Modern Control SystemEKT 308

Root Locus Method (contd…)

Root Locus Procedure(contd…)

Root Locus ProcedureStep 1 (review):

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

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

Step 2 (review): 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.

Magnitude and Angle Criterion

0)(1 equation, sticcharacteri theSuppose,

sKP

(1) ---- 0)(

)(1

1

1

n

jj

M

ii

ps

zsK

1)(...)()()(....)()(

Magnitude,21

21

n

M

pspspszszszs

K

integer.an is where,

360.180)(-....)(-)(-

)(...)( )( Angle,

21

21

l

lpspsps

zszszsoo

n

M

Magnitude and Angle Criterion (contd…)

432143

21

4433

2211

r.rr.r

:refresh Background

rrrr

1. figurein shown are )1(s toingcorrespond Angles

11 2

4.1.K-42-,

areeqn sitccharacteri theof roots theSay,

)02 ,( 0)2(

1 equation, sticCharacteri

Example,

1

21

2

K

Kss

KssorssK

Figure 1: Angle for s = s1

Note: Because complex roots appear as complex conjugate pairs, root loci must be symmetrical with respect to horizontal real axis.

Step 3:The loci proceed to the zeros at infinity along asymptotes centered at

Mn

zp

MnsPofzerossPofPoles

n

j

M

iij

A

AA

1 1

)()(

)()( centroid, Asymptote

where, angle with and

)1,.....(2,1,0 ,180.12

,asymptotes of angle And

MnkMn

k oA

Where n, the order of numerator polynomial and M is the order of denominator Polynomial

Example for step 3.

2 figin shown are axis real on the lociroot and zeros and Poles

0)4)(2(

)1(11

equation, sticcharacteri heConsider t

2

ssssKGH(s)

Figure 2: Root loci on real axis

ly.respective 2 and 1, 0,for 300 ,180 ,60

60).12(180.12

are, asymptotes of angles The

.339

14)1()4(2)2(

),asymptotes ofon intersecti called (also centroid Asymptote

ooo

k

kMn

k ooA

A

Asymptotes are shown in Figure 3

Figure 3: Asymptotes

Step 4:Determine where the locus crosses the imaginary axis (if it does so), using Routh-Hurwitz criterion. Hint: When root locus crosses the imaginary axis from left to right, the system moves from stability to instability.

Example: Complete first four steps of sketching root locus of the characteristic equation

0)44)(44)(4(

1

jsjsss

K

Step 1: Poles and zeros are shown in figure 4.

Figure 4: Poles and zeros

Step 2: There is a segment of root locus on the real axis between s=0 to s=-4 as shown in figure 4 above.

34

444

centroid, Asymptotes.315 ,225 ,135 ,45

3. 2, 1, 0,k ,180.4

12

are, asymptotes of Angles :3 Step

A

oo0

oA

oA

k

The asymptotes are drawn in figure 5.

Figure 5: Asymptotes

Step 4.

Kc

Kb

K

ss

RouthKssss

11

12812641

sss

array, 01286412 equation, sticCharacteri

0

1

2

3

4

234

568.8912/)128(33.53)128(33.5312

012)128(33.53 ,33.53

12)128(33.531 ,33.531

KK

KSo

Kcb

above. 5 figurein shown as 266.3at axisimaginary thecrosses locusroot the568.89, when So,

)266.3)(266.3(33.53))266.3((33.53

)67.10(33.5389.56853.33s

by,given areequation aux theof Roots

222

22

jsK

jsjsjs

s