Systems Analysis & Control: Steady State Errors

Overheads for ACS211
Systems analysis and control
Anthony Rossiter
Department of Automatic Control and
Systems Engineering
University of Sheffield
www.shef.ac.uk/acse

© University of Sheffield 2009 This work is licensed under a
Department of Automatic Control
Creative Commons Attribution 2.0 License. and Systems 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

and Systems Engineering

3

ACS211: Compensation
and steady-state errors

Anthony Rossiter
Dept. ACSE
Some of this is revision of ACS114


4
Final value theorem
(FVT) Convergent
signals only
For analysis of offset, students should be
able to prove the following (was done in
ACS114). lim y ( t ) lim sY ( s )
t s 0
Examples:
1 2t
Y (s) lim s 0
sY ( s ) 0; { y (t ) e }
s 2
1 t 2t
Y (s) lim s 0
sY ( s ) 0; { y (t ) e e }
(s 1)( s 2)
1 1 1 2t
Y (s) lim s 0
sY ( s ) ; { y (t ) (1 e )}
s(s 2) 2 2

5
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?
Think about this for a few minutes and tell me.


6
In these slides assume
following feedback
structure


7
Illustration of offset
Step Response
1 1.2
G 2
s 3s 2 1

0.8
G=tf(1,[1 3 2])

Amplitude
K=1
K=1; 0.6
K2=10 offset
target
K2=10; 0.4
figure(1);clf reset
0.2
step(feedback(G*K,1),feedba
ck(G*K2,1)); 0
0 0.5 1 1.5 2 2.5 3 3.5 4
hold on Time (sec)

plot([-0.1,0,0,4],[0,0,1,1],'r');
axis([-0.1 4 0 1.2]);
legend('K=1','K2=10','target');
Note how as gain increases, speed of
response improves, but always an
offset.

8
What about the inputs?
Outputs
Note how the input
1 K=1
amplitude depends K2=10

Amplitude
target
upon the error 0.5

(target –output) and
the gain K. 0
0 0.5 1 1.5 2 2.5 3 3.5 4
Time (sec)
A bigger gain means
Inputs
a bigger initial input 10

so a faster 5
Amplitude

response and a
0
smaller offset.
-5
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (sec)

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 of Automatic Control
Department

9
Gain insufficient
It is clear that gain alone is insufficient to
remove offset.
e(s) r (s) G ( s )u ( s )
e(s) r (s) G ( s ) K ( s )e( s )
u (s) K ( s )e( s )

(1 G ( s ) K ( s )) e ( s ) r (s)

1
e(s) r (s)
1 G (s)K (s)

1 1 1
Let r(s)=(1/s) lim s 0
se ( s ) lim s 0
s .
and use FVT. 1 G (s)K (s) s 1 G (0) K (0)


10
Examples
Find the steady state offset for the
following, system controller
pairs, assuming the target is a unit step.

1 1 1
G G G
2
s 2s 1 ; s
3
3s
2
3s 1 ; (s 10 )( s 2)
K 2 K 0 .5 K 8


11
Answers
1 1 1
G G G
s
2
2s 1 ; s
3
3s 3s
2
1 ; (s 10 )( s 2)
K 2 K 0 .5 K 8
ERRORS
1
System 1 1 1 1

A pu e
0.5

ml d
t
i
1 G (0) K (0) 1 2 3 0
0 1 2 3 4 5 6 7
Time(sec)
Step Response
1 1 2
System 2

Amplitude
0.9
1 G (0) K (0) 1 0 .5 3 0.8
0.7

0 1 2 3 4 5 6 7 8 9 10

StepmResponse
Ti e (sec)
System 3
Amplitude

0.9
0.8
1 1 20 0.7
0 . 71
1 G (0) K (0) 1 (8 ) 28 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
20 Time (sec)

12
MATLAB CODE
G1=tf(1,[1 2 1])
K1=2;
Note there are spaces
G2=tf(1,[1 3 3 1])
between these numbers,
K2=0.5;
so it is
G3=tf(1,[1 12 20]) [1 space 3 space 3 space
K3=8; 1]
figure(1);clf reset
subplot(311) You can use commas if
step(feedback(1,G1*K1)); that is easier to read
title('ERRORS','Fontsize',18)
subplot(312) You may be able to copy
step(feedback(1,G2*K2)); and past this code straight
subplot(313) into MATLAB from the
step(feedback(1,G3*K3)); PDF file.

13
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?


14
Offset free tracking
requirement (steps).
Begin from the FVT [Assume target is a unit
step or 1/s ]
1 1 1
lim t
e (t ) lim s 0
se ( s ) lim s 0
s .
1 G (s)K (s) s 1 G (0) K (0)

It is clear that e(t) tends to zero if and only if
the term G(0)K(0) is infinite.
Clearly with no
Zero offset requires either G(s) or K(s) cancelling
derivative term
to include an integrator that is a term
of the form (1/s). Department of Automatic Control

15
Integral action
Most practical control laws include an
integrator to ensure offset free tracking of
constant targets.
ERRORS
1
1
G 2 0.5
s 2s 1 Amplitude

0 .4 0

K
s -0.5
0 5 10 15 20 25
Time (sec)

1
G 3 2
1

s 3s 3s 1
0.5
0 .2
Amplitude

K 0
s
-0.5
0 5 10 15 20 25 30 35
Time (sec) Department of Automatic Control

16
MATLAB CODE
G1=tf(1,[1 2 1])
K1=tf(0.4,[1 0]);
Note the term [1 0] defines
G2=tf(1,[1 3 3 1])
(s+0)
K2=tf(0.2,[1 0]);
figure(2);clf reset i.e. coefficient of 1 on s
subplot(211) and coefficient of 0 on the
step(feedback(1,G1*K1)); constant
subplot(212)
step(feedback(1,G2*K2));
title(' ','Fontsize',18)


17
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.
1 1 (s 3)
G G G
s
2
2s 3 ; s
3
5s
2
3s 2 ; (s 1)( s 2 )( s 4)
K 4 K 0 .6 K 8
s 3 1
G 2
G 3 2
(s 3)
s 2s 1 s 5s 3s 2 G
; ; 3
s 5s
2
3s
4 (s 2)
K K 0 .6 K 0 .8
s s
One is not as expected. WHY? and Systems Engineering

18

ERRORS
1
Figures for
Amplitude

0.5

0
0 1 2 3 4 5 6
previous page
Time (sec)
Step Response
Amplitude

0.9
0.8
0.7

0 5 10 15 20 25
Time (sec)
Step Response
1
ERRORS
Amplitude

6
x 10
0.5 2

Amplitude
0
0 0.5 1 1.5 2 2.5 3 3.5 4 0
4.5
Time (sec)
-2
0 50 100 150
Time (sec)

1

Amplitude
0

-1
0 50 100 150 200 250 300 350
Time (sec)

1
Amplitude

0

-1
0 2 4 6 8 10 12 14
Time (sec)

19
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.

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

20
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.


21
Offset to ramps
Revert back to the FVT but now use a target of
r(s)=(1/s2). [or scaled variant as appropriate]
1 1 1
e(s) r (s) . 2
1 G (s)K (s) 1 G (s)K (s) s

1 1 1
lim t
e (t ) lim s 0
se ( s ) lim s 0
s . 2
or
1 G (s)K (s) s sG ( s ) K ( s )

Clearly, the error diverges to infinity unless there is
an integrator in G(s) or K(s).
1
lim t
e (t ) lim s 0
sG ( s ) K ( s )


22
Examples
Find the offset to a ramp r=(1/s2) for the
following compensator system pairs.
1 1
G G 3 2
2
s 2s 1 ; s 3s 3s 1
0 .4 0 .2 ( s 3)
K K
s (s 4)
1
G 3 2
s 3s 3s 1
0 .2
K
s


23
Examples
1 1
G G 3 2
2
s 2s 1 ; s 3s 3s 1
0 .4 0 .2 ( s 3)
K K
s (s 4)
1
G 3 2
s 3s 3s 1
0 .2
K
s


24
Solutions
1
G 2
s 2s 1 lim sG ( s ) K ( s ) 0 .4; e 2 .5
s 0
0 .4
K
s
1
G 3 2
s 3s 3s 1
lim s 0
sG ( s ) K ( s ) 0; e
0 .2 ( s 3)
K
(s 4)
1
G 3 2
s 3s 3s 1 lim sG ( s ) K ( s ) 0 .2; e 5
s 0
0 .2
K Department of Automatic Control
s and Systems Engineering

25
Figures for 1st and 3rd
pairs (errors only)
ramp=0:20;
time=ramp;
ERRORS
G1=tf(1,[1 2 1]) 20

K1=tf(0.4,[1 0]); 15

Amplitude
G2=tf(1,[1 3 3 1]) 10
K2=tf(0.2,[1 0]);
5
figure(2);clf reset
0
subplot(211) 0 2 4 6 8 10 12 14 16 18 20

lsim(feedback(1,G1*K1),time,ramp); Time (sec)

20
subplot(212)
lsim(feedback(1,G2*K2),time,ramp); 15
Amplitude

title(' ','Fontsize',18) 10

5

0
0 2 4 6 8 10 12 14 16 18 20
Time (sec)

26
Zero offset to ramps

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

Tell me what you think.


27
Zero offset to ramps
Clearly, zero offset requires two integrators
in G(s)K(s) so that:
2
M (s) 1 s
G (s)K (s) 2
lim s 0
lim s 0
0
s sG ( s ) K ( s ) sM ( s )

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.).

28
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.
(s 3)
1 1 G
G G (s 1)( s 2 )( s 4)
s
2
2s 3 ; s
3
5s
2
3s ;
2
K 4 K 0 .6 K
s
s 3 s 1
G 2
G 3 2
(s 3)
s 2s 1 s 5s 3s G
; ; 3
s 5s
2
3s
0 .4 (s 2)
K K 6 K 0 .8
s s

Tutorial questions for 29

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] Department of Automatic Control

30
Figures for offsets
ERRORS ERRORS
20 20
Amplitude

Amplitude
10 10

0 0
2 4 6 8 10 12 14 16 18 20 2 4 6 8 10 12 14 16 18 20
Time (sec) Time (sec)
Linear Simulation Results
20 20
Amplitude

Amplitude
10 0

0
2 4 6 8 10 12 14 16 18 20 -20
2 4 6 8 10 12 14 16 18 20
Time (sec)
Linear Simulation Results Time (sec)
20 20
Amplitude

Amplitude

10
10
0
2 4 6 8 10 12 14 16 18 20 0
2 4 6 8 10 12 14 16 18 20
Time (sec)
Time (sec)


31
Extra tutorial question:
validation
MATLAB can be used to G1=tf(1,[1 2 3])
validate your K1=4;
computations in several
easy ways: G2=tf(1,[1 5 3 0])

1. Compute and display K2=0.6;
closed-loop responses
as in these notes. Gc1=feedback(G1*K1,1)
2. Compute closed-loop gain1=bode(Gc1,0)
transfer function
Gc2=feedback(G2*K2,1)
explicitly and evaluate
steady-state gain. gain2=bode(Gc2,0)


32
Examples of closed-
loop transfer functions
Try these examples by hand and with
MATLAB:
1
G 4 4
2
s 2s 3 Gc 2
G c (0) 0 . 57
K 4 s 2s 7 7

1
G 0 .6
3
s 5s
2
3s Gc 3 2
G c (0) 1
K 0 .6 s 5s 3s 0 .6

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

33
Reminders
1. 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.

Given G (s)
2
; K (s) k
s 2
s 3 s

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.


Given G (s)
2
; K (s) k
s 2
s 3 s

1. Find the closed-loop transfer functions from
target to input and target to output.
GK 2k (s 2)
G cy ( s ) ;
1 GK 2k (s 2) s(s 3)
K 2k (s 2 )( s 3)
G cu ( s ) ;
1 GK 2k (s 2) s(s 3)

2. Find the open-loop and closed-loop poles for
k=1. Which are better and why?
2
pc 2k (s 2) s(s 3) s (2k 3) s 4k ;
2
k 1 pc s 5s 4 (s 1)( s 4 );
po s(s 3 );

Given G (s)
2
; K (s) k
s 2
s 3 s

• 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
1 1 3
e ( step ) 0; e ( ramp )
1 G (0) K (0) sGK k .2 .2
• Best value for k involves a compromise between
speed of response and input activity. See
MATLAB for illustration.


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');


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?


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© 2009 University of Sheffield

This work is licensed under a Creative Commons Attribution 2.0 License.
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