Controllability. Chapter Reachable States. This chapter develops the fundamental results about controllability and pole assignment.

Transcription

1 Chapter Controllability This chapter develops the fundamental results about controllability and pole assignment Reachable States We study the linear system ẋ = Ax + Bu, t, where x(t) R n and u(t) R m Thus A R n n and B R n m We begin our study of controllability with a question: What states can we get to from the origin by choice of input? Are there states in R n that are unreachable? This is a question of control authority For example, if B is the zero matrix, there is no control authority, and if we start x at the origin, we ll stay there By contrast, if B = I it will turn out that we can reach every state Fix a time t > We say a vector v in R n is reachable (at time t ) if there exists an input u( ) that steers the state from the origin at t =tov at t = t To characterize reachability we have to recall the integral form of the above di erential equation It was derived in ECE that x(t) =e ta x() + If x() = and t = t,then x(t )= Z t e (t Z t e (t )A Bu( )d )A Bu( )d Thus a vector v is reachable at time t i (9u( )) v = Z t e (t )A Bu( )d Let us introduce the space U of all signals u( ) defined on the time interval [,t ] The smoothness of u( ) is not really important, but to be specific, let s assume u( ) is continuous Also, let us introduce the reachability operator R : U!R n, Ru = Z t e (t )A Bu( ) d

2 8 CHAPTER CONTROLLABILITY In words, R is the linear transformation (LT) that maps the input signal to the state at time t starting from x() = The LT R is not the same as a matrix because U isn t finite dimensional But its image is well defined and Im R is a subspace of R n, as is easy to prove It s time now to introduce the controllability matrix W c = B AB A n B In the single-input case, B is n and W c is square, n n The importance of this matrix comes from the theorem to follow Note The object B is a matrix However, associated with it is an LT, namely the LT that maps a vector u to the vector Bu Instead of introducing more notation, we shall write Im B for the image of this LT That is, the symbol B will stand for the matrix or the LT depending on the context Likewise for other matrices such as A and W c Theorem Im R =ImW c, ie, the subspace of reachable states equals the column span of W c Let s postpone the proof and instead note the conclusion: A vector v is reachable at time t i it belongs to the column span of W c,ie, rank W c = rank W c v Notice that reachability turns out to be independent of t rank W c = n Also, every vector is reachable i Example Consider this setup of carts, forces: y y The equations are u u M M K M ÿ = u + K(y y ), M ÿ = u + K(y y ) Taking the state x =(y, ẏ,y, ẏ ) and M =,M =/,K = we have the state model A =, B = We compute using Scilab/MATLAB (or by hand) that W c is 8 and its rank equals Thus every state is reachable from the origin That is, every position and velocity of the carts can be

3 REACHABLE STATES 9 produced at any time by an appropriate open-loop control In this sense, two forces gives enough control authority Example Now consider carts, common force: Now u = B = y y u M M K u and u = u SoA is as before, while Then W c = The rank equals The set of reachable states is the -dimensional subspace spanned by the first two columns So we don t have complete control authority in this case Example Two pendula balanced on one hand: M M L L d The linearized equations of motion are M ( d + L ) = M g M ( d + L ) = M g

4 CHAPTER CONTROLLABILITY Take the state and input to be x =(,, ), u = d Find W c and show that every state is reachable i L = L The proof of the theorem requires the Cayley-Hamilton theorem, which we discuss now Consider the matrix A = Its characteristic polynomial is s + s + Substitute s = A into this polynomial, regarding the constant as s You get the matrix A + A + I Verify that this equals the zero matrix: A + A + I = Thus A is a linear combination of {I,A} Likewise for higher powers A,A etc Theorem If p(s) denotes the characteristic polynomial of A, then p(a) = Thus A n is a linear combination of lower powers of A Proof Here s a proof for n = ; the proof carries over for higher n Wehavetheidentity (si A) = p(s) N(s), where p(s) =det(si A) and N(s) is the adjoint of si A We can manipulate this to read p(s)i =(si A)N(s) Say p(s) =s + a s + a s + a ThenN(s) must have the form s I + sn + N,whereN,N are constant matrices Thus we have (s + a s + a s + a )I =(si A)(s I + sn + N ) Equating coe cients of powers of s, we get a I = N A a I = N AN a I = AN Multiply the first equation by A and the second by A, and then add all three: You get a A + a A + a I = A,

5 REACHABLE STATES or A + a A + a A + a I = Proof of Theorem We first show Im R Im W c Let v Im R Then 9u U such that v = Ru, ie, v = = = B Z t e (t Z t Z t )A Bu( ) d I +(t )A + (t ) A +! u( )d + AB Z t Thus v belongs to the column span of B AB A B (t )u( )d + Bu( ) d By the Cayley-Hamilton theorem, the column span terminates at B AB A n B Now we show Im W c Im R An equivalent condition is in terms of orthogonal complements: (Im R)? (Im W c )? Soletv be a vector orthogonal to Im R Thus for every u( ) That is, Z t Z t v T e (t )A Bu( )d = v T e (t )A Bu( )d = Since this is true for every u( ), it must be that v T e (t )A B = 8, t This implies that and hence that Thus v T e ta B = 8t, t t v T I + ta + t A + v T B =, v T AB =, B = 8t, t t

6 CHAPTER CONTROLLABILITY That is, v is orthogonal to every column of W c An auxiliary question is, if v is a reachable state, what control input steers the state from x() = to x(t )=v? That is, if v Im R, solve v = Ru for u There are an infinite number of u because the nullspace of R is nonzero We can get one input as follows We want to solve v = Z t e (t )A Bu( )d for the function u Without any motivation, let s look for a solution of the form u( ) =B T e (t )A T w, where w is a constant vector Then the equation to be solved is v = Z t e (t )A BB T e (t )A T wd Now w can be brought outside the integral: v = Z t e (t )A BB T e (t )A T d w In square brackets is a square matrix: L c = Z t e (t )A BB T e (t )A T d If L c is invertible, we re done, because w = Lc v It can be shown that L c is in fact invertible if W c has full rank Recap: The set of all states reachable starting from the origin is a subspace, the image (column span) of W c, the controllability matrix Thus, if this matrix has rank n, every state is reachable If a state is reachable at some time, then it s reachable at any time Of course you ll have to use a big control if the time is very short Finally, we say that the pair of matrices (A, B) isacontrollable pair if every state is reachable, equivalently, the rank of W c equals n Go over the examples in this section and see which are controllable Properties of Controllability Invariance under change of basis The state vector of a system is certainly not unique For example, if x is a state vector, so is Vx for any square invertible matrix V Suppose we have the state model ẋ = Ax + Bu y = Cx + Du

7 PROPERTIES OF CONTROLLABILITY and we define a new state vector, x = Vx The new equations are x = V AV x + VBu y = CV x + Du Under the change of state x! Vx,theA, B matrices change like this (A, B)! (V AV,VB) and the controllability matrix changes like this B AB A B! V B AB A B Thus, (A, B) is controllable i (V AV,VB) is controllable A transformation of the form A! V AV is called a similarity transformation; wesaya and V AV are similar Invariance under state feedback Consider applying the control law u = Fx+ v to the system ẋ = Ax + Bu Here F R m n and v is a new independent input The new state model is ẋ =(A + BF)x + Bv That is, if u = Fx+ v (or u! Fx+ u), then (A, B)! (A + BF,B) Notice that the implementation of such a control law requires that there be a sensor for every state variable To emphasize this, consider the maglev example Think of the sensors required to implement u = Fx+ v You can prove that under state feedback, the set of reachable states remains unchanged; controllability can be neither created nor destroyed by state feedback Decomposition If we have a model ẋ = Ax + Bu where (A, B) isnot controllable, it is natural to try to decompose the system into a controllable part and an uncontrollable part Let s do an example Example carts, force We had these matrices: A =, B = And the controllability matrix is W c =

8 CHAPTER CONTROLLABILITY The rank equals The set of reachable states is the -dimensional subspace spanned by the first two columns Let {e,e } denote these two columns; thus {e,e } is a basis for Im W c Add two more vectors to get a basis for R,say e =(,,, ), e =(,,, ) Now, the matrix A is the matrix representation of an LT in the standard basis Write the matrix of the same LT but in the basis {e,,e } As you recall, you proceed as follows: Write Ae in the new basis and stack up the coe cients as the first column: Ae = e So the first column of the new matrix is (,,, ) Repeat for the other basis vectors The result is the matrix There s a more streamlined way to describe this transformation Form the matrix V by putting {e,,e } as its columns: V = Now transform the state via x = V x ThenA, B transform to V AV, V B: V AV =, V B = These matrices have a very nice structure, indicated by the partition lines: V AV =, V B = Let us write the blocks like this: V AV = A A A A, V B = B B Thus, the state x has been transformed to x = V x, and the state equation ẋ = Ax + Bu to x = A x + A x + B u x = A x + A x + B u

9 THE PBH (POPOV-BELEVITCH-HAUTUS) TEST Note these key features: B =, A =, and (A,B ) is controllable With these, the model is actually x = A x + A x + B u x = A x In this form, x represents the uncontrollable part of the system the second equation has no input And x represents the controllable part There is coupling only from x to x Now we describe the general theory The matrix W c is the controllability matrix and the subspace Im W c is the set of reachable states It s convenient to rename this subspace as the controllable subspace Now, if we have a model ẋ = Ax + Bu where (A, B) isnot controllable, we can decompose the system into a controllable part and an uncontrollable part The decomposition construction follows the example Let {e,,e k } be a basis for Im W c Complement it to get a full basis for state space: {e,,e k,,e n } Let V denote the square matrix with columns e,,e n Then V AV = A A A, V B = B Furthermore, (A,B ) is controllable The lower-left block of the new A equals zero because Im W c is A-invariant; the lower block of the new B equals zero because Im B Im W c The PBH (Popov-Belevitch-Hautus) Test If we have the model ẋ = Ax + Bu and we want to check if (A, B) is controllable, that is, if there is enough control authority, we can check the rank of W c This section develops an alternative test that is frequently more insightful Observe that the n n matrix A I is invertible i is not an eigenvalue In other words rank(a I) = n () is not an eigenvalue The PBH test concerns the n (n + m) matrix A I B Theorem (A, B) is controllable i rank A I B = n 8 eigenvalues of A Proof (=)) Assume rank A I B <nfor some eigenvalue Then there exists x = (x will be complex if is) such that x A I B =

10 CHAPTER CONTROLLABILITY (ie, x? every column of A I B ), where * denotes complex-conjugate transpose Thus x A = x, x B = So x A = x A = x etc ) x A k = k x Thus x B AB A n B = x B x B n x B = So (A, B) is not controllable ((=) Assume (A, B) is not controllable As in the preceding section, there exists V such that (Ã, B) =(V AV, V B) have the form à = A A A, B = B Thus rank [à I B] <nfor an eigenvalue of A So rank [A I B] <nsince à I B = V A I B V I In view of this theorem, it makes sense to define an eigenvalue of A to be controllable if rank A I B = n Then (A, B) is controllable i every eigenvalue of A is controllable Example carts, force We had these matrices: A =, B = The eigenvalues of A are {,, ± p j} The controllable ones are {± p j}, which are the eigenvalues of A, the controllable part of A

11 CONTROLLABILITY FROM A SINGLE INPUT Controllability from a Single Input In this section we ask, when is a system controllable from a single input? That is, we consider (A, B) pairs where A is n n and B is n A matrix of the form a a a n a n is called a companion matrix Its characteristic polynomial is s n + a n s n + + a s + a Companion matrices arise naturally in going from a di erential equation model to a state model Example Consider the system modeled by y + a ÿ + a ẏ + a y = b ü + b u + b u The transfer function is b s + b s + b s + a s + a s + a and then the controllable realization is A =, B = a a a C = b b b, D = Notice that if A is a companion matrix, then there exists a vector B such that (A, B) is controllable, namely, B = Example The controllability matrix of A =, B = a a a

12 8 CHAPTER CONTROLLABILITY is W c = B AB A B = a a a + a The rank of W c equals, so (A, B) is controllable Now we ll see that if (A, B) is controllable and B is n, then A is similar to a companion matrix Theorem Suppose (A, B) is controllable and B is n Let the characteristic polynomial of A be s n + a n s n + + a Define à = a a a n a n, B = Then there exists a W such that W AW = Ã, W B = B Proof Assume n = to simplify the notation The characteristic poly of A is s + a s + a s + a By Cayley-Hamilton, Multiply by B: Hence A + a A + a A + a I = A B + a A B + a AB + a B = A B = a B a AB a A B Using this equation, you can verify that A B AB A B = B AB A B a a a ()

13 CONTROLLABILITY FROM A SINGLE INPUT 9 Define the controllability matrix W c := B AB A B and the new matrix M = a a a Note that M T is the companion matrix corresponding to the characteristic polynomial of A From () we have W c AW c = M Regarding B, wehave B = B AB A B so W c B =, () () Now define à = a a a, B = These are the matrices we want to transform A, B to Define their controllability matrix W c := B à B à B Then, as in () and (), W c W c à W c = M, (same M!) () B = () From (), () and (), (), c AW c = W c à W c, W c B = W c B W Define W = W c W c Then W AW = Ã, W B = B Let s summarize the steps to transform (A, B) to(ã, B) :

14 CHAPTER CONTROLLABILITY Procedure Step Find the characteristic poly of A: s n + a n s n + + a Step Define W c = B AB A n B, Wc = B Ã B Ã n B, W = Wc W c Then W AW = Ã, W B = B Example A = 9, B = char poly A = s s s + Ã =, B = W c = W = 8, Wc = Pole Assignment This section presents the most important result about controllability: That eigenvalues can be arbitrarily assigned by state feedback This result is very useful and is the key to state-space control design methods The result was first proved by WM Wonham, a professor in our own ECE department As we saw, the control law u = Fx+ v transforms (A, B) to(a + BF,B) This leads us to pose the pole assignment problem: Given A, B Design F so that the eigs of A + BF are in a desired location It is customary to call the eigenvalues of A its poles Of course this more properly refers to the poles of the relevant transfer function

15 POLE ASSIGNMENT We might like to specify the eigenvalues of A + BF exactly, or we might be satisfied to have them in a specific region in the complex plane, such as a truncated cone for fast transient response and good damping: desired poles region The fact is that you can arbitrarily assign the eigs of A + BF i (A, B) is controllable We ll prove this, and also get a procedure to design F Single-Input Case Example A =, B = This A is in companion form, and also B is as in Theorem Let us design F = F F F to place the eigs of A + BF at WehaveA + BF is a companion matrix with final row +F +F +F Thus its characteristic poly is s +( F )s +( F )s +( F ) But the desired char poly is (s + )(s + )(s + ) = s +s + s + Equating coe cients in these two equations, we get the unique F : F = The example extends to general case of A a companion matrix with final row a a n

16 CHAPTER CONTROLLABILITY and B = as follows: Let {,, n} be the desired set of eigenvalues, occurring in complex-conjugate pairs The matrix F has the form F = F F n,soa + BF is a companion matrix with final row a + F a n + F n Equate the coe cients of the two polynomials s n +(a n F n )s n + +(a F ), (s ) (s n), and solve for F,,F n Let us turn to the general case of (A, B) controllable, B is n The procedure to compute F to assign the eigenvalues of A + BF is as follows: Step Using Theorem, compute W so that à := W AW, B := W B have the form that à is a companion matrix and B = Step Compute F to assign the eigs of à + B F to the desired locations Step Set F = FW To see that A + BF and à + B F have the same eigs, simply note that W (A + BF)W = à + B F Example pendula We had x =, u = d

17 POLE ASSIGNMENT Taking L =, L =/, g =, we have A =, B = Suppose the desired eigs are {,, ± j} Step eigs A = n ± p, ± p o char poly A = s s + à = W c = W c =, B = W = W W c = Step Desired char poly is (s + ) (s + j)(s ++j) =s +s + s + s + Char poly of à + B F is s F s +( F )s F s + ( F ) Equate coe s and solve: F = 9 Step

18 CHAPTER CONTROLLABILITY F = FW = This result is the same as via the MATLAB command acker We have now proved su ciency of the single-input pole assignment theorem: Theorem The eigs of A + BF can be arbitrarily assigned i (A, B) is controllable The direction (=)) is left as an exercise It says, if (A, B) is not controllable, then some eigenvalues of A + BF are fixed, that is, they do not move as F varies Note that in the single-input case, F is uniquely determined by the set of desired eigenvalues Multi-Input Case Now we turn to the general case of (A, B) whereb is n m, m> As before, if (A, B) is not controllable, then some of the eigs of A + BF are fixed; if (A, B) is controllable, the eigs can be freely assigned To prove this, and construct F, we first see that the system can be made controllable from a single input by means of a preliminary feedback Lemma (Heymann 98) Assume (A, B) is controllable Let B be any nonzero column of B (could be the first one) Then 9F such that (A + BF,B ) is controllable Before proving the lemma, let s see how it s used Theorem (Wonham 9) The eigs of A+BF are freely assignable i (A, B) is controllable Proof (=)) For you to do ((=) Step Select, say, the first column B of B If(A, B ) is controllable, set F = and go to Step Step Choose F so that (A + B F,B ) is controllable Step Compute F so that A + B F + B F has the desired eigs Step Set F = F + F Then A + BF = A + B F + B F

19 POLE ASSIGNMENT Two comments: A random F will work in Step ; In general F is not uniquely determined by the desired eigenvalues Now we turn to the proof of the lemma Proof (Hautus ) The proof concerns an artificial discrete-time state model, namely, x(k + ) = Ax(k)+Bu(k), x() = B () Suppose we can prove this: There exists an input sequence {u(),,u(n )} such that the resulting states {x(),,x(n)} span R n, that is, they are linearly independent Then define F as follows: u(n) = anything F x() x(n) = u() u(n) () To see that (A + BF,B ) is controllable, note from () and () that so x(k + ) = (A + BF)x(k), x() = B, x() x(n) = B (A + BF)B (A + BF) n B Thus the controllability matrix of (A + BF,B ) has rank n So it su ces to show, with respect to (), that there exist {u(),,u(n )} such that the set {x(),,x(n)} is lin indep This is proved by induction First, {x()} = {B } is lin indep since B = For the induction hypothesis, suppose we ve chosen {u(),,u(k )} such that {x(),,x(k)} is lin indep Define V =Im x() x(k) If V = R n we re done So assume V= R n (8) We must now show there exists u(k) such that x(k + ) / V (Note that x(k + ) / Vis equivalent to {x(),,x(k + )} is lin indep) That is, we must show (from ()) (9u(k)) Ax(k) +Bu(k) / V (9) Suppose, to the contrary, that (8u R m ) Ax(k)+Bu V Setting u = we get Ax(k) V ()

20 CHAPTER CONTROLLABILITY Then we also get (8u R m ) Bu V () Now we will show that AV V, () ie, Ax(i) V, i =,,k This is true for i = k by (); for i<kwe have from () that Ax(i) = x(i + ) Bu(i) where x(i + ) V by definition V, and Bu(i) V by () This proves () Using () and () we get that 8u R m ABu AV V A Bu A V AV V etc Thus every column of the controllability matrix of (A, B) belongs to V Hence Im W c V But Im W c = R n, by controllability, so V = R n This contradicts (8), so (9) is true after all Finally, the following test for controllability is fairly good numerically: compute the eigs of A choose F at random compute the eigs of A + BF Then (A, B) is controllable () {eigs A} and {eigs A + BF} are disjoint Stabilizability Recall that A is stable if all its eigenvalues have negative real parts This is the same as saying that for ẋ = Ax, x(t)! as t!for every x() Under state feedback, u = Fx+ v, thesystem ẋ = Ax + Bu is transformed to ẋ =(A + BF)x + Bv We say (A, B) isstabilizable if 9F such that A + BF is stable By the pole assignment theorem (A, B) controllable =) (A, B) stabilizable, ie, controllability is su cient (a stronger property) What exactly is a test for stabilizability? The following theorem answers this Theorem The following three conditions are equivalent:

21 PROBLEMS (A, B) is stabilizable The uncontrollable part of A, A, is stable rank A I B = n for every eigenvalue of A with Re The carts, force example is not stabilizable, because the uncontrollable eigenvalues,,, aren t stable Suppose (A, B) is stabilizable but not controllable, and we want to find an F to stabilize A+BF The way is clear: Transform to see the controllable part and stabilize that Here are the details: Let {e,,e k } be a basis for Im W c Complement it to get a full basis for state space: {e,,e k,,e n } Let V denote the square matrix with columns e,,e n Then V AV = A A A, V B = B, where (A,B ) is controllable Choose F to stabilize A + B F Then A A A + B F A + B = F A A is stable Transform the feedback matrix back to the original coordinates: F = F V Problems Write a Scilab/MATLAB program to verify the Cayley-Hamilton theorem Run it for A = Consider the state model ẋ = Ax + Bu with A =, B = (a) Find a basis for the controllable subspace (b) Is the vector (,, ) reachable from the origin? (c) Find a nonzero vector that is orthogonal to every vector that is reachable from the origin Show that the matrix A = nonzero matrices B has the property that (A, B) is controllable for all

22 8 CHAPTER CONTROLLABILITY Consider the setup of two identical systems with a common input: u - S - S (An example is two identical pendula balanced on one hand) Let the upper and lower systems be modeled by, respectively, ẋ = Ax + Bu, ẋ = Ax + Bu Assume (A, B) is controllable (a) Find a state model for the overall system, taking the state to be x = (b) Find a basis for the controllable subspace of the overall system Consider the state model ẋ = Ax + Bu with A =, B = x x (a) Transform x, A, and B so that the controllable part of the system is exhibited explicitly (b) What are the controllable eigenvalues? Consider the system with A =, B = (a) What are the eigenvalues of A? (b) It is desired to design a state-feedback matrix F to place the eigenvalues of A + BF at {,, ± j} Is this possible? If so, do it (c) It is desired to design a state-feedback matrix F to place the eigenvalues of A + BF at {,, ± j} Is this possible? If so, do it Consider a pair (A, B) wherea is n n and B is n (single input) Show that (A, B) cannot be controllable if A has two linearly independent eigenvectors for the same eigenvalue 8 Consider the two pendula Find the controllable subspace when the lengths are equal and when they re not

23 PROBLEMS 9 9 Continue with the two-pendula problem Give numerical values to L = L and stabilize by state feedback Let A =, B = (a) Check that (A, B) is controllable but that (A, B i ) is not controllable for either column B i of B (b) Compute an F to assign the eigenvalues of A + BF to be ± j, ± j Let A be an n n real matrix in companion form Then (A, B) is controllable for a certain column vector B What does this imply about the Jordan form of A? Consider the system model ẋ = Ax + Bu, x() = with A =, B = 9 Does there exist an input such that x() = (,, )? Consider the system model ẋ = Ax + Bu with A =, B = (a) It is desired to design a state feedback matrix F so that the eigenvalues of A + BF all have negative real part Is this possible? (b) Is it possible to design F so that each eigenvalue of A + BF satisfies Re? Consider the following -cart system: y y u u M M K There are two inputs: a force u applied to the first cart; a force u applied to the two carts in opposite directions as shown Take the state variables to be y,y, ẏ, ẏ in that order, and take M =,M =,K =

24 CHAPTER CONTROLLABILITY (a) Find (A, B) in the state model (b) According to the Cayley-Hamilton theorem, A can be expressed as a linear combination of lower powers of A Derive this expression for the -cart system Many mechanical systems can be modeled by the equation M q + D q + Kq = u, where q is a vector of positions (such as joint angles on a robot), u is a vector of inputs, M is a symmetric positive definite matrix, and D and K are two other square matrices Find a q state model by taking x = What can you say about controllability of this state model? q (a) Let A =, B = Find F so that the eigenvalues of A + BF are ± j, ± j (b) Take the same A but B = Find the controllable part of the system Consider the system ẋ = x + u Is the vector (,, ) reachable from the origin? 8 Is the following (A, B) pair controllable? A = 8, B = 9 Let A =, B =

25 PROBLEMS (i) Check that (A, B) is controllable (ii) Find a feedback law u = Fx such that the closed-loop poles are all at Prove that (A, B) is controllable if and only if (A + BF,B) is controllable for some F Prove that (A, B) is controllable if and only if (A + BF,B) is controllable for every F