Pole Placement Design

Transcription

1 Department of Automatic Control LTH, Lund University

2 1 Introduction 2 Simple Examples 3 Polynomial Design 4 State Space Design 5 Robustness and Design Rules 6 Model Reduction 7 Oscillatory Systems 8 Summary Theme: Be aware where you place them!

3 Introduction A simple idea Strong impact on development of control theory The only constraint is reachability and observability The robustness debate Classic control vs State feedback Easy to apply for simple systems Polynomial equations notoriously badly conditioned: z n = 0 OK for low order systems, use matrix formulation for high order systems How to choose closed loop poles - The Million $ question How do the closed loop poles influence performance How do the closed loop poles influence robustness A bit of history - Mats Lilja s PhD thesis TFRT 1031 (1989) Insight into model reduction

4 1 Introduction 2 Simple Examples 3 Polynomial Design 4 State Space Design 5 Robustness and Design Rules 6 Model Reduction 7 Oscillatory Systems 8 Summary Theme: Be aware where you place them!

5 Control of First Order Systems State: variables required to characterize storage of mass, momentum and energy Many systems are approximately of first order The key is that the storage of mass, momentum and energy can be captured by one parameter Examples Velocity of car on the road Control of velocity of rotating system Electric systems where energy storage is essentially in one capacitor or one inductor Incompressible fluid flow in a pipe Level control of a tank Pressure control in gas tank Temperature in a body with essentially uniform temperature distribution (e.g. steam filled vessel)

6 PI Control of First Order Systems Process, controller and loop- transfer function P(s) = b s + a, C fb(s) = k p + k i s, C ff = βk p + k i s Closed loop transfer functions with error feedback G yr (s) = G yd (s) = Controller with set point weighting G yr (s) = PC 1 + PC = b(k p s + k i ) s 2 + (a + bk p )s + bk i P 1 + PC = bs s 2 + (a + bk p )s + bk i PC 1 + PC = b(βk p s + k i ) s 2 + (a + bk p )s + bk i Poles chosen by controller gains k p, k i, zero by set-point weight β

7 Block Diagram Representations Error feedback. Controller gives the zero k i /k p in G yr (s) r e u y Σ k p + k i/s P 1 Controller with two degrees of freedom with β = 0. Controller generates no zeros in G yr (s). r e u y Σ k i/s Σ P k p 1

8 Transfer function of Gang of Four S = Gang of Seven s(s + a) s 2 + (a + bk p )s + bk i, PS = CS = (k ps + k i )(s + a) s 2 + (a + bk p )s + bk i, T = Transfer functions from command signal b(βk p s + k i ) G yr (s) = s 2 + (a + bk p )s + bk i G ur (s) = (βk ps + k i )(s + a) s 2 + (a + bk p )s + bk i G er (s) = s(s + a + (1 β)bk p) s 2 + (a + bk p )s + bk i bs s 2 + (a + bk p )s + bk i b(k p s + k i ) s 2 + (a + bk p )s + bk i

9 Second Order Systems Two states because storage of mass, momentum and energy can be captured by two parameter Examples Position of car on the road Control of angle of rotating system Stabilization of satellites Electric systems where energy is stored in two elements (inductors or capacitors) Levels in two connected tanks Pressure in two connected vessels

10 PD Control of Second Order System Process and controller transfer functions P(s) = b s 2 + a 1 s + a 2, C(s) = k p + k d s Closed loop transfer function from reference to output G yr (s) = PC 1 + PC = b(k d s + k p ) s 2 + (a 1 + bk d )s + a 2 + bk p Closed loop system of second order, controller has two parameters. All closed loop poles can be chosen, but no integral action. With setpoint weighting G yr (s) = PC 1 + PC = b(γk d s + k p ) s 2 + (a 1 + bk d )s + a 2 + bk p

11 PI Control of Second Order Systems Process and controller transfer functions P(s) = b 1s + b 2 s 2 + a 1 s + a 2, C(s) = k p + k i s Closed loop transfer function from reference r to output y G yr (s) = PC 1 + PC = (k p s + k i )(b 1 s + b 2 ) s 3 + (a 1 + b 1 k p )s 2 + (a 2 + b 1 k i + b 2 k p )s + b 2 k i Closed loop system of third order, controller has only two parameters. Not enough degrees of freedom. A more complex controller is required to choose closed loop characteristic polynomial.

12 PID Control of Second Order System Process and controller transfer functions P(s) = b 1s + b 2 s 2 + a 1 s + a 2, C(s) = k p + k i s + k ds Closed loop transfer function from reference r to output y (b 1 s + b 2 )(k d s 2 + k p s + k i ) G yr (s) = (1 + b 1 k d )s 3 + (a 1 + b 1 k p + b 2 k d )s 2 + (a 2 + b 1 k i + b 2 k p )s + b 2 k i All closed loop poles can be chosen arbitrarily; with setpoint weighting (b 1 s + b 2 )(γk d s 2 + βk p s + k i ) G yr (s) = (1 + b 1 k d )s 3 + (a 1 + b 1 k p + b 2 k d )s 2 + (a 2 + b 1 k i + b 2 k p )s + b 2 k i some zeros of G yr can also be chosen arbitrarily weighting, but process zero remains.

13 1 Introduction 2 Simple Examples 3 Polynomial Design 4 State Space Design 5 Robustness and Design Rules 6 Model Reduction 7 Oscillatory Systems 8 Summary Theme: Be aware where you place them!

14 Polynomial Design Process and controller d p (s)y (s) = n p (s)u(s), d c (s)u(s) = n cff (s)r(s) n c (s)y (s) Closed loop transfer function G yr (s) = n p (s)n cff (s) d p (s)d c (s) + n p (s)n c (s) = n p(s)n cff (s) d cl (s) Determine d c (s) and n c (s) to give the desired closed loop polynomial d cl (s). The zeros can be partially influenced through n cff (s). Parameter count deg d c + deg n c + 1 = deg d cl Introduce unknown coefficients and solve linear equation

15 The Diophantine Equation The equation 3x + 2y = 1, where x and y are integers, has the solution: x = 1 and y = 1. Many other solutions can be obtained by adding 2 to x and subtracting 3 from y. The equation 6x + 4y = 1, cannot have a solution, because the left hand side is even and the right hand side is odd. The equation 6x + 4y = 2, has a solution, because we can divide by 2 and obtain the first equation.

16 Main Result Let a, b, c, x and y be integers, the equation ax + by = c has a solution if and only if the greatest common factor of a and b divides c. If the equation has a solution x 0 and y 0 then x = x 0 bn and y = y 0 + an, where n is an arbitrary integer, is also a solution. Integers and polynomials same algebra, add, subtract, divide with remainder (size replaced by degree) Euclid s algorithm holds for polynomials (the same algebra add & mult!)

17 Proof - Euclid s Algorithm Assume that the degree of a is greater than or equal to the degree of b. Let a 0 = a and b 0 = b. Iterate the equations a n+1 = b n, b n+1 = a n mod b n until b n+1 = 0. The greatest common divisor is then g = b n. If a and b are co-prime we have b n = 1. Back-tracking we find that ax + by = b n = g where the polynomials x and y can be found by keeping track of the quotients and the remainders in the iterations. When a and b are co-prime (g = 1) we get ax + by = 1 Multiplying x and y by c gives the original equation ax + by = c. When a and b have a common factor the largest common divisor of a and b must be a factor of c.

18 An Algorithm Let g be the greatest common divisor of a and b and let u and v be the minimal degree solutions to ax + by = 0. x y a 1 0 = g x y u v b u v Make row transformations to transform (Gaussian elimination) A (0) = a 1 0 b 0 1 to A (n) = g x y 0 u v It follows from Euclid s algorithm that g = A (n) 11 is the largest common divisor of a and b, and that a and b are co-prime if and only if A (n) 11 = 1. The equation ax + by = c has a solution if A (n) 11 is a factor of c.

19 Non-uniqueness I Closed loop characteristic polynomial d p d c + n p n c = d cl, C = n c /d c If d c0, n c0 is a solution then so is d c0 qn p, n c0 + qd p, where q is an arbitrary polynomial. Many different choices: Minimal numerator degree deg n c < deg d p, generically deg d c = deg n c = deg d p 1, deg d cl = 2 deg d p 1 (Luenberger observer) deg d c = deg d p, deg n c = deg d p 1, deg d cl = 2 deg d p (Kalman filter) Minimal denominator degree deg d c deg n p (controller has excess of zeros, derivative action). Generically deg d c = deg n p 1, deg n c = deg d p 1, deg d cl = deg d p + deg n p 1 deg n p = 0, d c = 1, deg n c = deg d p 1, deg d cl = deg d p Integral action: Add s as an extra factor of d p (s) solve for d c and n c and the controller is then C = n c (s)/(sd c (s)).

20 Non-uniqueness II Process and controller transfer functions P(s) = n p(s) d p (s), Closed loop characteristic equation C(s) = n c(s) d c (s) d p (s)d c (s) + n p (s)n c (s) = d cl (s) If C 0 = n c0 (s)/d c0 (s) is a controller that gives the closed loop characteristic polynomial d cl (s) then the controller C(s) = n c0(s) + q(s)d p (s) d c0 (s) q(s)n p (s) where q(s) is an arbitrary polynomial also gives char. pol d cl (s). d p (s) ( d c0 0(s) q(s)n p (s) ) + n p (s) ( n c0 + q(s)d p (s) ) = d p (s)d c0 (s) + n p (s)n c0 (s) = d cl (s)

21 Youla-Kucera Parametrization I Consider a process with a stable transfer function P. Let the desired transfer function from reference to output be T. Let the requirement be realized by feedforward with the transfer function Q, where T = P Q. Since Q must be stable, T and P should have the same zeros in the right half plane. The transfer function T can also be obtained by error feedback with the controller C = Q 1 PQ Σ Q P For any stable Q, GoF is linear in Q: P Σ T = PQ, S = 1 T = 1 PQ v PS = P(1 PQ), CS = Q 1 All stabilizing controllers C can be represented by some such Q.

22 Youla-Kučera Parametrization II Process transfer function P = B/A, where A and B are stable co-prime rational functions. Assume that the controller C 0 = G 0 /F 0 stabilizes P. All stabilizing controllers are given by C = G 0 + QA F 0 QB Q is an arbitrary stable rational transfer function. GoF: S = A(F 0 QB) AF 0 + BG 0, CS = A(G 0 + QA) AF 0 + BG 0, PS = B(F 0 QB) AF 0 + BG 0 T = B(G 0 + QA) AF 0 + BG 0 The system is stable since the rational function AF 0 + BG 0 has all its zeros in the left half plane and A, B, F 0, G 0 and Q are stable rational functions.

23 Block Diagram Interpretation Controller C = G 0 + QA F 0 QB, F 0U = G 0 Y + Q(BU AY ) v Q B Σ A G 0 Σ F 1 0 P Notice that the input to Q is nominally zero

24 Where to Place the Poles - Standard Forms Butterworth configurations ω 2 0 G(s) =, G(s) = s ω 0s + ω0 2 s 3 + 2ω 0s 2 + 2ω0 2s + ω3 0 ω 3 0 a nω0 n G 1(s) = s n + a 1ω 0s n 1 + a 2ω0 2sn a nω0 n a n 1ω n 1 0 s + a nω0 n G 2(s) = s n + a 1ω 0s n 1 + a 2ω0 2sn a nω0 n G 1 zero step error min ITAE, G 2 zero ramp error min ITAE G 1(s) G 2(s) n a 1 a 2 a 3 a 4 a 5 a 1 a 2 a 3 a 4 a

25 Summary - Pole Placement Simple useful method easy to use for low order systems The only requirement is that there are no common zeros in the numerator and denominator polynomials of the process transfer function, or equivalently that the system is reachable and observable. There are multivariable extensions based on matrix polynomials BUT polynomial computations are numerically poorly conditioned for high order systems s n = ǫ, s = ǫ 1/n, n = 10, ǫ = 10 8, ǫ 1/n = 0.16 State space approach gives more reliable computations and other insights

26 1 Introduction 2 Simple Examples 3 Polynomial Design 4 State Space Design 5 Robustness and Design Rules 6 Model Reduction 7 Oscillatory Systems 8 Summary Theme: Be aware where you place them!

27 To Add Integral action Interpretation 1: dynamics of a controller arises from the observer Nice interpretation of closed loop characteristic polynomial det(si A + BL) det(si A + KC) Emphasize different 2DOF architectures Can we separate a given polynomial? How to apply the design rules - where should cancellation take part L or K?

28 Combining State Feedback with an Observer Consider a system which is assumed to be observable and reachable dx dt = Ax + Bu, y = Cx Determine the state with an observer and use state feedback from the observed state. This gives the controller dˆx = Aˆx + Bu + K(y Cˆx) dt u = Lˆx + L r r The controller is a dynamical system whose dynamics is represented by the observer dynamics (the internal model principle). Controller complexity is given by model complexity.

29 Architecture of Basic Controller r u ẋ x y Σ B Σ C L r Process A K Σ B Σ ˆx C ŷ Observer A L The controller contains a model of the process and its environment

30 The Closed Loop System 1 Process Controller dˆx dt dx dt = Ax + Bu, y = Aˆx + Bu + K(y Cˆx), u = Cx. = Lˆx + L rr. Introduce the state x = x ˆx instead of ˆx. dx dt = Ax + Bu = Ax BLˆx + Bl rr = (A BL)x + BL x + Bl r r d x = (A KC) x dt If the system is observable the observer can be designed so that the observer error x goes to zero as fast as desired. The system behaves like a system with state feedback when x = 0

31 Closed Loop System 2 The closed loop system d x x = A BL BL x x + Bl r r dt 0 A KC 0 y = C 0 Observe Block triangular structure of dynamics matrix Closed loop eigenvalues are eigenvalues of A BL and A KC State feedback gain L and observer gain K can be designed independently The state x is not reachable from the reference. Intuitively: changes in the reference should not generate observer errors!

32 More Flexible Architecture with Full 2DOF r Model and Feedforward Generator x m Σ State Feedback u fb Σ u ff u Process y -ˆx Observer A nice two degree-of-freedom (2DOF) structure which decouples response to disturbances (handled by state feedback and observer) from response to reference signals (handled by reference model and feedforward) There are many ways to generate the signals x m and u ff

33 1 Introduction 2 Simple Examples 3 Polynomial Design 4 State Space Design 5 Robustness and Design Rules 6 Model Reduction 7 Oscillatory Systems 8 Summary Theme: Be aware where you place them!

34 Robustness Intuitively we may expect that well-damped closed loop poles would guarantee robustness Unfortunately this is not true!!! Always necessary to check robustness if it is not part of design process Always check requirements that are not explicit requirements in the design procedure, particularly if you optimize Looking at the Gang of Four is a good idea A long forgotten problem (Mats Lilja s PhD # ) Two examples that give insight Two simple design rules for placing poles properly

35 Example 1 Consider a first order system with PI control P(s) = b s + a = 1 s + 1, C(s) = k p + k i s where the controller parameters are chosen to give a closed loop system with the characteristic polynomial s 2 + ω 0 s + ω 2 0. Characteristic polynomial Controller parameters s(s + a) + b(k p s + k i ) = s 2 + ω 0 s + ω 2 0 k p = ω 0 a b = ω 0 1, k i = ω2 0 b = ω2 0 What is special about ω 0 = 1? What does it mean that k p is negative?

36 Nyquist Plot ω 0 /a = 0.1 (red), 1 and 10 (blue) imag L(iω) real L(iω)

37 Gang of Four ω 0 /a = 0.1 (red), 1 and 10 (blue) CS T S P S ω ω Looks OK for ω 0 /a = 1 and 10 BUT not for ω 0 = 0.1 (red curves)

38 Reflections We have made what looks like a perfectly reasonable pole placement design with nicely damped closed loop poles ζ 0 = 0.5. The results look good for ω 0 /a = 1 and 10 The design for ω 0 /a = 0.1 have very high sensitivities M s = 9.4 and M t = 10 It is apparently important where we place the poles Can we understand what goes on and fix it?

39 The Sensitivity Function We have for a = 1 and ω 0 = 0.1, M s = 9 (9.4) S = (s + a)s s 2 + ω 0 s + ω 2 0 = (s + 1)s s s = d p (s)d c (s) d p (s)d c (s) + n p (s)n c (s) 10 1 S(iω) ω What creates the peak? Start from high frequencies.

40 Generalization Transfer functions of process and controller P(s) = n p(s) d p (s), C(s) = n c(s) d c (s), Sensitivity functions S(s) = PC = d p (s)d c (s) d p (s)d c (s) + n p (s)n c (s) At high frequencies we have S 1. As the frequency decreases there will be a break-point at the process poles (zeros of d p. To avoid having high sensitivities high frequency process poles must be matched by corresponding closed loop poles. In the example there was a process pole at s = 1 but the closed loop poles were at 0.1.

41 Complementary Sensitivity Function We have for a = 1 and ω 0 = 0.1 T = (ω 0 1)s + ω 2 0 s 2 + ω 0 s + ω 2 0 = 0.9s s s = n p (s)n c (s) d p (s)d c (s) + n p (s)n c (s) Notice slow zero in the controller!! 10 1 T (iω) We have approximately M t 0.1 = 10 (10.04) 0.01 ω

42 Consider the process and controller Example 2 P (s) = b 1s + b 2 s 2, C(s) = g 0s + g 1 s 2 + f 1 s + f 2 Desired closed loop characteristic polynomial We have d cl (s) = (s 2 + 2ζ c ω c s + ω 2 c )(s2 + 2ζ o ω o s + ω 2 o ) s 2 (s 2 + f 1 s + f 2 ) + (b 1 s + b 2 )(g 0 s + g 1 ) = c(s) Identification of coefficients of equal powers of s gives f 1 = 2(ζ o ω o + ζ c ω c ) f 2 = ω 2 o + ω2 c + 4ζ 0ζ c ω o ω c b 1 g 0 g 0 = 2b 2(ζ o ω c + ζ c ω o )ω o ω c b 1 ω 2 0 ω2 c b 2 2 g 1 = ω 2 0 ω2 c /b 2

43 Another Simple Pole Placement Problem... With b 1 = 0.5, b 2 = 1, ω c = 10, ζ c = 0.707, ω o = 20 and ζ o = we get 11516s C(s) = s s L(iω) L(iω) Frequency ω [rad/s] Extremely poor robustness M s = 13 and M t = 12

44 Complementary Sensitivity Function P(s) = 0.5s + 1 s 2, C(s) = Notice slow zero in the controller!! 11516s s s T (iω) ω

45 Generalization Transfer functions of process and controller P(s) = n p(s) d p (s), C(s) = n c(s) d c (s), Sensitivity functions T (s) = PC 1 + PC = n p (s)n c (s) d p (s)d c (s) + n p (s)n c (s) At low frequencies we have T 1. As the frequency increases there will be breakpoints at the process zeros of (zeros of n p ). To avoid having high sensitivities low frequency process zeros must be matched by corresponding closed loop poles. In the example there was a process zero at s = 0.5 but the slowest closed loop poles were at s = 10, hence a peak of 10.

46 Design Rules Formally only reachability and observability are required To obtain robust closed loop systems the poles and zeros of the process must be taken into account. Design rule. Choose bandwidth ω b or dominating closed loop poles: classify poles and zeros as slow < ω b or fast > ω b Slow unstable zeros (time delays) and fast unstable poles restrict the choice of closed loop bandwidth ω b Design rule: Pick closed loop poles close to slow stable process zeros and fast stable process poles. Picking closed loop poles and zeros identical to slow stable zeros and fast stable poles give cancellations and simple calculations. Violating the design rule leads to closed loop systems that are not robust. Unstable poles and zeros cannot be canceled, slow unstable zeros and fast unstable poles therefore give fundamental limitations.

47 1 Introduction 2 Simple Examples 3 Polynomial Design 4 State Space Design 5 Robustness and Design Rules 6 Model Reduction 7 Oscillatory Systems 8 Summary Theme: Be aware where you place them!

48 Model Reduction - Fast Stable Poles Neglect fast modes in the model, explore a simple case P(s) = a s + a 1 Reduced order model is a static system, I controller sufficient C(s) = k i s Choose k i to give the closed loop pole a 0, hence k i = a 0, closed loop characteristic polynomial with true plant model s(s + a) + aa 0 = 0 s = a 0 a2 0 a..., s = a + a 0 + a2 0 a... Simple rule: neglect stable poles that are an order of magnitude larger than a 0 Notice a closed loop pole close to neglected pole a

49 Model Reduction - Slow Zeros For large ω 0 we can approximate the process model by neglecting slow zeros P(s) = s + a s 2 1 s Use a PI controller k p = 2ζω 0, k i = ω 2 0. Closed loop characteristic polynomial with true model s 3 + (s + a)(k p s + k i ) = s(s 2 + k p s + k i ) + a(k p s + k i ) = 0 For small a roots close to s 2 + k p s + k i = 0 For small s roots close to s + a = 0 Simple rule: neglect stable zeros that an order of magnitude smaller than ω 0

50 Design Rules Simple models are useful How to approximate? Start with desired bandwidth ω bw which is of the order of ω ms, ω mt, ω gc Neglect process poles and zeros that are an order of magnitude faster than ω bw unless they are highly oscillatory. Neglect process zeros that are an order of magnitude slower than ω bw, approximate slow poles by integrators. Cancel fast process poles and slow process zeros that cannot be neglected. Make the design based on the simplified model, add the canceled factors to the controller after the design. Add high frequency roll-off if necessary.

51 An Example Consider P(s) = K p (1 + st 4 )(1 + st 5 ) (1 + st 1 )(1 + st 2 )(1 + st 3 ) e sl where T 1 is significantly larger than L and the other time constants. For controller with very low performance, settling times slower than T 1, approximate the process by its gain and use an integrating controller. For slightly higher performance approximate by P(s) K p 1 + s(t 1 + T e ) e ste or P(s) where T e = (T 2 + T 3 + L T 4 T 5 )/2 and T e > 0 K p (1 st e ) (1 + st 1 )(1 + st e )

52 1 Introduction 2 Simple Examples 3 Polynomial Design 4 State Space Design 5 Robustness and Design Rules 6 Model Reduction 7 Oscillatory Systems 8 Summary Theme: Be aware where you place them!

53 Oscillatory Systems Truxal 1961: The design of feedback systems to effect satisfactorily the control of very lightly damped physical systems is perhaps the most basic of the difficult control problems. A simple prototype problem P(s) = ω 2 0 s 2 + 2ζ 0 ω 0 s + ω 2 0 with very low damping ζ = 0.01 (Q = 50) Integral (I) control PI control PID Control - Notch-filter design PID Control - Active damping of oscillatory modes Systems with many resonances

54 Integrating (I) Control Any stable system with P(0) 0 can be controlled by an integrating controller provided that requirements are modest. Approximate the process transfer function by neglecting the oscillatory mode P(s) = ω 2 0 s 2 + 2ζ 0 ω 0 s + ω With pure integral control we have C(s) = k i s, L(s) = k i ω0 2 s(s 2 + 2ζ 0 ω 0 s + ω0 2) k i s Requirements on the gain margin limits integral gain k i = ω gc g m = 1 L(i ω 0 ) = 2ζ 0ω 0 k i, k i = 2ζ 0ω 0 g m, ω gc = ω 0 Qg m

55 I Control Any stable system with P(0) 0 can be controlled by an integrating controller provided that requirements are modest. L(s) = k i ω 2 0 s(s 2 + 2ζ 0 ω 0 s + ω 2 0 ), k i = 2ζ 0ω 0 g m Numerical values: ζ 0 = 0.01, g m = 2.5 ( M s 1.7) gives k i = 0.008ω 0 L(iω) L(iω) ω/ω 0 ImL(iω) ReL(iω)

56 Gang of Four for Integral Control S(iω) P S(iω) CS(iω) T (iω) ω/ω 0 ω/ω 0 Good robustness M s = 1.7, M t = 1 low bandwidth ω gc = 0.008ω 0 extra sensitive to load disturbances at resonance frequency

57 I Control with Roll-off Controller transfer function with high-frequency roll-off C(s) = k i s(1 + st f + s 2 T 2 f /2) Loop transfer function and low frequency approximation L(s) = k i ω 2 0 s(s 2 T 2 f /2 + st f + 1)(s 2 + 2ζ 0 ω 0 s + ω 2 0 ) k i s (1 st f s 2ζ 0 ω 0 ) k i s k it f Choosing T f = 0.5/k i gives a Nyquist curve that is close to the robustness valley (Re L(iω) = 0.5) for small s, adjust k i to be as large as possible without sacrificing robustness, k i = 0.1ω 0 an order of magnitude larger than without a filter (k i = 0.008ω 0 ).

58 I Control with Roll-off... Nyquist Plot The filter shifts the red circle to a safer region (2nd order filter important!)

59 I Control with Roll-off... S(iω) P S(iω) CS(iω) T (iω) ω/ω 0 ω/ω 0 I Control: k i = ω 0 (red) I Control: with filter: k i = 0.1 ω 0 (blue) The filter gives a significant improvement at least 10X

60 Summary of I Control Gain crossover frequency is limited to k i = ω gc < 2ζ 0ω 0 g m = ω 0 Proportional action does not give any improvements Roll-off filter helps to increase k i significantly to 0.1ω 0 Notice bending of the loop to a safer region High resonant peak in P S remains Controller noise gains CS moderate

61 PID Control - Notch filter design The resonant poles must be considered in order to increase the bandwidth. If the bandwidth is less than ω 0 we must choose closed loop poles that are close to the process poles. A simple way to do this is to cancel the fast process poles by controller zeros. Controller (C(s)) and the loop transfer function (L(s)) become C(s) = k i s 2 + 2ζ 0 ω 0 s + ω 2 0 ω 2 0 s = k ds 2 + k p s + k i, L(s) = k i s s Add high frequency roll-off C(s) = k i(s 2 + 2ζ 0 ω 0 s + ω 2 0 ) ω 2 0 s(1 + st f + s 2 T 2 f /2), L(s) = k i s(1 + st f + s 2 T 2 f /2) Relate T f to gain crossover frequency ω gc k i, T f k i = 0.5

62 Gang of Four for PID Notch Filter Control S(iω) P S(iω) CS(iω) 10 0 T (iω) ω/ω 0 ω/ω 0 k i = 0.5ω 0, T f = 0.5/k i, M s = 1.6, M t = 1 (robustness valley!) Without filter red, with filter blue High peak in response to load disturbances. High frequency roll-off important

63 Parameter Variations 1% for k i = 0.2 ω S(iω) 10 0 P S(iω) CS(iω) T (iω)

64 Parameter Variations 1% for k i = ω S(iω) 10 0 P S(iω) CS(iω) T (iω)

65 Nyquist Plots for k i = 0.2 ω 0 and k i = ω 0 ω 0 = 0.99, k i = 0.2 ω 0 = 1.00, k i = 0.2 ω 0 = 1.01, k i = ω 0 = 0.99, k i = 1.0 ω 0 = 1.00, k i = 1.0 ω 0 = 1.01, k i = Underestimating ω 0 better than overestimating (ω design = 1.00)

66 Summary of Notch-filter Design Bandwidth can be increased significantly compared to I-Control Maximum sensitivities are OK. Small parameter variations give moderate changes for designs with ω b = k i = 0.2ω 0 but very high sensitivities for designs with k i = ω 0. High sensitivity to small time delay variations for designs with k i = ω 0. Still very sensitive to disturbances with energy at the resonance frequency, which are not attenuated by the feedback loop.

67 Active Damping of Resonant Mode A PID controller is sufficient. The loop transfer function is L(s) = (k ds 2 + k p s + k i )ω 2 0 s(s 2 + 2ζ 0 ω 0 s + ω 2 0 ). Closed loop characteristic polynomial s 3 + (k d ω ζ 0 ω 0 )s 2 + (k p + 1)ω 2 0s + k i ω 2 0. A general third order polynomial can be parameterized as (s + α c ω c )(s 2 + 2ζ c ω c s + ω 2 c ) = s 3 + (α c + 2ζ c )ω c s 2 + (1 + 2α c ζ c )ω 2 cs + α c ω 3 c. Identification of coefficients of equal powers of s give k d = (α c + 2ζ c )ω c 2ζ 0 ω 0 ω0 2, k p = (1 + 2α cζ c )ωc 2 ω0 2 1, k i = α cωc 3 ω0 2

68 Add high frequency roll-off Parameters: C(s) = The Controller k ds 2 + k p s + k i s(1 + st f + s 2 T 2 f /2) k d = (α c + 2ζ c )ω c 2ζ 0 ω 0 ω 2 0 k p = (1 + 2α cζ c )ω 2 c ω k i = α cω 3 c ω 2 0 T f = T d 10 = k d 10k p

69 Changing Integral Gain k i = 0.27ω 0, ω 0, 8 ω 0 S(iω) P S(iω) CS(iω) T (iω) ω/ω 0 ω/ω 0 k i = 0.27ω 0 (dash-dotted), k i = ω 0 (dashed) and k i = 8ω 0, (full). Notice dramatic increase of bandwidth and controller HF gain and bandwidth! Bandwidth increased while maintaining robustness M s = 1.7, M t = 1.6. But we have to look at parametric uncertainty too!

70 Admissible High Frequency Controller Gain To estimate allowable high frequency gains we must consider High frequency measurement noise Range of control signal Bandwidth of sensors and actuators Model accuracy Robustness

71 Robustness to Parameter Variations 20 ω 0 = ω 0 = ω 0 = ω 0 = ω 0 = ω 0 = Not very sensitive to parameter variations

72 Closed Loop Poles I and Ifilt Root Locus Root Locus Imaginary Axis (seconds -1 ) Imaginary Axis (seconds -1 ) Real Axis (seconds -1 ) Real Axis (seconds -1 )

73 Closed Loop Poles Notch and Damp Root Locus Root Locus Imaginary Axis (seconds -1 ) Imaginary Axis (seconds -1 ) Real Axis (seconds -1 ) Real Axis (seconds -1 )

74 Posicast Control A scheme for command signal control of oscillatory systems Proposed by Otto Smith who also invented the Smith predictor The idea: move a hanging load from one position to another

75 Time Response and Transfer Function 2 Transfer function interpretation G(s) = 1 ( 1 + e st ) p/2 2 Notice that G(iω) = 0 for ωt p = 2π + 4nπ u y Time t Posicast controller is an efficient notch filter Can be modified to handle damped oscillations Application to MEMS drives

76 I Control with I Posicast Notch with Filter Controller transfer function C(s) = k i ( γ + (1 γ)e st ) d s ( ) γ = 1 + e ζ 0π/ 1 ζ 1, 0 2 π Td = ω 0 1 ζ0 2 Loop transfer function L(s) = k ( ) i γ + (1 γ)e st d ω 2 0 s(s 2 + 2ζ 0 ω 0 s + ω0 2). Series expansion for small s L(s) k i s ( 1 (1 γ)t d s 2ζ 0s ω 0 ) = k i Pick k i is so that Re L(iω) = 0.5 for small s, ω 0 1 ζ 2 k i = (1 γ)π + 2ζ 0 1 ζ 2 0 s k (1 γ)π i 2ζ 0k i ω 0 1 ζ 2 0 ω 0

77 Bode and Nyquist Plots L(iω) L(iω) ω/ω 0 ImL(iω) ReL(iω) Integral gain k i = 0.32 ω 0, bandwidth ω B = 0.63 ω 0 maximum sensitivities: M s = 1.53, M t = 1.00, M ps = 50.0, M cs = 1.00.

78 Gang of Four S(iω) P S(iω) CS(iω) T (iω) ω/ω ω/ω 0 Integral gain k i = 0.32 ω 0, bandwidth ω B = 0.63 ω 0 maximum sensitivities: M s = 1.53, M t = 1.00, M ps = 50.0, M cs = 1.00

79 Summary I control: k i 2ζ 0 ω 0 /g m (k i = ω 0 ), high sensitivity for disturbances at resonance frequency I control with filter: k i = 0.1 ω 0, high sensitivity for disturbances at resonance frequency I control with posicast notch k i = 0.32 ω 0, high sensitivity to disturbance at resonance frequency PI: Proportional action gives no improvement PID Notch with roll-off: k i = 0.8 ω 0 high sensitivity for disturbances at resonance frequency, high sensitivity to variations in resonance frequency PID with active damping and roll-off: k i = 8 ω 0, sensitivity for disturbances at resonance frequency reduced, requires high gain of controller at high frequencies M cs = 100, and a wide band controller ω ccs = 200 ω 0

80 1 Introduction 2 Simple Examples 3 Polynomial Design 4 State Space Design 5 Robustness and Design Rules 6 Model Reduction 7 Oscillatory Systems 8 Summary Theme: Be aware where you place them!

81 Summary Simple design method for SISO systems Probably quickest way to introduce design Most importantly we get insight and design rules Insights through Euclid s algorithm and Youla parametrization Polynomials are bad numerically, matrix calculations much more robust There are multivariable versions but they are complicated

82 Reading Suggestions Åström Murray Feedback Systems - An Introduction for Scientists and Engineers, Princeton 2008 (also on Richards home page). GoF Ch 11. Use index for other things. Åström Hägglund. Advanced PID control - Second edition has full chapter on oscillatory systems - drafts available