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LINEAR CONTROL
SYSTEMS
Ali Karimpour
Associate Professor
Ferdowsi University of Mashhad
Lecture 12
Dr. Ali Karimpour Feb 2013
2
Lecture 12
Topics to be covered include:
v Different region of S plane.
u Constant natural frequency loci.
u Constant natural damped frequency loci.
u Constant damping factor loci.
u Constant damping ratio loci.
v Transient response of a position control system.
Time domain analysis of control systems
Lecture 12
Dr. Ali Karimpour Feb 2013
3
Introducing a prototype second order system.
2هعزفی یک سیستن نوونه درجه
+
-
c e r
)2(
2
n
n
ss
c r 22
2
2 nn
n
ss
10 if-1- :are Poles 2 dnn jj
frequency Naturaln فرکانس طبیعی
ratio Damping نسبت میرائی
frequency damped Natural1 2 nd
فرکانس طبیعی میرا شده
factor Dampingn ضریب میرائی
n-
21 nj
d
n
j
j
21
n
θ
Lecture 12
Dr. Ali Karimpour Feb 2013
4
v Constant natural frequency loci.
Sنواحی هختلف در صفحه
Different region of S plane.
n
v Constant natural damped frequency loci.
d
n
j
j
21
v Constant damping factor loci.
n-
v Constant damping ratio loci.
cos
Lecture 12
Dr. Ali Karimpour Feb 2013
5
هکاى فزکانس طبیعی ثابت
Constant natural frequency loci.
1 2
2 n
1 n
Lecture 12
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6
هکاى فزکانس طبیعی هیزا شذه ثابت
Constant natural damped frequency loci.
-1
1
-2
2
1d
2d
Lecture 12
Dr. Ali Karimpour Feb 2013
7
Constant damping factor loci. هکاى ضزیب هیزائی ثابت
1 n2 n
Lecture 12
Dr. Ali Karimpour Feb 2013
8
Constant damping ratio loci. هکاى نسبت هیزائی ثابت
45
707.0
60
5.0
30
866.0
Lecture 12
Dr. Ali Karimpour Feb 2013
9
Example 1: Find the loci of a system where damping
ratio is greater then 0.707 and damping factor greater
than 2
707.0
45
2 n
Lecture 12
Dr. Ali Karimpour Feb 2013
10
Example 2: Find the loci that the percent overshoot
is less than 16% and settling time is less than 1 sec
(according to 5% bound).
6.0
60
5.0%16.. OP
605.0
2.3
2.31
2.3
ns
n
s
t
t
2.3 n
Lecture 12
Dr. Ali Karimpour Feb 2013
11
Effect of roots loci on step response تاثیز هکاى ریشه ها بز پاسخ پله
-5 -2.5 2.5
10
5
Conjugate root is not shown ریشه هشدوج نشاى داده نشذه است
Lecture 12
Dr. Ali Karimpour Feb 2013
12
Example 3: Find the poles of the following systems and
its corresponding step response for k=75, 100, 200 and
1000
+
-
c e r
)20( ss
k
kss
k
sr
scsM
20)(
)()(
2
75k 15,5 21 pp
100k 10,10 21 pp
200k jpjp 1010,1010 21
1000k jpjp 3010,3010 21
Lecture 12
Dr. Ali Karimpour Feb 2013
13
+
- c e r
)20( ss
k
kss
k
sr
scsM
20)(
)()(
2
k=75 k=100
k=200
k=1000
0 0.2 0.4 0.6 0.8 1 1.20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
(sec)
0 0.5 1 1.50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
(sec)
0 0.1 0.2 0.3 0.4 0.5 0.60
0.2
0.4
0.6
0.8
1
1.2
1.4
(sec)
0 0.1 0.2 0.3 0.4 0.5 0.60
0.2
0.4
0.6
0.8
1
1.2
1.4
(sec)
0 0.5 1 1.50
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Am
plit
ude
k=75
k=100
k=200
k=1000
Lecture 12
Dr. Ali Karimpour Feb 2013
14
Dynamics of electromechanical system
دیناهیک سیستوهای الکتزوهکانیکی
Lecture 12
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15
A simplified aeroplane (position control system)
(سیستن کنتزل هوقعیت)یک هواپیوای ساده شذه
Position control
Lecture 12
Dr. Ali Karimpour Feb 2013
16
Block diagram of a position control system
بلوک دیاگزام یک سیستن کنتزل هوقعیت
Load r
c
cGear
DC Motor
Power Amplifier
with current
feedback
Ks
Tacometer
Lecture 12
Dr. Ali Karimpour Feb 2013
17
Block diagram of a position control system
بلوک دیاگزام یک سیستن کنتزل هوقعیت
tt sJB
1 m
ikaI
s
1 cN
l
bk
sk
r
1k
rI
2k
k
tk
Load
c
Gear
DC Motor
Power Amplifier
with current
feedback
Ks
Tacometer
t V
a a sL R
1
bk
Lecture 12
Dr. Ali Karimpour Feb 2013
18
Let ignore the velocity feedback.
.اس فیذبک سزعت اغواض کنیذ
tt sJB
1 m
ikaI
s
1 mN
c
bk
t V
sk
r
1k
rI
2k
k
a a sL R
1
tt sJB
1 m
ikaI
s
1 mN
c
bk
t V
sk
r
1k
rI
2k
k
tk
aa sLR
1
Lecture 12
Dr. Ali Karimpour Feb 2013
19
Simplification ساده ساسی
tt sJB
1 m
ikaI
s
1 mN
c
bk
t V
sk
r
1k
rI
2k
k
a a sL R
1
c
)(sG
r
))((1
))(()(
21
1
ttaa
bi
aa
ttaa
is
sJBsLR
kk
sLR
kk
sJBsLRs
Nkkkk
sG
bittttaa
is
kksJBkksJBsLRs
Nkkkk
)())(( 21
1
Lecture 12
Dr. Ali Karimpour Feb 2013
20
Simplification
ساده ساسی و عذد گذاری
c
)(sG
r
bittttaa
is
kksJBkksJBsLRs
NkkkksG
)())(()(
21
1
Let we ignore La با اغواض اسLa دارین
ta
bitta
ta
is
JkkR
kkBkkBRss
JkkR
Nkkkk
sG
)(
)()(
~
21
21
21
1
Lecture 12
Dr. Ali Karimpour Feb 2013
21
Simplification
ساده ساسی و عذد گذاری
c
)(sG
r
ta
bitta
ta
is
bittttaa
is
JkkR
kkBkkBRss
JkkR
Nkkkk
sG
kksJBkksJBsLRs
NkkkksG
)(
)()(
~
)())(()(
21
21
21
1
21
1
0001.001.0
005.011.003.05
0636.0195.010 21
ml
mlaa
bsi
JJ
BBNLR
kkkkk
0002.0100
01.00001.0
015.0100
1005.0
2
2
lmt
lmt
JNJJ
BNBB
Lecture 12
Dr. Ali Karimpour Feb 2013
22
Simplification ساده ساسی و عذد گذاری
c
)(sG
r
2.361
4500)(
~
)3008)(3.400(
105.1
12040003.3408
105.1)(
7
2
7
ss
ksG
sss
k
sss
ksG
0001.001.010005.0
1.0003.050636.0
195.010 21
mllm
aab
si
JJBB
NLRk
kkkk
0002.0100
01.00001.0
015.0100
1005.0
2
2
lmt
lmt
JNJJ
BNBB
ta
bitta
ta
is
bittttaa
is
JkkR
kkBkkBRss
JkkR
Nkkkk
sG
kksJBkksJBsLRs
NkkkksG
)(
)()(
~
)())(()(
21
21
21
1
21
1
Lecture 12
Dr. Ali Karimpour Feb 2013
23
Stability بزرسی پایذاری c
)(sG
r
2.361
4500)(
~
ss
ksG
)3008)(3.400(
105.1)(
7
sss
ksG
ksss
k
sG
sGsM
7
7
105.1)3008)(3.400(
105.1
)(1
)()(
0105.112040003.3408 723 ksss
ks
s
72
3
105.13.3408
12040001
ks
ks
70
7
105.1
03.3408
105.14100593200
Stable for 0<k<273.57
kss
ksM
45002.361
4500)(
2
045002.3612 kss
Stable for all k>0
Lecture 12
Dr. Ali Karimpour Feb 2013
24
Analysis تجشیه تحلیل c
)(sG
r
2.361
4500)(
~
ss
ksG kk nn 5304500
2
kkn
692.2
560
2.3612.3612
sn tOPppk ..21
dampedover is System81.12.30331222.2
damping critically is System16.1806.18025.7
unstable is System233842
unstable is System03610
022.0%3.4255707.06.1806.1805.14 j
022.0%601110163.010966.180274 j
045002.3612 kss
Lecture 12
Dr. Ali Karimpour Feb 2013
25
Step response پاسخ پله c
)(sG
r
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6Step Response
Time (sec)
Am
plit
ude
222.2k
25.7k
5.14k
274k
Lecture 12
Dr. Ali Karimpour Feb 2013
26
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
0.02
0.04
0.06
0.08
0.1
0.12Step Response
Time (sec)
Am
plit
ude
Velocity response c پاسخ شیب
)(sG
r
2.361
4500)(
~
ss
ksG
222.2k
25.7k
5.14k
274k
222.2
68.27
036.0
25.7
3.90
011.0
5.14
6.180
005.0
274
3414
00029.0
k
vk
sse
Lecture 12
Dr. Ali Karimpour Feb 2013
27
Analysis تجشیه تحلیل c
)(sG
r
..321 OPpppk n
unstable is System2342830042
0105.112040003.3408 723 ksss
)3008)(3.400(
105.1)(
7
sss
ksG
unstable is System040030080
dampedover is System303663012222.2
dampedover is System1562303022250.7
%6.42677.019218630355.14 j
%5968717.06771153178100 j
unstable is System10982.03408274 j
Lecture 12
Dr. Ali Karimpour Feb 2013
28
Step response پاسخ پله c
)(sG
r
)3008)(3.400(
105.1)(
7
sss
ksG
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6Step Response
Time (sec)
Am
plit
ude
5.14k 222.2k
25.7k
100k
Lecture 12
Dr. Ali Karimpour Feb 2013
29
Step response comparison هقایسه پاسخ پله
c
)(sG
r
)3008)(3.400(
105.1)(
7
sss
ksG
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1Step Response
Time (sec)
Am
plit
ude 22.2k
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1Step Response
Time (sec)
Am
plit
ude
22.2k
25.7k
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20
0.2
0.4
0.6
0.8
1
1.2
1.4Step Response
Time (sec)
Am
plit
ude
22.2k
25.7k
5.14k
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6Step Response
(sec)
22.2k
25.7k
5.14k
100k
)2.361(
4500)(
~
ss
ksG
Lecture 12
Dr. Ali Karimpour Feb 2013
30
Exercises توزینها
1- Find the roots of following system for -301<k<301 and show
them on the s plane.
2- Find the roots of following system for -301<k<301 and show
them on the s plane.
)3008)(3.400(
105.1)(
7
sss
ksG
)2.361(
4500)(
ss
ksG
Let k=-300,-280,-260,……….260,280,300
Let k=-300,-280,-260,……….260,280,300
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