Chapter 7

ROOT LOCUS

R + K GC

H

-

E

C

R

K G s

K G s H s====

++++

( )

( ) ( )1

1+ K G(s) H(s) = 0 is called the

characteristic equation of the system

and its roots are closed loop poles.

Root locus is the locus of the roots of

the characteristic equation as

parameter K is varied ( 0 < K < ∞∞∞∞)

+K

C

-

1

1s ++++

First order System

R

Closed loop Pole, s = - 1 - K

K ←←←←∞∞∞∞ K = 0

××××- 1

Root locus for G(s)H(s) = 1/(s+1)

�closed loop system is stable for all K

�As K is increased, closed loop pole is

moved farther away from imaginary axis;

hence the time constant reduces. i.e.

Transient response improves.

K ←←←←∞∞∞∞

K = 0××××1

s = 1 - K

K = 1

Root locus forG(s)H(s) = 1/(s-1)

From the plot ,

for K < 1, system is unstable

K G H = K

s - 1Ex:

+K

-)1+s(s

1

Second Order

CR

Ch Eq: s2 + s + K = 0

closed loop poles at2

41±-0.5

K -

For K = 0 : Roots at 0 and - 1

For 0 < K < 0.25 :

Roots are real and equidistant

from -0.5.

For K = 0.25 : Roots at -0.5

For K > 0.25: Roots are complex

conjugate with real part at - 0.5

0.5××××0××××

-1

>

>

>

>

Root locus forG(s)H(s) = 1/s(s+1)

• locus is always symmetric about the real axis

• closed loop system is stable for all K

• system under damped for K > 0.25

• damping factor reduces as K increases

+ K G

H

-

Ch Eq: 1 + K G H = 0

R C

For systems of order 3 & above, it is difficult

to solve for roots and hence we will work

out short cuts to plot the root locus.

K G H = - 1=

Magnitude of K G H = 1

→Magnitude Criterion

1∠180°

Phase angle of K G H

= Phase angle of G H, as ∠K = 0

= (2n+1) 180° where n = 0,1,2,….

→Angle Criterion

&

The method :

On the s - plane, check for the angle

of G H at a chosen (Test) point….

If the angle criterion is satisfied,

the point is on the locus.

Substitute in magnitude criterion to find K, at that point.

°=°°

=

180- )45-135 -∠(

pointat test (s) F∠

××××

.

-1××××0

j1

-2

Point is on locus

K GH = 1= K/s(s+2)

K/s(s+2) =1 at s= -1+j1

K =2

Ex:2) + s s(

K = F(s)

is -1 + j 1 lie on locus ? If yes, K=?

Ch. eq:1 + K G H = 0

0)...p+)(s p+ (s

)... z + (s ) z + (sK + 1

2 1

21 =or

or (s +p1)(s +p2)…+ K (s + z1)(s + z2)… = 0

for K = 0, s = - p1, - p2, ...

K = 0, open loop poles are closed loop poles.

i.e. From every open loop pole, a branch of the root locus STARTS .

0)...p+)(s p+ (s

)... z + (s ) z + (sK + 1

2 1

21 =

(s +p1)(s +p2)…+ K (s + z1)(s + z2)…= 0

or (1/K)(s +p1)(s +p2)…+ (s +z1)(s +z2)..= 0

for K = ∞, s = - z1,- z2, ...

Open loop zeroes are closed loop poles

i.e. Each branch of the root locus ENDSat a open loop zero

Ex:1) + s (

1 G(s)H(s)=

s = - 1, is the open loop pole

Hence the root locus STARTS at s = - 1

There is no open loop zero at a finite point. However, the function goes to zero as ∞∞∞∞→→→→s

========1) + s (

1G(s)

s1+1

s1

0 = G(s) , s As ∞→

i.e. The open loop zero is at infinity.

. .××××- 1T2 T1

At T1

angle = 0°°°°, T1

is not on locus

At T2

angle = 180°°°°, T2

is on locus

K ←←←←∞∞∞∞ K = 0××××

- 1

Ex

××××

.T2

×××××××××××× . .T3T4

Example

- 2 0

T1 .

T1 - not on locusT2 - on locusT3 - on locusT4 - not on locus.

∴Locus on real axis

is from 0 to - 2

Rule : Real axis locus

Point on real axis is part of locusif Number of open loop zeroes & poles to

right of test point are ODD.

Example :2) 1 +s (

1 =H(s) )s(G

××××K= 0

-1××××

No locus on real axis

××××

T

-1××××

.

θθθθ

At any point ‘T’,Angle = - 2 θθθθ

= - ( 2n + 1 ) 180 °°°°∴∴∴∴ θθθθ = 90 °°°° , 270 °°°°

××××××××-1

Root locus For 2) 1 +s (

1 = )s(H)s(G

××××

××××-1

j 1

-j 1

Root locus for ?

××××××××-1

××××

Example :3) 1 +s (

1 =H(s) )s(G

real axis locus from - 1 to ∞∞∞∞

.

××××

T

××××θ

××××

At any point T, -3θθθθ = - ( 2n + 1 ) 180 °°°°∴∴∴∴ θθθθ = 60 °°°° , 180 °°°° 300°°°°

××××××××××××3) 1 +s (

1 = )s(H)s(G

Root locus

for

Ex.2) (s 1) (s s

1 GH

++++++++====

1. Locus starts at s = 0, -1, -2, 2

2. All three branches go to infinity

3. Real axis locus from 0 to -1 & -2 to - ∞∞∞∞4. Break away point between 0 and -1

×××××××× ××××-2 -1 0

> <<

break away point, dK / ds = 0

3 s2 + 6s + 2 = 0

31 1 - s ±±±±====

= - 0 .42 or - 1. 58

s = - 0 .42 is the break away point

5. Imaginary axis crossing: Ch. Equation :

s3 + 3s + 2s + K = 0

s3

s2

s1

s0

1 2

3

K - 6

K

3K

Closed loop system is stable for 0 < K < 6

K = 6, A(s) = 3s2 + 6 = 0 , 2 j s ±±±±====

The locus crosses imaginary axes at

2 j s ±±±±==== with K = 6

Asymptotes :

G H = 1

( s + ) A

n - mσσσσ

Consider GH with n poles & m zeroes m n ≥≥≥≥

Suppose all poles are at point A - = s σ

i.e. (n - m) branches go to infinity at angle

.... ) m -n (

) 1 + 2q ( 180 =

°±θ

When n poles and m zeroes are distributed ,

G H

i

n=

s + z )

( s + p )

i=1

m

i

= 1i

∏∏∏∏

∏∏∏∏

(

=

( s + p ) ( s + z )= 1

i= 1

i

1

i

n

i

m

∏∏∏∏ ∏∏∏∏

... + s )z -p ( + s

1 =

1 - m -n

ii

m -n ΣΣ

Considering very large values for s, we take first 2 terms

When s takes very large values , all poles and zeroes may be considered

to be at the same point.

) + s (

1

) p + s (

) z + s (

= H m -n

Ai

1 =

i

m

1=i

σ=∴

∏

∏n

i

G

1 - m -n

ii

m -n m -n

A s )z -p ( + s = ) + s ( ΣΣσ∴

Or

) m-n (

z -p = - ii

A

ΣΣ−σ∴

-σσσσ A is known as centroid

zeroes.) of no. - poles of (no.

zeroes) ofpart (real - poles) ofpart real (

ΣΣ=

×××× ××××<××××-2 0

<-1

>

2j

asymptotes

While two branches cross Imaginary axis and go

to ∞∞∞∞, the third branch from - 2 goes to - ∞∞∞∞

The third order system can be approximated as a

second order system with dominant pole pairs

located on the complex conjugate parts the third

closed loop pole is on the negative real axis far

away from imaginary axis.

P1 - P2 dominant pole pair

×××× ××××< ×××× <>

..

P1

.P2

P3

°°°°====

====

45

0.707,

θθθθ

ζζζζFor

(θθθθ

K = 0

××××- 1

××××

××××

××××- 2

K = 0

K = 0

K = 0

0

- j1

j1

Example:

) 2 2s 2s ( 2) s ( s

1 HG

++++++++++++====

• 4 branches go to infinity

• Real axis locus from 0 to -2

°±°±=+°±

=θ 135 ,45 4

)1n2(180

A

1- 4

1 - 1 - 2 - 0 = A

=σ−

1- = s 0, = 31)+(s 0 = 1+3s+23s+3sor 0= 4+12s+212s+34s

0 = ds/dK →

j1 = s,5K ±=

Imaginary axis crossing :

Break Away Pt

Angle of departure from a complex conjugate pole :

××××

××××××××××××

P θθθθ

Let P be the point in the neighbourhood of

s = - 1 +j 1(open loop pole)

°θ

°−=−−−θ−

90 - =

180 45 135 90

××××

××××

××××××××-2 0

-j1

j1

> <

<

<<

<

<

<

Root Contours

Locus of closed loop

poles as any other

parameter also is

varied.

‘K‘ and ‘a’ are varied

+K

- a)s(s

1

+

Example :

Characteristic Equation :

s2 + a s + K = 0 ⇒⇒⇒⇒ ( s2 + K )+a s = 0

0 K s

s . a 1

2=

++

This equation is similar to 1+KGH = 0

With ‘a’ varying, we plot the locus with

open loop zero at s = 0 and open loop

poles at K j s ±=

××××

××××a = 0

∞∞∞∞====a

0

a = 0

.

a ←∞<

>

< >

××××

××××K = 1

0

j 2

..

<

>

< >

××××

××××<

>

>

j 1K = 4

.

.

.

.

Root Contours

<