Modern Control SystemEKT 308

Aspects of State Equation &

Root Locus Method

Transfer Function from State Equation

.....(2) (1) ..... Given,

DuxCyBuxAx

....(4) )()()( (3) .... )()()(

condition) zero (initial Transform, Laplace Taking

sDUsXCsYsBUsXAsXs

1

1

)( ere, wh

)()()()(

)( )()()( (3), From

(sI-A)s

sBUssBU (sI-A)sX

sBU) (sI-A)X(ssBUsAXssX

Transfer Function from State Equation (contd…)

DBsCsUsYsG

sUDBsCsYsDUsBUsC

sDUsXCsY

)()()()(

function, transfer theTherefore,)()( )(

)()()( )()()(

(4),Equation From

)()()( function, transfer theFind

0] [5y 10

4310

equations,matrix thehas system SISOA :

sUsYsG

x

uxx

Example

Answer:

22

221

43433

431

434

431

)(

431

4310

1001

sss

ss

sssss

ss

s

ss

sAsI

345

10

43433

431

434

05

)()(

2

22

22

ss

sss

ss

sssss

DBsCsG

Concept of Root Locus

)()(

)()()(

sqsp

sRsYsT

Characteristic equation (above fig), KsKG 0,0)(1

01)(/)(1)(

jsKGsKGsKG

,.......2,1,0

360.180)(/,1)(

k

ksKGsKG oo

Fig 1

Concept of Root Locus (contd…)

The root locus is the path of the roots of the characteristic equation traced out in the s-plane as the system parameter varies from zero to infinity.

Root Locus ProcedureStep 1:

KsPKsF

0 where0)(1 as, arrange-reThen 0)(1equation, sticcharacteri theWrite

(1) ---- 0)(

)(1

follows, as zeros and poles of form in the writeRe

1

1

n

jj

M

ii

ps

zsK

Root Locus Procedure (contd…)

Locate poles and zeros in the s-plane (‘x’ for poles, ‘o’ for zeros

as, written becan (1) Eq

(2) ... 0)()(11

M

ii

n

jj zsKps

Let us vary K from 0 to infinity. When K = 0,

n

n

jj ppppsps ,......,,, i.e. ,0)( 321

1

So, when K=0, values of s coincide with poles of P(s).

0)()(

,When

11

M

ii

n

jjK

zsKpsLt

K

Note:

M

M

ii

M

ii

n

jjK

zzzzszs

zspsK

Ltei

,......,, giving ,0)(

)()(1..

3211

11

The locus of the roots of the characteristic equation 1+KP(s)=0 begins at the poles of P(s) and ends at the zeros of P(s) as K increases from 0 to infinity. If n > M, (n-M) branches of root locus approach the (n-M) zeros at infinity.

Step 2: Locate the segments of the real axis that are root loci. The root locus on the real axis lies in a segment of the real axis to the left of an odd number of poles and zeros.

Example fo step 1 an 2.

0

41

121

.1)(1

equation sticcharacteri following thehas which system afor locusroot Find

2

ss

sKsGH

2- : eros Z4- 0, :Poles

)4()2(2

4)2(2)(

0)(14

)2(21 eq, sticCharacteri :1 Step

:

2

2

sss

ssssP

sKPss

sK

Ans