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4 Root-Locus Method

4.1 Root-Locus Analysis

(cf. Chapter 6 of the textbook)

Feedback System:

+

KG(s)

R YE

where K >0 and

G(s) = b(s)

a(s)

=

mi=1(s zi)

n

i=1(s pi)

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Example 4.1 Find a graphical representation

of all closed-loop poles with respect toK when

G(s) = 1

s(s + 1)

Solution.As the closed-loop transfer function is

K

s2 + s + K

the closed-loop poles can be found to be

p1, p2= 12 1 4K

2

As K varies from 0 to, the two poles forma cross as shown in the following figure.

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Re(s)0

Im(s)

1

1

1

Definition:The graph of all possible closed-loop poles rel-

ative to some particular system parameter is

called the root locus.

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Angle and Magnitude Conditions:

For a given K, a complex number p is a closed-

loop pole if and only if it satisfies

G(p)= 180(2k + 1) (k = 0, 1, 2 )

|KG(p)|= 1

Characterization of Root Loci:

The root-locus consists of all points satisfying

G(s)= 180(2k + 1) (k = 0, 1, 2 )

Main Steps for Sketching Root Loci:

Step 1 Mark the open-loop poles with and zeros

with in the s-plane.

Step 2 Find the real axis portions of the locus:

The test point of any real axis portion must

be to the left of odd number of poles +zeros on the real axis.

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)1(...,,1,0,360180

mnk

mn

kk

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q is the multiplicity of pj or zj, and k is

chosen so that

|

dep| 180 and

|arr

| 180

Remark 4.1 Every branch of a root locus always

starts at an open-loop pole and ends at an open-

loop zero or infinity.

Remark 4.2 A root locus is symmetric about the

real axis.

Remark 4.3 The points where the locus crosses

the imaginary axis can be found by solving

a(j) + Kb(j) = 0

Remark 4.4 The locations of multiple closed-loop

poles on the real axis can be determined by solving

a(s) + Kb(s) = 0a(s) + Kb(s) = 0

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Example 4.2 For

G(s) =

1

s(s + 1)(s + 2)

sketch the root-locus plot and then determine the

value of K such that the damping ratio of a pair

of complex-conjugate closed-loop poles is 0.5.

Solution.

To find the breakaway point, we solve

s(s + 1)(s + 2 ) + K= 0

[s(s + 1)(s + 2 ) + K] = 0

The two solutions are

s= 0.4226 and s= 1.5774The breakaway point is seen to be s = 0.4226 as

it is between 0 and1.

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Example 4.3 For

G(s) = s + 2s2 + 2s + 3

sketch the root-locus plot.

Solution.

To find the breakin point, we solve

s2 + 2s + 3 + K(s + 2) = 0

[s2 + 2s + 3 + K(s + 2)]= 0

The two solutions are

s= 3.7320 and s= 0.2680The breakin point is seen to be s =3.7320 asthe other point is not on the root locus.

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4.2 Root-Locus Synthesis

(cf. Chapter 7 of the textbook)

Dominant Poles: A pair of complex conjugatepoles is said to be dominant if all the other

poles are to the left of the pair.

Lead compensators: A lead compensator is asystem whose steady-state response to any si-

nusoidal input has a phase lead. A first order

lead compensator is of the form

C(s) =Ks + b

s + a, a > b >0

Lag Compensators: A lead compensator is asystem whose steady-state response to any si-

nusoidal input has a phase lag. A first order

lag compensator is of the form

C(s) =Ks + bs + a

, 0< a < b

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Lag-Lead Compensators: A lag-lead compen-sator is a system whose steady-state response

to a sinusoidal input has a phase lag in some

frequency region and a phase lead in another.

Role of Compensation: A compensator is tocompensate for deficit performance of the orig-

inal system.

Realization of Compensators: electronic net-works and mechanical systems.

+u

+y

R1

R2

C

Lead network.

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+

u

+

y

R1

R2

C

Lag network.

Compensation technique:Consider the following feedback system dia-

gram:

+

C(s) P(s)

R Y

Feedback control system.

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whereP(s) is the transfer function of the plant

to be controlled and C(s) is a first-order com-

pensator of the form

C(s) =Ks + b

s + a

As the open-loop transfer function isP(s)C(s),

the angle and magnitude conditions for the

root locus are

P(s)C(s) = 180(2k+ 1) (k= 0, 1, 2 )|P(s)C(s)| = 1

i.e.

[(s + b)(s + a)] + P(s) =

180

(2k+ 1)

|C(s)||P(s)| = 1The basic compensation technique consists of

two steps:

Determine the location of the pole and zero

of C(S) from the angle condition.

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Compute the gain of C(s) from the magni-

tude condition.

The complete procedure for designing a com-pensator:

Step 1 Determine the desired location for the dom-

inant closed-loop poles from the performance

specifications.

Step 2 Ascertain whether or not the gain adjust-

ment alone can yield the desired closed-loop

poles by e.g. drawing the root-locus plot.

If not, go to the next step.

Step 3 If not, determine the angle contribution of

the compensator from the angle of P(s) at

a dominant closed-loop pole.

Step 4 Determine the location of the pole and zeroof the compensator.

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ROOT-LOCUS METHOD

Step 5 Determine the gain of the compensator from

the magnitude condition.

Step 6 Verify that all performance specifications have

been met.

Example 4.4 Consider the system shown as

on page 88, where

P(s)=4

s(s+ 2)

Design a lead compensator C(s) so that the

closed-loop system has poles at

jp 322

and has a static velocity error as small aspos-

sible.

Solution.

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Step 1 Note from the root-locus plot shown below

that a proportional controller cannot place

the closed-loop poles at the desired loca-

tion.

Re(s)0

Im(s)

2

2

2

Step 2 As the angle of P(s) at p= 2 + 23j is

4

s(s + 2)

s=2+23j

= 210o

the angle condition will be satisfied whenthe angle contribution of the compensator

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is 30o, i.e.,

= (p + b) (p + a) = 300 (4.1)

Step 3 As the static velocity error constant of the

closed-loop system is

Kv =K(b/a) lims0 sP(s) (4.2)

a and b should be chosen so that (4.1) is

satisfied and b/a is maximized. The follow-

ing figure explains how to calculate such apair which is given by

a= 5.4 and b= 2.9

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Re

Im

0

p

b

a

2

2

Step 4 From the magnitude condition, the gain of

the compensator is found to be

K= 4.68

It follows from (4.2) that the static velocity

error constant is

Kv = 5.02

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Step 5 The resulting closed-loop poles are

p= 2 2

3j and 3.4The following figure exhibits the unit-step

and unit-ramp responses of the closed-loop

systems associated with

C(s) = 1 and C(s) = 4.68s + 2.9

s + 5.4

respectively.

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0 0.5 1 1.5 2 2.5 3 3.5 40

0.2

0.4

0.6

0.8

1

1.2

1.4UnitStep Responses of Compensated and Uncompensated Systems

t Sec

Output

Compensated System

Uncompensated System

Unit-step responses of compensated and

uncompensated systems.

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0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

3

3.5

4UnitRamp Responses of Compensated and Uncompensated Systems

t Sec

Output

Uncompensated System

Compensated System

UnitRamp Input

Unit-ramp responses of compensated and

uncompensated systems.

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