feedback control system the root-locus design method dr.-ing. erwin sitompul chapter 5

Feedback Control System

THE ROOT-LOCUS DESIGN METHOD

Dr.-Ing. Erwin Sitompul

Chapter 5

http://zitompul.wordpress.com

6/2Erwin Sitompul Feedback Control System

Root Locus: Illustrative ExampleExamine the following closed-loop system, with unity negative feedback.

2

( )

( )

Y s K

R s s s K

1,2

1 1 4

2

Ks

The closed-loop transfer function is given as:

The roots of the characteristic equation are:

• The characteristic equation• The denominator of the

closed-loop transfer function


Real Axis

Imag

inar

y A

xis

-2 -1 0 1-2

-1

0

1

2

1,2

1 1 4 1, 0

2 2 4

1 4 1 1,

2 2 4

KK

sK

j K

Root Locus: Illustrative Example

K=0K=0

K=1/4

K=∞

K=∞

: the poles of open-loop transfer function


Real Axis

Imag

inar

y A

xis

-2 -1 0 1-1.5

-1

-0.5

0

0.5

1

1.5

Root Locus: Illustrative Example

Where are the location of the closed-loop roots when K=1?

2 2

( ) 1

( ) 1

Y s K

R s s s K s s

2 1 0.5n

2 1 1n n

0.5n 21 0.866d n

( ) ( 0.5 0.866)dj j

K=1

K=1

–0.5

–0.866

0.866

There is a relation between gain K and the position of closed-loop poles, which also affects the dynamic properties of the system (ζ and ωd)


Root Locus of a Basic Feedback System

The closed-loop transfer function of the basic feedback system above is:

( ) ( ) ( )( )

( ) 1 ( ) ( ) ( )

Y s D s G sT s

R s D s G s H s

The characteristic equation, whose roots are the poles of this transfer function, is:

1 ( ) ( ) ( ) 0D s G s H s


Root Locus of a Basic Feedback SystemTo put the characteristic equation in a form suitable for study of the roots as a parameter changes, it is rewritten as:

1 ( ) 0KL s

where( ) ( ) ( ) ( )KL s D s G s H s

( )( )

( )

b sL s

a s

K is the gain of controller-plant-sensor combination. K is selected as the parameter of interest. W. R. Evans (in 1948, at the age of 28) suggested to plot

the locus (location) of all possible roots of the characteristic equation as K varies from zero to infinity root locus plot.

The resulting plot is to be used as an aid in selecting the best value of K.


1 ( ) 0KL s

Root Locus of a Basic Feedback SystemThe root locus problem shall now be expressed in several equivalent but useful ways.

( )1 0

( )

b sKa s

( ) ( ) 0a s Kb s 1

( )L sK

The equations above are sometimes referred to as “the root locus form of a characteristic equation.”

The root locus is the set of values of s for which the above equations hold for some positive real value of K.


Root Locus of a Basic Feedback System

Explicit solutions are difficult to obtain for higher-order system General rules for the construction of a root locus were developed by Evans.

With the availability of MATLAB, plotting a root locus becomes very easy, using the command “rlocus(num,den)”.

However, in control design we are also interested in how to modify the dynamic response so that a system can meet the specifications for good control performance.

For this purpose, it is very useful to be able to roughly sketch a root locus which will be used to examine a system and to evaluate the consequences of possible compensation alternatives.

Also, it is important to be able to quickly evaluate the correctness of a MATLAB-generated locus to verify that what is plotted is in fact what was meant to be plotted.


Guideines for Sketching a Root Locus

1 ( ) 0KL s 1

( )L sK

( ) 1

( )

b s

a s K

Deriving using the root locus form of characteristic equation,

1 2

1 2

( )( ) ( ) 1

( )( ) ( )m

n

s z s z s z

s p s p s p K

Taking the polynomial a(s) and b(s) to be monic, i.e., the coefficient of the highest power of s equals1, they can be factorized as:

If any s = s0 fulfills the equation above, then s0 is said to be on the root locus.


0

1 2

1 2

( ) ( ) ( ) 1 1

( ) ( ) ( )m

n s s

s z s z s z

s p s p s p K K

0

1 2

1 2

( )( ) ( ) 1180

( )( ) ( )m

n s s

s z s z s z

s p s p s p K

Phase Condition

Magnitude Condition

Guideines for Sketching a Root LocusThe magnitude condition implies:

The phase condition implies:

Defining and , the phase condition can be rewritten as:

( )i is z ( )i is p

1 1

180 360 ( 1), 1, 2,3,m n

i ii i

l l



The root locus is the set of values of s for which 1 + KL(s) = 0 is satisfied as the real parameter K varies from 0 to ∞. Typically, 1 + KL(s) = 0 is the characteristic equation of the system, and in this case the roots on the locus are the closed-loop poles of that system.

“

”The root locus of L(s) is the set of points in the s-plane where the phase of L(s) is 180°. If the angle to a test point from a zero is defined as ψi and the angle to a test point from a pole as Φi, then the root locus of L(s) is expressed as those points in the s-plane where, for integer l, Σψi – ΣΦi = 180° + 360°(l–1).

“

”



2

1( )

( 5) ( 2) 4

sL s

s s s

Consider the following example.

: the poles of L(s): the zero of L(s)

0 1 2s j Test point

1 90 1

1 tan (2 1) 116.6

2 0 1

3 tan (4 1) 76.0 1

4 tan (2 4) 26.6

1 1 2 3 4i i 90 116.6 0 76.0 26.6

129.2 180 360 ( 1)l s0 is not on the root locus


Rules for Plotting a Root Locus

RULE 1:The n branches of the locus start at the poles of L(s) and m of these branches end on the zeros of L(s), while n–m branches terminate at infinity along asymptotes.

( ) 1

( )

b s

a s K

Recollecting( ) 1

( )

b s

a s K

1 2

01 2

( ) ( ) ( )1lim lim

( ) ( ) ( )i

m

K s pn

s z s z s z

K s p s p s p

1 2

1 2

( ) ( ) ( ) 1

( ) ( ) ( )m

n

s z s z s z

s p s p s p K

1 2

1 2

( ) ( ) ( )1lim 0 lim

( ) ( ) ( )i

m

K s zn

s z s z s z

K s p s p s p

The root locus starts at K = 0 at the poles of L(s) and ends at K = ∞ on the zeros of L(s)



RULE 2:On the real axis, the loci (plural of locus) are to the left of an odd number of poles and zeros.

12345: The root locus

1

1234

Angles from real poles or zeros are 0° if the test point is to the right and 180° if the test point is to the left of a given pole or zero.


2

1( )

( 4) 16L s

s s


The rule is now applied to obtain the root locus of:

2,3 4 4p j 1 0p



1 2 3 0

180 360 ( 1)l

For any test point s0 on the real axis, the angles φ1 and φ2 of two complex conjugate poles cancel each other, as would the angles of two complex conjugate zeros (see figure below).

11 tan (2 4) 26.6

12 tan ( 2 4) 26.6

1 2 0

3 0 The pair does not give contribution to the phase condition

s0 is not on the root locus

Now, check the phase condition of s1!

s1


RULE 3:For large K and s, n–m of the loci are asymptotic to lines at angles Φl radiating out from the point s = α on the real axis, where:


180 360 ( 1), 1, 2, ,3l

ll

n m

i ip z

n m

Center of Asymptotes

Angles of Asymptotes



2

1( )

( 4) 16L s

s s

For , we obtain

1 2,30, 4 4p p j 3, 0n m

180 360 ( 1)

3 0

l

i ip z

n m

60 120 ( 1)l 60 ,180 ,300

180 360 ( 1)l

l

n m

( 4 4) ( 4 4) 0

3 0

j j

8

2.673

–2.67

60°

180°

300°


RULE 4:The angle of departure of a branch of a locus from a pole is given by:


,dep 180 360 ( 1)l i ii l

l

and the angle of arrival of a branch of a locus to a zero is given by:

,arr 180 360 ( 1)l i ii l

l



For the example, the root loci must depart with certain angles from the complex conjugate poles at –4 ± j4, and go to the zero at ∞ with the angles of asymptotes 60° and 300°.



22

180 360 ( 1)i ii

l

2 1 3 180 360 ( 1)l

13 tan (4 4) 135 1 90

2 90 135 180 2 405 45

From the figure,

But

Thus

By the complex conjugate symmetry of the roots, the angle of departure of the locus from –4 – j4 will be +45°.



So, the root loci will start their journey from –4 ± j4 towards ∞ with the direction of 45°.

±



RULE 5:The locus crosses the jω axis (imaginary axis) at points where: The Routh criterion shows a transition from roots in the left

half-plane to roots in the right half-plane. This transition means that the closed-loop system is

becoming unstable. This fact can be tested by Routh’s stability criterion, with

K as the parameter, where an incremental change of K will cause the sign change of an element in the first column of Routh’s array.

The values of s = ± jω0 are the solution of the characteristic equation in root locus form, 1 + KL(s) = 0. The points ± jω0 are the points of cross-over on the

imaginary axis.



For the example, the characteristic equation can be written as:

1 ( ) 0KL s

2

11 0

( 4) 16Ks s

3 28 32 0s s s K

3

2

1

0

: 1 32

: 8

8 32:

8:

s

s K

Ks

s K

The closed-loop system is stable for K > 0 and K < 256 for 0 < K < 256.

For K > 256 there are 2 roots in the RHP (two sign changes in the first column).

For K = 256 the roots must be on the imaginary axis.


3 28 32 256 0s s s

The characteristic equation is now solved using K = 256.

1

2,3 0

85.66

ss j j Points of Cross-over

Another way to solve for ω0 is by simply replacing any s with jω0 without finding the value of K first.

3 20 0 0( ) 8( ) 32( ) 0j j j K

3 20 0 08 32 0j j K

2 30 0 08 (32 ) 0K j

≡ 0 ≡ 0

30 0

20

0

32 32 5.66

208

8 32 256

KKK


Same results for K and ω0


5.66

–5.66


The points of cross-over are now inserted to the plot.



The complete root locus plot can be shown as:



RULE 6:The locus will have multiple roots at points on the locus where:

( ) ( )( ) ( ) 0da s db s

b s a sds ds

The branches will approach and depart a point of q roots at angles separated by:

180 360 ( 1)l

q



A special case of point of multiple roots is the intersection point of 2 roots that lies on the real axis.

If the branches is leaving the real axis and entering the complex plane, the point is called the break-away point.

If the branches is leaving the complex plane and entering the real axis, the point is called the break-in point.

Realaxis

Imagaxis

Break-awaypoint

Realaxis

Imagaxis

Break-inpoint

90 90


Example 1: Plotting a Root Locus

Draw the root locus plot of the system shown below.

1( )

( 1)( 2)L s

s s s

1 2 30, 1, 2p p p 3, 0n m 3 zeros at infinity

RULE 1

( ) ( 1)( 2)( ) 1

( 1)( 2)

KY s s s s

KR ss s s

1 ( )KL s



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

RULE 2

1 2 30, 1, 2p p p



180 360 ( 1)l

l

n m

180 360 ( 1)

3 0

l

i ip z

n m

60 120 ( 1)l 60 ,180 ,300

0 1 2

3 0

1

1 2 30, 1, 2p p p

Center of Asymptotes

Angles of Asymptotes



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

RULE 3

60°

180°

300°

60 ,180 ,300l

1

Not applicable. The angles of departure or the angles of arrival must be calculated only if there are any complex poles or zeros.

RULE 4



1 ( ) 0KL s 1

1 0( 1)( 2)

Ks s s

3 23 2 0s s s K

Replacing s with jω0,3 2

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

0 0 03 2 0j j K 2 3

0 0 03 (2 ) 0K j

≡ 0 ≡ 0

30 0

20

0

2 2 1.414

203

3 2 6

KKK

Points of Cross-over



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

RULE 51.414

–1.414

0 1.414



1 ( ) 0KL s

3 2

( ) 1 1( )

( ) ( 1)( 2) 3 2

b sL s

a s s s s s s s

( ) ( )( ) ( ) 0da s db s

b s a sds ds

The root locus must have a break-away point, which can be found by solving:

2 3 21 (3 6 2) ( 3 2 ) 0 0s s s s s 23 6 2 0s s

1 21.577, 0.423s s



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

RULE 61.414

–1.414

1 21.577, 0.423s s

On the root locusThe break-away

point

Not on the root locus

–0.423


Example 1: Plotting a Root LocusAfter examining RULE 1 up to RULE 6, now there is enough information to draw the root locus plot.

Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

1.414

–1.414

–0.423

90



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

1.414

–1.414

–0.423

The final sketch, with direction of root movements as K increases from 0 to ∞ can be shown as:

Final Result

Determine the locus of all roots when K = 6!



a) Draw the root locus plot of the system.b) Define the value of K where the system is stable.c) Find the value of K so that the system has a root at s = –2.



1( ) ( 4)

1( ) 1 ( 2)( 4)

KY s sR s K s

s

2( )

4

sL s

s

1 14, 2p z 1, 1n m

1 ( )KL s

RULE 1

There is one branch, starts from the pole and approaches the zero



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

RULE 2

1 14, 2p z

Not applicable, since n = m.

RULE 3

Not applicable. The angles of departure or the angles of arrival must be calculated only if there are any complex poles or zeros.

RULE 4



1 ( ) 0KL s 2

1 04

sKs

4 ( 2) 0s K s

Replacing s with jω0,

0 0( ) 4 ( 2) 0j K j

04 2 (1 ) 0K j K

≡ 0 ≡ 0

4 2 2

KK

0

0

(1 ) 0

K

Points of Cross-over



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

The point of cross-over, as can readily be guessed, is at s = 0.

RULE 5

Not applicable. There is no break-in or break-away point.

RULE 6

K = 2



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

The final sketch, with direction of root movements as K increases from 0 to ∞ can be shown as:

Final Result

a) Draw the root locus plot of the system.



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

b) Define the value of K where the system is stable.

System is stable when the root of the characteristic equation is on the LHP, that is when 0 ≤ K < 2.

K = 2K = 0 K = ∞



Realaxis

Imagaxis

1 20–1–2–3–4

1

2

–1

–2

3

–3

K = 2K = 0 K = ∞

c) Find the value of K so that the system has a root at s = –2.

Inserting the value of s = –2 in the characteristic equation,

204 ( 2)

ss K s

2 4 ( 2 2) 0K ( 2 2) 2 4 0.5K

K = 0.5


Homework 6 No.1, FPE (5th Ed.), 5.2.Hint: Easier way is to assign reasonable values for the zeros and poles in

each figure. Later, use MATLAB to draw the root locus.

No.3Sketch the root locus diagram of the following closed-loop

system as accurate as possible.

No.2, FPE (5th Ed.), 5.7.(b)Hint: After completing the hand sketch, verify your result using MATLAB. Try

to play around with Data Cursor.

feedback control system the root-locus design method dr.-ing. erwin sitompul chapter 5

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