8/3/2019 Control Lect9
1/17
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
Control Engineering
Root Locus Analysis II
Dr. Ayman A. El-Badawy
Department of Mechatronics Engineering
Faculty of Engineering and Material Science
German University in Cairo
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
8/3/2019 Control Lect9
2/17
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
8/3/2019 Control Lect9
3/17
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
# 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
8/3/2019 Control Lect9
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Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
003 KK
Hurwitz Criteria:
02900058
04250
14
5800
31450
5800580014
504514
2
31
3141
31
KK
KKKKK
KKK
Routh criteria:
8/3/2019 Control Lect9
5/17
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
8/3/2019 Control Lect9
6/17
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
270,902
180
5.12
58
Departure angle from real axis90
180
q
No. of branches departing from the pt. (since,2poles 2 branches )
8/3/2019 Control Lect9
7/17
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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.
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
8/3/2019 Control Lect9
8/17
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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.
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
variable?controltheisifWhat rK
8/3/2019 Control Lect9
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Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
8/3/2019 Control Lect9
10/17
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
8/3/2019 Control Lect9
11/17
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
8/3/2019 Control Lect9
12/17
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
8/3/2019 Control Lect9
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Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
Selection of the zero and pole of a Lead Compensator
2(c)
102(b)
202(a)
:11,with
casesfor threelociRoot
ssD
sssD
sssD
sssG
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
Example
10
2
first trywillWets.requiremenesatisfy thwill2.725.0
8.1of
frequencynaturalaand0.5ofratiodampingathatestimateWe
sec.0.25thanmorenooftimeriseand20%thanmorenoof
overshootprovidethat will11foroncompensatiaFind
s
sKsD
sssG
n
Solution
8/3/2019 Control Lect9
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Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
Step Response for Example 2
Remember it is not really a second-order system !! Can use RLTOOL in Matlab
8/3/2019 Control Lect9
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Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
8/3/2019 Control Lect9
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Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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
8/3/2019 Control Lect9
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Dr. Ayman A. El-Badawy
Department of Mechatronics EngineeringFaculty of Engineering and Material Science
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.
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