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State estimation Propagating the model:
C
A
v z− 1 + B
6
ˆy(k) xˆ(k)
“Model:” Pˆ (z)
C
A
v z− 1 + B
6
y(k) x(k)
Plant: P (z)
v
u(k)
To calculate an estimated state, ˆx(k), we must choose an initial estimated state, ˆx(0), and run it through out model. Note that our controller will generate u(k) so we know what this is for times, k = 0,... , k − 1.
State Estimation State Estimation State feedback design assumes that we can measure the complete state. What do we do if we cannot? Estimate it. Approach: create a “model” of the system and use its state instead of the measured state.
C
A
v z− 1 + B
6
ˆy(k) xˆ(k)
“Model:” Pˆ (z)
C
A
v z− 1 + B
6
y(k) x(k)
Plant: P (z)
v
u(k)
Roy Smith: ECE 147b 12 : 1
State estimation Error properties: Look at the solution to the state equations:
x(k) = Ak^ x(0) + ∑^ k j=
A(k−j)B u(j)
xˆ(k) = Ak^ xˆ(0) + ∑^ k j=
A(k−j)B u(j)
Subtracting these gives, x˜(k) = Ak^ x˜(0) ←− the estimation error depends only on the initial error. Again, it’s easy to see that if A is stable the transient caused by the initial estimation error will decay to zero. Can we do better? Make use of the measurement, y(k).
State estimation Error properties: Define the state estimation error: x˜(k) := x(k) − xˆ(k). Applying the state equation for both the model and the plant gives,
x˜(k + 1) = x(k + 1) − ˆx(k + 1) = A x(k) − A ˆx(k) = A (x(k) − xˆ(k)) = A ˜x(k).
So the dynamics of the error, ˜x(k), are the same as the open-loop dynamics of the plant. If the plant is open-loop unstable, the state estimation error, ˜x(k), will blow up.
Roy Smith: ECE 147b 12 : 3
Estimator design Designing L Define the “observability matrix”,
O :=
C
CA.
CAn−^1
and now L = γo(A) O−^1
This can have an analogous problem to the controllability matrix, O may not be invertible. Observability The system is “observable” if and only if O is full rank. Observability means that the states can be estimated from measurements of the output.
Example: Consider f = md
(^2) x dt^2 ( a double integrator). Good choices of state are x (position), and v (velocity). Is the system observable by measuring x? Is the system observable by measuring v?
State estimation Designing an estimator This comes down to choosing an L matrix. Notice that the dynamics of the error are given by A − LC, so we can view this as a pole placement problem. Ackermann’s formula still works. Note that eig(A − LC) = eig(A − LC)T^ = eig(AT^ − CT^ LT^ ), and this is exactly the same as the state feedback pole placement problem: A − BK. Ackermann’s formula for L Select pole positions for the error: η 1 , η 2 , · · · , ηn. Specify these as the roots of a polynomial, γo(z) = (z − η 1 )(z − η 2 ) · · · (z − ηn). We will again use this polynomial with the A matrix as the variable, γo(A) = (A − η 1 I)(A − η 2 I) · · · (A − ηnI). Roy Smith: ECE 147b 12 : 7
Output feedback The big picture: using both K and L to create C(z)
C
A
z−^1 B
L
v +
6
ˆy(k)
ˆx(k)
6
Controller: C(z)
C
A
z−^1 B
−K
v +
6
v
?
y(k) x(k)
Plant: P (z)
u(k)
Output feedback State feedback with estimated states We can now put the two pieces together: State feedback: u(k) = −K x(k) ←− this is based on the true state. and, Estimator: ˆx(k + 1) = (A − LC) ˆx(k) + B u(k) + L y(k).
To do this we use the estimated state to calculate the feedback: u(k) = −K xˆ(k).
This idea is also referred to as “certainty equivalence”; using our best estimate in place of reality. But will it work? What is the effect of using ˆx(k) in place of x(k)? How will this affect the closed-loop poles of the plant?
Roy Smith: ECE 147b 12 : 9
Tradeoffs in designing L How should we design L? If we make the poles of A − LC fast the estimator error transient will decay quickly. What is the tradeoff here though? What prevents us from putting the poles as close to zero as we want? Is there a problem if L is large?
Estimator/state feedback controller What does the controller look like?
C
A
z−^1 B
L
v +
6
y ˆ(k)
xˆ(k)
6
Controller: C(z)
−K
y(k) -
v -
-^ u(k)
ˆx(k + 1) = A xˆ(k) + B u(k) + L(y(k) − C ˆx(k)) = (A − LC) ˆx(k) + B u(k) + L y(k) = (A − LC − BK) ˆx(k) + L y(k) ←− dynamics are specified by L and K and u(k) = −K ˆx(k)
Roy Smith: ECE 147b 12 : 13
Designing L Tradeoffs in the design of L Noise enters the estimator equations, xˆ(k + 1) = (A − LC − BK) ˆx(k) + L y(k) + L n(k), and corrupts the estimate. The error is now given by, x˜(k + 1) = (A − LC) ˜x(k) − L n(k), so, although it is still stable, it doesn’t decay to zero if n(k) 6 = 0. Note that the larger the size of L, the more the noise affects the error. The complete closed-loop system is: [x(k + 1) x˜(k + 1)
]
[A − BK BK
0 A − LC
] [x(k) ˜x(k)
]
[ 0
−L
]
n(k).
Rule of Thumb: The estimator error poles should be placed 2 to 4 times faster than the closed-loop poles. This is a trade-off between estimator error transients and noise magnification.
Estimator design Noise!
C
A
z−^1 B
L
v +
6
ˆy(k)
ˆx(k)
6
Controller: C(z)
n(k) - +
?
C
A
z−^1 B
−K
v +
6
v
?
y(k) x(k)
Plant: P (z)
u(k)
Roy Smith: ECE 147b 12 : 15
