Csl11 11 f15

ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION

Control Systems

B.Madhuri


Second order system

H(s)

G(s)R(s) C(s)+ -

𝐺 (𝑠)=𝜔𝑛

2

𝑠 (𝑠+2𝜉 𝜔𝑛) 22

2

2)(

)(

nn

n

sssR

sC

If H(s)=1


Steady-State Error Definition

Steady-state error is the difference between the

input and the output for a prescribed test input as time approaches infinity


Steady-State Error - Test Input

1.Step input2.Ramp input3.Parabolic input


Steady-State Error – Introduction


Evaluating Steady-State Error

Steady-state error analysis only applicable when the

system response is stable. Unstable system cannot be analyzed for steady-state

error


Evaluating Steady-State Error – Step input

Error = 0

Error = constant


Evaluating Steady-State Error – Ramp input

Error = 0

Error = constant

Error = ∞


Evaluating Steady-State Error – General Block Diagram

E(s) = R(s) – C(s)

E(s) = R(s) – C(s)

General Representation (unity and non-unity feedback)

Unity feedback System

Error Calculation


Evaluating Steady-State Error – Sources of Error

1.Non-linear elements2.System configuration3.Type of applied input


Evaluating Steady-State Error – Sources of Error by System Configuration

Since K is nonzero and positive, e(t) will never be zero for a step input

Kte

Kte

KsKs

K

sRsE

sRsKEsE

sKEsC

sCsRsE

t

1

1)(

1

1)(

)1(

1

1

1

1

)()(

)()()(

)()(

)()()(

The pure gain configuration will never

produce zero steady-state error.


Evaluating Steady-State Error – Sources of Error by System Configuration

Since K is nonzero and positive, e(t) will always approach zero for a step input.

The integrator configuration will always produce zero steady-state error. 0)(

)(

11)(

1

)()(

)()()(

)()(

)()()(

t

Kt

te

ete

Kss

Kss

s

KssR

s

KsR

sE

sRsEs

KsE

sEs

KsC

sCsRsE


Steady-State Error for Unity Feedback System Using Closed Loop Transfer Function

)(1)()(

)()()()(

)()()(

)()()(

sTsRsE

sTsRsRsE

sTsRsC

sCsRsE

Since we are interested to find the steady-state error (not the transient error), we can use final

value theorem


Steady-State Error for Unity Feedback System Using Closed Loop Transfer Function

)(lim)(lim)(0

ssEteest

Final Value Theorem

)(1)(lim)(lim)(00

sTssRssEess


Steady-State Error for Unity Feedback System Using Closed Loop Transfer Function - Example

ssR

sssT

1)(

107

5)(

2

Find steady-state error?

5.0107

51)

1(lim)(

)(1)()(lim)(

20

0

sssse

sTssRssEe

s

s


Steady-State Error for Unity Feedback System Using Forward Path Transfer Function

)(1

)()(

)()()()(

)()()(

)()()(

sG

sRsE

sRsGsEsE

sGsEsC

sCsRsE

By final value theorem,

)(1

)(lim)(lim)(00 sG

ssRssEe

ss

No need to evaluate

T(s)


Steady-State Error for Unity Feedback System – Step Input

)(lim1

1

)(1

1

lim)(lim)(

)(1

)(lim)(lim)(

000

00

sGsGss

ssEe

sG

ssRssEe

sss

ss

In order to have zero steady-state error,

)(lim0

sGs

1)(lim

))((

))(()(

0

21

21

nsG

pspss

zszssG

s

n

In order to have zero steady-state error, n must be equal or greater than one. There must exist at least one pole at origin. If n = 0, we have

21

21

0

21

21

)(lim

))((

))(()(

pp

zzsG

psps

zszssG

s

There is finite steady-state error


Steady-State Error for Unity Feedback System – Ramp Input

)(lim

1

)(lim

1

)(lim1

/1

)(1

1

lim)(1

)(lim)(lim)(

000

2

000 ssGssGssG

s

sGss

sG

ssRssEe

ssssss


)(lim0

ssGs

2)(lim

))((

))(()(

0

21

21

nssG

pspss

zszssG

s

n

In order to have zero steady-state error, n must be equal or greater than two. There must exist at least two poles at origin.


Steady-State Error for Unity Feedback System – Ramp Input

If n = 1, we have

21

21

21

21

00

21

21

))((

))((lim)(lim

))((

))(()(

pp

zz

pspss

zszssssG

pspss

zszssG

ss

There is finite steady-state error

If n = 0, we have

0))((

))((lim)(lim

))((

))(()(

21

21

00

21

21

psps

zszssssG

psps

zszssG

ss

There is infinite steady-state error

)(lim

1)(

0ssG

e

s


Steady-State Error for Unity Feedback System – Parabolic Input

)(lim

1

)(lim

1

)(1

1

lim)(1

)(lim)(lim)(

2

0

2

0

2

3

000 sGssGsssGss

sG

ssRssEe

sssss


)(lim 2

0sGs

s

3)(lim

))((

))(()(

2

0

21

21

nsGs

pspss

zszssG

s

n In order to have zero steady-

state error, n must be equal or greater than three. There must exist at least three poles at origin.

If n = 2, then finite errorIf n = 1 or less, then infinite error


Static Error Constants

)(lim1

1)(

0sG

e

s

step

Step input

)(lim

1)(

0ssG

e

s

ramp

Ramp input

)(lim

1)(

2

0sGs

e

s

parabola

Parabolic input

Static Error ConstantSteady-state error

)(lim0

sGKs

p

Position error constant

Velocity error constant

Acceleration error constant

)(lim 2

0sGsK

sa

)(lim0

ssGKs

v

April 14, 2023


Static Error Constants

22

)(lim0

sGKs

p

The static error constants can assume three values:1.Zero2.Finite constant3.Infinity

)(lim 2

0sGsK

sa

)(lim

0ssGK

sv

The value of steady-state error decreases as the value of static error constant increases

pstep Ke

1

1)(

vramp Ke

1)(

aparabola Ke

1)(

April 14, 2023

ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATIONApril 14, 2023

Static Error Constants - Example

Evaluate steady-state error constants and the steady state errors for step, ramp and parabolic inputs?

208.5)12)(10)(8(

)5)(2(500

)12)(10)(8(

)5)(2(500lim)(lim00

sss

sssGK

ssp

161.0208.51

1

1

1)(

pstep Ke

0)12)(10)(8(

)5)(2(500lim)(lim00

sss

sssssGK

ssv

0)12)(10)(8(

)5)(2(500lim)(lim 2

0

2

0

sss

ssssGsK

ssa

0

11)(

aparabola Ke

0

11)(

vramp Ke

Step Input




)12)(10)(8)(0(

)6)(5)(2(500

)12)(10)(8(

)6)(5)(2(500lim)(lim00 ssss

ssssGK

ssp

01

1

1

1)(

pstep Ke

25.31)12)(10)(8(

)6)(5)(2(500lim)(lim00

ssss

ssssssGK

ssv

0)12)(10)(8(

)6)(5)(2(500lim)(lim 2

0

2

0

ssss

sssssGsK

ssa

0

11)(

aparabola Ke

032.025.31

11)(

vramp Ke




)12)(10)(8(

)7)(6)(5)(4)(2(500lim)(lim

200 ssss

ssssssGK

ssp

01

1

1

1)(

pstep Ke

)12)(10)(8(

)7)(6)(5)(4)(2(500lim)(lim

200 ssss

ssssssssGK

ssv

875)12)(10)(8(

)7)(6)(5)(4)(2(500lim)(lim

22

0

2

0

ssss

sssssssGsK

ssa

31014.1875

11)(

aparabola Ke

011

)(

v

ramp Ke

April 14, 2023


Steady-State Error for Unity Feedback System – Example

Find steady-state error for input 5u(t), 5tu(t) and 5t2u(t).


Steady-State Error for Unity Feedback System – Example

Find steady-state error for input 5u(t), 5tu(t) and 5t2u(t).


H(s)

G(s)R(s) C(s)+ -

Steady state error for non-unity feedback system

𝑒𝑠𝑠=lim𝑠→0

𝑠𝑅 (𝑠)1+𝐺 (𝑠 )𝐻 (𝑠)


April 14, 2023

System Types

System type is the value of n.n = 0, System type 0n = 1, System type 1n = 2, system type 2Etc.


System Types


System Types - Example


Check for stability first

)9)(7(

)8(1000)(

ss

ssG

2s1s0s

1 8063

1016 0

8063

No sign changes.Therefore, all poles are in the left-half plane.Therefore, the closed-loop system is stable.

Engineering

Csl11 11 f15