Control Systems Engineering( Chapter 7. Steady-State Errors )

Prof. Kwang-Chun Hokwangho@hansung.ac.krTel: 02-760-4253 Fax:02-760-4435

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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

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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

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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!

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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

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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

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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

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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

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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

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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

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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

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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))

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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

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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)

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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)!!

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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 !)

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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)

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SSE for Test Inputs

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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

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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

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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)

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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)

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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

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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

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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

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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

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Homework Assignment #7

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Homework Assignment #7

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Homework Assignment #7

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Homework Assignment #7