1. s to Z-Domain Transfer Function

1. s to Z-Domain Transfer Function

1. s to Z-Domain Transfer Function Discrete ZOH Signals

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

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. 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 ).

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).

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

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

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

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 ]

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

2. Important Result from Differentiation

2. Important Result from Differentiation Recall 1(n)a n Differentiating w.r.t. a, z z a = n=0 a n z n,

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

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

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

3. ZOH Equivalence of 1/s

3. ZOH Equivalence of 1/s The step response of 1/s is

3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2.

3. ZOH Equivalence of 1/s The step response of 1/s is 1/s 2. In time domain, it is,

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

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

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,

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 ) =

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

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

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

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

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

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

4. ZOH Equivalence of 1/s 2

4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is

4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is 1/s 3.

4. ZOH Equivalence of 1/s 2 The step response of 1/s 2 is 1/s 3. In time domain, it is,

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 =

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.

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

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

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

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

5. ZOH Equivalent First Order Transfer Function

5. ZOH Equivalent First Order Transfer Function Find the ZOH equivalent of K/(τ s + 1).

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 ]

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

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

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 ]

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 )

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

6. ZOH Equivalent First Order Transfer Function - Example

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

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:

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]);

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));

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,

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) = 0.9546 z 0.9048

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) = 0.9546 z 0.9048 = 10(1 e 0.1 ) z e 0.1

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) = 0.9546 z 0.9048 = 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

7. Discrete Integration

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

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

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

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

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

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)]

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:

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) ]

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:

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) ]

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

8. Transfer Function for Discrete Integration

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

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 1 + z 1 1 z 1U(z)

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 2 1 + z 1 1 z 1U(z) z + 1 z 1 U(z)

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 2 1 + z 1 1 z 1U(z) z + 1 z 1 U(z) Integrator has a transfer function,

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 2 1 + 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

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 2 1 + 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

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 2 1 + 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)

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 2 1 + 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 + 1 2 z 1 Digital Control 8 Kannan M. Moudgalya, Autumn 2007

9. Derivative Mode

9. Derivative Mode Integral Mode: 1 s T s 2 z + 1 z 1

9. Derivative Mode Integral Mode: 1 s T s z + 1 2 z 1 Derivative Mode: s 2 z 1 T s z + 1

9. Derivative Mode Integral Mode: 1 s T s z + 1 2 z 1 Derivative Mode: s 2 z 1 T s z + 1 High pass filter

9. Derivative Mode 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.

9. Derivative Mode 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

9. Derivative Mode 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

9. Derivative Mode 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. Digital Control 9 Kannan M. Moudgalya, Autumn 2007

10. Derivative Mode - Other Approximations

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

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)

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

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)

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

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

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)

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)

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)

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)

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)

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

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,

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,

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

11. PID Controller

11. PID Controller Proportional Mode: Most popular control mode.

11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in

11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset

11. PID Controller Proportional Mode: Most popular control mode. Increase in proportional mode generally results in Decreased steady state offset and increased oscillations

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.

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

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

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

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.

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

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

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

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

12. PID Controller - Basic Design

12. PID Controller - Basic Design Let input to controller by E(z)

12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z).

12. PID Controller - Basic Design Let input to controller by E(z) and output from it be U(z). If gain is K,

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

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,

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

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)

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)

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.

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:

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 ( 1 + 1 τ i s + τ ds 1 + τ ds N ) e(t)

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 ( 1 + 1 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

13. Reaction Curve Method - Ziegler Nichols Tuning

13. Reaction Curve Method - Ziegler Nichols Tuning Applicable only to stable systems

13. Reaction Curve Method - Ziegler Nichols Tuning Applicable only to stable systems Give a unit step input to a stable system and get

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),

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

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 )

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

14. Reaction Curve Method - Ziegler Nichols Tuning

14. Reaction Curve Method - Ziegler Nichols Tuning K R = K/τ L τ

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.

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:

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

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

15. Stability Method - Ziegler Nichols Tuning

15. Stability Method - Ziegler Nichols Tuning Another way of finding the PID tuning parameters is as follows.

15. Stability Method - Ziegler Nichols Tuning Another way of finding the PID tuning parameters is as follows. Close the loop with a proportional controller

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

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 )

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:

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

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

16. Design Procedure

16. Design Procedure A common procedure to design discrete PID controller:

16. Design Procedure A common procedure to design discrete PID controller: Tune continuous PID controller by any popular technique

16. Design Procedure A common procedure to design discrete PID controller: Tune continuous PID controller by any popular technique Get continuous PID settings

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

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

17. 2-DOF Controller

17. 2-DOF Controller r T c u y R c G = B A S c R c

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

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.

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

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

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

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

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

18. Offset-Free Tracking of Steps with Integral

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)

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

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

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

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

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

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

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

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

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

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

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