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s to Z-Domain Transfer Function 1. s to Z-Domain Transfer Function 1. Discrete ZOH Signals s to Z-Domain Transfer Function 1. Discrete ZOH Signals 1. Get step response of continuous trans- fer function y s ( t ) . s to Z-Domain

  1. ZOH Equivalent First Order Transfer Function 5. Find the ZOH equivalent of K/ ( τs + 1) . � � Y s ( s ) = 1 K 1 1 τs + 1 = K s − s + 1 s τ � 1 − e − t/τ � y s ( t ) = K , t ≥ 0 � 1 − e − nT s /τ � y s ( nT s ) = K , n ≥ 0 Kz (1 − e − T s /τ ) z z � � Y s ( z ) = K = z − 1 − z − e − T s /τ ( z − 1)( z − e − T s /τ )

  2. ZOH Equivalent First Order Transfer Function 5. Find the ZOH equivalent of K/ ( τs + 1) . � � Y s ( s ) = 1 K 1 1 τs + 1 = K s − s + 1 s τ � 1 − e − t/τ � y s ( t ) = K , t ≥ 0 � 1 − e − nT s /τ � y s ( nT s ) = K , n ≥ 0 Kz (1 − e − T s /τ ) z z � � Y s ( z ) = K = z − 1 − z − e − T s /τ ( z − 1)( z − e − T s /τ ) Dividing by z/ ( z − 1) , we get G ( z ) = K (1 − e − T s /τ ) z − e − T s /τ 5 Digital Control Kannan M. Moudgalya, Autumn 2007

  3. ZOH Equivalent First Order Transfer Function 6. - Example

  4. ZOH Equivalent First Order Transfer Function 6. - Example Sample at T s = 0 . 5 and find ZOH equivalent trans. function of 10 G a ( s ) = 5 s + 1

  5. ZOH Equivalent First Order Transfer Function 6. - Example Sample at T s = 0 . 5 and find ZOH equivalent trans. function of 10 G a ( s ) = 5 s + 1 Scilab Code:

  6. ZOH Equivalent First Order Transfer Function 6. - Example Sample at T s = 0 . 5 and find ZOH equivalent trans. function of 10 G a ( s ) = 5 s + 1 Scilab Code: Ga = tf(10,[5 1]);

  7. ZOH Equivalent First Order Transfer Function 6. - Example Sample at T s = 0 . 5 and find ZOH equivalent trans. function of 10 G a ( s ) = 5 s + 1 Scilab Code: Ga = tf(10,[5 1]); G = ss2tf(dscr(Ga,0.5));

  8. ZOH Equivalent First Order Transfer Function 6. - Example Sample at T s = 0 . 5 Scilab output is, and find ZOH equivalent trans. function of 10 G a ( s ) = 5 s + 1 Scilab Code: Ga = tf(10,[5 1]); G = ss2tf(dscr(Ga,0.5));

  9. ZOH Equivalent First Order Transfer Function 6. - Example Sample at T s = 0 . 5 Scilab output is, and find ZOH equivalent 0 . 9546 G ( z ) = trans. function of z − 0 . 9048 10 G a ( s ) = 5 s + 1 Scilab Code: Ga = tf(10,[5 1]); G = ss2tf(dscr(Ga,0.5));

  10. ZOH Equivalent First Order Transfer Function 6. - Example Sample at T s = 0 . 5 Scilab output is, and find ZOH equivalent 0 . 9546 G ( z ) = trans. function of z − 0 . 9048 10 = 10(1 − e − 0 . 1 ) G a ( s ) = 5 s + 1 z − e − 0 . 1 Scilab Code: Ga = tf(10,[5 1]); G = ss2tf(dscr(Ga,0.5));

  11. ZOH Equivalent First Order Transfer Function 6. - Example Sample at T s = 0 . 5 Scilab output is, and find ZOH equivalent 0 . 9546 G ( z ) = trans. function of z − 0 . 9048 10 = 10(1 − e − 0 . 1 ) G a ( s ) = 5 s + 1 z − e − 0 . 1 Scilab Code: In agreement with the Ga = tf(10,[5 1]); formula in the previous G = ss2tf(dscr(Ga,0.5)); slide 6 Digital Control Kannan M. Moudgalya, Autumn 2007

  12. Discrete Integration 7.

  13. Discrete Integration 7. u ( n ) u ( k − 1) u ( k ) n

  14. Discrete Integration 7. y ( k ) = blue shaded area u ( n ) u ( k − 1) u ( k ) n

  15. Discrete Integration 7. y ( k ) = blue shaded area u ( n ) + red shaded area u ( k − 1) u ( k ) n

  16. Discrete Integration 7. y ( k ) = blue shaded area u ( n ) + red shaded area u ( k − 1) y ( k ) = y ( k − 1) u ( k ) n

  17. Discrete Integration 7. y ( k ) = blue shaded area u ( n ) + red shaded area u ( k − 1) y ( k ) = y ( k − 1) + red shaded area u ( k ) n

  18. Discrete Integration 7. y ( k ) = blue shaded area u ( n ) + red shaded area u ( k − 1) y ( k ) = y ( k − 1) + red shaded area u ( k ) y ( k ) = y ( k − 1) + T s 2 [ u ( k ) + u ( k − 1)] n

  19. Discrete Integration 7. y ( k ) = blue shaded area u ( n ) + red shaded area u ( k − 1) y ( k ) = y ( k − 1) + red shaded area u ( k ) y ( k ) = y ( k − 1) + T s 2 [ u ( k ) + u ( k − 1)] n Take Z-transform:

  20. Discrete Integration 7. y ( k ) = blue shaded area u ( n ) + red shaded area u ( k − 1) y ( k ) = y ( k − 1) + red shaded area u ( k ) y ( k ) = y ( k − 1) + T s 2 [ u ( k ) + u ( k − 1)] n Take Z-transform: Y ( z ) = z − 1 Y ( z ) + T s U ( z ) + z − 1 U ( z ) � � 2

  21. Discrete Integration 7. y ( k ) = blue shaded area u ( n ) + red shaded area u ( k − 1) y ( k ) = y ( k − 1) + red shaded area u ( k ) y ( k ) = y ( k − 1) + T s 2 [ u ( k ) + u ( k − 1)] n Take Z-transform: Y ( z ) = z − 1 Y ( z ) + T s U ( z ) + z − 1 U ( z ) � � 2 Bring all Y to left side:

  22. Discrete Integration 7. y ( k ) = blue shaded area u ( n ) + red shaded area u ( k − 1) y ( k ) = y ( k − 1) + red shaded area u ( k ) y ( k ) = y ( k − 1) + T s 2 [ u ( k ) + u ( k − 1)] n Take Z-transform: Y ( z ) = z − 1 Y ( z ) + T s U ( z ) + z − 1 U ( z ) � � 2 Bring all Y to left side: Y ( z ) − z − 1 Y ( z ) = T s U ( z ) + z − 1 U ( z ) � � 2

  23. Discrete Integration 7. y ( k ) = blue shaded area u ( n ) + red shaded area u ( k − 1) y ( k ) = y ( k − 1) + red shaded area u ( k ) y ( k ) = y ( k − 1) + T s 2 [ u ( k ) + u ( k − 1)] n Take Z-transform: Y ( z ) = z − 1 Y ( z ) + T s U ( z ) + z − 1 U ( z ) � � 2 Bring all Y to left side: Y ( z ) − z − 1 Y ( z ) = T s U ( z ) + z − 1 U ( z ) � � 2 (1 − z − 1 ) Y ( z ) = T s 2 (1 + z − 1 ) U ( z ) 7 Digital Control Kannan M. Moudgalya, Autumn 2007

  24. Transfer Function for Discrete Integration 8.

  25. Transfer Function for Discrete Integration 8. Recall from previous slide (1 − z − 1 ) Y ( z ) = T s 2 (1 + z − 1 ) U ( z )

  26. Transfer Function for Discrete Integration 8. Recall from previous slide (1 − z − 1 ) Y ( z ) = T s 2 (1 + z − 1 ) U ( z ) 1 + z − 1 Y ( z ) = T s 1 − z − 1 U ( z ) 2

  27. Transfer Function for Discrete Integration 8. Recall from previous slide (1 − z − 1 ) Y ( z ) = T s 2 (1 + z − 1 ) U ( z ) 1 + z − 1 Y ( z ) = T s 1 − z − 1 U ( z ) 2 = T s z + 1 z − 1 U ( z ) 2

  28. Transfer Function for Discrete Integration 8. Recall from previous slide (1 − z − 1 ) Y ( z ) = T s 2 (1 + z − 1 ) U ( z ) 1 + z − 1 Y ( z ) = T s 1 − z − 1 U ( z ) 2 = T s z + 1 z − 1 U ( z ) 2 Integrator has a transfer function,

  29. Transfer Function for Discrete Integration 8. Recall from previous slide (1 − z − 1 ) Y ( z ) = T s 2 (1 + z − 1 ) U ( z ) 1 + z − 1 Y ( z ) = T s 1 − z − 1 U ( z ) 2 = T s z + 1 z − 1 U ( z ) 2 Integrator has a transfer function, G I ( z ) = T s z + 1 2 z − 1

  30. Transfer Function for Discrete Integration 8. Recall from previous slide (1 − z − 1 ) Y ( z ) = T s 2 (1 + z − 1 ) U ( z ) 1 + z − 1 Y ( z ) = T s 1 − z − 1 U ( z ) 2 = T s z + 1 z − 1 U ( z ) 2 Integrator has a transfer function, G I ( z ) = T s z + 1 2 z − 1 A low pass filter!

  31. Transfer Function for Discrete Integration 8. Recall from previous slide Im ( z ) (1 − z − 1 ) Y ( z ) = T s 2 (1 + z − 1 ) U ( z ) × Re ( z ) 1 + z − 1 Y ( z ) = T s 1 − z − 1 U ( z ) 2 = T s z + 1 z − 1 U ( z ) 2 Integrator has a transfer function, G I ( z ) = T s z + 1 2 z − 1 A low pass filter!

  32. Transfer Function for Discrete Integration 8. Recall from previous slide Im ( z ) (1 − z − 1 ) Y ( z ) = T s 2 (1 + z − 1 ) U ( z ) × Re ( z ) 1 + z − 1 Y ( z ) = T s 1 − z − 1 U ( z ) 2 1 s ↔ T s z + 1 = T s z + 1 z − 1 U ( z ) 2 z − 1 2 Integrator has a transfer function, G I ( z ) = T s z + 1 2 z − 1 A low pass filter! 8 Digital Control Kannan M. Moudgalya, Autumn 2007

  33. Derivative Mode 9.

  34. Derivative Mode 9. • Integral Mode: 1 s ↔ T s z + 1 2 z − 1

  35. Derivative Mode 9. • Integral Mode: 1 s ↔ T s z + 1 2 z − 1 • Derivative Mode: s ↔ 2 z − 1 T s z + 1

  36. Derivative Mode 9. • Integral Mode: 1 s ↔ T s z + 1 2 z − 1 • Derivative Mode: s ↔ 2 z − 1 T s z + 1 • High pass filter

  37. Derivative Mode 9. • Integral Mode: 1 s ↔ T s z + 1 2 z − 1 • Derivative Mode: s ↔ 2 z − 1 T s z + 1 • High pass filter • Has a pole at z = − 1 .

  38. Derivative Mode 9. • Integral Mode: 1 s ↔ T s z + 1 2 z − 1 • Derivative Mode: s ↔ 2 z − 1 T s z + 1 • High pass filter • Has a pole at z = − 1 . Hence produces in partial fraction expansion, a term of the form

  39. Derivative Mode 9. • Integral Mode: 1 s ↔ T s z + 1 2 z − 1 • Derivative Mode: s ↔ 2 z − 1 T s z + 1 • High pass filter • Has a pole at z = − 1 . Hence produces in partial fraction expansion, a term of the form z z + 1 ↔ ( − 1) n

  40. Derivative Mode 9. • Integral Mode: 1 s ↔ T s z + 1 2 z − 1 • Derivative Mode: s ↔ 2 z − 1 T s z + 1 • High pass filter • Has a pole at z = − 1 . Hence produces in partial fraction expansion, a term of the form z z + 1 ↔ ( − 1) n • Results in wildly oscillating control effort. 9 Digital Control Kannan M. Moudgalya, Autumn 2007

  41. Derivative Mode - Other Approximations 10.

  42. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k )

  43. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z )

  44. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z ) 1 Y ( z ) = T s 1 − z − 1

  45. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z ) 1 z Y ( z ) = T s 1 − z − 1 = T s z − 1 U ( z )

  46. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z ) 1 z Y ( z ) = T s 1 − z − 1 = T s z − 1 U ( z ) 1 s ↔

  47. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z ) 1 z Y ( z ) = T s 1 − z − 1 = T s z − 1 U ( z ) 1 z s ↔ T s z − 1

  48. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z ) 1 z Y ( z ) = T s 1 − z − 1 = T s z − 1 U ( z ) 1 z s ↔ T s z − 1 Forward difference: y ( k ) = y ( k − 1) + T s u ( k − 1)

  49. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z ) 1 z Y ( z ) = T s 1 − z − 1 = T s z − 1 U ( z ) 1 z s ↔ T s z − 1 Forward difference: y ( k ) = y ( k − 1) + T s u ( k − 1) (1 − z − 1 ) Y ( z ) = T s z − 1 U ( z )

  50. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z ) 1 z Y ( z ) = T s 1 − z − 1 = T s z − 1 U ( z ) 1 z s ↔ T s z − 1 Forward difference: y ( k ) = y ( k − 1) + T s u ( k − 1) (1 − z − 1 ) Y ( z ) = T s z − 1 U ( z ) z − 1 Y ( z ) = T s 1 − z − 1 U ( z )

  51. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z ) 1 z Y ( z ) = T s 1 − z − 1 = T s z − 1 U ( z ) 1 z s ↔ T s z − 1 Forward difference: y ( k ) = y ( k − 1) + T s u ( k − 1) (1 − z − 1 ) Y ( z ) = T s z − 1 U ( z ) z − 1 T s Y ( z ) = T s 1 − z − 1 U ( z ) = z − 1 U ( z )

  52. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z ) 1 z Y ( z ) = T s 1 − z − 1 = T s z − 1 U ( z ) 1 z s ↔ T s z − 1 Forward difference: y ( k ) = y ( k − 1) + T s u ( k − 1) (1 − z − 1 ) Y ( z ) = T s z − 1 U ( z ) z − 1 T s Y ( z ) = T s 1 − z − 1 U ( z ) = z − 1 U ( z ) 1 s ↔

  53. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z ) 1 z Y ( z ) = T s 1 − z − 1 = T s z − 1 U ( z ) 1 z s ↔ T s z − 1 Forward difference: y ( k ) = y ( k − 1) + T s u ( k − 1) (1 − z − 1 ) Y ( z ) = T s z − 1 U ( z ) z − 1 T s Y ( z ) = T s 1 − z − 1 U ( z ) = z − 1 U ( z ) 1 T s s ↔ z − 1

  54. Derivative Mode - Other Approximations 10. Backward difference: y ( k ) = y ( k − 1) + T s u ( k ) (1 − z − 1 ) Y ( z ) = T s U ( z ) 1 z Y ( z ) = T s 1 − z − 1 = T s z − 1 U ( z ) 1 z s ↔ T s z − 1 Forward difference: y ( k ) = y ( k − 1) + T s u ( k − 1) (1 − z − 1 ) Y ( z ) = T s z − 1 U ( z ) z − 1 T s Y ( z ) = T s 1 − z − 1 U ( z ) = z − 1 U ( z ) 1 T s s ↔ z − 1 Both derivative modes are high pass,

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