Sec 8b -Root Locus

8/12/2019 Sec 8b -Root Locus

1/19

E&CE 380 Root Locus Analysis

Root Locus and Lead Controllers Consider the previous example

)

2

1)(

2

1(

)

4()

(js

js

s

s

Ks

GH-

++

+

+=

X O

jw

sX

4

X

O2

2j

2j

+1

p2

p3

p1

z1

X

Angle of departure fromp2is

7.1

+ 90

tan

-1

(2/5) =75.3

The new asymptote intersects

the real axis at

[(0116)(41)]/(42) =1.5

Add a pole/zero combination

ats=6 ,1.


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jw

sXX

24

4j

4j

8

Root Locus Lead Design

Consider again the

system defined by

)2()(

+=

ssKsGH s

1Design Point

s-4

settling time 1.0 s

0.707

Control design

specifications:


3/19


jw

sXX

24

4j

4j

8


Can we force the root

locus to go through the

design point,s1?s

1

Design Point

Consider the effect of

adding apoleand zero

combination (a leadcompenstor).

X


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Three effects of the pole

zero combination:

modifies the real axis

segments

shifts the asymptote real

axis intersection to the

left

modifies the angle

criterion by

=z-p

jw

sXX

24

4j

4j

8

s1

Design Point

X

zp


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Design steps:

place the zero below the

design point,s1(approx.)

jw

sXX

24

4j

4j

p

s1

Design Point

X

p

locate the pole location

from the angle, pandthe location of the zero

determine the angle

difference,=90 -pthat will satisfy the angle

criterion for the design

point,s1 .


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calculate p

tan(18.4) =4 / (-4 -p)

p =-4/0.333 -4 =-16

X

jw

sXX

24

4j

4j

8

s1

Design Point

-135 -116.6 =180

=431.6 or 71.6

p=90 -71.6 =18.4135116.618.4

calculate the pole location

16


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Root Locus Lead Design The asymptote intersection:

K=5.56 4.47 12.65 / 4 = 78.6

Determine the gain at thedesign point:

X

jw

sXX

24

4j

4j

8

s1Design Point

16

sa=(-0 -2 -16) -(-4)

3 -1

sa = -7


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Root Locus Lead Design The final compensator is

The compensated open-

loops system is

16

46.78

+

+=

s

sGc

)2(

1

16

46.78

++

+=

sss

sGG pc

4.3146.11018

46.78

23 +++

+=

sss

sG

The final closed-loop

systems isNote: the lead compensator

=16/4 =4


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Time (sec.)

Amplitude

Closed-Loop System Step Responses

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0

0.5

1

1.5

)2(

1)(

sssGH

)2(20)(ss

sGH

)2(

1

)16(

)4(6.78)(

sss

ssGH


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Root Locus Lag Design

Pole/zero placement:

The lag compensator is

represented by a pole nearthe imaginary axis and a

zero further to the left.

jw

s

1X

pz

Lag compensator

ps

zs

+

+

||

||

p

z=

Low frequency gain:

The compensator lowfrequency gain is


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s1

jw

sX

1 pz

Lag compensator

ps

zs

+

+

2


Desired effects of the lag:

increase the low frequency gain

to achieve desired steady-stateerror specifications, the gain

increase is .

2)()( 11 +-+ pszs

Introduce very little effect onthe path of the root locus, ie.


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Relate the angle requirement

and low frequency gain: (note

2 = 0.035 rad.)By similar triangles

2

jw

sX

s1

s1

pz

jw1

L

1035.0

||||

w

L

L

pz

-

1

2035.0)1(||

w

L

p =-

)1(

)(035.0||

1

2

1

2

1

-

+=

w

wsp

then


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Design steps:

Determine a point,s1on the

uncompensated loci that

satisfies the dynamicrequirements.

Find the gain at s1and then

the low frequency gainof the

system.

Determine the low frequency

gain, of the compensator

required to meet the system

steady-state error requirement.

)1(

)(035.0||

1

2

1

2

1

-

+=

w

wsp

Calculate the compensator

pole location usings1 and

in the relationship

Finally, calculate the

compensator zero from

||

||

p

z=


14/19


60

Root Locus Lag Design Example

Consider again the

system defined by

)2()(

+=

ssKsGH

ess5% for a ramp

input.

0.50

Control design

specifications:

K= 1

jw

sXX

2

1.73j

s1

K = 2 2 /1 = 4

K= 4


15/19


Root Locus Lag Design Example

The velocity steady-

state error constant is

Foress0.05,Kv20

2)2(

4

lim0 =+= sssK sv

Therefore the compensatorlow-frequency gain must

be 10 .

The compensator pole

magnitude is

)1()(035.0||

2

-

+=

101.731.731p

= 0.0128

The compensator zero

magnitude is

|z| = |p| = 0.128


16/19


Root Locus Lag Design Example The final compensator is

The compensated open-

loops system is

0.0128

0.1280.4

+

+=

s

sGc

The final closed-loop

systems is

)2(

1

0128.0

128.0

0.4 ++

+=

sss

s

GG pc

512.0026.4013.2128.04 23 ++

+=sss

sG

The roots of the final

closed-loop systems ares= -0.9386 1.7000i

s= -0.1358


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-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Real Axis

ImagAxis

Root Locus of Final System

x

-0.25 -0.2 -0.15 -0.1 -0.05 0 0.05

-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

Real Axis

ImagAxis

x x


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Step Responses

Time (sec.)

Amplitude

Step Response

0 2 4 6 8 10 12 14 16 18 20

0

0.2

0.4

0.6

0.8

1

1.2

uncompensated

compensated


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Ramp Response

Time (sec.)

Amplitude

Ramp Response

0 1 2 3 4 5 6 7 8 9 10

0

1

2

3

4

5

6

7

8

9

10

uncompensated

compensatedunit ramp

Documents

Sec 8b -Root Locus