control lect9

8/3/2019 Control Lect9

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Dr. Ayman A. El-Badawy

Department of Mechatronics EngineeringFaculty of Engineering and Material Science

Control Engineering

Root Locus Analysis II


Department of Mechatronics Engineering

Faculty of Engineering and Material Science

German University in Cairo



Example:An airline is mechanizing a pitch control auto pilot system. The system is

shown below.

Draw the Root-Locus diagram.

Indicate what happens to the system response as Kchanges.

Find values of Kfor which the system is stable.

r

pM

+_

54

32

ss

s

1

s

1

10s

KeM

++

Aircraft dynamicsElevator servo


2/17



0rSet

Step 1:

Find T.F.

pM

543

2

sss

s

10s

K

543

2

sss

s

10s

K

pM

35410

103

1054

31

54

3

2

2

2

sKssss

ss

s

K

sss

s

sss

s

Mp

C/C equation=0



035410 2 sKssss

5410

31

2

ssss

sK

Step 2:Put T.F. in proper form;

We must write this as:

5410 2 ssss

by

01 sKGp

Step 3:

Draw Root-Locus Diagram jsjsss

sKGop

2210

3


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# of branches =4=#of poles

Asymptote:

3

11ReRe

mn

zeropole

3003

1802

6011360180

l

l

l

mn

l

Departure Angle ?

5.25

1801

1

3

1

d

i

i

i

id

250.5 45


4/17



03504514

035410

035410

234

22

2

KsKsss

KKsssss

sKssss

1

2

3

4

s

s

s

s

51

41

31

14

1

Step 4:

Find the point where the branches moves to R.H.P

(i.e. system becomes unstable)We start with the C/C equation:

0

3

50

45

42

32

K

K 0

3K

KK

KK

KK

K

303

31450

314

01314

14

5014514

41

3141

51

31

31

41

32

31



003 KK

Hurwitz Criteria:

02900058

04250

14

5800

31450

5800580014

504514

2

31

3141

31

KK

KKKKK

KKK

Routh criteria:


5/17



07.2017.143029000582 KKKK

7.20107.201

7.14307.143

KK

KK

7.20107.201

7.14307.143

KK

KK

Two Cases are

Possible for the

Inequality to hold

OR

Violates 0K Therefore,

Cant exist

003051 KK



7.1430 K

01.4317.1934514

07.14337.143504514

234

234

ssss

ssss

7.143K

Putting all of the conditions underlined above together gives:

At

78.2

22.11

72.30

4

3

2,1

s

s

js

C/C equation:

72.3js

We solve the poly.

Using an equation

Solver such as roots

In MatLab.

C/C roots

or

Poles

The system crosses the imaginary axis at


6/17



Example:

An important element of an Intelligent Highway System (IHS) is controlling

the spacing between vehicles on a guideway. Assuming the dynamics of an

automated guiding system can be described by;

r

VK

+_ 8

12 ss

y

5s

Desired

spacing

Actual spacing

Between vehicles

Estimate the R.L. Diagram for the systemSolution:

Form suitable for R.L. diagram 8

52

ss

sK



270,902

180

5.12

58

Departure angle from real axis90

180

q

No. of branches departing from the pt. (since,2poles 2 branches )


7/17



Example:

Consider the following servo-mechanism with a PD controller:

r

VK

+

_ 11

ss

y

1sK

Kr

A. Put the C/C equation in a form suitable for R.L.

C.L.T.F:

KsKss

K

K

KsK

ss

K

ss

K

sFrr

1

11

1

C/C equation.



rK

opG

0

11

Kss

sKr

C/C equation :

0

1

11

sKssK

r

01 KsKss r

If is the control variable:

KIf is the control variable:

B. Estimate the R.L. Diagrams

Control Variable: 01110111 rr KssK

sKssKK


8/17



2,270,90

12

1

2

010

mn

KK

rr

K

Zeros : None

2 poles : 2 branches

Asymptote :

System initially over damped, then it becomes under damped.

rK Poles move further apart from each other, and it takes largerValues of K to make system underdamped.



variable?controltheisifWhat rK


9/17



Selected Illustrative Root Loci(P control)

r

V

+_ 2

1

s

ypk

Double integrator TF:Ex: The control ofAttitude of a satellite

The root locus with respect to controller gain is: 01

12

skp

Rule 1. The locus has two branches that start at s= 0

Rule 2. There are no parts of the locus on the real axis.

Rule 3. The two asymptotes have origin at s= 0 and are

at the angles of +/- 90

Rule 4. The loci depart from s= 0 at the angles of +/- 90.

Rule 5. The loci remain on the imaginary axis for all values of kp

Hence the transient would be oscillatory for any value of kp



Root Locus with PD Control

01

1

formlocusrootin theresultswhich,1/asratiogainselect theyarbitraril

momentfor theand,defineweform,locusrootinequationput theTo

01

1

:iscontrolPDithequation wsticcharacteriThe

2

2

s

sK

kk

kK

sskk

Dp

D

Dp

The addition of the zero has pulled the locus into the

left half-plane, a point of general importance in

constructing a compensation


10/17



Lead Compensator

The physical operation of differentiation is not practical and in practice

PD control is approximated by

01

011

iscontrollerthisplant with/1for theequationisticchararcterThe

thatso/

anddefiningbyformlocusrootinputbecanwhich

1

2

2

psszsK

sKLsGsD

s

ps

zsKsD

Kpkz

pkkK

ps

skksD

p

Dp

Dp



Lead Compensation Example

12,1 pz4,1 pz

9,1 pz

An additional pole movingin from the far left tends to

push the locus branches to

the right as it approaches a

given locus


11/17



Design Using Dynamic Compensation

Lead compensation approximates the function of PD control and

acts mainly to speed up a response by lowering rise time anddecreasing the transient overshoot.

Lag compensation approximates the function of PI control and is

usually used to improve the steady-state accuracy of the system.

pzpz

ps

zsKsD

ifoncompensatilagandifoncompensatileadcalledis

formtheoffunctiontransferaon withCompensati



Design Using Dynamic Compensation

Lead compensation approximates the function of PD control and

acts mainly to speed up a response by lowering rise time and

decreasing the transient overshoot.

Lag compensation approximates the function of PI control and is

usually used to improve the steady-state accuracy of the system.

pzpz

ps

zsKsD

ifoncompensatilagandifoncompensatileadcalledis

formtheoffunctiontransferaon withCompensati


12/17



Design Using Lead Compensation

If we apply this compensation to a second-order position control

system with normalized TF

1

1

sssG

linesdashed2controlPDwithand

linessolidoncompensatiwith

,01

forlociRoot

sKsD

KsD

sGsD



Design Using Lead Compensation Selecting exact values of zand pis usually done by trial and error.

In general, the zero is placed in the neighborhood of the closed-loop

wn , as determined by rise-time or settling time requirements, and the

pole is located at a distance 5 to 20 times the value of the zero

location.

The choice of the exact pole location is a compromise between the

conflicting effects of noise suppression, for which one wants a small

value for p, and compensation effectiveness for which one wants a

large p.

pz

js

pz

ps

zsKsD

11-tantan

bygivenisatT.Fthisofphasetheexample,Forlead.phaseimpartTFsthese

signals,sinusoidaltofact thattheofreflectionaisitsinceoncompensatileadcalledisitthenif

For


13/17



Selection of the zero and pole of a Lead Compensator

2(c)

102(b)

202(a)

:11,with

casesfor threelociRoot

ssD

sssD

sssD

sssG



Example

10

2

first trywillWets.requiremenesatisfy thwill2.725.0

8.1of

frequencynaturalaand0.5ofratiodampingathatestimateWe

sec.0.25thanmorenooftimeriseand20%thanmorenoof

overshootprovidethat will11foroncompensatiaFind

s

sKsD

sssG

n

Solution


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Example 2

ions.specificatmeet theseon tocompensatileadaDesign

20.nlonger thanobepoleleadthat therequiretsrequiremennsuppressionoiseThe

.35.35.3atpoleahavetosystemloop-closedtherequireweSuppose 0 jr

The root-locus angle condition will be satisfied if the angle from the lead zero is 72.6. The location of the zero

is found to be z= -5.4 at a gain of 127. Thus 20

4.5127s

ssD



Step Response for Example 2

Remember it is not really a second-order system !! Can use RLTOOL in Matlab


15/17



Step response

To achieve better damping in order to reduce the overshoot in the transient response,

move the pole of the lead compensator more to the left in order to pull the locus inthat direction, and selecting K= 91.

Step response for K= 91 and L(s) =

13

291

s

ssD



Design Using Lag Compensation

Once satisfactory dynamic response has been obtained, we may discover that

the low-frequency gain the value of the relevant steady-state error constant,

such as kv is still too low.

As we saw, the system type, which determined the degree of the polynomial

the system is capable of following, is determined by the order of the pole of the

TF.D(s)G(s) at s = 0

If the system is type 1, the velocity-error constant, which determines the

magnitude of the error to a ramp input, is given by

In order to increase this constant, it is necessary to do so in a way that does not

upset the already satisfactory dynamic response.

sGssDs 0lim


16/17



Thus, we want an expression forD(s) that will yield a significant gain at s = 0

to raise kv (or some other steady-state error constant) but is nearly unity (no

effect) at the higher frequency n, where the dynamic response is determined.

The result is

boosting)requires

gainstate-steadywhich theextent toon thedependingvalue(the10to3

0yet,withcomparedsmallareandofvaluesthewhere

,

pzDpz

pzps

zssD

n



Example

.01.005.0Thus

.7arounddynamicsdominantthengrepresentilocustheofportionson the

effectlittlehavewouldthatsosmallveryandbothofvaluesthekeepswhich

,05.0atzeroaand01.0atpoleawithedaccomplishbecanThis

5.offactorabyconstantvelocitytheincreaseorder toin5with

oncompensatilagarequirewes,obtain thiTo.70thatrequireweSuppose

.1413

291

1

1

13

291lim

lim

isconstantvelocitytheThus

.13291oncompensatileadtheincludingand1

1For

2

2

0

10

1

sssD

sDpz

zp

pz

K

sss

ss

GsKDK

sssKDss

sG

n

v

s

sv


17/17



Example continued

Root locus with both lead and lag compensations: (a) whole locus; (b) portion of part (a) expanded

to show the root locus near the lag compensation.

control lect9

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