ROOT LOCUSKHALID SAEED AL BADRI

A system to automatically track a subject in a visual image can be modeled as follows

A root locus plot is simply a plot of the s zero values and the s poles on a graph with real and imaginary coordinates.

• This method is very powerful graphical technique for investigating the effects of the variation of a system parameter on the locations of the closed loop poles.

The root locus of an (open-loop) transfer function G(s) is a plot of the locations of all possible closed loop poles and zeros.

Poles and zeros

)())((

)())(()(

21

21

n

m

pspsps

zszszsksF

n

m

ppp

zzz

,,

,,

21

21 zerospoles

axisRe

axisIm

pole

zero

Root Locus Sketching RulesRule 1:Find the number

of poles Np and zeros Nz

Rule 2:Root locus is symmetric about the real axis.

Rule 3: Along the real axis, the root locus includes all segments that are to the left of an odd number of poles and zeros.

Root Locus Sketching RulesRule4: find the asymptotic location at 0 and angles k.

Example 1Sketch the root locus of the closed-loop system.

qEi

Plant G(s)Controller

qDV 16

0 0174 1s s( . )KP0.03

qV

qD

0.03

1 0

0.48

1 00.0174 1

p

N s

p

D s

K G s H s

Ks s

Example 1Step 1:Transform the closed-loop

characteristic equation into the standard forms:

Step 2:Find the open-loop zeros, zi , and the open-loop poles, pi :

1

1 27.58 057.47N s

p

D s

Ks s

No open-loop zeros

open-loop poles 47.57,0 21 pp

K

Example 1Step 3:Determine the real axis segments

that are to be included in the root locus.

1 0p 2 57.47p

Example 1Step 4: Determine the asymptotes s0 and

angles qk .

(2 1) [rad]kP Z

kN N

2

2

0i i

P Z

p z

N N

57.47

28.742

Example 1Step 5: Determine the break-away and break-in

points .

( ) ( )0 or 0,

( ) ( )

d N s d D s

ds D s ds N s

0.0174 10,0.0348 1 0, 28.74

0.48

s sds s

ds

Example 1

-60 -50 -40 -30 -20 -10 0Real Axis

-30

-20

-10

0

10

20

30

Img. Axis

-57.47-28.74

Example 2A positioning feedback control system is proposed. The corresponding block diagram is:

Sketch the root locus.

Y(s)U(s)

Plant G(s)Controller

R(s) 16

0 0174 1s s( . )K(s + 80)

+

Example 2

122

-140 -120 -100 -80 -60 -40 -20 0

-40

-30

-20

-10

0

10

20

30

40

Real Axis

Imag Axis

1 0p 2 57.47p 1 80z 37.6

Example 3A feedback control system is proposed. The corresponding block diagram is:

Sketch the root locus.

Y(s)U(s)

Plant G(s)Controller

R(s) 1

4 202s s s( ) +

K

s( ) 4

2

1 0

11 0

4 4 20

cG s G s H s

K

s s s s

Find closed-loop characteristic equation:

Example 3

Step 1:

Step 2:

Step 3:

2

1

1 04 20 4

N s

D s

Ks s s s

1 2 3,40, 4, 2 4p p p j open-loop zeros

open-loop poles

No open-loop zeros

1 0p 2 4p

Example 3Step 4:

Step 5:

(2 1) [rad]kP Z

kN N

43

45

47

4

0i i

P Z

p z

N N

0 4 2 4 2 4

24 0

j j

( ) ( )0 or 0,

( ) ( )

d N s d D s

ds D s ds N s

2

4 3 2

3 2

8 36 801

4 24 72 80

4 20 4

0

D sd d ds s s s

ds N s ds d

s s s s

s

s s s

1 2,32, 2 2.45s s j

Example 3Step 6:Determine the imaginary axis crossings

2

11 0

4 20 4Ks s s s

2

4 3 2

4 20 4 0

8 36 80 0

s s s s K

s s s s K

s j

4 2 336 8 80 0K j

4 221

31 2

260036 0,

08 80 0 10 3.16

KKK

Example 3

-6 -5 -4 -3 -2 -1 0

-4

-3

-2

-1

0

1

2

3

4

Real Axis

Imag Axis

Example 1/ MATLAB

1

1 27.58 057.47N s

p

D s

Ks s

num=[0 1];den=[1 57.47 0];rlocus(num,den)

Example 2/ MATLAB

num=[1 80];den=[1 57.47 0];rlocus(num,den)

16 80 80

1 1 920 00.0174 1 57.47

N s N s

D s D s

s s

K Ks s s s

Example 3/ MATLAB

num=[0 1];den=[1 8 36 80 0];rlocus(num,den)

2

11 0

4 20 4Ks s s s

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