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

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

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

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

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,

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

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

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 =

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

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 =

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

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

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 ()

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) :

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

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

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

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

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

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 ()

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:

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

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

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 =

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 =

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