سیستمهای کنترل خطی پاییز 1389 بسم ا... الرحمن الرحيم دکتر...

سیستمهای کنترل خطی

1389پاییز

بسم ا... الرحمن الرحيم

دکتر حسين بلندي- دکتر سید مجید اسما عیل زاده

Recap.

• Optimal Systems,• Index of Performance,• Introduction to Ess.

Steady-State Tracking & System Types

In General:

• Unity feedback control:

G(s) C(s)+

-r(s) e y(s)

Go.l.(s)+

-r(s) e

y(s)ol

T.F. get & open, loop cut i.e.

yto e from T.F. loop open the is )()(

)(.. sG

s.an cancelcan otherwise ,0 need ,0 If

)1()1)(1(

:into factored be alwayscan

asasasasas

sTsTsTs

sTsTsTKG

psloss

ssRsseete

sysRse

1step to

1)(:input stepFor

)()(lim)(

: trackingstate-steady

1)()()(:error tracking

)()(:loop-closed

r, respect to with 0" type" called is system the,0 If

)K control alproportion with confused be not to p, small use here(

1step toThen

const.error position static called

)0()(lim denote

1step to

,2 or type ,1 typecalled is system

larger or ,2or ,1 If

. zero-non input with step acan track system 0 type

higheror 1 for type 0

0 for type 1

then,)(:unitnot is step If

higheror 1 for type 0

0 for type 1

:input stepunit For

. zero input with step acan track higher or 1 typeof systemA

const errorvelocity static called

:denote

ramp to

:ramp unit is If

ssGssGs

signal. input ramp a track not can system 0 type

ramp to

system, 0 type For

bsbsbsK

. error statesteady zero-non

withramp tracks system 1 type

finiteramp to

finite,

1: type For

bsbsbsK

!0,0,0,1

. no withinput ramp a

track can system higher or 2 type

ramp to

higher, or 2 type For

factor a as s has still ones cancel

asasasas

bsbsbsK

1type if :then

unit, not is ramp If

2type if

1type if

0type if

:input ramp unit For

input onaccelerati unit

KsGssGss

lim)(1

acc to

bsbsbsK

1acc to

0 den.in offactor no system, 0 For type

constant.error on accelerati theis )(lim

sig. acc. tract tcan' system 1 or 0 type

acc to

bu i.e.

den. in of factor one i.e.

. in of factor one system, 1 type For

asasas

bsbsbsK

error. s.s. finite withsig.acc tract can system 2 type

acc to

den. in of factors two or,

in of factors two :2 type

bsbsbsK

error. s.s. no withsig.acc tract can syst. higher or 3 type

acc to

higher or 3 type

bsbsbsK

stabilize. to difficult are system higher or 2 type but

tracting. bettertyper larger like seems

:Caution

by A. multiplied be to needs then,

:rather acc, unit not If

3type if

2type if

0,1type if

acc to

:input acc. forsummary

r(t)=R·1(t)r(s)=R/s

r(t)=R·t·1(t)r(s)=R/s2

r(t)=R·1/2·t2

r(s)=R/s3

type 0(N=0 a0≠0)

Kp=b0/a0

ess=R/(1+Kp)

ess=∞

type 1(N=1 a0=0 a1≠0 b0≠0 )

Kp= ∞

Kv=b0/a1

ess=R/Kv

ess=∞

type 2, N=2a0=a1=0

a2≠0,b0≠0

Kp= ∞

Kv= ∞

Kp=b0/a2

ess=R/Ka

type≥3, N ≥ 3a0=a1=a2=0

b0≠0

Kp= ∞

Kv= ∞

Ka= ∞

sys.type

ref.input

Example of tank

KKRKGK

RKsHsCsG

KsCRAs

1acc to

1ramp to

1step to

0 type0,1

)()()(

:control

KRKssGKK

RKsKsHsCsG

0,)(lim,

)()()()(

acc to

ramp to

step to

type den, in of factor one

same :control PI

the to loop the following from path the in #

T.F. loop open the in #i.e.

in # is w.r.t.type sys.

type. sys. is tracking statesteady toKey

Kps+KI

r(s) e ωn2

s(s+2ξ ωn)

w.r.t.2 type 2 :path in #count

example

1type one is there

:default Take

specified. dist. or input No :Note

acc. ramp. step. to error statesteady &

constants error type, system find

)5.0)(5.1(

)15.3()(

G(s)r(s) e(s) y(s)

2.45.05.1

15.3)(lim

ramp to

1 type for acc to

1 type for step to

1 type for

15)(lim

)5)(12(

)1(5)(

acc to

typeramp to

step to

type ,#

:Example

The stability of linear feedback systems.

The concept of stability.

The stability of a cone.

Tacoma Narrows Bridge

As oscillation begins

At catastrophic failureon November 7, 1940.

Bounded Input - Bounded Output criterion

An unconstrained linear system is stable if the output response is bounded for all bounded inputs.

Note: By a bounded input, we mean an input variable that stays within upper and lower limits for all values of time

Example: BI-BO criterion

Q.If the step function is applied at the input of continuous system and the output remains below a certain level for all time, is the system stable?

A.The system is not necessarily stable since the output must be bounded for every bounded input.

A bounded output to one specific bounded input does not ensure stability.

Q. If the step function is applied at the input of continuous system and the output is of the form y = t, is the system stable or unstable?

A. This system is unstable since a bounded input produced an unbounded output.

Characteristic root locations criterion

A system is stable if all the poles of the transfer function have negative real parts.

Stability in the s-plane.

Stable system

A necessary and sufficient condition for a feedback system to be stable is that all the poles of the transfer function have negative real parts.

This means that all the poles are in the left-hand s-plane.

• Determine in each case if the set of roots represents, stable, marginally stable, or unstable systems.

a) -1, -2 e) -2 + j, -2 - j, 2j, -2jb) -1, +1 f) 2,-1,-3c) -3,-2,0 g) -6,-4,7d) -1 + j, -1- j h) -2 + 3j, -2 -3j, -2

• Determine in each case if the set of roots represents, stable, marginally stable, or unstable systems.

a) -1, -2 e) -2 + j, -2 - j, 2j, -2jb) -1, +1 f) 2,-1,-3c) -3,-2,0 g) -6,-4,7d) -1 + j, -1- j h) -2 + 3j, -2 -3j, -2

Marginally stable

If there are any poles on the jω-axis (with none on the right hand side), then the system is considered marginally stable.

Unstable

If there are one or more poles in the right-hand s-plane or there are repeated roots on the jω-axis, the system is unstable.

Example: Poles and zeros

Q. A system has poles at -1 and -5 and zeros at 1 and -2. Is the system stable?

Example: Poles and zeros

Q. A system has poles at -1 and -5 and zeros at 1 and -2. Is the system stable?

A. The system is stable since the poles are roots of the system characteristic equation which have negative real parts.

The fact that the system has a zero with a positive real part does not affect its stability.

Example: Characteristic equation

Q. Determine if the system with the following characteristic equation is stable:

(s+1)(s+2)(s-3) = 0

Example: Characteristic equation

Q. Determine if the system with the following characteristic equation is stable:

(s+1)(s+2)(s-3) = 0

A. This characteristic equation has the roots

-1,-2, and 3 and therefore represent an unstable systems since there is a positive real root.

Example: integrator

Q. The differential equation of an integrator may be written as follows:

dy/dt = x

Determine if an integrator is stable.

Example: integrator

y(t) x(t)dt0

Y(s) 1

Y(s)X(s)

s s1 0 0 j0

Example: integrator

A. The characteristic equation of his system is s = 0. Since the root does not have a negative real part, an integrator is not stable.

Since it has no roots with positive real parts, integrator is marginally stable

Stability in the s-plane.

NOT STABLESTABLE

The Routh-Hurwitz Stability Criterion.

Introduction

• Goal: Determining whether the system is stable or unstable from a characteristic equation in polynomial form without actually solving for the roots.

• Routh’s stability criterion is useful for determining the ranges of coefficients of polynomials for stability, especially when the coefficients are in symbolic (nonnumerical) form.

A necessary condition for Routh Stability

• A necessary condition for stability of the system is that all of the roots of its characteristic equation have negative real parts, which in turn requires that all the coefficients be positive.

A necessary (but not sufficient) condition for stability is that all the coefficients of the polynomial

characteristic equation be positive.

A necessary and sufficient condition for Stability

• Routh’s formulation requires the computation of a triangular array that is a function of the coefficients of the polynomial characteristic equation.

A system is stable if and only if all the elements of

the first column of the Routh array are positive

Characteristic Equation

• Consider an nth-order system whose the characteristic equation (which is also the denominator of the transfer function) is:

q(s) ansn an 1s

n 1 an 2sn 2 a1s a0

Determining the Routh array

• Consider the characteristic equation:

• First arrange the coefficients of the characteristic equation in two rows, beginning with the first and second coefficients and followed by the even-numbered and odd-numbered coefficients:

q(s) an sn an 1sn 1 an 2s

n 2 an 3sn 3 a1s a0

sn : an an 2 an 4 sn 1 : an 1 an 3 an 5

Routh array: method (cont’d)

• Then add subsequent rows to complete the Routh array:

sn : an an 2 an 4 sn 1 : an 1 an 3 an 5 sn 2 : bn 1 bn 3 bn 5 sn 3 : cn 1 cn 3 cn 5

s0 : hn 1

Routh array: method (cont’d)

• Compute elements for the 3rd row:

sn : an an 2 an 4 sn 1 : an 1 an 3 an 5 sn 2 : bn 1 bn 3 bn 5 sn 3 : cn 1 cn 3 cn 5

s2 : * *

s1 : *

s0 : *

bn 1 1

an an 2

an 1 an 3

bn 3 1

an an 4

an 1 an 5

cn 1 1

an 1 an 3

bn 1 bn 3

Routh-Hurwitz Criterion

• The Routh-Hurwitz criterion states that the number of roots of q(s) with positive real parts is equal to the number of changes in sign of the first column of the Routh array.

Three distinct cases of the first column array

1. No element in the first column is zero.

2. There is a zero in the first column and in this row.

3. There is a zero in the first column and the other zero

in this row.

Case1. No element in the first column is zero

2)( asasasq

2011 0

Second-order system

2)( asasasq

The requirement for a stable second-order system is that all the coefficients be positive or all coefficients be negative.

Third-order system

3)( asasasasq

Third-order system

30121 a

3)( asasasasq

For system to be stable, it is necessary and sufficient

that the coefficients a0 ,a1, a2, a3 be positive

and a2a1 > a0a3.

Unstable system

Two roots of q(s) lie in the right-hand s-plane.

q(s) s3 s2 2s 24

q(s) (s 1 j 7)(s 1 j 7)

Two changes in sign

Case 2: There is a zero in the first column and in this row.

Ksssssq 234)(

The system is unstable for all values of gain K.

Case 3:A zero in the first column and the other zero in this row.

q(s) s3 2s2 4sKFor a stable system:

U(s) = 2s2 + 8 = 2 (s + j2)(s - j2)

q(s) (s 2)(s j2)(s j2)

For a marginal stability: K=8

0 < K < 8

s3 1 4

s2 2 K

s1 8 K

s0 K 0

Ex. A six-legged micro robot

It is equipped with sensor network that includes 150 sensors of 12 different types. The legs are instrumented so that the robot can determine the lay of the terrain, the surface texture, hardness, and color.

Micro robot

633244)( 2345 ssssssq

)3)(3(21)3(216321)( 22 jsjssssU

The auxiliary polynomial:

The auxiliary polynomial

• The equation that immediately precedes the zero entry in the Routh array.

Micro robot

)( 232

s 1 j 6

The robot is using flexible legs with high-gaincontrollers that may become unstable and oscillate.

Example :

Given the characteristic equation,

is the system described by this characteristic equation stable?

Answer:• One coefficient (-2) is negative.

• Therefore, the system does not satisfy the necessary condition for stability.

4s4ss2s3s4s)s(a 23456

Example :

Given the characteristic equation,

is the system described by this characteristic equation stable?

Answer:• All the coefficients are positive and nonzero.• Therefore, the system satisfies the necessary condition for

stability.• We should determine whether any of the coefficients of

the first column of the Routh array are negative.

44234)( 23456 sssssssa

Example (cont’d): Routh array

44234)( 23456 sssssssa

0424:s

4131:s

44234)( 23456 sssssssa

05122:

44234)( 23456 sssssssa

2)2(25

01576:

05122:

44234)( 23456 sssssssa

The elements of the 1st column are not all positive:

the system is unstable

Example : Stability versus Parameter Range

Consider a feedback system such as:

The stability properties of this system are a function of the proportional feedback gain K. Determine the range of K over which the system is stable.

Example (cont’d)

0)6)(1(

Example (cont’d)

• Expressing the characteristic equation in polynomial form, we obtain:

0)6(5 23 KsKss

0)6)(1(

Example (cont’d)

• The corresponding Routh array is:

• Therefore, the system is stable if and only if

q(s) s3 5s2 (K 6)sK

5)304(:

5 0 and K 0

K 7.5 and K 0

Example (cont’d)

• Solving for the roots gives:

-5 and ±1.22j for K=7.5=> The system is unstable (or critically stable) for K=7.5

-4.06 and –0.47 ±1.7j for K=13-1.90 and –1.54 ±3.27j for K=25

=> The system is stable for both K=13 and K=25

KsKsssa )6(5)( 23

Example : Stability versus Two Parameter Range

Consider a Proportional-Integral (PI) control such as:

Find the range of the controller gains so that the PI feedback system is stable.

)K,K( 1

Example (cont’d)

• The characteristic equation for the system is given by:

)s)(s(s

Example (cont’d)

• Expressing the characteristic equation in polynomial form, we obtain:

023 123 Ks)K(ss

)s)(s(s

Example 4 (cont’d)

• The corresponding Routh array is:

• For stability, we must have:

KsKsssa )6(5)( 23

)KK(:s

KKandK

The Routh-Hurwitz criterion

Relative Stability

• Further characterization of the degree of stability of a stable closed loop system.

• Can be measured by the relative real part of each root or pair of roots.

Root r2 is relatively more stable than roots r1 and r1’.

Axis shift

464)( 23 ssssq

14)1(6)1(4)1( 2323 nnnnnn ssssss

)1)(1())((1)( 2 jsjsjsjsssU nnnn

EX. Flow graph diagram

12 3 sL

L3 Ks 2

)3()3(1)(1 221121321

sKsssLLLLL

s2 2s (K 3) 0 K 3 for stability

Ex. Block diagram model.

T(s) KG1(s)G2(s)

1 KG1(s)G2(s)

(s) 1KG1(s)G2(s) 0

0)3(2)3)(1()( 2 KssKsss

G2(s) 1

K 3 for stability

Tracked vehicle turning control.

Select K and a so that the system is stable and ess≤ 24%of a ramp command.

Characteristic equation:

1Gc (s)G(s) 0

1K(s a)

s(s1)(s 2)(s 5)0

s4 8s3 17s2 (K 10)sKa 0

Routh array

b1 126 K

b1(K 10) 8Ka

b1 126 K

b1(K 10) 8Ka

Ramp response for a=0.6 and K=70

The steady-state error

E(s) 1

1Gc (s)G(s)R(s)

E(s) s(s1)(s 2)(s 5)

s(s1)(s 2)(s 5) K(s a)

ess(t) limsE (s) 10A

aKfor s 0

The closed-loop T(s)

T(s) 70s 42

s4 8s3 17s2 80s 42s 7.08

s 0.58

s 0.17 3.2 j

The stable region

(K 10)(126 K) 64Ka 0

ess AKa

K 70 a 0.6

Absolute Stability

• A closed loop feedback system which could be characterized as either stable or not stable.

Stable System

• A stable system is a dynamic system with a bounded response to a bounded input.

Stability of S.S. Eqn.

xx A0)( xAI

0)det( AI

Example: Closed epidemic system

det)det(

det)det( AI

]))([()det( 2 AI

0)]()([()det( 22 AI

سیستمهای کنترل خطی پاییز 1389 بسم ا... الرحمن الرحيم دکتر...

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