自动控制原理Lecture22
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Controllability
Example
State Feedback Control
Examples
Summary
Xn = A Xn−1 + B un−1 = An−1 Bu0 + . . . + A Bun−2 + Bun−1 .
Controllability of a linear system (3)
Examples
Summary
Controllability of a linear system
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Control III Lecture 22
A system is said to be completely state controllable if it is possible to transfer the system from any arbitrary initial state to any arbitrary desired state. This is an important criteria for state feedback design since if any state variable is independent of the control signal then it is impossible to control the system. Why do we care? Controllability is an important part of pole placement in the state space design process.
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In matrix form this becomes: Xn = [ B AB A2 B · · · ] An−1 B un − 1 un − 2 . . . u0 The controllability matrix C is defined to be the n × n matrix [ ] C = B AB A2 B · · · An−1 B . (1) The system is controllable if and only if C has rank n, ie C −1 exists. .
Controllability
Example
State Feedback Control
Examples
Summary
Controllability of a linear system (2)
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Control III Lecture 22
Recall: the recursive solution to the state equation Xk +1 = A Xk + B uk , is X1 = A X0 + B u0 = B u0 X2 = A X1 + B u1 = A B u0 + B u1 X3 = A X2 + B u2 = A2 B u0 + A B u1 + B u2 . . . X0 = 0
Learning Objectives :
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Control III Lecture 22
Controllability
Example
State Feedback Control
Controllability of a linear system. State feedback control design
State Feedback Control Design (3)
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Control III Lecture 22
How do we use Q (z )? Recall from the last lecture that the transfer function of the system can be written as Q (z ) = D + C (zI − A)−1 B that is, in terms of state space matrices. Let’s begin by assuming we have a strictly causal system ⇒ D = 0.
Controllability
Example
State Feedback Control
Examples
Summary
State Feedback Control Design (2)
Output feedback controller:
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Control III Lecture 22
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Control III Lecture 22
Controllability
Example
State Feedback Control
Examples
Summary
Example: Non-controllable linear system (2)
Exploring this further: the TF is not minimal because there is a pole and a zero at z = −0.3723. Cancelling the common factor (z + 0.3723) yields the minimal TF 3.186 Gm (z ) = z − 5.372 with canonical state space representation A = 5.372, B = 1, C = 3.186, D = 0. Since we cancelled the factor (z + 0.3723) the system is not completely state controllable. However even though this system is unstable in open loop, it can still be stabilised by suitable feedback. This example demonstrates the importance of the system being minimal.
We now explore the idea of designing system dynamics through state feedback. Until now, you have used the idea of output feedback where you used system output yk as a measurement to determine the control design. Recall that with output feedback you would construct an error signal ek such that ek = rk − yk where yk is the output signal and the desired value of yk was represented by the reference input rk .
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Control III Lecture 22
Controllability
Example
State Feedback Control
Examples
Summary
Controllability under transformation?
Recall Xk = T Zk where T is an n × n nonsingular matrix. Let [A, B , C , D ] describe the state Xk and [F , G , H , J ] describe the transformed state Zk then [ ] B AB A2 B · · · An−1 B Cx = [ ] G FG F 2 G · · · F n−1 G . Cz = However from previous lectures we know F = T −1 AT and G = T −1 B , thus [ −1 ] T B T −1 ATT −1 B · · · T −1 An−1 TT −1 B Cz = [ −1 ] T B T −1 AB T −1 A2 B · · · T −1 An−1 B = = T −1 Cx . Thus a change of state by a nonsingular linear transformation does not change controllability.
y k
Controllability
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Control III Lecture 22
CoState Feedback Control
Examples
Summary
State Feedback Control Design
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Controllability
Example
State Feedback Control
Examples
Summary
State Feedback Control Design (4)
Thus
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ek
Control III Lecture 22
C(z)
uk
Q(z)
Control III Lecture 22
