✫ ✪ Chapter 7 Compensator Design Compensator Structure Compensator = Control Law + Estimator -K ˙ ˆ x = A ˆ x + B 2 u +L(y - C 2 ˆ x) ˙ x = Ax + B 2 u +B 1 w y = C 2 x + v x w v u Control law Estimator y Process Compensator ˆ x ESAT–SCD–SISTA CACSD pag. 190
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Chapter 7
Compensator Design
Compensator Structure
Compensator = Control Law + Estimator
−K˙x = Ax+ B2u
+L(y − C2x)
x = Ax+B2u
+B1wy = C2x+ v
x
w v
u
Control law Estimator
y
Process
Compensator
x
ESAT–SCD–SISTA CACSD pag. 190
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Standard Plant Formulation
In control theory the following general system description,
called standard plant, is often used
+ + +
}
uk
vk
yk
wk
zk
B1
B2
C1
C2
D1
D2
z−1
A
xk+1 = Axk +B1wk +B2uk
yk = C2xk +D2uk + vk
zk =
[
C1xk
D1uk
]
where xk is the state vector, zk is the regulated output
vector, yk is the measurement vector, uk is the control in-
put, wk and vk are process noise and measurement noise
respectively.
ESAT–SCD–SISTA CACSD pag. 191
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Clearly there are 2 types of inputs :
• the actuator or control input uk : inputs manipulated
by the compensator
• the exogeneous input wk : all other input signals
and 2 types of outputs :
• measured output yk : inputs to the compensator
• regulated output zk : all outputs that are of interest for
control. These outputs may be virtual and not mea-
sured. They typically define the state feedback cost
function
JN =N∑
k=0
zTk zk =N∑
k=0
(xTkCT1 C1xk + uTkD
T1D1uk)
so, using the LQR formalism, by definition of zk,
Q = CT1 C1 and R = DT
1D1.
ESAT–SCD–SISTA CACSD pag. 192
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State-space Description
Plant :
xk+1 = Axk +B2uk + B1wk,
zk =
[
C1xk
D1uk
]
,
yk = C2xk +D2uk + vk.
Estimator :
xk+1 = Axk + L(yk − C2xk −D2uk) + B2uk.
Control law :
uk = −Kxk.
Compensator :
xk+1 = (A−B2K − LC2 + LD2K)xk + Lyk,
uk = −Kxk.
ESAT–SCD–SISTA CACSD pag. 193
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Closed-Loop System
A state-space model for plant + compensator is
[
xk+1
xk+1
]
=
[
A −B2K
LC2 A− B2K − LC2
][
xk
xk
]
+
[
B1 0
0 L
][
wk
vk
]
,
yk =[
C2 −D2K][
xk
xk
]
+[
0 I][
wk
vk
]
.
Using the error system (estimator) dynamics
xk+1 = (A− LC2)xk +B1wk − Lvk,
we obtain
[
xk+1
xk+1
]
=
[
A−B2K B2K
0 A− LC2
][
xk
xk
]
+
[
B1 0
B1 −L
][
wk
vk
]
,
yk =[
C2 −D2K D2K][
xk
xk
]
+[
0 I][
wk
vk
]
.
ESAT–SCD–SISTA CACSD pag. 194
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Separation Principle:Pole Placement
The characteristic equation of the closed-loop system
det
[
sI − A +B2K −B2K
0 sI − A + LC2
]
= 0
The poles of the closed-loop system can be obtained from
det(sI − A + BK) det(sI − A + LC) = αc(s)αe(s) = 0
⇒ Separation Principle (for Pole Placement):
The set of poles of the
closed-loop system consists
of the union of the control
poles and estimator poles.
ESAT–SCD–SISTA CACSD pag. 195
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Design procedure :
• Choose the desired poles for the state feedback control
loop. Determine the control law K via pole placement.
• Choose the desired poles for the estimator. Determine
the estimator feedback gain L via pole placement.
• Combine the estimator and the control law to get the
final compensator.
ESAT–SCD–SISTA CACSD pag. 196
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Linear Quadratic Gaussian Control
Measurement feedback controlLet w and v be Gaussian zero mean white noises.
Our aim is to find an output feedback controller u = Ky
such that
limT→∞
{1
T
∫ T
0
zTzdt
}
or limK→∞
{
1
K
K∑
k=0
zTk zk
}
is minimized.
This problem is called the Linear Quadratic Gaussian
(LQG) control problem.
The difference between LQG and LQR is that LQR uses
state feedback and hence K is just a constant matrix while
LQG uses measurement or output feedback and hence K
is a dynamic system.
ESAT–SCD–SISTA CACSD pag. 197
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Stochastic Separation Principle
It can be proven that the solution
to the LQG optimal control prob-
lem consists in using the optimal
state estimator (Kalman filter) to-
gether with the same feedback as
would have been applied had the
states been measured (LQR). This
principle relies on the assumption
of linearity, the Gaussian charac-
ter of the noise and the use of a
quadratic optimization criterion.
ESAT–SCD–SISTA CACSD pag. 198
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Design procedure (summary) :
• LQR design: find an optimal state feedback control law
uk = −Kxk for the system
xk+1 = Axk +B1wk +B2uk,
zk =
[
C1xk
D1uk
]
.
where K = (DT1D1 + BT
2 SB2)−1BT
2 SA, in which S is
a solution for the Riccati equation :
S = AT[S − SB2(D
T1D1 +BT
2 SB2)−1BT
2 S]A+CT
1 C1.
• Kalman filter design : find an optimal estimator
xk+1 = Axk + L(yk − C2xk −D2uk) +B2uk
for the system
xk+1 = Axk +B1wk +B2uk,
yk = C2xk +D2uk + vk,
where L = APCT2 (R + C2PC
T2 )−1, in which P is a
solution for the Riccati equation :
P = B1QBT1 +APAT−APCT
2 (R+C2PCT2 )−1C2PA
T .
with R = cov(v) and Q = cov(w).
ESAT–SCD–SISTA CACSD pag. 199
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• Combination of the LQR control law and the Kalman
filter : the final controller K is given as follows :
xk+1 = (A−B2K − LC2 + LD2K)xk + Lyk,
uk = −Kxk.
Design strategy for LQG :
Trade-off between speed of time responses and
attenuation of noises by choosing the covariance
matrices of w and v.
Similarly,
Trade-off between the energy of the states and energy
of the control effort by choosing the weighting matrices
CT1 C1 and DT
1D1.
ESAT–SCD–SISTA CACSD pag. 200
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Compensator Dynamics
Compensator :
xk+1 = (A−B2K − LC2 + LD2K)xk + Lyk,
uk = −Kxk.
Characteristic equation of the compensator :
det(zI − A +B2K + LC2 − LD2K) = 0
⇒The poles of the compensator are not necessary in the unit
circle, the compensator might be unstable.
Transfer function of the compensator :
K(z) = −K(zI − A +B2K + LC2 − LD2K)−1L.
ESAT–SCD–SISTA CACSD pag. 201
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Summary of State Space Control
Design
LQG
compensator design
system modeling
LQR Kalman filter pole placementpole placement
pole placement
control law design estimator design
ESAT–SCD–SISTA CACSD pag. 202
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Examples of Compensator Design
Example Boeing 747 Aircraft control - Compensator de-
sign
The complete model of the aircraft is
x = Ax +Bu +Bw,
z =
[
q1/2Cx
u
]
,
y = Cx + v.
where q is the weighting factor and w and v are zero
mean white noises with covariance Rw and Rv respectively.
(A,B,C,D) is the nominal aircraft model (D = 0).
ESAT–SCD–SISTA CACSD pag. 203
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First, we design a compensator via pole placement. From
results of pole placement design examples in chapter 3 and
chapter 5, we know that the state feedback gain
K =[
1.06 −0.19 −2.32 0.10 0.04 0.49]
places the poles of the control loop at
−0.0051, −0.468, −1.106, −9.89, −0.279± 0.628i
and the estimator gain
L =
2.5047e + 01
−2.0517e + 03
−5.1935e + 03
−2.4851e + 04
−4.0914e + 04
−1.5728e + 04
places the estimator poles at
−0.0255, −2.34 ,−5.53, −49.45, −1.395± 3.14i.
The compensator then is
˙x = (A−BK − LC)x + Ly,
u = −Kx.
ESAT–SCD–SISTA CACSD pag. 204
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The poles of the compensator are then at
−51.35, −4.074± 10.12i, −0.8372± 0.6711i, −0.027.
So the compensator itself is stable, which is preferred since
otherwise an accident resulting in a breakdown of the
control loop will cause the control output to go to infinity.