Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
1
ROOT-LOCUS ANALYSIS
Lecture 11: Root Locus Plot
Consider a general feedback control system with a variable gain K.
Root-Locus is a plot of the loci of the poles of the closed-loop transfer
function when one of the system’s parameters (K) is varied. Root locus
allows us to visualize the changes in the closed-loop poles as the
parameter K is increase from 0 to infinity.
K G(s)Y(s)
H(s)
R(s)−
+
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
2
A distinction is usually made between the following categories of
root-locus:
(a) Root-Locus Plot:
The plot of the root loci when one parameter varies in positive values;
this parameter is usually the forward gain K,
i.e. 0 ≤ K < ∞
(b) Complementary Root-Locus Plot:
The plot is obtained for negative values of K,
i.e. −∞ < K ≤ 0
(c) Root Contours Plot:
The plot is obtained when more than one parameter varies.
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
3
Example: Consider the system.
Suppose that , find the roots of the C.E. for
0 ≤ K < ∞ and plot these roots in the s-plane.
1( )
( 2 )G s
s s=
+
2
( ) ( )
( ) 1 ( ) 2
Y s K G s K
R s K G s s s K= =
+ + +
The C.E. is s2 + 2s + K = 0. The roots of the C.E. are:
1 2
1 1 for 1 1 1 for 1;
1 1 for 1 1 1 for 1
K K K Ks s
j K K j K K
− + − ≤ − − − ≤ = =
− + − > − − − >
KY(s)R(s)
−+ G(s)
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
4
The effect of K on the roots s1 and s2 is shown by the table below:
K s1 s2 Remark
0
0.5
1.0
2.0
3.0
4.0
:
:
→∞
0
−0.293
−1.0
−1 + j1
−1 + j1.414
−1 + j1.732
:
:
- 1 + j∞
−2.0
−1.707
−1.0
−1 – j1
−1 – j1.414
−1 – j1.732
:
:
−1 − j∞
breakaway
point
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
5
The root-locus is plotted as shown:
K = 0
jωK→∞
j2
j1
K = 0
σ
−j1
−j2
K = 2
+∞↑
K = 3
K = 1
−2−3 −1 0
breakaway
point
s-plane
K = 2
K = 3
↓−∞
K→∞OLTF poles
→ root-locus as K is increased
We shall develop a procedure to sketch the root-locus without
computing the roots of the C.E. point by point.
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
6
( 3)1 ( ) 1 0
( 2)p
K sG s
s s
++ = + =
+1 ( ) 1 0
( 2)( 3)p
KG s
s s s+ = + =
+ +The root loci plots of and
are given below. Take note of the asymptotes and break-away (in) points.jω
K→∞
σ−2−3 0
OLTF poles
OLTF zero
break-in
point
K→∞
break-away
pointjω
K→∞
σ−2−3 0
s-plane
K→∞
K→∞
break-away
point
centroid of
asymptotes
asymptotes
root loci
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
7
Mathematical Definition of Root-Locus
Consider the closed-loop transfer function
The Characteristic Equation (C.E.) is
1 + KG(s)H(s) = 0
i.e. KG(s)H(s) = −1
Root-locus is defined by the conditions:
(I) Magnitude Condition:
(II) Angle Condition:
where n = 0, 1, 2, …… (an integer)
( ) ( )
( ) 1 ( ) ( )
Y s K G s
R s K G s H s=
+
( ) ( ) 1 K G s H s =
( ) ( ) (2 1)KG s H s n π∠ =± +
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
8
The open-loop transfer function may be expressed as
where -zi’s are the zeros of OLTF
-pi’s are the poles of OLTF
Hence, the C.E. is given by
When K = 0, the roots of the C.E. are:
−p1, −p2, ……, −pn
When K → ∞, the roots of the C.E. are:
−z1, −z2, ……, −zm
In other words, as K varies from 0 to ∞, the roots of the C.E.
traverse from the poles of OLTF to the zeros of OLTF.
1 2
1 2
( )( ) ( )( ) ( )
( )( ) ( )
m
n
K s z s z s zKG s H s
s p s p s p
+ + ⋅⋅⋅ +=
+ + ⋅⋅⋅ +
0)())((
)())((
21
21
=+⋅⋅⋅++
++⋅⋅⋅++
m
n
zszszsK
pspsps
1+ ( ) ( ) 0 KG s H s =
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
9
1 2
1 2
( )( ) ( )( ) ( )
( )( ) ( )
m
n
K s z s z s zKG s H s
s p s p s p
+ + ⋅⋅⋅ +=
+ + ⋅⋅⋅ +
1
1
1( ) ( )
m
i
i
n
i
i
s z
G s H sK
s p
=
=
+= =
+
∏
∏
1 1
( ) ( ) ( ) ( )
(2 1) ; is an integer (2)
m n
i ii i
G s H s s z s p
j jπ= =
∠ = ∠ + − ∠ +
= ± +
∑ ∑
With the OLTF KG(s)H(s) given by
Magnitude Condition:
Angle condition:
Equations (1) and (2) define the root-locus. The root-locus of a system
is a plot of all the values of s which satisfy eqns (1) and (2).
(1)
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
10
Thus, given the pole-zero plot of G(s)H(s), the root-locus plot of the
closed-loop system involves:
Example: Consider 1
2 3
( )( ) ( ) ; 0
( )( )
K s zKG s H s K
s s p s p
+= ≥
+ +
s1
jω
B
C
θp2
−p2
θp1
σ−p1
A
θz1
−z1 D
θp3−p3
s1 is an arbitrary
point
open loop poles
open loop zeros
The poles and zero are
shown in the diagram, and
we want to check whether
s1 is on the root locus.
(a) A search for points in the s-plane which satisfy equation (2).
(b) The determination of the values of K on the root-locus using
equation (1).
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
11
Kpspss
zs 1
31211
11 =++
+
( )1 1 1 1 2 1 3( ) ( ) ( )
(2 1)
s z s s p s p
j π
∠ + − ∠ + ∠ + + ∠ +
= ± +
1 1 2 3( ) (2 1)
z p p pjθ θ θ θ π⇒ − + + = ± +
A
DCBK
⋅⋅=
If s1 is a point on the Root-Locus, it must satisfy the following two
conditions:
Magnitude Condition:
Angle Condition:
If s1 satisfies equation (4), then the value of K at that point is obtained
from eqn (3), i.e.
Next, repeat for new value of s1 and so on.
(3)
(4)
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
12
Rules for Constructing the Root-Locus
The OLTF is given by
The closed-loop C.E. is
i.e. n = number of finite poles of G(s)H(s):
m = number of finite zeros of G(s)H(s):
Rule 1 : K = 0, points on the root-locus are at the finite poles of G(s)H(s).
Rule 2 : K → +∞, points on the root-locus are at the finite zeros of
G(s)H(s) and infinity.
Rule 3 : The number of branches of the root loci is equal to n.
Rule 4 : The root loci are symmetrical with respect to the real-axis of the
s-plane.
1 2
1 2
( )( ) ( )( ) ( ) ; ( )
( )( ) ( )
m
n
K s z s z s zKG s H s n m
s p s p s p
+ + ⋅⋅⋅ += ≥
+ + ⋅⋅⋅ +
1 2
1 2
( )( ) ( )
( )( ) ( ) 0
n
m
s p s p s p
K s z s z s z
+ + ⋅⋅⋅ + +
+ + ⋅⋅⋅ + =
; 1,2,i
z i m− = …
; 1,2,ip i n− = …
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
13
(2 1) ; 0,1, ,( 1)
j
jj n m
n m
πθ
+= = − −
−…
(finite_poles_of_ ( ) ( )) (finite_zeros_of_ ( ) ( ))
( )
G s H s G s H s
cn m
σ∑ ∑−
=−
1 1i.e.
( ) ( )
( )
n m
i ii i
c
p z
n mσ = ==
− − −
−
∑ ∑
(The explanations for the following rules 5-8, and 10 are given in the
Appendix.)
Rule 5: There are |n – m| asymptotes. As |s| → ∞ the root loci are
asymptotic to straight lines (asymptotes) with angle with the real-axis
given by
Rule 6: The |n – m| asymptotes intersect on the real axis at the point
given by:
Rule 7: Root loci are found on a section of the real axis only if the
total number of real poles and zeros of G(s)H(s) to the right of the
section is ODD.
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
14
Rule 8:
(a) The angle of departure from complex poles are given by
θd = 180° + θ′
where θ′ is the angle of G(s)H(s) at that pole with the angle contribution
from the pole itself ignored.
(b) The angles of arrival at complex zeros are given by:
θa = 180° − θ″
where θ″ is the angle of G(s)H(s) at that zero with the angle contribution
from the zero itself ignored.
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
15
Rule 9: The points where the root loci intersect the imaginary axis are
obtained using the Routh-Hurwitz criterion.
Rule 10: The breakaway/breakin points are points where two or more
branches of the root locus depart from or arrive at a main branch.
The angle of departure/arrival of breakaway/breakin point is given by
(2 1)180 ; 0,1, , 1ba
jj k
kθ
+ °= = −⋯
where k = number of loci leaving or approaching the point.
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
16
( ) 1 ( ) ( ) 1 0
( )
B sKG s H s K
A s+ = + =
)(
)(
sB
sAK −=
)(
)()()()(2 sB
sBsAsAsB
ds
dK ′−′−=
0=ds
dK
0)()()()( =′−′ sBsAsBsA
Suppose that the C.E. is expressed as
Then,
(6)
The breakaway/breakin points are given by
i.e. the breakaway/breakin point are given by the roots of
NB: NOT all roots of equation (8) correspond to the actual
breakaway/breakin point.
(5)
(8)
(7)
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
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( )
( )
AK
B
σσ
= −
σ
K
σ2 σ1
With condition (7), we can use eqn. (6) to derive a numerical method to
obtain the breakaway/breakin points. For breakaway(in) points on the
real axis, s = σ, so we have
We plot a graph of K for various values of σ between the selected points
where the breakaway(in) point is expected.
Then σ at which K is maximum (or minimum) is the breakaway (or
breakin) point.
σ1 is the breakaway point.
σ2 is the breakin point.
(9)
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
18
1
1 1
1
1
1
1
1
( ) ( )
n
j
j
m
j
j
KG s H s
s p
s z
=
=
=
+
=+
∏
∏
1
product_of_vector_lengths_from_OLTF_poles
product_of_vector_lengths_from_OLTF_zerosK =
Rule 11: The value of K at a point s1 on the root locus is obtained by
applying the Magnitude Condition and is obtained from
i.e.
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
19
Appendix: Explanations of rules 5-8, and 10.
Rule 5: As K → +∞, |n – m| loci will approach infinity.
i.e. There are |n – m| asymptotes.
If |s1| → ∞⇒ θj ≈ θ1 ≈ θ2 ≈ θ3 ≈ θ4 ≈ θ5 ≈ θ6
By the Angle Condition:
1 1( ) ( ) ( ) (2 1)j
G s H s n m jθ π∠ = − ≡ +(2 1)
; 0,1, , ( 1)j
jj n m
n m
πθ
+∴ = = − −
−…
σ
jω
s1
∞
θj
θ1
0
θ2
θ3θ4θ5θ6
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
20
1
1 1
1
1 1
1 2
1 2
( )( ) ( )( ) ( )
( )( ) ( )
mmm m
i ii i
nnn n
j jj j
K s z s z
m
s p s pn
K s z s z s zKG s H s
s p s p s p
−
= −
−
= −
+ + +
+ + +
∑ ∏
∑ ∏
+ + ⋅⋅⋅ += =
+ + ⋅⋅⋅ +
⋯⋯
⋯⋯
1
1 1
( ) ( )n m
n m n m
j i
j i
KKG s H s
s p z s− − −
= =
≈
+ − ∑ ∑
( )
1
( )
( )
n m
c
n m n m
c
KKP s
s
K
s n m s
σ
σ
−
− − −
=−
=− − +⋯⋯
Rule 6: From the OLTF
Divide the denominator and the numerator by the numerator term,
Consider the following function
(A2)
(A1)
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
21
)1(,,1,0;)12(
−−=−
+= mnj
mn
jj …
πθ
4n m− =
(2 1); 0,1, ,( 1).
j
jj n m
n m
πθ
+= = − −
−…
jω
σσc
(4 poles here)
45o
45o
The C.E. 1 + KP(s) = 0 has (n – m) root locus branches which are
straight lines passing through the point s = σc and having angles,
The case of is illustrated below. It can be readily verified that
every point on the 4 straight lines satisfies the angle condition.
That is, the root locus of 1 + KP(s) = 0 is a
set of (n – m) lines drawn from s = σc at
angles
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
22
• The 1+KG(s)H(s) behaves in the same manner as 1+KP(s) for values
of s → ∞, because the first two higher order terms of their
denominators are identical.
• 1+KG(s)H(s) approaches 1+KP(s) for values of s → ∞. Therefore,
the branches of 1 + KG(s)H(s) = 0 which tend to infinity approach the
straight line root locus branches of 1 + KP(s) = 0.
• The straight line root loci of 1 + KP(s) = 0 act as asymptotes to the (n
– m) branches of 1+KG(s)H(s) = 0.
• From eqns (A1) and (A2)
1 1
( )n m
c j i
j i
n m p zσ= =
− − = −∑ ∑
1 1
( ) ( )
( )
n m
j i
j i
c
p z
n mσ = =
− − −
=−
∑ ∑i.e.
- the centroid of the asymptotes.
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
23
Rule 7: Consider the following case:jω
∠(so + p1)
σ− p1− p2 − z1
∠(so + z1)∠( so + p2)∠( so + z2)
-θ
so
θ
− p3 − z2
∠(so + p3)
mR : Number of zeros on the right of so.
nR : Number of poles on the right of so.
• The poles and zeros on the real axis to the right of this point so
contribute an angle of 180° each.
• The poles and zeros to the left of this point so contribute an angle of
0° each.
• The net angle contribution of a complex conjugate pole or zero pair
is always zero.
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
24
( ) ( ) ( )180 (2 1)180R RG s H s m n j∠ = − ° = ± + °
R Rm n− R Rm n+
jω
σ
p
θ1
so
θ3
θ2
θp → θd (so → p)
θ4θ5
So,
i.e. (and hence
Rule 8: Consider the following case:
) must be an odd number.
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
25
• At point so, the net angle contribution of all other poles and zeros at
this point is:
( ) 53214 θθθθθθ +++−=′
• In the limit as the point so on the root locus approaches p, θP equals
the angle of departure of the root locus from the pole p, (θP → θd).
From the Angle Condition:
θ′ − θd = −(2j + 1)180° = −180°or θd = 180° + θ′.
Similar explanation applies to angle of arrival.
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
26
[ ] ' 1
1 1
1 '
1 1
1 ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
r r
r
dKG s H s s b A s A s r s b
ds
s b rA s s b A s
−
−
+ = + + +
= + + +
[ ]1 ( ) ( ) 0 d
KG s H sds
+ =
Rule 10: The Characteristic Equation is
1 + KG(s)H(s) = 0
If C.E. has a multiple root at s = −b of multiplicity r, i.e.
1 + KG(s)H(s) = (s + b)r A1(s)
where A1(s) does not contain the factor (s + b), then
At s = −b, RHS = 0 (only if r ≥ 2)
(A3)⇒
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
27
( ) 1 ( ) ( ) 1 0
( )
B sKG s H s K
A s+ = + =
[ ] 2
( ) ( ) ( ) ( )1 ( ) ( )
( )
d A s B s A s B sKG s H s K
ds A s
′ ′−+ =
0)()()()( =′−′ sBsAsBsA
)(
)(
sB
sAK −=
The C.E. can be expressed as
So,
From eqn (A3), the roots are the breakaway points, therefore the
breakaway points are also given by the roots of
From eqn (A4),
(A6)
(A4)
(A5)
Lecture 11: Root Locus Plot
Dr. Kalyana Veluvolu
28
)(
)()()()(2
sB
sBsAsAsB
ds
dK ′−′−=
)(
)()()()(2 sB
sAsBsBsA ′−′=
0=ds
dK
Hence
With eqn (A5), we have
That is, the breakaway(in) points can also be computed from the roots
of (A7), or through a numerical method.
Warning: NOT all roots of equation (A5) (or (A7) correspond to the
actual breakaway point. (Verify with Angle Condition!!!)
(A7)
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