UCSI University Kuala Lumpur, Malaysia Faculty of Engineering Department of Mechatronics Lecture 18 State Space Design Mohd Sulhi bin Azman Lecturer Department of Mechatronics UCSI University sulhi@ucsi.edu.my 1 August 2011 Page : 1 Contents Open loop and closed loop system representation Controller and observer design via poleplacement method Ackermann s formula for computing the gain, K Page : 2 1
State Space Representation Open-loop control system representation: Closed-loop control system representation with feedback: Page : 3 Pole Placement Design Method The pole-placement design method is one of the many method used to design the controller and observer for a system. The problem of placing the poles at the desired location (on the s-plane) is called the poleplacement design method. We shall look at two ways of designing the controller and observer via pole-placement method. Page : 4 2
Controller Design via Pole Placement Method To design a controller with pole-placement method, the system considered must be completely controllable. Furthermore, all state variables are assumed measureable and available for feedback, and then only the poles of the closed-loop system maybe placed at any desired locations by means of state feedback through an appropriate state feedback gain. Page : 5 Controller Design via Pole Placement Method Consider the following state equations: xɺ = Ax + Bu y = Cx The open-loop state space representation of the plant is given by: Page : 6 3
Controller Design via Pole Placement Method The closed-loop representation is given as: In pole placement method, another feedback is introduced to feed the state back to the control (input) signal. Page : 7 Controller Design via Pole Placement Method Now, the control signal is: u = Kx Substituting gives: A B( K ) ( A BK ) x = ( ) x xɺ = x + x xɺ = xɺ ( t) A BK ( t) The solution of the above equation is: ( ) A BK t xɺ ( t) = e x(0) Page : 8 4
Controller Design via Pole Placement Method Thus, in state space, given the pair (A,B), we can always determine the K (gain), to place all the system closed-loop poles in the left-half of the plane if and only if the system is controllable that is, if and only if the controllability matrix C M is of full rank. And for a full rank C M in a SISO system, it implies that the C M matrix is invertible. Page : 9 Observer Design via Pole Placement Method In designing the observer using the poleplacement method, all state variables are assumed to be available for direct measurement and feedback. However, not all state variables are available. Therefore, the state variable needed to be estimated through the state observer. Page : 10 5
Observer Design via Pole Placement Method Open-loop observer: Page : 11 Observer Design via Pole Placement Method Closed-loop observer: Page : 12 6
Observer Design via Pole Placement Method Exploded view of a closed-loop observer, showing feedback arrangement to reduce statevariable estimation error: Page : 13 Observer Design via Pole Placement Method Full state feedback observer: Page : 14 7
Observer Design via Pole Placement Method Consider a system defined by: xɺ = Ax + Bu y = Cx (i) And another estimator or observer is defined as: Subtracting gives: xɺɶ = Axɶ + Bu yɶ = Cxɶ error ( ɶ ) ( x xɶ ) xɺ xɺɶ = A x x y yɶ = C (ii) (iii) Page : 15 Observer Design via Pole Placement Method Note that the design of the observer is different from the design of the controller. Similar to the design of the controller vector K, the design of the observer consists of evaluating the constant vector L, so that the transient response of an observer is faster than the response of the controlled loop in order to yield a rapidly updated estimate of the state vector. The vector (matrix) L is the observer gain matrix and is to be determined as part of the observer design procedure. Page : 16 8
Observer Design via Pole Placement Method Let us now consider the following figure : The state equations are: ( ɶ ) xɺɶ = Axɶ + Bu + L y y y = Cx (iv) Page : 17 Observer Design via Pole Placement Method Now, subtracting equation (iv) from (i), results in: Re-arranging gives: Simplifying gives: ( ) ( ) ( x xɶ ) xɺ xɺɶ = A x xɶ L y yɶ y yɶ = C ( ) LC ( ) ( x xɶ ) xɺ xɺɶ = A x xɶ x xɶ y yɶ = C ( A LC )( ɶ ) ( x xɶ ) xɺ xɺɶ = x x y yɶ = C Page : 18 9
Observer Design via Pole Placement Method Let the estimation error signal be denoted as e, hence: eɺ x ( A LC) = y yɶ = Ce Take note that the dynamic behaviour of the error vector is determined by the eigenvalues of the matrix (A LC). x e x Also note that if the system is completely observable, then it is possible to choose matrix L such that (A LC) has arbitrarily eigenvalues. Page : 19 Approaches to State Space Design There are three methods to find the gain, K: 1. Direct approach. 2. Direct substitution approach. 3. Ackermann s formula. We shall now examine each and every step of this approach. Page : 20 10
Ackermann s Formula Consider the following system: xɺ = Ax + Bu y = Cx We assume that the system is completely state controllable. And furthermore, we also assume that the desired closed-loop poles are at s=µ 1, µ 2,, µ n. The control law says that: u = Kx. Which then gives, upon substitution: ( A BK ) xɺ = x Page : 21 Next, let us define: Ackermann s Formula ( ) Aɶ = A BK, hence: x ɺ ɶ = A ɶ x Thus, the eigenvalues are: ( ) ( µ )( µ ) ( µ ) si Aɶ = si A + BK = s s s 1 2 n Remember that the Cayley-Hamilton theorem says that every square matrix A satisfies its own characteristic equation. Then: n 1 n 2 ( ) φ Aɶ = Aɶ + α Aɶ + α Aɶ + + α Aɶ + α I = n 1 2 n 1 n 0 (1) We are going to use the above equation to derive Ackermann s formula, and we choose n=3 for simplicity. Page : 22 11
Ackermann s Formula Consider the following identities: I = I Aɶ = A BK ( ) 2 2 2 2 2 2 Aɶ = A BK = A ABK ABK + B K = A ABK + BKAɶ ( ) factorize Aɶ = A BK = A A BK ABKAɶ BKAɶ 3 3 3 2 2 Let us multiply the above identity in order by α3, α2, α1, α0 wherein α 0 = 1, respectively and let us add all the results together to obtain: 2 3 ( Aɶ ) = 3I + 2A ɶ + 1A ɶ + 0A ɶ φ α α α α Page : 23 Ackermann s Formula On expansion, we obtain: ( Aɶ ) = ( ) ( ) ( ) 3I + 2 A BK + 1 A BK + 0 A BK φ α α α α On substitution, we obtain: 2 3 ( Aɶ ) = 3I + 2A ɶ + 1A ɶ + 0A ɶ = 3I + 2 ( A BK) 2 3 2 2 + α1 ( A ABK + BKAɶ ) + ( A A BK ABKAɶ BKAɶ ) φ α α α α α α Simplifying gives: 2 3 = 1 ( ɶ 2 3 ) φ A = α I + α A + α A + A α BK α ABK α BKAɶ 3 2 1 2 1 1 A BK ABKAɶ BKAɶ 2 2 (2) Page : 24 12
Ackermann s Formula We refer back to the equation (1) given in slide 19. This equation was obtain as a result of Cayley-Hamilton theorem and for n=3. We reproduce the equation here: And also: ( ) 3 2 φ Aɶ = Aɶ + α Aɶ + α Aɶ + α I = ( ɶ 3 2 ) 1 2 3 0 φ A = A + α A + α A + α I 1 2 3 0 (3) (4) We next substitute equations (3) and (4) into equations (2), to get: 2 3 φ ( Aɶ ) = α3i + α2a + α1a + A α 2BK α1abk 1 φ ( Aɶ ) α BKAɶ A BK ABKAɶ BKAɶ 2 2 Page : 25 Ackermann s Formula Simplifying leads us to: 2 2 ( A ɶ ) = ( A ɶ ) 2BK 1ABK 1BKA ɶ A BK ABKA ɶ BKA ɶ φ φ α α α Since φ ( A ɶ ) = 0, the above equation is reduced to: 2 2 ( A ɶ ) = 2BK + 1ABK + 1BKA ɶ + A BK + ABKA ɶ + BKA ɶ φ α α α Next, we factorize the above expression to get: ( ɶ ) ( ɶ ɶ 2 1 ) ( ɶ 1 ) 2 2 A = B K + KA + KA + AB K KA + A BK φ α α α (5) Page : 26 13
Ackermann s Formula And because this is a matrix, then we can write equation (5) in the previous slide as: 2 α 2K + α1kaɶ + KAɶ 2 φ ( Aɶ ) = B AB A B α1k KAɶ K controllability matrix Note that since the system is completely controllable, then the inverse matrix exists. Pre-multiplying equation (5) with the inverse gives: 2 α 2K + α1kaɶ + KAɶ 2 1 B AB A B φ ( A) = α1k KAɶ K (6) Page : 27 Ackermann s Formula Now, note that the Ackermann s formula is used to find the system gain, K. So, to find K, we pre-multiply equation (6) by [0 0 1]. 2 α 2K + α 1KA ɶ + KAɶ B AB A B A = α1k KAɶ = K K 2 1 [ 0 0 1] φ ( ) [ 0 0 1] Which actually, if re-written, gives: 2 1 [ 0 0 1] φ ( ) K = B AB A B A (7) Page : 28 14
Ackermann s Formula Ladies and gentlemen, for an arbitrary integer n, equation (7) can be generally written as: 2 n 1 1 [ 0 0 0 1] φ ( ) K = B AB A B A B A And this is what we call the Ackermann s formula for evaluating the system gain in designing the controller in state space. Page : 29 Example 1 : Controller Design Page : 30 15
Example 2 : Observer Design Page : 31 Next Step Textbook reference : Chapter 12. Homework 17 has been posted on the course website. Attempt them. You do not have to submit Homework 17 as it will not be graded. Thank You. Page : 32 16