Root Locus

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Chapter 5 Root Locus Method

Contents §1 Introduction

1.1 Time Domain Methods 1.2 Requirements of Anal. and Des. Methods 1.3 Major Indirect Methods

§2 Root Locus Plots 2.1 An Example 2.2 Angle and Magnitude Condition 2.3 Constant gain Locus

§3 General Rules for Constructing RL 3.1~3.9

§4 RL Analysis of Control Systems 4.1 Effect of poles and Zeros 4.2 Comparison Study 4.3 Conditionally Stable Systems 4.4 NM Phase Systems and +ve F/B Systems 4.5 Systems with Pure Time Delay 4.6 Root Contours 4.7 Roots of a Polynomial

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§1 Introduction

1.1 Direct Analysis Method —Time Domain Solution

lBest for judging performance lToo complicated lDifficult to predict the performance as parameters change lTime consuming

1.2 Requirements of Analysis Methods lSimple to implement lEasy to predict the system performance lPossible to indicate better parameters

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1.3 Major Indirect Methods

(1) Algebraic Stability Criterion lRouth Stability Criterion ¬Absolute stability ¬No proper measure of relative stability ¬Difficult to judge that if the CL poles reside in a given region ¬No clue to parameter changes

(2) Frequency Response Method lNyquist plot, Bode diagrams, Nichols charts are available lEasy to predict the CL performances lEasy to improve the CL performance by modifying OL frequency response lNot able to know the relation between the CL poles and the OL parameters

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(3) Root Locus Method

lWhat is a root locus?

A plot of the Roots of

of the characteristic equation

of the Closed Loop

as the function of Gain

lWhy to study the root locus?

Easy to predict the CL performances

K = 0 K = 0

K = 0

K = 0

lPossible to study the root locus?

YES. W R Evans, 1948, 1950

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§2 Root Locus Plots

2.1 An Example

Example 2.1 Draw the CL root locus for the given OL transfer function

G sK

s s( )

( )=

+1 G s( )

R s( ) C s( )+−

Solution: CL TF is

G sC sR s

G sG s

Ks s K

CL ( )( )( )

( )( )

= =+

=+ +1 2

K varies à CL poles change. CL characteristic equation:

s s K2 0+ + = CL poles are:

s K1 212

12

1 4, = − ± −

(i) K ≥ 0

K = 0, s1 212

12

0 1, ,= − ± = − , OL poles

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K <14

, sK

1 212

1 42, = − ±−

K =14

, s1 212

012

12, ,= − ± = − −

K >14

, sK

1 212

4 12, = − ±

−j

K = 0 K = 0

K =14

K → ∞

K → ∞

− 1

CL system damping K=0 0.25 ∞ Overdamped Underdamped (ii) K ≤ 0 K = 0, s1 2 0 1, ,= − , OL poles

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K < 0

sK

1 212

1 42, = − ±

+

s1 1∈ −∞ −( , ]

s2 0∈ +∞[ , )

K = 0 K = 0K → −∞

−1

K → −∞

o

2.2 Angle and Magnitude Conditions lAssumption: K ≥ 0 lRoot locus conditions All CL poles satisfy G s F s( ) ( ) = −1 i.e. Magnitude condition: G s F s( ) ( ) = 1

Angle condition:

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[ ]arg ( ) ( ) ( )G s F s k= ± ° +180 2 1

k = 0 1 2, , ,L

Example 2.2 Condition check for Example 2.1

Solution: Take arbitrarily s112

2= − + j

(i) Angle condition

− 1

θ2 θ1

σ

jωs1

arg ( ) arg( )

G ss s1

1 1

11

=+

= − − +arg( ) arg( )s s1 1 1

= − −θ θ1 2

= −−

−arctan.

arctan.

205

205

= − °− °= − °104 04 7596 180. .

(2) Magnitude condition

GK

.( . )

.− ± =

− + −05 2

05 0 5j

j2 + j2 +1

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= =417

1K

i.e. K =174

⇒ CL poles s112

2= − ± j o

lAngle condition ALONE gives root locus ⇑ All CL poles satisfy angle condition without knowing K lMagnitude condition gives K for a given CL pole

2.3 Constant Gain Locus lLocus of complex s satisfying G s F s( ) ( ) = 1 with constant K

lMagnitude condition ALONE gives constant gain locus

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Example 2.3 Draw the constant gain locus for

G s F sK

s s( ) ( )

( )=

+ 1

Solution:

−1 σ

jω

K = 1K = 5

K = 20

o

lIntersecting points of root locus and constant gain locus are the CL poles

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lRoot locus and constant gain locus are always orthogonal In G s F s( ) ( ) plane ¬ [ ]arg ( ) ( ) ( )G s F s k= ± ° +180 2 1 and

G s F s( ) ( ) = 1 are conformal mappings

of root locus and constant gain locus ¬ [ ]arg ( ) ( ) ( )G s F s k= ± ° +180 2 1 and

G s F s( ) ( ) = 1 are orthogonal

So in s plane Root locus and constant gain locus are orthogonal

σ

jω

s plane ⇒

conformal mapping

Re

Im

G s F s( ) ( ) plane

G s F s( ) ( ) = 1

arg ( ) ( )( )

G s F sk= ± ° +180 2 1

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§3 General rules for Constructing Root Locus

lAssumptions: ¬Argument:

σ

jω

Positive real axis—0°

counterclockwise — +ve direction

¬Negative F/B—K ≥ 0

¬Only CL root locus on the upper s plane are calculated

3.1 Write Char. Eqn. in the Form

1 1 1+ = + = +G s F s KW s KB sA s

( ) ( ) ( )( )( )

= +− −

− − −=1 01

1 2

K s z s zs p s p s p

m

n

( ) ( )( )( ) ( )

LL

And LOCATE the OL poles and zeros in the s plane.

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3.2 Find Starting/Terminating Points and Number of Branches lStarting points: ( )K = 0

KW s( ) = 1 ⇒ W sK

( ) =1

lim ( ) limK K

W sK→ →

= = ∞0 0

1

It is equivalent to ( )( ) ( )s p s p s pn− − − =1 2 0L

¬ All OL poles are starting points lTerminating points: ( )K = ∞

lim ( ) limK K

W sK→∞ →∞

= =1

0

It is equivalent to

( )( ) ( )s z s z s zs n m

m− − − =→ ∞ >

1 2 0Lif

¬ All OL zeros are (finite) terminating points

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¬ If n m> , there are n m− infinite terminating points lNumber of Branches n branches ← n starting points where m branches to m finite zeros n m− branches to infinity NB: If n m< ,

n branches from finite OL poles

m n− branches from infinity

All branches terminate at OL zeros

3.3 Loci on the Real Axis l Depends only on the real poles & zeros l Conjugate poles make no contributions to the angle condition

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θ2

θ1

σ

jωs1

s2

θ θ1 2 360+ = ° l Odd number of real poles and zeros to the right of loci ¬ Real axis point to the right of poles/zeros

Phase angle ≡ 0° ¬Real axis point to the left of poles/zeros

Phase angle ≡ 180°

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¬If 2k poles/zeros on the right to a real axis point

- 360°

0°

Phase angle= ± °×180 2k which does not satisfy Angle Condition

¬Draw real axis root loci

Rightest pole / zero

3.4 Asymptotes of loci as s → ∞ lAngle of asymptotes

γ

As s → ∞ All phase angle are the same—γ

[ ]arg ( ) ( )G s F s s→∞

= −( )m n γ

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From [ ]arg ( ) ( ) ( )G s F s k= ± ° +180 2 1

we have γ =° +−

m180 2 1( )kn m

, k = 0 1 2, , ,L

ln m− distinct asymptotes Angle of one asymptote with the real axis

γ =− ° +

−180 2 1( )k

n m, k n m= − −0 1 1, , ,L

lReal axis intersect of asymptotes

γ

− σ a

− =

−

−= =∑ ∑

σa

jj

n

ii

n

p z

n m1 1

(See Appendix I for the proof)

3.5 Breakaway and break-in points lMultiple CL poles case

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lReal axis breakaway/break-in points K=0 K=0

σ

σK

Break-in point K takes a maximum

K =∞ K = ∞

K

σ

σ

Breakaway point K takes a minimum

NB: (i) There may be more than ONE breakaway/break-in points (ii) Breakaway/break-in points are not necessarily on the real axis

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lCalculation CL poles satisfy f s A s KB s( ) ( ) ( )= + = 0 (1) Multiple CL poles satisfy (at least)

d

df s

sA s KB s

( )( ) ( )= ′ + ′ = 0

KA sB s

= −′′( )( )

(2)

Substituting Eqn (2) into Eqn (1) gives

A sA sB s

B s( )( )( )

( )−′′

= 0

i.e A s B s A s B s( ) ( ) ( ) ( )′ − ′ = 0 (3)

Since KA sB s

= −( )( )

,

so ddKs

A s B s A s B sB s

=′ − ′( ) ( ) ( ) ( )

( )2 (4)

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Comparing Eqns (3) and (4) gives the necessary condition:

ddKs

= 0 (5)

NB: Breakaway/break-in points are only those points satisfying K > 0

Example 3.1 Given

G s F sK

s s s( ) ( )

( )( )=

+ +1 2

To find the breakaway/break-in points. Solution: The characteristic equation is

K s s s s s s= − + + = − − −( )( )1 2 3 23 2

ddKs

= 0 → − − − =3 6 2 02s s

s = − −0 423 1577. , .

For s = −0 423. K s s s s= − + + = >=−( )( ) ..1 2 0 385 00 423

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For s = −1577. K s s s s= − + + = − <=−( )( ) ..1 2 0 385 01 577

So -0.423 is the breakaway point. NB: (i) -1.557 is not on the root locus, ⇒ it is not a breakaway/ break-in point. (ii) When B s( ) = 1, the necessary condition is ′ =A s( ) 0. o (See Appendix II for further discussions on multiple breakaway/break-in points)

3.6 Angle of Departure from Complex poles and Angle of Arrival at Complex Zeros lThese angles are important for drawing

ω 1 1, K

ω 2 2, K

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lA test point near the complex pole/zero is used

pr

ϕpr

s

ϕzr

zr

s

lAngle of departure from a complex pole pr

A test point s should satisfy

arg ( ) ( ) arg( )G s F s p zr i pri

m= − −

=∑ ϕ

1

− −=≠

∑arg( )p pr jjj r

n

1

= ± ° +180 2 1( )k

ϕpr r jjj r

nk p p= ± ° + − −

=≠

∑180 2 11

( ) arg( )

+ −=∑arg( )p zr ii

m

1

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Example 3.2 Find the angle of departure for

G s F sK s

s s( ) ( )

( )=

++ +

22 32

Solution: OL poles are p1 2 1 2, = − ± j

σ

jωϕpr

j 2p1

p1

55°

90°

z1

−2 −1

ϕp k1 180 2 1= ± ° +( )

− −arg( )p p1 2 + −arg( )p z1 1 = ± ° +180 2 1( )k

− °+ °90 55 = °145

ϕp2 145= − ° (symmetry) o

lAngle of arrival at a complex zero zr

A test point s should satisfy

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arg ( ) ( ) arg( )G s F s z zr i zrii r

m= − +

=≠

∑ ϕ1

− −=∑arg( )z pr jj

n

1

= ± ° +180 2 1( )k

ϕzr r jj

nk z p= ± ° + + −

=∑180 2 1

1

( ) arg( )

− −=≠

∑arg( )z zr iii r

m

1

(See Appendix III for the angle of departure from multiple OL poles)

3.7 Imaginary Axis Crossing Points lCrossing of imag. axis by root locus ⇒ change of stability status lThe crossing points indicate critical K and oscillation frequency

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lMethods (i) Routh stability criterion (ii) Let s = jω . Solve

[ ][ ]

Re ( ) ( )

Im ( ) ( )

1 0

1 0

+ =

+ =

G F

G F

j j

j j

ω ω

ω ω

for ω and K. (iii) Trial and error approach

Example 3.3 Find the crossing points for

G s F sK

s s s( ) ( )

( )( )=

+ +1 2

Solution:

Char eqn: f s s s s K( ) = + + +3 23 2

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(i) Routh criterion Routhian array

s3 1 2 6 0− =K

s2 3 K K = 6

s1 ( )6 3− K Aux. Eqn from line s2

s0 K 3 02s K+ = s = ± j 2

(ii) f K( ) ( ) ( ) ( )j j j jω ω ω ω= + + + =3 23 2 0

K − =

− =

3 0

2 0

2

3

ω

ω ω

ω

ω

= ±

= =

2

3 62K o

3.8 Conservation of the Sum of System Poles l Sum of CL poles = Sum of OL poles if n m− ≥ 2 f s A s KB s( ) ( ) ( )= +

= − +−

=∑s p sn

in

i

n1

1

L

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= − +

−

=∑K s z sm

jm

j

n1

1L

When n m− ≥ 2, i.e. m n≤ − 2 .

f s s p sni

n

i

n( ) = − −

=∑ 1

1

+ ∑ − − −terms of (s s s sn n m m2 3 1, , , , , )L L

Suppose the CL poles are λ λ λ1 2, , , ,L n

then f s s ii

n( ) ( )= −

=∏ λ

1

= − +−

=∑s sn

in

i

nλ 1

1

L

So λii

n

ii

np

= =∑ ∑=

1 1

lPart of CL poles may be found according to the rule

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3.9 Illustrative Examples Example 3.4 Sketch the root locus for

G sK s

s s( )

( )=

++ +

22 32 , F s( ) = 1

Solution:

(i) G s F sK s

s s( ) ( )

( )=

++ +

22 32

OL poles: p1 2 1 2, = − ± j

OL zeros: z1 2= −

σ

jω

j 2

− 2 − 1

Char eqn: f s s s K s( ) ( )= + + + +2 2 3 2

= + + + + =s K s K2 2 3 2 0( ) ( )

KA sB s

s ss

= − = −+ +

+( )( )

2 2 32

(ii) Starting points: − ±1 2j

Terminating points: − ∞2,

Number of branches: 2 (iii) Real axis locus: ( , )− ∞ − 2

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(iv) Asymptotes

γ =± ° +

−=

± ° +180 2 1 180 2 11

( ) ( )kn m

k

= − °180 (k = 0) No need to calculate − σa

(v) Breakaway points

ddKs

= 0 ′′

=A sB s

A sB s

( )( )

( )( )

2 2

12 3

2

2s s ss

+=

+ ++

s s2 4 1 0+ + = s1 2 2 3 3732 0 268, . , .= − ± = − −

− 3732. is on the root locus and is the breakaway point

NB: Ks s

ss

= −+ +

+= >

=−

2

3 732

2 32

5 4641 0.

.

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(vi) Angle of departure

ϕp1 180= ± °− −arg( )p p1 2 + −arg( )p z1 1

= ± °− °+ °= °180 90 45 145 or − °215

ϕp2 145= − °

(vii) Imaginary axis crossing points

σ

jω

j 2

− 2 − 1

ζ = 0 7.

No crossing point

(viii) A simple application: To find the CL poles and corresponding gain for ζ = 0 7. .

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Let s n n= − ± −ζω ω ζj 1 2

= − ± −σ σ ζζ

j1 2

= − ±σ σj10202.

then G s( ) = K( . )( . )( . )

− + +− + + + − + + −

σ σσ σ σ σ

jj j 2 j j 2

10202 210202 1 10202 1

arg ( ) ( )G s F s = arctan.102022

σσ−

−+

−arctan

.10202 21

σσ

σ

jω

j 2

− 2

ζ = 07.ϕ1

ϕ2

θ

σ1

−−

−arctan

.10202 21

σσ

= − − = − °θ ϕ ϕ1 2 180 θ ϕ ϕ+ °= +180 1 2

tantan tan

tan tanθ ϕ ϕ

ϕ ϕ= +

−1 2

1 21

2 0408 4 1 02. σ σ− + = σ1 1666= . , σ 2 0 294= .

Take σ1 1666= . (See root locus for the reason)

then s = − ±167 1 70. .j

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According to magnitude condition

Ks

ss

=+ +

+=

=− ±

( ).

. .

1 22

1342

1 67 1 70j

K = 134. sd = − ±167 170. .j ζ = 0 7. (See Appendix IV for more discussions)

o

Example 3.5 Sketch the root locus for

G s F sK

s s s s( ) ( )

( . )( )=

+ + +2 73 2 22

Solution: (i) OL poles: 0, − 2 73. , − ±1 j

K s s s s= − + + +( . )( )2 73 2 22

= − + + +( . . . )s s s s4 3 24 73 7 46 5 46 (ii) Starting points: 0, − 2 73. , − ±1 j

Terminating points: ∝ Number of branches: 4

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(iii) Real axis locus: (-2.73, 0)

σ

jω

− 2 −1

-1183.

-2 0565. j1

− 2 73.

(iv) Asymptotes

γ =± ° +

−=

± ° +180 2 1 180 2 14

( ) ( )kn m

k

= ± ° ± °45 135,

− =

−

−= =∑ ∑

σa

jj

n

ii

m

p z

n m1 1

=− − + − −0 2 73 1 1

4. j1 j1

= −1183.

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(v) Break-in points

ddKs

= 0, 4 14 19 14 92 546 03 2s s s+ + + =. . .

One real root s = −2 0565. (on the root locus) K A s s= − = >=−( ) ..2 0565 2 931 0

(vi) Angle of departure

σ

jω

−2 −1

j1135°

90°

30°

p1p2

p3

p4

ϕp3 180= ° − −arg( )p p3 1

− −arg( )p p3 2 − −arg( )p p3 4

= °− °− °− °180 135 30 90 = − °75

ϕp4 75= °

(vii) Imaginary crossing points

f s s s s s K( ) . . .= + + + + =4 3 24 73 7 46 5 46 0

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Let s = jω

ω ω ω ω4 3 24 73 7 46 5 46 0− − + + =j j. . . K

K + − =

− =

ω ω

ω ω

4 2

3

7 46 0

546 4 73 0

.

. .

ω = ±10744. , K = 7 28.

o

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§4 Root Locus Analysis of Control Systems

4.1 Effect of Zeros and Poles (1) Addition of Zeros and Poles

Example 4.1 Addition of a zero

G s F sK

s p s p( ) ( )

( )( )=

+ +1 2

σ

jω

− p1 − p2

p p1 2 0> >

σ

jω

− p2− p1− z

z p p> >1 2

σ

jω

− p 2− p1 − z

p p z1 2> >

σ

jω

− p2− p1 − z

p z p1 2> >

★ A zero moves the root locus to the left

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Example 4.2 Addition of a pole

G s F sK s z

s p s p( ) ( )

( )( )( )

=+

+ +1 2

σ

jω

− p2− p1− z

z p p> >1 2

σ

jω

− p2− p1− z − p3

p p3 2<

σ

jω

− p2− p1− z− p3

p z3 >

★A pole moves the root locus to the right

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(2) Movements of Poles and Zeros with a Parameter

Example 4.3 Given OL transfer function

G s F sK ss s

( ) ( )( )( )

=++

12 a

σ−1−a

It is important to check if there are break-in/breakaway points in ( , )− −a 1

KA sB s

s ss

= − = −+

+( )( )

( )2

1a

ddKs

= 0 ⇒ A s B s A s B s( ) ( ) ( ) ( )′ = ′

s s s s s2 23 2 1( ) ( )( )+ = + +a a

2 3 2 02s s+ + + =( )a a

s =− + ± − +( )a a a3 10 9

4

2

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Break-in/breakaway points exist if

a a2 10 9 0− + ≥ , i.e. a ≥ 9, a ≤ 1 l a = ≥10 9 s1 2 4 2 5, , .= − −

l a = 9 s1 2 3 3, ,= − −

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l a = 8 No break-in & breakaway points

l a = 3

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l a = 1 G s F sKs

( ) ( ) = 2

l a = 0 5.

★Minor change in the OL poles may

cause dramatic change in the CL root loci

NB: − 1 must be a closed- loop pole.

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4.2 Comparison Study:

Example 4.4 Derivative control and velocity feedback in a positional servemechanism

G ss s

( )( )

=+

15 1

(i) Three Loop Configurations lP-control

−+R s( ) C s( )

G s( )5

G s F sI I( ) ( ) =+

55 1s s( )

=+1

0 2s s( . )

lPD-control

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

−5 1 08( . )+ s

G s F sII II( ) ( ) =+

+5 1 0 8

5 1( . )( )

ss s

=++

0 8 1250 2

. ( . )( . )s

s s

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43

lVelocity F/B

−+ C s( )G s( )5

R s( )

−1 0 8+ . s

G s F sIII III( ) ( ) =+

+5 1 0 8

5 1( . )( )

ss s

=++

0 8 1250 2

. ( . )( . )s

s s

(ii) Root Loci

σ

jω

− 0 2.

σ

jω

− 02.−125.

Remarks: System I s1 2 01 0 995, . .= − ± j , ζ = 01.

Heavy oscillation, slow decaying rate

System II,III s1 2 05 0866, . .= − ± j , ζ = 05.

Better performance could be expected

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44

(iii) Time response lUnit impulse response

★The zero causes the difference

between systems II and III (See Appendix V for impulse response equations)

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45

lUnit step response

¬ II derivative of position error, fast response. III velocity F/B, correction before error occurs, smaller overshoot. (See Appendix V for step response equations)

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46

4.3 Conditionally Stable Systems

Example 4.5 Given OL transfer function

)14.1)(6)(4(

)42()(

2

2

++++++

=sssss

ssKsG

with F s( ) = 1. Find the stability ranges

of gain K .

Solution:

(i) OL poles: 0 4 6 0 7 0 714, , , . .− − − ± j

OL zeros: − ±1 17321j .

Breakaway point: 3557.2−

(ii) Imag axis crossing points:

CL char equation:

f s s s s k s( ) . ( . )= + + + +5 4 3 2114 39 43 6

+ + + =( )24 2 4 0K s K

=++−

=++−

)2(0)]224(39[

)1(04)6.43(4.1124

24

K

KK

ωωω

ωω

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K = − + −05 195 124 2. .ω ω (from (2))

ω ω ω6 4 2202 92 9 96 0− + − =. . ω1 12115= . ω2 21545= . ω3 37538= .

with K1 1554= . K2 64 74= . K3 16351= .

(iii) Stability ranges K < 1554. , 64 74 16351. .< <K

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48

4.4 Non-minimum Phase Systems and Positive Feedback

(1) Root Locus Condition for Non-minimum Phase Systems

−+ G s( )R s( ) C s( )

where G sK T ss Ts

( )( )( )

=−+

11

1 ,T1 0> ,K > 0,T > 0

Angle condition:

arg ( ) arg( )( )

G sK T ss Ts

=−+

11

1

= −−+

arg( )( )

K T ss Ts

1 11

= °+−+

18011

1arg( )( )

K T ss Ts

So arg( )( )

K T ss Ts

k1 11

360−+

= ± °× , k = 01 2, , ,L

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49

(2) Root Locus Condition for a System Under Positive Feedback

++ G s( )

R s( ) C s( )E s( )

[ ]C s G s E s G s R s C s( ) ( ) ( ) ( ) ( ) ( )= = +

= +G s R s G s C s( ) ( ) ( ) ( )

C sR s

G sG s

( )( )

( )( )

=−1

Char eqn of CL system: 1 0− =G s( )

G s( ) = 1 Angle condition: arg ( )G s k= ± °×360 , k = 01 2, , ,L

Root Locus

50

(3) Complementary Root Loci and Complete Root Loci

Let G s F sK s z s z

s p s p s pm

n( ) ( )

( ) ( )( )( ) ( )

=− −

− − −1

1 2

LL

0 < < ∞K : Root loci − ∞ < <K 0: Complementary root loci − ∞ < < ∞K : Complete root loci

(4) Plotting of Complementary Root Loci for Negative Feedback with K < 0 Char eqn G s F s KW s( ) ( ) ( )= = −1 (K < 0) Let ′ = −K K , then ′ =K W s( ) 1 ( ′ >K 0) Root locus conditions:

′ =

′ = ± °× =

K W s

K W s k k

( )

arg ( ) , , , ,

1

360 0 1 2 L

★All rules related to the angle

condition should be modified.

Root Locus

51

lReal axis locus

Even number of real poles/zeros lAsymptotes

γ =± °×

−= − −

3600 1 2 1

kn m

k n m, , , , ,L

lAngle of departure and arrival

ϕpr r ii

mk p z= ± °× + −

=∑360

1

arg( )

− −=≠

∑arg( )p pr jjj r

n

1

ϕzr r iii r

mk z z= ± °× − −

=≠

∑3601arg( )

+ −=∑arg( )z pr jj

n

1

Root Locus

52

Example 4.6 Sketch root locus plot for

G s F sK T ss Ts

K T T( ) ( )( )( )

, , ,=−

+> > >

11

0 0 011

Solution: Write

+

−′

=+

−−=

Tss

TsK

TsssTK

sFsG1

1

)1()1(

)()( 11

where 01 <−

=′TKT

K

★The root locus for 0 < < ∞K is the complementary root locus for 0<′<∞− K .

σ

jω

K = 0 + ∞0=′K − ∞

−1

1T−1T

o

Root Locus

53

Example 4.7 Sketch the complete root locus plot for given system

G sK

s s sF s( )

( )( ), ( )=

+ +=

1 21

Solution:

o

Root Locus

54

4.5 Systems with Pure Time Delay

TssKWsFsG −= e)()()(

where W s( ) is a rational function.

Let s = +σ ωj , then ωσ TTTs jee −−− = CL characteristic equation:

0e)(1 =+ −TssKW

i.e. 1e)( j −=−− ωσ TTsKW

Since

[ ] [ ]

−== −

ω

σ

TsKWsFsGsKWsFsG T

)(arg)()(arge)()()(

We have root locus conditions

++±==−

TkW(s)KW(s) T

ωπ

σ

)12(arg1e

= ± ° + + °×180 2 1 57 3( ) .k Tω k = 0 1 2, , L (1) Starting points ( )K = 0

lim ( ) limK

T

KW s e

K→

−

→= = ∞

0 0

1σ

Root Locus

55

★All poles of W s( ) and σ → −∞

(2) Terminating points ( )K → ∞

lim ( ) limK

T

KW s e

K→∞

−

→∞= =σ 1

0

★All zeros of W s( ) and σ → +∞

(3) Asymptotes ★s → ∞ when K → 0 and K → ∞

arg ( ) argW sss n m→∞ −=

1

= − −( )arctann mωσ

lσ → +∞

arctanωσ

→ 0 σ

jω

arg ( ) ( )W s k T= ± + + =π ω2 1 0

ωπ

=± +

=± ° +( ) ( )

.2 1 180 2 1

57 3k

TkT

k = 0 1 2, , ,L

Root Locus

56

lσ → −∞

arctanωσ

π→ σ

jω

arg ( ) ( ) ( )W s k T n m= ± + + = − −π ω π2 1

ω π= −− ± +( ) ( )n m k

T2 1

¬If n m− is odd

ωπ

=2kT

, k = ± ± ±0 1 2, , ,L

¬If n m− is even

ωπ

=+(2 )kT

1, k = ± ± ±0 1 2, , ,L

★All asymptotes are parallel to the real axis (4) Number of branches: infinite (5) Real axis root locus ★Same as that of G s F s KW s( ) ( ) ( )= since ω = 0

Root Locus

57

(6) Breakaway points

ddKs

= 0, dd

es W s

Ts−

=

( )0

Example 4.8 Sketch the root locus plot for

G sKs

Ts( ) =

+

−e1

, F s( ) = 1, T=1

Solution:

(i) Write W ss

( ) =+1

1, K s Ts= − +( )1 e

(ii) Asymptotes: σ → +∞ : ω π π π= ± ± ±, , ,3 5 L σ → −∞ : n-m=1 is odd ω π π π= ± ± ±0 2 4 6, , , ,L (iii) Real axis locus: ( , )−∞ −1

Root Locus

58

(iv) Breakaway point

[ ]dd

e e es

s T sTs Ts Ts− + = − + −( ) ( )1 1

= eTs Ts T( )− − − =1 0

sT

T= −

+= −

12

σ

jω

ω ==

2 032 26

.

.K

ω ==

7 9838 04

.

.K

24.1421.14

==

Kω

− 1

− 2jπ

j3π

j5π

o ★Delay causes instability. (See Appendix VI for detailed calculations)

Root Locus

59

4.6 Root Contour Characteristic equation is P s K Q s K Q s( ) ( ) ( )+ + =1 1 2 2 0

where P s( ),Q s1( ),Q s2( ) are polynomials.

(1) Let K2 0= , then P s K Q s( ) ( )+ =1 1 0

1 01 1+ =K Q s

P s( )

( )

corresponding OL TF

G s F sK Q s

P s1 11 1( ) ( )

( )( )

=

★The roots of P s K Q s( ) ( )+ =1 1 0 are on

the root locus for K1 0= → ∞

(2) Write the char eqn as

1 02 2

1 1+

+=

K Q sP s K Q s

( )( ) ( )

Root Locus

60

corresponding OL TF

G s F sK Q s

P s K Q s2 22 2

1 1( ) ( )

( )( ) ( )

=+

★OL poles of G s F s2 2( ) ( ) are

on the root locus of G s F s1 1( ) ( )

★Root locus of G s F s2 2( ) ( ) starts from

the root locus of G s F s1 1( ) ( ) for

a given K1

★ A set of root loci is formed as K1 varies Example 4.9 Given

G sK

s s( )

( )=

+ a, F s( ) = 1

Sketch the root contour for varying K and a. Solution: CL char eqn:

s s K2 0+ + =a (i) Let a = 0, then s K2 0+ =

Root Locus

61

Sketch root locus of 211 )()(sK

sFsG =

Root Locus

62

(ii) Set K = const , we have

1 02++

=as

s K

Sketch the root locus of G s F ss

s K( ) ( ) =

+a

2

OL poles: ± j K

o ★Root locus for a varying parameter

other than gain can be constructed in the similar way

Root Locus

63

4.7 Roots of a Polynomial

Example 4.10 To find approximately the roots of the polynomial

162421103)( 234 −+++= sssssf

Solution: (i) Rewrite 0)( =sf as

021103

16241 234 =

++−

+sss

s

so 23411

21103

1624)()(

sss

ssFsG

++

−=

)21103(

)23(822 ++

−=sss

s

++

−

=7

310

32

22 sss

sK

(ii) Sketch root locus for )()( 11 sFsG :

OL poles: 0, 0, 055.2j667.1 ±−

OL zeros: 32

Asymptotes: °−°°= 60,180,60γ ,

− = −σa 4 3

Root Locus

64

Imaginary axis crossing point: ± j3037.

(We may find by trial and error s1 0 4482= . , s2 2 2627= − . ,

and s3 4 0 7459 21638, . .= − ± j )

(iii) Rewrite 0)( =sf as

121 24 16

3 100

2

4 3++ −

+=

s ss s

so 34

2

22103

162421)()(

ss

sssFsG

+

−+=

Root Locus

65

+

−+

=

310

2116

78

7

3

2

ss

ss

(iv) Sketch root locus for )()( 22 sFsG

OL poles: 0, 0, 0, − 10 3

OL zeros: 0 4719. , − 16147.

− = −σa 219.

★Crossing points of the two root locus plots give the roots of the polynomial. o