R7003E - Automatic Control Lesson 11staff.damvar/Classes/R7003E-2015-LP2/Lessons/L11_annotated.pdfTable of Contents 1 Dead beat controllers 2 Control of non-fully controllable systems

R7003E - Automatic ControlLesson 11

1 December 2015

1

Table of Contents

1 Dead beat controllers

2 Control of non-fully controllable systems

3 Estimation of non-fully observable systems

4 Design for systems with pure time delays - 7.13

5 Digital control - 8

6 z-Transform – 8.2.1

7 Relationships between s and z - 8.2.3

8 Aliasing and anti-aliasing filters – 8.4.3

2

Dead beat controllers

these controllers are definedonly for discrete time systems

3


Assumption:

{ x(k + 1) = Ax(k) +Bu(k)y(k) = Cx(k)

fully controllable, i.e.,

C = �B AB . . . An−1B� full rank

full controllability≡can choose K so that eigenvalues of (A −BK) are arbitrary

4


Assumption:

{ x(k + 1) = Ax(k) +Bu(k)y(k) = Cx(k)

fully controllable, i.e.,

C = �B AB . . . An−1B� full rank

full controllability≡can choose K so that eigenvalues of (A −BK) are arbitrary

4


factcan choose K so that eigenvalues of (A −BK) are arbitrary⇓let’s choose K so that all the eigenvalues of (A −BK) are 0

factif all the eigenvalues of (A −BK) are 0 then ∃T s.t.

(A −BK) = T −1

⎡⎢⎢⎢⎢⎢⎢⎣J1 ⋱

Jn′

⎤⎥⎥⎥⎥⎥⎥⎦T with Ji =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1⋱ ⋱⋱ 10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦5



factif all the eigenvalues of (A −BK) are 0 then ∃T s.t.

(A −BK) = T −1

⎡⎢⎢⎢⎢⎢⎢⎣J1 ⋱

Jn′

⎤⎥⎥⎥⎥⎥⎥⎦T with Ji =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1⋱ ⋱⋱ 10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦5

Dead beat controllersImportant: ⎡⎢⎢⎢⎢⎢⎢⎣

J1 ⋱Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

k

=⎡⎢⎢⎢⎢⎢⎢⎣Jk

1 ⋱Jk

n′

⎤⎥⎥⎥⎥⎥⎥⎦∀k ∈ N

Examples: ⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

2

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 1 00 0 0 10 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

3

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 10 0 0 00 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦6


J1 ⋱Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

k

=⎡⎢⎢⎢⎢⎢⎢⎣Jk

1 ⋱Jk

n′

⎤⎥⎥⎥⎥⎥⎥⎦∀k ∈ N


0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

2

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 1 00 0 0 10 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

3

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 10 0 0 00 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦6


J1 ⋱Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

k

=⎡⎢⎢⎢⎢⎢⎢⎣Jk

1 ⋱Jk

n′

⎤⎥⎥⎥⎥⎥⎥⎦∀k ∈ N


0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

2

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 1 00 0 0 10 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

3

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 10 0 0 00 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦6


J1 ⋱Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

k

=⎡⎢⎢⎢⎢⎢⎢⎣Jk

1 ⋱Jk

n′

⎤⎥⎥⎥⎥⎥⎥⎦∀k ∈ N


0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

2

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 1 00 0 0 10 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

3

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 10 0 0 00 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦6


J1 ⋱Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

k

=⎡⎢⎢⎢⎢⎢⎢⎣Jk

1 ⋱Jk

n′

⎤⎥⎥⎥⎥⎥⎥⎦∀k ∈ N


0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

2

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 1 00 0 0 10 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

3

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 10 0 0 00 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦6


⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

4

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 00 0 0 00 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦Generalizing,

Ji ∈ Rν×ν ⇒ Jνi = 0ν×ν

Thus ⎡⎢⎢⎢⎢⎢⎢⎣J1 ⋱

Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

n

=⎡⎢⎢⎢⎢⎢⎢⎣Jn

1 ⋱Jn

n′

⎤⎥⎥⎥⎥⎥⎥⎦= 0n×n

7


⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

4

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 00 0 0 00 0 0 00 0 0 0



Thus ⎡⎢⎢⎢⎢⎢⎢⎣J1 ⋱

Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

n

=⎡⎢⎢⎢⎢⎢⎢⎣Jn

1 ⋱Jn

n′

⎤⎥⎥⎥⎥⎥⎥⎦= 0n×n

7


⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

4

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 00 0 0 00 0 0 00 0 0 0



Thus ⎡⎢⎢⎢⎢⎢⎢⎣J1 ⋱

Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

n

=⎡⎢⎢⎢⎢⎢⎢⎣Jn

1 ⋱Jn

n′

⎤⎥⎥⎥⎥⎥⎥⎦= 0n×n

7


⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

4

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 00 0 0 00 0 0 00 0 0 0



Thus ⎡⎢⎢⎢⎢⎢⎢⎣J1 ⋱

Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

n

=⎡⎢⎢⎢⎢⎢⎢⎣Jn

1 ⋱Jn

n′

⎤⎥⎥⎥⎥⎥⎥⎦= 0n×n

7


⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 1 0 00 0 1 00 0 0 10 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

4

=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 00 0 0 00 0 0 00 0 0 0



Thus ⎡⎢⎢⎢⎢⎢⎢⎣J1 ⋱

Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

n

=⎡⎢⎢⎢⎢⎢⎢⎣Jn

1 ⋱Jn

n′

⎤⎥⎥⎥⎥⎥⎥⎦= 0n×n

7



Implications:

(A −BK)n = T−1

⎡⎢⎢⎢⎢⎢⎢⎣J1 ⋱

Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

n

T = T−10T = 0

8



Implications:

(A −BK)n = T−1

⎡⎢⎢⎢⎢⎢⎢⎣J1 ⋱

Jn′

⎤⎥⎥⎥⎥⎥⎥⎦

n

T = T−10T = 0

8


What does (A −BK)n = 0 mean?

Open-loop system:

{ x(k + 1) = Ax(k) +Bu(k)y(k) = Cx(k)

Closed-loop system:

{ x(k + 1) = (A −BK)x(k)y(k) = Cx(k)

the closed loop x(k) is 0 after at most n steps!9



Open-loop system:

{ x(k + 1) = Ax(k) +Bu(k)y(k) = Cx(k)

Closed-loop system:

{ x(k + 1) = (A −BK)x(k)y(k) = Cx(k)




Open-loop system:

{ x(k + 1) = Ax(k) +Bu(k)y(k) = Cx(k)

Closed-loop system:

{ x(k + 1) = (A −BK)x(k)y(k) = Cx(k)



the closed loop x(k) is 0 after at most n steps!

10


Definition (d.b.c.)A K s.t. (A −BK) has all eigenvalues equal to zero is called adead beat controller

Q) is it necessary to have full controllability to be able to find ad.b.c.?

yes no

Example:

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

�x1(k + 1)x2(k + 1)� = �1 0

0 0� �x1(k)x2(k)� + �10�u(k)

y(k) = �1 1� �x1(k)x2(k)�

11




yes no

Example:

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

�x1(k + 1)x2(k + 1)� = �1 0

0 0� �x1(k)x2(k)� + �10�u(k)

y(k) = �1 1� �x1(k)x2(k)�

11




yes no

Example:

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

�x1(k + 1)x2(k + 1)� = �1 0

0 0� �x1(k)x2(k)� + �10�u(k)

y(k) = �1 1� �x1(k)x2(k)�

11


⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

�x1(k + 1)x2(k + 1)� = �1 0

0 0� �x1(k)x2(k)� + �10�u(k)

y(k) = �1 1� �x1(k)x2(k)�

Controllability matrix:

C = � �10� �1 00 0� �10� � = �1 1

0 0�At the same time, K = �K1 K2� implies

A −BK = �1 00 0� − �10� �K1 K2� = �1 −K1 −K2

0 0 �12


⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

�x1(k + 1)x2(k + 1)� = �1 0

0 0� �x1(k)x2(k)� + �10�u(k)

y(k) = �1 1� �x1(k)x2(k)�

Controllability matrix:

C = � �10� �1 00 0� �10� � = �1 1

0 0�At the same time, K = �K1 K2� implies

A −BK = �1 00 0� − �10� �K1 K2� = �1 −K1 −K2

0 0 �12


⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

�x1(k + 1)x2(k + 1)� = �1 0

0 0� �x1(k)x2(k)� + �10�u(k)

y(k) = �1 1� �x1(k)x2(k)�

non controllable

A −BK = �1 −K1 −K20 0 � closed loop

K1 = 1 K2 = 0 ⇒ A −BK = �0 00 0�

i.e., K = d.b.c.

13


⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

�x1(k + 1)x2(k + 1)� = �1 0

0 0� �x1(k)x2(k)� + �10�u(k)

y(k) = �1 1� �x1(k)x2(k)�

non controllable

A −BK = �1 −K1 −K20 0 � closed loop

K1 = 1 K2 = 0 ⇒ A −BK = �0 00 0�

i.e., K = d.b.c.

13


(A, B) fully controllable ⇒ exists d.b.c.

(A, B) fully controllable /⇐ exists d.b.c.

14

Dead beat observers

(d.b.c.)d = d.b.o.

Caveat: dead beat corresponds in continuous time to haveinfinitely fast poles, thus

d.b.c. = use the actuators a lotd.b.o. = amplify a lot the noise

15

Dead beat observers

(d.b.c.)d = d.b.o.

Caveat: dead beat corresponds in continuous time to haveinfinitely fast poles, thus

d.b.c. = use the actuators a lotd.b.o. = amplify a lot the noise

15

?

16

Control of non-fully controllable systems

Assumption:

��A11 A120 A22

� , �B10 � , [C1, C2] , D�⇓

x2 non controllable

Potential cases:A22 has only stable eigenvalues ⇒ lim

t→+∞x2 = 0

A22 has some unstable eigenvalues ⇒ in general settingslim

t→+∞ ∥x2∥ = +∞

17


Assumption:

��A11 A120 A22

� , �B10 � , [C1, C2] , D�⇓

x2 non controllable


t→+∞x2 = 0


t→+∞ ∥x2∥ = +∞

17


Assumption:

��A11 A120 A22

� , �B10 � , [C1, C2] , D�⇓

x2 non controllable


t→+∞x2 = 0


t→+∞ ∥x2∥ = +∞

17


Thus

��A11 A120 A22

� , �B10 � , [C1, C2] , D� ” + ” A22 unstable

⇓the system in general diverges

in this case you are ******only way to control the system is to add some actuators

(i.e., change B)

18


Thus

��A11 A120 A22

� , �B10 � , [C1, C2] , D� ” + ” A22 unstable

⇓the system in general diverges

in this case you are ******only way to control the system is to add some actuators

(i.e., change B)

18


Case

��A11 A120 A22

� , �B10 � , [C1, C2] , D� ” + ” A22 stable

Fact:poles relative to A11 can be allocated arbitrarilypoles relative to A22 can not be allocated

what you can control is x1

19


Case

��A11 A120 A22

� , �B10 � , [C1, C2] , D� ” + ” A22 stable



19


Case

��A11 A120 A22

� , �B10 � , [C1, C2] , D� ” + ” A22 stable



19

Control of non-fully controllable systemsAlgorithm:

1 assume A11 ∈ Rn1×n1 , A22 ∈ Rn2×n2

2 consider (A11, B1)3 design the new poles positions for this subsystem, say αc(s)4 design K ∈ R1×n1 so that det (sI −A11 +B1K) = αc(s)

(acker!)5 set u = −�K 01×n2�x = −Kx1

6 closed loop system:

x = ��A11 A120 A22

� − �B1K

0 ��x = ��A11 −B1K A120 A22

��x

discrete time systems = the very same

20


1 assume A11 ∈ Rn1×n1 , A22 ∈ Rn2×n2


(acker!)5 set u = −�K 01×n2�x = −Kx1


x = ��A11 A120 A22

� − �B1K

0 ��x = ��A11 −B1K A120 A22

��x


20


1 assume A11 ∈ Rn1×n1 , A22 ∈ Rn2×n2


(acker!)5 set u = −�K 01×n2�x = −Kx1


x = ��A11 A120 A22

� − �B1K

0 ��x = ��A11 −B1K A120 A22

��x


20


1 assume A11 ∈ Rn1×n1 , A22 ∈ Rn2×n2


(acker!)5 set u = −�K 01×n2�x = −Kx1


x = ��A11 A120 A22

� − �B1K

0 ��x = ��A11 −B1K A120 A22

��x


20


1 assume A11 ∈ Rn1×n1 , A22 ∈ Rn2×n2


(acker!)5 set u = −�K 01×n2�x = −Kx1


x = ��A11 A120 A22

� − �B1K

0 ��x = ��A11 −B1K A120 A22

��x


20


1 assume A11 ∈ Rn1×n1 , A22 ∈ Rn2×n2


(acker!)5 set u = −�K 01×n2�x = −Kx1


x = ��A11 A120 A22

� − �B1K

0 ��x = ��A11 −B1K A120 A22

��x


20


1 assume A11 ∈ Rn1×n1 , A22 ∈ Rn2×n2


(acker!)5 set u = −�K 01×n2�x = −Kx1


x = ��A11 A120 A22

� − �B1K

0 ��x = ��A11 −B1K A120 A22

��x


20

Estimation of non-fully observable systems

�estimation of non-fully observable systems�≡�control of non-fully controllable systems�d

21

?

22

Design for systems with pure time delays

this is only for continuous time systems

23

Design for systems with pure time delaysTime delay: Dτ [f(t)] = f(t − τ)Laplace-transform:

L(Dτ [f(t)]) (s) = ∫ +∞−∞ Dτ [f(t)] e−stdt

= ∫ +∞−∞ f(t − τ)e−stdt

= ∫ +∞−∞ f(T )e−s(T+τ)dT

= ∫ +∞−∞ f(T )e−sT e−sτ dT

= e−sτ ∫ +∞−∞ f(T )e−sT dT

= e−sτL(f(t)) (s)24



= ∫ +∞−∞ f(t − τ)e−stdt

= ∫ +∞−∞ f(T )e−s(T+τ)dT

= ∫ +∞−∞ f(T )e−sT e−sτ dT

= e−sτ ∫ +∞−∞ f(T )e−sT dT

= e−sτL(f(t)) (s)24



= ∫ +∞−∞ f(t − τ)e−stdt

= ∫ +∞−∞ f(T )e−s(T+τ)dT

= ∫ +∞−∞ f(T )e−sT e−sτ dT

= e−sτ ∫ +∞−∞ f(T )e−sT dT

= e−sτL(f(t)) (s)24



= ∫ +∞−∞ f(t − τ)e−stdt

= ∫ +∞−∞ f(T )e−s(T+τ)dT

= ∫ +∞−∞ f(T )e−sT e−sτ dT

= e−sτ ∫ +∞−∞ f(T )e−sT dT

= e−sτL(f(t)) (s)24



= ∫ +∞−∞ f(t − τ)e−stdt

= ∫ +∞−∞ f(T )e−s(T+τ)dT

= ∫ +∞−∞ f(T )e−sT e−sτ dT

= e−sτ ∫ +∞−∞ f(T )e−sT dT

= e−sτL(f(t)) (s)24



= ∫ +∞−∞ f(t − τ)e−stdt

= ∫ +∞−∞ f(T )e−s(T+τ)dT

= ∫ +∞−∞ f(T )e−sT e−sτ dT

= e−sτ ∫ +∞−∞ f(T )e−sT dT

= e−sτL(f(t)) (s)24



= ∫ +∞−∞ f(t − τ)e−stdt

= ∫ +∞−∞ f(T )e−s(T+τ)dT

= ∫ +∞−∞ f(T )e−sT e−sτ dT

= e−sτ ∫ +∞−∞ f(T )e−sT dT

= e−sτL(f(t)) (s)24

Design for systems with pure time delaysIn time:

u(t) Dτ G(s) y(t)u(t − τ)

In Laplace:

U(s) Dτ G(s) Y (s)e−sτ U(s)

Overall transfer function:Y (s)U(s) = e−sτ G(s)

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In Laplace:



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In Laplace:



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big problem: e−sτ is not rational!!

e−sτ = 1est= 1

1 + st + (st)2/2! + (st)3/3! + . . .

Implication: there is no finite state description of e−sτ

what shall we do?

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e−sτ = 1est= 1

1 + st + (st)2/2! + (st)3/3! + . . .


what shall we do?

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e−sτ = 1est= 1

1 + st + (st)2/2! + (st)3/3! + . . .


what shall we do?

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Design for systems with pure time delaysSmith compensator

Algorithm:1 design a controller D(s) considering simply G(s)2 complete the controller by putting it in this way:

Drawback: must know G(s) and e−sτ very accurately!

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Design for systems with pure time delaysSmith compensator

Algorithm:1 design a controller D(s) considering simply G(s)2 complete the controller by putting it in this way:

Drawback: must know G(s) and e−sτ very accurately!

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Design for systems with pure time delays – discrete timecase

Case τ = multiple of the sampling period Δ

example: τ = 3Δ

y(k + 1) = b0 + b1z−1

1 + a1z−1 + a2z−2 u(k − 3)= b0z−3 + b1z−4

1 + a1z−1 + a2z−2 u(k)

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example: τ = 3Δ

y(k + 1) = b0 + b1z−1

1 + a1z−1 + a2z−2 u(k − 3)= b0z−3 + b1z−4

1 + a1z−1 + a2z−2 u(k)

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example: τ = 3Δ

y(k + 1) = b0 + b1z−1

1 + a1z−1 + a2z−2 u(k − 3)= b0z−3 + b1z−4

1 + a1z−1 + a2z−2 u(k)

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Case τ ≠ multiple of the sampling period Δ

example: τ = 3.4Δ1 Δ≪ bandwidth of the system (∼10 to 20 times) + control

system requirements not too tight ⇒ may ignore the .4 andtreat the 3 as before

2 otherwise ⇒ treat the 3 as before + do a Smith compensatorfor the .4

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Digital control

Continuous time

y(t) = b(s)a(s)u(t)

Discrete time

y(k) = b(z)a(z)u(k)

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Digital control

Continuous time

y(t) = b(s)a(s)u(t)

Discrete time

y(k) = b(z)a(z)u(k)

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Digital control

Continuous time

Discrete time

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Digital controlWorkflow

physics-based considerations system identification procedures

{ x = Ax +Bu

y = Cx{ x(k + 1) = Ax(k) +Bu(k)

y(k) = Cx(k)

continuous controller discrete controller

implementation

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{ x = Ax +Bu

y = Cx{ x(k + 1) = Ax(k) +Bu(k)

y(k) = Cx(k)


implementationdiscrete equivalent

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{ x = Ax +Bu

y = Cx{ x(k + 1) = Ax(k) +Bu(k)

y(k) = Cx(k)


implementationdiscrete design

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Table of Contents

1 Dead beat controllers

2 Control of non-fully controllable systems

3 Estimation of non-fully observable systems

4 Design for systems with pure time delays - 7.13

5 Digital control - 8

6 z-Transform – 8.2.1

7 Relationships between s and z - 8.2.3

8 Aliasing and anti-aliasing filters – 8.4.3

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Documents

R7003E - Automatic Control Lesson 11staff.damvar/Classes/R7003E-2015-LP2/Lessons/L11_annotated.pdfTable of Contents 1 Dead beat controllers 2 Control of non-fully controllable systems