Discrete Time Systems:

Transcription

1 Discrete Time Systems: Finding the control sequence {u(k): k =, 1, n} that will drive a discrete time system from some initial condition x() to some terminal condition x(k). Given the discrete time system (DTS) whose state equation is given by: x(k+1) = Ax(k) + Bu(k): (1) with the solution given as: k 1 k k j 1 xk ( ) = Ax( ) + A Buj ( ) j= (2) we wish to determine how to find the sequence {u(l): l =, 1, k} that drives the system from an arbitrary initial condition x() to a terminal condition x(k). Equation (2) contains both the transient and forced response of the system. If we define an initial condition x() and a terminal condition x(k) and expand the convolution summation for some final value of k, the equation may be written in vector matrix form as: [x(k) - A k x()] = [B, AB, A 2 B,. A K-1 B] [u(k-1) u(k-2) u(1) u()] T () Copyright Dr. Thomas Chmielewski 27 Page 1 of 1

2 The input or forcing function sequence {u(k)} may be found by solving vector matrix equation (). A key question to be answered is how big does k have to be to drive the system from some initial condition to a terminal condition. Alternatively we may ask if there are any limits to the size of k. At this point it may be obvious that the answer lies in the controllability test matrix : C T = [B, AB, A 2 B, A k-1 B] (4) Note answering the first question also allows us to understand controllability in a better sense. Rewrite Eq. (2) as follows: [B, AB, A 2 B,. A K-1 B] [u(k-1); u(k-2); u(1) u()] = [x(k) - A k x()] (2a) consider Eq. (2a) to be in the form C T U = Y (5) where: C T : n by m U: m by 1 Y: n by 1 Copyright Dr. Thomas Chmielewski 27 Page 2 of 1

3 From basic linear algebra, there exists a solution for U (possibly more than one), if and only if Y can be expressed as a linear combination of the columns of C T. The equivalent rank calculation for the existence of a solution of U is given as Rank [C T Y] = Rank [C T ] (6) that is by forming an augmented system, the addition of the column y does not change the rank. Since in our situation Y corresponds to an n by 1 vector (since x() and x(k) are n by 1 and a must be square and of order n) then a solution for U only exists if the partitioned matrix has rank n. Alternatively there must be n linearly independent columns out of C T. This allows us to put a bound on the size of C T and therefore the size of the vector U. C T contains n rows, based on the number of states, but could have as many columns as desired (by Eq. 2 and ). By the Cayley Hamilton theorem powers of a matrix A whose order is n by n can be expressed as A m = f(i, A, A 2, A n-1 ) where m n. Thus the additional partitions BA N provide no new linearly independent columns. In summary, if the rank of Equation (4) equals the order of the system, n, then the system is completely controllable and a control sequence {u(k): k =, 1, n-1} exists that will drive the system from some arbitrary initial condition to the origin or any other desired terminal state. Copyright Dr. Thomas Chmielewski 27 Page of 1

4 Example 1 : driving a system to the origin in k = n steps Given the system described by the matrices A = [ 1; 2] B = [ ;] x2 = A = B = 1 2 with initial condition: x= [6;] x = 6 and terminal condition x2 = [ ;] The solution is computed from the relationship: [x(2) - A 2 x()] = [B AB] [u(1); u()] yielding [u(1); u()] = [B AB] -1 [x(2) - A 2 x()] Using Matlab to evaluate the expression yields: u = inv([b, A*B])*(x2 - A^2*x) u = -2 Copyright Dr. Thomas Chmielewski 27 Page 4 of 1

5 Note that u() = -2 and u(1) = (this is the reverse order from the natural indexing for the vector u) The best way to validate the result is by recursive substitution at k = x1 = A x + Bu() x1 = A*x + B*(-2) x1 = at k = 1 x2 = A x1 + Bu(1) x2 = A*x1 + B*() x2 = which is the desired result. Copyright Dr. Thomas Chmielewski 27 Page 5 of 1

6 Example 2 : driving a system to an arbitrary location in k = n steps Now let us consider driving the system to the terminal condition x2 = [1;1] x2 = 1 1 The solution is the same with the new numerical value for x2 u = inv([b, A*B])*(x2 - A^2*x) u = Remember to modify the vector u so that the indices match the discrete time equation, Validating the solution: at k = x1 = A x + Bu() x1 = A*x + B*1. x1 = at k = 1 x2 = A x1 + Bu(1) x2 = A*x1 + B*(-.) x2 = Copyright Dr. Thomas Chmielewski 27 Page 6 of 1

7 which is the desired result within the precision of the numbers used 12 It is interesting to plot the trajectory in state space. We define a time history of the states in a trajectory matrix whose columns correspond to the state at a given time. traj = [x, x1, x2] traj = now we will plot the second state (i.e., second row of traj) versus the first state, using arrows to show the direction of the trajectory. plot(traj(1, : ), traj(2, : ), '->') axis([, 12,, 12]) The trajectory plot shows the two transitions necessary to drive the initial state to the terminal state Copyright Dr. Thomas Chmielewski 27 Page 7 of 1

8 Using more than n steps to drive the system from x() to x(k) Consider the case for k > n steps, the equation to find {u(k)} i is the same as before: [x(k) - A k x()] = [B, AB, A 2 B,. A K-1 B] [u(k-1); u(k-2); u(1) u()] (2) here the controllability test matrix C T = [B, AB, A 2 B,. A K-1 B] is no longer terminated at a power of n-1 but contains the number of partitions equals that of the desired number of steps. By the Cayley Hamilton Theorem, the extra columns are linearly dependent on the first n columns With these additional columns, C T is no longer square and the inverse is undefined. This is called the underspecified case. The solution of Eq. (5) and utilizes a least squares solution. C T U = Y (5) Since there are more unknowns k than equations, n, there are many solutions, we will specify the solution, U, that has the minimum norm. The minimum norm solution uses the pseudo inverse defined for a general non square matrix as: [C T T (C T C T T ) -1 ]. Hence U = [C T T (C T C T T ) -1 ] Y (7) The sizes of the matrices in Eq. (7), these are given in the ordinal position of the matrices of Eq. (7) as:(k by 1) = [(k by n) [(n by k)(k by n)] (n by 1) Copyright Dr. Thomas Chmielewski 27 Page 8 of 1

9 Example : driving a system to the origin in k = steps Using the system of Example 1, we wish to drive the system to the origin in three steps. Computing the pseudo inverse yieids: CT = [B, A*B, A^2*B] A = [ 1; 2] B = [ ; ] x= [6;] x = [ ;] A = B = x = x = CT = P_INV = CT'*inv(CT*CT') P_INV = Solving for vector U yields : U = P_INV*(x - A^*x) U = Copyright Dr. Thomas Chmielewski 27 Page 9 of 1

10 Once again we will verify via recursive substitution: x1 = A*x + B*(-1.6) x1 =. 1.2 x2 = A*x1 + B*(-.8) x2 = traj = [x, x1, x2, x] traj = Plotting the trajectory as before: plot(traj(1, : ), traj(2, : ), '-*') axis([-1, 7, -1, 7]) x = A*x2 + B*() x = 1.e-14 * x is essentially zero x = [;] x = Copyright Dr. Thomas Chmielewski 27 Page 1 of 1

11 Example 4 : driving a system to the origin in k = 5 steps Here we find 5 inputs that will drive the system from the initial condition to the origin CT = [B, A*B, A^2*B, A^*B, A^4*B] CT = x5 = [;] x5 = P_INV = CT' * inv(ct * CT') P_INV = Copyright Dr. Thomas Chmielewski 27 Page 11 of 1

12 U = P_INV*(x5 - A^5*x) U = x1 = A*x + B*(-1.559) x1 = x2 = A*x1 + B*(-.7529) x2 = x = A*x2 + B*(-.765) x = x4 = A*x + B*(-.1882) x4 =.282. x5 = A*x4 + B*() x5 = 1.e-14 * so again x5 is essentially zero, the small numbers are due to floating point implementation used in Matlab traj = [x, x1, x2, x, x4, x5] traj = Copyright Dr. Thomas Chmielewski 27 Page 12 of 1

13 plot(traj(1,: ), traj(2, : ), '-*') axis([-1, 7, -1, 7]) Copyright Dr. Thomas Chmielewski 27 Page 1 of 1