View
3
Download
0
Category
Preview:
Citation preview
Control Systems Engineering( Chapter 7. Steady-State Errors )
Prof. Kwang-Chun Hokwangho@hansung.ac.krTel: 02-760-4253 Fax:02-760-4435
2Dept. Electronics and Information Eng.
Introduction
In this lesson, you will learn the following :How to find the steady-state error for a unity feedback
systemHow to specify a system’s steady-state error
performanceHow to design system parameters to meet steady-state
error performance specifications
3Dept. Electronics and Information Eng.
Introduction
In chapter 1, we learnt about 3 requirements needed when designing a control system Transient responseStabilitySteady-state errors (SSE)
Up until now, we only covered until transient response and stabilityWe learned in chapter 4, there are 4 types of transient response for a 2nd-order systemOverdamped, Underdamped, Undamped, Critically
damped
4Dept. Electronics and Information Eng.
Introduction
An example of elevator response
The transient response for elevator can be considered as overdamped
The system is stable but has steady-state error!
5Dept. Electronics and Information Eng.
Introduction
What is steady-state error?Steady-state error is the difference between the input
and output for a certain test input as time approaches infinity( )
The concept of stead-state error is limited to system that are stable
In these systems, the natural response approaches zero as time approaches infinity
Test input used for steady-state error analysis and design areStep, Ramp, Parabola
t
lim ( ) lim ( ( ) ( ))t t
e t r t c t
6Dept. Electronics and Information Eng.
Introduction
Test waveforms for evaluating steady-state errors of position control systems
Test inputs for steady-state error analysis and design vary with target type
7Dept. Electronics and Information Eng.
Introduction
Example of systems tested using the test signalTargeting system:
Targeting a static target. (e.g. a stopping car) We test the system using step input because the position of the car is
in constant position Targeting a car moving with constant velocity
We test the system using ramp input because the car is moving in constant velocity
Targeting an accelerating car We test the system using parabola input because the car is
accelerating
8Dept. Electronics and Information Eng.
SSE for unity feedback system
Unity feedback system can be represented as
Steady state error can be calculated from a system’s closed-loop transfer function, T(s), or the open-loop transfer function, G(s), for unity feedback systems
SSE in terms of T(s)Considering figure (a), we find the SSE E(s) between the input,
R(s) and the output C(s)( ) ( ) ( )( ) ( )[1 ( )] ( ) ( ) ( )
E s R s C sE s R s T s C s R s T s
9Dept. Electronics and Information Eng.
SSE for unity feedback system
We can find final value of the error, in terms of T(s) using
We can only use this equation if T(s) is stable, E(s) has no poles in the right-half plane or poles on the imaginary axis other than the origin
Math Ref.: Initial Value Theorem
If the function x(t) and its first derivative are Laplace transformable and x(t) Has the Laplace transform X(s), and the exists, then
This is particularly useful in circuits and systems to find out the initial condition in the time domain
( )e
0 0( ) lim ( ) lim ( ) lim ( )[1 ( )]
t s se e t sE s sR s T s
0lim ( ) lim ( ) (0)s t
sX s x t x
lim ( )s
sX s
10Dept. Electronics and Information Eng.
SSE for unity feedback system
Final Value Theorem
This is particularly useful in circuits and systems to find out the the final value of x(t) in the time domain
Proof:
0lim ( ) lim ( ) ( )s t
sX s x t x
11Dept. Electronics and Information Eng.
SSE for unity feedback system
Example:Find the steady state-error for a unity feedback system that has
T(s) = 5/(s2+7s+10) and the input is a unit stepSolution:
R(s) =unit step = 1/s T(s) = 5/(s2+7s+10), we must check the stability of T(s) using Routh
table or poles If T(s) is stable, E(s) is then
12Dept. Electronics and Information Eng.
SSE for unity feedback system
Before calculating the final value of the error, we must check the position of E(s) poles The poles for E(s) are at (0,0), (-2,0) and (-5,0) Since all the poles are not on the right half plane or the imaginary
axis we can use the equation to calculate final error value in terms of T(s)
Finally, SSE is given by
20 0
1 5lim 1 lim 17 10
5 5 1110 10 2
s se sR s T s s
s s s
2
2 2 2
2 2 2
2 2
1 5 1 7 10 5( ) 17 10 7 10 7 10
1 7 5 7 5 7 57 10 2 57 10
s sE ss s s s s s s s
s s s s s ss s s s s ss s s
(similar to Output 2 of Fig. 7.2(a))
13Dept. Electronics and Information Eng.
SSE for unity feedback system
SSE in terms of G(s)Considering figure (b), we find the SSE E(s) between the input,
R(s) and the output C(s)
Applying the final value theorem and verifying that the system is stable, we have
( ) ( ) ( )( )( ) ( ) ( ) ( )
1 ( )
E s R s C sR sE s C s E s G sG s
0 0
( )( ) lim ( ) lim1 ( )s s
sR se sE sG s
14Dept. Electronics and Information Eng.
SSE for Test Inputs
We are going to use three types of input R(s) : step, ramp and parabolaSo, the final value of the error for this types of input can be described as
(Step Input)0
0
(1/ ) 1( ) lim1 ( ) 1 lim ( )s
s
s seG s G s
)(lim1
)(1)/1(lim)(
0
2
0 ssGsGsse
ss
(Ramp Input)
)(lim1
1)/1(lim)(
2
0
3
0 )( sGssse
ss sG
(Parabolic Input)
15Dept. Electronics and Information Eng.
SSE for Test Inputs
Example: Find the steady-state errors for inputs of 5u(t), 5tu(t),
and 5t2u(t) to the system with no integration below
Solution:
0
2
0
23 2
0
5 5 5 55 ( ) : ( )1 lim ( ) 1 20 21
5 5 55 ( ) : ( )lim ( ) 0
10 10 105 ( ) ( )lim ( ) 0
step
ramp
parabola
s
s
s
u t es G s
tu t es sG s
t u t es s G s
(similar to Output 2 of Fig. 7.2(a))
(similar to Output 3 of Fig. 7.2(b))
Can get the same result by T(s)!!
16Dept. Electronics and Information Eng.
SSE for Test Inputs
Example: Find the steady-state errors for inputs of 5u(t), 5tu(t),
and 5t2u(t) to the system with one integration below
Solution:
0
2
0
23 2
0
5 5 55 ( ) : ( ) 01 lim ( )
5 5 5 15 ( ) : ( )lim ( ) 100 20
10 10 105 ( ) ( )lim ( ) 0
step
ramp
parabola
s
s
s
u t es G s
tu t es sG s
t u t es s G s
(similar to Output 1 of Fig. 7.2(a))
(similar to Output 2 of Fig. 7.2(b))Improvement for a ramp input !
(No improvement for a parabolic input !)
17Dept. Electronics and Information Eng.
SSE for Test Inputs
Now let’s define the static error constants
Example: Find SSE via static error constantsSolution:
(Position Constant)0
0
(1/ ) 1 1( ) lim1 ( ) 1 lim ( ) 1 p
ss
s seG s G s K
2
00
(1/ ) 1 1( ) lim1 ( ) lim ( ) v
ss
s seG s sG s K
(Velocity Constant)
3
200
(1/ ) 1 1( ) lim1 ( ) lim ( ) a
ss
s seG s s G s K
(Acceleration Constant)
18Dept. Electronics and Information Eng.
SSE for Test Inputs
19Dept. Electronics and Information Eng.
SSE for Test Inputs
First step is to calculate the static error constants
Next step is to calculate the final error value
0 0
0 0
22
0 0
500( 2)( 5)( 6) 500(0 2)(0 5)(0 6)lim lim 5.208( 8)( 10)( 12) (0 8)(0 10)(0 12)(500)( 2)( 5)( 6)lim lim 0( 8)( 10)( 12)(500)( 2)( 5)( 6)lim lim(
p s s
v s s
a s s
s s sK G ss s s s
s s s sK sG ss s s
s s s sK s G ss
0
8)( 10)( 12)s s
1 1 1( ) 0.161, ( ) , ( )1step ramp parabola
p v a
e e eK K K
20Dept. Electronics and Information Eng.
SSE for Test Inputs
SSE via system typeWe are still focusing on unity negative feedback systemBelow is a feedback control system for defining system
type
We define the system type to be the value of n in the denominator Type 0 when n = 0 Type 1 when n = 1 Type 2 when n = 2
21Dept. Electronics and Information Eng.
SSE for Test Inputs
Relationship between input, system type, static error constant, and steady-state errors can be summarized as
(n=0) (n=1) (n=2)
22Dept. Electronics and Information Eng.
SSE Specification
We can use the static error constants to represent the steady-state error characteristic of our systemExample:Gain design to meet a steady-state error specification
(Find the value of K so that there is 10% error in the steady state)
Solution:
0
1 5( ) 0.1 10 lim ( )6 7 8
672
v sv
Ke K sG sK
K
(The system is of Type 1, since only Type 1 have Kv that are finite constant)
23Dept. Electronics and Information Eng.
SSE for Disturbance
Feedback control system are used to compensate for disturbances (or unwanted inputs) that enter a systemConsider feedback control system showing disturbance
SSE isUsing )()()()()()( 221 sGsDsGsGsEsC
)()()( sCsRsE
)()()(1
)()()()(1
1)(21
2
21
sDsGsG
sGsRsGsG
sE
24Dept. Electronics and Information Eng.
SSE for Disturbance
Assuming a step disturbance D(s)=1/s
The steady-state error produced by a step disturbance can be reduced by increased the dc gain of G1(s) or decreasing the dc gain of G2(s)
0 01 2
2
01 2
( ) lim ( ) lim ( )1 ( ) ( )
( )lim ( ) ( ) ( )1 ( ) ( )
s s
R Ds
se sE s R sG s G s
sG s D s e eG s G s
10 02
1( ) 1lim lim ( )( )
D
s s
eG s
G s
25Dept. Electronics and Information Eng.
SSE with State Space
Using the final value theorem, one can find the steady-state error for single-input, single-output systems in state-space formConsider the closed-loop system represented in state
space
The Laplace transform of the error is
usingApplying the final value theorem, we have
,r yx = Ax +B = Cx
( ) ( ) ( ) ( ) 1E s R s Y s R s s -1= C( I A) B
( ) ( )Y s R s s -1C( I A) B
0 0( ) lim ( ) lim ( ) 1
s se sE s sR s s
-1C( I A) B
26Dept. Electronics and Information Eng.
SSE with State Space
Example:Evaluate the SSE for the system
Solution:Using
For the unit-step, R(s)=1/s,For the unit-ramp, R(s)=1/s2,Notice that the system behaves like a Type 0 system
0 0( ) lim ( ) lim ( ) 1
s se sE s sR s s
-1C( I A) B
( ) 4 / 5e ( )e
27Dept. Electronics and Information Eng.
Homework Assignment #7
28Dept. Electronics and Information Eng.
Homework Assignment #7
29Dept. Electronics and Information Eng.
Homework Assignment #7
30Dept. Electronics and Information Eng.
Homework Assignment #7
Recommended