Lecture 7

Lecture 7Lecture 7

Poles and ZerosPoles and Zeros

StabilityStability

Poles and ZerosPoles and Zeros

Transfer Function ModelsTransfer Function Models

General RepresentationGeneral Representation

whwhere zi are the zeros pi are the poles n ≥ m to have a physically realizable system

The dynamic behavior of a transfer function model can be characterized by the numerical value of its poles and zeros

)())((

)())(()(

21

21

0

0

mn

mmn

i

ii

m

i

ii

pspspsa

zszszsb

sa

sbsG

Transfer Function ModelsTransfer Function Models

Transfer Function can be expressed in Transfer Function can be expressed in gaingain//time time constant formconstant form, i.e. factoring out , i.e. factoring out bbmm and and aann

Pole–Zero cancellation happens when the Pole–Zero cancellation happens when the numerator and denominator terms cancels each numerator and denominator terms cancels each otherother

)1)(1(

)1)(1(

)())((

)())(()(

21

21

21

ss

ssK

pspspsa

zszszsbsG

ba

mn

mm

PolePole

The factor/s of the denominator of the transfer function

It is the value wherein the transfer function approaches infinity as the value of s approaches the pole

pole of order n where as

sgsf x

i

ni

i

1

)(

)()(

Example 1Example 1

Pole of order 1 or simple pole at Pole of order 1 or simple pole at ss = 0 = 0

Pole of order 2 at Pole of order 2 at ss = 5 and pole of order 3 = 5 and pole of order 3 at at ss = – 7 = – 7

32 )7()5(

2)(

ss

ssf

ssf

3)(

ZeroZero

The factor/s of the numerator of the transfer function

It is the value wherein the transfer function approaches 0 as the value of s approaches the zero

pole of order m where sg

assf

y

i

mi

i

)(

)()( 1

Complex Plane PlotComplex Plane Plot

Graphical representation of a rational transfer Graphical representation of a rational transfer function in the complex plane which helps to function in the complex plane which helps to convey certain properties of the systemconvey certain properties of the system

x–axis is real part

y–axis is imaginary part

)12)(1()(

222

21

ssss

KsG

Pole–Zero PlotPole–Zero Plot

ZeroZeross = – 2 = – 2

o in ploto in plot PolePole

ss = ±0.5 = ±0.5ii

x in plotx in plot

412

2)(

s

ssf

Effects of Pole LocationEffects of Pole Location

Left Half Plane (LHP) results in a stable system, i.e. stable response

Right Half Plane (RHP) results in unstable system, i.e. unstable step response

x

x

x

Real axis

Imaginary axis

x

x

x

Real axis

Imaginary axis


Faster response when pole is farther from imaginary axis

Complex pole results to oscillatory response

p = a + bj where j = √– 1

Real axis

Imaginary axis

x

x

x → complex poles


More oscillatory transient response when pole is farther from real axis

Pole at the origin, i.e. 1/s term in Transfer Function Model, results in an integrating element/process

Effects of Zero LocationEffects of Zero Location

Zeros have no effect on system stability Zero in Right Half Plane (RHP) results in

inverse response to a step change in the input

x y

t

inverse response

Real

axis

Imaginary axis

Effects of Zero LocationEffects of Zero Location

Zero in Left Half Plane (LHP) results in overshoot during a step response

StabilityStability

StabilityStability

Most industrial processes are stable without Most industrial processes are stable without feedback control and are said to be feedback control and are said to be open–loop open–loop stablestable or or self–regulatingself–regulating

An openAn open––loop stable process will return to the loop stable process will return to the original steady state after a transient disturbance, original steady state after a transient disturbance, i.e. one that is not sustained, occursi.e. one that is not sustained, occurs

By contrast there are a few processes, such as By contrast there are a few processes, such as exothermic chemical reactors, that can be exothermic chemical reactors, that can be open-open-loop unstableloop unstable

StabilityStability

DefinitionDefinitionAn unconstrained linear system is said to be stable if the An unconstrained linear system is said to be stable if the output response is bounded for all bounded inputs or output response is bounded for all bounded inputs or BIBO stability. Otherwise it is said to be unstable.BIBO stability. Otherwise it is said to be unstable.

ApplicabilityApplicability– Any linear control system comprised of linear Any linear control system comprised of linear

elementselements– Nonlinear systems operating near the point of Nonlinear systems operating near the point of

linearizationlinearization

StabilityStability

Asymptotically stability is when the variables of the stable control system always decrease from their initial value and do not show permanent oscillations– Permanent oscillations occur when a pole has a real part exactly

equal to zero (in the continuous time case) or a modulus equal to one (in the discrete time case)

Marginally stability is when a simply stable system response neither decays nor grows over time, and has no oscillations– System transfer function has non–repeated poles at complex

plane origin, i.e. their real and complex component is zero in the continuous time case

Closed Loop SystemClosed Loop System

Transfer FunctionTransfer Function

Km

Gm

PYsp

Ym

Y’sp E–

+

G*d

GpGvGc

U+

+

D

YV

pvcmOL

spOL

pvcm

OL

pd

sppvcm

pvcm

pvcm

pd

GGGGG where

YG

GGGKD

G

GG

YGGGG

GGGKD

GGGG

GGY

11

11*

*

Closed Loop System

Characteristic Equation of Closed Loop System– Is the denominator of the transfer function

– Can be simplified to 1 + GOL where GOL is the open loop transfer function

– Used to solve for poles by equation to zero, i.e. 1 + GOL = 0

Stability for Closed Loop Stability for Closed Loop SystemSystemGeneral Stability CriterionGeneral Stability Criterion

The feedback control system is stable if and only if The feedback control system is stable if and only if allall roots of the characteristic equation are negative or have roots of the characteristic equation are negative or have negative real parts. Otherwise, the system is unstable.negative real parts. Otherwise, the system is unstable.

AssumptionsAssumptions– Set–point changes rather than disturbance changes Set–point changes rather than disturbance changes

were consideredwere considered– Closed–loop transfer function was a ratio of Closed–loop transfer function was a ratio of

polynomialspolynomials– Poles are all distinctPoles are all distinct

Stability Region in Complex Plane

A system is stable if the poles of the transfer function lie strictly in the closed left half of the complex plane, i.e. the real part of all the poles is less than zero

Closed Loop Response

Stable System

Closed Loop Response

Unstable System

Example 1Example 1

Determine whether system is stable or not givenDetermine whether system is stable or not given

Solution:Solution:

Solving for 1 + Solving for 1 + GGOLOL

Equating 1 + Equating 1 + GGOLOL = 0 = 0

)12(

)15.0(10

ss

sGOL

)12(

53

)12(

)15.0(1011

2

ss

ss

ss

sGOL

0532 ss

Example 1Example 1

Solving for Solving for ss

Inference on systemInference on system– System is stable since real part of pole/s is negativeSystem is stable since real part of pole/s is negative– Behavior is oscillatory due to the presence of Behavior is oscillatory due to the presence of

imaginary termimaginary term

2

113

2

113

2

113

2

2093

21

s j

s

or

js

Example 2Example 2


1 + 1 + GGOLOL = = ss + 0.2 + 0.2KKcc – 1 – 1

Solution:Solution:


ss + 0.2 + 0.2KKcc – 1 = 0 – 1 = 0

ss = 1 – 0.2 = 1 – 0.2KKcc

For system to be stable For system to be stable ss should be less than 0, should be less than 0, i.e. i.e. KKcc > > 5 5

Example 3Example 3


Solution:Solution:



)15)(12(

ss

KG c

OL

)15)(12(

1710

)15)(12(11

2

ss

Kss

ss

KG cc

OL

01710 2 cKss

Example 3Example 3

Solving for Solving for ss

For system to be stable, For system to be stable, ss should be less than 0, should be less than 0, i.e. 40(i.e. 40(KKcc + 1) > 0 or + 1) > 0 or KKcc > – 1 > – 1

20

)1(40497 cK

s

Stability Test

Direct Substitution Method Routh Stability Criterion Root Locus Diagram Bode Stability Criterion Nyquist Stability Criterion

Direct Substitution MethodDirect Substitution Method

Imaginary axis divides the complex plane into stable and unstable regions for the roots of characteristic equation

On the imaginary axis, the real part of s is zero, and thus we can write s = j. Substituting s = j into the characteristic equation allows us to find a stability limit such as the maximum value of Kc

As the gain Kc is increased, the roots of the characteristic equation cross the imaginary axis when Kc = Kcm

Direct Substitution MethodDirect Substitution Method

Methodology– s = j is substituted in the characteristic

equation

– Kc is equated to Kcm

– Both the real part and imaginary part is equated to 0 and value of and Kcm is computed

– Stability region is determined for Kc

Example 4Example 4

Determine whether system is stable or not givenDetermine whether system is stable or not given1010ss33 + 17 + 17ss22 + 8 + 8s + s + 11 + + KKcc = 0 = 0

Solution:Solution:Substituting Substituting s = j and Kc = Kcm in the characteristic equation

– – 1010jj33 – 17 – 1722 + 8 + 8jj + + 11 + + KKcmcm = 0 = 0oror

(1(1 + + KKcm cm – 17– 1722) + ) + jj (8 (8 – 10 – 1033)) = 0= 0

Example 4Example 4

Equating both real and imaginary part to zeroEquating both real and imaginary part to zero11 + + KKcm cm – 17– 1722 = 0 = 0

88 – 10 – 1033 = 0= 0

Solving Solving from 88 – 10 – 1033 = 0= 0 results to = 0 → KKcmcm = – 1 = – 1

= ±0.894 → KKcmcm = 12.6 = 12.6

Region of stability would beRegion of stability would be – – 1 < 1 < KKcc < 12.6 < 12.6

Example 5Example 5


Solution:Solution:

Solving for 1 + Solving for 1 + GGOL OL = 0= 0

1

31

21

sss

KG c

OL

0

13

12

111

ss

s

KG c

OL

Example 5Example 5

ss33 + 6 + 6ss22 + 11 + 11s s + 6(1+ 6(1 + + KKcc) = 0) = 0

Substituting Substituting s = j and Kc = Kcm in the characteristic equation

– – jj33 – 6 – 622 + 11 + 11jj + + 6(1 6(1 + + KKcmcm) = 0) = 0

oror

(6(6 + 6+ 6KKcm cm – 6– 622) + ) + jj (11 (11 – – 33)) = 0= 0

0

13

12

1

)1(6116 23

ss

s

Ksss c

Example 5Example 5

Equating both real and imaginary part to zeroEquating both real and imaginary part to zero66 + 6+ 6KKcm cm – 6– 622 = 0 = 0

1111 – – 33 = 0= 0

Solving Solving from 11 – – 33 = 0= 0 results to = 0 → KKcmcm = – 1 = – 1

= ±3.32 → KKcmcm = 10 = 10

Region of stability would beRegion of stability would be – – 1 < 1 < KKcc < 10 < 10

Routh Stability Criterion

Developed by E J Routh in 1905 a.k.a.

– Routh–Hurwitz Stability Criterion

– Routh Test Purely algebraic method Used to establish stability in single input single

output (SISO) linear time invariant control system


Applied to systems with characteristic equation that has a polynomial form.

Hence, it can not be used to systems with time delays or transport lag, i.e. e –s term

For system with e –s term, Padé approximation is done on the time delay term/s

positive is a where

asa sasa

n

nn

nn 001

11


It is necessary (but not sufficient) that all the coefficients of the characteristic equation, i.e. an, an – 1, . . . , a1 and a0, be positive else the system is unstable. Hence, no need to perform the Routh Test

Flow Process for Performing Flow Process for Performing Stability AnalysisStability Analysis

Example 6Example 6

Determine the stability of system that has Determine the stability of system that has characteristic equationcharacteristic equation

ss44 + 5 + 5ss33 + 3 + 3ss22 + 1 + 1 = 0= 0

Solution:Solution:Since the Since the s term is missing, its coefficient is zero

Thus, the system is unstable. Recall that a necessary condition for stability is that all of the coefficients in the characteristic equation must be positive.

Routh Array Generation

Given a polynomial of the form:

ansn + an–1sn–1 + ● ● ● + a1s + a0 = 0

an–1

anan–3

an–2

z1

b1

Row1

● ● ● an–5

an–4 ● ● ●

b3b2 ● ● ●

c1 c2 ● ● ●

● ●

●

3

2

4

n + 1

● ●

●

a

aaaab

a

aaaab

n

nnnn

n

nnnn

1

5412

1

3211

b

baabc

b

baabc

nn

nn

1

33512

1

21311

Theorems on Routh TestTheorems on Routh Test

Theorem 1Theorem 1A necessary and sufficient condition for all the roots A necessary and sufficient condition for all the roots of the characteristic equation to have a negative real of the characteristic equation to have a negative real parts (or stable system) is that all the elements of the parts (or stable system) is that all the elements of the first column in the Routh array be positive and first column in the Routh array be positive and nonzerononzero

Theorem 2Theorem 2If some of the elements in the first column are If some of the elements in the first column are negative, the number of roots with a positive real negative, the number of roots with a positive real part, i.e. in the right hand plane, is equal to the part, i.e. in the right hand plane, is equal to the number of sign changes in the first columnnumber of sign changes in the first column

Theorems on Routh TestTheorems on Routh Test

Theorem 3Theorem 3If one pair of the roots is on the imaginary axis, If one pair of the roots is on the imaginary axis, equidistant from the origin, and all other roots are in equidistant from the origin, and all other roots are in the left half plane, all the elements of the nthe left half plane, all the elements of the nthth row will row will vanish and none of the elements preceding row will vanish and none of the elements preceding row will vanish. The location of the pair of imaginary roots vanish. The location of the pair of imaginary roots can be found by solving the equationcan be found by solving the equation

CCss22 + D = 0 + D = 0

where C and D are the elements of the where C and D are the elements of the ((nn–1)–1)thth row row read left to right, respectivelyread left to right, respectively

Example 7Example 7


ss44 + 3 + 3ss33 + 5 + 5ss22 + 4 + 4ss + 2 + 2 = 0= 0Solution:Solution:

3

1

4

5

2

11/3

Row1 2

226/11

3

2

4

5System is stable since all terms in the 1st column is positive (Theorem 1)

Example 8Example 8


ss66 + + s s55 + + 44ss44 + 3 + 3ss33 + 2 + 2ss22 + 4 + 4ss + 2 + 2 = 0= 0Solution:Solution:

11

34

–12/5

1

Row1 2

–2

5

3

2

4

5

42

2

2

2–74/126

27

System is unstable since not all terms in the 1st column is positive

2 roots are in the right half plane due to 2 sign changes

Example 9Example 9

Determine value of Determine value of KKcc to have a stable system to have a stable system

Solution:Solution:


1

31

2)1(

sss

KG c

OL

1

31

2)1(

)1(6116

13

12

)1(11

23

sss

Ksssss

s

KG cc

OL

Example 9Example 9

Characteristic equationCharacteristic equationss33 + 6 + 6ss22 + 11 + 11ss + 6(1 + + 6(1 + KKcc)) = 0= 0

6

1

6(1 + Kc)

11

10 – Kc

Row1

6(1 + Kc)

3

2

4

For system to be stable, all terms in the 1st column should be greater than zero (Theorem 1)

10 – Kc > 0 → Kc < 106 (1 + Kc) > 0 → Kc > –1

Region of stability –1 < Kc < 10

Example 9Example 9

When When KKcc = 10, system is on the verge of = 10, system is on the verge of instability. The Routh array becomesinstability. The Routh array becomes

6

1

66

11

0

Row1

66

3

2

4

According to Theorem 3, the (n–1)th row is the coefficient C and D used in solving the imaginary roots.

6s2 + 66 = 0

s = ±√11

Example 10Example 10

Determine value of Determine value of KKcc to have a stable system to have a stable system

GGOLOL = 5s +2 = 5s +2KKccee–s–s

Solution:Solution:


1 + 1 + GGOLOL = 1 + 5s +2 = 1 + 5s +2KKccee–s–s

Using 1/1 PadUsing 1/1 Padéé approximation approximation

s

se s

5.01

5.01


Characteristic equationCharacteristic equation

2.52.5ss22 + (5.5 – + (5.5 – KKcc))ss + (1 + 2 + (1 + 2KKcc)) = 0= 0

5.5 – Kc

2.5 1 + 2Kc

1 + 2Kc

Row1

3

2

For system to be stable, all terms in the 1st column should be greater than zero (Theorem 1)

5.5 – Kc > 0 → Kc < 5.5 1 + 2Kc > 0 → Kc > –0.5

Region of stability –1 < Kc < 10

05.01

5.01251

s

sKs c

Root Locus

The locus of the roots of the characteristic equation of the closed loop transfer function as the loop gain, Kc, of the feedback system is increased from zero to infinty

It is a useful tool for analyzing the transient response, as well as the stability of a single input single output dynamic systems

A system is stable if all of its poles are in the left hand side of the s–plane

Graphical procedure for finding roots of the characteristic equation, 1 + GOL = 0

Root LocusRoot Locus

MethodologyMethodology– Obtain characteristic equationObtain characteristic equation

– Vary value of Vary value of KKcc

– Solve for the roots of the equationSolve for the roots of the equation

– Plot the roots based on specific Plot the roots based on specific KKcc values values

– Connect the points based on increasing Connect the points based on increasing KKcc

valuesvalues


Plot the Root Locus Diagram of the characteristic Plot the Root Locus Diagram of the characteristic equationequation

ss33 + 6 + 6ss22 + 11 + 11ss + 6(1 + + 6(1 + KKcc)) = 0= 0

Solution:Solution:Rearranging equation toRearranging equation to

((ss + 1) ( + 1) (ss + 2) ( + 2) (ss + 3) + + 3) + KK = 0 = 0where where KK = 6 = 6KKcc


KK = 6 = 6KKcc Root 1Root 1 Root 2Root 2 Root 3Root 3

––3030 1.21 1.21 – 3.61 – 2.60– 3.61 – 2.60jj – 3.61 + 2.60– 3.61 + 2.60jj––66 0 0 – 3.00 – 1.41– 3.00 – 1.41jj – 3.00 + 1.41– 3.00 + 1.41jj 00 – 3– 3 – 2– 2 – 1– 10.230.23 – 3.1– 3.1 – 1.75– 1.75 – 1.15– 1.150.390.39 – 3.16– 3.16 – 1.42– 1.42 – 1.42– 1.421.581.58 – 3.45– 3.45 – 1.28 – 0.75– 1.28 – 0.75jj – 1.28 + 0.75– 1.28 + 0.75jj 6.66.6 – 4.11– 4.11 – 0.95 – 1.50– 0.95 – 1.50jj – 0.95 + 1.50– 0.95 + 1.50jj26.526.5 – 5.10– 5.10 – 0.45 – 2.50– 0.45 – 2.50jj – 0.45 + 2.50– 0.45 + 2.50jj 6060 – 6.00– 6.00 – 3.32 – 3.32jj + 3.32+ 3.32jj100100 – 6.72– 6.72 0.35 – 4.00 0.35 – 4.00jj 0.35 + 4.000.35 + 4.00jj

-5

-4

-3

-2

-1

0

1

2

3

4

5

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2


-5

-4

-3

-2

-1

0

1

2

3

4

5

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2X XX

X

X

Kc = 0

Kc = 10

Kc = 10

Kc = 10

X

Kc = 0.07

Kc = –5X

Kc = –5

Kc = –5

Kc = –1

Kc = –1

Kc = –1

X

X


Consider a feedback control system that has the Consider a feedback control system that has the open loop transfer functionopen loop transfer function

Plot the root locus for 0 Plot the root locus for 0 ≤≤ KKcc ≤≤ 20 20

Solution:Solution:

The characteristic equation 1 + The characteristic equation 1 + GGOL OL = 0 or= 0 or

((ss + 1)( + 1)(ss + 2)( + 2)(ss + 3) + 4 + 3) + 4KKcc = 0 = 0

)3)(2)(1(

4

sss

KG c

OL


When Kc = 0, the roots are merely the poles of the open loop transfer function, i.e. – 1, – 2 and – 3

X denotes an open loop pole

Dots denote locations of the closed loop poles for different values of Kc

ReferencesReferences

1.1. Coughanowr, Donald R. Coughanowr, Donald R. Process Systems Process Systems Analysis and Control. Analysis and Control. 22ndnd ed. New York: ed. New York: McGraw–Hill, Inc, 1991.McGraw–Hill, Inc, 1991.

2.2. Seborg, Dale E. Seborg, Dale E. et al. et al. Process Dynamics and Process Dynamics and Control. Control. 22ndnd ed. New York: John Wiley & ed. New York: John Wiley & Sons, Inc, 2004.Sons, Inc, 2004.

3.3. http://www.wikipedia.orghttp://www.wikipedia.org

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