Lecture 7 and 8. Fall EE 105, Feedback Control Systems (Prof. Khan) September 30 and October 05, 2015

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1 1 Lecture 7 and 8 Fall EE 105, Feedback Control Systems (Prof Khan) September 30 and October 05, 2015 I CONTROLLABILITY OF AN DT-LTI SYSTEM IN k TIME-STEPS The DT-LTI system is given by the following n-dimensional state-space: where: x k+1 = Ax k + Bu k, (1) The state vector, x k, lies in R n, ie, there are n state variables; The input vector, u k, lies in R m, ie, there are m inputs; The above naturally lead to A R n n and B R n m ; The matrix A is called the system matrix and the matrix B is called the input matrix Definition 1 (Controllability): A DT-LTI system is said to be controllable in n time-steps when there exists a sequence of inputs, u 1,, u n 1, such that x n = 0 regardless of the initial condition, x 0, where 0 is a vector with n zeros In other words, using the control inputs, u 1,, u n 1, we can force any arbitrary initial condition, x 0, to go to 0 in n time-steps In the following, let us assume the extreme case when there is only 1 control input: u k R (a lowercase italic represents a scalar) Naturally this forces the input matrix, B, to lie in R n 1 A Controllability in n time-steps The n-dimensional DT-LTI state-space is given by: x k+1 = Ax k + Bu k, k = A k+1 x 0 + A m Bu k m (2) We are interested in studying the controllability in n time-steps, ie, we would like x n = 0, where 0 is a vector with n zeros From Eq (2), we have 0 = x n = A n x 0 + A n x 0 = n 1 m=0 m=0 n 1 m=0 A m Bu n 1 m A m Bu n 1 m,

2 2 Subsequently, A n x 0 = n 1 m=0 A m Bu n 1 m, = Bu n 1 + AB + A 2 Bu + + A n 2 Bu 1 + A n 1 B u n 1 u = B AB A 2 B A n 2 B A n 1 B, (3) }{{} C n }{{} u 0,,n 1 = C n u 0,,n 1 (4) Let us consider the controllability matrix, C n The first element B is n 1, the second element AB is also n 1, and the last (nth) element is also n 1 Hence, controllability matrix, C, is n n Similarly, the sequence of n control inputs, u 1,, u n 1, is compactly written as the vector, u 0,,n 1 ; note that it is n 1 From the linear system of equations, it is clear that the following system of equations A n x 0 = C n u 0,,n 1, (5) has a solution when C n is invertible However, it is easy to verify that for an arbitrary input matrix, B R n m, ie, with m control inputs, the controllability matrix, C n has the dimensions n mn being a square matrix only when m = 1 What we require is to guarantee the existence of some u 0:,n 1 such that Eq (5) holds Clearly, if A n x 0 R(C n ), then there exists a solution, where R(C n ) is the range space of C n Because we are allowing arbitrary values for both A and x 0, the only way to guarantee the existence condition is for R(C n ) = R n, or, equivalently, for rank(c n ) = n Hence, the necessary condition for Eq (5) to have a solution is rank(c n ) = n (6) This is because an n mn matrix cannot have a rank greater than n The above exposition can be summarized as the following Conclusion: For any DT-LTI system with one control input, there exists some sequence of n control inputs, u 1,, u n 1, such that any arbitrary initial condition, x 0, can be forced to 0 in n time-steps; when ( ) rank(c n ) = rank B AB A n 1 B = n (7) Such a system is said to be controllable by Def 1 u 1

3 3 B Controllability in less than n time-steps For a DT-LTI system with one control input, recall from the previous section (Eq (3)) that u n 1 u A n x 0 = B AB A 2 B A n 2 B A n 1 B u 1 If we want the system to be controllable in n 1 time steps, then the above equation leads to A n 1 x 0 = u B AB A 2 B A n 2 B, }{{} u C 1 n 1 }{{} u 0,,n 2 ie, with n 1 control inputs, u 1,, The new matrix, C n 1, has n rows but only n 1 columns So this system of linear equations does not have a solution since the rank cannot exceed that minimum of the number of rows and columns, ie, rank(c n 1 ) min(n, n 1) = n 1 < n (8) Conclusion: Any DT-LTI system is not controllable with one control input in (strictly) less than n time-steps

4 4 C Controllability in more than n time-steps For a DT-LTI system with one control input, recall from the previous section (Eq (3)) that u n 1 u A n x 0 = B AB A 2 B A n 2 B A n 1 B u 1 If we want the system to be controllable in n + 1 time steps, ie, with n + 1 control inputs, then the above equation leads to A n+1 x 0 = B AB A 2 B A n 2 B A n 1 B A n B }{{} C n+1 u n u u 1 } {{ } u 0,,n Controllability in n + 1 time-steps is only desirable if the system is not controllable in n timesteps When a system is not controllable in n time-steps, we must have ( ) rank(c n ) = rank B AB A n 1 B < n In other words, in the hope of improving the rank of C n, we are appending an additional column, A n B, to C n ie, C n+1 = C n A n B We now claim that (See Appendix 1) In fact, the following is true rank(c n ) = rank(c n+1 ) rank(c n ) = rank(c n+1 ) = rank(c n+2 ) = Before justifying this claim, let us understand the consequence of the above statement In words, it is saying that adding additional columns of the form A n B, A n+1 B, A n+2 B,, to the matrix, C n, do not improve the rank Recall that adding additional such columns is equivalent to using more control inputs, u n, u n+1, u n+2, Conclusion: In short, if a system (with one control input) is not controllable in n time-steps, then it is not controllable in any number of time-steps greater than n

5 5 II TESTS FOR CONTROLLABILITY A few notable tests for controllability are as follows: (i) Fundamental rank test: rank(c) = n (ii) Popov-Belevitch-Hautus (PBH) test 1: An LTI system is not controllable if and only if a non-zero left eigenvector, w, of the system matrix, A, such that w B = 0 Note that if A = V DV 1, then the left eigenvectors are the rows of V 1 ; what are the eigenvalues corresponding to the left eigenvectors? For real-valued symmetric system matrices, ie A = A, note that the left eigenvector is a simple transpose of the right eigenvector with the same eigenvalue, is it obvious? (iii) Popov-Belevitch-Hautus (PBH) test 2: An LTI system is controllable if and only if for all s C rank(si A B) = n, Refer to the lecture for examples Building on the fact that a control input is required for each state in a diagonal system, PBH test 1 follows rather immediately using the state transformation: x k = V z k where A = V DV 1

6 6 III STATE FEEDBACK CONTROL Let us consider single input systems and use the state-feedback control law, ie, With the state-feedback control, the system evolution is u k = Kx k (9) x k+1 = Ax k + Bu k = Ax k BKx k = (A BK)x k = (A BK) k+1 x 0 If a system is controllable then any arbitrary initial condition, x 0, can be forced to zero in n time steps Recall that x n = A n x 0 + C n u 0,,n 1, u n 1 u = A n x 0 + C n = A n x 0 C n u 1 K(A BK) n 1 K(A BK) n 2 K(A BK) 0 = x n = A n x 0 C n x 0, K(A BK) K, K(A BK) n 1 K(A BK) n 2 K(A BK) A n x 0 = C n x 0, K(A BK) K, K(A BK) n 1 K(A BK) n 2 K(A BK) C 1 n A n x 0 = K(A BK) K, x 0 K(A BK) n 1 K(A BK) n 2 K(A BK) K(A BK) K, x 0,

7 7 Clearly, the above is only true when C n is invertible Since the above has to be true for all x 0, we have K = Cn 1 A n, (10) where there are n 1 zeros in the above Conclusion: For a single input controllable system, ie, C n is invertible, Eq (10) takes any initial condition, x 0, to 0 in n time-steps

8 8 APPENDIX 1: CAYLEY HAMILTON THEOREM The well-known Cayley Hamilton theorem states that 1 Every real-valued square matrix, A, satisfies its own characteristic polynomial The characteristic polynomial of a matrix, A R n n is defined as χ(λ) det(λi A), = λ n + a n 1 λ n 1 + a n 2 λ n a 1 λ + a 0, for some a i s From the theorem, we have the the matrix satisfies its own characteristic polynomial, which is a fancy way of saying that: From this equation, or, χ(a) = 0 (11) A n + a n 1 A n 1 + a n 2 A n a 1 A + a 0 I = 0 n n, (12) A n = a n 1 A n 1 a n 2 A n 2 a 1 A a 0 I, (13) ie, A n is linear combination of A 0, A 1,, A n 1 Multiplying both sides by A, we get A n+1 = a n 1 A n a n 2 A n 1 a 1 A 2 a 0 A, = a n 1 ( a n 1 A n 1 a n 2 A n 2 a 1 A a 0 I) a n 2 A n 1 a 1 A 2 a 0 A, ie, A n+1 is also linear combination of A 0, A 1,, A n 1, and so on for any integer power of A strictly greater than n 1 Since A n, A n+1, are linear combinations of A 0, A 1,, A n 1, we can easily conclude that rank(c n ) = rank(c n+1 ) = rank(c n+2 ) = 1 The statement is more general but the following suffices

9 9 Example 1 (Cayley Hamilton Mechanics): Let a 3 3 square matrix be given by A = Verify that From CH theorem, we must have χ(λ) = det(λi A) = λ 3 8λ λ 6 A 3 8A A 6I = 0 3 3, A 3 = 8A 2 13A + 6I This can be verified in Matlab A = eig(a) ans = poly(6 1 1) ans = Aˆ3 ans = *Aˆ2-13*A + 6*eye(3) ans =