staff.damvar/Classes/R7003E-2015-LP2/Lessons/L09_a… · Introduction of the reference input How do we design the zeros of the controller? controller zeros: det sI−(A−BK−LC)−

R7003E - Automatic ControlLesson 9

25 November 2015

1

Introduction of the reference inputMost generic strategy

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

{ x = Ax +Bu

y = Cxplant

˙x = (A −BK −LC) x +Ly +Mr observer

u = −Kx +Nr reference

2

Introduction of the reference input

Dynamics of the plant:

{ x = Ax +Bu

y = Cx

Transfer function:Y (s) = Gu→y(s)U(s)

3

Introduction of the reference inputDynamics of the controller:

⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Ly +Mr

u = −Kx +Nr

inputs = y and r

outputs = u

Transfer function(s):

U(s) = Gy→u(s)Y (s) +Gr→u(s)R(s)4


⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Ly +Mr

u = −Kx +Nr

inputs = y and r

outputs = u




⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Ly +Mr

u = −Kx +Nr

inputs = y and r

outputs = u




Transfer functions for the whole system:

⎧⎪⎪⎨⎪⎪⎩Y (s) = Gu→y(s)U(s)U(s) = Gy→u(s)Y (s) +Gr→u(s)R(s)

poles = fixed by fixing K and L

zeros = ???

Y (s) = Gu→y(s)U(s)↓Y (s) = Gu→y(s)�Gy→u(s)Y (s) +Gr→u(s)R(s)�

↓�I −Gu→y(s)Gy→u(s)�Y (s) = Gu→y(s)Gr→u(s)R(s)

5





zeros = ???



5





zeros = ???



5





zeros = ???



5





zeros = ???



5





zeros = ???



5


Y (s) = �I −Gu→y(s)Gy→u(s)�−1Gu→y(s)Gr→u(s)R(s)

= Gy→y(s)Gr→u(s)R(s)Thus

Gr→u(s) = 0 ⇒ Gy→y(s)Gr→u(s) = 0

zeros from r to u are also zeros from r to y

zeros for the closed loop are the zerosfrom r to y and the zeros of the plant

6




Gr→u(s) = 0 ⇒ Gy→y(s)Gr→u(s) = 0



6




Gr→u(s) = 0 ⇒ Gy→y(s)Gr→u(s) = 0



6




Gr→u(s) = 0 ⇒ Gy→y(s)Gr→u(s) = 0



6




Gr→u(s) = 0 ⇒ Gy→y(s)Gr→u(s) = 0



6



I.e., we are interested only in r → u!!

7

Introduction of the reference inputWe are interested only in r → u!!

U(s) = Gy→u(s)Y (s) +Gr→u(s)R(s)↧

U(s) = Gr→u(s)R(s)Dynamics of the controller:

⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Ly +Mr

u = −Kx +Nr

↧⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Mr

u = −Kx +Nr

8

Introduction of the reference inputWe are interested only in r → u!!

U(s) = Gy→u(s)Y (s) +Gr→u(s)R(s)↧

U(s) = Gr→u(s)R(s)Dynamics of the controller:

⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Ly +Mr

u = −Kx +Nr

↧⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Mr

u = −Kx +Nr

8

Introduction of the reference inputWhat are the zeros of the controller?

⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Mr

u = −Kx +Nr

Transfer function:

Gr→u(s) =det �sI − (A −BK −LC) −M−K N

�det �sI − (A −BK −LC)�

9


det �sI − (A −BK −LC) −M−K N� = 0

⇓det⎡⎢⎢⎢⎢⎣sI − (A −BK −LC) −M

N−K 1

⎤⎥⎥⎥⎥⎦ = 0

⇓det⎡⎢⎢⎢⎢⎣sI − (A −BK −LC) − M

NK −M

N0 1

⎤⎥⎥⎥⎥⎦ = 0

⇓det �sI − (A −BK −LC) − M

NK� = 0

10


det �sI − (A −BK −LC) −M−K N� = 0

⇓det⎡⎢⎢⎢⎢⎣sI − (A −BK −LC) −M

N−K 1

⎤⎥⎥⎥⎥⎦ = 0

⇓det⎡⎢⎢⎢⎢⎣sI − (A −BK −LC) − M

NK −M

N0 1

⎤⎥⎥⎥⎥⎦ = 0


NK� = 0

10


det �sI − (A −BK −LC) −M−K N� = 0

⇓det⎡⎢⎢⎢⎢⎣sI − (A −BK −LC) −M

N−K 1

⎤⎥⎥⎥⎥⎦ = 0

⇓det⎡⎢⎢⎢⎢⎣sI − (A −BK −LC) − M

NK −M

N0 1

⎤⎥⎥⎥⎥⎦ = 0


NK� = 0

10


det �sI − (A −BK −LC) −M−K N� = 0

⇓det⎡⎢⎢⎢⎢⎣sI − (A −BK −LC) −M

N−K 1

⎤⎥⎥⎥⎥⎦ = 0

⇓det⎡⎢⎢⎢⎢⎣sI − (A −BK −LC) − M

NK −M

N0 1

⎤⎥⎥⎥⎥⎦ = 0


NK� = 0

10

Introduction of the reference inputHow do we design the zeros of the controller?

det �sI − (A −BK −LC) − M

NK� = 0

designing M

N= same problem as

choosing L for designing an observer

⇒ use acker!

How do we select αzeros(s)? → Chapters 3 and 4

11



NK� = 0

designing M

N= same problem as


⇒ use acker!


11



NK� = 0

designing M

N= same problem as


⇒ use acker!


11


designing M

Nplaces the closed loop zeros in


NK� b(s) = 0,

but what about the gain of the closed loop system?12


controller zeros: det �sI − (A −BK −LC) − M

NK� = 0

How do we select N , when M

Nhas already been selected? → must

ensure the closed loop gain to be 1

System equations:

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

�x˙x� = �A −BK BK

0 A −LC� �x

x� + � B

B −M�Nr

y = �C 0� �xx�

13


controller zeros: det �sI − (A −BK −LC) − M

NK� = 0

How do we select N , when M

Nhas already been selected? → must

ensure the closed loop gain to be 1

System equations:

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩


0 A −LC� �x

x� + � B

B −M�Nr

y = �C 0� �xx�

13


System equations:⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩


0 A −LC� �x

x� + � B

B −M�Nr

y = �C 0� �xx�

DC gain = closed loop TF (s = 0)

Algorithm:

��C 0��0 ⋅ I − �A −BK BK

0 A −LC��−1 �� B

B −M�N� = 1

14




0 A −LC� �x

x� + � B

B −M�Nr

y = �C 0� �xx�


Algorithm:

��C 0��0 ⋅ I − �A −BK BK

0 A −LC��−1 �� B

B −M�N� = 1

14




0 A −LC� �x

x� + � B

B −M�Nr

y = �C 0� �xx�


Algorithm:

��C 0��0 ⋅ I − �A −BK BK

0 A −LC��−1 �� B

B −M�N� = 1

14

?

15


are we finished?

yes no

need to talk about some practical cases

16


are we finished?

yes no

need to talk about some practical cases

16


what are the dynamics of x = x − x?

{ x = Ax +Bu

y = Cxplant

⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Ly +Mr

u = −Kx +Nrcontroller

⇒ ˙x = x − ˙x

= Ax −BKx +BNr−Ax +BKx +LCx −LCx −Mr

= (A −LC) x + (BN −M) r17



{ x = Ax +Bu

y = Cxplant

⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Ly +Mr


⇒ ˙x = x − ˙x


= (A −LC) x + (BN −M) r17



{ x = Ax +Bu

y = Cxplant

⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Ly +Mr


⇒ ˙x = x − ˙x


= (A −LC) x + (BN −M) r17



{ x = Ax +Bu

y = Cxplant

⎧⎪⎪⎨⎪⎪⎩˙x = (A −BK −LC) x +Ly +Mr


⇒ ˙x = x − ˙x


= (A −LC) x + (BN −M) r17

Introduction of the reference inputPractical case 1

˙x = (A −LC) x + (BN −M) r

we may want the dynamics of the error to be independent of r

(intuition: x should be free from external excitations)

18


˙x = (A −LC) x + (BN −M) r

we may want the dynamics of the error to be independent of r

(intuition: x should be free from external excitations)

18


Dynamics of the error independent of r implies˙x = (A −LC) x + (BN −M) r

⇓M = BN

19


What happens to the controller zeros when M = BN?


NK� = 0

↧det �sI − (A −LC)� = 0

20


What happens to the controller zeros when M = BN?


NK� = 0

↧det �sI − (A −LC)� = 0

20


Closed loop zeros:

det �sI − (A −LC)� b(s) = 0

Closed loop poles (principle of separation):

det �sI − (A −LC)�det �sI − (A −BK)� = 0

⇒ cancellations!21


˙x = (A −LC) x + (BN −M) r

sometimes we do not have y and r but only y − r

(examples: radars, thermostats, . . . )

22


˙x = (A −LC) x + (BN −M) r

sometimes we do not have y and r but only y − r

(examples: radars, thermostats, . . . )

22


˙x = (A −BK −LC) x +L(y − r)⇓

M = −L N = 0

23


What happens to the controller zeros when M = −L, N = 0?

det �sI − (A −BK −LC) −M−K N� = 0 ↦ det �sI − (A −BK −LC) L−K 0� = 0

↦ det �sI −A L−K 0� = 0

24


open loop zeros: det �sI −A −B

C D� = b(s) = 0

closed loop zeros: det �sI −A L−K 0� b(s) = 0

25

Introduction of the reference inputSummary

Closed loop transfer function:

G(s) = Y (s)R(s) =K

γ(s)b(s)αc(s)αe(s)

αc(s) = det [sI −A +BK] = controller polesαe(s) = det [sI −A +LC] = estimator poles

γ(s) = det �sI − (A −BK −LC) − M

NK� = reference zeros

b(s) = plant zerosK = overall gain (usually = 1)

26

?

27

summary of what has been done up to now

28

Integral control and robust tracking

To be controlled system:

{ x = Ax +Bu

y = Cx

Control design workflow:1 design the controller K

2 design the estimator L

3 design the reference M, N

4 ensure that the gain from r to y is 1

is this robust?

29


To be controlled system:

{ x = Ax +Bu

y = Cx

Control design workflow:1 design the controller K

2 design the estimator L

3 design the reference M, N

4 ensure that the gain from r to y is 1

is this robust?

29


if

{ x = Ax +Bu +B1w

y = Cx

then doing as before does not en-sure the gain from r to y to be 1!

30


Assumptions

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

{ x = Ax +Bu +B1w

y = Cxperturbed plant

r + α1r + α2r = 0 reference

w + α1w + α2w = 0 disturbance

? = r0 ≠ w0 =? unknown initial conditions

Control requirementse ∶= y − r should be s.t. lim

t→+∞ e(t) = 0 even under not preciseknowledge of (A, B, C)

31


Assumptions

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

{ x = Ax +Bu +B1w

y = Cxperturbed plant

r + α1r + α2r = 0 reference

w + α1w + α2w = 0 disturbance

? = r0 ≠ w0 =? unknown initial conditions

Control requirementse ∶= y − r should be s.t. lim

t→+∞ e(t) = 0 even under not preciseknowledge of (A, B, C)

31

Integral control and robust trackingExamples of what can be modeled with w + α1w + α2w = 0

w = 0⇒ step disturbance

w = 0⇒ ramp disturbance

w +w = 0⇒ sinusoidal disturbance

32

Integral control and robust trackingWorkflow

1 rewrite the state equations

{ x = Ax +Bu

y = Cx

so to include the dynamics of e = y − r

2 find a controller for the novel state equations

3 rephrase this controller in terms of the original state equations

33


r + α1r + α2r = 0 ” + ” e = y − r

⇓e + α1e + α2e = y + α1y + α2y

= Cx +Cα1x +Cα2x

= C (x + α1x + α2x)= Cξ

we choose ξ as a new state ⇒ must compute ξ

34


r + α1r + α2r = 0 ” + ” e = y − r

⇓e + α1e + α2e = y + α1y + α2y

= Cx +Cα1x +Cα2x

= C (x + α1x + α2x)= Cξ


34


r + α1r + α2r = 0 ” + ” e = y − r

⇓e + α1e + α2e = y + α1y + α2y

= Cx +Cα1x +Cα2x

= C (x + α1x + α2x)= Cξ


34


r + α1r + α2r = 0 ” + ” e = y − r

⇓e + α1e + α2e = y + α1y + α2y

= Cx +Cα1x +Cα2x

= C (x + α1x + α2x)= Cξ


34


r + α1r + α2r = 0 ” + ” e = y − r

⇓e + α1e + α2e = y + α1y + α2y

= Cx +Cα1x +Cα2x

= C (x + α1x + α2x)= Cξ


34


r + α1r + α2r = 0 ” + ” e = y − r

⇓e + α1e + α2e = y + α1y + α2y

= Cx +Cα1x +Cα2x

= C (x + α1x + α2x)= Cξ


34

Immediately:ξ ∶= x + α1x + α2x

⇓ξ = ...

x + α1x + α2x

From the original system:x = Ax +Bu +B1w

x = Ax +Bu +B1w...x = Ax +Bu +B1w

Thusξ = ...

x + α1x + α2x

= (Ax +Bu +B1w)+ α1 (Ax +Bu +B1w)+ α2 (Ax +Bu +B1w)= Aξ +B (u + α1u + α2u) + 0

indeed w + α1w + α2w = 0 by assumption!35


⇓ξ = ...

x + α1x + α2x



Thusξ = ...

x + α1x + α2x




⇓ξ = ...

x + α1x + α2x



Thusξ = ...

x + α1x + α2x




⇓ξ = ...

x + α1x + α2x



Thusξ = ...

x + α1x + α2x




⇓ξ = ...

x + α1x + α2x



Thusξ = ...

x + α1x + α2x



New definition:µ ∶= u + α1u + α2u

Thusξ = Aξ +B (u + α1u + α2u) = Aξ +Bµ

Summary of the equations:ξ = Aξ +Bµ

e + α1e + α2e = Cξ

⇒⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

0 1 0−α2 −α1 C

0 0 A

⎤⎥⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦+⎡⎢⎢⎢⎢⎢⎢⎣

00B

⎤⎥⎥⎥⎥⎥⎥⎦µ ↦ z = Aez +Beµ

(Ae, Be) ∶= error system

36




e + α1e + α2e = Cξ

⇒⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

0 1 0−α2 −α1 C

0 0 A

⎤⎥⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦+⎡⎢⎢⎢⎢⎢⎢⎣

00B

⎤⎥⎥⎥⎥⎥⎥⎦µ ↦ z = Aez +Beµ


36




e + α1e + α2e = Cξ

⇒⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

0 1 0−α2 −α1 C

0 0 A

⎤⎥⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦+⎡⎢⎢⎢⎢⎢⎢⎣

00B

⎤⎥⎥⎥⎥⎥⎥⎦µ ↦ z = Aez +Beµ


36




e + α1e + α2e = Cξ

⇒⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

0 1 0−α2 −α1 C

0 0 A

⎤⎥⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦+⎡⎢⎢⎢⎢⎢⎢⎣

00B

⎤⎥⎥⎥⎥⎥⎥⎦µ ↦ z = Aez +Beµ


36


Fact: if the roots of s2 + α1s + α2 are not zeros of (A, B, C, 0)then (Ae, Be) is fully controllable

Important:

(Ae, Be) fully controllable

⇓∃Ke = �K2 K1 K0� s.t.(Ae −BeKe) has arbitrary dynamics

through the feedback µ = −Kez

37


how do we go back to x and u from (Ae, Be) and Ke?

µ = −Kez = −�K2 K1 K0�⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦= −K2e −K1e −K0ξ

Old definitions:

µ ∶= u + α1u + α2u ξ ∶= x + α1x + α2x

Thus

u + α1u + α2u = −K2e −K1e −K0x − α1K0x − α2K0x≡(u +K0x) + α1 (u +K0x) + α2 (u +K0x) = −K2e −K1e

38



µ = −Kez = −�K2 K1 K0�⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦= −K2e −K1e −K0ξ

Old definitions:

µ ∶= u + α1u + α2u ξ ∶= x + α1x + α2x

Thus


38



µ = −Kez = −�K2 K1 K0�⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦= −K2e −K1e −K0ξ

Old definitions:

µ ∶= u + α1u + α2u ξ ∶= x + α1x + α2x

Thus


38



µ = −Kez = −�K2 K1 K0�⎡⎢⎢⎢⎢⎢⎢⎣e

e

ξ

⎤⎥⎥⎥⎥⎥⎥⎦= −K2e −K1e −K0ξ

Old definitions:

µ ∶= u + α1u + α2u ξ ∶= x + α1x + α2x

Thus


38


How to implement

(u +K0x) + α1 (u +K0x) + α2 (u +K0x) = −K2e −K1e ?

integrate!

(u +K0x) + α1 �1s

u + K0s

x� + α2 � 1s2 u + K0

s2 x� = −K2s2 e − K1

se

≡�1 + α1

s+ α2

s2 �u +K0 �1 + α1s+ α2

s2 �x = −�K1s+ K2

s2 � e

39


How to implement

(u +K0x) + α1 (u +K0x) + α2 (u +K0x) = −K2e −K1e ?

integrate!

(u +K0x) + α1 �1s

u + K0s

x� + α2 � 1s2 u + K0

s2 x� = −K2s2 e − K1

se

≡�1 + α1

s+ α2

s2 �u +K0 �1 + α1s+ α2

s2 �x = −�K1s+ K2

s2 � e

39

Integral control and robust trackingBlock schematic implementation

�1 + α1s+ α2

s2 �u +K0 �1 + α1s+ α2

s2 �x = −�K1s+ K2

s2 � e

40

Integral control and robust trackingPractical example: constant disturbance

r = 0 ⇒ u +K0x = −K1s

e

41

?

42

Table of Contents

1 Introduction of the reference input

2 Summary

3 Integral control and robust tracking – 7.10

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Documents

staff.damvar/Classes/R7003E-2015-LP2/Lessons/L09_a… · Introduction of the reference input How do we design the zeros of the controller? controller zeros: det sI−(A−BK−LC)−