ACS Week733b

Lecture 11: Root Locus Plot

Dr. Kalyana Veluvolu

1

ROOT-LOCUS ANALYSIS


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)−

+



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.



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)



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



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.



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



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∠ =± +



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 =



9

1 2

1 2

( )( ) ( )( ) ( )

( )( ) ( )

m

n


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)



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).



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)



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− = …



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.



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.



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.



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)



17

( )

( )

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)



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.



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



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


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)



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



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.



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.



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.



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.



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)⇒



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)



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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