1. Cancellation of Left Half Plane Poles with Zeros
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1 1. Cancellation of Left Half Plane Poles with Zeros
2 1. Cancellation of Left Half Plane Poles with Zeros Control of DC Motor (Astrom & Wittenmark) Velocity of motor shaft - x 1, position - x 2.
3 1. Cancellation of Left Half Plane Poles with Zeros Control of DC Motor (Astrom & Wittenmark) Velocity of motor shaft - x 1, position - x 2. [ ] [ ] [ ] [ ] d x1 1 0 x1 1 = + u dt x x 2 0 y = [ 0 1 ] [ ] x 1 x 2
4 1. Cancellation of Left Half Plane Poles with Zeros Control of DC Motor (Astrom & Wittenmark) Velocity of motor shaft - x 1, position - x 2. [ ] [ ] [ ] [ ] d x1 1 0 x1 1 = + u dt x x 2 0 y = [ 0 1 ] [ ] x 1 x 2 Step response requirements: rise time of 3 seconds
5 1. Cancellation of Left Half Plane Poles with Zeros Control of DC Motor (Astrom & Wittenmark) Velocity of motor shaft - x 1, position - x 2. [ ] [ ] [ ] [ ] d x1 1 0 x1 1 = + u dt x x 2 0 y = [ 0 1 ] [ ] x 1 x 2 Step response requirements: rise time of 3 seconds overshoot of not more than 0.05
6 1. Cancellation of Left Half Plane Poles with Zeros Control of DC Motor (Astrom & Wittenmark) Velocity of motor shaft - x 1, position - x 2. [ ] [ ] [ ] [ ] d x1 1 0 x1 1 = + u dt x x 2 0 y = [ 0 1 ] [ ] x 1 x 2 Step response requirements: rise time of 3 seconds overshoot of not more than 0.05 Choose T s = 0.25s
7 1. Cancellation of Left Half Plane Poles with Zeros Control of DC Motor (Astrom & Wittenmark) Velocity of motor shaft - x 1, position - x 2. [ ] [ ] [ ] [ ] d x1 1 0 x1 1 = + u dt x x 2 0 y = [ 0 1 ] [ ] x 1 x 2 Step response requirements: rise time of 3 seconds overshoot of not more than 0.05 Choose T s = 0.25s G(z) = z z z z 2 Digital Control 1 Kannan M. Moudgalya, Autumn 2007
8 2. Cancellation of Left Half Plane Poles with Zeros
9 2. Cancellation of Left Half Plane Poles with Zeros We get the following factorizations: G(z) = z z z z 2
10 2. Cancellation of Left Half Plane Poles with Zeros We get the following factorizations: G(z) = z z z z 2 A b = 1 z 1 A g = z 1
11 2. Cancellation of Left Half Plane Poles with Zeros We get the following factorizations: G(z) = z z z z 2 A b = 1 z 1 A g = z 1 B b = 1 B g = = ( z 1 )
12 2. Cancellation of Left Half Plane Poles with Zeros We get the following factorizations: G(z) = z z z z 2 A b = 1 z 1 A g = z 1 B b = 1 B g = = ( z 1 ) φ cl = z z 2
13 2. Cancellation of Left Half Plane Poles with Zeros We get the following factorizations: G(z) = z z z z 2 A b = 1 z 1 A g = z 1 B b = 1 B g = = ( z 1 ) φ cl = z z 2 R 1 = z 1 S 1 = Digital Control 2 Kannan M. Moudgalya, Autumn 2007
14 3. Performance of Pole Placement Controller On DC Motor
15 3. Performance of Pole Placement Controller On DC Motor u(n) Time offset: Time
16 3. Performance of Pole Placement Controller On DC Motor u(n) Time offset: 0 Oscillations Time
17 3. Performance of Pole Placement Controller On DC Motor u(n) Time offset: Time Oscillations - due to cancellation of pole-zero in Left Half Plane Digital Control 3 Kannan M. Moudgalya, Autumn 2007
18 4. Why is Control Effort Oscillatory?
19 4. Why is Control Effort Oscillatory? Recall controller: R c (z)u(n) = γt c (z)r(n) S c (z)y(n)
20 4. Why is Control Effort Oscillatory? Recall controller: R c (z)u(n) = γt c (z)r(n) S c (z)y(n) u(n) = γ T c(z) R c (z) r(n) S c(z) R c (z) y(n)
21 4. Why is Control Effort Oscillatory? Recall controller: Recall, R c (z)u(n) = γt c (z)r(n) S c (z)y(n) u(n) = γ T c(z) R c (z) r(n) S c(z) R c (z) y(n) R c = B g R 1, T c = A g T 1, S c = A g S 1
22 4. Why is Control Effort Oscillatory? Recall controller: Recall, R c (z)u(n) = γt c (z)r(n) S c (z)y(n) u(n) = γ T c(z) R c (z) r(n) S c(z) R c (z) y(n) R c = B g R 1, T c = A g T 1, S c = A g S 1 Substituting, the control law becomes u(n) = γ Ag T 1 B g R 1 r(n) Ag S 1 B g R 1 y(n)
23 4. Why is Control Effort Oscillatory? Recall controller: Recall, R c (z)u(n) = γt c (z)r(n) S c (z)y(n) u(n) = γ T c(z) R c (z) r(n) S c(z) R c (z) y(n) R c = B g R 1, T c = A g T 1, S c = A g S 1 Substituting, the control law becomes u(n) = γ Ag T 1 B g R 1 r(n) Ag S 1 B g R 1 y(n) If B g has root with negative real part (e.g ), as in the antenna control problem, it shows up as a pole of the controller. Digital Control 4 Kannan M. Moudgalya, Autumn 2007
24 5. Recall: Poles on Negative Real Axis
25 5. Recall: Poles on Negative Real Axis Im(z) n n n Re(z) n n n
26 5. Recall: Poles on Negative Real Axis Im(z) n n n Re(z) n n n Poles on negative real axis
27 5. Recall: Poles on Negative Real Axis Im(z) n n n Re(z) n n n Poles on negative real axis produce oscillations! Digital Control 5 Kannan M. Moudgalya, Autumn 2007
28 6. Redefining Good and Bad Polynomials
29 6. Redefining Good and Bad Polynomials If B with negative real part is defined as good, it shows up as controller pole
30 6. Redefining Good and Bad Polynomials If B with negative real part is defined as good, it shows up as controller pole Whatever is cancelled, shows up in the controller expression
31 6. Redefining Good and Bad Polynomials If B with negative real part is defined as good, it shows up as controller pole Whatever is cancelled, shows up in the controller expression Preferably, bad factors should lie outside shaded region
32 6. Redefining Good and Bad Polynomials If B with negative real part is defined as good, it shows up as controller pole Whatever is cancelled, shows up in the controller expression Preferably, bad factors should lie outside shaded region For ease of calculation, negative roots are taken as bad
33 6. Redefining Good and Bad Polynomials If B with negative real part is defined as good, it shows up as controller pole Whatever is cancelled, shows up in the controller expression Preferably, bad factors should lie outside shaded region For ease of calculation, negative roots are taken as bad - so don t get cancelled
34 6. Redefining Good and Bad Polynomials If B with negative real part is defined as good, it shows up as controller pole Whatever is cancelled, shows up in the controller expression Preferably, bad factors should lie outside shaded region For ease of calculation, negative roots are taken as bad - so don t get cancelled A b, A g, same as before
35 6. Redefining Good and Bad Polynomials If B with negative real part is defined as good, it shows up as controller pole Whatever is cancelled, shows up in the controller expression Preferably, bad factors should lie outside shaded region For ease of calculation, negative roots are taken as bad - so don t get cancelled A b, A g, same as before B b = z 1 B g =
36 6. Redefining Good and Bad Polynomials If B with negative real part is defined as good, it shows up as controller pole Whatever is cancelled, shows up in the controller expression Preferably, bad factors should lie outside shaded region For ease of calculation, negative roots are taken as bad - so don t get cancelled A b, A g, same as before B b = z 1 B g = φ cl = z z 2
37 6. Redefining Good and Bad Polynomials If B with negative real part is defined as good, it shows up as controller pole Whatever is cancelled, shows up in the controller expression Preferably, bad factors should lie outside shaded region For ease of calculation, negative roots are taken as bad - so don t get cancelled A b, A g, same as before B b = z 1 B g = φ cl = z z 2 R 1 = z 1, S 1 = Digital Control 6 Kannan M. Moudgalya, Autumn 2007
38 7. Redefining Good and Bad Polynomials
39 7. Redefining Good and Bad Polynomials y(n) t (s) u(n) t (s)
40 7. Redefining Good and Bad Polynomials y(n) u(n) t (s) t (s) Rise time can be improved by shortening the specified rise time from 3s to 2s Digital Control 7 Kannan M. Moudgalya, Autumn 2007
41 8. 2-DOF PP Control of IBM Lotus Domino Server
42 8. 2-DOF PP Control of IBM Lotus Domino Server T.F. between Max users and no. of remote procedure calls: G(z) = 0.47z z 1
43 8. 2-DOF PP Control of IBM Lotus Domino Server T.F. between Max users and no. of remote procedure calls: G(z) = 0.47z z 1 Track step inputs with Rise time 10, overshoot ε 0.1.
44 8. 2-DOF PP Control of IBM Lotus Domino Server T.F. between Max users and no. of remote procedure calls: G(z) = 0.47z z 1 Track step inputs with Rise time 10, overshoot ε 0.1. A g = z 1 A b = 1 B g = 0.47 B b = 1 k = 1
45 8. 2-DOF PP Control of IBM Lotus Domino Server T.F. between Max users and no. of remote procedure calls: G(z) = 0.47z z 1 Track step inputs with Rise time 10, overshoot ε 0.1. A g = z 1 A b = 1 B g = 0.47 B b = 1 k = 1 T s = 1 N r = 10 ω = ρ = φ cl = z z 2 A b R 1 + z k B b S 1 = φ cl Digital Control 8 Kannan M. Moudgalya, Autumn 2007
46 9. 2-DOF PP Control of an IBM Lotus Domino Server
47 9. 2-DOF PP Control of an IBM Lotus Domino Server (1 z 1 )R 1 + z 1 S 1 = z z 2 The solution is given by, R 1 = z 1 S 1 = and the 2-DOF pole placement controller is given by, R c = z z 2 S c = z 1 T c = z 1 γ = Digital Control 9 Kannan M. Moudgalya, Autumn 2007
48 10. ibm pp.sce
49 10. ibm pp.sce 1 / / U p d a t e d ( ) 2 / / / / C o n t r o l o f I B M l o t u s d o m i n o s e r v e r 4 / / T r a n s f e r f u n c t i o n 5 6 B = ; A = [ ]; k = 1 ; 7 [ zk, dzk ] = zpowk ( k ) ; 8 9 / / T r a n s i e n t s p e c i f i c a t i o n s 10 r i s e = 10; e p s i l o n = ; Ts = 1 ; 11 p hi = d e s i r e d ( Ts, r i s e, e p s i l o n ) ; / / C o n t r o l l e r d e s i g n 14 Delta = [ 1 1]; / / i n t e r n a l m o d e l o f s t e p u s e d 15 [ Rc, Sc, Tc,gamm] = pp im (B, A, k, phi, Delta ) ; 16 Digital Control 10 Kannan M. Moudgalya, Autumn 2007
50 17 / / S i m u l a t i o n p a r a m e t e r s f o r s t b d i s c. c o s 18 s t = 1 ; / / d e s i r e d c h a n g e 19 t i n i t = 0 ; / / s i m u l a t i o n s t a r t t i m e 20 t f i n a l = 40; / / s i m u l a t i o n e n d t i m e 21 C = 0 ; D = 1 ; N var = 0 ; [ Tcp1, Tcp2 ] = c o s f i l i p (Tc, 1 ) ; / / T c / 1 24 [ Rcp1, Rcp2 ] = c o s f i l i p (1, Rc ) ; / / 1 / R c 25 [ Scp1, Scp2 ] = c o s f i l i p ( Sc, 1 ) ; / / S c / 1 26 [ Bp, Ap ] = c o s f i l i p (B,A ) ; / / B / A 27 [ zkp1, zkp2 ] = c o s f i l i p ( zk, 1 ) ; / / z k / 1 28 [ Cp, Dp ] = c o s f i l i p (C,D) ; / / C / D Digital Control 11 Kannan M. Moudgalya, Autumn 2007
51 11. stb disc.cos
52 11. stb disc.cos stb_disc u1 random generator num(z) den(z) Filter 1 Step switch 1 num(z) den(z) +! Feedforward Filter num(z) den(z) Filter 2 num(z) den(z) Plant num(z) den(z) Delay + + y1 Ramp num(z) den(z) Feedback Filter Digital Control 12 Kannan M. Moudgalya, Autumn 2007
53 12. Saturation Block
54 12. Saturation Block v r γt c 1 u u sat R c z kb A y S c Digital Control 13 Kannan M. Moudgalya, Autumn 2007
55 13. Saturation Block Simulation Results u(n) 0 u(n) n n u sat (n) 0 y(n) n n Digital Control 14 Kannan M. Moudgalya, Autumn 2007
56 14. AWC - When Within Limits
57 14. AWC - When Within Limits v r 1 γp T c F u u sat z kb A y E F P S c
58 14. AWC - When Within Limits v r 1 γp T c F u u sat z kb A y E F P S c When within limits,
59 14. AWC - When Within Limits v r 1 γp T c F u u sat z kb A y E F P S c When within limits, v r γp T c 1 F E z kb A y P S c
60 14. AWC - When Within Limits v r 1 γp T c F u u sat z kb A y E F P S c When within limits, v r γp T c 1 F E z kb A y P S c Choose F E = P R c Digital Control 15 Kannan M. Moudgalya, Autumn 2007
61 15. AWC
62 15. AWC v r γt c 1 u u sat R c z kb A y S c Digital Control 16 Kannan M. Moudgalya, Autumn 2007
63 16. AWC - When Limits are Exceeded
64 16. AWC - When Limits are Exceeded δ r 1 γp T c F u u sat z kb A y E F P S c
65 16. AWC - When Limits are Exceeded δ r 1 γp T c F u u sat z kb A y E F P S c δ r γp T c 1 F z kb A y Ez k A B P S c Digital Control 17 Kannan M. Moudgalya, Autumn 2007
66 17. Performance of AWC
67 17. Performance of AWC δ r γp T c 1 F z kb A y Ez k A B P S c T δy =
68 17. Performance of AWC δ r γp T c 1 F z kb A y Ez k A B P S c T δy = 1 + z k B A z k B A ( P Sc Ez k A B) 1 F
69 17. Performance of AWC δ r γp T c 1 F z kb A y Ez k A B P S c T δy = = 1 + z k B A z k B A ( P Sc Ez k A z k BF B) 1 F F A + (z k BP S c EA)
70 17. Performance of AWC δ r γp T c 1 F z kb A y Ez k A B P S c T δy = = 1 + z k B A z k B A ( P Sc Ez k A z k BF B) 1 F F A + (z k BP S c EA) Denom: (F E)A + z k BP S c.
71 17. Performance of AWC δ r γp T c 1 F z kb A y Ez k A B P S c T δy = = 1 + z k B A z k B A ( P Sc Ez k A z k BF B) 1 F F A + (z k BP S c EA) Denom: (F E)A + z k BP S c. Using F E = P R c,
72 17. Performance of AWC δ r γp T c 1 F z kb A y Ez k A B P S c T δy = = 1 + z k B A z k B A ( P Sc Ez k A z k BF B) 1 F F A + (z k BP S c EA) Denom: (F E)A + z k BP S c. Using F E = P R c, z k BF T δy = P R c A + z k BP S c Digital Control 18 Kannan M. Moudgalya, Autumn 2007
73 18. Performance of AWC
74 18. Performance of AWC T δy = z k BF P R c A + z k BP S c If we choose F = AR c + z k BS c
75 18. Performance of AWC z k BF T δy = P R c A + z k BP S c If we choose F = AR c + z k BS c = φ cl A g B g
76 18. Performance of AWC z k BF T δy = P R c A + z k BP S c If we choose F = AR c + z k BS c = φ cl A g B g T δy = z kb P
77 18. Performance of AWC z k BF T δy = P R c A + z k BP S c If we choose F = AR c + z k BS c = φ cl A g B g T δy = z kb P If P well behaved,
78 18. Performance of AWC z k BF T δy = P R c A + z k BP S c If we choose F = AR c + z k BS c = φ cl A g B g T δy = z kb P If P well behaved, the effect of T δy will diminish with time.
79 18. Performance of AWC z k BF T δy = P R c A + z k BP S c If we choose F = AR c + z k BS c = φ cl A g B g T δy = z kb P If P well behaved, the effect of T δy will diminish with time. A popular choice,
80 18. Performance of AWC z k BF T δy = P R c A + z k BP S c If we choose F = AR c + z k BS c = φ cl A g B g T δy = z kb P If P well behaved, the effect of T δy will diminish with time. A popular choice, when A is Hurwitz is
81 18. Performance of AWC z k BF T δy = P R c A + z k BP S c If we choose F = AR c + z k BS c = φ cl A g B g T δy = z kb P If P well behaved, the effect of T δy will diminish with time. A popular choice, when A is Hurwitz is P = A Digital Control 19 Kannan M. Moudgalya, Autumn 2007
82 19. Saturation Block Simulation Results u(n) 0 u(n) n n u sat (n) 0 y(n) n n Digital Control 20 Kannan M. Moudgalya, Autumn 2007
CL Digital Control
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