nuaa-control system engineering chapter 4 root-locus technique

NUAA-Control System Engineering

Chapter 4

Root-locus Technique

Control System Engineering-2008

2

4-1 Introduction4-2 Root-locus equation4-3 Rules to draw regular root loci4-4 Generalized root loci4-5 Analysis of control system by RL

method4-6 Control system design by the root-

locus method

Contents in Chapter 4


3

Introduction

Stability and the characteristics of transient response of closed-loop systems

Locations of the closed-loop poles

Problems to solve characteristic equation : 1. Difficult for a system of third or higher order. 2. Tedious for varying parameters.


4

Varying the loop gain K

In many systems, simple gain adjustment may move the closed-loop poles to desired locations.

Then the design problem may become the selection of an appropriate gain value.

It is important to know how the closed-loop poles move in the s plane as the loop gain K is varied.

The open loop gain K is an important parameter that can affect the performance of a system

R(s)K

Y(s)

H(s)

G (s)－


5

A simple method for finding the roots of the characteristic equation has been developed by W.R.Evans. This method is called root locus method.

Graphical Analysis of Control System, AIEE Trans. Part II,67(1948),pp.547-551. Control System Synthesis by Root Locus Method, AIEE Trans. Part II,69(1950),pp.66-69

Root locus method


6

Root Locus: the locus of roots of the characteristic equation of

the closed-loop system as a specific parameter (usually, gain K) is varied form 0 to ∞.

The advantages of RL approach: 1. Avoiding tedious and complex roots-solving

calculation2. Clearly showing the contributions of each loop poles or zeros to the location of the closed-loop poles. 3. Indicating the manner in which the loop poles and zeros should be modified so that the response meets system performance specifications.


7

4-2 Root-locus equation


8

Consider a second-order system shown as follows:

)1( sskR(

s) -

Y(s)

Start from an example

The roots of CE change as the value of k changes.

Closed-loop TF:

kss

ks

2)(

Characteristic equation (CE): 02 kss

Roots of CE: kk

s 412

1

2

1

2

4112,1

When k changes from 0 to ∞, how will the locus of the roots of CE move?


9

1,2

1 11 4 , : 0

2 2s k k

1/ 4k＝ 2/121 ss

01 s 1s2 k＝ 0

0 1/ 4k As the value of k increases, the two negative real roots move closer to each other.

A pair of complex-conjugate roots leave the negative real-axis and move upwards and downwards following the line s=-1/2.

1/ 4 k

－ 1/2－ 1 0

On the s plane, using arrows to denote the direction of characteristic roots move when k increases, by numerical value to denote the gain at the poles.

0k 0k

1

4k

k

k


10

By Root loci, we can analyze the system behaviors

(2)Steady-state performance:

there ’s an open-loop pole at s=0, so the system is a type

I system. The steady-state error is

0 under step input signal

v0/Kv under ramp signal v0t

∞ under parabolic signal.

(1)Stability: when Root loci are on the left half plane, then the system is definitely stable for all k>0.


(3)Transient performance: there’s a close relationship between root loci and system

behavior

on the real-axis: k<0.25 underdamped;

k=0.25 critically damped

k>0.25 underdamped.

However, it’s difficult to draw root loci directly by closed-

loop characteristic roots-solving method.

The idea of root loci : by loop transfer function, draw closed-loop root loci directly.


12

Relationship between zeros and poles of G(s)H(s) and closed-loop ones

G(s)

H(s)

-

R(S) Y(s)

Forward path TF: Closed-loop TF:( )G s

)()(1)(

)(sHsG

sGs

Feedback path TF: ( )H s


Write G(s) and H(s) into zero-pole-gain (zpk) form:

1

1

( )( )

( )

a

G iib

ii

K s ZG S

s P

1

1

( )

( )( )

c

H jj

d

jj

K s Z

H Ss P

GK RL gain of forward path

HK RL gain of feedback path

iZ Zeros of forward path TF

iP Poles of forward path TF

jZ Zeros of feedback path TF

jP Poles of feedback path TF


1 1

1 1

( ) ( )

( ) ( ) ,( ) ( )

a c

i ji j

b d

i ji j

K s Z s Z

G S H Ss P s P

1 1

*

1 1 1 1

( ) ( )( )

( )1 ( ) ( )

( ) ( ) ( ) ( )

a d

G i ji j

b d a c

i j i ji j i j

K s Z s PG s

sG s H s

s P s P K s Z s Z

1.The closed-loop zeros = feed forward path zeros + feedback path

poles.

2. The closed-loop poles are related to poles and zeros of G(s)H(s)

and loop gain K.

3.The closed-loop root loci gain = the feed forward path root loci

gain .

GK

Loop gain:

G HK K K


15

Suppose that G(s)H(s) has m zeros (Zi) and n poles （ Pi), the above equation can be re-written as

1

1

( )( ) ( ) 1

( )

m

iin

ii

K s ZG s H s

s P

K* varies

from 0 to ∞

Root-locus equation

To draw Closed-loop root locus is to solve the CE

1 ( ) ( ) 0G s H s That is

( ) ( ) 1G s H s RL equation


16

Magnitude equation (ME)

1

1

1

m

iin

ii

K s Z

s P

Angle equation (AE)Angle equation (AE)

1 1

( ) ( )

(2 1) , 0, 1, 2,

m n

i ii i

s Z s P

l l

Since G(s)H(s) is function of a complex variable s, the root locus equation can be described by the following two equations:

1

1

( )( ) ( ) 1

( )

m

iin

ii

K s ZG s H s

s P

Magnitude equation is related not only to zeros and poles of G(s)H(s), but also to RL gain ；

Angel equation is only related to zero and poles of G(s)H(s).

Use AE to draw root loci and use ME to determine the value of K on root loci.

RL equation:


17

2 21 2 2

'

1

( 1)( ) ( )

( 1)( 2 1)

( 1/ )

( 1/ )( )( )n d n d

K sG s H s

s T s T s T s

K s

s s T s j s j

' 21 2/K K TT

21/n T 2

2d T1

/11 z

1 4

1 11 1

1 1 2 3 4

( ) ( ) ( ) ( )

( )

i ii i

G s H s s z s p

Poles of G(s)H(s) （ × ）

Zeros of G(s)H(s) （〇）

Angel is in the direction of anti-clockwise

For a point s1 on the root loci, use AE

Use ME

11

4131211'

zs

pspspssK

Example 1

2 11/p T dn4,3 jp 1 0p


18

Unity-feedback transfer function:s

K)s(G

Test a point s1 on the negative real-axis

11 1

1 1

( ) ( ) |

180

m n

i i si i

s z s p

s p

Test a point outside the negative real-axis s2=-1-j 2

1 1

2 1

( ) ( ) |

135

m n

i i si i

s z s p

s p

All the points on the negative real-axis are on RL.

All the points outside the negative real-axis are not on RL.

Example 2

One pole of G(s)H(s): 1 0p

No zero.


1. Find all the points that satisfy the Angle Equation on the s-plane, and then link all these points into a smooth curve, thus we have the system root locus when k* changes from 0 to ∞;2. As for the given k, find the points that satisfy the Magnitude Equation on the root locus, then these points are required closed-loop poles.

However, it’s unrealistic to apply such “probe by each point” method. W.R. Evans (1948) proposed a set of root loci drawing rules which simplify the our drawing work.

Probe by each point:


20

4-3 Rules to draw regular root loci

(suppose the varying parameter is open-loop gain K )


21

2 Number of Branches on the RL

3 Symmetry of the RL

4 Root Loci on the real-axis

5 Asymptotes of the RL

6 Breakaway points on the RL

7 Departure angle and arrival angle of RL

8 Intersection of the RL with the imaginary axis

9 The sum of the roots and the product of the roots of the closed-loop characteristic equation

Properties of Root Loci1 and points of Root Loci0K K


22

Root loci originate on the poles of G(s)H(s) (for K=0) and terminates on the zeros of G(s)H(s) (as K=∞).

MagnitudeEquation:

1

1

n

iim

ii

s PK

s Z

＝

0K Root loci start from poles of G(s)H(s)

K Root loci end at zeros of G(s)H(s).

1

1

1

m

iin

ii

K s Z

s P

1 and points0K K

is P

is Z

1

1

( )( ) ( ) 1

( )

m

iin

ii

K s ZG s H s

s P

RLEquation:


23

2 Number of branches on the RLnth-order system, RL have n starting points and RL have n branches

For a real physical system, the number of poles of G(s)H(s) are more than zeros ， i.e. n > m.

m root loci end at open-loop zeros （ finite zeros) ；（ n － m ） root loci end at (n － m) infinite zeros.

The order of the characteristic equation is n as K varies from 0 to ∞ ,n roots changen root loci.

1

1

( )( ) ( ) 1

( )

m

iin

ii

K s ZG s H s

s P

RLEquation:

n root loci end at open-loop zeros （ finite zeros) ；


24

3 Symmetry of the RL

The RL are symmetrical with respect to the real axis of the s-plane.

The roots of characteristic equation are real or complex-conjugate.

Therefore, we only need to draw the RL on the up half s-plane and on the real-axis, the rest can be obtained by plotting its mirror image.


25

On a given section of the real axis, RL for k>0 are found in the section only if the total number of poles and zeros of G(s)H(s) to the right of the section is odd.

4 RL on the Real Axis

zero ： z1 poles ： p1 、 p2 、 p3 、 p4 、 p5

Pick a test point s1 on [p2 ， p3]

The sum of angles provided by every pair of complex conjugate poles are 360°;

The angle provided by all the poles and zeros on the left of s1 is 0°.

The angle provided by all the poles and zeros on the right of s1 is 180°;

1 5

1 1 1 11 1

( ) ( ) ( ) ( )i ii i

G s H s s z s p

1 1( ) ( ) (2 1)180G s H s l ＝？


jw

26

Consider the following loop transfer function

2 2

( 1)( 4)( 6)( ) ( )

( 2)( 3)

K s s sG s H s

s s s

On the right of [-2 ， -1] the number of real zeros and poles=3.

On the right of [-6 ， -4], the number of real zeros and poles=7.

Example

Determine its root loci on the real axis.

Repeated poles:S-plane

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

Poles: Zeros:


27

5 Asymptotes of RL

(2 1)180a

i

n m

mn

zpm

1ii

n

1ii

a

＝

When n ≠ m, there will be 2|n-m| asymptotes that describe the behavior of the RL at |s|=∞.

The angles between the asymptotes and the real axis are（ i= 0 ， 1 ， 2 ，… ,n-m-1 ）：

The asymptotes intersect the real axis at ：

1

1

( )( ) ( ) 1

( )

m

iin

ii

K s ZG s H s

s P

RLEquation:


28

*(0.25 1) ( 4)( ) ( )

( 1)(0.2 1) ( 1)( 5)

K s K sG s H s

s s s s s s

The angles between the asymptotes and the real axis are

3 poles ： 0 、 -1 、 -5

1 zero ： -4n-m = 3 -1 = 2

The asymptotes intersect the real axis at

113

)4()5()1()0(

mn

zpm

1ii

n

1ii

a

＝

(2 1)180, 0,1a

ii

n m

90 ,270a

Example Consider the following loop transfer function

Determine the asymptotes of its root loci.


29

2

( 1)( ) ( )

( 4)( 2 2)

K sG s H s

s s s s

4 poles ： 0 、 -1+j 、 -1-j 、 -4

1 zero ： -1

n-m=4-1=3

The asymptotes intersect the real axis at

1 1

(0) ( 1 ) ( 1 ) ( 4) ( 1) 5

4 1 3

n m

i ii i

a

p z

n mj j

＝

The angles between the asymptotes and the real axis are

(2 1)180, 0,1,2a

ii

n m

60 ,180 ,300a


Determine the asymptotes of its root loci.


30

6 Breakaway points on the RLBreakaway points on the RL correspond to multi-order roots of the RL equation.

j

-4 -3 -2 -1 0

1d 2d

分离点

j

4p

3p

1p2p

A

B

0

[s]

The breakaway points on the RL are determined by finding the roots of dK/ds=0 or dG(s)H(s)=0.


31

Method － 1

n

1i ib

m

1i ib p

1

z

1

1

1

( )( ) ( )

( )

m

iin

ii

s zG s H s K

s p

1 1

( ) 1 ( ) ( ) ( ) ( ) 0n m

i ii i

D s G s H s s p K s z

Suppose the multi-order root is s1 ， then we have

Proof ：

1 1 11 1

( ) ( ) ( ) 0n m

i ii i

D s s p K s z

1 1 11 11 1

( ) ( ) ( ) 0n m

i ii i

d dD s s p K s z

ds ds

1 11 1

( ) ( )n m

i ii i

s p K s z

1 11 11 1

( ) ( )n m

i ii i

d ds p K s z

ds ds

The breakaway point b is the solution of


32

1 11 1

( ) ( ) (1)n m

i ii i

s p K s z

1 11 11 1

( ) ( ) (2)n m

i ii i

d ds p K s z

ds ds

m

ii

m

ii

n

ii

n

ii

zs

zsds

d

ps

psds

d

11

11

1

11

11

1

)(

)(

)(

)(

m

ii

n

ii zs

ds

dps

ds

d

11

111

1

)(ln)(ln

(2) divided by (1) yields

m

ii

n

ii zs

ds

dps

ds

d

11

111

1

)ln()ln(

Or

n

i i

m

i i pszs 1 11 1

11

Thus

Solutions1 is the breakaway point b


33

The poles and zeros of G(s)H(s) are shown in the following figure, determine its root loci.

Rule 1 、 2 、 3RL have three branches ， starting from poles 0 、－ 2 、－ 3 ， ending at on finite zero － 1 and two infinite zeros. The RL are symmetrical with respect to the real axis.

Ruel 4The intersections [-1,0] and[-3,-2] on the real axis are RL.

Rule 5The RL have two asymptotes(n － m ＝ 2)

(2 1)18090 ,270

0,1

a

i

n mi

22

)1()3()2(0

mn

zpm

1ii

n

1ii

a

＝

Rule 6The RL have breakaway points on the real axis (within[-3,-2])

3

1

2

1

0

1

1

1

＋＋

＋＋＝

＋ bbbb 47.2b

0-1-2-3

jω

σ

Example


34

Method － 2The breakaway point b is the solution of

1

1

( )0

( )

n

iim

ii

s pdK d

ds dss z

－

Proof ：Suppose s moves from p2 to p1, if we

increase K from zero ， K reaches its maximum when s reaches b. Then K decreases and reaches zero at p1.

K corresponding to the breakaway point has the maximum value. Thus

1

1

( )0

( )

n

iim

ii

s pdK d

ds dss z

－


35

2( ) ( )

( 4)( 4 20)

KG s H s

s s s s

Rule 1 、 2 、 3 、 4 n=4,m=0the RL are symmetrical with respect to the real axis;the RL have four branches which start from poles 0,-4 and -2±j4 ;the RL end at infinite zeros;the intersection [-4,0] on the real-axis is RL

0-2

-j4

-4

jω

σ

j4

(2 1)180( 0,1,2,3)

45 135 , 225 315

a

ll

n m

，，

Rule 5 The RL have four asymptotes.

24

)2()2()4(0

mn

zpm

1ii

n

1ii

a

＋

＝

Example


36

21 ( ) ( ) 1 0

( 4)( 4 20)

KG s H s

s s s s

2

4 3 2

( 4)( 4 20)

( 8 36 80 )

K s s s s

s s s s

3 2(4 24 72 80) 0dK

s s sds

Rule 6the breakaway point of the RL

21 b

45.223,2 jb

2( ) ( )

( 4)( 4 20)

KG s H s

s s s s

0-2

-j4

-4

jω

σ

j4

Example


37

7 Angles of departure and angles of arrival of the RL

The angle of departure or arrival of a root locus at a pole or zero, respectively, of G(s)H(s) denotes the angle of the tangent to the locus near the point.


38

Angle of Departure:

1 1

(2 1) ( ) ( ), 0, 1,m n

pj j i j ii i

i j

l p z p p l

＝＋

Pick up a point s1 that is close to p1

Applying Angle Equation (AE)

1 1 1 1 1 2 1 3( ) ( ) ( ) ( )

(2 1)

s z s p s p s p

l

s1p1 )( 11 ps angle of departure θp1

1 1 1 1 2 1 3(2 1) ( ) ( ) ( )p l p z p p p p ＝＋


39

1 1

(2 1) ( ) ( ), 0, 1,n m

zj j i j ii i

i j

l z p z z l

＝＋

Angle of Arrival:


40

2( ) ( )1 1

KsG s H s

s s

3 poles P1,2=-1(repeated poles) P3=1 ； 1 zero Z1=0 ， n-m=2 。3 branches ， 2 asymptotes

1 1 1 1 1 00.5

3 1

n m

i ii i

a

P Z

n m

(2 1) 3,

2 2a

l

n m

3(2 1) ,

2 2 2pl l

Angle of departure:


Determine its RL when K varies from 0 to ∞.

-1

j

-0.5 1


41

8 Intersection of the RL with the Imaginary Axis

Intersection of the RL with Im-axis?

Method 1 Use Routh’s criterion to obtain the value of K when the system is marginally stable, the get ω from K.

The characteristic equation have roots on the Im-axis and the system is marginally stable.

Method 2

js 1 ( ) ( ) 0G j H j

0)()(1Im

0)()(1Re

jHjG

jHjG

1 ( ) ( ) 0G s H s


42

( ) ( )( 1)( 2)

KG s H s

s s s

Closed-loop CE:3 2( 1)( 2) 3 2 0s s s K s s s K

3

2

1

0

1 2

3

60

3

s

s K

Ks

s K

Marginally stable: K =6

Auxiliary equation:

063 2 s

2js Intersection point:


Determine the intersection of the RL with Im-axis.

Routh’s Tabulation:

Method 1


43

3 2 2 3( ) 3( ) 2( ) ( 3 ) (2 ) 0j j j K K j

2 3 0K Real part: ＝

3Imaginary part: 2 0 ＝2 6K

( ) ( )( 1)( 2)

KG s H s

s s s

js 1 ( ) ( ) 0G j H j → Closed-loop CE


Determine the intersection of the RL with Im-axis.

Method 2

Closed-loop CE:3 2( 1)( 2) 3 2 0s s s K s s s K


Summary of General Rules for Constructing Root Loci

First, obtain the characteristic equation (CE)

1 ( ) ( ) 0G s H s

Then, rearrange the equation so that the parameters of interest appear as the multiplying factor in the form

1 2

1 2

( )( ) ( )1 0

( )( ) ( )m

n

K s z s z s z

s p s p s p

We sketch the Root Loci when K>0 varies from 0 to ∞.

G(s)

H(s)

-

R(S) Y(s)


Step 1: Locate the poles and zeros of G(s)H(s) on the s-plane. The RL start from poles of G(s)H(s) and terminate at zeros (finite zeros or zeros at infinity)

Step 2: Determine the RL on the real-axis. Choose a test point on the real-axis, if the total number of real poles and real zeros to the right of the test is odd, then the test point is on the RL.

Step 3: Determine the asymptotes of RL.

(2 1)180a

i

n m

The angles between the asymptotes and the real axis are（ i= 0 ， 1 ， 2 ，… ,n-m-1 ）：

The asymptotes intersect the real-axis at ：

1 1

n m

i ii i

a

p z

n m

＝


Step 4: Determine the breakaway and break-in points.

The breakaway and break-in points can be determined by finding the roots of

( ) ( )0 0

dK dG s H sor

ds ds

Step 5: Determine the angle of departure (angle of arrival) of the RL from a complex pole (at a complex zero)

Angle of departure from a complex pole pj=

1, 1

180 ( ) ( )i i

n mo

j ji i j i

p p p z

Angle of arrival at a complex zero zk=

1 1,

180 ( ) ( )i i

n mo

k ki i i k

z p z z


Step 6: Find the points where the RL may cross the imaginary-axis.

Method 1 Use Routh’s criterion to obtain the value of K when the system is marginally stable, the get ω from K.

Method 2

js 1 ( ) ( ) 0G j H j

0)()(1Im

0)()(1Re

jHjG

jHjG

1 ( ) ( ) 0G s H s


48

Consider a unity feedback system with the open-loop transfer function as:

2( )

( 3)( 2 2)

KG s

s s s s

loop poles: p1=0, p2=-3, p3,4=-1±j, n=4; no loop zeros: m=0;

2( ) ( )

( 3)( 2 2)

KG s H s

s s s s

loop TF:

Example

Sketch RL of the closed-loop system when K varies from 0 to ∞.

Solution.

Step 1: Locate the poles and zeros of G(s)H(s) on the s-plane.


49

2( ) ( )

( 3)( 2 2)

KG s H s

s s s s

-0.5

0.5

1

1.5

0-1-1.5-2-2.5-3

Step 2: Determine the RL on the real-axis. The section [0,-3] of the real axis are RL.

Step 3: Determine the asymptotes of RL

(2 1)180

45 ,135 5 ,315

a

i

n m

, 22

The angles between the asymptotes and the real axis are （ i= 0 ， 1 ， 2 ，… ,n-m-1 ）：

The asymptotes intersect the real-axis at ： 1 1 1.25

n m

i ii i

a

p z

n m


50

2( ) ( )

( 3)( 2 2)

KG s H s

s s s s

-0.5

0.5

1

1.5

0-1-1.5-2-2.5-3

1

10, i.e.,

1 1 1 10

3 1 1

n

i id P

d d d j d j

2.3d

Step 4: Determine the breakaway and break-in points.


51

3 31

313

180 ( )

( )

m

p ii

n

iii

p z

p p

－

2( ) ( )

( 3)( 2 2)

KG s H s

s s s s

3 71.6p

-0.5

0.5

1

1.5

0-1-1.5-2-2.5-3

3p

Using measurement to estimate

Step 5: Determine the angle of departure (angle of arrival) of the RL from a complex pole (at a complex zero)


52

2( ) ( )

( 3)( 2 2)

KG s H s

s s s s

-0.5

0.5

1

1.5

0-1-1.5-2-2.5-3

3p

Substituting S=jw into the above equation yields:

4 2

3

8 0

5 6 0

K

4 3 2( ) 5 8 6 0D s s s s s K

The characteristic equation:

1.1

8.16K

Step 6: Find the points where the RL may cross the imaginary-axis.


53

2( ) ( )

( 3)( 2 2)

KG s H s

s s s s

-0.5

0.5

1

1.5

0-1-1.5-2-2.5-3

3p

0K＝ 0K＝

K ＝

K ＝

K ＝

K ＝

8 16K＝ .

The complete RL:

0K＝

0K＝8 16K＝ .


RL Demo 1

j

0

j

0

j

0

j

0

j

0

j

0 0

j

0

j

0

j

j

00

j


RL Demo 2j

0

j

0

j

00

j j

0

j

0

j

0

j

00

jj

00

j j

0


56

Draw RL using MATLAB

2 4 3 2

1( ) ( )

( 3)( 2 2) 5 8 6

KG s H s K

s s s s s s s s

Loop function:

>>n=1;

>> d=[1 5 8 6 0];

>>sys=tf(n,d);

>>rlocus(sys)


Every point on the RL represents a closed-loop pole corresponding to a certain value of K which can be determined according to the Magnitude Equation 。

Determination of closed-loop poles

Magnitude Equation :

1

1

1

m

iin

ii

K s Z

s P


Determine the closed-loop poles when K =4.

Example

2( )

( 3)( 2 2)

KG s

s s s s

-0.5

0.5

1

1.5

0-1-1.5-2-2.5-3

d ＝－ 2.3

8 16K＝ .

jw ＝－ 1.1

Given the RL of a unity feedback system whose open-loop transfer function is:

Control System Engineering-2008 Determine the closed-loop pole according to ME:

Magnitude Equation :

1

1

1

m

iin

ii

K s Z

s P

4 poles ， no zero

2( ) ( )

( 3)( 2 2)

KG s H s

s s s s

4

1

2

4

( 3)( 2 2) 4

ii

K s P

s s s s

Pick two points within [ － 3 ， -2.3] and [-2.3,0] , using trial-and-error to find the closed-loop poles on the real-axis.

1 22, 2.51s s

So there are exist the two real closed-loop poles locating at two sides of the breakaway point corresponding to K=4

At the breakaway point d=-2.3, its corresponding K=4.3

K increases from 0 to 4.3


Closed-loop CE: 4 3 25 8 6 0s s s s K

Using the ‘Synthetic Division’ to get

23 4( )( ) 0.48 0.79s s s s s s

4 3 23 4( 2)( 2.52)( )( ) 5 8 6 4s s s s s s s s s s

2 4 3 23 4( 4.52 5.04)( )( ) 5 8 6 4s s s s s s s s s s

1 2using 2, 2.51 and 4s s K

3

4

0.24 0.86

0.24 0.86

s j

s j


So when K=4, the closed-loop transfer function is

4( )

( 2)( 2.51)( 0.24 0.86)( 0.24 0.86)s

s s s j s j


Effects of Adding Poles and Zeros to G(s)H(s)

Example

( ) ( )( 4)

KG s H s

s s

0-4 -2

Breakaway point:

Poles: -4, 0

( ) ( )0 2

dG s H ss

ds

Control System Engineering-20081. Addition of poles of G(s)H(s)

Example

( ) ( )( 4)

KG s H s

s s

Add a pole at s=-5, we have

( ) ( )( 4)( 5)

KG s H s

s s s

0-4-5

Poles: -5,-4, 0 no zeros, n=3,m=0

Breakaway point:

( ) ( )0 1.47

dG s H ss

ds

RL on real-axis: (-∞,-5] and [-4,0]

Angle of asymptotes= 60°,180° ,300°

Intersection of asymptotes with real-axis = -3

-3

Intersection of RL with Im-axis = ±4.47j

0-4 -2

Adding a pole to G(s)H(s) has the effect of pushing the root loci toward the right-half plane.

Control System Engineering-20082. Addition of zeros of G(s)H(s)

Example

( ) ( )( 4)

KG s H s

s s

Add two zeros at s=-5±3j, we have2( 10 34)

( ) ( )( 4)

K s sG s H s

s s

0-4-5

Poles: -4, 0 zero: -5±3j, n=2,m=2

Breakaway point:

( ) ( ).930 2

dG s H ss

ds

RL on real-axis: [-4,0] -3

Adding left-half plane zeros to G(s)H(s) generally has the effect of moving and bending the root loci toward the left-half plane.

3

-30-4 -2


Example 1. Given a system’s structure diagram, where the parameter is speed feedback constant.

Problem: Draw RL when varies from 0 to ∞.tK

)2(10ss

Y(s)

1+ s

-

R(s)2)s(s

10

tK

RL for the case of varying zeros of G(s)H(s)

tK


The RL loop transfer function is

)2(

)1(10)()(

ss

sKsHsG t

The closed-loop transfer function is ：

2

( ) 10( )

1 ( ) ( ) 2 10 10 t

G ss

G s H s s s K s

)2(10ss

Y(s)

1+ s

-

R(s)2)s(s

10

tK


2

101 0

2 10tK s

s s

2 2 10 10 0ts s K s

It can be re-written as

Define

The characteristic equation is:

The is the equivalent RL loop transfer function whose RL loop gain is

)(1 SG10 tK

*

1 2

*

10( )

2 10 ( 1 3)( 1 3)

10

t t

t t

K s K sG S

s s s j s j

K K


The new system is shown as follows:

The closed-loop transfer function ：

Note: The same denominator, but different numerator, compared with the original system.

2

10

2 10tK s

s s --

Y(s)Y(s)R(s)

sKss

sK

sssK

sssK

st

t

t

t

10102

10

10210

1

10210

)(2

2

2


2 poles, 1 zeros n=2, m=1

1 2 1

1 2

l

The breakaway point :

1/( - ) 1/ ( ) 1/ ( )

3.16, 3.16 (deleted)

180=

d

d p d p d z

d d

l

1 2 11 3, 1 3, 0p j p j z

1

2

From 1 (s) 0, the corresponding gain is:

-(s 2s 10)/10s 0.43

t

t

G K

K

*

1 2

*

( )2 10

( 10 )

t

t t

K sG S

s s

K K

jω

σ-1-2-3-4

1

2

3

2 branches, one approaches to zero, the other approaches to infThere are RL on the real axis [-∞, 0]

0.43tK

For the new system:


RemarkWith the root loci drew by equivalent open-loop transfer function, we can only determine system closed-loop poles, we still need to determine the system closed-loop zeros to analyze system dynamic behavior.

Analysis of the influence of variation of Kt on the system

)2(10ss

Y(s)

1+ s

-

R(s)2)s(s

10

tK

When Kt varies from 0 to infinity, an zero is added to the loop function G(s)H(s) and the zero moves from the negative infinity to the origin of s-plane.


When Kt is small, a pair of closed-loop complex-conjugate poles are close to the Im-axis, the overshoot of system step response is large and the oscillation is strong. When Kt is small, the system velocity feedback signal is weak and under-damped.When Kt increases, the system damp is enhanced, the oscillation is weaken and the overshoot decreases, the system behavior is enhanced.The breakaway point: Kt ＝ 0.43 ， the system is critically damped.When Kt>0.43, the two closed-loop poles are negative and real , the system is over-damped.

jω

σ-1-2-3-4

1

2

3

0.43tK

Adding left-half plane zeros to open-loop transfer function generally has the effect of moving and bending the root loci toward the left-half s-plane.


Example 2: Consider a unity feedback system whose open-loop transfer function is

)1)(1()()(

sTssk

SHSG

where K and T are given constants,while the parameterτ is to be determined. Problem: Draw RL whenτvaries from 0 to infinity.

RL for the case of varying poles of G(s)H(s)


The characteristic equation ： 1+G(S)H(s)=0

2 ( 1) ( 1) 0s Ts s Ts k

)1)(1()()(

sTssk

SHSG

2 ( 1)1 0

( 1)

s Ts

s Ts k

The equivalent open loop transfer function G1(s) ： 2 2

1 2

( 1) ( 1)( )

( 1)

s Ts s TsG S

s Ts k Ts s k

SolvingT

TksksTs

2

411 yields 0 2,1

2

Suppose that 4KT>1, then two complex poles of G(S)are:

22

21 )

2

1(

2

1,)

2

1(

2

1

TT

Kj

TP

TT

kj

TP


2 poles, 3 zeros Z1=Z2=0,Z3=-1/T

So number of branches of the RL is equal to 3.

According to rule 8, the interaction point with the imaginary axis is determined by 1+G1(jw)=0, that is

0

00

)1()1()(

13

222

T

Tk

kjwTjwjwTjw

c

cc

It follows that 1,

1

KTT

TKT

c

2

11 2

( 1)( )

( )( )

s TsG S

s P s P

2

22

1 )2

1(

2

1,)

2

1(

2

1

TT

Kj

TP

TT

kj

TP


RL whenτvaries from 0 to infinity is shown as follows:

Adding a pole to open-loop transfer function has The effect of pushing the RL toward the right-half plane

j

p1

p2

-1/T 1/ (-2T)

0c c the system is unstable.

Control System Engineering-2008Example

( 1)( 2)( )

( 1)

K s sG s

s s

2 21.37, 139d K

Consider the RL of a unity-feedback system

j

0 1-1-2

d1=0.366k1=0.0718

d2= -1.37

k2=13.9

s=0.7jk1=0.332

1 Determine the closed-loop TF when its CE has multi-order roots.

2 Determine the steady-state error with a ramp input for underdamped closed-loop system

Solution. 1. RL loop function:The CE of closed-loop system has multi-order roots at two

breakaway points which are actually two closed-loop roots

1 10.366, 0.0718d K 2

( ) 0.067( 1)( 2)( )

1 ( ) ( 0.366)

G s s ss

G s s

2

( ) 0.933( 1)( 2)( )

1 ( ) ( 1.37)

G s s ss

G s s

2 21.37, 13.9d K


( 1)( 2)( )

( 1)

K s sG s

s s

2 21.37, 139d K

Consider the RL of a unity-feedback system

j

0 1-1-2

d1=0.366k1=0.0718

d2= -1.37

k2=13.9

s=0.7jk1=0.332

1 Determine the closed-loop TF when its CE has multi-order roots.

2 Determine the steady-state error with a ramp input for underdamped closed-loop system

Solution. 2. RL loop function:

Type-1 system:

For a ramp input, the ramp error constant: 0

lim ( ) 2vs

K sG s K

Closed-loop system is underdamped:

0.332 13.9K

The steady-state error:1

ssv

eK

1.5 0.036sse

Example


An example of a control system analysis and design utilizing Root Locus technique

Control System Engineering-2008An automatic self-balancing scale


We desire to accomplish the following:

Select the parameters of the systemDetermine the specificationsObtain a model represent the systemDesign the gain K using root locusDetermine the dominant mode of response


Parameters

2

Counterweight:

Inertia of the beam:

Length of the weight to the pivot:

Lengt

2

0.05

5

20h of the beam to the viscous damper:

Damping constant 10 of the viscous damper: 3 / /

c

w

i

W N

I kgm

l cm

l cm

b kg m s

Lead screw gain:

Input potentiometer gain:

Feedback pote

1 /

400

ntiome

0 4800 /

ter gai 4 0 /n 0:

s

i

f

K m rad

K V m

K V m


SpecificationsA rapid and accurate response resulting in a small

steady-state weight measurement error is desired.

Specifications

Steady-state error (for a step input) Kp=∞, ess=0

Underdamped response ζ=0.5

Settling time (error band Δ<2%) ts<2sec


Mathematical model

i

y

l

The motion of the beam about the pivot is represented by the torque equation:

2

2

dI torquesdt

For a small deviations from balance, the deviation angle is

22

2 w c i

d dI l W xW l bdt dt


Mathematical model

( )m i fv t K y K x

The transfer function of the motor is:

( )

( ) ( 1)m m m

m

s K K

V s s s s

The input voltage to the motor is

where τis negligible with respect to the time constants of the overall system and θm is the output shaft rotation.


Block diagram

Signal-flow


Using Mason’s formula: 1

( )

( )

Nout k k

kin

y MX s

W s y

211 21

331

,i m s f

i m s f c

L l b Is L K K K s

L l K K K W Is

31 1, 1w i i m sM l l K K K Is Forward path:

Loop path:

2 212 i m s fL l bK K K IsNontouching loops:

11 21 31 121 ( )L L L L

2

( )( )

( ) ( )( )w i i m s

i m s f c m s i i

l l K K KX ss

W s s Is l b s K K K W K K K l

The closed-loop transfer function:

Characteristic equation:2( )( ) 0i m s f c m s i is Is l b s K K K W K K K l

The steady-state gain of the system is:

0

( ) 0.05lim 0.025 /

( ) 2w i i m s w

sc m s i i c

l l K K K lX sm kg

W s W K K K l W


Substituting the selected parameter into the CE yields:

968 3 0

10 10m mK K

s s s

To rewrite it into root locus form, first isolate Km as follows:

2 968 3 8 3 0

10 10m mK K

s s s s


Rewrite it into root locus form:

2

( 6.93 6.93)( 6.93 6.93)( ) ( )

( 13.856 )4

s j s jG s H s

s s

2

1 ( ) ( )

( 10 )( 6.93 6.93)( 6.93 6.93)1 0

( 8 3)m

KG s H s

K s j s j

s s

10mK K


We draw the root locus as K varies from 0 to ∞

With MATLAB:

2

( 6.93 6.93)( 6.93 6.93)1 0

( 13.8564)

s j s jK

s s

> >z=[-6.93-6.93i, -6.93+6.93i];

>>rlocus(sys)>>sys=zpk(z,p,k);>>k=1;>>p=[0 0 -13.8564];

The dominant closed-loop poles can be placed at:

0.5, 25.3K

To achieve this gain:

/10 795m

rad sK K

volt

Closed-loop CE:


Let us observe the time-domain step response of the closed-loop system with the designed Km in MATLAB to see if it satisfied the desired specifications.

Specifications

Steady-state error (for a step input) Kp=∞, ess=0

Underdamped response ζ=0.5

Settling time (error band Δ<2%) ts<2sec


Parameter Design by Root Locus Technique


Originally the root locus method was developed in determine the loci of roots of the characteristic equation as the system gain, K, varies from 0 to ∞.

However, as we have see, the effect of other system parameters may be readily investigated by using the root locus technique.

For example, a third-order equation of interest might be

3 2(3 ) 3 6 0s s s

To ascertain the effect of the parameter α, we isolate the parameter and rewrite the equation in root locus form 2

3 21 0

3 3 6

s

s s s

Control System Engineering-2008If there are more than one (m>1) parameters we want to investigate, we must repeat the root locus approach m times.

3 2 0s s s

3 21 0

s

s s

The denominator of the above RL equation is the characteristic equation of the system with . 0

For example, a third-order characteristic equation with and as parameters

The effect of varying from 0 to ∞ is determined from the root locus equation


3 2 0s s Rewritten as RL form:

21 0

( 1)s s

Then upon evaluating the effect of , a value of is selected and used to evaluate the effect of .

Therefore, we must first evaluate the effect of varying from 0 to ∞ by utilizing

When more than one parameters varies continuously from 0 to ∞, the family of root loci are referred to as the root contours(RC).


Example

3 23 2 0s s s s

nuaa-control system engineering chapter 4 root-locus technique

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