Chapter #6 EEE State Space Analysis and Controller Design

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1 hapter #6 EEE3-83 State Space nalsis and ontroller Design State feedack Linear Quadratic Regulator (LQR) Open and closed loop estimators Reduced Order Estimators (not assessed) racking Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /33

2 State feedack Until now we have seen how to determine the ehaviour of the sstem either solving the ODE or checking its eigenvalues. If this ehaviour is not the desired one then we have to properl control it. In this chapter we will see how we can control a sstem that is modelled in state space. he D of the state space model is: u dt D he open loop lock diagram of a F model is: U G(s) Y lassical control strateg: Output feedack R G c (s) U G(s) Y Or: Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /33

3 R Gc (s) R U G(s) Y G c (s) If the controller is a pure gain: R K R U G(s) Y K nd in the more general case: R K R U G(s) Y K We will do eactl the same with the state space model. In state space sstems we can have two kinds of feedack; the output and the state feedack (we will onl stud the state feedack method): Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 3/33

4 r ontroller u D dt he task of the controller is to produce the appropriate control signal u that will insure that =r. Let s assume the simplest form of control: u t K r t K t, where K and K are appropriatel selected gain vectors/matrices. he eact dimensions of these vectors will e defined later. Using the standard state space equation: u t K r t K t t t u t t t K r t K t t K t K r t t t Du t DK t DK rt ut the closed loop sstem is a state space model, so it must e descried a state equation: L L t t r t t t D u t L L where r= the sstem is called regulator. L K L K L DK DL DK are the closed loop matrices. If Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 4/33

5 he L state matri is a function of K, therefore appropriate changing K we can change the eigenvalues of L ) which means that we can improve s performance, i.e. to make it faster/stale. his method is called pole placement. WE MUS HEK IF HE SYSEM IS ONROLLLE. Eample 4.: ssume that we have an RL circuit: L R I di dt L he sstem is descried the equation: V ir. We want to control the sstem in such a wa that changing V we will achieve a satisfactor value of i (stead state). lso we want to control the dnamics of the sstem (oscillations, settling time, pole location ). di di R dt L dt L L he state space model: V ir i V, this implies that u, where =-R/L, =i, =/L and u=v. Oviousl the eigenvalue is R/L and the sstem will eponentiall converge to zero when () is not zero (assume that this eigenvalue is.5) and V=: V Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 5/33

6 .8 (t).6.4. If we want to make the sstem faster then we can use a state feedack control strateg. he new signal u is -K R K R K his implies that L K. he new sstem has an L L L eigenvalue at R K L R K L R 6 K 5R : L time, s. If we want to place the eigenvalue at -6R/L then.5.4 Open loop losed Loop ime, s Eample 4.: ssume 3 u (unstale sstem), ()=. We use a state feedack controller u k which implies that the losed Loop (L) state equation is 3 k, hence the eigenvalue of the L sstem is 3-k. If we want the L eigenvalues at - (stale and ver fast) then k=3: Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 6/33

7 .5 Open loop losed Loop ime, s Eample 4.3: t t u 3 4 t t Find the eigenvalues of create a controller that will stailise the sstem.. If the sstem is unstale he eigenvalues are: -.373, Hence we need to create a feedack controller to stailise the sstem. efore that we need to see if we can influence oth states, i.e. if the sstem is controllale: , since the controllailit matri has LI vectors, the sstem is controllale. o check that we can also use the determinant of the controllailit matri: 3 3. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 7/33

8 So the sstem is controllale and hence we can use a pole placement strateg. he new L state matri is: k k L k k he eigenvalues of the closed loop sstem are: k k k k eig ssume that we want to place the L poles at - and -: k k k k k 3 k 54 4k 6 3k 4 k k k k k 3 k 8 5k 6 3k 5 4k 3k 48 So a sstem: k 6 5k 3k 74 so k 7. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 8/33

9 5 4 3 Open loop losed Loop, ime, s Linear Quadratic Regulator (LQR) Previousl we saw that the necessar gains for the sstem: ( t) ( t) ut, ( t) ( t) 3 4 are 6 and 7. his will place the poles at - and - and we have a nice stale response:, time, s Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 9/33

10 Oviousl in that sstem we want to converge to zero as fast as possile. In order to quantif the phrase converge to zero as fast as possile we can use various performance indees like: I dt, Question: Wh do we need to take the asolute values? he last inde effectivel tries to minimize the area under the curves produced, : time, s time, s Oviousl the faster the sstem, the smaller the aove areas and hence the smaller the performance inde, in this case the inde has as a final value.9383 and its graph is:.8.6 I time, s Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /33

11 nother wa to quantif the speed is to use the following inde: I dt his is preferred as the graphs will e smoother: time, s time, s In this case we have I =.43:.5 I time, s In order to see that indeed this inde quantifies the speed, we see that this is reduced if we have a faster sstem. o do that we choose the pole location at: -5 -: Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /33

12 time, s time, s nd the inde I =.7.5 I time, s So someone could state that the smaller the desired eigenvalues the etter the sstem. his is not true as the smaller the eigenvalues, the higher the gains that are required. Which effectivel means that the signal u will e higher in the second case: 5 u=-k -5 - case case time, s Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /33

13 his practicall means that we need to use more energ, and more epensive equipment (for eample a higher rating dc converter to drive a D motor). In order to quantif this energ we use a similar inde as efore I u u dt : 6 I u ase ase time, s hus we see that the est sstem is the one that minimises oth indees: I dt and u I u dt I u dt or If we have a sstem where the speed is more important than the energ then we use: I q u dt In general we can use I q ru dt and we chose the positive gains q and r to denote the importance of the speed or of the energ. In general we use: I t Q t u t Ru t dt Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 3/33

14 Where Q and R are positive definite matrices (the equivalent of positive numers in matrices, the are square smmetric matrices with positive eigenvalues and other properties like Q for all nonzero ). In general Q and R can e an positive definite matrices ut we will use onl diagonal matrices where all the elements are equal. Our task: Design a controller (called Linear Quadratic Regulator - LQR) u K for u such as it is going to minimise: J Q u Rudt, where Q and R are positive definite matrices. Q: Importance of the error, R: Importance of the energ that we use. he optimum controller can e found: Not assessed material K (assume that K is stale) So: J Q K R K dt J Q K RK dt ssume that P is positive definite. d dt Q K RK P P P Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 4/33

15 replacing K and K : Q K RK K P P K K P P K and since the previous equation holds for all : Q K RK K P P K his equation can e solved onl if -K is stale and it can e shown that K R P replacing K R P into Q K RK K P P K => P P PR P Q he last equation is called Reduced Riccati Equation. Steps to design an LQR controller:. Solve P P PR P Q to find the optimum P.. Use K R P to find the optimum gain K. In general it is difficult to solve the Reduced Riccati Equation and for that reason we need to use Matla. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 5/33

16 Eample 4.4: sstem is given, Find K, the eigenvalues of -K, and the response of the sstem for R= and Q=ee() and Q=*ee() (()=[ ]). It is given that the matri P in the first case is P and in the second P We know that K R P, hence: K and K hus we see that in the nd case where we have a higher importance in the sstem s error we have higher gains. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 6/33

17 Estimating techniques Until now we assumed that we can measure all states for our state feedack laws. ut usuall we onl have and not at our disposal. If we had a perfect model of the sstem then we can use a mathematical representation of the actual sstem: u u dt Plant u dt Estimator Where, are the estimated (or calculated) state and output vector respectivel. hen we can use in the state feedack the estimated states instead of the actual ones. he error etween the estimated and real state is e t t t e t t t t u t t t u t e t e t Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 7/33

18 herefore the error is descried a homogeneous ODE. If e and has unstale eigenvalues, then the error will diverge to infinit. Even if is stale the error will depend on the dnamics of. If is slow then the error will slowl converge to zero. Oviousl one wa to quantif the estimation is to monitor the error etween the actual and the estimated state vector. Unfortunatel we cannot monitor the actual state vector (if we could then we would not need the estimator) so we have to use the error etween the actual and the estimated output. We feed this signal into the estimator: u u dt - Plant G u - dt Estimator he gain G will e used so that we have a fast and stale estimator: et G t G t G et Hence appropriatel choosing G we can force the error to converge to zero ver fast. ut the sstem must e oservale Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 8/33

19 State estimators can e designed if and onl if the oservailit condition is satisfied. he estimator design is to find G such that G is stale. his is still a pole placement prolem ut this time these poles are the estimator poles. Hence the eigenvalues of G are the poles of the closed loop estimator. Usuall the oserver poles are chosen around 5 to times higher than the closed-loop sstem, so that the state estimation is good as earl as possile. his is quite important to avoid that the oserver makes the closed-loop sstem slower. Eample 4.5: Design a closed loop estimator for the sstem represented in state space :,, ssume that the estimator poles are located at -, -5. First we need to check oservailit!! For oservailit: M O his matri is non-singular with rank and determinant is -3 hence the sstem is oservale. hen we can design an oserver for that sstem. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 9/33

20 g et Ge G g G g g g 3 4 g 3 4 g g he characteristic equation of the closed loop estimator is 3 g 4g 3 g g I G : eig 4 g g g 6 4g he closed loop poles are placed at - and -5: 3 g 4g 3 g 4g 3 g 6 4g 3 g 6 4g 3 g 6 4g 4g 3 g 74 6 g 45 g 5 3 g 4g 5 3 g 4g 3 g 5 6 4g 3 g 9 4g 5 3 g 9 4g 4g 3 g g 57 g g =.93 and g = -.45 G Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /33

21 Effect of addition of the estimator on the closed loop sstem In the pole-placement design process, we assumed that the actual state (t) was availale for feedack. In practice, however, the actual state (t) ma not e measurale, so we will need to design an estimator and use the oserved state tfor feedack. he design process, therefore, ecomes a two-stage process, the first stage eing the determination of the feedack gain matri K to ield the desired characteristic equation and the second stage eing the determination of the oserver gain matri G, to ield the desired oserver characteristic equation. Let us now investigate the effects of the use of the oserved state t, rather than the actual state (t), on the characteristic equation of a closed-loop control sstem. u u dt - Plant G u - dt Estimator -K lock diagram of the overall sstem with the controller and the closed loop estimator Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /33

22 onsider the completel state controllale and oservale state space sstem: ( t) ( t) u( t) ( t) ( t) Du( t). he state feedack control using the estimated state vector is: ut With this control the state equation ecomes: K t. where the error et ( t) ( t) K t K ( t) K ( t) t was defined as the difference etween actual and estimated state et t t. Hence ( t) K ( t) K ( t) t K ( t) Ke t he estimator error equation is also given : et Ge. Hence the dnamics of the oserved-state feedack control sstem can e descried : t K K t e t G e t he characteristic equation of the overall sstem is: si ( K) K si G si ( K) si G he closed loop poles of the oserved-state feedack control sstem consist of the poles of the pole placement controller alone and the poles of the Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk /33

23 estimator alone. his means that the pole placement design and the estimator design are independent of each other. hen the can e designed separatel and comined to form the oserved-state feedack control sstem. Reduced Order Estimators (ROE) Not assessed Material In the previous case the estimator was estimating all the states of the controller. On the other hand, we usuall want to estimate just some of the states since the others can e measured. With this wa we have a simpler oserver and the estimation process is etter. he estimator or oserver that is estimating some states is called Reduced Ordered Estimator. ssume that we can onl measure one state (as it is usuall the case), then we can use a different approach for the estimator design. Imagine that the onl state that can e measured is a. he rest of the state vector X cannot e measured: X a. partitioning the state space X equations we can have: a aa a a a U X a X a Y X where aa is a scalar numer a is (n-) vector a is (n-) vector is (n-) (n-) matri a is scalar is (n-) vector hen a new state space model can e defined the non-measurale state: X X aa U In the last eqn. the matri is the new state matri and a a U is the new known input. he known state is: a Y aa a a X a U Y aa a a U a X Y n n X n a X Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 3/33

24 Y t the last eqn. the term aa a au is a known measurement. Hence the new state space model will have states: Xn=X, state matri n=, Input Un= a a U, n= a and the output is Y n Y aa a a U X( t) X( t) U( t) X Y( t) X( t) Y Hence the estimator eqn. is: ~ ~ ~ X X U GY Y ~ ~ Y X ~ X ~ ~ X a a n n n n X X n n U GY U aa a n X Y a n U X Y a ~ X aa or: X G X G G U G Y a ~ ~ X a a nd hence the estimating error: ~ a G X G Y G a E aa aa a a a U GY E G o avoid differentiating the Y we can define as a new state: X hence the state eqn. is: ~ X c G X G Y G U a Notice that the term is: ~ a aa a a a ~ c X a a U GY G Y is not at the LHS. he lock diagram of this ROE U a X Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 4/33

25 U U X dt X Y X Plant Y Y a -G aa G -G a ~ G Y X X c dt Xc X ~ Estimator -G a With this method the gain matri G is found from: G=place(, a,pe) G=G Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 5/33

26 Revision & racking he lock diagram of a state space sstem is: u u dt With equations: u, he qualitative properties of the sstem depend on the eigenvalues of. In realit we can onl access/measure the output vector. So unless the output matri is the identit matri we need to estimate the state vector (either for monitoring or if we want to stailise the sstem using a state feedack controller). he simplest case is an open estimator, i.e. a mathematical model of the sstem. We assume that we have a linear sstem and we perfectl know the sstem s matrices (, and ): Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 6/33

27 u u dt Plant u dt Estimator Estimated State Vector We have proved that the error dnamics etween the actual state vector and the estimated state vector are given : e e. Hence if e and the state matri has unstale eigenvalues the error will diverge to infinit. o overcome this prolem we create a closed loop estimator (the sstem under stud is still open loop): u u dt - Plant G u - dt Estimator Estimated State Vector Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 7/33

28 hen the error dnamics of the estimator will e given : e Ge Note: If the sstem is unstale then the response of the estimator will also e unstale ut at the same rate as the original sstem such as their difference converges to zero. Now, we can design a pole placement controller to stailise the sstem: u u dt - Plant G u - dt Estimator -K So in this case the estimator must e a lot faster than the original sstem s dnamics (governed -K and hence K) so that the error etween the actual state vector and estimated state vector will converge to zero ver fast and hence the estimated state vector that is eing used to control the actual sstem and the estimator (the have the same input) will e the correct one. his means that the original sstem has dnamics: K ut if ver fast then we have K Note: o stailise the sstem we can also use an LQR controller. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 8/33

29 Now let s assume that we want to converge to a nonzero value. If we just change the diagram to: r ss u u dt - Plant G u - dt Estimator -K hen while the sstem will e stale (from the state feedack controller) it will e proal not converge to rss. o avoid that and assuming that we have a D input (for SISO sstems):,, u u ss ss ss hus we can sa that if u u K ss ss then ss u u as ss. (the gain matri K can e found pole placement or LQR). ss uss he state equation at the stead state is: ss ss Now we can define new vectors: N r, u N r and hence our prolem is transformed into find these vectors. ss ss ss u ss Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 9/33

30 ss uss N rss Nurss N Nu ss ss rss N rss N N N N u N u lso u u K N r K N r K N KN r K K r ss ss u ss ss u ss ss nd thus the lock diagram is: r ss K u u dt - Plant G u - dt Estimator -K Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 3/33

31 Eample 4.6: Previousl we saw that the necessar gains for the sstem: ( t) ( t) ut, ( t) ( t) 3 4 are K=[6, 7] If we assume that =[ ] then the matri is: N Hence... N u , therefore: N, Nu K Nu KN Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 3/33

32 utorial 4. sstem is descried,, 5 i. Is it stale? ii. Find the sstem s response (()=, ()=) and hence crosscheck our answer. iii. Stailise the sstem using a pole placement controller.. Repeat eercise for,, sstem is descried,, 5 i. Is it stale? ii. Find the sstem s response (()=, ()=) and hence crosscheck our answer. iii. Use a state feedack controller that will force the sstem to a stead state in ~.5s. iv. Use a state feedack controller that will force the sstem to a stead state in ~.5s. v. Plot the signal K in the previous cases. What do ou notice? vi. For the previous sstem design an LQR controller that will force: a) he speed to e times more important than the energ used. ) he speed to e times less important than the energ used. 4. Repeat question 3 for,, sstem is descried,, i. Find the sstem s response (()=, ()=) and comment on the output signal.. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 3/33

33 ii. Use an open loop estimator to estimate the states when the initial error etween the actual and the estimated states is.. Plot the estimation error. Under what conditions can we use the estimated states in a pole placement control law? Simulate a case where we can use them and a case where we cannot use them Repeat the previous question when,, 5 7. Use a closed loop estimator for,,.. Place the poles of the estimator to [- -]. 8. reate different pole placement control strategies and then use the aove estimator: i. One with desired closed loop poles at [-5-6] ii. One with desired closed loop poles at [-5-6] Repeat question 3 when,,. Module Leader: Dr Damian Giaouris - damian.giaouris@ncl.ac.uk 33/33

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