automatic control systems - ???????? -...
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AUTOMATIC AUTOMATIC CONTROLCONTROLSYSTEMSSYSTEMS
Ali KarimpourAssociate Professor
Ferdowsi University of Mashhad
Lecture 8
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Lecture 8
Time domain analysis of control systemsTopics to be covered include: Time domain analysis. Transient response of a prototype second order system Different region of S plane. Transient response of a position control system.
Lecture 8
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Introducing a prototype second order system.2معرفی یک سیستم نمونه درجه
+-
cer)2(
2
n
n
ss
cr22
2
2 nn
n
ss
)(2
)( 22
2
sRss
sCnn
n
)2()( 22
2
nn
n
ssssC
Step response
10 if1- :are Poles 2 nn j
222
222 1)(1)(
1)(
nn
n
nn
n
ss
ss
sC
)1sin(
111)()( 2
2
tetutc n
tn 1cos
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Step response پاسخ پله
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Step Response
Time (sec)
Ampl
itude
13 n 8.0,13 n
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Step Response
Time (sec)
Ampl
itude
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Step Response
Time (sec)
Ampl
itude
6.0,8.0,13 n
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Step Response
Time (sec)
Ampl
itude
4.0,6.0,8.0,13 n
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Step Response
Time (sec)
Ampl
itude
2.0,4.0,6.0,8.0,13 n
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Step Response
Time (sec)
Ampl
itude
0,2.0,4.0,6.0,8.0,13 n
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Step response پاسخ پله13.0 n
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Step Response
Time (sec)
Ampl
itude
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Step Response
Time (sec)
Ampl
itude
2,13.0 n
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Step Response
Time (sec)
Ampl
itude
3,2,13.0 n
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Step Response
Time (sec)
Ampl
itude
4,3,2,13.0 n
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Step Response
Time (sec)
Ampl
itude
28.6,4,3,2,13.0 n
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Rise time, Settling time, Overshoot, Under shoot and Peak time زمان صعود، زمان نشست، فراجهش، فروجهش و زمان پیک
Lecture 8
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Introducing a prototype second order system. 2معرفی یک سیستم نمونه درجه
+-
cer)2(
2
n
n
ss
cr22
2
2 nn
n
ss
10 if-1- :are Poles 2 dnnn jj
n-
21 nj
d
n
jj
21
n
frequency Naturaln فرکانس طبیعی
ratio Damping نسبت میرائیfrequency damped Natural1 2 nd
فرکانس طبیعی میرا شده
factor Dampingn ضریب میرائی
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Peak time لحظه پیک
ntn 21
21
n
pt
21
n
nt
Let n=1
n=1
n=3
n=4
n=5
n=2
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Percent Overshootدرصد فراجهش
21100%..
eOP ..OP0 100%
0.100 73%0.200 53%0.300 37%0.400 25%0.500 16%0.707 4.3%
1 0%0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
10
20
30
40
50
60
70
80
90
100
Per
cent
Ove
rsho
ot
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Rise time زمان صعود
rt
How can we findrise time ?
چگونه زمان صعودرا بیابیم؟
yktc r)(
ytc 9.0)(Let
nrt
5.28.0 10917.24167.01 2
nrt
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Settling time زمان نشست
st
How can we findsettling time ?
چگونه زمان نشسترا بیابیم؟
ytcy )(
%5for 2.3
nst %2for 4
n
st 7.00
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Constant natural frequency loci.
Sنواحی مختلف در صفحه Different region of S plane.
n
Constant natural damped frequency loci.
d
n
jj
21
Constant damping factor loci.
n-
Constant damping ratio loci.
cos
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مکان فرکانس طبیعی ثابتConstant natural frequency loci.
12
2 n1 n
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مکان فرکانس طبیعی میرا شده ثابتConstant natural damped frequency loci.
-1
1
-2
2
1d
2d
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Constant damping factor loci.مکان ضریب میرائی ثابت
1 n2 n
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Constant damping ratio loci.مکان نسبت میرائی ثابت
45707.0
605.0
30866.0
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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
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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
ns
t
t
2.3 n
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Effect of roots loci on step responseتاثیر مکان ریشه ها بر پاسخ پله
-5 -2.5 2.5
10
5
Conjugate root is not shown ریشه مزدوج نشان داده نشده است
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Example 3: Find the poles of the following systems and its corresponding step response for k=75, 100, 200 and 1000
+-
cer)20( ss
k
kssk
srscsM
20)()()( 2
75k 15,5 21 pp
100k 10,10 21 pp
200k jpjp 1010,1010 21
1000k jpjp 3010,3010 21
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+-
cer)20( ss
k
kssk
srscsM
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)
Ampl
itude
k=75k=100k=200k=1000
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Dynamics of electromechanical systemدینامیک سیستمهاي الکترومکانیکی
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A simplified aeroplane (position control system))سیستم کنترل موقعیت(یک هواپیماي ساده شده
Position control
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Block diagram of a position control systemبلوك دیاگرام یک سیستم کنترل موقعیت
Loadr
c
cGearDC MotorPower Amplifier
with current feedback
Ks
Tacometer
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Stability بررسی پایداريc
)(sG
r
2.3614500)(~
ss
ksG)3008)(3.400(105.1)(
7
sssksG
ksssk
sGsGsM 7
7
105.1)3008)(3.400(105.1
)(1)()(
0105.112040003.3408 723 ksss
kss
72
3
105.13.340812040001
ks
ks
70
7
105.1
03.3408
105.14100593200
Stable for 0<k<273.57
kssksM
45002.3614500)( 2
045002.3612 kss
Stable for all k>0
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Analysis تجزیه تحلیلc
)(sG
r
2.3614500)(~
ss
ksG kk nn 53045002
kkn692.2
5602.3612.3612
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
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Step response پاسخ پلهc
)(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)
Ampl
itude
222.2k25.7k
5.14k
274k
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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)
Ampl
itude
Velocity response cپاسخ شیب
)(sGr
2.3614500)(~
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
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Analysis تجزیه تحلیلc
)(sG
r
..321 OPpppk n
unstable is System2342830042
0105.112040003.3408 723 ksss
)3008)(3.400(105.1)(
7
sssksG
unstable is System040030080
dampedover is System303663012222.2
dampedover is System1562303022250.7
%6.42677.019218630355.14 j
%5968717.06771153178100 j
unstable is System10982.03408274 j
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Step response پاسخ پلهc
)(sG
r
)3008)(3.400(105.1)(
7
sssksG
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)
Ampl
itude
5.14k 222.2k
25.7k
100k
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Step response comparison مقایسه پاسخ پله
c
)(sGr
)3008)(3.400(105.1)(
7
sssksG
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)
Ampl
itude 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)
Ampl
itude
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)
Ampl
itude
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
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Real dominant poles of transfer function.قطبهاي غالب حقیقی در تابع انتقال
Region of insignificant poles
Region of dominant
poles UnstableRegion
dD
What about D? D > 5 or 10 times of d.
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Complex dominant poles of transfer function.قطبهاي غالب مختلط در تابع انتقال
Region of insignificant poles
Region of dominant
poles UnstableRegion
D
What about D? D > 5 or 10 times of d.
d
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Design procedure. روش طراحی
Region of insignificant poles
Region of dominant
poles
D
45
45
For design purposes, such as in the pole placement design we try to put poles on: . در طراحی ها، مثلا در جاگذاري قطب سعی می کنیم قطبها در مکان زیر قرار گیرند
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Example 4: What are the dominant poles of the following transfer function.
.قطبهاي غالب در تابع انتقال زیر را بیابید: 4مثال
)16)(2(32)(
sssM
16143.0
2143.11)(
ssssC
)16)(2(32)()()(
ssssRsMsC
Step response
)()143.0143.11()( 162 tueetc tt
dominant
insignificant
insignificant
0
dominant
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Example 5: Simplify the system in example 1.
.را ساده کنید 1سیستم مثال : 5مثال
)2)(16(32)(~
sssM
)2(32)()()(~
ss
sRsMsC)2(
1616
ss
)()1616()(~ 2 tuetc t
Step response
)2(2)(~
ss
sC
)()11()(~ 2 tuetc t
)2(32
s
Step response)2)(116/(
16/32)2)(16(
32)(~
ssss
sM
)2)(16(32)(
sssM )()143.0143.11()( 162 tueetc tt
22
)2)(116/(16/32
)2)(16(32)(~
ssssssM
211
)2(2)(~
sssssC
غ ق ق
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Example 6: What are the dominant poles of the following transfer function.
.قطبهاي غالب در تابع انتقال زیر را بیابید: 6مثال
)2)(16()1.2(24.15)(
ss
ssM
169457.0
20544.01)(
ssssC
)2)(16()1.2(24.15)()()(
sss
ssRsMsC
Step response
)()9457.00544.01()( 162 tueetc tt
insignificant
0
dominant
The zero near to pole makes it insignificant
dominant insignificant
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Example 7: Discuss the dominant poles of following system.
.در مورد قطبهاي غالب سیستم زیر بحث کنید: 7مثال
)40010)((400)( 2
ssas
asM
Simplification
If (a-5) > 5 ∙ 20=100
dominant-5-a
20
)40010)(1/(400)(~2
ssas
sM
)40010(400)(~
2
sssM for a > 105
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Example 8: Compare the step response of M and its approximation for different values of k.
.بدست آورید aو تقریب آن را براي مقادیر مختلف Mپاسخ پله : 8مثال
)40010)((400)( 2
ssas
asM)40010(
400)(~2
ss
sM
step(400,[1 10 400]);hold on
a=10;step(400*a,conv([1 a],[1 10 400])
a=20;step(400*a,conv([1 a],[1 10 400]))
a=50;step(400*a,conv([1 a],[1 10 400]))
a=100;step(400*a,conv([1 a],[1 10 400]))0 0.2 0.4 0.6 0.8 1 1.2
0
0.5
1
1.5
Step Response
Time (sec)
Ampl
itude
Second orderThird order a=10Third order a=20Third order a=50Third order a=100
2nd order
a=10
a=20
a=50
a=100
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Exercises تمرینها
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
sssksG
)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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Exercises تمرینها
2 Derive a suitable second order system for following system.
1 Consider following system. Find the dominant poles and insignificant poles of system.
)3)(2)(16(96)(
ssssM
)3)(2)(16(96)(
ssssM
3 Compare the step response of the system in problem 2 and step response of its second order approximation.
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Exercises تمرینها
5 Derive a suitable second order system for following system.
4 Consider following system. Find the dominant poles and insignificant poles of system.
)1)(3)(20)(160()05.3(3150)(
ssss
ssM
6 Compare the step response of the system in problem 5 and step response of its second order approximation.
)1)(3)(20)(160()05.3(3150)(
ssss
ssM