lect 6 stability2

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Control Systems

Lect.6 Stability

Basil Hamed


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Chapter 6After completing this chapter the student will be able to:

Make and interpret a basic Routh table todetermine the stability of a system (Sections 6.16.!"

Make and interpret a Routh table #here either the$rst element of a ro# is %ero or an entire ro# is%ero (Sections 6.&6.'"

se a Routh table to determine the stability of asystem represented in state space (Section 6.)"

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Stability A Simple Example

*e #ant the mass to stay at + , - but #ind /a0esome initial speed (f(t" , -". *hat #ill happen

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How to characterize different behaviors with TF?


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Stability Importance The most basic and important specification in control analysis

and synthesis!

Unstable systems have to be stabilized by feedback.

Unstable closed-loop systems are useless. What happens if a system is unstable?

may hit mechanical/electrical stops (saturation)

may break down or burn out

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StabilityWhat Will Happen to UnstableSystems?

Tacoma Narrows Bridge (July 1-Nov.7, 1940)

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Wind-induced vibration Collapsed!

2008


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6.1 Introduction

Three requirements enter into the design of a control system:transient response, stability, and steady-state errors. Thus far

we have covered transient response, which we will revisit in

Chapter 8. We are now ready to discuss the net requirement,

stability. !tability is the most important system specification. "f a

system is unstable, transient response and steady-state errors

are moot points.

#n unstable system cannot be designed for a specific transientresponse or steady-state error requirement.

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6.1 Introduction

*hat is stability

2here are many de$nitions for stability dependin/upon the kind of system or the point of 0ie#. 3n thissection #e limit oursel0es to linear timein0ariant

systems.

4 system is stable if e0ery bounded input yields abounded output. *e call this statement the

boundedinput boundedoutput (B3B5" de$nition ofstability.

4 system is unstable if any bounded input yields an

unbounded output.Basil Hamed


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Stability DefinitionBIBO(Bounded-Input-Bounded-Output)stability:Any boundedinput generates a bounded output

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Asymptotic stability:

4ny 3Cs /enerates y(t" con0er/in/ to %ero.


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6.1 Introductionwe present the following definitions of stability, instability, andmarginal stability:

# linear, time-invariant system is stable if the natural responseapproaches $ero as time approaches infinity.

# linear, time-invariant system is unstable if the naturalresponse grows without bound as time approaches infinity.

# linear, time-invariant system is marginally stable if thenatural response neither decays nor grows but remains

constant or oscillates as time approaches infinity.

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6.1 Introduction

%ow do we determine if a system is stable& 'et us focus on the

natural response definitions of stability

!table systems have closed-loop transfer functions with polesonly in the left half-plane.

"f the closed-loop system poles are in the right half of the s-plane and hence have a positive real part, the system is

unstable.

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6.1 Introduction

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6.1 Introduction

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Stability s Domain Stability

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For a system represented by a transferfunction G(s),

system is BIBO stable

All the poles of G(s) are in the openleft half of the complex plane

system is asymptotically stable


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Stability Remarks on Stability Definition

For a general system (nonlinear etc.), BIBO stabilitycondition and asymptotic stability condition are different.

Forlinear time-invariant (LTI) systems(to which we canuse Laplace transform and we can obtain a transferfunction), the conditions happen to be the same.

In this course, we are interested in only LTI systems, weuse simply stable to mean both BIBO and asymptoticstability.

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Stability Examples

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Stability Summary

Stability for LTI systems

(BIBO and asymptotically) stable, marginally stable,

unstable

Stability forG (s" is determined by poles ofG.

Next

Routh-Hurwitz stability criterionto determine stability

without explicitly computing the poles of a system.

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6.2 Routh-Hurwitz Criterion

"n this section, we learn a method that yields stability

information without the need to solve for the closed-loop

system poles.

(sing this method, we can tell how many closed-loop systempoles are in the left half-plane, in the right half-plane, and on

the )w-ais. *+otice that we say how many, not where.

The method requires two steps:

* enerate a data table called a /outh table

*0 interpret the /outh table to tell how many closed-loop

system poles are in the '%1, the /%1, and on the jw-axis.

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Generating a Basic Routh Table

'oo2 at the equivalent closed-loop transfer function shown in 3ig

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!ince we are interested in the system poles, we focus our

attention on the denominator.

"n order the above characteristic eq. does not have roots in /%1,it is necessary but not sufficient that the following hold4

. #ll the coefficient of the polynomial have the same sign.

0. +one of the coefficient vanishes.


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We first create the /outh table shown in Table below


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5nly the $rst ! ro#s of the array are obtained fromthe characteristic e9. the remainin/ are calculatedas follo#s:

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Stability Routh-Hurwitz Example

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;ind the stability of the characteristic e9:


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Example 6.1 P. 306

1/56'7: a2e the /outh table for the system shown in

3igure

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Example 6.1 P. 306

!5'(T"5+: The first step is to find the equivalent closed-loop

system because we want to test the denominator of this function.

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System is unstable and has ! poles in

RH


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=+ample

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Consider the equation

Solution 6ecause the equation has no missing terms and the coefficientsare all of the same sign, it satisfies the necessary condition

!ystem is unstable

because there are two

sign changes in the first

column of the tabulation,the equation has two

roots in the right half of

the .y-plane.


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6.3 Routh!ur"it# $riterion% &pecial

$ases

Two special cases can occur:

* The /outh table sometimes will have a $ero only in thefirst column of a row, or

*0 The /outh table sometimes will have an entire row thatconsists of $eros

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'ero (nl) in the *irst $olumn

"f the first element of a row is $ero, division by $ero wouldbe required to form the net row.

To avoid this phenomenon, an epsilon, 9, is assigned to

replace the $ero in the first column.

The value 9 is then allowed to approach $ero from eitherthe positive or the negative side, after which the signs of

the entries in the first column can be determined.

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Example 6.+ P.30,

?etermine the stability of the closedlooptransfer function

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# sign change from the s3

row to the s2 row, and therewill be another sign change

from thes2row to the srow.

%ence, the system is

unstable and has two poles

in the right half-plane.


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Entire Ro" is 'ero

We now loo2 at the second special case. !ometimes while ma2ing a

/outh table, we find that an entire row consists of $eros.

/eason for entire row is $ero4 it indicates one or more of the following

eist:

* The roots are symmetrical and real,*0 the roots are symmetrical and

imaginary, or

* the roots are quadrantal.

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Entire Ro" is 'ero

To continue with /outh;s tabulation when a row of $ero appears,

we conduct the following steps:

. 3orm the auiliary equation 1*s < = by using the coefficients from

the row )ust preceding the row of $eros.

0. Ta2e the derivative of the auiliary equation with respect to s4 thisgives d1*s>ds < =.

. /eplace the row of $eros with the coefficients of d1*s>ds .

?. Continue with /outh;s tabulation in the usual manner with the newly

formed row of coefficients replacing the row of $eros.@. "nterpret the change of signs, if any, of the coefficients in the first

column of the /outh;s tabulation in the usual manner.

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Example 6.- P. 310

?etermine the number of ri/hthalfplanepoles in the closedloop transfer function

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Solution


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Example 6.- P. 310

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6.- Routh!ur"it# $riterion% dditional

Examples

=+ample 6.6 ;ind the stability of the systemsho#n.

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!5'(T"5+: 3irst, find the closed-loop transfer function as

The system is unstable, since it

has two right-half plane poles

and two left-half-plane poles.


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Example 6./ P.31-

;ind the stability of the system

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!5'(T"5+: The closed-loop transfer function is

there are two sign changes, and thesystem is unstable, with two poles in

the right half-plane. The remaining

poles are in the left half-plane.


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Example 6., P.31-

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;ind the stability of the system

!5'(T"5+: The closed-loop

transfer function for the system

The closed-loop system is unstable

because of the right-half-plane poles.


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Example 6. P. 31,

1/56'7: 3ind the range of gain, K, for the system shown, that willcause the system to be stable, unstable, and marginally stable. #ssume

K A =.

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!5'(T"5+: 3irst find the closed-loop transfer function as

" ;or stable System -@k@1&76i" ;or nstable System k A1&76ii" ;or mar/inal stable k, 1&76


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=+ample

Consider that the characteristic equation of a closed-loop control

system is

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!olution: "t is desired to find the range ofK so that the system is stable.

3rom the row, the condition of stability isK > =, and from the row, thecondition of stability is


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Stability Control Example

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Determine the range ofK that stabilize the closed-loop system


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Characteristic equation

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IfK = 35, oscillation frequency is obtained by the auxiliary equation


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6. &tabilit) in &tate &pace

(p to this point we have eamined stability from the s-planeviewpoint. +ow we loo2 at stability from the perspective of state

space.

!tability of !tate !pace depends on eigenvalues of matri #.6ecause the values of the system;s poles are equal to the eigenvaluesof the system matri, #.

eigenvalues of matri # is found by det*s" - # < =

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Example 6.11 P. 3+1

1/56'7: iven the system

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3ind out how many poles are in the left half-plane, in the right half-

plane, and on the )w-ais.

!5'(T"5+: 3irst form *s" - #:


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Example 6.11 P. 3+1

+ow find the det*s" B #:

!ince there is one sign

change in the first

column, the system has

one right half- plane

pole and two left-half-plane poles. "t is

therefore unstable.

lect 6 stability2

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