# Control Systems Design

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1 ELEC4410 Control Systems Design Lecture 18: State Feedback Tracking and State Estimation Julio H. Braslavsky School of Electrical Engineering and Computer Science Lecture 18: State Feedback Tracking and State Estimation p. 1

2 Outline Regulation and Tracking Robust Tracking: Integral Action State Estimation A tip for Lab 2 Lecture 18: State Feedback Tracking and State Estimation p. 2

3 State Feedback In the last lecture we introduced state feedback as a technique for eigenvalue placement. Briefly, given the open-loop state equation ẋ(t) = Ax(t) + Bu(t) y(t) = Cx(t), we apply the control law u(t) = Nr(t) Kx(t) and obtain the closed-loop state equation ẋ(t) = (A BK)x(t) + BNr(t) y(t) = Cx(t), If the system is controllable, by appropriately designing K, we are able to place the eigenvalues of (A BK) at any desired locations. Lecture 18: State Feedback Tracking and State Estimation p. 3

4 Regulation and Tracking The associated block diagram is the following Two typical control problems of interest: The regulator problem, in which lim and we aim to keep (i.e., a pure stabilisation problem) The tracking problem, in which is specified to track. When, constant, the regulator and tracking problems are essentially the same. Tracking a nonconstant reference difficult problem, called the servomechanism problem. is a more Lecture 18: State Feedback Tracking and State Estimation p. 4

5 Regulation and Tracking We review the state feedback design procedure with an example. Example (Speed control of a DC motor). We consider a DC motor described by the state equations h 5 i ω + V i The input to the DC motor is the voltage V(t), and the states are the current i(t) and rotor velocity ω(t). We assume that we can measure both, and take ω(t) as the output of interest. Lecture 18: State Feedback Tracking and State Estimation p. 5

6 Regulation and Tracking Example (continuation). ➀ The open-loop characteristic polynomial is (s) = det(si A) = det [ s s+2 ] = s s which has two stable roots at s = and s = The motor open-loop step response is Step Response 0.12 The system takes about 3s to reach steady-state. speed is about the amplitude of the voltage step. The final Amplitude Time (sec) We would like to design a state feedback control to make the motor response faster and obtain tracking of reference inputs. to constant Lecture 18: State Feedback Tracking and State Estimation p. 6

7 Regulation and Tracking Example (continuation). To design the state feedback gain, we next ➁ compute the controllability matrix [ ] C = B AB = which is full rank the system is controllable. Also, from the open-loop characteristic polynomial we form controllability matrix in x coordinates is C = 1 α = = Lecture 18: State Feedback Tracking and State Estimation p. 7

8 Regulation and Tracking Example (continuation). We now ➂ propose a desired characteristic polynomial. Suppose that we we would like the closed-loop eigenvalues to be at s = 5 ± j, which yield a step response with 0.1% overshoot and about 1s settling time. The desired (closed-loop) characteristic polynomial is then K (s) = (s + 5 j)(s j) = s s + 26 With K (s) and (s) we determine the state feedback gain in x coordinates [ ] [ ] K = (ᾱ 1 α 1 ) (ᾱ 2 α 2 ) = (10 12) ( ) [ ] = Lecture 18: State Feedback Tracking and State Estimation p. 8

9 Regulation and Tracking Example (continuation). Finally, ➃ we obtain the state feedback gain K in the original coordinates using Bass-Gura formula, K = K CC 1 = [ ] [ [ ] = ] [ 0 2 ] As can be verified, the eigenvalues of (A BK) are as desired Step Response The closed-loop step response, as desired, settles in significant overshoot. s, with no Amplitude Note, however, that we still have steady-state error ( ). To fix it, we use the feedforward gain Time (sec) Lecture 18: State Feedback Tracking and State Estimation p. 9

10 Regulation and Tracking Example (continuation). The system transfer function does not have a zero at s = 0, which would prevent tracking of constant references (as we can see in the step response, which otherwise, would asymptotically go to 0). Step Response 1.2 Thus, ➄ we determine 1 formula with the 0.8 Amplitude Time (sec) and achieve zero steady-state error in the closed-loop step response. Lecture 18: State Feedback Tracking and State Estimation p. 10

11 Regulation and Tracking Example (continuation). We have designed a state feedback controller for the speed of the DC motor. However, the tracking achieved by feedforward precompensation would not tolerate (it is not robust to) to uncertainties in the plant model. 1.2 Step Response To see this, suppose the real matrix in the system is slightly different from the one we used to compute, Amplitude The closed-loop step response given by the designed gains N, K (based on a different A-matrix) doesn t yield tracking. Time (sec) Lecture 18: State Feedback Tracking and State Estimation p. 11

12 Outline Regulation and Tracking Robust Tracking: Integral Action State Estimation Lecture 18: State Feedback Tracking and State Estimation p. 12

13 Robust Tracking: Integral Action We now introduce a robust approach to achieve constant reference tracking by state feedback. This approach consists in the addition of integral action to the state feedback, so that the error ε(t) = r y(t) will approach 0 as t, and this property will be preserved under moderate uncertainties in the plant model under constant input or output disturbance signals. Lecture 18: State Feedback Tracking and State Estimation p. 13

14 Robust Tracking: Integral Action The State Feedback with Integral Action scheme: Lecture 18: State Feedback Tracking and State Estimation p. 14

15 Robust Tracking: Integral Action The State Feedback with Integral Action scheme: Lecture 18: State Feedback Tracking and State Estimation p. 14

16 Robust Tracking: Integral Action The State Feedback with Integral Action scheme: Lecture 18: State Feedback Tracking and State Estimation p. 14

17 Robust Tracking: Integral Action The State Feedback with Integral Action scheme: Lecture 18: State Feedback Tracking and State Estimation p. 14

18 Robust Tracking: Integral Action The State Feedback with Integral Action scheme: The main idea in the addition of integral action is to augment the plant with an extra state: the integral of the tracking error, (IA1) The control law for the augmented plant is then h i (IA2) Lecture 18: State Feedback Tracking and State Estimation p. 14

19 Robust Tracking: Integral Action The closed-loop state equation with the state feedback control u(t) given by (IA1) and (IA2) is ẋ(t) = A 0 x(t) ] [K k z x(t) + ż(t) C 0 } {{ } A a z(t) B 0 }{{} B a } {{ } K a z(t) 0 r 1 Lecture 18: State Feedback Tracking and State Estimation p. 15

20 Robust Tracking: Integral Action The closed-loop state equation with the state feedback control u(t) given by (IA1) and (IA2) is ẋ(t) = A 0 x(t) ] [K k z x(t) ż(t) C 0 } {{ } A a = (A a B a K a ) B z(t) 0 }{{} [ ] x(t) z(t) B a + [ 0 1 } {{ } K a ] r + 0 r z(t) 1 Lecture 18: State Feedback Tracking and State Estimation p. 15

21 Robust Tracking: Integral Action The closed-loop state equation with the state feedback control u(t) given by (IA1) and (IA2) is ẋ(t) = A 0 x(t) ] [K k z x(t) ż(t) C 0 } {{ } A a = (A a B a K a ) B z(t) 0 }{{} [ ] x(t) z(t) B a + [ 0 1 } {{ } K a ] r + 0 r z(t) 1 The state feedback design with integral action can be done as a normal state feedback design for the augmented plant Lecture 18: State Feedback Tracking and State Estimation p. 15

22 Robust Tracking: Integral Action The closed-loop state equation with the state feedback control u(t) given by (IA1) and (IA2) is ẋ(t) = A 0 x(t) ] [K k z x(t) ż(t) C 0 } {{ } A a = (A a B a K a ) B z(t) 0 }{{} [ ] x(t) z(t) B a + [ 0 1 } {{ } K a ] r + 0 r z(t) 1 The state feedback design with integral action can be done as a normal state feedback design for the augmented plant If K a is designed such that the closed-loop augmented matrix (A a B a K a ) is rendered Hurwitz, then necessarily in steady-state lim t ż(t) = 0 lim t y(t) = r, achieving tracking. Lecture 18: State Feedback Tracking and State Estimation p. 15

23 Integral Action How does it work? Lecture 18: State Feedback Tracking and State Estimation p. 16

24 Integral Action How does it work? Lecture 18: State Feedback Tracking and State Estimation p. 16

25 Integral Action How does it work? Lecture 18: State Feedback Tracking and State Estimation p. 16

26 Integral Action How does it work? The block diagram of the closed-loop system controlled by state feedback with integral action thus collapses to where G K (s) is a BIBO stable transfer function, by design, and also the overall closed-loop is BIBO stable. Lecture 18: State Feedback Tracking and State Estimation p. 17

27 Integral Action How does it work? The block diagram of the closed-loop system controlled by state feedback with integral action thus collapses to where G K (s) is a BIBO stable transfer function, by design, and also the overall closed-loop is BIBO stable. Let s express G K (s) in terms of its numerator and denominator polynomials, G K (s) = N(s) D(s) Then, from the block diagram above, Y(s) = ( k z )N(s) sd(s) 1 + ( k z)n(s) sd(s) R(s)+ N(s) D(s) 1 + ( k z)n(s) sd(s) D i (s) ( k z)n(s) sd(s) D o (s) Lecture 18: State Feedback Tracking and State Estimation p. 17

28 Integral Action How does it work? In other words, Lecture 18: State Feedback Tracking and State Estimation p. 18

29 Integral Action How does it work? In other words, and, If the reference and the disturbances are constants, say then, because the closed-loop is BIBO stable, the steady state value of, is determined by the 3 transfer functions above evaluated at lim Lecture 18: State Feedback Tracking and State Estimation p. 18

30 Integral Action How does it work? In other words, If the reference and the disturbances are constants, say, and then, because the closed-loop is BIBO stable, the steady state value of is determined by the 3 transfer functions above evaluated at, lim That is: the output will asymptotically track constant references and reject constant disturbances irrespective of the values,. and Lecture 18: State Feedback Tracking and State Estimation p. 18

31 Robust Tracking Example Example (Robust speed tracking in a DC motor). Let s go back to the DC motor example and design a state feedback with integral action to achieve robust speed tracking. We have to redesign the state feedback gain K it must be computed together with k z for the augmented plant given by [ ] ] ] ] A a = A 0 =, B C 0 a = = [ If we compute the characteristic polynomial of the augmented matrix A a we find [ B 0 a (s) = s s s = s (s). Unsurprisingly, it just adds a root at s = 0 to the original. [ Lecture 18: State Feedback Tracking and State Estimation p. 19

32 Robust Tracking Example Example (continuation). We now move on to compute C a and C a for the augmented pair (A a, B a ), C a = [ ] , C a = [ ] 1 = [ ] Note that the augmented pair (A a, B a ) will always be controllable as long as the original pair (A, B) is controllable and the system has no zeros at s = 0. We select as desired augmented characteristic polynomial Ka (s) = (s + 6) K (s) = s s s (We keep the original desired characteristic polynomial and let the extra pole be faster, to keep the same specifications in the response of y(t).) Lecture 18: State Feedback Tracking and State Estimation p. 20

33 Robust Tracking Example Example (continuation). As before, from a (s) and ak (s) we compute K a and now obtain [ ] [ ] K a = (16 12) ( ) (156 0) = and finally, K a = K a C a C 1 = [ [ ] } {{ } K [ ]] 78 }{{} k z Note that the first two elements of the augmented K a correspond ] to the new state feedback gain K for the motor state, while the last element is the state feedback gain k z for [ ω(t) i(t) the augmented state z(t). Lecture 18: State Feedback Tracking and State Estimation p. 21

34 Robust Tracking Example Example (continuation). A block diagram for implementation: And a SIMULINK diagram (including disturbances): d_i d_o r 1 s Integrator1 78 Gain2 B* u Matrix Gain1 1 s Integrator A* u C* u Matrix Gain2 Scope Gain 2 em Matrix Gain Gain1 Lecture 18: State Feedback Tracking and State Estimation p. 22

35 Robust Tracking Example Example (continuation). We simulate the closed loop system to a unit step input applied at t = 0, an input disturbance d = 0.5 applied at t = 8s, and an output disturbance d o = 0.3 applied at t = 4s y(t) time [s] We can see the asymptotic tracking of the reference despite the disturbances. Lecture 18: State Feedback Tracking and State Estimation p. 23

36 Outline Regulation and Tracking Robust Tracking: Integral Action State Estimation Lecture 18: State Feedback Tracking and State Estimation p. 24

37 State Estimation State feedback requires measuring the states but, normally, we do not have direct access to all the states. Then, how do we implement state feedback? If the system is observable the states can be estimated by an observer. Plant Observer State feedback from estimated states An observer is a dynamic system that estimates the states of the plant based on the measurement of its inputs and outputs. Lecture 18: State Feedback Tracking and State Estimation p. 25

38 State Estimation: A Naive Observer How would we build an observer? One intuitive way would be to reproduce a model of the plant and run it simultaneously to obtain a state estimate ^x(t). Plant copy A naive design of an observer. The problem with this naive design is that if the plant and its copy in the observer have different initial conditions, the estimates will generally not converge to the true values. Lecture 18: State Feedback Tracking and State Estimation p. 26

39 State Estimation: A Feedback Observer A better observer structure includes error feedback correction. error A self-corrected (feedback) design of an observer. By appropriately designing the matrix gain L, we could adjust the observer to give a state estimate that will asymptotically converge to the true state. Lecture 18: State Feedback Tracking and State Estimation p. 27

40 State Estimation: Final Observer Structure By rearranging the previous block diagram, we get to the final structure of the observer If the system is observable, we can then choose the gain L to ascribe the eigenvalues of (A LC) arbitrarily. Observer We certainly want the observer to be stable! From the block diagram, the observer equations are Observer structure ^x(t) = A^x(t) + Bu(t) + L [y(t) C^x(t)] = (A LC)^x(t) + Bu(t) + Ly(t) (O) Lecture 18: State Feedback Tracking and State Estimation p. 28

41 State Estimation From the observer state equation (O), and the plant state equation ẋ = Ax + Bu y = Cx We can obtain a state equation for the estimation error ε = x ^x ε = ẋ ^x = Ax + Bu A^x Bu LC(x ^x) = A(x ^x) LC(x ^x) = (A LC)ε ε(t) = e (A LC)t ε(0). Thus, we see that for the error to asymptotically converge to zero ε(t) 0, (and so ^x(t) x(t)) we need (A LC) to be Hurwitz. Lecture 18: State Feedback Tracking and State Estimation p. 29

42 Observer Design In summary, to build an observer, we use the matrices A, B and C from the plant and form the state equation = (A LC)^x(t) + Bu(t) + Ly(t) where L is such that the eigenvalues of (A LC) have negative real part. How to choose L? We can use, by duality, the same procedure that we already know for the design of a state feedback gain K to render A BK Hurwitz. Notice that the matrix transpose (A LC) T = A T C T L T = A dual B dual K dual We choose K dual to make A dual B dual K dual Hurwitz, and finally L = K T dual Lecture 18: State Feedback Tracking and State Estimation p. 30

43 State Estimation Example Example (Current estimation in a DC motor). We revisit the DC motor example seen earlier. Before we used state feedback to achieve robust reference tracking and disturbance rejection. This required measurement of both states: current i(t) and velocity ω(t). Suppose now that we don t measure current, but only the motor speed. We will construct an observer to estimate i(t). Recall the plant equations d ω(t) = 10 1 ω(t) + dt i(t) i(t) [ ] y = 1 0 ω(t) i(t) 0 V(t) 2 Lecture 18: State Feedback Tracking and State Estimation p. 31

44 State Estimation Example Example (continuation). We first check for observability (otherwise, we won t be able to build the observer) O = C = 1 0 CA 10 1 which is full rank, and hence the system is observable. By duality, in the procedure to compute K dual, the role of C is played by C dual = O T, and C dual is the same C dual = C = 1 α = since C only depends on the characteristic polynomial of A, which is the same as that of A T. Lecture 18: State Feedback Tracking and State Estimation p. 32

45 State Estimation Example Example (continuation). Say that the desired eigenvalues for the observer are s = 6 ± j2, (slightly faster than those set for the closed-loop plant, which is standard) which yields Kdual = s s + 40 Thus, from the coefficients of Kdual and those of (s) we have K dual = [ ] (12 12) ( ) = [ ] We now return to the original coordinates and get K dual, ] K dual = K dual CC 1 dual [0, = 19.98, by chance, the same as K dual Finally, L = K T dual = [ ]. It can be checked with MATLAB that L effectively place the eigenvalues of (A LC) at the desired locations. Lecture 18: State Feedback Tracking and State Estimation p. 33

46 State Estimation Example Example (continuation). We simulated the observer with SIMULINK, setting some initial conditions to the plant (and none to the observer) r B* u Matrix Gain1 1 s Integrator A* u C* u States Matrix Gain estimation error in speed estimation error in current Matrix Gain B* u 0.2 Estimated states 1 s L* u A L*C* u time [s] We can see how the estimation errors effectively converge to zero in about 1s. Lecture 18: State Feedback Tracking and State Estimation p. 34

47 State Estimation Example Example (Feedback from Estimated States). The observer can be combined with the previous state feedback design; we just need to replace the state measurements by the estimated states. d_i d_o r 1 78 s 1 B* u s Gain2 C* u Scope A* u B* u em 1 s L* u Gain 2 Gain1 A L*C* u Note that for the integral action part we still need to measure the real output (its estimate won t work). Lecture 18: State Feedback Tracking and State Estimation p. 35

48 State Estimation Example Example (continuation). The figure below shows the results of simulating the complete closed-loop system, with feedback from estimated states and integral action for robust reference tracking and disturbance rejection y(t) time [s] Lecture 18: State Feedback Tracking and State Estimation p. 36

49 One Tip for Lab 2 Although we will further discuss state feedback and observers, we have now all the ingredients necessary for Laboratory 2 and Assignment 3. One trick that may come handy in the state design for Lab 2 is plant augmentation. It consists of prefiltering the plant before carrying out the state feedback design. The system for Lab 2 has a transfer function of the form G(s) = k s(s + a)(s 2 + 2ζωs + ω 2 ) When the damping ζ is small, the plant has a resonance at ω, as is the case in the system of Lab 2. Lecture 18: State Feedback Tracking and State Estimation p. 37

50 One Tip for Lab 2 Control design for a system with resonances is tricky. One way to deal with them is prefilter the plant with a notch filter of the form F(s) = s2 + 2ζωs + ω 2 s ζωs + ω 2 with better damping ζ = 0.7 (say) and the same natural frequency. The notch filter cancels the resonant poles and replaces them by a pair of more damped poles. Of course, this can only be done for stable poles. The augmented plant then becomes G a (s) = G(s)F(s) = k s(s + a)(s ζωs + ω 2 ) We then do state feedback design for this augmented plant. Lecture 18: State Feedback Tracking and State Estimation p. 38

51 Summary For a plant with a controllable state-space description, it is possible to design a state feedback controller u(t) = Kx(t) that will place the closed-loop poles in any desired location. Lecture 18: State Feedback Tracking and State Estimation p. 39

52 Summary For a plant with a controllable state-space description, it is possible to design a state feedback controller u(t) = Kx(t) that will place the closed-loop poles in any desired location. State feedback control can incorporate integral action, which yields robust reference tracking and disturbance rejection for constant references and disturbances. Lecture 18: State Feedback Tracking and State Estimation p. 39

53 Summary For a plant with a controllable state-space description, it is possible to design a state feedback controller u(t) = Kx(t) that will place the closed-loop poles in any desired location. State feedback control can incorporate integral action, which yields robust reference tracking and disturbance rejection for constant references and disturbances. If the states are not all measurable, but the state-space description of the plant is also observable, then it is possible to estimate the states with an observer. Lecture 18: State Feedback Tracking and State Estimation p. 39

54 Summary For a plant with a controllable state-space description, it is possible to design a state feedback controller u(t) = Kx(t) that will place the closed-loop poles in any desired location. State feedback control can incorporate integral action, which yields robust reference tracking and disturbance rejection for constant references and disturbances. If the states are not all measurable, but the state-space description of the plant is also observable, then it is possible to estimate the states with an observer. The observer can be used in conjunction with the state feedback by feeding back estimated estates. Lecture 18: State Feedback Tracking and State Estimation p. 39

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