1. s to Z-Domain Transfer Function
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1 1. s to Z-Domain Transfer Function
2 1. s to Z-Domain Transfer Function Discrete ZOH Signals
3 1. s to Z-Domain Transfer Function Discrete ZOH Signals 1. Get step response of continuous transfer function y s (t).
4 1. s to Z-Domain Transfer Function Discrete ZOH Signals 1. Get step response of continuous transfer function y s (t). 2. Discretize step response: y s (nt s ).
5 1. s to Z-Domain Transfer Function Discrete ZOH Signals 1. Z-transform the step response to obtain Y s (z). 1. Get step response of continuous transfer function y s (t). 2. Discretize step response: y s (nt s ).
6 1. s to Z-Domain Transfer Function Discrete ZOH Signals 1. Get step response of continuous transfer function y s (t). 2. Discretize step response: y s (nt s ). 1. Z-transform the step response to obtain Y s (z). 2. Divide the result from above by Z-transform of a step, namely, z/(z 1).
7 1. s to Z-Domain Transfer Function Discrete ZOH Signals 1. Get step response of continuous transfer function y s (t). 2. Discretize step response: y s (nt s ). 1. Z-transform the step response to obtain Y s (z). 2. Divide the result from above by Z-transform of a step, namely, z/(z 1). G a (s): function Laplace transfer
8 1. s to Z-Domain Transfer Function Discrete ZOH Signals 1. Get step response of continuous transfer function y s (t). 2. Discretize step response: y s (nt s ). 1. Z-transform the step response to obtain Y s (z). 2. Divide the result from above by Z-transform of a step, namely, z/(z 1). G a (s): function Laplace transfer G(z): Z-transfer function
9 1. s to Z-Domain Transfer Function Discrete ZOH Signals 1. Get step response of continuous transfer function y s (t). 2. Discretize step response: y s (nt s ). G(z) = z 1 z Z 1. Z-transform the step response to obtain Y s (z). 2. Divide the result from above by Z-transform of a step, namely, z/(z 1). G a (s): function Laplace transfer G(z): Z-transfer function [ L 1G ] a(s) s
10 1. s to Z-Domain Transfer Function Discrete ZOH Signals 1. Get step response of continuous transfer function y s (t). 2. Discretize step response: y s (nt s ). G(z) = z 1 z Z Step Response Equivalence 1. Z-transform the step response to obtain Y s (z). 2. Divide the result from above by Z-transform of a step, namely, z/(z 1). G a (s): function Laplace transfer G(z): Z-transfer function [ L 1G a(s) s ]
11 1. s to Z-Domain Transfer Function Discrete ZOH Signals 1. Get step response of continuous transfer function y s (t). 2. Discretize step response: y s (nt s ). 1. Z-transform the step response to obtain Y s (z). 2. Divide the result from above by Z-transform of a step, namely, z/(z 1). G a (s): function Laplace transfer G(z): Z-transfer function G(z) = z 1 [ z Z L 1G ] a(s) s Step Response Equivalence = ZOH Equivalence Digital Control 1 Kannan M. Moudgalya, Autumn 2007
12 2. Important Result from Differentiation
13 2. Important Result from Differentiation Recall 1(n)a n Differentiating w.r.t. a, z z a = n=0 a n z n,
14 2. Important Result from Differentiation Recall 1(n)a n Differentiating w.r.t. a, z z a = z (z a) 2 = a n z n, n=0 na n 1 z n n=0
15 2. Important Result from Differentiation Recall 1(n)a n Differentiating w.r.t. a, z z a = z (z a) 2 = na n 1 1(n) n=0 n=0 a n z n, na n 1 z n z (z a) 2
16 2. Important Result from Differentiation Recall 1(n)a n Differentiating w.r.t. a, z z a = z (z a) 2 = na n 1 1(n) n(n 1)a n 2 1(n) n=0 n=0 a n z n, na n 1 z n z (z a) 2 2z (z a) 3 Digital Control 2 Kannan M. Moudgalya, Autumn 2007
17 3. ZOH Equivalence of 1/s
18 3. ZOH Equivalence of 1/s The step response of 1/s is
19 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2.
20 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is,
21 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is, y s (t) = L 1 1 s 2
22 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is, y s (t) = L 1 1 s 2 = t
23 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is, y s (t) = L 1 1 s 2 = t Sampling it with a period of T s,
24 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is, y s (t) = L 1 1 s 2 = t Sampling it with a period of T s, y s (nt s ) =
25 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is, y s (t) = L 1 1 s 2 = t Sampling it with a period of T s, y s (nt s ) = nt s
26 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is, Taking Z-transforms Y s (z) = y s (t) = L 1 1 s 2 = t Sampling it with a period of T s, y s (nt s ) = nt s
27 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is, Taking Z-transforms Y s (z) = T sz (z 1) 2 y s (t) = L 1 1 s 2 = t Sampling it with a period of T s, y s (nt s ) = nt s
28 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is, y s (t) = L 1 1 s 2 = t Taking Z-transforms Y s (z) = T sz (z 1) 2 Divide by z/(z 1), Sampling it with a period of T s, y s (nt s ) = nt s
29 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is, y s (t) = L 1 1 s 2 = t Sampling it with a period of T s, Taking Z-transforms Y s (z) = T sz (z 1) 2 Divide by z/(z 1), to get the ZOH equivalent discrete domain transfer function y s (nt s ) = nt s
30 3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is, y s (t) = L 1 1 s 2 = t Sampling it with a period of T s, y s (nt s ) = nt s Taking Z-transforms Y s (z) = T sz (z 1) 2 Divide by z/(z 1), to get the ZOH equivalent discrete domain transfer function G(z) = T s z 1 Digital Control 3 Kannan M. Moudgalya, Autumn 2007
31 4. ZOH Equivalence of 1/s 2
32 4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is
33 4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is 1/s 3.
34 4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is 1/s 3. In time domain, it is,
35 4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is 1/s 3. In time domain, it is, y s (t) = L 1 1 s 3 =
36 4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is 1/s 3. In time domain, it is, y s (t) = L 1 1 s 3 = 1 2 t2.
37 4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is 1/s 3. In time domain, it is, y s (t) = L 1 1 s 3 = 1 2 t2. Sampling it with a period of T s, y s (nt s ) = 1 2 n2 T 2 s
38 4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is 1/s 3. In time domain, it is, Take Z-transform y s (t) = L 1 1 s 3 = 1 2 t2. Sampling it with a period of T s, y s (nt s ) = 1 2 n2 T 2 s
39 4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is 1/s 3. In time domain, it is, y s (t) = L 1 1 s 3 = 1 2 t2. Take Z-transform Y s (z) = T s 2 z(z + 1) 2(z 1) 3 Sampling it with a period of T s, y s (nt s ) = 1 2 n2 T 2 s
40 4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is 1/s 3. In time domain, it is, y s (t) = L 1 1 s 3 = 1 2 t2. Sampling it with a period of T s, y s (nt s ) = 1 2 n2 T 2 s Take Z-transform Y s (z) = T s 2 z(z + 1) 2(z 1) 3 Dividing by z/(z 1), we get G(z) = T s 2 (z + 1) 2(z 1) 2 Digital Control 4 Kannan M. Moudgalya, Autumn 2007
41 5. ZOH Equivalent First Order Transfer Function
42 5. ZOH Equivalent First Order Transfer Function Find the ZOH equivalent of K/(τ s + 1).
43 5. ZOH Equivalent First Order Transfer Function Find the ZOH equivalent of K/(τ s + 1). Y s (s) = 1 s K τ s + 1 = K [ 1 s 1 s + 1 τ ]
44 5. ZOH Equivalent First Order Transfer Function Find the ZOH equivalent of K/(τ s + 1). Y s (s) = 1 [ K 1 s τ s + 1 = K s 1 ] s + 1 ] τ y s (t) = K [1 e t/τ, t 0
45 5. ZOH Equivalent First Order Transfer Function Find the ZOH equivalent of K/(τ s + 1). Y s (s) = 1 [ K 1 s τ s + 1 = K s 1 ] s + 1 ] τ y s (t) = K [1 e t/τ, t 0 [ ] y s (nt s ) = K 1 e nt s/τ, n 0
46 5. ZOH Equivalent First Order Transfer Function Find the ZOH equivalent of K/(τ s + 1). Y s (s) = 1 [ K 1 s τ s + 1 = K s 1 ] s + 1 ] τ y s (t) = K [1 e t/τ, t 0 [ ] y s (nt s ) = K 1 e nt s/τ, n 0 Y s (z) = K [ z z 1 z z e T s/τ ]
47 5. ZOH Equivalent First Order Transfer Function Find the ZOH equivalent of K/(τ s + 1). Y s (s) = 1 [ K 1 s τ s + 1 = K s 1 ] s + 1 ] τ y s (t) = K [1 e t/τ, t 0 [ ] y s (nt s ) = K 1 e nt s/τ, n 0 Y s (z) = K [ z z 1 z z e T s/τ ] = Kz(1 e T s/τ ) (z 1)(z e T s/τ )
48 5. ZOH Equivalent First Order Transfer Function Find the ZOH equivalent of K/(τ s + 1). Y s (s) = 1 [ K 1 s τ s + 1 = K s 1 ] s + 1 ] τ y s (t) = K [1 e t/τ, t 0 [ ] y s (nt s ) = K 1 e nt s/τ, n 0 Y s (z) = K [ z z 1 z 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 ] = Kz(1 e T s/τ ) (z 1)(z e T s/τ ) Digital Control 5 Kannan M. Moudgalya, Autumn 2007
49 6. ZOH Equivalent First Order Transfer Function - Example
50 6. ZOH Equivalent First Order Transfer Function - Example Sample at T s = 0.5 and find ZOH equivalent trans. function of G a (s) = 10 5s + 1
51 6. ZOH Equivalent First Order Transfer Function - Example Sample at T s = 0.5 and find ZOH equivalent trans. function of G a (s) = 10 5s + 1 Scilab Code:
52 6. ZOH Equivalent First Order Transfer Function - Example Sample at T s = 0.5 and find ZOH equivalent trans. function of G a (s) = 10 5s + 1 Scilab Code: Ga = tf(10,[5 1]);
53 6. ZOH Equivalent First Order Transfer Function - Example Sample at T s = 0.5 and find ZOH equivalent trans. function of G a (s) = 10 5s + 1 Scilab Code: Ga = tf(10,[5 1]); G = ss2tf(dscr(ga,0.5));
54 6. ZOH Equivalent First Order Transfer Function - Example Sample at T s = 0.5 and find ZOH equivalent trans. function of G a (s) = 10 5s + 1 Scilab Code: Ga = tf(10,[5 1]); G = ss2tf(dscr(ga,0.5)); Scilab output is,
55 6. ZOH Equivalent First Order Transfer Function - Example Sample at T s = 0.5 and find ZOH equivalent trans. function of G a (s) = 10 5s + 1 Scilab Code: Ga = tf(10,[5 1]); G = ss2tf(dscr(ga,0.5)); Scilab output is, G(z) = z
56 6. ZOH Equivalent First Order Transfer Function - Example Sample at T s = 0.5 and find ZOH equivalent trans. function of G a (s) = 10 5s + 1 Scilab Code: Ga = tf(10,[5 1]); G = ss2tf(dscr(ga,0.5)); Scilab output is, G(z) = z = 10(1 e 0.1 ) z e 0.1
57 6. ZOH Equivalent First Order Transfer Function - Example Sample at T s = 0.5 and find ZOH equivalent trans. function of G a (s) = 10 5s + 1 Scilab Code: Ga = tf(10,[5 1]); G = ss2tf(dscr(ga,0.5)); Scilab output is, G(z) = z = 10(1 e 0.1 ) z e 0.1 In agreement with the formula in the previous slide Digital Control 6 Kannan M. Moudgalya, Autumn 2007
58 7. Discrete Integration
59 7. Discrete Integration u(n) u(k 1) u(k) n
60 7. Discrete Integration u(n) y(k) = blue shaded area u(k 1) u(k) n
61 7. Discrete Integration u(n) u(k 1) y(k) = blue shaded area + red shaded area u(k) n
62 7. Discrete Integration u(n) u(k 1) u(k) y(k) = blue shaded area + red shaded area y(k) = y(k 1) n
63 7. Discrete Integration u(n) u(k 1) u(k) y(k) = blue shaded area + red shaded area y(k) = y(k 1) + red shaded area n
64 7. Discrete Integration u(n) u(k 1) u(k) n y(k) = blue shaded area + red shaded area y(k) = y(k 1) + red shaded area y(k) = y(k 1) + T s 2 [u(k) + u(k 1)]
65 7. Discrete Integration u(n) u(k 1) u(k) n y(k) = blue shaded area + red shaded area y(k) = y(k 1) + red shaded area y(k) = y(k 1) + T s 2 [u(k) + u(k 1)] Take Z-transform:
66 7. Discrete Integration u(n) u(k 1) u(k) n y(k) = blue shaded area + red shaded area y(k) = y(k 1) + red shaded area y(k) = y(k 1) + T s 2 [u(k) + u(k 1)] Take Z-transform: Y (z) = z 1 Y (z) + T s 2 [ U(z) + z 1 U(z) ]
67 7. Discrete Integration u(n) u(k 1) u(k) n y(k) = blue shaded area + red shaded area y(k) = y(k 1) + red shaded area y(k) = y(k 1) + T s 2 [u(k) + u(k 1)] Take Z-transform: Y (z) = z 1 Y (z) + T s [ U(z) + z 1 U(z) ] 2 Bring all Y to left side:
68 7. Discrete Integration u(n) u(k 1) u(k) n y(k) = blue shaded area + red shaded area y(k) = y(k 1) + red shaded area y(k) = y(k 1) + T s 2 [u(k) + u(k 1)] 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 2 [ U(z) + z 1 U(z) ]
69 7. Discrete Integration u(n) u(k 1) u(k) n y(k) = blue shaded area + red shaded area y(k) = y(k 1) + red shaded area y(k) = y(k 1) + T s 2 [u(k) + u(k 1)] 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 2 [ U(z) + z 1 U(z) ] (1 z 1 )Y (z) = T s 2 (1 + z 1 )U(z) Digital Control 7 Kannan M. Moudgalya, Autumn 2007
70 8. Transfer Function for Discrete Integration
71 8. Transfer Function for Discrete Integration Recall from previous slide (1 z 1 )Y (z) = T s 2 (1 + z 1 )U(z)
72 8. Transfer Function for Discrete Integration Recall from previous slide (1 z 1 )Y (z) = T s 2 (1 + z 1 )U(z) Y (z) = T s z 1 1 z 1U(z)
73 8. Transfer Function for Discrete Integration Recall from previous slide (1 z 1 )Y (z) = T s 2 (1 + z 1 )U(z) Y (z) = T s 2 = T s z 1 1 z 1U(z) z + 1 z 1 U(z)
74 8. Transfer Function for Discrete Integration Recall from previous slide (1 z 1 )Y (z) = T s 2 (1 + z 1 )U(z) Y (z) = T s 2 = T s z 1 1 z 1U(z) z + 1 z 1 U(z) Integrator has a transfer function,
75 8. Transfer Function for Discrete Integration Recall from previous slide (1 z 1 )Y (z) = T s 2 (1 + z 1 )U(z) Y (z) = T s 2 = T s z 1 1 z 1U(z) z + 1 z 1 U(z) Integrator has a transfer function, G I (z) = T s 2 z + 1 z 1
76 8. Transfer Function for Discrete Integration Recall from previous slide (1 z 1 )Y (z) = T s 2 (1 + z 1 )U(z) Y (z) = T s 2 = T s z 1 1 z 1U(z) z + 1 z 1 U(z) Integrator has a transfer function, A low pass filter! G I (z) = T s 2 z + 1 z 1
77 8. Transfer Function for Discrete Integration Recall from previous slide (1 z 1 )Y (z) = T s 2 (1 + z 1 )U(z) Y (z) = T s 2 = T s z 1 1 z 1U(z) z + 1 z 1 U(z) Integrator has a transfer function, A low pass filter! G I (z) = T s 2 z + 1 z 1 Im(z) Re(z)
78 8. Transfer Function for Discrete Integration Recall from previous slide (1 z 1 )Y (z) = T s 2 (1 + z 1 )U(z) Y (z) = T s 2 = T s z 1 1 z 1U(z) z + 1 z 1 U(z) Integrator has a transfer function, A low pass filter! G I (z) = T s 2 z + 1 z 1 Im(z) Re(z) 1 s T s z z 1 Digital Control 8 Kannan M. Moudgalya, Autumn 2007
79 9. Derivative Mode
80 9. Derivative Mode Integral Mode: 1 s T s 2 z + 1 z 1
81 9. Derivative Mode Integral Mode: 1 s T s z z 1 Derivative Mode: s 2 z 1 T s z + 1
82 9. Derivative Mode Integral Mode: 1 s T s z z 1 Derivative Mode: s 2 z 1 T s z + 1 High pass filter
83 9. Derivative Mode Integral Mode: 1 s T s z z 1 Derivative Mode: s 2 z 1 T s z + 1 High pass filter Has a pole at z = 1.
84 9. Derivative Mode Integral Mode: 1 s T s z 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
85 9. Derivative Mode Integral Mode: 1 s T s z 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
86 9. Derivative Mode Integral Mode: 1 s T s z 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. Digital Control 9 Kannan M. Moudgalya, Autumn 2007
87 10. Derivative Mode - Other Approximations
88 10. Derivative Mode - Other Approximations Backward difference: y(k) = y(k 1) + T s u(k)
89 10. Derivative Mode - Other Approximations Backward difference: y(k) = y(k 1) + T s u(k) (1 z 1 )Y (z) = T s U(z)
90 10. Derivative Mode - Other Approximations 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
91 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z)
92 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z) 1 s
93 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z) 1 s T z s z 1
94 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z) 1 s T z s z 1 Forward difference: y(k) = y(k 1) + T s u(k 1)
95 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z) 1 s T z 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)
96 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z) 1 s T z 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 1U(z)
97 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z) 1 s T z 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) Y (z) = T s z 1 1 z 1U(z) = T s z 1 U(z)
98 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z) 1 s T z 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) Y (z) = T s 1 s z 1 1 z 1U(z) = T s z 1 U(z)
99 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z) 1 s T z 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 1U(z) = T s z 1 U(z) 1 s T s z 1
100 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z) 1 s T z 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 1U(z) = T s z 1 U(z) 1 s T s z 1 Both derivative modes are high pass,
101 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z) 1 s T z 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 1U(z) = T s z 1 U(z) 1 s T s z 1 Both derivative modes are high pass, no oscillations,
102 10. Derivative Mode - Other Approximations 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 = T z 1 s z 1 U(z) 1 s T z 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 1U(z) = T s z 1 U(z) 1 s T s z 1 Both derivative modes are high pass, no oscillations, same gains Digital Control 10 Kannan M. Moudgalya, Autumn 2007
103 11. PID Controller
104 11. PID Controller Proportional Mode: Most popular control mode.
105 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in
106 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset
107 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset and increased oscillations
108 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset and increased oscillations Integral Mode: Used to remove steady state offset.
109 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset and increased oscillations Integral Mode: Used to remove steady state offset. Increase in integral mode generally results in
110 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset and increased oscillations Integral Mode: Used to remove steady state offset. Increase in integral mode generally results in Zero steady state offset
111 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset and increased oscillations Integral Mode: Used to remove steady state offset. Increase in integral mode generally results in Zero steady state offset Increased oscillations
112 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset and increased oscillations Integral Mode: Used to remove steady state offset. Increase in integral mode generally results in Zero steady state offset Increased oscillations Derivative Mode: Mainly used for prediction purposes.
113 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset and increased oscillations Integral Mode: Used to remove steady state offset. Increase in integral mode generally results in Zero steady state offset Increased oscillations Derivative Mode: Mainly used for prediction purposes. Increase in derivative mode generally results in
114 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset and increased oscillations Integral Mode: Used to remove steady state offset. Increase in integral mode generally results in Zero steady state offset Increased oscillations Derivative Mode: Mainly used for prediction purposes. Increase in derivative mode generally results in Decreased oscillations and improved stability
115 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset and increased oscillations Integral Mode: Used to remove steady state offset. Increase in integral mode generally results in Zero steady state offset Increased oscillations Derivative Mode: Mainly used for prediction purposes. Increase in derivative mode generally results in Decreased oscillations and improved stability Sensitive to noise
116 11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset and increased oscillations Integral Mode: Used to remove steady state offset. Increase in integral mode generally results in Zero steady state offset Increased oscillations Derivative Mode: Mainly used for prediction purposes. Increase in derivative mode generally results in Decreased oscillations and improved stability Sensitive to noise The most popular controller in industry. Digital Control 11 Kannan M. Moudgalya, Autumn 2007
117 12. PID Controller - Basic Design
118 12. PID Controller - Basic Design Let input to controller by E(z)
119 12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z).
120 12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z). If gain is K,
121 12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z). If gain is K, τ i is integral time
122 12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z). If gain is K, τ i is integral time and τ d is derivative time,
123 12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z). If gain is K, τ i is integral time and τ d is derivative time, u(t) = K [e(t) + 1τi t ] de(t) e(t)dt + τ d dt 0
124 12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z). If gain is K, τ i is integral time and τ d is derivative time, u(t) = K [e(t) + 1τi t ] de(t) e(t)dt + τ d 0 dt U(s) = K(1 + 1 τ i s + τ ds)e(s)
125 12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z). If gain is K, τ i is integral time and τ d is derivative time, u(t) = K [e(t) + 1τi t ] de(t) e(t)dt + τ d 0 dt U(s) = K(1 + 1 τ i s + τ ds)e(s) U(s) = S c(s) R c (s) E(s)
126 12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z). If gain is K, τ i is integral time and τ d is derivative time, u(t) = K [e(t) + 1τi t ] de(t) e(t)dt + τ d 0 dt U(s) = K(1 + 1 τ i s + τ ds)e(s) U(s) = S c(s) R c (s) E(s) If integral mode is present, R c (0) = 0.
127 12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z). If gain is K, τ i is integral time and τ d is derivative time, u(t) = K [e(t) + 1τi t ] de(t) e(t)dt + τ d 0 dt U(s) = K(1 + 1 τ i s + τ ds)e(s) U(s) = S c(s) R c (s) E(s) If integral mode is present, R c (0) = 0. Filtered derivative mode:
128 12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z). If gain is K, τ i is integral time and τ d is derivative time, u(t) = K [e(t) + 1τi t ] de(t) e(t)dt + τ d 0 dt U(s) = K(1 + 1 τ i s + τ ds)e(s) U(s) = S c(s) R c (s) E(s) If integral mode is present, R c (0) = 0. Filtered derivative mode: u(t) = K ( τ i s + τ ds 1 + τ ds N ) e(t)
129 12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z). If gain is K, τ i is integral time and τ d is derivative time, u(t) = K [e(t) + 1τi t ] de(t) e(t)dt + τ d 0 dt U(s) = K(1 + 1 τ i s + τ ds)e(s) U(s) = S c(s) R c (s) E(s) If integral mode is present, R c (0) = 0. Filtered derivative mode: u(t) = K ( τ i s + τ ds 1 + τ ds N ) e(t) N is a large number, of the order of 100. Digital Control 12 Kannan M. Moudgalya, Autumn 2007
130 13. Reaction Curve Method - Ziegler Nichols Tuning
131 13. Reaction Curve Method - Ziegler Nichols Tuning Applicable only to stable systems
132 13. Reaction Curve Method - Ziegler Nichols Tuning Applicable only to stable systems Give a unit step input to a stable system and get
133 13. Reaction Curve Method - Ziegler Nichols Tuning Applicable only to stable systems Give a unit step input to a stable system and get 1. the time lag after which the system starts responding (L),
134 13. Reaction Curve Method - Ziegler Nichols Tuning Applicable only to stable systems Give a unit step input to a stable system and get 1. the time lag after which the system starts responding (L), 2. the steady state gain (K) and
135 13. Reaction Curve Method - Ziegler Nichols Tuning Applicable only to stable systems Give a unit step input to a stable system and get 1. the time lag after which the system starts responding (L), 2. the steady state gain (K) and 3. the time the output takes to reach the steady state, after it starts responding (τ )
136 13. Reaction Curve Method - Ziegler Nichols Tuning Applicable only to stable systems Give a unit step input to a stable system and get 1. the time lag after which the system starts responding (L), 2. the steady state gain (K) and 3. the time the output takes to reach the steady state, after it starts responding (τ ) K R = K/τ L τ Digital Control 13 Kannan M. Moudgalya, Autumn 2007
137 14. Reaction Curve Method - Ziegler Nichols Tuning
138 14. Reaction Curve Method - Ziegler Nichols Tuning K R = K/τ L τ
139 14. Reaction Curve Method - Ziegler Nichols Tuning K R = K/τ L τ Let the slope of the response be calculated as R = K τ.
140 14. Reaction Curve Method - Ziegler Nichols Tuning K R = K/τ L τ Let the slope of the response be calculated as R = K τ. Then the PID settings are given below:
141 14. Reaction Curve Method - Ziegler Nichols Tuning K R = K/τ L τ Let the slope of the response be calculated as R = K τ. Then the PID settings are given below: K p τ i τ d P 1/RL PI 0.9/RL 3L PID 1.2/RL 2L 0.5L
142 14. Reaction Curve Method - Ziegler Nichols Tuning K R = K/τ L τ Let the slope of the response be calculated as R = K τ. Then the PID settings are given below: K p τ i τ d P 1/RL PI 0.9/RL 3L PID 1.2/RL 2L 0.5L Consistent units should be used Digital Control 14 Kannan M. Moudgalya, Autumn 2007
143 15. Stability Method - Ziegler Nichols Tuning
144 15. Stability Method - Ziegler Nichols Tuning Another way of finding the PID tuning parameters is as follows.
145 15. Stability Method - Ziegler Nichols Tuning Another way of finding the PID tuning parameters is as follows. Close the loop with a proportional controller
146 15. Stability Method - Ziegler Nichols Tuning Another way of finding the PID tuning parameters is as follows. Close the loop with a proportional controller Gain of controller is increased until the closed loop system becomes unstable
147 15. Stability Method - Ziegler Nichols Tuning Another way of finding the PID tuning parameters is as follows. Close the loop with a proportional controller Gain of controller is increased until the closed loop system becomes unstable At the verge of instability, note down the gain of the controller (K u ) and the period of oscillation (P u )
148 15. Stability Method - Ziegler Nichols Tuning Another way of finding the PID tuning parameters is as follows. Close the loop with a proportional controller Gain of controller is increased until the closed loop system becomes unstable At the verge of instability, note down the gain of the controller (K u ) and the period of oscillation (P u ) PID settings are given below:
149 15. Stability Method - Ziegler Nichols Tuning Another way of finding the PID tuning parameters is as follows. Close the loop with a proportional controller Gain of controller is increased until the closed loop system becomes unstable At the verge of instability, note down the gain of the controller (K u ) and the period of oscillation (P u ) PID settings are given below: K p τ i τ d P 0.5K u PI 0.45K u P u /1.2 PID 0.6K u P u /2 P u /8
150 15. Stability Method - Ziegler Nichols Tuning Another way of finding the PID tuning parameters is as follows. Close the loop with a proportional controller Gain of controller is increased until the closed loop system becomes unstable At the verge of instability, note down the gain of the controller (K u ) and the period of oscillation (P u ) PID settings are given below: K p τ i τ d P 0.5K u PI 0.45K u P u /1.2 PID 0.6K u P u /2 P u /8 Consistent units should be used Digital Control 15 Kannan M. Moudgalya, Autumn 2007
151 16. Design Procedure
152 16. Design Procedure A common procedure to design discrete PID controller:
153 16. Design Procedure A common procedure to design discrete PID controller: Tune continuous PID controller by any popular technique
154 16. Design Procedure A common procedure to design discrete PID controller: Tune continuous PID controller by any popular technique Get continuous PID settings
155 16. Design Procedure A common procedure to design discrete PID controller: Tune continuous PID controller by any popular technique Get continuous PID settings Discretize using the method discussed now or the ZOH equivalent method discussed earlier
156 16. Design Procedure A common procedure to design discrete PID controller: Tune continuous PID controller by any popular technique Get continuous PID settings Discretize using the method discussed now or the ZOH equivalent method discussed earlier Direct digital design techniques Digital Control 16 Kannan M. Moudgalya, Autumn 2007
157 17. 2-DOF Controller
158 17. 2-DOF Controller r T c u y R c G = B A S c R c
159 17. 2-DOF Controller r T c u y R c G = B A S c R c u = T c R c r S c R c y
160 17. 2-DOF Controller r T c R c u G = B A y S c R c u = T c r S c y R c R c It is easy to arrive at the following relation between r and y.
161 17. 2-DOF Controller r T c R c u G = B A y S c R c u = T c r S c y R c R c It is easy to arrive at the following relation between r and y. y = T c B/A r R c 1 + BS c /AR c
162 17. 2-DOF Controller r T c R c u G = B A y S c R c u = T c r S c y R c R c It is easy to arrive at the following relation between r and y. y = T c B/A r = R c 1 + BS c /AR c BT c AR c + BS c r
163 17. 2-DOF Controller r T c R c u G = B A y S c R c u = T c r S c y R c R c It is easy to arrive at the following relation between r and y. y = T c B/A BT c r = r R c 1 + BS c /AR c AR c + BS c Error e, given by r y is given by
164 17. 2-DOF Controller r T c R c u G = B A y S c R c u = T c r S c y R c R c It is easy to arrive at the following relation between r and y. y = T c B/A r = R c 1 + BS c /AR c Error e, given by r y is given by e = ( 1 ) BT c r AR c + BS c BT c AR c + BS c r
165 17. 2-DOF Controller r T c R c u G = B A y S c R c u = T c r S c y R c R c It is easy to arrive at the following relation between r and y. y = T c B/A r = R c 1 + BS c /AR c Error e, given by r y is given by e = ( 1 BT c AR c + BS c ) BT c AR c + BS c r r = AR c + BS c BT c r AR c + BS c Digital Control 17 Kannan M. Moudgalya, Autumn 2007
166 18. Offset-Free Tracking of Steps with Integral
167 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z)
168 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z) lim e(n) = n
169 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z) lim e(n) = lim n z 1 z 1 A(z)R c (z) + B(z)S c (z) B(z)T c (z) z A(z)R c (z) + B(z)S c (z) z z 1
170 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z) lim e(n) = lim n z 1 z 1 A(z)R c (z) + B(z)S c (z) B(z)T c (z) z A(z)R c (z) + B(z)S c (z) Because the controller has an integral action, z z 1
171 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z) lim e(n) = lim n z 1 z 1 A(z)R c (z) + B(z)S c (z) B(z)T c (z) z A(z)R c (z) + B(z)S c (z) Because the controller has an integral action, R c (1) = 0: z z 1
172 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z) lim e(n) = lim n z 1 z 1 A(z)R c (z) + B(z)S c (z) B(z)T c (z) z A(z)R c (z) + B(z)S c (z) Because the controller has an integral action, R c (1) = 0: e( ) = z z 1
173 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z) lim e(n) = lim n z 1 z 1 A(z)R c (z) + B(z)S c (z) B(z)T c (z) z A(z)R c (z) + B(z)S c (z) Because the controller has an integral action, R c (1) = 0: e( ) = S c(z) T c (z) S c (z) z=1 z z 1
174 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z) lim e(n) = lim n z 1 z 1 A(z)R c (z) + B(z)S c (z) B(z)T c (z) z A(z)R c (z) + B(z)S c (z) Because the controller has an integral action, R c (1) = 0: e( ) = S c(z) T c (z) S c (z) = S c(1) T c (1) z=1 S c (1) z z 1
175 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z) lim e(n) = lim n z 1 z 1 A(z)R c (z) + B(z)S c (z) B(z)T c (z) z A(z)R c (z) + B(z)S c (z) Because the controller has an integral action, R c (1) = 0: e( ) = S c(z) T c (z) S c (z) = S c(1) T c (1) z=1 S c (1) This condition can be satisfied if one of the following is met: z z 1
176 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z) lim e(n) = lim n z 1 z 1 A(z)R c (z) + B(z)S c (z) B(z)T c (z) z A(z)R c (z) + B(z)S c (z) Because the controller has an integral action, R c (1) = 0: e( ) = S c(z) T c (z) S c (z) = S c(1) T c (1) z=1 S c (1) This condition can be satisfied if one of the following is met: T c = S c z z 1
177 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z) lim e(n) = lim n z 1 z 1 A(z)R c (z) + B(z)S c (z) B(z)T c (z) z A(z)R c (z) + B(z)S c (z) Because the controller has an integral action, R c (1) = 0: e( ) = S c(z) T c (z) S c (z) = S c(1) T c (1) z=1 S c (1) This condition can be satisfied if one of the following is met: T c = S c T c = S c (1) z z 1
178 18. Offset-Free Tracking of Steps with Integral E(z) = A(z)R c(z) + B(z)S c (z) B(z)T c (z) R(z) A(z)R c (z) + B(z)S c (z) lim e(n) = lim n z 1 z 1 A(z)R c (z) + B(z)S c (z) B(z)T c (z) z A(z)R c (z) + B(z)S c (z) Because the controller has an integral action, R c (1) = 0: e( ) = S c(z) T c (z) S c (z) = S c(1) T c (1) z=1 S c (1) This condition can be satisfied if one of the following is met: T c = S c T c = S c (1) T c (1) = S c (1) z z 1 Digital Control 18 Kannan M. Moudgalya, Autumn 2007
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