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LINEAR CONTROLLINEAR CONTROL SYSTEMS SYSTEMS
Ali Karimpour
Associate Professor
Ferdowsi University of Mashhad
Lecture 16
Dr. Ali Karimpour May 2013
2
Lecture 16
Topics to be covered include:
Design of controller in time domain. Various controller configurations.Different kind of controllers. Controller realization.
Time domain design of the PID controllers. Design of PID controllers. Design of PD controllers. Design of PI controllers.
Time domain design of control systems
Lecture 16
Dr. Ali Karimpour May 2013
3
Various controller configurations.ساختارهای کنترلی
متفاوت
)(sGp)(sGc
+
-
r(t) e(t) u(t) c(t)1 Series or cascade compensation. ساختار کنترلی سری1
2 State-feedback control. کنترل فیدبک حالت2
K
+
-
r(t) u(t) x(t))(sGp
c(t)D
3 Forward compensation with series compensation. (Two degree of freedom) جبران سازی پیش رو با جبران سازی سری )دو درجه آزادی(3
+
-
r(t) e(t) u(t) c(t))(sGp
)(1 sGc)(2 sGc
Lecture 16
Dr. Ali Karimpour May 2013
44
Various controller configurations.ساختارهای کنترلی
متفاوت
4 Feed forward compensation.(Two degree of freedom)
کنترلر پیش خور 4)دو درجه آزادی(
)(2 sGc
+r(t) e(t) u(t) c(t))(sGp+
+)(1 sGc
-
Controller
Controller
Controlledprocess
Lecture 16
Dr. Ali Karimpour May 20135
Series Compensation Structure
In this Course we consider the series or cascade compensation.
جبران سازی سری
Controller
In This course
Different kind of
controllers
PID controllers.
Lead lag controllers
H2 Controllers
H∞ Controllers
Intelligent Controllers
………….. Adaptive Controllers
NN Controllers…………..
In Graduate courses
Lecture 16
Dr. Ali Karimpour May 2013
66
PID Controllers
PID has become almost universally used in industrial control.
These controllers have proven to be robust and extremely beneficial in the control of many important applications.
PID stands for: P (Proportional)
I (Integral)
D (Derivative)
The standard form PID are:
An Alternative form for PID
کنترلر PID
Proportional only:
Proportional plus Integral:
Proportional plus derivative:
Proportional, integral and derivative:
sKs
KKsG d
ipPID )(
s
KKsG i
pPI )(
sKKG dpPD
pP KsG )(
1sK
sK
d
d
11)(
sK
sK
s
KKsG
d
dipPIDseries
Lecture 16
Dr. Ali Karimpour May 2013
77
Lead-lag Compensators
Closely related to PID control is the idea of lead-lag compensation. The transfer function of these compensators is of the form:
If a<1 , then this is a lag network. Or (z>p) in other form.
If a>1 this is a lead network. Or (z<p) in other form.
کنترلر پیش فاز پس فاز
1
1)(
s
saksG
ps
zsksG
)( or
Lecture 16
Dr. Ali Karimpour May 2013
PID and Operational Amplifiers کنترلرPID و تقویت کننده عملیاتی
88)(
)(
)(
)()(
sZ
sZ
sV
sVsG
i
f
i
o
Lecture 16
Dr. Ali Karimpour May 2013
99
PID and Operational Amplifiers کنترلرPID و تقویت کننده عملیاتی
pin
f
in
out KR
R
sV
sVsG
)(
)()(
sKRCsCs
R
sV
sVsG D
in
out /1)(
)()(
sKRCsR
Cs
sV
sVsG i
in
out //1/1
)(
)()(
)(
)(
)(
)()(
sZ
sZ
sV
sVsG
i
f
i
o
Lecture 16
Dr. Ali Karimpour May 2013
1010
PID and Operational Amplifiers کنترلرPID و تقویت کننده عملیاتی
)(
)(
)(
)()(
sZ
sZ
sV
sVsG
i
f
i
o
sCR
sCR
sV
sVsG
i
o
11
22
11
1
)(
)()(
sCRsCRCR
CRCRsG 12
2121
1122 1)()(
Lecture 16
Dr. Ali Karimpour May 2013
11
0 5 10 15 20 25 300
1
2Step Response
Time (sec)
Ampli
tude
0 5 10 15 20 25 30-1
0
1Step Response
Time (sec)
Ampli
tude
0 5 10 15 20 25 30-1
-0.5
0
0.5Impulse Response
Time (sec)
Ampli
tude
)(tc
)(te
)(te
Effects of the PD control on the time response.
بر پاسخ زمانیPDتاثیر کنترلر
)2(
2
n
n
ss
)(sC
PK
sKD
+
+
+
-
)(sR )(sE
PD controller
Derivative part can improve the oscillation.
جمله مشتق می تواند رفتار گذرا را بهبود بخشد.
Lecture 16
Dr. Ali Karimpour May 2013
12
Effects of the PI control on the time response.
بر پاسخ زمانیPIتاثیر کنترلر
)2(
2
n
n
ss
)(sC
PK
s
K I
+
++ -
)(sR )(sE
PI controller
Loop transfer function without controller
)2()(
2
n
n
sssG
)2(
)()(2
n
nIPc sss
KsKsGsG
Loop transfer function with controller
PI controller can improve error by increases the type of system by one
Lecture 16
Dr. Ali Karimpour May 2013
13
Tuning of PID Controllers
Because of their widespread use in practice, we present below several methods for tuning PID controllers. Actually these methods are quite old and date back to the 1950’s. Nonetheless, they remain in widespread use today.In particular, we will study.
Ziegler-Nichols Oscillation Method Ziegler-Nichols Reaction Curve Method Cohen-Coon Reaction Curve Method Time domain design Frequency domain design
PIDتنظیم کنترلرهای
Lecture 16
Dr. Ali Karimpour May 2013
14
Ziegler-Nichols Design
طراحی زیگلر نیکولز
This procedure is only valid for open loop stable plants. Open-Loop Tuning
Closed-Loop Tuning
According to Ziegler and Nichols, the open-loop transfer function of a system can be approximated with time delay and single-order system, i.e.
where TD is the system time delay and T1 is the time constant.
Lecture 16
Dr. Ali Karimpour May 2013
15
Ziegler-Nichols Reaction Curve Method(Open-Loop Case)
For open-loop tuning, we first find the plant parameters by applying a step input to the open-loop system.
The plant parameters K, TD and T1 are then found from the result of the step test as shown in Figure.
طراحی زیگلر نیکولز حالت حلقه باز
Lecture 16
Dr. Ali Karimpour May 2013
16
PIDKT
T19.0
Kp Ki Kd
2127.0
DKT
T
PIDDKT
T12.1
P
طراحی زیگلر نیکولز حالت حلقه باز
DKT
T1
216.0
DKT
T
K
T16.0
Ziegler-Nichols Reaction Curve Method(Open-Loop Case)
Lecture 16
Dr. Ali Karimpour May 2013
17
Numerical Example
Consider step response of an open-loop system as:
مثال عددی
s
esGTTCK
DsT
D 201
40)(sec20sec,5,40 :So 1
Lecture 16
Dr. Ali Karimpour May 2013
18
PI 09.09.0 1
DKT
T
Kp Ki Kd
0054.027.0
21
DKT
T
PID 12.02.1 1
DKT
T
P 1.01 DKT
T
012.06.0
21
DKT
T3.0
6.0 1 K
T
s
esGTTCK
DsT
D 201
40)(sec20sec,5,40 :So 1
1.0)( sKP
ssKPI
0054.009.0)(
ss
sKPID 3.0012.0
12.0)(
Numerical Exampleمثال عددی
Lecture 16
Dr. Ali Karimpour May 2013
19
Ziegler-Nichols Oscillation Method(Closed-loop)
This procedure is only valid for open loop stable plants and it is carried out through the following steps
Set the true plant under proportional control, with a very small gain.
Increase the gain until the loop starts oscillating. Note that linear oscillation is required and that it should be detected at the controller output.
Record the controller critical gain Kc and the oscillation period of the controller output, T.
Adjust the controller parameters according to Table
طراحی زیگلر نیکولز بروش نوسانی)حلقه بسته(
Lecture 16
Dr. Ali Karimpour May 2013
20
PI cK45.0
Kp Ki Kd
T
Kc54.0
PID T
Kc2.1 TKc075.0cK6.0
P cK5.0
Ziegler-Nichols Oscillation Method(Closed-loop) طراحی زیگلر نیکولز بروش نوسانی)حلقهبسته(
Lecture 16
Dr. Ali Karimpour May 2013
Consider a plant with a model given by
Find the parameters of a PID controller using the Z-N oscillation method. Obtain a graph of the response to a unit step input reference.
21
Numerical Exampleمثال عددی
Lecture 16
Dr. Ali Karimpour May 2013
22
Solution
Applying the procedure we find:
Kc = 8 and ωc = 3. T=3.62
Hence, from Table, we have
The closed loop response to a unit step in the reference at t = 0 is shown in the next figure.
حل
17.2075.065.22.18.46.0 TKKT
KKKK cd
cicp
Lecture 16
Dr. Ali Karimpour May 2013
23
Response to step reference
0 5 10 150
0.5
1
1.5Step response for PID control
Time (sec)
Am
plitu
de
پاسخ سیستم به پله
ss
sCPID 17.265.2
8.4)(
117.201.0
17.265.28.4)(
s
s
ssCPID
Lecture 16
Dr. Ali Karimpour May 2013
24
Time domain design طراحی حوزه زمانی
)25(
40
ss
)(sCk
+
-
)(sR )(sE
را بگونه ای تنظیم کرد که kدر سیستم زیر آیا می توان گردد؟0.707نسبت میرائی قطبهای مختلط سیستم
Is it possible to set the value of k such that the damping ratio of complex poles be 0.707?
-25 0
k
k
0k0k
45
45
?40 k25.12
8125.7 k
Yes
بله
25.12
1
Lecture 16
Dr. Ali Karimpour May 2013
25
Time domain design طراحی حوزه زمانی
)25(
40
ss
)(sCk
+
-
)(sR )(sE
را بگونه ای تنظیم کرد که ثابت kدر سیستم زیر آیا می توان گردد؟100خطای شیب معادل
Is it possible to set the value of k such that ramp error constant be 100?
100vk
100)25(
40lim
0
ss
ks
s
250040 k
5.62 k
Yes
بله
Lecture 16
Dr. Ali Karimpour May 2013
26
Time domain design طراحی حوزه زمانی
)25(
40
ss
)(sCk
+
-
)(sR )(sE
را بگونه ای تنظیم کرد که kدر سیستم زیر آیا می توان ثابت خطای و 0.707نسبت میرائی قطبهای مختلط سیستم
؟ گردد100شیب معادل
Is it possible to set the value of k such that the damping ratio of complex poles be 0.707 and ramp error constant be 100 ?
8125.7k
5.62k
Clearly the design is not possible
???!!!???Other controllers
Lecture 16
Dr. Ali Karimpour May 2013
27
نسبت در سیستم ضرایب کنترلر را بگونه ای تنظیم کنید که ثابت خطای شیب و 0.707میرائی قطبهای مختلط سیستم
؟ گردد100معادل
Determine the controller coefficient such that the damping ratio of complex poles be 0.707 and ramp error constant be 100 ?
)25(
40
ss
)(sCskk DP
+
-
)(sR )(sE
10025
40
)25(
40)(lim
0
P
sDPv
k
ssskksk 5.62 Pk
0250025
401
2
ss
skD
Tuning PD controller طراحی کنترلر PD
Lecture 16
Dr. Ali Karimpour May 2013
28
)25(
40
ss
)(sCskk DP
+
-
)(sR )(sE
0250025
401
2
ss
skD
Tuning kD by graphical method.
Tuning PD controller طراحی کنترلر PD
Dk
-12.5
48.4
0Dk
0Dk
45Dk40-50 -35
22 4.135.22 22 4.835.22 22 3535
1426.1Dk
Lecture 16
Dr. Ali Karimpour May 2013
29
)25(
40
ss
)(sCskk DP
+
-
)(sR )(sE
0250025
401
2
ss
skD
Tuning kD by mathematical method
0250025
401
2
ss
skD 02500)4025(2 sks D
02: withCompare 22 nnss 5025002 nn
1425.140
7.457.7040252 DDn kk
Tuning PD controller طراحی کنترلر PD
Lecture 16
Dr. Ali Karimpour May 2013
30
)25(
40
ss
)(sCskk DP
+
-
)(sR )(sE1425.15.62 DP kk
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.160
0.2
0.4
0.6
0.8
1
1.2
1.4Step Response
Time (sec)
Am
plitu
de
100Clearly vk
Tuning PD controller طراحی کنترلر PD
Why P.O. > 4.3%
25007.70
)5.621425.1(40
)(
)(2
ss
s
sR
sC
Lecture 16
Dr. Ali Karimpour May 2013
31
Tuning PI controller طراحی کنترلر PI
)25(
40
ss
)(sC
Pk
s
kI
+
+
+
-
)(sR )(sE
)25(
40
ss
)(sC
s
ksk IP +
-
)(sR )(sE
Clearly type of system is 2 so: vk
نسبت در سیستم ضرایب کنترلر را بگونه ای تنظیم کنید که ثابت خطای شیب و 0.707میرائی قطبهای مختلط سیستم
؟ گردد100معادل
Determine the controller coefficient such that the damping ratio of complex poles be 0.707 and ramp error constant be 100 ?
Lecture 16
Dr. Ali Karimpour May 2013
32
Tuning PI controller (Continue) )ادامه(PIطراحی کنترلر
)25(
40
ss
)(sC
s
ksk IP +
-
)(sR )(sE
نسبت میرائی قطبهای مختلط سیستم حال نیاز داریم که گردد.0.707
We now need damping ratio of complex poles be 0.707.
Root loci with proportional controller
k
k
0k0k 45
45
0)25(
401
sskP
0)25(
40)/(1
sss
kksk PI
P
5.12/Let PI kk
lociroot in changeimportant No -25 0
2/ PI kk
Pk4025.12 25.12
45.8Pk
22 5.125.10
25.12
Lecture 16
Dr. Ali Karimpour May 2013
33
Tuning PI controller (Continue) )ادامه(PIطراحی کنترلر
را تعیین کنیم. kPحال نیاز داریم که
We now need to set kP.
Root loci with PI controller
k
k
0k0k45
45
0)25(
40)2(1
sss
skP
-25 0
-10.9 -4.6-11.5
Lecture 16
Dr. Ali Karimpour May 2013
34
)25(
40
ss
)(sC
s
kk I
P +
-
)(sR )(sE9.16245.8 PIP kkk
vkClearly
Tuning PI controller طراحی کنترلر PI
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60
0.2
0.4
0.6
0.8
1
1.2
1.4Step Response
Time (sec)
Am
plitu
de
Why P.O. > 4.3%
67633825
)2(338
)(
)(23
sss
s
sR
sC
j5.123.11,37.2 :are Poles
Lecture 16
Dr. Ali Karimpour May 2013
35
Compare PI and PD controllers مقایسه کنترلرهایPI PDو
)25(
40
ss
)(sC
s
kk I
P +
-
)(sR )(sE
9.1645.8 IP kk
)25(
40
ss
)(sCskk DP
+
-
)(sR )(sE
1425.15.62 DP kk
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60
0.2
0.4
0.6
0.8
1
1.2
1.4Step Response
Time (sec)
Am
plitu
de
With PI controller
With PD controller
Lecture 16
Dr. Ali Karimpour May 2013
36
Exercises تمرینها
1- In the following system design a PID controller with Ziegler-Nichols Oscillation Method
)5.48(
40
ss
)(sC)(sGPID
+
-
)(sR )(sE
2- In the following system design a PID controller with Ziegler-Nichols Oscillation Method
)10)(5(
4
sss
)(sC)(sGPID
+
-
)(sR )(sE
Lecture 16
Dr. Ali Karimpour May 2013
37
Exercises تمرینها
3 In the following system design a PD controller such that that the damping ratio of complex poles be 0.6 and ramp error constant be 80.
)5.48(
40
ss
)(sC)(sGPID
+
-
)(sR )(sE
)10)(5(
4
sss
)(sC)(sGPID
+
-
)(sR )(sE
4 In the following system design a PD controller such that that the damping ratio of complex poles be 0.6 and ramp error constant be 80.
Lecture 16
Dr. Ali Karimpour May 2013
38
Exercises تمرینها
5 In the following system design a PI controller such that that the damping ratio of complex poles be 0.6 and ramp error constant be 80.
)5.48(
40
ss
)(sC)(sGPID
+
-
)(sR )(sE
)10)(5(
4
sss
)(sC)(sGPID
+
-
)(sR )(sE
6 In the following system design a PI controller such that that the damping ratio of complex poles be 0.6 and ramp error constant be 80.
Lecture 16
Dr. Ali Karimpour May 2013
39
Exercises تمرینها
Let the input impedance be generated by a resistor R2 be in series with a resistor R1 and a capacitor C1) that are in parallel, and let the feedback impedance be generated by a resistor Rf and a capacitor C f .a) Show that this choices lead to form a PID controller with high frequency gain limit as;
7 Consider following structure:
b) Derive the parameters in the controller.