vgu control theory

Control Theory

Mechatronic and Sensor Systems Technology VGU

Prof. Dr. Frieder Keller

Karlsruhe University of Applied Sciences

[email protected]

Prof. Dr.-Ing. F. Keller VGU-Course: Control Theory VGU_Control_Theory_new.DOC Karlsruhe University of Applied Sciences page 1

Contents

1 Recommended Literature

2 Examples of Control Systems

3 General Structure of a Control Loop

4 Block Diagrams

5 Mathematical Description of Dynamic Systems

6 Differential Equations

7 Laplace-Transform

8 Frequency-Response Analysis

9 Basic Functional Elements

10 Analysis of Control Systems

11 Stability Analysis of Control Systems

12 Tuning-Rules for Control-Parameters

13 Modifications of the Standard Control Loop


Recommended Literature [1] Nise, Norman S. Control systems engineering, John Wiley, 2000. [2] Ogata, Katsuhiko Modern Control Engineering, Prentice Hall [3] Ogata, Katsuhiko System Dynamics, Pearson, Prentice Hall It is also recommended to visit the Internet-page: [4] http://www.engin.umich.edu/group/ctm/ which provides a Control Systems Tutorial You can also access this tutorial offline by clicking on index.html on the subdirectory tutorial in the public directory. This tutorial covers much more aspects than the lec-ture.


Examples of Control Systems

control knob

heater

temperaturesensor

Temperature Control

inertia

motoru

nc

nr

speedsensor

nc

controller

Speed Control

spindle

motoru

xc

xr

controller

positionsensor

xc

Position Control

qin

pump

u

levelsensor

controller

hc

hr

hc

Liquid Level Control

coil

iron ball

solarcell

lamp

Position Control (suspended ball)


Structure of a Control Loop

controller motor-

nc

x Ks

nrx Ks

u(nr)

Speed Control System

controller motor-

ncnrx Ks

Equivalent System

controller process-

r ce y

General structure of a Control Loop

r: request signal, input signal, reference input e: error signal y: controller output c: controlled variable, plant output


Controlled variable:

The controlled variable is the output of the process. It is measured by a sensor element and controlled.

Reference input:

The reference input is the desired value for the controlled variable. It may be a constant value or may be a time-dependent signal.

Manipulated variable:

The manipulated variable is the output of the controller and the input of the process. The goal is to affect the value of the controlled variable to minimize the deviation of the controlled variable from the desired value.

Process:

Any operation to be controlled is called a process. Disturbance:

A disturbance influences the process and tends to affect the value of the con-trolled variable.

Feedback control or closed loop control:

A system where the controlled variable is measured and compared with the desired value. Any deviation leads to a controller output which reduces the dif-ference between the desired value and the controlled variable.

Open loop control:

In an open-loop control system the output is not measured and thus not com-pared with the input signal. The effect of disturbances cannot be compen-sated.


Goals in Control Engineering:

Stability, Dynamics, Accuracy, Overshoot

0 5 10 15 20 25 30 0

0.5

1

1.5

t

r,c

accuracyovershoot

dynamics

Dynamics, Overshoot, Accuracy

0 5 10 15 20 25 30-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

t

r,c

Step Response of an unstable System


Signals and Block Diagrams

t

(t)

1

t

r(t)

1

00 1

t

(t)

1

0 t

approximationfor (t)

0

1/

UNIT STEP UNIT RAMP

DIRAC PULSE

Basic functions

systeminput signal output signal

Functional Block

yG1(s)

-G2(s)

x

summing point branch point

Block Diagram


Mathematical Description of Dynamic Systems - Differential Equations,

- Laplace-Transformation

LTI-

systeminput signal output signal

x(t) y(t)

Linear, time-invariant system

a d y tdt

a d y tdt

a dy tdt

a y t b d x tdt

b d x tdt

b x tnn

n n

n

n m

m

m m

m

m( ) ( ) ... ( ) ( ) ( ) ( ) ... ( )+ + + + = + + +

1

1

1 1 0 1

1

1 0

for technical systems: n m

Differential Equation

1) Given: x(t), inital conditions Obtain: y(t) solution of the differentail equation

2) Given: x(t)=(t), inital conditions = 0 Obtain: y(t) = h(t) step response 3) Given: )sin()( 0 txtx = Obtain: y(t) for steady state

frequency response 4) Given: x(t) and y(t) Obtain: coefficients ai and bi

system identification 5) Check the stability of the system

stability test

Standard Problems


Definition of the Laplace-Transform: ( ){ }L f t F s f t e dts t= = ( ) ( )0

Some rules:

( ){ } ( ){ } ( ){ }( ){ } ( ){ }

( ){ }( ) { }

L L L

L L

L L

L L

f t g t f t g t

a f t a f tdf tdt

s f t f t

f ds

f tt

+ = + =

= =

=

( )

( )( )

( )

0

1

0

Small Laplace Transform Table:

( )t

1s

e a t 1s a+

t 12s

( )1 1a e at ( )1s s a + ( )sin at a

s a2 2+ ( )cos at s

s a2 2+

fr d1: ( )1 111 2

2

1 21 2

21 2 + = T

T Te T

T Te mit T T d d

tT

tT

,

12 12 2( )T s d T s s + +


Some other rules:

( ){ } ( ){ }( ){ }

0)at t sderivativeor their impulses no are there(if )(lim)0(

exists) )f(t (if )(lim)()(

0

==+=

+==

ssFf

ssFtfasFtfe

tfeTtf

s

s

at

sTD

D

L

LL

In practice Laplace-transform is performed using a table and the Laplace-theorems or a computer program like MAPLE or MATLAB with the symbolic toolbox. Inverse Laplace-transform also uses tables and computer programs. Sometimes the partial-fraction expansion method can be used to split a complicated expression in a sum of simpler expressions. Consider a LTI-system which is characterized by the following differential equation:

a d y tdt

a d y tdt

a dy tdt

a y t b d x tdt

b d x tdt

b x tnn

n n

n

n m

m

m m

m

m( ) ( ) ... ( ) ( ) ( ) ( ) ... ( )+ + + + = + + +

1

1

1 1 0 1

1

1 0

Laplace-Transformation gives:

a s Y s a s Y s a Y s b s X s b s X s b X sa s a s a Y s b s b s b X sY sX s

b s b s ba s

nn

nn

mm

mm

nn

nn

mm

mm

mm

mm

nn

( ) ( ) ... ( ) ( ) ( ) ... ( )( ... ) ( ) ( ... ) ( )

( )( )

...

+ + + = + + + + + = + + + = + + ++

11

0 11

0

11

0 11

0

11

0

a s aG s

nn

+ + =1 1 0... ( )

G(s) is the transfer-function. To obtain the output y(t) with given input x(t) and initial conditions (here considered to be 0) the following recipe can be used:

1. Obtain G(s) 2. Find the Laplace-transform of x(t) 3. Obtain Y(s)=G(s)X(s) 4. Find the inverse Laplace-transform of Y(s)


Example: Obtain the step response of a 1st order lowpass with initial condition =0. Solution:

differential equation: xyTy =+ & input signal: )()( ttx = initial condition: 0)0( ==ty

to obtain output signal: )(ty

Laplace-Transformation gives:

)()()( sXsYsTsY =+

This equation may be solved for Y(s)!

sT

sTssTsX

sTsY 11

1111

1)(1

1)( +

=+=+=

We have got the solution (but it is in Laplace-domain).

With the help of our Laplace-table we obtain the inverse Laplace-transform:

Tt

ety= 1)(

This is the well-known formula for the step response of a first order lowpass.


Example: Solution of a differential equation of 2nd order with nonzero initial condi-tions.

differential equation: xyTy =+ &&2 input signal: 0)( =tx initial conditions: 00 )0()0( vtyyty ==== &

to obtain signal: )(ty Solution:

Preliminary considerations: The Laplace-Transform of a second derivative is:

{ } { } { }( )

)0()0()()0()0()()0()((

2 ========

tytyssYstytytysstytysty

&&&&&& LLL

This way the initial conditions are included in the solution process. Now Laplace-Transformation of the differential equation gives:

0)()( 02

0222 =+ vTysTsYsTsY

and we can solve for Y(s):

( )0202221 1)( vTysTsTsY ++= 0

22

0

22

022

2

022

2

1

1

111)( v

TsTTy

Ts

svsT

TysTsTsY

++

+=+++

=

With the help of our Laplace-table we obtain:

+

=TtTv

Ttyty sincos)( 00

This describes for example the undamped oscillation of a spring-mass-system.


Example: Obtain the response of a 1st order lowpass with initial condition =0 for an input signal ( )txtx sin)( 0 = . Solution:

differential equation: xyTy =+ & input signal: ( )txtx sin)( 0 = initial condition: 0)0( ==ty with Laplace-transformation:

TssxsY ++= 1

1)( 220

written with partial fractions

TsC

sBAs

TssxsY +++

+=++= 111)( 22220

The coefficients A,B,C are determined as follows:

( ) CT

xCTss

BAss

x =+

++++=+ 22022220 1

1

( ) BAjTj

xBAjTj

xsTsCBAs

Tsx +=+=+++++=+ 1

1;1

111

100

220

( )( ) ( )( )22

0

0000

00

1

112

11112

11

11

11;

11

TxB

TjTjx

TjTjTjTjxB

Tjx

Tjx

BAjTj

xBAjTj

x

+=+=+

++==++

+=+=+

( )( ) ( )( )22

0000

00

1

112

11112

11

11

11;

11

TTA

TjTjTjx

TjTjTjTjxAj

Tjx

Tjx

BAjTj

xBAjTj

x

+=

+=+

==+

+=+=+


With the coefficients A,B,C from the above calculations we can write Y(s) as:

++++++++

=sTT

TsTs

sTTxsY

/11

111

1)( 22222222220

Inverse Laplace-transform gives:

+++++

= Tt

eTTt

Tt

TTxty 2222220 1

)sin(1

1)cos(1

)(

( )[ ]

++=

Tt

eTTTtVxty 220 1

arctansin)(

with

( ) 2222222

11

1

1

TT

TV +=++=

y(t) has two parts: the steady state solution ysteady(t) and the transient solution ytransient(t)

( )[ ]Tt

transient

steady

eTTxty

TtVxty

+=

=

220

0

1)(

arctansin)(

The following figure shows x(t) and y(t) for T1= :


0 0.5 1 1.5 2 2.5 3-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

t/T

2

/4

ysteady and ytransient are depicted in the next figure:

0 0.5 1 1.5 2 2.5 3-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

t/T

x

ysteady

ytransient


For this example it can be seen that the amplification factor V is given by:

)( jGV =

and the phase shift is: { }{ } )()(Re

)(Imarctan jGjGjG ==


Frequency-Response Analysis The two above relations are valid for all stable LTI-systems. This is shown in the following. We consider a system with input signal ( )txtx sin)( 0 = and transfer function G(s). The output signal y(t) has the Laplace-transform:

)()( 220 sGsxsY +=

With partial fraction expansion we have:

.........

...

...)()( 22220 +++++=+=

s

BAssGs

xsY

In this expression all the terms with ... belong to the transient response which tend towards 0 for large values of t. This is only the case if the system under consideration is stable. For the steady state response it is only necessary to determine the values A and B. Multiplying both sides by 22 +s we have:

( )220 ...............)( + ++++= sBAssGx With s=j and s=-j we have:

BAjBAjjGx

BAjBAjjGx

+=

++++=

+=

++++=

0.........

...

...)(

0.........

...

...)(

0

0

These are two equations for the two unknowns A and B. It can be easily seen that:

{ }{ })(Im

)(Re

0

0

jGxAjGxB

==

So we have: { } { } .....)(Re)(Im)( 220220 ++

++=

sjGx

ssjGxsY


With the inverse Laplace-transform we obtain: { } ( ) { } ( )[ ]

{ }{ } ....)(Re

)(Imarctansin)()(

....sin)(Recos)(Im)(

0

0

+

+=

=++=

jGjGtjGxty

tjGtjGxty

This shows that for a stable LTI-system gain V and phase shift are given by:

)( jGV =

and: { }{ } )()(Re

)(Imarctan jGjGjG ==

For a sinusoidal input x(t) a stable LTI-system in steady state is characterized by a magnitude response and a phase response. Both, magnitude and phase depend on the frequency. Frequency response is obtained as follows:

1. Replace s by j. in G(s) to obtain G(j) 2. The magnitude is given by G(j) 3. The phase is given by G(j)

Note that s may only be substituted by j if the system is stable, otherwise a steady state response does not exist.


Bode diagram and Nyquist plot Bode diagrams and Nyquist plots are a graphical representation of )( jG . The Bode plot consists of two graphs: the magnitude and the phase

)()( jGandjG The horizontal axis in the Bode plots show the frequency in logarithmic scale. The magnitude is expressed in decibels (dB):

)(lg20 jGadB =

Some special values are given in the following table:

Gain 100

1 101

21

22 1 2 10 100

adB -40 -20 -6 -3 0 3 20 40 The phase is shown in linear scale. The following figure shows the Bode diagram for a first order system with

sTwithTs

sG 11

1)( =+=1

Frequency (rad/sec)

Pha

se (

deg)

; M

agni

tude

(dB

)

Bode Diagrams

-20

-15

-10

-5

0From: U(1)

10-1 100 101-100

-80

-60

-40

-20

0

To:

Y(1

)

1 The plot is generated with MATLAB using the command bode(tf([1],[1 1]),'k')


The Nyquist plot shows the real and imaginary parts of )( jG when is varied. Usually is varied between 0 and . The following figure is generated with MATLAB using the Nyquist-command2. MATLAB varies in the range -


10-3

10-2

10-1

100

101

102

103

-80

-60

-40

-20

0

20

40

60

80

f/f0

a/dB

10-3

10-2

10-1

100

101

102

103

-200

-150

-100

-50

0

50

100

150

200

f/f0

/

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4

real part

imag

inar

y pa

rt


Basic functional elements

The proportional element: Equation: xKy = . K is called the gain. Transfer function: KsG =)( Step response of the proportional element: )()( tKth = Bode-diagram:

10-3

10-2

10-1

100

101

102

103

-80

-60

-40

-20

0

20

40

60

80

f/f0

a/dB

10-3

10-2

10-1

100

101

102

103

-200

-150

-100

-50

0

50

100

150

200

f/f0

/

Nyquist-plot:

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4

Examples:

Mechanical beam Input signal is the position of the left end, output signal is the position of the beams right end

Voltage divider using a potentiometer as position sensor. input: position of the wiper, output: output voltage


The integrator: Equation: = xdtKy I . Transfer function:

sKsG I=)(

Step response: )()( trKth I = (the ramp function) Bode-diagram:

10-3

10-2

10-1

100

101

102

103

-80

-60

-40

-20

0

20

40

60

80

f/f0

a/dB

10-3

10-2

10-1

100

101

102

103

-200

-150

-100

-50

0

50

100

150

200

f/f0

/

Nyquist-plot:

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4

Examples:

- Water tank Input: flow rate, output: liquid level

qin

h

out

valve

q

- Relation between velocity and distance

Input: velocity, output: distance


The differentiator: Equation: xKy D &= .. Transfer function: sKsG D =)( Step response: )()( tKth D = (the Dirac pulse) Bode-diagram:

10-3

10-2

10-1

100

101

102

103

-80

-60

-40

-20

0

20

40

60

80

f/f0

a/dB

10-3

10-2

10-1

100

101

102

103

-200

-150

-100

-50

0

50

100

150

200

f/f0

/

Nyquist-plot:

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4

Examples:

Relation between velocity and acceleration input: velocity, output: acceleration

Electronic differentiator (Opamp-circuit)


The first order lowpass: Equation: xKyyT P =+ & Transfer function:

TsKsG P+= 1)(

Step response:

= T

t

P eKth 1)(

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t/T

y

At t=T the step-response reaches 63% of the final value. The tangent at t=0 intersects the line of the final value at t=T. Bode-diagram (for KP=1):

10-3

10-2

10-1

100

101

102

103

-80

-60

-40

-20

0

20

40

60

80

f/f0

a/dB

10-3

10-2

10-1

100

101

102

103

-200

-150

-100

-50

0

50

100

150

200

f/f0

/

Nyquist-plot (for KP=1):

-1 -0.5 0 0.5 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1


Examples:

RC-lowpass filter input: input voltage, output: output voltage

U2RC

U1

Electrical motor input: voltage u, output: speed

Motorui

inertia

Inductance with resistance

input: voltage, output: current Exercise: Draw the block diagram of an electrical motor. input: voltage, output: angular position

Motorui

inertia


The first order highpass: Equation: xKyyT D && =+ Transfer function:

TssKsG D+= 1)(

Step response: Tt

D eTKth

=)( Drawing for KD=T:

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t/T

y

Bode-diagram (for KD=T):

10-3

10-2

10-1

100

101

102

103

-80

-60

-40

-20

0

20

40

60

80

f/f0

a/dB

10-3

10-2

10-1

100

101

102

103

-200

-150

-100

-50

0

50

100

150

200

f/f0

/

Nyquist-plot (for KD=T):

-1 -0.5 0 0.5 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1


Examples:

RC-highpass filter input: input voltage, output: output voltage

U2R

CU1

Modified differentiator (exercise: draw the circuit)

RC-lowpassfilter (exercise: find the appropriate input and output signal)

Exercise: Represent the highpass by a series connection of a lowpass and a differentiator.


The second order lowpass: Equation: xKyyTdyT P =++ &&& 22 d is called the damping ratio.

Transfer function: 2221)(

sTdTsKsG P++=

Step response:

d>1

( )11)( 22,121

2

21

1 21 =

+= ddTTmite

TTTe

TTTKth

Tt

Tt

P

d=1

+= Tt

P eTtKth 11)(

d


Bode-diagram (for KP=1):

10-2

10-1

100

101

102

-80

-70

-60

-50

-40

-30

-20

-10

0

10

20

/ 0

|G| i

n dB

Asymptote fr 0

Asymptote fr

d=0.2

d=0.6

d=1.0 d=1.4

d=1.8

10-2

10-1

100

101

102

-200

-180

-160

-140

-120

-100

-80

-60

-40

-20

0

/ 0

G

Asymptote fr 0

Asymptote fr

d=0.2

d=0.6 d=1.0

d=1.4

d=1.8


-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

Realteil

Imag

inr

teil

d=0.2

d=0.6

d=1.0

d=1.4 d=1.8


Examples:

RLC-filter input: input voltage u1, output: output voltage u2

U2RC

U1 L

Spring-mass-system input: force, output: position of the mass

md

cF

y

Exercise: Describe the above RLC-filter with a block diagram containing two integra-tors

Exercise: Replace the series connection of two identical lowpasses of first order by a lowpass of 2nd order.


The delay element: Equation: )()( DP TtxKty = TD is called the delay-time. Transfer function: sTP DeKsG

=)( Step response: )()( DP TtKth = Example: Conveyor belt:

h1 h2

l

v

Bode-diagram (for KP=1):

10-3

10-2

10-1

100

101

102

103

-80

-60

-40

-20

0

20

40

60

80

f/f0

a/dB

10-3

10-2

10-1

100

101

102

103

-200

-150

-100

-50

0

50

100

150

200

f/f0

/


-1 -0.5 0 0.5 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1


Analysis of Control Systems Closed loop transferfunction:

cG1(s)

-G2(s)

r

controller process

The transfer function of a closed loop is easily obtained. [ ]

)()(1)()(

)()(

)()()()()(

21

21_

21

sGsGsGsG

sRsCG

sCsRsGsGsC

loopclosed +==

=

The PID-controller: The equation of a controller with proportional-plus-integral-plus-derivative control ac-tion is:

++= eTedtTeKy diP &

1

An electronic circuit or PID-control-action is shown in the next figure:

uin

uout

Rp

R1

R1

R1

R1

R1

Rd

Cd

Ri

Ci


The transfer function is:

sTsTTsTKsG

sTsT

KsG

i

diiP

di

P

++=

=

++=

21)(

11)(

With the help of variable resistors RP, Ri and Rd the parameters KP, Ti and Td can be adjusted. Variation of a resistor does only affect one of the control parameters.A PID-controller may also be realized in a circuit with fewer components:

uin uout

R1

CiR2CdR3

However, varying one element affects not only one control parameter. So we prefer the first circuit for experimental tuning of control parameters and the second for reali-zation with reduced number of components. Exercise: Establish the transfer function of the above circuits and find the relation between the control and component parameters. Hint: The resistor R3 realizes modified differen-tiation.


Control of an integrating process

r

-

ce uPK Ts

1

sKT

sTK

sTK

sRsC

PP

P

+=

+=

1

111

1

)()(

From this transfer function we conclude that - the control loop is stable (first order lowpass is stable) - the control loop has no overshoot (because a first order system cannot have an

overshoot) - the control loop is accurate (because for s=0 we obtain a gain of 1) - the dynamics can be adjusted by the variation of KP (because the time constant

is T/KP)

Exercise: Obtain the transfer function)()(sRsU . Sketch u(t) for r=(t).

Exercise: Analyze the following control loop, where z is a disturbance z=z0(t). Is the control loop still accurate?

r

-

ce uPK Ts

1

z


P-Control of a first order lowpass system

r

-

ce uPK Ts

K+1

sKKTKKKK

sTKKKK

sTKKsT

KK

sRsC

P

P

P

P

PP

+++

=++

=++

+=1

1

11

11

1)()(

From this transfer function we conclude that - the control loop is stable (first order lowpass is stable) - the control loop has no overshoot (because a first order system cannot have an

overshoot) - the control loop is not accurate (because for s=0 we obtain a gain of

11


Solution: We compare the transfer function with the standard form of a 2nd order lowpass.

222 ''21

1

1

1sTsdTs

KTTs

KT ii ++=++

Now we find:

KTTT i =' and

TKT

TTKKT

KTTd

KTdT i

i

iii

====

22'2'2

An overshoot exists if d


Control of a 2nd order lowpass system From the above we may conclude that the strategy of pole compensation is also suit-able for a second order process. However, in this case a PID-controller is needed.

r

-ce u

2221 sTdTsK

++sTi1

PK

sTd

sKK

TsTKKsTKK

sTdTsK

sTsTTsTK

sTdTsK

sTsTTsTK

sTdTsKsT

sTK

sTdTsKsT

sTK

sRsC

P

i

i

P

i

P

i

idiP

i

idiP

di

P

di

P

+=

+

=++

+++++

++

=++

+++

++

++=

1

1

121

11

211

21111

2111

)()(

22

2

22

2

22

22

We have chosen: dT

TTTanddTTi

di 22

2

=== This leads to the same transfer function as in the previous example. However, in practice we have to take care for the output of the controllers differenti-ator not to go in saturation. This can be achieved by rate limiting of the input signal and/or modified differentiation. The realization of a rate limiter with an Opamp-circuit is shown in the following figure:

R

C

R1

R1

R2

R3

uout

uin

Note that we usually choose a high gain of the non-inverting amplifier, so it may also go in saturation. Let umax and umin denote the maximum and minimum output voltage


of the amplifier. Then RC

udtduout minmax/

minmax/= . With the negative feedback it is ensured

that the output voltage always follows the input voltage. Note that the output is in-verted, so one needs an additional inverter to have the correct sign. Control of a 3rd order lowpass (academic example) Consider the following control loop:

r cuTs+1

1PK Ts+1

1Ts+1

1-

e

Exercise: Find the transfer function of the closed loop. What is the steady state error for KP=1, KP=3 and KP=9? Here we find that the control loop is unstable if the controller gain KP exceeds a cer-tain threshold (Kcritical=8). For KP


Stability Analysis

We will introduce two stability criterions. The first is based on the closed loops transfer function and its pole locations in the complex plane. The second is the Nyquist criterion. It analyzes the open loops Nyquist curve. The pole location criterion states: A system is stable if all poles of its transfer function are located in the left half of the complex plane if all poles have a negative real part. Note that this stability criterion can only be applied if the system does not contain any delay elements. For the above example we may evaluate the pole locations depending on the pa-rameter K. Often the pole locations are visualized in the so-called root-locus-diagram. The corresponding MATLAB-command is rlocus. The following graph is obtained with this procedure:

-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5-2

-1.5

-1

-0.5

0

0.5

1

1.5

2Root Locus

Real Axis

Imag

inar

y Axi

s

There are three poles. As K is increasing one pole (blue line) is moving to the left, the other poles however are approaching the imaginary axis and for K>8 they are enter-ing the right half plane and the closed loop gets unstable. As another example consider P-control of a 2nd order lowpass. The corresponding root locus plot looks like this:

-0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0-2

-1.5

-1

-0.5

0

0.5

1

1.5

2Root Locus

Real Axis

Imag

inar

y Axi

s

From this we conclude that the control loop cannot get unstable.


A powerful and complex theory has been developed on the basis of root locus plots including also the design of controllers. However these details are not treated in this lecture. The Rouths criterion allows us to determine whether the system is stable without explicitly evaluating the poles. Rouths stability criterion is outlined in the following: Write the polynomial (denominator of the closed loop) in the following form (it is as-sumed that 0na ):

n

nnnn asasasasa +++++ ...3322110

Check if all the coefficients are nonzero and have the same sign. If not, the system is not stable or there are roots which are imaginary. However this is only a necessary but not a sufficient condition.

Arrange the coefficients in rows and columns according to the following scheme:

sn a0 a2 a4 a6 . . .

sn-1 a1 a3 a5 a7 . . . sn-2 b1 b2 b3 b4 . . . sn-3 c1 c2 c3 c4 . . .

. . . .

.

. s0 g1

The coefficients ai have to be filled in from the polynomial. The coefficients bi, ci, are evaluated according to the following rule:

1

30211 a

aaaab = 1

50412 a

aaaab =

1

21311 b

baabc = 1

31512 b

baabc = The table is filled until we have completed the line with label s0. The system is stable if all coefficients in the first column have the same sign.


Exercise: Find the necessary and sufficient conditions for a system with transfer-

function 32

21

30

1)(asasasa

sG +++= to be stable. Exercise: Apply the Routh criterion to the academic example The Nyquist criterion is based on the open loop:

processu c

controllerr

-

e

If we use a sinusoidal input (e.g. from a function generator) for the controller the steady state output c(t) will also be a phase-shifted sine-function (provided the open loop is stable). In the case of our academic example it is found that the phase shift

may be -1800 (for a frequency T3= ) and the gain may be 1 (for a controller gain

8=PK ). This may be shown by analytical calculation and verified by simulation (see the following SIMULINK-model).

1

tau*tau*tau.s +3*tau*taus +3*taus+13 2

Transfer Fcn

Subtract

8

SliderGain2

Sine Wave

Scope

Manual SwitchGround

If we now toggle the switch and close the control loop the oscillation continues with-out changing the amplitude. This means the control loop is at the verge to instability. For higher values of the gain however the amplitude will increase: the control loop is unstable. For gains less than 8 the oscillation is damped and goes towards 0: the control loop is stable.


This example illustrates the simplified Nyquist criterion:

A closed loop system is stable if the point (-1+0.j) of the complex plane is al-ways on the left side of the open loops Nyquist curve when is varied from 0 to . The application of this simplified form requires that the open-loop system is stable. However 2 poles are allowed at s=0.

For a more detailed explanation of the general Nyquist criterion please refer to [2].

There are two standard measures for the distance of the Nyqusit plot from the critical point: gain margin and phase margin. They are illustrated in the following figure:

Im

Re

phase margin

representing thegain margin

unit circle

1

The gain and phase margin is easily obtained with the MATLAB-command margin. The next figure shows the result of margin when applied to the academic example.

-120

-100

-80

-60

-40

-20

0

Mag

nitu

de (dB

)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

-45

0

Pha

se (de

g)

Bode DiagramGm = 18.1 dB (at 1.73 rad/sec) , Pm = -180 deg (at 0 rad/sec)

Frequency (rad/sec)


The gain margin is stated to be 18.1 dB. This corresponds to the well-known maxi-mum gain of 8, for which the control loop is still stable. As the Nyquist plot touches the unit circle at s=1, the phase margin is 180o.

Tuning Rules for Control Parameters In the literature one finds a lot of different methods and recipes for tuning the pa-rameters of a controller. Some of them are based on the theory of differential equa-tions and use mathematical methods. Other methods propose a procedure for the experimental tuning of the parameters.

In the following some methods are listed.

Compensation of Poles This method was already mentioned. With a PID-controller it is possible to compen-sate up to 2 poles of the process transfer function. As it was shown earlier the pre-ferred transfer function of the open loop is an integrator resulting in a first order low-pass behaviour of the closed loop. If the process has more than two poles we may choose to compensate the dominating poles, these are the poles which are closest to the imaginary axis. Adjustment of the Phase Margin This method is based on the frequency response of the open loop. From the Nyquist criterion it is obvious that a control loops step response has increasing oscillations the closer it gets to the critical point -1. The recommended distance is a phase mar-gin of approx. 600. Note that the compensation of poles will prefers a phase margin of 90o, a phase margin which is less will usually result in an overshoot.

Tuning Rule according to Ziegler-Nichols This rule was developed for processes with a transport delay in series with a first or-der element. But often the rule is also applied to other types of processes. Recipe:

Start with P-control. Increase the amplification to Kcritical , where the control loop is marginally stable. Measure the period Tcritical of the oscillation. Adjust the parameters according to the following table:


Controller K TI TD P-Controller 0.5.Kcritical - - PI-Controller 0.45. Kcritical 0.83.Tcritical - PID-Controller 0.6. Kcritical 0.50.Tcritical 0.125.Tcritical

These parameter values may not give satisfactory results. But they may be good ini-tial values for further improving the dynamic behaviour of the control loop experimen-tally. Note that increasing the integral control action usually produces more oscilla-tions and increasing the differential control action has a damping effect. The reason for this is best seen in the Nyquist plot. Tuning Rule according to Tietze, Schenk This tuning rule may be used to find the control parameters for higher order proc-esses. - Start with P-control and increase the controller gain until the step response

shows a damped oscillation. On a scope one should see 4-5 periods of oscilla-tion.

- Continue with PD-control. Increase the differential control action to get rid of the oscillations.

- Activate the integral control action and increase its weight until settling time is found to be good.


Modifications of the Standard Control Loop Cascaded Control Loops

In many applications not only one signal may be monitored by a sensor element. Cascading control loops then may be a powerful method for improving the perform-ance. The general structure is shown in the following figure:

processpart 1

ucontroller1

r

-controller

2 -processpart 2

c

The tuning of the controller is of course done starting with the parameters of control-ler 1, and then adjusting the parameters of the outer control loop.

Example: Position control with an inner control loop for speed control.

Linearization of nonlinear processes

If one has to deal with nonlinear processes linearization may be realized (in some but not all cases as for example saturation effects) as shown in the following figure.

processu c

controllerr

-

ef(u)f-1(u)

f-1(u) is the inverse of f(u) and must be realized as part of the controller. In analog technique this is often very difficult, in digital controllers it may be easily done with the help of lookup-tables.

Compensation of disturbances

Sometimes an information about the disturbance z is available and it may be com-pensated as shown in the next figure. With this method we may efficiently suppress the effect of the disturbance before it affects the error signal and has to be compen-


sated by the controller. In the ideal case Gz(s) is the inverse of the transfer function of process part 1, however an approximate realization of Gz(s) also may show signifi-cant effect.

processpart 1

r

-controller processpart 2

c-

z

ucontroller

Gz(s)

Rate limiting the reference input to avoid the windup phenomenon

If the controller includes an integral control action overshoot of the controlled variable may occur due to the saturation of the controller output. In those cases rate limiting the reference input is a good countermeasure.

processu c

controllerr

-

eratelimiter

vgu control theory

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