chapter 7 compensator design compensator structure · 2014-01-21 · chapter 7 compensator design...

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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.


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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.


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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.


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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

]

.


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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.


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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.


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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.


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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.


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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).


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• Combination of the LQR control law and the Kalman

filter : the final controller K is given as follows :


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.


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Compensator Dynamics

Compensator :


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.


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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


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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).


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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.


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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.

The transfer function of the compensator K(s) is

K(s) = −K(sI − A +BK + LC)−1L

= 102 −8.42s5 − 92.3s4 − 72.1s3 + 74.9s2 + 113s + 2.55

s6 + 61.2s5 + 640s4 + 7086s3 + 11029s2 + 7316s+ 194.


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The initial condition response with β0 = 1◦ :

0 5 10 15 20 25 30 35 40 45 50−4

−2

0

2

4

6

8

10

12x 10

−3

Time (secs)

r(ra

d/s

ec)

The response is almost the same as with state feedback

control but with a slightly larger peak value and with the

slowly decaying error (see the plot on page 97) filtered out

here.


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Now consider LQG design. Let q = 9.527, Rw = 0.7 and

Rv = 1. From the results of LQR and Kalman filter design

examples in chapter 4 and 6, we obtained a state feedback

gain K

K =[

1.06 −0.19 −2.32 0.10 0.04 0.49]

and an estimator gain

L =

−1.5465e− 02

4.9686e− 02

2.2539e− 01

−7.3199e− 01

−3.0200e− 01

−8.2157e− 15

.

The poles of the compensator with the K and L above are

at:

−9.845, −1.405, −0.2616± 0.5571i, −0.4756, −0.0049.

Again the compensator is stable.


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0 5 10 15 20 25 30 35 40 45 50−8

−6

−4

−2

0

2

4

6

8

10

12x 10

−3

Time (secs)

r(ra

d/s

ec)

This initial condition response shows that the Kalman filter

is too slow compared with the state feedback control loop.


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The reason for this result is that the measurement noise v

is large compared with the process noise w, so that the

Kalman filter has to reduce the measurement noise in-

fluence using a small estimator L, which slows down the

Kalman filter. Now increase the process noise covariance

to Rw = 700. The Kalman filter gain is now

L =

−8.0289e + 00

−3.3434e− 01

8.2913e + 00

−4.8082e + 00

−3.4682e + 00

3.5982e− 11

The poles of the compensator are at

−9.845, −1.405, −0.2616± 0.5571i, −0.4756, −0.0049.


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It’s not clear whether or not the Kalman filter is speeding

up from these poles location. But the following initial

condition response shows that the result is much better.


0 5 10 15 20 25 30 35 40 45 50−3

−2

−1

0

1

2

3

4

5

6

7x 10

−3

Time (secs)

r(ra

d/s

ec)

This response is even better than the one from pole place-

ment and with a better noise attenuation.


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Example Tape drive control - LQG design

The complete model of the tape drive is

x = Ax +Bu +Bw,

z =

[

Q1/2Cx

u

]

,

y = Cx + v.

where Q is a weighting matrix, w and v are zero mean

white noises with covariance Rw and Rv respectively.

Pole placement design:

From the results of the pole placement design examples for

control law and estimator design in chapter 3 and 5, we

know that the state feedback gain K

K =

[

0.55 1.58 0.32 0.56 0.67 0.05

0.60 0.60 0.68 3.24 −0.21 1.74

]

places the poles of the control loop at

−0.451± 0.937i, −0.947± 0.581i, −1.16, −1.16


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and the estimator gain

L =

5.3833e + 01 −4.2084e + 01

2.0377e + 02 −1.5615e + 02

3.9038e + 01 −7.9226e + 00

4.0009e + 02 −3.0809e + 02

6.5103e + 02 −5.8228e + 02

1.7691e + 03 −1.5000e + 03

places the estimator poles at

−2.255± 4.685i, −4.735± 2.905i, −5.800, −5.800.

The compensator then is

˙x = (A−BK − LC)x + Ly,

u = −Kx.


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The compensator has poles at

−1.696± 8.531, −12.09, −5.847, −3.335± 0.2469i.

The initial condition response with x1(0) = x3(0) = 1 looks

like

0 1 2 3 4 5 6 7 8 9 10−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Time (secs)

Am

plitu

de

p3

T


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Now consider LQG design.

Let

Q = 2I2, Rw = 0.001I4, Rv =

[

4 0

0 0.0001

]

.

From the results of the LQR and Kalman filter design ex-

amples in chapter 4 and 6, we obtain the optimal state

feedback gain K

K =

[

0.55 1.78 0.45 1.14 0.86 0.46

0.45 1.14 0.55 1.78 0.46 0.86

]

and the optimal estimator gain

L =

5.56e− 02 −1.67e + 00

7.74e− 04 −5.71e− 01

5.56e− 02 1.67e + 00

7.74e− 04 5.71e− 01

4.32e− 05 −6.02e− 01

4.32e− 05 6.02e− 01

.


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The poles of the compensator with the K and L above are

at

−0.7204± 1.093i, −0.6294± 0.90502i, −1.308, −1.365.

These poles are obviously slower than those for pole

placement estimator design. As a consequence, the time

response is slower.

The initial condition response with x1(0) = x3(0) = 1 :

0 10 20 30 40 50 60 70 80 90 100−0.2

0

0.2

0.4

0.6

0.8

1

Time (secs)

Am

plitu

de

p3

T


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Matlab Functions

initial

kalman

lqg

lqgreg

lqr

reg


chapter 7 compensator design compensator structure · 2014-01-21 · chapter 7 compensator design...

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