Systems Analysis & Control: Steady State Errors

Department of Automatic Control and Systems

Engineering

Overheads for ACS211 Systems analysis and control Anthony Rossiter

Department of Automatic Control and Systems Engineering

University of Sheffieldwww.shef.ac.uk/acse

© University of Sheffield 2009 This work is licensed under a

Creative Commons Attribution 2.0 License.

http://creativecommons.org/licenses/by/2.0/uk/



Engineering

Contents• ACS211: Compensation and steady-state errors• Final value theorem (FVT)• Fast response• Illustration of offset• What about the inputs?• Gain insufficient• Examples• MATLAB CODE• Offset free tracking• Offset free tracking requirement (steps).• Integral action• MATLAB CODE• Tutorial questions (at home)• Tracking ramps• Offset to ramps• Solutions• Figures for 1st and 3rd pairs (errors only)• Zero offset to ramps• Tutorial questions (at home)• Figures for offsets• Extra tutorial question: validation• Examples of closed-loop transfer functions• Reminders• Given • MATLAB code to investigate k• Credits


Engineering

ACS211: Compensation and steady-state errors

Anthony Rossiter

Dept. ACSE

3

Some of this is revision of ACS114


Engineering

Final value theorem (FVT)

For analysis of offset, students should be able to prove the following (was done in ACS114).

Examples:

4

)(lim)(lim 0 ssYty st

)}1(2

1)({;

2

1)(lim

)2(

1)(

})({;0)(lim)2)(1(

1)(

})({;0)(lim2

1)(

20

20

20

ts

tts

ts

etyssYss

sY

eetyssYss

sY

etyssYs

sY

Convergent signals only


Engineering

Fast response

• How do we get a system to respond fast to an error between target and output?

• What is the mathematical expression for this?

• Why is this law insufficient to give offset free tracking in general?

5

Think about this for a few minutes and tell me.


Engineering

In these slides assume following feedback structure

6


Engineering

Illustration of offset

G=tf(1,[1 3 2])

K=1;

K2=10;

figure(1);clf reset

step(feedback(G*K,1),feedback(G*K2,1));

hold on

plot([-0.1,0,0,4],[0,0,1,1],'r');

axis([-0.1 4 0 1.2]);

legend('K=1','K2=10','target');

7

Note how as gain increases, speed of response improves, but always an offset.

0 0.5 1 1.5 2 2.5 3 3.5 40

0.2

0.4

0.6

0.8

1

1.2

Step Response

Time (sec)

Am

plitu

de K=1

K2=10

target

offset

23

12

ss

G


Engineering

What about the inputs?

Note how the input amplitude depends upon the error (target –output) and the gain K.

A bigger gain means a bigger initial input so a faster response and a smaller offset.

8

0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

Outputs

Time (sec)

Am

plitu

de

K=1

K2=10

target

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-5

0

5

10Inputs

Time (sec)

Am

plitu

de

In steady-state we have consistency from:y=G(0)u, u=K(r-y) which gives y=[G(0)K/(1+G(0)K)] r


Engineering

Gain insufficientIt is clear that gain alone is insufficient

to remove offset.

9

)()()(1

1)(

)()())()(1(

)()()()()()()()(

)()()()(

srsKsG

se

srsesKsG

sesKsGsrsesesKsu

susGsrse

Let r(s)=(1/s) and use FVT. )0()0(1

11.)()(1

1lim)(lim 00 KGssKsG

ssse ss


Engineering

Examples

Find the steady state offset for the following, system controller pairs, assuming the target is a unit step.

10

8)2)(10(

1;

5.0133

1;

212

1232

Kss

G

Ksss

G

Kss

G


Engineering

Answers

System 1

System 2

System 3

11

3

1

21

1

)0()0(1

1

KG

8)2)(10(

1;

5.0133

1;

212

1232

Kss

G

Ksss

G

Kss

G

3

2

5.01

1

)0()0(1

1

KG

71.028

20

)208(1

1

)0()0(1

1

KG

0 1 2 3 4 5 6 70

0.5

1ERRORS

Time (sec)

Amplit

ude

0 1 2 3 4 5 6 7 8 9 10

0.70.80.9

Step Response

Time (sec)Am

plitud

e

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

0.70.80.9

Step Response

Time (sec)

Amplit

ude


Engineering

MATLAB CODEG1=tf(1,[1 2 1])

K1=2;

G2=tf(1,[1 3 3 1])

K2=0.5;

G3=tf(1,[1 12 20])

K3=8;

figure(1);clf reset

subplot(311)

step(feedback(1,G1*K1));

title('ERRORS','Fontsize',18)

subplot(312)


subplot(313)


12

Note there are spaces between these numbers, so it is [1 space 3 space 3 space 1]

You can use commas if that is easier to read

You may be able to copy and past this code straight into MATLAB from the PDF file.


Engineering

Offset free tracking

• What conditions are required for offset free tracking of steady targets? how do we control speed in a car? what are the key steps and concepts? what form of control law is this?

• Having understood the key concept, can we prove this mathematically?

13


Engineering

Offset free tracking requirement (steps).

Begin from the FVT [Assume target is a unit step or 1/s ]

It is clear that e(t) tends to zero if and only if the term G(0)K(0) is infinite.

14

)0()0(1

11.)()(1

1lim)(lim)(lim 00 KGssKsG

sssete sst

Zero offset requires either G(s) or K(s) to include an integrator that is a term of the form (1/s).

Clearly with no cancelling derivative term


Engineering

Integral actionMost practical control laws include an

integrator to ensure offset free tracking of constant targets.

15

sK

sssG

sK

ssG

2.0133

1

4.012

1

23

2

0 5 10 15 20 25-0.5

0

0.5

1ERRORS

Time (sec)

Am

plitu

de

0 5 10 15 20 25 30 35-0.5

0

0.5

1

Time (sec)

Am

plitu

de


Engineering

MATLAB CODEG1=tf(1,[1 2 1])

K1=tf(0.4,[1 0]);

G2=tf(1,[1 3 3 1])

K2=tf(0.2,[1 0]);

figure(2);clf reset

subplot(211)


title('ERRORS','Fontsize',18)

subplot(212)


title(' ','Fontsize',18)

16

Note the term [1 0] defines (s+0)

i.e. coefficient of 1 on s and coefficient of 0 on the constant


Engineering

Tutorial questions (at home)

Find the offset for a unit step target for the following pairs of compensators and systems.

Validate all your answers with MATLAB.

17

8.035

)3(;

)2(6.0

235

1

;4

12

3

8)4)(2)(1(

)3(;

6.0235

1;

432

1

23232

232

Ksss

sG

s

sK

sssG

sK

ss

sG

Ksss

sG

Ksss

G

Kss

G

One is not as expected. WHY?


Engineering

18

0 1 2 3 4 5 60

0.5

1ERRORS

Time (sec)

Am

plitu

de

0 5 10 15 20 25

0.70.80.9

Step Response

Time (sec)

Am

plitu

de

0 0.5 1 1.5 2 2.5 3 3.5 4 4.50

0.5

1Step Response

Time (sec)

Am

plitu

de

Figures for previous page

0 50 100 150-2

0

2x 10

6 ERRORS

Time (sec)

Am

plitu

de

0 50 100 150 200 250 300 350-1

0

1

Time (sec)

Am

plitu

de

0 2 4 6 8 10 12 14-1

0

1

Time (sec)

Am

plitu

de


Engineering

Tracking ramps

1. Why would you want to track a ramp given this diverges to infinity?

2. How would you compute the offset to a ramp target?

3. How will an actuator supply an input big enough to track a ramp in general?

Discuss this with you neighbours briefly.

19

WARNING: When using MATLAB, remember only to compute convergent signals, hence you cannot easily plot ramp responses which go to infinity.


Engineering

Targets that are ramps

May operate over limited time scales.

Circular motion (DVD players, etc.).

As input energy is limited, usually ramps can only be tracked asymptotically where the system model itself contains an integrator.

20


Engineering

Offset to rampsRevert back to the FVT but now use a target

of r(s)=(1/s2). [or scaled variant as appropriate]

Clearly, the error diverges to infinity unless there is an integrator in G(s) or K(s).

21

2

1.)()(1

1)(

)()(1

1)(

ssKsGsr

sKsGse

)()(

11.)()(1

1lim)(lim)(lim

200 sKssGor

ssKsGsssete sst

)()(

1lim)(lim 0 sKssG

te st


Engineering

ExamplesFind the offset to a ramp r=(1/s2) for

the following compensator system pairs.

22

sK

sssG

s

sK

sssG

sK

ssG

2.0133

1

)4(

)3(2.0133

1

;4.012

1

23

232


Engineering

ExamplesFind the offset to a ramp r=(1/s2) for


23

sK

sssG

s

sK

sssG

sK

ssG

2.0133

1

)4(

)3(2.0133

1

;4.012

1

23

232


Engineering

SolutionsFind the offset to a ramp r=(1/s2) for


24

5;2.0)()(lim2.0

133

1

;0)()(lim

)4(

)3(2.0133

1

5.2;4.0)()(lim4.012

1

0

23

0

23

0

2

esKssG

sK

sssG

esKssG

s

sK

sssG

esKssG

sK

ssG

s

s

s


Engineering

Figures for 1st and 3rd pairs (errors only)

25

0 2 4 6 8 10 12 14 16 18 200

5

10

15

20ERRORS

Time (sec)

Am

plitu

de

0 2 4 6 8 10 12 14 16 18 200

5

10

15

20

Time (sec)

Am

plitu

de

ramp=0:20;time=ramp; G1=tf(1,[1 2 1])K1=tf(0.4,[1 0]);G2=tf(1,[1 3 3 1])K2=tf(0.2,[1 0]);figure(2);clf resetsubplot(211)lsim(feedback(1,G1*K1),time,ramp);title('ERRORS','Fontsize',18)subplot(212)lsim(feedback(1,G2*K2),time,ramp);title(' ','Fontsize',18)


Engineering

Zero offset to ramps

Look at the FVT and determine how to get zero offset for a ramp target.

Tell me what you think.

26


Engineering

Zero offset to ramps

Clearly, zero offset requires two integrators in G(s)K(s) so that:

Warning: Throughout we have assumed a simple feedback structure. You must modify all rules if the feedback structure is different (e.g. more compensators, something in return path, etc.).

27

0)(

lim)()(

1lim

)()()(

2

002 ssM

s

sKssGs

sMsKsG ss


Engineering

Tutorial questions (at home)

Find the offset for a ramp target r=(1/s2) for the following pairs of compensators and systems.

Validate all your answers with MATLAB.

28

8.035

)3(;

)2(6

35

1

;4.012

3

2)4)(2)(1(

)3(

;6.035

1;

432

1

23232

232

Ksss

sG

s

sK

sss

sG

sK

ss

sG

sK

sss

sG

Ksss

G

Kss

G


Engineering

Tutorial questions for home

1. How might you deal with targets that are parabola?

2. Why in general would a controller containing proportional and integral be a good idea? [Usually denoted PI and the most common structure in industry].

3. Industry often uses PID. What does the D stand for and why does this help? What problems can it introduce? [Look at some books if you are stuck]

29


Engineering

Figures for offsets30

2 4 6 8 10 12 14 16 18 200

10

20ERRORS

Time (sec)

Amplit

ude

2 4 6 8 10 12 14 16 18 200

10

20Linear Simulation Results

Time (sec)

Amplit

ude

2 4 6 8 10 12 14 16 18 200

10

20Linear Simulation Results

Time (sec)

Amplit

ude

2 4 6 8 10 12 14 16 18 200

10

20ERRORS

Time (sec)

Ampl

itude

2 4 6 8 10 12 14 16 18 20-20

0

20

Time (sec)

Ampl

itude

2 4 6 8 10 12 14 16 18 200

10

20

Time (sec)

Ampl

itude


Engineering

Extra tutorial question: validation

MATLAB can be used to validate your computations in several easy ways:

1. Compute and display closed-loop responses as in these notes.

2. Compute closed-loop transfer function explicitly and evaluate steady-state gain.

G1=tf(1,[1 2 3])

K1=4;

G2=tf(1,[1 5 3 0])

K2=0.6;

Gc1=feedback(G1*K1,1)

gain1=bode(Gc1,0)

Gc2=feedback(G2*K2,1)

gain2=bode(Gc2,0)

31


Engineering

Examples of closed-loop transfer functionsTry these examples by hand and with

MATLAB:

Gc(0) gives steady-state output for a step input r(s)=1/s.

32

1)0(6.035

6.0

6.035

1

57.07

4)0(

72

4

432

1

2323

22

cc

cc

Gsss

GK

sssG

Gss

GK

ssG


Engineering

Reminders1. The closed-loop must be stable or the FVT does

not apply.

2. Next we will consider how to determine whether a closed-loop is stable or not and also whether performance is expected to be good or bad.

3. In previous years the lecturer spent some time on state space models. I have omitted this year to avoid overload, however awareness of state space is essential in the longer term. Spend a few hours looking at this in the text books. The lecturer of ACS214/206 will assume some familiarity!

4. State space is introduced formally in year 3.

33


Engineering

Given s

sksK

ssG

2)(;

3

2)(

1. Find the closed-loop transfer functions from target to input and target to output.

2. Find the open-loop and closed-loop poles for k=1. Which are better and why?

3. Determine the open-loop (usual to ignore K(s) for open loop) and closed-loop offset for a unit step and unit ramp.

4. Propose a value for k. Give a justification.


Engineering

Given

1. Find the closed-loop transfer functions from target to input and target to output.

2. Find the open-loop and closed-loop poles for k=1. Which are better and why?

;)3()2(2

)3)(2(2

1)(

;)3()2(2

)2(2

1)(

sssk

ssk

GK

KsG

sssk

sk

GK

GKsG

cu

cy

s

sksK

ssG

2)(;

3

2)(

);3(

);4)(1(451

;4)32()3()2(22

2

ssp

sssspk

ksksssskp

o

c

c


Engineering

Given s

sksK

ssG

2)(;

3

2)(

• Determine the open-loop (usual to ignore K(s) for open loop) and closed-loop offset for a unit step and unit ramp.

Open-loop steady-state for a step is 2/3. Offset for a ramp will be infinite.

Closed-loop steady state errors are

• Best value for k involves a compromise between speed of response and input activity. See MATLAB for illustration.

2.2.

31)(;0

)0()0(1

1)(

ksGKrampe

KGstepe


Engineering

MATLAB code to investigate k

G=tf(2,[1 3]); K=tf([1 2],[1 0]);

Gcy=feedback(G*K,1); Gcu=feedback(K,G);

figure(1);clf reset

pzmap(Gcy);

step(Gcy,Gcu);legend('Output','Input');

figure(1);clf reset

k=1:.5:4;clear u y t

t=linspace(0,2,200);

for kk=1:7;

[y(:,kk),tt]=step(feedback(G*K*k(kk),1),t);

[u(:,kk),tt] = step(feedback(K*k(kk),G),t);

end

subplot(211); plot(t,y);

subplot(212);plot(tt,u);

legend('k=1','k=1.5','k=2','k=2.5','k=3','k=3.5','k=4');


Engineering

How fast can response get?

Use MATLAB to try a very large value of k

step(feedback(G*K*1000,1),t)

Where do poles go as k tends to infinity?


Engineering

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