objective determination of root from the characteristic equation by using graphical solution. ...

OBJECTIVE

Determination of root from the characteristic equation by using graphical solution. Rules on sketching the root locus. Analysis of closed-loop using root locus

4.0 ROOT LOCUS

Root locus conceptRoot locus concept

Consider

Open-loop transfer functionAnd

WhereIf

And

KG(s)

H(s)

R(s) + E(s) Y(s)

B(s)

)()( sHsKG

)()()(

1

1

sPsZsG

m

iim )zs()zs)....(zs)(zs()s(Z

1211

n

jjn pspspspssP

1211 )())....()(()(

Closed-loop transfer function

)()()(

)()(

11

1

sKZsPsKZ

sRsY

~ Number and position of zeros for open-loop and closed-loop are the same

~ Position of poles for the closed-loop depend on the position of poles, zeros and K.

Characteristic equation is 0)()( 11 sKZsP

.

If 1)( sHLet

)()()(

1

1

sPsZsG and

)()()(

2

2

sPsZsH

Its open-loop transfer function

)()()()(

)()(21

21

sPsPsZsZ

KsHsKG

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

)()(

2121

21

sZsKZsPsPsPsKZ

sRsY

and its closed-loop as

~ Position of poles for the closed-loop depend on the position of poles, zeros and K of the open-loop. transfer function.


Where H(s)=1

As K varies the closed-loop poles follow similarly and form a locus. For that we define Root locus is a locus of characteristic equation as K varies from 0 to .

Referring to a characteristic equation0)()(1 sHsKG

Let say

n

jj

m

ii

ps

zsKsHsKG

1

1

)(

)()()(

1)()( sHsKGRe-arrange

-1 is a complex numberjre11

where ....,,,r 531


Magnitude conditionMagnitude condition1)()( sHsKG

Re-arrange

m

nm

ii

n

jj

zszszspspsps

zs

psK

............

21

21

1

1

where npspsps ......21 is the magnitude from a test point to open-loop poles

nppp ......, 21and

mzszszs ......21 the magnitude from a test point to open-loop zeros nzzz ......, 21

From the magnitude condition, we can determine K. Let s be the test point, the magnitude are

nnn ,.....1 mmm .....1 and from open-loop zeros and poles respectively.

Hence the magnitude condition is

1..........

1

1

m

n

mmnn

j

s-plane

n1nn mm

m1

s

Magnitude conditionMagnitude condition

Angle conditionAngle condition Revisiting the complex number of -1

jre11Its angle ,5,3,1,)()()()(

11

rrpszssHsKGn

jj

m

ii

Expand

.....,5,3,1)(.....)()(

(.....)()(

21

21

rrpspspszszszs

n

m

where mzszszs ,....,, 21 is the angle from a test point to open-loop zeros

nz......z,z 21

and

where npspsps ,.....,, 21 is the angle from a test point to open-loop poles

nppp ......, 21

From the angle condition, we can determine the angle contribution by s the test

point, the angle are n ....1 m ....1 and from open-loop zeros and poles respectively.

Hence the angle condition is

rm

n

.....

.....

1

1

Angle conditionAngle condition

n

1 1 2

s

3

j

m

s-plane

Angle conditionAngle condition

ExampleExampleGive the unity feedback system that has the forward

transfer function.

a) Calculate the angle of G(s) at the point (-2+j0.707) by finding the algebraic sum of angles of the vectors drawn from the zeros and poles of G(s) to the given point.

b) Determine if the point specified in a) is on the root locus.

c) If the point specified in a) is on the root locus, find the gain, K, using the lengths of the vectors.

21

43

ssssK)s(G

ExampleExample

1+2-3-4=19.471o+35.264o-90o-144.736o=-180o

b) Point is on the root locus.c)

-4 -3 -2 -1

j0.707

j

1

2

3

4

330221122

22122

21

43 ...

.

LLLLK

s

Procedure for drawing a Root Procedure for drawing a Root LocusLocus

(2) The locus is symmetrical at the real axis.

ConsiderKasssR

sC

42

5)()(

2

Characteristic equation is

0422 Kass

If 1,3 aK

» K=3;a=1;roots([1 2*a 4+K])

ans =

-1.0000 + 2.4495i

-1.0000 - 2.4495i

(1) Start with the characteristic equation.

j

That shows the locus is symmetrical at -axis.

s-plane


3) The number of locus depends on the number of open loop poles

Open-loop transfer function Closed-loop transfer function

Its characteristic equation

Number of locus is reflected by the order of characteristic equation which is the same as the number of open-loop poles of the system.

K )()(

1

1

sPsZ

)()(

2

2

sPsZ

-

)()()()(

)(21

21

sPsPsZsZ

KsGo

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

)()(

2121

21

sZsKZsPsPsPsKZ

sRsY

02121 )s(Z)s(KZ)s(P)s(P


+

4) Root locus begin from open loop poles and end at open loop zeros or symmetrical lines

Consider closed-loop transfer function,

If K=0, then the characteristic equation becomes P1(s)P2(s)=0 which are open-loop poles.

If K0, characteristic equation is

If K→∞, the equation becomes Z1(s)Z2(s)≈0 , which are open-loop zeros.

Locus will begin from the open-loop poles, K=0, and end at the open-loop zeros, K→∞.

If the number of open-loop zeros are less than the open-loop poles, then the locus will end at asymptote lines given by angle

)s(Z)s(KZ)s(P)s(P)s(P)s(KZ

)s(R)s(Y

2121

21

KsPsP

sZsZ)()(

)()( 2121 .

mnr

180


where m and n are the order of numerator and denominator of the open-loop transfer function respectively.

And r=1, 3, 5, 7,…..And the line intersect the real-axis at

where and are the sum of all poles and zeros of open-loop transfer function respectively.

mn

ZPm

ii

n

jj

11

n

jjP

1

m

iiZ

1


pK)8(

2sss

51s

)(sY)(sR-

ExampleExample

Consider a P-compensator

Find the open-loop zero and poles. Obtain the closed-loop poles at Kp=0 and Kp→∞.

Consequently, shows where the locus end.

Solution

which give an open-loop zero at –2 open-loop poles at 0, –5 and –8 Its characteristic equation

As we know that the locus start at Kp=0 substitute this to the characteristic equation s3+13s+s(40)=s(s+5)(s+8)=0 , give closed-loop poles at 0, –5 and –8, which is the same as open-loop poles.

)8)(5()2(

51.

)8(2.)(

sssKs

ssssKsG P

PO

0)(1 sGO

024013 23 PP K)K(sss

ExampleExample

To find the closed-loop poles at pKWe rearrange the characteristic equation

becomes05

1.)8(

2.1

ssssK P

PKssss )8)(5(2

If pK , the root of the characteristic equation is –2, which is the same as the open-loop zero.

As there is 3 open-loop poles, there will be 3 loci, one will end at open-loop zero of -2 and the other two will end at the asymptote lines, with angles at

9002

180180 rrmn

r

where ,......5,3,1 r

with intersection of the real-axis at

5.513

)2()850(

ExampleExample

-8 -5.5 -5 -2

j

s-plane

ExampleExample

Consider the following closed-loop transfer function

KasssRsC

425

)()(

2

5a 40,21,10,3,0KIf , find the closed-loop poles at and 100.

Finally, trace the loci for pK0

Solution:

As the characteristic equation is 04102 Kss . By substituting the above values to the equation

will give the following table:

ExampleExample

Gain, K Closed-loop pole

0-9.5826 -0.4174

3 -9.2426 -0.7574

10 -8.3166 -1.6834

21

40

100

-5

-5

-5.0000 + 4.3589i

-5.0000 - 4.3589i

-5.0000 + 8.8882i -5.0000 - 8.8882i

ExampleExample

21K

100K

40K

0K

-5.0 40K

100K

j s -plane

0K

- j8.8882

-j4.3589

j4.3589

j8.8882

1000 K

ExampleExample

K0For we can trace the locus as it move from 0 to ∞ as shown below

K

K

0K 0K

js-plane

-j4.3589

- j8.8882

j4.3589

j8.8882

ExampleExample

(5) Locus on the real-axis is It should fulfill the angle condition.

Consider

H G E D C B A

j

s-plane


F

Take a test point at A-H and use the angle condition

)()( sHsKGAngle condition

A 0 No

B -180 Yes

C -180-180 = -360 No

D -180-180+180 = -180 Yes

E -180-180+180-180 = -360 No

F -180-180+180-180+180 = -180 Yes

G -180-180+180-180+180-180 = -360 No

H -180-180+180-180+180-180-180 = -540 Yes

Test point


(6) Breakaway point of locus from the real-axis

It is the point where the root locus originating from the open poles meet at the real-axis and breakaway

Consider back the closed-loop transfer function of

If we plot a graph of the closed-loop position against the gain K. We notice that at the point of -5,

before the closed-loop poles become complex, the gain is maximum for any real poles. After this the

locus will breakaway from the real-axis

We can determine this by taking the differential of the characteristic equation,

KasssRsC

425

)()(

2

-9.6 -9.2 -8.3 -5.0

2110

3

K

-1.7 -0.8 -0.4

0dsdK


A unity feedback system with an open-loop system as given below

Determine the maximum gain K before the system starts to oscillate.

Solution:

The characteristic equation is

Rearrange

Differentiate to obtain the maximum gain

)8)(2()(

sssKsGO

0)8)(2(

1

sss

K

01610 23 Ksss

sssK 1610 23

016203 2 ssdsdK

ExampleExample

which gives

Out of the two values, we choose -0.93 as –5.74 is not on the locus. Use the magnitude condition

Using s=-0.93, maximum K before the starts to oscillate

74593032

16342020 2

21 .,.)(

))((s ,

821 sssK

04.7

893.0293.093.01

82193.093.0

ss

sssK

ExampleExample

0)()(1Re sHsKG

(7) Stability boundary of the locus(7) Stability boundary of the locus

We use the Routh-Hurwitz criteria determine the maximum gain before instability and its frequency. This can be obtained through


(8) Angle of departure and arrival(8) Angle of departure and arrival

For complex poles, the locus will leave the poles from an angle called angle of departure. By selecting a test point very near to the open-loop pole, -py the angle of departure, is given by

or = (Sum of angle from all open-loop zeros to test point - (Sum of angle from all

open-loop poles to test point ) -r

r)ps(......)ps()ps(zs(.....)zs()zs( nm 2121


Example:

Find the angle of departure of the complex poles

1 1 2 3

j

90

s-plane

ExampleExample

As for the above diagram, the angle of departure from the complex pole is 18090 3211 r)()(

while the conjugate pair, its angle of departure is

For complex open-loop zero xz , the locus will end here and the angle of arrival, , given by an angle condition as

)ps(......)ps()ps(zs(.....)zs()zs(r nm 2121

or

= r - (Sum of angle from all open-loop zeros to xz ) + (Sum of angle from all open-loop

poles to xz )

ExampleExample

ExampleExample

Given a unity feedback system as given below. Find the angle of departure from the complex pole

-

+R(s)

C(s)

)ss)(s(

sK223

22

ExampleExample

4 3

1j

s-plane

2

18021

1190 11

14321

tantan

410862511 ..

Solution:

-3 -2 -1

j1

-j1

A plant of transfer function is feedback with a unity feedback. Sketch a root locus for 0<K<∞.

Solution Its characteristic equation is

As the number of open-loop poles is 3, there will be 3 loci and the loci start at 0,-1 dan –2. The locus on the real-axis

)2)(1( sssK

0)2)(1()2)(1(

1

Kssssss

K

ExampleExample

-1-2

j

Satah-s

As there is no zeros, the locus will end at asymptotes lines,

where 6003

180180 rrmn

r

,......5,3,1 r

This gives asymptote angles at 300,180,60 , with real-axis crossing at

-103

(0)-0)-1-2- (11

mn

ZPm

ii

n

jj

ExampleExample

-2 -1

s-plane

j

The loci will meet and depart at breakaway point

023 23 Ksss

Rearrange

sssK 23 23

Differentiate for the maximum value of K

0263 2 ssdsdK

ExampleExample

Its root

6.1,4.0)3(2

)2)(3(466 2

2,1

s

The probable root is –0.4 as it is on the locusThe crossing at the boundary of stability

3s2s1s

36 K

0s

1 2

3 K

K

03

6

K 0K and

Range of gain for stability 60 K . The natural frequency is determined by taking

6K njs 063 2 n 1.2 sradn, replacing to gives

.

ExampleExample

» num=[1];den=[1 3 2 0]; %Takrifkan rangkap pindah gelung buka

» rlocus(num,den); %KerahanWe can determine the gain at particular points through» [k,p]=rlocfind(num,den)

ExampleExample

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

-3

-2

-1

0

1

2

3

40.160.340.50.640.76

0.86

0.94

0.985

0.160.340.50.640.760.86

0.94

0.985

123456

Root Locus

Real Axis

Imag

inar

y Ax

is

Determination of gainDetermination of gain

Can be determined using magnitude condition.

O

s

js-plane

m1 n1 m2 m3

Zero and poles magnitude from test point

Take a test point, s the gain K is

1

321

1

1

nmmm

zs

psK m

ii

n

jj

Pole DeterminationPole Determination

If the order of the characteristic equation is n, there will be n poles. If n-1 poles are known we know the last pole by comparing the coefficient.

Consider

0..... 011

1 asasas n

nn

Which can be factored to 0))....()(( 21 npspsps where its poles are

nn pppp ,....,,, 121 . Expanding the factor form

0)....(.....)....( 211

21 n

nn

n pppsppps

Equating the coefficient

1121 .... nnn apppp

will give the n pole as

1211 .... nnn pppap

Example

For a third order system with 5K , two of the poles are –2 and –3, determine its third pole.

1116 23 sssK+

-

Solution:

Characteristic equation

06116 23 sss

Comparing 0..... 011

1 asasas n

nn

and 0)....(.....)....( 211

21 n

nn

n pppsppps

which give 6321 ppp hence the third pole is 13 p

ExampleExample

objective determination of root from the characteristic equation by using graphical solution. ...

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