chapter 8 reference introduction – integral control reference …€¦ · chapter 8 reference...

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Chapter 8

Reference Introduction – Integral

Control

Reference Input – Zero Design

Motivation

A controller obtained by combining a control law with an

estimator is essentially a regulator design : the charac-

teristic equations of the controller and the estimator are

basically chosen for good disturbance rejection. However,

it does not lead to tracking, which is evidenced by a good

transient response of the combined system to command

changes. A good tracking performance is obtained by

properly introducing the reference input into the system.

This is equivalent to design proper zeros from the reference

input to the output.

ESAT–SCD–SISTA CACSD pag. 217

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Reference input – full state feedback

Discrete-time :The reference signal rk is typically the signal that the out-

put yk is supposed to follow. To ensure zero steady-state

error to a step input rk, the feedback control law has to be

modified.

Modification of the control law :

• Calculate the steady-state values xss and uss of the state

xk and the output yk for the step reference rss (=the

steady-state of step reference rk) :

xss = Axss +Buss

rss = Cxss +Duss

Let xss = Nxrss and uss = Nurss, then[

A− I B

C D

][

Nx

Nu

]

=

[

0

I

]

⇒ [

Nx

Nu

]

=

[

A− I B

C D

]−1 [

0

I

]


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• Modify the control law:

uk = Nurk−K(xk−Nxrk) = −Kxk+(Nu +KNx)︸︷︷︸

N

rk

In this way the steady-state error to a step input will be 0.

Proof :

1. Verify that the closed-loop system from rk to yk is given

by[

xk+1

xk+1

]

=

[

A −BKLC A−BK − LC

][

xk

xk

]

+

[

BN

BN

]

rk,

yk =[

C −DK][

xk

xk

]

+DNrk.

2. If |eig(A−BK)| < 1 and |eig(A−LC)| < 1 we obtain

the following steady-state equations :

xss = Axss −BKxss + BNrss

xss = xss

yss = (C −DK)xss +DNrss

uss = −Kxss + Nrss

⇓yss = Cxss +D(−Kxss + Nrss) = Cxss +Duss = rss


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So the transfer matrix relating y and r is a unity matrix

at DC ⇒ zero steady–state tracking error, steady–state

decoupling.

Note that:

• rk is an exogenous signal, the reference introduction will

NOT affect the poles of the closed-loop system.

•[

A− I B

C D

]−1

must exist, and thus for MIMO

number of references = number of outputs

• also for MIMO, reference introduction implies a steady-

state decoupling between different reference and output

pairs. This means that yss = rss.

• some properties of this controller are discussed on page

227.

Continuous-time :Try to verify that in this case

[

Nx

Nu

]

=

[

A B

C D

]−1 [

0

I

]


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There are two types of interconnections for reference input

introduction with full state-feedback :

Type I: uk = Nurk −K(xk −Nxrk)

Type II: uk = −Kxk + (Nu +KNx)︸︷︷︸

N

rk

-

+-

+

-

+

EstimatorEstimator

PlantPlant

xx

rr uu y y

K K

Nu

Nx

N

For a type II interconnection, the control lawK used in the

feedback (uk = −Kxk) and in the reference feedforward

(N = Nu + KNx) should be exactly the same, otherwise

there is a steady-state error. There is no such problem in

type I.

⇒Type I is more ROBUST to parameter errors than Type

II.


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Reference Input - General Compensator

Plant and compensator model :

Plant : xk+1 = Axk +Buk,

yk = Cxk +Duk;

Compensator : xk+1 = (A−BK − LC + LDK)xk

+Lyk,

uk = −KxkThe structure of a general compensator with reference in-

put r :

N

−K

M

yu

r +

+

xEstimator

Process


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The general compensator is defined by the following closed-

loop equations from rk to yk :[

xk+1

xk+1

]

=

[


][

xk

xk

]

+

[

BN

M

]

rk,

yk =[

C −DK][

xk

xk

]

+DNrk.

Hence, the equations defining the compensator are

xk+1 = (A−BK − LC + LDK)xk + Lyk

+(M − LDN)rk,

uk = −Kxk + Nrk

where M ∈ Rn×m and N ∈ R

p×m.

The estimator error dynamics are

xk+1 = (A− LC)xk +BNrk −Mrk.


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Poles:

Characteristic equation:

det

(

zI −[


])

= 0.

This is the same characteristic equation as without ref-

erence introduction. So introducing references will NOT

change the poles.

Zeros :

The equations for a transmission zero are (see page 82)

det

ζI − A BK −BN−LC ζI − A +BK + LC −MC −DK DN

= 0

⇔

ζI − A BK −BN−LC ζI − A +BK + LC −MC −DK DN

u

v

w

︸︷︷︸6=0

= 0

⇔


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det

[

ζI − A −BC D

]

det

[

ζI − A +BK + LC −M−K N

]

= 0

The first term determines the transmission zeros of the open

loop system while the second term corresponds to the trans-

mission zeros of the compensator from rk to uk:

xk+1 = (A−BK − LC + LDK)xk + (M − LDN)rk,

uk = −Kxk + Nrk

These transmission zeros are designed via reference intro-

duction.


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Autonomous estimator (cfr. pag. 218-220) :

Select M and N such that the state estimator error equa-

tion is independent of r ⇒

M = BN

where N is determined by the method for introducing the

reference input with full state feedback.

−K

y

r +

xEstimator

Process

+

u

N


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Zeros :

The transmission zeros from rk to uk in this case are deter-

mined by

det(ζI − A + LC) = 0

which is the characteristic equation for the estimator, hence

the transmission zeros from rk to uk cancel out the poles

of the state estimator.

Properties :

• The compensator is in the feedback path. The refer-

ence signal rk goes directly into both the plant and the

estimator.

• Because of the pole-zero cancelation which causes “un-

controllability” of the estimator modes, the poles of the

transfer function from rk to yk consist only of the state

feedback controller poles (the roots of det(sI − A +

BK) = 0).

• The nonlinearity in the input (saturation) cancels out in

the estimator since in this case the state estimator error

equation is independent of u (xk+1 = (A− LC)xk)


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Tracking–error estimator

Select M and N such that only the tracking error, ek =

(rk − yk), is used in the controller.

⇒ N = 0, M = −L

−K

y

xEstimator

Processu

r+

−−e

The control designer is sometimes forced to use a tracking–

error estimator, for instance when the sensor measures only

the output error. For example, some radar tracking sys-

tems have a reading that is proportional to the pointing

error, and this error signal alone must be used for feedback

control.


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Zeros :

The transmission zeros from rk to yk are determined by

det

[

ζI − A −BC D

]

det

[

ζI − A +BK + LC L

−K 0

]

= 0

⇔

det

[

ζI − A −BC D

]

det

[

ζI − A L

−K 0

]

= 0.

Once K and L are fixed by the control and estimator de-

sign, so are the zeros. So there is no way to choose the

zeros.

Properties :

• The compensator is in the feedforward path. The ref-

erence signal r enters the estimator directly only. The

closed-loop poles corresponding to the response from rk

to yk are the control poles AND the estimator poles (the

roots of det(sI − A + BK) det(sI − A + LC) = 0).

• In general for a step response there will be a steady-state

error and there will exist a static coupling between the

input-output pairs.

• Used when only the output error ek is available.


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Zero-assignment estimator (SISO) :

Select M and N such that n of the zeros of the overall

transfer function are placed at desired positions. This

method provides the designer with the maximum flexibil-

ity in satisfying transient-response and steady-state gain

constraints. The previous two methods are special cases of

this method.

Zeros of the system from rk to uk:

det

[

ζI − A +BK + LC −M−K N

]

= 0

⇓ M∆= MN−1

λ(ζ)∆= det(ζI − A + BK + LC − MK) = 0


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Solution :

Determine M using a estimator pole-placement strategy

for “system” (Az, Cz), with

Az = A− BK − LC, Cz = K,

N is determined such that the DC gain from rk to yk is

unity.

For instance, in the case of a SISO system in continuous

time, for which D = 0

N = − 1

C(A−BK)−1B[1−K(A− LC)−1(B − M)]

and finally M = MN .


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Example Tape drive control - reference introduction

Autonomous Estimator :

Consider the model of the tape drive on page 39. From the

pole-placement design example on page 114, K is known.[

A B

C D

]−1 [

0

I

]

=

[

1 0 1 0 0 0 0 0

−2.5 0 2.5 0 −0.67 0.67 −0.67 0.67

]T

.

Thus,

Nx =

1 −2.5

0 0

1 2.5

0 0

0 −0.67

0 0.67

, Nu =

[

0 −0.67

0 0.67

]

.


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Let M = BN . The control law is

u = −Kx + (Nu +KNx)r

= −Kx +

[

0.8666 −1.6514

1.2779 2.1706

]

r.

Let L be the matrix from the pole placement estimator

design example on page 172. Then the closed-loop system

from r to y is[

x˙x

]

=

[

A −BKLC A− BK − LC

][

x

x

]

+

[

BN

BN

]

r,

y =[

C 0][

x

x

]

.

This system is NOT controllable, as was expected (Try to

prove it!).


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Step responses from the reference r to the output y.

0 2 4 6 8 10 12 14 16 18 20−0.5

0

0.5

1

1.5

Time (secs)

Am

plitu

de

0 2 4 6 8 10 12 14 16 18 20−0.5

0

0.5

1

1.5

Time (secs)

Am

plitu

de

T

T

p3

p3

Step input to reference 1

Step input to reference 2

Output 1 (p3) follows a step input to reference 1 while

output 2 (T ) is zero in steady state.

Output 2 (T ) follows a step input to reference 2 while

output 1 (p3) is zero in steady state.

⇒ steady-state decoupling.


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Integral Control and Robust Tracking

Motivation:The choice of N will result in a step response with a zero

steady-state error (see page 218). But the result is not

robust because any change in the parameters will cause the

error to be nonzero. Integral control is needed to obtain

robust tracking of step inputs.

A more general method for robust tracking, called the error

space approach (see page 238), can solve a broader class of

tracking problems, i.e. tracking signals that do not go to

zero in steady-state (a step, ramp, or sinusoidal signal).


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Integral control

Augment the plant

xk+1 = Axk +Buk,

yk = Cxk +Duk

with extra states integrating the output error ek = yk− rkxIk+1

= xIk+ Cxk +Duk − rk︸︷︷︸

ek

.

The augmented state equations become[

xIk+1

xk+1

]

=

[

I C

0 A

][

xIk

xk

]

+

[

D

B

]

uk −[

I

0

]

rk.

What are the equivalent equations in continuous-time ?

We now close the loop to stabilize the system. The feedback

law is

uk = −[

K1 K0

]

︸︷︷︸K

[

xIk

xk

]

.


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Use pole placement or LQR methods to design the control

feedback gain K. Once the closed-loop is stable, the track-

ing error e goes to zero even if some parameters change.

Processr y

+

−+

+−K1u

x

1z−1

−K0

The states of the plant xk are estimated using a state es-

timator. The estimator gain L is determined using pole

placement or Kalman filtering techniques. The integrator

states xIk need not to be estimated as they are being com-

puted explicitly.

What will be the closed-loop response from rk to yk ? Try

to derive a state-space model.

Note that pole placement or LQR might not work since the

augmented system is NOT always stabilizable and in this

case integral control can not be used.


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Tracking control - the error-space approach

Integral control is limited to step response tracking. A

more general approach, the error-space approach, gives a

control system the ability to track a non-decaying or even

a growing input such as a step, a ramp, or a sinusoid.

Suppose the external signal, the reference, is generated by a

certain dynamic system. By including the dynamic system

as a part of the formulation and solving the control problem

in an error space, the error approaches zero.


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Given the plant dynamics

xk+1 = Axk +Buk,

yk = Cxk +Duk

and the reference dynamics

rk+2 + α1rk+1 + α2rk = 0,

the tracking error is defined as

ek = yk − rk.


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Define the error-space state:

ξk∆= xk+2 + α1xk+1 + α2xk,

and the error-space control:

µk = uk+2 + α1uk+1 + α2uk.

Then

ek+2 + α1ek+1 + α2ek = Cξk +Dµk,

and the state equation for ξk becomes

ξk+1 = Aξk +Bµk

Combining these two equations, the final error system is

zk+1 = Aezk +Beµk

where

zk =

ek

ek+1

ξk

, Ae =

0 I 0

−α2I −α1I C

0 0 A

, Be =

0

D

B

.


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Controllability of the error system:

If (A,B) is controllable and has no zero at the roots of

αe(z) = z2 + α1z + α2,

then (Ae, Be) is controllable.

Control design: Pole-placement or LQR

µk = −[

K2 K1 K0

]

ek

ek+1

ξk

= −Kzk

The actual control uk is determined by the following inter-

nal model:

(u +K0x)k+2 +2∑

i=1

αi(u +K0x)k+2−i = −2∑

i=1

Kiek+2−i.

Once the closed-loop is stable, ek and ek+1 go to zero even

if some parameters change.


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Disturbance rejection

by disturbance estimation

Motivation

If the state is not available then −Kx can be replaced by

the estimate−Kx where x comes from the state estimator.

The disturbance rejection problem consists in designing an

estimator such that the error x = x− x goes to zero even

when there is a disturbance signal with known dynamics.

Suppose that the disturbance is generated by a certain

known dynamic system. The method consists in augment-

ing the estimator with the disturbance system in a way to

cancel out the disturbance effects in the estimator output.


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Augmenting the disturbance system to the plant

Given a plant with a disturbance input:

xk+1 = Axk +B(uk + wk),

yk = Cxk +Duk

and the disturbance dynamics (suppose 2nd order):

wk+2 + α1wk+1 + α2wk = 0.

The final error system is

zk+1 = Adzk +Bduk

where

z =

wk

wk+1

xk

, Ad =

0 I 0

−α2I −α1I 0

B 0 A

, Bd =

0

0

B

,

Cd =[

D 0 C]

, Dd = D.


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Observability:

If the plant (A,C) is observable and has no zero at any

roots of

αd(z) = z2 + α1z + α2,

then (Ad, Cd) is observable.

Estimator for the error system:

zk+1 = Adzk + Bduk + L(yk − Cdzk −Dduk).

The output uk:

uk = −Kxk + Nrk︸︷︷︸

introduce reference

− wk︸︷︷︸cancel disturbance

.

Final closed-loop system:

xk+1 = (A− BK)xk + BNrk +BKxk + Bwk.

where xk = xk − xk and wk = wk − wk.

Stable estimator ⇒ xk → 0 and wk → 0. The final state

is NOT affected by the disturbance.


chapter 8 reference introduction – integral control reference …€¦ · chapter 8 reference...

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