Manchester School of Computer Science C omp 3 0 291 : Digital Media Processing Section 5 z-transforms

University of Manchester School of Computer Science C omp 3 0 291 : Digital Media Processing Section 5 z-transforms & IIR-type digital filters. Introduction . General causal digital filter has difference equation:. Order is maximum of N and M.

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University of Manchester School of Computer Science C omp 3 0 291 : Digital Media Processing Section 5 z-transforms

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  1. University of Manchester School of Computer Science Comp 30291 : Digital Media Processing Section 5 z-transforms & IIR-type digital filters Comp30291 DMP Section 5

  2. Introduction • General causal digital filter has difference equation: • Order is maximum of N and M. • Recursive if any b j is non-zero. • A 2nd order recursive filter has the difference-equation: • y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] - b 1 y[n-1] - b 2 y[n-2] Comp30291 DMP Section 5

  3. y[n] x[n] + + a0 z-1 z-1 z-1 z-1 + + a1 -b1 a2 -b2 Signal-flow-graph for 2nd order recursive diff equn y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] - b 1 y[n-1] - b 2 y[n-2] Comp30291 DMP Section 5

  4. Derivation of ‘z-transform’ If {x[n]} with x[n] = zn is applied to an LTI system with impulse-response {h[n]}, output would be, by convolution : Comp30291 DMP Section 5

  5. ‘z-transform’ of {h[n]} • If input is {zn}, output is {H(z).zn} • z may be any real or complex number for which series converges. • For causal & stable IIR, the series converges when |z|  1. • Replacing z by ej gives frequency-response: Comp30291 DMP Section 5

  6. Imaginary part of z z = ej  1 Real part of z Visualising H(z) On Argand diagram (‘z-plane’), z = ej lies on unit circle. Comp30291 DMP Section 5

  7. Example 5.1 Find H(z) for the non-recursive difference equation: y[n] = x[n] + x[n-1] Solution: {h[n]} = { ... , 0, 1, 1, 0, ... }, therefore H(z) = 1 + z - 1 Comp30291 DMP Section 5

  8. Example 5.2 • Find H(z) for the recursive difference equation: • y[n] = a 0 x[n] + a 1 x[n-1] - b 1 y[n-1] • Solution:If x[n] = z n then y[n] = H(z) z n , • y[n-1] = H(z) z n - 1 • Substitute to obtain: • H(z) z n = a 0 z n + a 1 z n - 1 - b 1 H(z) z n - 1 • H(z) = a 0 + a 1 z - 1 - b 1 H(z) z - 1 • (1 + b1z-1) H(z) = a 0 + a 1 z - 1 • When z = -b1, H(z) =  Comp30291 DMP Section 5

  9. Find H(z) for the 2nd order difference-equation y[n] = a0 x[n] + a1x[n-1] + a2x[n-2] - b1 y[n-1] - b2 y[n-2] The same method gives: H(z) z n = a0zn + a1zn - 1 +a2zn - 2 - b1H(z) zn - 1 - b2H(z)zn-2 a 0 + a 1 z - 1 + a 2 z - 2  H(z) =  b 0 + b 1 z - 1 + b 2 z - 2 with b0 = 1. Comp30291 DMP Section 5

  10. Consider difference-equation for general digital filter • N M • y[n] =  a i x[n  i]  b j y[n  j] • i=0 j=1 • The same method gives: • This is the ‘system function’ of the digital filter. • Also referred to as ‘transfer function’ Comp30291 DMP Section 5

  11. System function • Ratio of 2 polynomials in z-1 • Equal to z-transform of impulse-response when this converges. • Easily derived from difference-equation & signal-flow graph. • Replacing z by ej gives frequency-response Comp30291 DMP Section 5

  12. z-1 z-1 z-1 z-1 y[n] x[n] + + a0 + + a1 -b1 -b2 a2 Example 5.3 Give a signal-flow graph for the system function: Solution: The difference-equation is: y[n] = a0 x[n] + a1 x[n-1] + a2 x[n-2] - b1 y[n-1] - b2 y[n-2] Represented by the signal-flow graph below: 2nd order or ‘bi-quadratic’ IIR section in ‘direct form 1’. Comp30291 DMP Section 5

  13. Y + + X a0 z-1 z-1 z-1 z-1 X1 + + Y1 a1 -b1 X2 Y2 a2 -b2 To implement ‘direct form 1’ biquad as a program Label ‘z-1’ box inputs & outputs as shown: Comp30291 DMP Section 5

  14. In MATLAB using floating point arithmetic X1=0; X2=0; Y1=0; Y2=0; while 1 X = input(‘X=’); Y = a0*X + a1*X1 + a2*X2 - b1*Y1 - b2*Y2; disp(sprintf(‘Y = %f’ , Y) ) ; % output Y X2 = X1; X1 = X ; Y2 = Y1; Y1 = Y ; end; Comp30291 DMP Section 5

  15. In MATLAB using Signal Processing toolbox: y = filter([a0 a1 a2], [b0 b1 b2], x ); Comp30291 DMP Section 5

  16. In MATLAB using fixed point arith & shifting K=1024; A0=round(a0*K); A1=round(a1*K); A2=round(a2*K); B1=round(b1*K); B2=round(b2*K); X1=0; X2=0; Y1=0; Y2=0; while 1 X = input(‘X=’) ; Y = A0*X + A1*X1 + A2*X2 - A1*Y1 - A2*Y2 ; Y = round(Y/K); % Divide by arith right shift disp(sprint(‘Y=%f’,Y)); % Output Y X2 = X1; X1 = X ; % Prepare for next time Y2 = Y1; Y1 = Y ; end; Comp30291 DMP Section 5

  17. y[n] x[n] + + a0 + + a1 -b1 z-1 z-1 z-1 z-1 -b2 a2 • Look again at ‘Direct Form 1’ signal-flow-graph • It may be thought of as two signal-flow-graphs: Non-recursive part Recursive part x[n] y[n] Comp30291 DMP Section 5

  18. L1 L2 x[n] y[n] Re-ordering LTI systems • It may be shown than if we have two LTI systems as shown: then re-ordering L1 & L2 does not change the behaviour of the overall system. L2 L1 x[n] y[n] • Only guaranteed to work for LTI systems Comp30291 DMP Section 5

  19. + + a0 + + a1 -b1 z-1 z-1 z-1 z-1 z-1 z-1 z-1 z-1 z-1 z-1 -b2 a2 + + + + x[n] x[n] y[n] y[n] a0 a0 + + + + a1 -b1 -b1 a1 a2 a2 -b2 -b2 Alternative signal-flow-graph Look again at ‘Direct Form 1’: y[n] x[n] Re-order the two ‘halves’ & then simplify to ‘direct form 2’: Comp30291 DMP Section 5

  20. W W1 z-1 z-1 W2 + + x[n] y[n] a0 + + -b1 a1 a2 -b2 ‘Direct Form II’ signal-flow-graph • Direct form II economises on ‘delay boxes’. • Notice labels: W,W1 & W2 It is a 2nd order (bi-quad) section whose system function is: Its difference equation is: y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] - b 1 y[n-1] - b 2 y[n-2] i.e. exactly the same as Direct Form 1 Comp30291 DMP Section 5

  21. Program to implement Direct Form II using normal arithmetic W1 = 0; W2 = 0; %For delay boxes while 1 X = input(‘X=’) ; % Input to X W =X - b1*W1 - b2*W2; % Recursive part Y = W*a0 + W1*a1 + W2*a2; % Non-rec. part W2 = W1; W1 = W; % For next time disp(sprintf(‘Y=%f’,Y)); % Output Y using disp end;% Back for next sample Comp30291 DMP Section 5

  22. Direct Form II in fixed point arithmetic with shifting K=1024; A0=round(a0*K); A1=round(a1*K); A2=round(a2*K); B1=round(b1*K); B2=round(b2*K); W1 = 0; W2 = 0; %For delay boxes while 1 X = input(‘X=’) ; % Assign X to input W =K*X - B1*W1 - B2*W2; % Recursive part W =round( W / K); % By arith right-shift Y = W*A0+W1*A1+W2*A2; % Non-rec. part W2 = W1; W1 = W; %For next time Y = round(Y/K); %By arith right-shift disp(sprintf( ‘Y=%f’, Y)); %Output Y using disp end;% Back for next sample Comp30291 DMP Section 5

  23. Poles & zeros of H(z) • For a discrete time filter: • Re-express as: • Now factorise numerator & denominator: Comp30291 DMP Section 5

  24. Poles & zeros of H(z) continued • z 1 , z 2 ,..., z N, are ‘zeros’. p 1 ,p 2 ,..., pN, are ‘poles’. • H(z) generally infinite with z equal to a pole. • H(z) generally zerowith z equal to a zero. • For a causal stable system all poles must satisfy  p i < 1. • i.e. on Argand diagram: poles must lie inside unit circle. • No restriction on the positions of zeros. Comp30291 DMP Section 5

  25. Design of IIR ‘notch’ filter • Design a 4th order 'notch' filter to eliminate an unwanted sinusoid at 800 Hz without severely affecting rest of signal. The sampling rate is FS = 10 kHz. • One way is to use the MATLAB function ‘butter’ as follows: • FL = 800 – 25 ; FU = 800+25; • [a b] = butter(2, [FL FU]/(FS/2),’stop’); • a = [0.98 -3.43 4.96 -3.43 0.98] • b= [ 1 -3.47 4.96 -3.39 0.96] • freqz(a, b); • freqz(a, b, 512, FS); % Better graph • axis([0 FS/2 -50 5]); % Scales axes • Notch has -3 dB frequency band: 25 + 25 = 50 Hz. MATLAB response Comp30291 DMP Section 5

  26. Gain/phase response of notch filter Comp30291 DMP Section 5

  27. 0 Magnitude (dB) -20 -40 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 Frequency (Hz) 0 -100 Phase (degrees) -200 -300 -400 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 Frequency (Hz) Gain/phase responses of notch filter Comp30291 DMP Section 5

  28. Details • How sharp is the notch? • Can answer this question by specifying notch’s -3 dB bandwidth. • Have just designed what Barry calls a 4th order band-stop filter. • MATLAB calls it a 2nd order band-stop filter. • For a sharper notch, decrease -3 dB bandwidth • But this will decrease its ‘depth’, i.e. the attenuation at the ‘notch’ frequency. • If necessary, increase the order to 6 (3) or 8 (4). Comp30291 DMP Section 5

  29. Gain 0 dB 1 0.5 -3 dB F 0 FS/2 800 -  800 800 +  • Sketch the same gain-response • The -3dB frequencies are at ( 800 +  ) and ( 800 -  ) Hz. • Given  (= 25 Hz say) can sketch gain-response: Comp30291 DMP Section 5

  30. Implement the 4th order ‘notch’ filter MATLAB function gave us: a = [0.98 -3.43 4.96 -3.43 0.98] b= [ 1 -3.47 4.96 -3.39 0.96] Transfer (System) Function is: Comp30291 DMP Section 5

  31. + x[n] y[n] 0.98 3.47 z-1 z-1 z-1 z-1 + + + + + + + -3.43 -4.96 4.96 3.39 -3.43 -0.96 0.0.98 A direct Form 2 implementation of the 4th order IIR notch filter Comp30291 DMP Section 5

  32. Problems with ‘direct form’ IIR implementations • Implementation on previous slide works fine in MATLAB. • But ‘direct form’ IIR implementations of order >2 are rarely used. • Sensitivity to round-off error in coeff values will be high. • Also range of ‘intermediate’ signals in z-1 boxes is high. • High wordlength floating point arithmetic hides this problem • But in fixed point arithmetic, great difficulty occurs. • Instead we use ‘cascaded biquad sections’ Comp30291 DMP Section 5

  33. x[n] y[n] H1(z) H2(z) H(z) G x[n] y[n] Using biquad (2nd order) IIR sections Given 4th order H(z), instead of: we prefer to arrange two biquad sectns as follows: Comp30291 DMP Section 5

  34. Converting to the new implementation • Get a & b for 4th order H(z) as before: [a b] = butter(2, [FL FU]/(FS/2),’stop’); • Then execute: [SOS G] = tf2sos(a,b) • MATLAB responds with: SOS = 1 -1.753 1 1 -1.722 0.9776 1 -1.753 1 1 -1.744 0.9785 G = 0.978 Transfer function to 2nd order sectns First sectn 2nd sectn Comp30291 DMP Section 5

  35. x[n] y[n] 0.978 -1.753 1.722 -1.753 1.744 -0.978 -0.979 Fourth order IIR notch filter realised as two biquad (SOS) sections H(z) may now be realised as: Comp30291 DMP Section 5

  36. Example A digital filter with a sampling rate of 200 Hz is required to eliminate an unwanted 50 Hz sinusoidal component of an input signal without affecting the magnitudes of other components too severely. Design a 4th order "notch" filter for this purpose whose 3dB bandwidth is not greater than 3.2 Hz. Solution method: FS=200; FL=50-1.6; FU=50+1.6; [a b]=butter(2,[FL,FU]/(FS/2), ‘stop’); [SOS G] = tf2sos(a,b) Comp30291 DMP Section 5

  37. IIR digital filter design by bilinear transformation • Many design techniques for IIR digital filters have adopted ideas of analogue filters. • Can transform analogue ‘prototype’ transfer function Ha(s) into H(z) for an IIRdigital filter. • Analogue filters have infinite impulse-responses. • Many gain-response approximations exist which are realisable by analogue filters • e.g. Butterworth low-pass approximation which can be transformed to high-pass, band-pass & band-stop. Comp30291 DMP Section 5

  38. Butterworth low-pass gain approximation of order n At C , Gain  0.71 i.e -3 dB Comp30291 DMP Section 5

  39. Transformations from Ha(s) to H(z) for IIR digital filter • Can transform Ha(s), with gain-response Ga(), to H(z) for an IIR digital filter with similar gain-response G(). • Many ways exist. • Most famous is ‘bilinear transformation’. • Replaces s by 2(z-1)/(z+1) to transform Ha(s) to H(z). • Fortunately MATLAB does all this for us. Comp30291 DMP Section 5

  40. Properties of bilinear transformation • (i) Order of H(z) = order of Ha(s) • (ii) If Ha(s) is causal & stable, so is H(z). • (iii) G() = Ga() where  = 2 tan(/2) • So gain of analog filter at  radians/second becomes gain of digital filter at  radians/sample where  = 2tan( /2). Comp30291 DMP Section 5

  41. Frequency warping By (iii),  from - to  mapped to  in range - to .   Comp30291 DMP Section 5

  42. Frequency warping (cont) • Shape of G() will change under the transformation. • If Ga() is Butterworth, G() will not have exactly the same shape, but we still call it Butterworth. • Mapping approx linear for  in the range -2 to 2. • As  increases above 2 , a given increase in  produces smaller and smaller increases in . Comp30291 DMP Section 5

  43. G() Ga() W w p/2 p p (a): Analogue gain response (b): Effect of bilinear transformation Comparing Ga() with G() • G() becomes more and more compressed as . • Illustrate for an analog gain-response with ripples: Comp30291 DMP Section 5

  44. ‘Prototype’ analogue transfer function • Although the shape changes, we would like G() at its cut off C to the same as Ga() at its cut-off frequency. • If Ga() is Butterworth, it is -3dB at its cut-off freq • So we would like G() to be -3 dB at its cut-off C. • Achieved if analogue prototype is designed to have its cut-off frequency at C = 2 tan(C/2). • C is the ‘pre-warped’ cut-off frequency. • Designing analog prototype with cut-off freq 2 tan(C/2) guarantees that the digital filter will have its cut-off at C. Comp30291 DMP Section 5

  45. Design 2nd order IIR lowpass digital filter by bilinear transfm • Let required cut-off frequency C = /4 radians/sample. • Need prototype transfer fn Ha(s) for 2nd order Butt low-pass filter with 3 dB cut-off at 2tan(C/2) = 2 tan(/8) radians/second. • C = 2 tan(/8) = 0.828 • I happen to remember that the transfer fn for a 2nd order Butt low-pass filter with cut-off C is: • If you don’t believe me, check that replacing s by j and taking the modulus gives G() = 1/[1+(/C)2n] with n=2. • Set C = 0.828 in this formula, then replace s by 2(z-1)/(z+1). • Gives us H(z) for an IIR digital filter. That’s it! Comp30291 DMP Section 5

  46. x[n] y[n] 0.098 2 0.94 -0.33 - Resulting IIR digital filter Comp30291 DMP Section 5

  47. Design of 2nd order IIR low-pass digital filter by bilinear transform using MATLAB • If required cut-off freq is /4 radians/sample, type: • [a b] = butter(2, 0.25) • MATLAB gives us: • a = [0.098 0.196 0.098] • b = [1 -0.94 0.33] • The required expression for H(z) is therefore: To save multipliers it is a good idea to re-express this as: Comp30291 DMP Section 5

  48. x[n] y[n] 0.09 8 2 0.94 - 0.33 Realise by ‘direct form 2’ signal-flow graph Comp30291 DMP Section 5

  49. Higher order IIR digital filters Example: Design 4th order Butterwth-type IIR low-pass digital filter with 3 dB c/o at fS / 16. . Solution: Relative cut-off frequency is /8. Typing: [a b] = butter(4, 0.125) gives the response: a = 0.0009 0.0037 0.0056 0.0037 0.0009 b = 1 -2.9768 3.4223 -1.7861 0.3556 Comp30291 DMP Section 5

  50. + x[n] y[n] 0.00093 2.977 z-1 z-1 z-1 z-1 + + + + + + + 0.0037 -3.422 0.0056 1.79 0.0037 -0.356 0.00093 A direct Form 2 implementation of 4th order IIR filter Comp30291 DMP Section 5

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