Modelling and Control of Dynamic Systems Controllability and - - PowerPoint PPT Presentation

Modelling and Control of Dynamic Systems Controllability and Observability

Sven Laur University of Tartu

Closed-loop Controllers: Basic Structure

Closed-loop control with state estimation

y[k] u[k] System Controller r[k] ˆ G[z]

State Estimator

The instability of a open loop controller is caused by gradual accumulation of

• disturbances. The control signal becomes unsynchronised with the system.

⊲ If we can estimate the system state from the output, we can circumvent such synchronisation errors. The system must be observable for that. ⊲ Even if we know the system state, it might be impossible to track the reference signal. In brief, the system must be controllable.

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Controllability

The state equation is controllable if for any input state x0 and for any final state x1 there exists an input u that transfers x0 to x1 in a finite time. ⊲ T6. The n dimensional pair (A, B) is controllable iff the controllability matrix

• B

AB A2B An−1B

• has the maximal rank n.

The desired input can be computed from the system of linear equations x1 = Anx0 + An−1Bu[0] + An−2Bu[1] + · · · + Bu[n − 1] . Although the theorem T6 gives an explicit method for controlling the system, it is an off-line algorithm with a time lag n. A good controller design should yield a faster and more robust method.

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Observability

The state equation is observable if for any input state x0 and for any input signal u, finite the output y sequence determines uniquely x0. ⊲ T7. The pair (A, C) is observable iff the observability matrix     C CA . . . CAn−1     has the maximal rank n.

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Offline state estimation algorithm

Again, the input state x0 can be computed from a system of linear equations              y[0] = Cx0 + Du[0] y[1] = CAx0 + CBu[0] + Du[1] · · · y[n − 1] = CAn−1x0 + · · · + CBu[n − 2] + Du[n − 1] Although this equation allows us to find out the state of the system, it is

• ffline algorithm, which provides a state estimation with a big time lag.

A good state space estimator must be fast and robust.

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General structure of state estimators

y[k] u[k] System State Estimator ˆ x[k]

x[k] = Ax[k − 1] + Bu[k − 1] y[k] = Cx[k] + Du[k]

A state estimate ˆ x[k] is updated according to u[k] and y[k]: ⊲ Update rules are based on linear operations. ⊲ The state estimator must converge quickly to the true value x[k] ⊲ The state estimator must tolerate noise in the inputs u[k] and y[k].

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Example

Consider a canonical realisation of the transfer function ˆ g[z] = 0.5z+0.5

z2−0.25

A =

• 0.25

1

• B =
• 1
• C = [0.5

0.5] D = 0 Then a possible stable state estimator is following ˆ x[k + 1] = Aˆ x[k] + Bu[k] + 1(y[k] − C ˆ x[k]

ˆ y[k]

) . In general, the feedback vector 1 can be replaced with any other vector to increase the stability of a state estimator.

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Kalman Decomposition

Equivalent state equations

Two different state descriptions of linear systems

• x[k + 1] = A1x[k] + B1u[k]

y[k] = C1x[k] + D1u[k]

• x[k + 1] = A2x[k] + B2u[k]

y[k] = C2x[k] + D2u[k] are equivalent if for any initial state x0 there exists an initial state x0 such that for any input u both systems yield the same output and vice versa. ⊲ T8. Two state descriptions are (algebraically) equivalent if there exists an invertible matrix P such that A2 = P A1P −1 B2 = P B1 C2 = C1P −1 D2 = D1 .

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Linear state space transformations

The basis e1 = (0, 1) and e2 = (1, 0) is a canonical base in R2. However a basis a1 = (1, 1) and a2 = (1, −1) is also a basis. Now any state x = x1e1 + x2e2 can be represented as x = x1a1 + x2a2 and vice versa. The latter is known as a basis transformation:      x1 = x1 + x2 2 x2 = x1 − x2 2

• x1 = x1 + x2

x2 = x1 − x2 For obvious reasons, we can do all computations wrt the basis {a1, a2} so that the underlying behaviour does not change. The same equivalence of state descriptions hold for other bases and larger state spaces, as well.

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Kalman decomposition

⊲ T9. Every state space equation can be transformed into an equivalent description to a canonical form     xco[k + 1] xco[k + 1] xco[k + 1] xco[k + 1]     =     Aco A13 A21 Aco A23 A24 Aco A43 Aco         xco[k] xco[k] xco[k] xco[k]     +     Bco Bco     u[k] y[k] = [Cco Cco 0] x[k] + Du[k] where ⋄ xco is controllable and observable ⋄ xco is controllable but not observable ⋄ xco is observable but not controllable ⋄ xco is neither controllable nor observable

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Minimal realisation

⊲ T10. All minimal realisations are controllable and observable. A realisation of a proper transfer function ˆ g[z] = N(z)/D(z) is minimal iff it state space dimension dim(x0) = deg D(z).

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Closed-loop Controllers: Design Principles

General setting

According to Kalman decomposition theorem, the state variables can be divided into four classes depending on controllability and observability. ⊲ We cannot do anything with non-controllable state variables. ⊲ Non-observable variables can be controlled only if they are marginally

• stable. We can do it with an open-loop controller.

⊲ For controllable and observable state variables, we can build effective closed-loop controllers. Simplifying assumptions ⊲ From now on, we assume that we always want to control state variables that are both controllable and directly observable: y[k] = x[k]. ⊲ If this is not the case, then we must use state estimators to get an estimate of x. The latter just complicates the analysis.

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Unity-feedback configuration

y[k] u[k] System Controller r[k] ˆ g[z] ˆ C[z] p +

• 1

Design tasks ⊲ Find a compensator ˆ c[z] such that system becomes stable. ⊲ Find a proper value of p such that system starts to track reference signal. It is sometimes impossible to find p such that y[k] ≈ r[k]. ⊲ The latter is impossible if ˆ g[1] = 0, then the output y[k] just dies out.

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Overall transfer function

Now note that ˆ y = ˆ g[z]ˆ c[z](pˆ r − ˆ y) ⇐ ⇒ ˆ y = pˆ g[z]ˆ c[z] 1 + ˆ g[z]ˆ c[z]ˆ r and thus the configuration is stable when the new transfer function ˆ g◦[z] = pˆ g[z]ˆ c[z] 1 + ˆ g[z]ˆ c[z] has poles lying in the unit circle. Now observe ˆ g[z] = N(z) D(z), ˆ c[z] = B(z) A(z) = ⇒ ˆ g◦[z] = pB(z)N(z) A(z)D(z) + B(z)N(z)

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Pole placement

We can control the denominator of the new transfer function ˆ g◦[z]. Let F(z) = A(z)D(z) + B(z)N(z) be a desired new denominator. Then there exists polynomials A(z) and B(z) for every polynomial F(z) provided that D(z) and N(z) are coprime. The degrees of the polynomials satisfy deg B(z) ≥ deg F(z) − deg D(z).

ℑ(s) ℜ(s) BIBO stable ℜ(z) ℑ(z) BIBO stable

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Signal tracking properties

Let ˆ g◦[z] be the overall transfer function. ⊲ The system stabilises if ˆ g◦[z] is BIBO stable. ⊲ The system can track a constant signal r[k] ≡ a if ˆ g◦[1] = 1. ⊲ The system can track a ramp signal r[k] ≡ ak if ˆ g◦[1] = 1 and ˆ g′

• [1] = 0.

The latter follows from the asymptotic convergence y[k] → aˆ g′

• [1] + kaˆ

g◦[1] for an input signal r[k] = ak. Moreover, let ˆ g◦ = N(z)/D(z). Then ˆ g◦[1] = 1 ∧ ˆ g′

• [1] = 0

⇐ ⇒ N(1) = D(1) ∧ N ′(1) = D′(1) .

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Trade-offs in pole placement

A pole placement is a trade-off between three design criteria: ⊲ response time ⊲ overshoot ratio ⊲ maximal strength of the input signal There is no general recipe for pole placement. Rules of thumb are given in ⊲ C.-T. Chen. Linear System Theory and Design. page 238.

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Illustrative example

Consider a feedback loop with the transfer function g[z] =

1 z−2. Then we

can try many different pole placements F(z) ∈

• z − 0.5, z + 0.5, z2 − 0.25, z2 + z + 0.25, z2 − z + 0.25
• The corresponding compensators are

ˆ c[z] ∈ 3 2, 5 2, 15 4z + 8, 24 4z + 12, 9 4z + 4

• ,

p ∈ 1 3, 3 5, 1 5, , 9 25 , 1 9

• .

They can be found systematically by solving a system of linear equations.

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Robust signal tracking

Sometimes the system changes during the operation. The latter can be modelled as an additional additive term w[k] in the input signal. If we know the poles of reference signal r[k] and w[k] ahead, then we can design a compensator that filters out the error signal w[k]. For instance, if w[k] is a constant bias, then adding an extra pole

1 z−1

cancels out the effect of bias. See pages 277–283 for further examples. In our example, the robust compensator for F[z] = z2 − 0.25 is ˆ c[z] = 12z − 9 4z − 4 p = 1 .

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Model matching

Find a feedback configuration such that ˆ g◦ is BIBO stable and ˆ g◦[1] = 1.

y[k] u[k] System r[k] ˆ g[z] Controller ˆ c1[z] + ˆ c2[z]

Feedback

⊲ The two-parameter configuration described above provides more flexibility and it is possible to cancel out zeroes that prohibit tracking. ⊲ There are many alternative configurations. A controller design is acceptable if every closed-loop configuration is BIBO stable.

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