Analysis of Stability & Steady -State Errorswebstaff.kmutt.ac.th/~sarawan.won/AE/INC341/Lec4.pdf ·...

INC 341 – Feedback Control Systems

Analysis of Stability &

Steady-State ErrorsSteady-State Errors

S Wongsa

[email protected]

Stability & Steady-State Error

Summary from previous class

� First-order & second order systems response

22

2

2)(

nn

n

ssG

ωζωω

++=

1)(

+=s

KsG

τ

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

c(t)

ζ=0

ζ=0.2ζ=0.5

ζ=1

ζ=1.5

10,100%)1/( 2

<≤⋅= −− ζζζπeOST

8.1≅

� Block diagrams

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Time(sec)

d

pT ωπ

=

10,100%)1/( <≤⋅= −− ζζζπ

eOS

n

sT ζω4

=

n

rT ω8.1

≅

-18 -16 -14 -12 -10 -8 -6 -4 -2 0-8

-6

-4

-2

0

2

4

6

8

Cascade Form Parallel Form Feedback Form


Today’s goal

� Stability analysis

- Definition of stability

- Stability conditions

- Routh-Hurwitz criterion

� Steady- state error analysis


Part I : Stability analysis


Definition of stability

LTI

systemx(t)

Initial conditions

y(t) = ynatural(t) + yforced (t)

Example: unit step response of a first-order system G(s)u(t)

tetysss

s

ssY 5

5

3

5

2)(

5

5/35/2

)5(

)2(1)( −+=→

++=

++

=

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time(sec)

y(t)

force respone: 2/5u(t)

natural response:3/5exp(-5t)

A pole of the input

function

(forced response)

A pole of the

transfer function

(natural response)


Definition of stability

LTI

systemx(t)

Initial conditions

y(t) = ynatural(t) + yforced (t)

� A system is stable if

- the natural response approaches zero as t →∞- for every bounded input the output is also bounded as t →∞- for every bounded input the output is also bounded as t →∞

� A system is unstable if

- the natural response approaches infinity as t →∞- for every bounded input the output is unbounded as t →∞

� A system is marginally stable if

- the natural response remains constant or oscillates

- there is at least one bounded input for which the output oscillates


� A LTI system is said to be stable if all the poles are in the LHP.

� A LTI system is said to be unstable if the system has any poles in the RHP.

� A LTI system is said to be marginally stable if the system has nonrepeated jωωωω axis poles.

Stability conditions

Source: R.C. Dorf & R.H. Bishop, Modern Control Systems, 9th Ed..


Is this closed-loop system stable ?


Routh Table

Routh-Hurwitz criterion


� Routh table can tell how many system poles are in the LHP, in the RHP, and on the jωaxis.

� The number of roots of the polynomial that are in the RHP is equal to the number of

sign changes in the first column.

Routh-Hurwitz criterion

� Any row of the Routh table can be multiplied by a positive constant without changing

the values of the rows below.


Example5950.87068.1,4146.13at are poles js ±−=

Try Skill-assessment Exercise 6.1


Zero only in the first column

� Replace the zero with an epsilon, ε, then analyse the sign changes by taking

ε�0-/0+.

10)( =sT

35632)(

2345 +++++=

ssssssT


Zero only in the first column

� Replace the zero with an epsilon, ε, then analyse the sign changes by taking

ε�0-/0+.

10)( =sT

35632)(

2345 +++++=

ssssssT

0.7020 j 0.5088-

1.6681-

1.5083 j 0.3429

at are poles

±

±=s


Entire row is zero

� An entire row of zeros tells us that there is an even polynomial as a factor of the original

polynomial.

� Even polynomials only have roots that are symmetrical about the origin.


Example

)7)(86(7

10

5684267

10)(

242345 +++=

+++++=

sssssssssT

A factor of T(s)

86)( 24 ++= sssP

0124)( 3 ++= ss

ds

sdP


Example

)7)(86(7

10

5684267

10)(

242345 +++=

+++++=

sssssssssT

Used to consider the remaining roots

4 (jω),

1 (LHP)

Used to consider only the roots of the even polynomial (s4)


Example20384859392212

20)(

2345678 ++++++++=

sssssssssT


Example128192128964824103

128)(

2345678 ++++++++=

sssssssssT



Review Questions

1. Where do system poles have to be to ensure that a system is not unstable?

2. What does the Routh-Hurwitz criterion tell us?

3. What causes an entire row of zeros to show up in the Routh table?

4. If a Routh table has two sign changes above the even polynomial and five sign changes

below the even polynomial, how many right-half-plane poles does the system have?

5. If a seven-order system has a row of zeros at the s3 row and two sign changes

below the s4 row, how many jω poles does the system have?


Part II : Steady-State Error Analysis


Definition of Steady-State Error

� Steady-state error is the difference between the input and the output for a prescribed

test input as t � ∞.


Steady-state error for unity feedback systems

Applying the final value theorem gives

)(1

)(

)()()(

sG

sR

sCsRsE

+=

−=

G(s)1

R(s)slim

)(lim)(

0

0

+=

=∞

→

→

s

s ssEe

Applying the final value theorem gives


Steady-state error vs. input types

)(1

)(lim)( 0

sG

ssRe s +

=∞ →

p

s

s

K

sGssG

se

+=

+=⋅

+=∞

→→

1

1

)(lim1

11

)(1lim)(

0

0

vs

ss

KssG

sGsssG

se

1

)(lim

1

))(1(

1lim

1

)(1lim)(

0

020

==

+=⋅

+=∞

→

→→

as

ss

KsGs

sGsssG

se

1

)(lim

1

))(1(

1lim

1

)(1lim)(

2

0

2030

==

+=⋅

+=∞

→

→→


System type

System type is the value of n in the denominator, i.e. the number of pure

integrations in the forward path.

n=0 � Type 0

n=1 � Type 1

n=2 � Type 2

and so on


Steady-state error vs. system types

where is know as the position constant.

is know as the velocity constant.

is know as the acceleration constant.

)(lim 0 sGK sp →=

)(lim 0 ssGK sv →=

)(lim 2

0 sGsK sa →=



Example

Find the value of K so that there is 10% error in the steady state.


Since the system is of Type 1, the finite error is possible only for a ramp input.

1.01

)(lim

1)(

0

===∞→ vs KssG

e

876

5)(lim10 0 ⋅⋅=== →

KssGK sv

Thus, K=672.


What if the feedback gain is not unity?

• Well, make it unity then.

See Examples 7.8 & 7.9 and Try Skill-assessment Exercise 7.5

)()()( 21 sGsGsG =

)(/)()( 11 sGsHsH =


Review Questions

1. Name of the test inputs used to evaluate steady-state error.

3. How many integrations in the forward path are required in order for there to be zero

steady-state error for each of the test inputs listed in Q.1?

2. Define system type.

steady-state error for each of the test inputs listed in Q.1?

4. Increasing system gain has what effect upon the steady-state error?


Summary

� Stability analysis

Routh Table

� Steady-state error analysis

Analysis of Stability & Steady -State Errorswebstaff.kmutt.ac.th/~sarawan.won/AE/INC341/Lec4.pdf ·...

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