Lecture 18 - Summary of the Course

Preparations for Final Examination

1. Dynamics

K. J. strm

1. Dynamics
2. PID Control
3. The Simple Feedback Loop

Most problems require knowledge about dynamical systems

Problem 6 - Physical modeling and observability
Problem 5 - Linearization
Problem 3 - Stability

2. PID Control

4. Feedforward

Problem 1, Problem 2, Problem 3

5. State Feedback and Observers

3. The Simple Feedback Loop

6. Summary

Problem 4, Problem 1, Problem 2

Theme: Reviewing what we learned and filling some gaps.

4. Feedforward
Problem 10, Problem 1, Problem 2
5. State Feedback and Observers
Problem 7, Problem 8, Problem 9

Homework Problems and Midterm

1. Dynamics

1. State models

Homework 1, Homework 2, Homework 3, Midterm

2. PID Control
Homework 2, Midterm
Homework 2, Homework 4, Homework 5, Midterm
Homework 6, Midterm
5. State Feedback and Observers
Homework 6, Homework 7

Linearization
Linear time invariant systems
2. Input-output models

3. The Simple Feedback Loop

4. Feedforward

1. Dynamics

Transient response, step response, impulse response

Transfer function, poles, zeros
Frequency response, Bode and Nyquist plots
3. Controllability, observability, system structure
4. Relations between representations
Coordinate changes, standard forms
5. Stability
6. Modeling (from physics and from data)

c K. J. strm August, 2001

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State Models

Linearization

State a collection of variables that summarize the past for

the purpose of predicting the future, e.g.
dx
= f ( x , u)
dt
y = g( x, u)

dx
= f ( x, u), y = g( x, u)
dt
The stationary solution x0 , u0 is obtained by solving the
equation f ( x0 , u0 ) = 0. For small deviations x = x0 + x,
u = u0 + u, y = y0 + x around the equilibrium we have
d x
= A x + B u
dt

x state, u control signal, y measured signal

The states describe storage of mass energy and momentum. Constitutive equations are also needed to obtain the
model.

y = C x + D u

dx
= Ax + Bu
dt
y = Cx + Du

Synonyms: internal descriptions, white boxes

Linear Time Invariant Systems

Input-Output Models
Input

x( t) = e At x(0) +

e A( t s) Bu( s) ds

y( t) = Ce x(0) + C
At

A ( t s )

Vf
( x0 , u0 )
Vu
Vg
D=
( x0 , u0 )
Vu
B=

Change notation: x x, u u, y y

Key problem: How accurately do we need to describe the

storage?

dx
= Ax + Bu
dt
y = Cx + Du

Vf
( x0 , u0 )
Vx
Vg
C=
( x0 , u0 )
Vx
A=

System

Output

Table over all input-output pairs

Bu( s) ds + Du( t)

Superposition: if (u1, y1) and (u2 , y2) are i-o pairs so is

(au1 + bu2, a y1 + b y2)

Think in scalars interpret as vectors and matrices!

If the matrix A can be diagonalized A = T T 1 we have
eAt = Tet T 1 . The elements of eAt is a sum of terms e i t . In
general when the
multiple eigenvalues the elements
P matrix has
t
i
ki Pi (t) e where i is an eigenvalue and P(t)
have the form
is a polynomial

Table can be simplified drastically if initial conditions

neglected
Rt
Impulse response y(t) = 0 g(t s)u(s)ds
Transfer function Y (s) = G (s) U (s)
Frequency response

c K. J. strm August, 2001

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Laplace Transforms

Some Properties of the Laplace Transform

Consider a function f defined on 0 t < and a real number

> 0. Assume that f grows slower than e t for large t. The
Laplace transform F = L f of f is defined as
Z
L f = F ( s) =
est f (t)dt
0

Example 1:

f (t) = 1,

F ( s) =

Example 2:
at

f (t) = e

Z
,

F ( s) =

st

1 st  1
dt = e  =
s
0
s

e(s+a)t dt =

1 st
1
e  =
s+a
0
s+a

est f (t)dt

Transform of the derivative

Z
df
L =
est f P (t)dt = f (0) + sL f
dt
0
= sL f f (0)
Differentiation: L df
dt
Rt
Integration: L 0 f ( )d = 1s L f
Rt
Convolution: L 0 f (t )g( )d = F (s) G (s)
Final value Theorem: lim s0 sF (s) = lim t f (t)
Initial value Theorem: lims sF (s) = lim t0 f (t)

Shape of e t for Real and Complex

Poles and Zeros

<0

The transfer function of an LTI system is a rational function

2.5
0.8

The denominator polynomial A(s) = det(sI A) is the

characteristic polynomial of the matrix A. (Notice a slight
notational conflict!)

0.6

1.5

0.4

0.2

0.5

= 0.25

Poles are the zeros of the polynomial A(s), they are also the
eigenvalues of the matrix A. A pole means that the response
of the system has the component e t .
y
c K. J. strm August, 2001
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0.4

0.6

0.8

= 0.25

20

0.2

10

0.2

10

0.4

0.2

30

0.4

Zeros are the zeros of the polynomial B (s), they depend on

all matrices in the state description and reflect how the states
are coupled to the inputs and outputs. A zero means that
transmission of the signal e t is blocked.

40

0.6

B ( s)
b1 sn1 + b2 sn2 + . . . + bn
= n
A(s)
s + a1 s n 1 + a2 s n 2 + . . . + a n

>0

G (s) = C(sI A)1 B =

L f = F ( s) =

Linearity: L (a f + bg) = aL f + bL g
e

10

15

20

10

15

Frequency Response

Frequency Response ...

The steady state transmission of sinusoids is given by G (i )

0.1

h G (i )h gain, amplitude ratio

0
0

10

15

arg G (i ), phase

Graphical representations of G (i )

1
0

10

Nyquist plot

15

An LTI system is completely characterized by its steady state

response to sinusoidal signals.
u( t) = u0 sin t
y( t) = u0 h G ( i )h sin t + arg G ( i )

Bode plots

Bode Plot

The Nyquist Curve

Represent the complex number G (i ) as a point in the complex plane. The Nyquist curve is the locus of the point as frequency changes.
Im G(i )

Interpretation

Interpretation

Ultimate point

Critical point
Positive feedback
Negative feedback

Represent the function G (i ) as two curves, one for the

magnitude h G (i )h (the gain curve or amplitude ratio) and one
for the phase arg G (i ) (the phase curve). Use logarithmic
scales for frequency and gain and linear scales for the phase.

Re G(i )
a

Asymptotes

Gain

10

10

Minimum phase
Relation between curves

Stability

10

10

10

10

10
w

10

10

Phase
0

50

d log h G (i )h
arg G (i ) 
2
d log
c K. J. strm August, 2001
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100

150
1

10

10

10
w

10

10

Non-minimum Phase
Right half plane zero
as
Transfer function G (s) =
,
a+s

dx
= Ax + Bu
dt
y = C x + Du

h G (i )h = 1 and

arg G (i ) = 2 arctan
Time delay
Transfer function G (s) = esT ,

Controllability and Observability

The system can be steered from one point to another if the controllability matrix Wc has full rank

Wc = B AB . . . An1 B

h G (i )h = 1 and

arg G (i ) = T
Right half plane pole

Transfer function G (s) =

s+b
,
sb

The state can be computed from the output if the observability

matrix Wo has full rank
C

CA

Wo = ..

h G (i )h = 1 and

arg G (i ) = 2 arctan

= + 2 arctan

C An1

System Structure

Relations

An LTI system can be split into four subsystems. The transfer

function is uniquely given by the subsystem Sco
Soc observable and
controllable
Soc
not observable but
controllable
Soc observable but not
controllable
Soc neither observable
nor controllable

Soc

dx
= Ax + Bu
dt
y = Cx + Du

Input-output relation
Z
y( t) = Ce At x(0) + Du( t) + C

Soc

Soc-

= Ce x(0) +
At

e A( t ) Bu( ) d

0
t

g( t )u( ) d

-Soc

Impulse response
g(t) = D (t) + CeAt B

c K. J. strm August, 2001

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Relations ...

Coordinate Changes

dx
= Ax + Bu
dt
y = Cx + Du

dx
= Ax + Bu
dt
y = Cx + Du

Change coordinates by the transformation z = T x, x = T 1 z

Transfer function
b0 sn + b1 sn1 + b2 sn2 + . . . + bn
G ( s) = D + C[sI A]1 B = n
s + a1 sn1 + a2 sn2 + . . . + an
Z
=
e stg( t) dt

dz
+ Bu

= T AT 1 z + T Bu = Az
dt
+ Du
y = CT 1 z + Du = Cz

Same form but different matrices. Some relations

or equivalently

At
B
g( t) = D ( t) + Ce At B = D ( t) + Ce
1
G ( s) = D + C( sI A) B = D + C ( sI A )1 B

dn y
d n1
dnu
d n1 u
+
a
+
.
.
.
+
a
y
=
b
+
b
+ . . . + bnu
1
n
0
1
dtn
dtn1
dtn
dtn1

c = T Wc ,
W

The Diagonal Form

o = Wo T 1 X S
W

Controllable Canonical Form

If the controllability matrix Wc has full rank we have

If the matrix A can be diagonalized we have

a a . . . a
1
1
2
n1 an

1
0
0
0
0

dz

0
1
0
0
0

=
z
+
u

dt

.
.

.
.

.
.

0
0
1
0
0

y = b1 b2 . . . bn1 bn z + Du

1
1
0

2
2

dz

=
z
+
u

.
.

.
.
dt

.
.

0
n
n

y = 1 2 . . . n z + Du

Transfer function

The transfer function is

1 1
2 2
n n
G ( s) = D +
+
+ ... +
s 1
s 2
s n
The mode zi is observable if i = 0 and controllable if i = 0

G ( s) = D +

b 1 s n1 + b 2 s n2 + . . . + b n
sn + a1 sn1 + a2 sn2 + . . . + an

The characteristic polynomial is

c K. J. strm August, 2001

&

sn + a1 sn1 + a2 sn2 + . . . + an
6

Observable Canonical Form

a1
1

a2

dz

.
..
=

dt

an1 0

an
0

y = 1 0 0...

...

0
1

0
0

From Transfer Functions to State Equations

0
b1

0
b2

.
z
+
u

0
bn

Y ( s)
1
1
=
= 4
4
3
U ( s)
(s + 1)
s + 4s + 6s2 + 4s + 1

Hence

(s4 + 4s3 + 6s2 + 4s + 1) Y (s) = U (s)

Switch to time domain

0 z + Du

d4 y
d3 y
d2 y
dy
+4 3 +6 2 +4
+y=u
4
dt
dt
dt
dt

Transfer function
G ( s) = D +

b1 sn1 + b2 sn2 + . . . + bn
s n + a1 s n 1 + a2 s n 2 + . . . + a n

Characteristic polynomial
s n + a1 s n 1 + a2 s n 2 + . . . + a n

From Transfer Functions to State Equations ...

From Transfer Functions to State Equations

d4 y
d3 y
d2 y
dy
+
4
+
6
+4
+y=u
4
3
2
dt
dt
dt
dt

Y ( s)
b1 s2 + b2 s + b3
= 3
U ( s)
s + a1 s2 + a2 s + a3

Introduce
x1 =

d y
,
dt3
dx1
dt
dx2
dt
dx3
dt
dx4
dt

x2 =

d y
,
dt2

x3 =

dy
,
dt

x4 = y

= 4x1 6x2 4x3 x4 + u

= x1
= x2

s2
U
s3 + a1 s2 + a2 s + a3
s
1
X2 = 3
U = X1
s + a1 s2 + a2 s + a3
s
1
1
1
X3 = 3
U = X2 = 2 X1
2
s + a1 s + a2 s + a3
s
s
1
1
s X 1 = a1 X 1 a2 X 1 a3 2 X 1 + U
s
s
= a1 X 1 a2 X 2 a3 X 3 + U
X1 =

s X2 = X1
s X3 = X2

= x3

Y = b1 X 1 + b2 X 2 + b3 X 3
c K. J. strm August, 2001
&

From Transfer Functions to State Equations ...

From Transfer Functions to State Equations

Y ( s)
b1 s2 + b2 s + b3
= 3
U ( s)
s + a1 s2 + a2 s + a3

1
1
sX 1 = a1 X 1 a2 X 1 a3 2 X 1 + U
s
s
= a1 X 1 a2 X 2 a3 X 3 + U

s3 X 1 + a1 s2 X 1 + a2 s X 1 + a3 X 1 = b1 s2 U + b2 sU + b3 U
1
1
1

s X 1 = a1 X 1 + b1 U +
a2 X 1 a3 X 1 + b2 U + b3 U
s
s
s
1
1
1

X 2 = 2 a2 X 1 a3 X 1 + b2 U + b3 U
s
s
s
s X 1 = a1 X 1 + b1 U + X 2 = a1 X 1 + X 2 + b1 U


1
s X 2 = a2 X 1 + b2 U +
a3 X 1 + b3 U
s


1
a2 X 1 a3 X 1 + b3 U
X3 =
s
s X 2 = a2 X 1 + b2 U + X 3 = a2 X 1 + X 3 + b2 U

sX 2 = X 1
sX 3 = X 2
Y = b1 X 1 + b2 X 2 + b3 X 3

Back to time domain

dx1
= a1 x1 a2 x a3 x3 + u
dt
dx2
= x1
dt
dx3
= x2
dt
y = b1 x1 + b2 x2 + b3 x3

s X 3 = a3 X 1 + b3 U

From Transfer Functions to State Equations ...

sX 1 = a1 X 1 + b1 U + X 2 = a1 X 1 + X 2 + b1 U
sX 2 = a2 X 1 + b2 U + X 3 = a2 X 1 + X 3 + b2 U
sX 3 = a3 X 1 + b3 U

In the time domain

dx1
= a1 x1 + x2 + b1 u
dt
dx2
= a2 x1 + x3 + b2 u
dt
dx3
= a3 x1 + b3 u
dt
y = x1

Stability
For the nonlinear system
dx
= f ( x)
dt

only stability of a solution.

For LTI systems stability
of one solution guarantees
stability of all
dx
= Ax
dt

Stable
1

Unstable
1

0.5

0.5

0
0

10

20

30

40

50

Stable
1

10

20

30

40

50

30

40

50

Unstable
1

1
0

0
0

1
10

20

30

40

50

10

20

An LTI system is stable if all eigenvalues of A are in the left half

plande or, equivalently, if the characteristic equation
det (sI A) = sn + a1 sn1 + a2 sn2 + . . . + an = 0

has all roots in the left half plane.

c K. J. strm August, 2001
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The Nyquist Criterion

Stability Margins
The margins express the distance from the Nyquist curve to
the critical point in different ways. Two classical measures: gain
margin gm and phase margin m. An alternative is the shortest
distance to the critical point 1 which is 1/ Ms , where Ms is the
maximum sensitivity.

5
5

Investigate propagation of sinusoids through the closed loop.

An oscillation can be maintained if L(i 0 ) = 1. Notice that
the feedback is positive for the frequency 0 .
The point 1 is the critical point!
Stability based on encirclements of critical point.

Sensitivity to Process Variations

Closed loop transfer function with error feedback
T=

Modeling
Modeling from physics
Use knowledge of physics and engineering to derive
differential equations

PC
1 + PC

Try to understand what factors are relevant for the problem

Small variations in the process

Useful to think of frequencies and magnitudes

dh T h
1
dh P h
dh P h
=
=S
hT h
h1 + PCh h Ph
h Ph

States describe storage of mass momentum and energy

Modeling from data

Large variations stable if

Introduce perturbations (steps, pulses, sinusoids) to the

process and observe its response

1
h Ph
<
h Ph
hT h

Maximum sensitivities Ms = max h S (i )h, Mt = max hT (i )h

Fit LTI models (system identification)

Often useful to combine the approaches

c K. J. strm August, 2001

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2. PID Control

PI Control of First Order System

Process: P(s) =

b
s+a

ki
ki
, Cf f (s) = k +
s
s
Closed loop transfer function from reference r to output y

Controller: Cf b (s) = k +

Y ( s)
PC
b( ks + ki)
=
= 2
R( s )
1 + PC
s + (a + bk)s + bki






1
sTd
Y ( s)
U (s) = k R(s) Y (s)) +
R( s ) Y ( s ) +
sTi
1 + sTd/ N
Notice

All poles can be assigned. A good characteristic polynomial is

s2 + 2 0 s + 02

Identification of coefficients of equal powers of s gives

Set point weighting (Cheap 2 DOF)

k = (2 0 a)/b,

Filtering of derivative

PD Control of Second Order System

Process: P(s) =

s2

b
+ a1 s + a2

Controller: C(s) = k + kd s,

PI Control of Second Order System

Process: P(s) =

Cf f (s) = k + kds

s2

b
+ a1 s + a2

ki
s
Closed loop transfer function from reference r to output y with
error feedback

Controller: C(s) = k +

Closed loop transfer function from reference r to output y

Y ( s)
PC
b( kds + k)
=
= 2
R( s )
1 + PC
s + (a1 + bkd)s + a2 + bk

Y ( s)
PC
b( ks + ki )
=
=
R( s )
1 + PC
s( s2 + a1 s + a2 ) + b( ks + ki )
b( ks + ki )
= 3
s + a1 s2 + ( a2 + bk) s + bki

All poles can be assigned. A good characteristic polynomial is

s2 + 2 0 s + 02

Identification of coefficients of equal powers of s gives

k = ( 02 a2 )/b,

ki = 02 /b

Not enough parameters to assign all poles.

kd = (2 0 a1 )/b
c K. J. strm August, 2001
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10

PI Control of Second Order System

PID Control of Second Order System

A compromise

Process: P(s) =

Y ( s)
b( ks + ki)
= 3
2
R( s )
s + a1 s + (a2 + bk)s + bki

ki
+ kds
s
Closed loop transfer function from reference r to output y

s3 + 2 0 s2 + 2 02 s + 03

Y ( s)
PC
b( kd s2 + ks + ki )
=
=
2
R( s )
1 + PC
s( s + a1 s + a2 ) + b( kd s2 + ks + ki )

and also choose 0 (response speed). Hence

0 =

a1
,
2

k=

2 a2
=
b

a21 /2

a2
,
b

ki =

b
+ a1 s + a2

Controller: C(s) = k +

Make characteristic polynomial equal to

2
0

s2

3
0

a31
8b

s3

b( kd s2 + ks + ki )
+ ( a1 + bkd ) s2 + ( a2 + bk) s + bki

All poles can be assigned.

Pole pattern can be chosen but the response speed is given by

process parameter a1 .

PID Control of Second Order System ...

Y ( s)
PC
b( kds2 + ks + ki)
=
= 3
R( s )
1 + PC
s + (a1 + bkd)s2 + (a2 + bk)s + bki
A good characteristic polynomial is
s3 + 2s2 0 + 2 02 s + 03

Identification of coefficients of equal powers of s gives

2 02 a2
k=
,
b

ki =

03
b

2 0 a1
kd =
b

Controllers with Error Feedback

Error feedback
U (s) = C(s)( R(s) Y (s)) = ( k +

ki
+ kds)( R(s) Y (s))
s

gives the closed loop transfer function from reference r to

output y
Y ( s)
PC
b( kds2 + ks + ki)
=
= 3
R( s )
1 + PC
s + (a1 + bkd)s2 + (a2 + bk)s + bki

All poles can be placed arbitrarily. Can something be done

about the zeros?

c K. J. strm August, 2001

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11

Controllers with Set Point Weighting

Controller with set point weighting (a kind of feedforward)
ki
( R( s) Y ( s)) + kd s( R( s) Y ( s))
s
ki
ki
= ( k +
+ kd s) R( s) ( k +
+ kd s) Y ( s)
s
s
= Cr ( s) R( s) C( s) Y ( s)

U ( s) = k( R( s) C( s)) +

Saturation Windup and Windup Protection

The feedback loop is broken if signals saturates. Problem
with unstable modes in the process and the controller. The
integrator in a PI is a typical example.
y

KT d s

e = r y

K
Ti

1
s

es

1
Tt

The numerator can be modified by choosing the parameters

and .

Actuator

gives the closed loop transfer function

Y ( s)
Cr (s) P(s)
b( kds2 + ks + ki)
=
= 3
R( s )
1 + P ( s) C ( s)
s + (a1 + bkd)s2 + (a2 + bk)s + bki

Actuator
model

Look for the error. Assume system works.

Controller Tuning

Ziegler-Nichols Step Response Method

Model based: Make a model and apply some control

design method e.g. the methods just discussed

Switch controller to manual.

Make simple experiments on the process and adjust

Log process output. Normalize the curve so that it corresponds to a unit step.

Make a step in the control variable.

Ziegler and Nichols methods

Determine parameters a and L. The controller parameters

are obtained from a table.

Good idea but systems are not well damped and too
little process information is used
Open loop experiments (bump test)
Closed loop experiments

Improved tuning rules

A0

a
L

c K. J. strm August, 2001

&

Controller

Ti

P
PI

1/a
0.9/a

3L

PID

1.2/a

2L

Td

Tp
4L
5.7L

L/2

3.4L

12

Ziegler-Nichols Frequency Response Method

Switch the controller to pure proportional.

Properties

Adjust the gain so that the closed loop system is at the

stability boundary.
Determine the gain ku (the ultimate gain) and the period
Tu (the ultimate period) of the oscillation.
Suitable controller parameters are obtained from a table.
Im G(i )
Ultimate point

Re G(i )

Reg.

0.5ku

PI

0.4ku

0.8Tu

PID

0.6ku

0.5Tu

Ti

Properties of Ziegler and Nichols Rules

+ Very common
- The closed loop system obtained too oscillatory  0.2.
Part of the criterion (quarter amplitude damping)
- Too large overshoot

Tp

- Sensitive to process variations

Tu

- Can only be used on a restricted class of processes

1.4Tu

A good idea to base tuning on simple process experiments.

0.85Tu

Large scope for improvements.

Td

0.125Tu

+ Easy to explain and use

More process information needed.

A Ziegler Nichols Replacement

Computer Implementation

ZN is basically a sound method, BUT

Sampling, aliasing and anti-aliasing filters

Damping specified is too low < = 0.2

Too little process information (Two parameters ku, Tu )
An additional parameter give a big improvement

Approximate derivatives by differences

Use backward differences
Organization of the code

= h G ( 180 )/ G (0)h
K/Ku vs kappa

p( tk) = k b( ysp( tk ) y( tk))

Ti/Tu vs kappa

10

e( tk) = ysp( tk ) y( tk)




Td
d( t k ) =
d( tk1) kN y( tk) y( tk1)
Td + Nh
v= p( tk) + i( tk ) + d( tk )

10

10

10

u( tk )= sat(v)
2

10

i( tk+1) = i( tk ) +

0.2

0.4

0.6

0.8

10

0.2

0.4

0.6

0.8

c K. J. strm August, 2001

&

kh
kh
e( t k ) +
uv
Ti
Tr
13

3. The Basic Feedback Loop

Some Observations
n

d
r

A system based on error feedback is characterized by four

transfer functions (The Gang of Four)
The basic feedback system with two degrees of freedom is
characterized by six transfer function (The Gang of Six)
To fully understand a system it is necessary to look at all
systems

Six transfer functions are required to describe the system

P
PC
PCF
D
N+
R
1 + PC
1 + PC
1 + PC
P
1
PCF
D+
N+
R
Y=
1 + PC
1 + PC
1 + PC
PC
C
CF
U=
D
N+
R
1 + PC
1 + PC
1 + PC
X =

It may be strongly misleading to only show properties of

a few systems for example the response of the output to
command signals. This is a common error in the literature.
The properties of the different transfer functions can be
illustrated by transient or frequency responses.

A Possible Choice

The Sensitivity Functions

Six transfer functions are required to show the properties of a

basic feedback loop. Four characterize the response to load
disturbances and measurement noise.
PC
1 + PC
C
1 + PC

P
1 + PC
1
1 + PC

Two more are required to describe the response to set point

changes.
CF
PC F
1 + PC
1 + PC

1
L
, T=
, L = PC,
1+ L
1+ L
Disturbance reduction
Ycl(s)
=S
Yol (s)
Sensitivity to small process variations
S=

S+T =1

dh T h
dh P h
= hSh
hT h
h Ph

Large variations do not give instability if

c K. J. strm August, 2001

&

1
h Ph
<
h Ph
hT h

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Maximum Sensitivities
1
1
=
,
1+ L
1 + PC
Maximum sensitivities
S=

Ms = max h S (i )h,

T=

The Sensitivity Functions

L
PC
=
1+ L
1 + PC

Mt = max hT (i )h

The value 1/ Ms is the shortest distance from the Nyquist curve

of the loop transfer function L(i ) to the critical point 1.

h Ph
1
L
Ycl (s)
1
= S,
<
, T=
,
1+ L
1+ L
Yol (s)
h Ph
hT h
A compromise: Good disturbance rejection |S| small! Good
robustness hT h small, but S + T = 1 and
S=

Bodes integrals: sL(s) goes to zero as s

Z

The condition

1
h P(i )h
<
h P(i )h
h Mt h
guarantees that the perturbation P does not give instability.

Feedforward

Closed loop

Open loop

Market Driven

Planning

Robust to model errors

h S h < 1 for some

Not robust to model

errors h S h = 1 for all

Risk for instability

No risk for instability

Reduce the effect of measurable disturbances

log h

Acts before deviations

show up

Improve response to reference signals

System with Two Degrees of Freedom

Acts only when there

are deviations

Feedforward is used to

log h S( i )h d =

The water bed effect.

4. Feedforward
Feedback

X
1
h d =
pi
1 + L( i )
Z 0
Z0
X1
L(1/ i )
h d =
log h T (1/ i )hd =
log h
1 + L(1/ i )
zi
0
0

The system is said to have two degrees of freedom (2DOF)

because there are separate signal paths from set-point to u
and from measured signal to u.
Set point weighting in PI control is a simple version of 2DOF
Z t
u(t) = k( r(t) y(t)) +
(r( ) y( ))d

c K. J. strm August, 2001

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Many Versions of 2DOF

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d
r

C u

Response to Reference Signals

P x

My

Mu
My

Mu

um
ym

uf b
C

um
ym

uf b

y
P

Notice that Mu and M y can be viewed as generators of the

desired output ym and the inputs um which corresponds to ym.

Horowitz: "Some structures have been presented as fundamentally different from the others ... all 2DOF configurations
have basically the same properties and potentials."

Assume that process P, controller C and desired response M y

are given are given. Then Mu = M y / P. Does not depend on C!

FC = Mu + Mu C

Attenuation of Measurable Disturbance

d

5. State Feedback and Observers

Basic idea: PID controller predicts by linear extrapolation

G f f

ep = e + Td
u

P1

P2

Can we predict better?

Transfer function from disturbance to process output with

combined feedback and feedforward
Gd =

de
dt

Y ( s)
P2 (1 P1 G f f )
=
= P2 (1 P1 G f f ) S
D ( s)
1 + PC

The state captures all information about the past that is

relevant to predict the future.
Find a way to compute the state from measurements of
input and output
A nice structuring of the control problem

Notice combined effect of feedback and feedforward. Ideal

feedforward compensator G f f = P11 = P yQ/ P yu.
c K. J. strm August, 2001
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Determinants, Inverses, and Eigenvalues

A vector e is an eigenvector if Ae = se and the number s is

called an eigenvalue. We have

Consider the following matrices and their transpose



a11 a12
A=
a21

Determinants, Inverses, and Eigenvalues

Ae = se

a22

( sI A) e = 0
det ( sI A) = 0

The determinant is
det A = a11 a22 a12 a21

If the determinant is different from zero the matrix has an

inverse which is given by


1
a22 a12
1
A =
det A a21 a11

The eigenvalues are thus given by the characteristic equation



s a11
a12
= s2(a11 +a22)s+a11 a22 a12 a21
det (sI A) = det
a21

s a22

The roots of this equation are the eigenvalues of the matrix.

State Feedback

The Closed Loop System

System and linear controller

dx
= Ax + Bu = ( A B L) x + B lr r
dt
y = Cx

dx
= Ax + Bu
dt
u = Lx + lr r

u = Lx + lr r

The closed loop system

Transfer function from reference r to output y

dx
= Ax + Bu = Ax + B ( Lx + lr r) = ( A B L) x + Blr r
dt

Y ( s)
= C(sI A + B L)1 Blr
R( s )

If the system is controllable; i.e. if controllability matrix

Wc = ( B AB . . . An1 B )

is regular, then the matrix L can be chosen to give arbitrary

eigenvalues of A B L.
c K. J. strm August, 2001
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Computing the State Feedback Matrix L

Start by determining controllability!

Observers

Simple problems, determine closed loop characteristic polynomial, pick L to match desired polynomial
Easy if system is in controllable canonical form

a1 a2 . . . an
1

1
0
0
0

dz

0
1
0
0
=
z
+
u

dt
.
.

.
.

.
.
0

Observer

CA

Wo =

..

n1

System

dx
= Ax + Bu
dt
y = Cx

CA

The state can be computed from the output if system observable; i.e. if the observability matrix Wo is regular. Many different
ways, differentiation, simulation of model, A compromise:
d x
d x
= ( A K C) x

= A x + Bu + K ( y C x )
dt
dt
Find a matrix K so that A K C has desired eigenvalues.

c = T Wc .
Transform to controllable canonical form W

Use Matlab for numerical solutions.

Design of Observers

State Feedback from Observed State

Basic compromise: Trust models or measurements?

uc

Use results for feedback design A K C = ( A C K ) ,

compare with state feedback A B L
T

T T

Model and
Feedforward
xm
Generator

Simple problems, determine closed loop characteristic polynomial, pick L to match desired polynomial

u ff
u fb

Process

Observer

o = Wo T 1
Easy if system is in observable canonical form W

a1
1

a
0

dz

.
=

..

dt

an1 0

an
0

y = 1 0 0...

...

0
0

0 z

0
b1

0
b2

.
z
+
u

0
bn

dx
= Ax + Bu
dt
y = Cx
u = u m + L( xm x )
d x
= A x + Bu + K ( y y )
dt
y = C x
c K. J. strm August, 2001
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The Closed Loop System

The Closed Loop System ...

dx
= Ax + Bu
dt
y = Cx
u = L x + lr r

dx
= ( A B L) x + B L x + B lr r
dt
d x
= ( A K C) x
dt

d x
= A x + Bu + K ( y C x )
dt

Characteristic equation
det (sI A + B L) det (sI A + K C) = 0

Introduce the state x = x x instead of x .

dx
= Ax + Bu = Ax B L x + B lr r
dt
= Ax B L( x x ) + B lr r = ( A B L) x + B L x + B lr r
d x
= ( A K C) x
dt

Transfer function from reference r to output y

Y ( s)
= C(sI A + B L)1 Blr
R( s )

Integral Action
Meaning of the model

An alternative
dx
= Ax + B (u + v)
dt
dv
=0
dt

dx
= Ax + Bu
dt
dv
=0
dt

Feedback from estimated states x and v

d
dt

  
x
A
=
v
0

   


x
B
K
+
u+

v
0
0
Kv
= ( y C x )
u = L x v
B

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