State variable feedback

Transcription

1 State variable feedbak We have previosly disssed systems desribed by the linear state-spae eqations Ax B y Cx n m with xt () R the internal state, t () R the ontrol inpt, and yt () R the measred otpt. Matrix A is the system or plant matrix, B is the ontrol inpt matrix, C is the otpt or measrement matrix, and is the diret feed matrix. We learned that the open-loop poles are given by the roots of ( s) si A and the open-loop transfer fntion is given by H ( s) C( si A) B. In previos letres we disssed how feedbak ontrollers (whih old be of PI-type) an be designed for the system given in transfer fntion form, sing the Root-Los method. This method allowed allation of the parameters of a PI-type ontroller sh that the poles of the losed loop system are plaed in desired positions in the s-plane orresponding to the desired losed loop performanes. It has to be mentioned here that this is a general method whih old be sed in onjntion with any sort of ontroller having a transfer fntion representation. In the following we will look a one method whih allows derivation of feedbak ontrollers sing the state desription (A,B,C,) of a system. The STATE-VARIABLE FEEBACK (SVFB) ontrol law is a basi ontrol sheme whih is based on the assmption that ALL the states of the system an be measred as otpts. Ths we wold like to determine a ontroller in the form of a gain matrix K whih allates the ontrol inpt to be sent to the system based on the measred state. Kx v For the moment we will onsider that the signal v is zero. This signal plays the role of the referene for the losed loop system, and hoosing it as 0 only means that the desired performane for the system is one of stabilization (i.e. the states of the system need to be transferred to the eqilibrim point x0). The feedbak matrix K is m n so that there are now mn ontrol loops. Assming SVFB, with Kx the losed-loop system is given as ( A BK) x Ax y ( C K) x Cx where A is the CLOSE-LOOP SYSTEM MATRIX and C is the losed-loop otpt matrix. p

2 The losed-loop poles are given by the roots of ( s) si A si ( A BK) One asks now the very good qestion: oes sh matrix K, whih stabilizes the system states, exist? Or, in other words, does this problem of finding the matrix K whih stabilizes the system dynamis have a soltion? To answer this let s remember the notion of reahability of a state variable system: The system (A,B,C) is alled reahable if the ontrol inpt an be seleted to drive any initial state to any desired final state at some final time. Ths we an solve the stabilization problem if and only if the system is reahable, sine only then the ontroller wold be able to drive any initial state of the system to the eqilibrim point. The following provides a test for reahability. A system is reahable if and only if the reahability matrix n U [ B AB... A B] has fll rank n. We also remember that reahability is eqivalent to the absene of inpt-deopling zeros. At this point, provided that the system is reahable, we wold like to proeed to deriving a methodology whih allows the allation of a matrix K whih plaes the poles of the losed loop system in desired loations given by the desired performanes of the losed loop system. SVFB Pole Plaement with Akermann's Formla In the ase of SVFB the otpt y(t) plays no role. This means that only matries A and B will be important in SVFB. We wold like to hoose the feedbak gain K so that the losed-loop harateristi polynomial ( s) si A si ( A BK) has presribed roots (whih the engineer determines based on the given desired performanes). This is alled the POLE-PLACEMENT problem. An important theorem says that the poles may be plaed arbitrarily as desired iff (A,B) is reahable. If the system is reahable, there are many tehniqes to find a sitable K that garantees stability and/or plaes the poles. One tehniqe that works for the single inpt ase, i.e. m, is ACKERMANN'S FORMULA K e U n ( A), where [ ] e n is the last row of the n n identity matrix, and (s ) is the ESIRE harateristi polynomial for the losed loop system. Note that (A ) is a MATRIX POLYNOMIAL.

3 erivation of Akermann s formla We wold like to hoose the feedbak gain K so that the losed-loop harateristi polynomial ( s) si A si ( A BK) has presribed roots, i.e. () s () s. Ths we wold like that the harateristi polynomial of the losed loop system matrix is () s (again we mention here that this polynomial is speified by the engineer as it gives the desired loations for the poles of the losed loop system). We remember Cayley-Hamilton theorem whih says that a matrix satisfies its own harateristi polynomial ( A ) 0. Starting from this ondition Akermann determined his famos formla. As a nie exerise we develop in the following the derivation of Akermann s formla for the ase of a system with 3 states. Say that the desired harateristi polynomial for the losed loop system (whih has as soltions the desired vales for the poles of the losed loop system) is given by 3 0 () s s qs qs q. Then the losed loop system matrix A A BK mst satisfy ( A) 0. Ths we an write 3 ( A) A qa qa qi 0 A ( A BK) AA ( BK) BK( A BK) A AB K BK( A BK) 3 3 ( ) [ ( )]( ) 3 A A BK A ABK BK A BK A BK A ABK ABK( A BK) BK( A BK) 3 ( A ) A ABK ABK( A BK) BK( A BK) q [ A AB K BK( A BK)] q ( A BK) qi 0 We an rewrite and arrange to obtain Bt 3 0 ( A ) A qa qa qi 3 ABK ABK( A BK) BK( A BK) A qa qa qi 0 ( A) ABqK BqK( A BK) BqK 3

4 Ths we an write qk qk ( A BK) K( A BK) ( A ) ( A) B AB AB qk K( A BK) K Remember that we started with the assmption that ( A ) 0 (i.e. the ontroller that we are looking for plaes the poles of the system in the positions given by the roots of the eqation () s 0 ) then we get qk qk ( A BK) K( A BK) ( A ) B AB AB qk K ( A BK ) 0 K whih is qk qk ( A BK) K( A BK) ( A ) B AB AB qk K ( A BK ) K We also see that B AB AB is the ontrollability matrix U B AB ABand we an write qk qk ( A BK) K( A BK) ( A ) U qk K ( A BK ) K Now, provided that the system is ontrollable (i.e. U has an inverse) we an write qk qk ( A BK) K( A BK) U ( A ) qk K ( A BK ). K Here we notie that the matrix U ( A ) has on the last line exatly the ontroller that we were set to determine. Ths determining the ontroller gain K redes to simply extrating the last line of this matrix: K [0 0 ] U ( A). And this is exatly Akermann s formla for the ase of three states. We an easily extend this to the ase of n states and see that n K [ ] U ( A) where U [ B AB... A B]. 4

5 Example The angle sbsystem of the inverted pendlm is given for some speifi vales of parameters by 0 0 Ax B x 9 0 where the state is x [ θ θ ] (angle and anglar veloity). The design speifiations are to design a state variable feedbak gain K whih will give a losed-loop POV of 4% with a settling time of τ s se. This will make the rod of the inverted pendlm balane pright. The open-loop poles are given by the roots of s ( s ) si A s 9 ( s 3)( s 3), 9 s whih are s-3, s3. The system is nstable, with natral modes e 3 t, e 3t. The design speifiations allow one to ompte the desired losed-loop poles. In fat, sine πζ / ζ τ s 5 τ 5/ α and POV 00e one may find the reqired real-part and damping ratio of the losed loop poles to be α 5, ζ / Ths, the natral freqeny is ω n α / ζ 7.07 so that the desired harateristi polynomial is s α s ω n s 0s50. To se Akermann's formla, one first verifies reahability by ompting the reahability matrix 0 U [ B AB] 0 and heking that it is indeed nonsinglar. Compting the qantities needed for Akermann's formla now yields U ( A) A 0A 50I Sbstitting now into Akermann's formla yields the reqired SVFB of K enu ( A) [ 0 ] [ 9.5 5] This solves the problem. To hek the design, one shold ompte the atal losed-loop poles sing A A BK [ 9.5 5]

6 s ( s) si A s 0s s 0 This is indeed eqal to (s). Note that for this problem where the meaning of the state variables is angle and anglar veloity one may write the P SVFB as θ Kx [ 9.5 5] x [ k k] kθ kθ θ, so that in effet a proportional-pls-derivative (P) ontrol is proded, with the proportional gain given by k and the derivative gain by k. In this ase the otpt of the system is the angle whih is desired to beome 0 (i.e. the pendlm shold be in the pright position). One now sees that this method allows allation of both the proportional and the derivative gains at the same time, whih allows plaing the poles of the system in any loation in the s-plane in a single design step. Ths state variable feedbak design offers a major advantage ompared with a root los method of design. Bt do not forget that this method reqires measrement of all the system states. One now asks another very good qestion: What do we do in the ase in whih we only have available a transfer fntion model for the system? Can we obtain a state spae model whih will allow s to se this nie method of design? The answer is yes. Even more, when going from transfer fntion model to state spae model of a system, one an obtain an infinite nmber of representations in the state-spae for the same single transfer fntion. In what follows we will look at three poplar methods of determining a state spae representation for a given transfer fntion. 6

7 Realization and anonial forms The problem of finding a state variable or blok diagram representation given a presribed transfer fntion is alled the realization problem. A transfer fntion an be realized as a blok diagram in series form or parallel form. We now introde two series forms that are very onvenient for solving the blok diagram realization problem for single-inpt/single-otpt (SISO) systems. We will then introde a parallel form realization.. Blok iagram realization of a transfer fntion series forms Consider for example a general desription of a third-order transfer fntion bs bs b0 Gs () 3 s as as a 0 For realization, it is important to ensre that highest order term in the denominator has a oeffiient of. If this is not tre then divide the nmerator and denominator by this oeffiient to pt the transfer fntion in the desired form. The transfer fntion mst also have relative degree of or more. If the relative degree is zero (e.g. same power of s in the nmerator as the denominator), then divide the denominator by the nmerator in one step of long division to write H(s) as a onstant term pls a term whose relative degree is at least one. The onstant term is a diret feedthrogh term, and the proedres below may be arried ot to realize the remainder term. A transfer fntion is said to be proper if its relative degree is greater than or eqal to zero, and stritly proper if the relative degree is greater than or eqal to one. We se a third-order system to illstrate the approah, whih works for any n-th order rational, moni, stritly proper transfer fntion. To find a B realization of H(s), divide by the highest power of s to obtain 3 3 bs bs bs 0 bs bs bs 0 Gs () 3 3 as as as ( as as as ) 0 0 Now think of Mason's Formla. To draw a B we an se three feedforward paths and three loops if we selet the orret transmissions and loop strtre. We give two series forms that have a onvenient strtre for realizing SISO systems. Note partilarly that Mason's Formla is very easy to se if there are no disjoint loops, and all loops toh all feedforward paths. Then, the determinant (s) is simply mins the sm of the loop gains, and all ofators are eqal to one. 7

8 .A. Reahable Canonial Form (RCF) Note that all loops and all feedforward paths have the left-hand integrator in ommon, so all ofators are eqal to and the determinant has no higher-order terms. Applying Mason's Formla to this B gives the transfer fntion that we onsidered. Eah integrator otpt is labeled as a state. The rle sed in this orse for labeling states will be: Label the states from right to left, from top to bottom. We will see some examples of this to larify it. With the states labeled as shown, one may write down diretly the state eqations 3 ax 3 ax ax 0 x3 x x y bx 3 bx bx 0 whih an now be arranged in the nie form 0 0 x x 0 3 a0 a a x3 x y [ b0 b b ] x x3 As an exerise, one may find the transfer fntion Gs () CsI ( A) B and verify that it is the same as the one we started with. 8

9 This development gives a very easy way to realize SISO system in state variable form. One notes that it is easy to write down the state spae representation diretly from the transfer fntion withot having to draw the blok diagram. In fat, simply take the denominator of H(s), trn the oeffiients bakwards, make them negative, and plae them into the bottom row of the A matrix. Take the oeffiients of the nmerator, trn them bakwards, and plae them into the C matrix. The A matrix in (3) is known as a bottom ompanion matrix for the harateristi polynomial The sperdiagonal 's in A and the lower in B mean simply that the three integrators are onneted in series. Look at: [ A B] a0 a a Example. Realize the given transfer fntion as reahable state variable system s s Gs () 3 s s 3s 4 The SV eqations are diretly written down as 0 0 x x x3 x y [ ] x x3 Now one may analyze the system inlding simlation, finding otpt given an inpt and initial onditions, et. 9

10 .B. Observable Canonial Form (OCF) A B satisfying this ondition is drawn below. Note that all loops and all feed forward paths have the right-hand integrator in ommon, so all ofators are eqal to and the determinant has no higher-order terms. Applying Mason's Formla to this B gives the transfer fntion that we started with. With the states labeled from right to left as shown, one may write down diretly the state eqations ax x b ax x3 b x 3 ax 0 b 0 y x whih an now be arranged in the nie form a 0x b a 0 x b 3 a0 0 0 x3 b0 x y [ 0 0 ] x x3 As an exerise, one may find the transfer fntion Gs () CsI ( A) B and verify that it is the same as the one we started with. Again, one notes that it is easy to write down the state spae desription diretly from the transfer fntion, withot having to draw the blok diagram. In fat, simply take the denominator of H(s), stak the oeffiients on end, make them negative, and plae them into the first olmn of the A matrix. Take the oeffiients of the nmerator, stak them on end, and plae them into the B matrix. 0

11 Note that this OCF state-spae form is not the same as RCF, thogh both have the same transfer fntion. In fat, RCF and OCF are related by a state-spae transformation, whih we shall not disss in this orse (it is disssed in Linear Systems). In fat the states of one representation an be written as a linear ombination of the states of the other representation. The A matrix in this ase is known as a left ompanion matrix for the harateristi polynomial. The sperdiagonal 's in A and the left-hand in C mean simply that the three integrators are onneted in series. Example 3. Realize Transfer Fntion as OCF SV System Let there be given the same transfer fntion for the system s s Gs () 3 s s 3s 4 The state variable eqations are diretly written down as 0x 3 0 x x3 x y [ 0 0 ] x x3. Blok diagram realization of transfer fntions in parallel form To realize a system in parallel form, one performs a partial fration expansion of the transfer fntion to obtain bs bs b0 k k k3 Gs () 3 s as as a0 s p s p s p3 where the poles are at p, p, p 3 and the resides are k, k, k 3. Now note that a single term of this form an be realized sing the simple blok diagram shown in the next figre. The omplete transfer fntion with three parallel paths an be realized as shown in the next figre.

12 This realization is known as parallel form. If there are repeated poles, then the transfer fntion has higher-order poles in the partial fration expansion. In this event, some parallel paths will ontain mltiple integrators. A system whih has a PFE with no higher-order poles is alled simple. The parallel form is known as Jordan Normal Form in mathematis. The ase of higher-order pole fators, orresponding to mltiple integrators in some paths, orresponds to what is known as eigenvetor hains in those paths. With the states labeled from top to bottom as shown in the figre, one may write down diretly the state eqations With the states labeled from right to left as shown, one may write down diretly the state eqations px px x 3 px 3 3 y kx kx kx 3 3 whih an now be arranged in the nie form p 0 0 x 0 p 0 x p3 x3 x y [ k k k 3] x x3 The resides an be plaed on the inpt paths in the figre above. In fat, one an split the resides between inpt and otpt paths and ths one an have

13 where p 0 0 x b 0 p 0 x b p3 x3 b3 x y [ 3] x x3 k b, i,3 i i i Note that this parallel state-spae form is not the same as RCF or OCF, thogh all three have the same transfer fntion. RCF, OCF, and the Jordan form are related by state-spae transformations, whih we shall not disss in this orse. Example 4. etermine state feedbak ontrollers sing Akermann s formla for the system s s Gs () onsidering the reahable anonial form and observable anonial form 3 s s 3s 4 realizations, sh that the poles of the losed loop system are the soltions of -5 i, -5-i and -5. The poles of the system to be ontrolled are , i and i. a. reahable anonial form 0 0 x x x3 x y [ ] x x3 K r [ ] b. observable anonial form 0x 3 0 x x3 x y [ 0 0 ] x x3 K [ ] 3

14 Notie that depending on the anonial form that was sed, a different state feedbak ontroller was determined. In order to make the state feedbak ontrol shemes work there are different states that need to be measred in eah ase. The bad news is that often times the states of the system an not be measred diretly from the system. And withot the measrements of the states one an not implement the state feedbak ontroller. The good news is that we have a model of the system available. And this model old give the vales of the states in response to the same inpt. The bad news is that even if we have a model, to obtain the orret vales for the state (at all times) we also need to know the exat vales of the initial vales of the states. And these vales may not always be known. For this reason a state estimation sheme needs to be developed sh that the estimated state gets lose to the real state of the system even in the ase of a wrong initialization of the state of the model system. This way one an implement the state feedbak ontroller sing the vales of the estimated state. 4