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Root Locus
1
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
Root Locus
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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
Root Locus
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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
Root Locus
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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
Root Locus
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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
Root Locus
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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:
Root Locus
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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
Root Locus
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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
Root Locus
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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.
Root Locus
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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
Root Locus
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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
Root Locus
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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°
Root Locus
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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
Root Locus
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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
Root Locus
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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)
Root Locus
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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
Root Locus
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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
Root Locus
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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
Root Locus
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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
Root Locus
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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
Root Locus
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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
Root Locus
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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
Root Locus
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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
Root Locus
28
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
Root Locus
29
(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.
.
Root Locus
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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. .
Root Locus
31
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
Root Locus
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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
Root Locus
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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.
Root Locus
34
(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
Root Locus
35
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
Root Locus
36
§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
Root Locus
37
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
Root Locus
38
(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
Root Locus
39
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, ,= − −
Root Locus
41
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.
Root Locus
42
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
Root Locus
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
Root Locus
44
(iii) Time response lUnit impulse response
★The zero causes the difference
between systems II and III (See Appendix V for impulse response equations)
Root Locus
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)
Root Locus
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
Root Locus
47
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
Root Locus
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
Root Locus
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
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
+
−+=