• Author / Uploaded
• Tushar Gupta

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Chapter 7

ROOT LOCUS

R +

E

-

K

G

H C K G (s) = R 1 + K G (s)H (s)

C

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 < ∞)

First order System

R

+ -

K

1 s + 1

Closed loop Pole, s = - 1 - K

C

∞ ←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.

Ex:

K KGH = s-1

s = 1-K ∞ ←K

K=0 K=1

Root locus for G(s)H(s) = 1/(s-1) From the plot , for K < 1, system is unstable

×

1

Second Order

R

+ -

K

1 s (s + 1)

C

Ch Eq: s2 + s + K = 0 closed loop poles at

1- 4 K -0.5 ± 2

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

×>

-1

> 0.5

>

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

×

0

>

• • • •

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

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.

R +

-

K

G H

Ch Eq: 1 + K G H = 0

C

K G H = - 1= 1∠180° Magnitude of K G H = 1 →Magnitude Criterion & 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.

Ex:

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

K F(s)= s( s + 2)

. ×

j1

×

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

∠ F (s) at test point = ∠( - 135° - 45°) = - 180°

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

-2

-1

Point is on locus

0

K =2

Ch. eq:1 + K G H = 0 (s+ z 1) (s+ z 2)... or 1+ K =0 (s+ p 1)(s+ p 2 )... 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 .

(s + z 1) (s + z 2 )... 1+ K =0 (s + p 1)(s + p 2 )... (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 ENDS at a open loop zero

Ex:

1 G(s)H(s) = ( s + 1)

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 G(s) = = ( s + 1)

1 s 1+1 s

As s → ∞ , G(s) = 0 i.e. The open loop zero is at infinity.

Ex

. T2

.

×

T1

-1

At T1 angle = 0°°, T1 is not on locus At T2 angle = 180°°, T2 is on locus

∞←K

K=0

×

-1

Example

× T4 .

×

-2 T1 T2 T3 T4

T2 .

T.3

- not on locus - on locus - on locus - not on locus.

×

T. 1

× 0

∴Locus on real axis is from 0 to - 2

Rule : Real axis locus Point on real axis is part of locus if Number of open loop zeroes & poles to right of test point are ODD.

Example :

1 G(s) H(s) = ( s + 1 )2

K= 0

× × -1

No locus on real axis

At any point ‘T’, Angle = - 2 θ = - ( 2n + 1 ) 180 ° T ∴ θ = 90 ° , 270 °

.

× ×-1

θ

Root locus For

1 G ( s ) H (s ) = 2 ( s +1)

× × -1

Root locus for ?

×

j1

-1

×

-j 1

Example :

1 G(s) H(s) = 3 ( s +1)

× ×

×-1

real axis locus from - 1 to ∞

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

.T × × ×

θ

1 Root locus G (s ) H (s ) = for ( s + 1 )3

× × ×

Ex. 1. 2. 3. 4.

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

Locus starts at s = 0, -1, -2, 2 All three branches go to infinity Real axis locus from 0 to -1 & -2 to - ∞ Break away point between 0 and -1


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