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Lecture 7
Dr. Ali Karimpour Dec 2017
LINEAR CONTROL
SYSTEMS
Ali Karimpour
Associate Professor
Ferdowsi University of Mashhad
Lecture 7
Dr. Ali Karimpour Dec 2017
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Lecture 7
Topics to be covered include:
Introduction. Various controller configurations.
Different kind of controllers.
Controller realization.
Controller design in time domain. PID controller design.
PD controller design.
PI controller design.
Lag controller design.
Lead controller design.
Time domain design of control systems
Introduction. Various controller configurations.
Different kind of controllers.
Controller realization.
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Introduction
1 Control structure design.
2 Controller or compensator type.
3 Controller or compensator parameter tuning.
4 Checking the compensated system.
• Feedforward controller.
• Feedback or parallel compensation.
• Series or cascade compensation.
• Feedforward and series compensation.
• Series parallel compensation or cascade system.
• Series or cascade compensation.
• Lead controller or PD. • Lag controller or PI.
• Lead/lag controller or PID.
• Analytic controller design. • Graphic controller design.
• Controller design by table.
• Graphic controller design.• Analytic controller design.
Lecture 7
Dr. Ali Karimpour Dec 20174
Series Compensation Structure
جبران سازی سریIn this Course we consider the
series or cascade compensation.
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 7
Dr. Ali Karimpour Dec 2017
PID and Operational Amplifiers
و تقویت کننده عملیاتی PIDکنترلر
55
)(
)(
)(
)()(
sZ
sZ
sV
sVsG
i
f
i
o
Lecture 7
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Introduction. Various controller configurations.
Different kind of controllers.
Controller realization.
Controller design in time domain. PID controller design.
PD controller design.
PI controller design.
Lag controller design.
Lead controller design.
Time domain design of control systems
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Time domain design طراحی حوزه زمانی
Cruise control of a vehicle.
Linear system is:
Transfer function ?
Closed loop and its specification?
We need zero steady state and settling time less than 1 sec?
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Time domain design طراحی حوزه زمانی
)25(
40
ss
)(sCk+
-
)(sR )(sE در سیستم مقابل آیا می توان کنترلر تناسبی یافت: 2مثال .کمتر گردد0/01که خطای شیب واحد در حالت دائم از
Example 2: Is it possible to design a P controller such that the steady
state error to unit ramp be less than 0.01?
)25(
40
ss
)(sCk+
-
)(sR )(sE
Example 3: Is it possible to design a P controller such that the steady
state error to unit ramp be less than 0.01 and damping ratio of complex
poles be 0.707 ?
What about other controller?
در سیستم مقابل آیا میتوان کنترلر تناسبی یافت که: 1مثال و کمتر0.707نسبت میرائی قطبهای مختلط سیستم بیشتر از
.از یک گردد
Example 1: Is it possible to design a P controller such that the damping
ratio of complex poles be larger than 0.707 and less than 1?
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Time domain design طراحی حوزه زمانی
Example 4: Design a controller such that the steady state error to unit
ramp be less than 0.01 and damping ratio of complex poles be 0.707?
در سیستم مقابل کنترلری طراحی کنید که ثابت خطای شیب: 4مثال گردد؟0.707و نسبت میرائی قطبهای مختلط سیستم 0/01واحد کمتر از
PD controller …..
PI controller …..
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Compare PI and PD controllers
PDو PIمقایسه کنترلرهای
)25(
40
ss
)(sC
s
kk I
P +
-
)(sR )(sE
9.1645.8 IP kk
)25(
40
ss
)(sC
skk 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
plit
ude
With PI controller
With PD controller
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Lag controller design procedure
رویه طراحی کنترلر پس فاز
)(sG)(sC
ps
zsk
+
-
)(sR )(sE
pz
1- Obtain the root-locus (without controller) and
determine the gain k0 to satisfy the desired
damping ratio or …...
2- Find the gain k to satisfy the desired steady
state error (without controller). If k is in
conflict with k0 continue.
3- Evaluate the needed controller gain
...or ratio damping desired hesatisfy t Gain to
error state-steady desired hesatisfy t Gain togain needed
0
k
k
4- Choose pole and zero of controller near origin such that:0
gain neededk
k
p
z
What is near?5 - Choose the controller as:ps
zsksGc
0)(
6 - Check the controller.
Why?
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Time domain design طراحی حوزه زمانی
Example 5: Design a controller such that the steady state error to unit
ramp be less than 0.01 and damping ratio of complex poles be 0.707?
و نسبت 0/01در حالت دائم کمتر از در سیستم مقابل کنترلری طراحی کنید که خطای شیب واحد: 5مثال .گردد0.707میرائی قطبهای مختلط سیستم
Lag controller …..
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Designing lag controller and its step response
)25(
40
ss
)(sC
1.0
8.08125.7
s
s+
-
)(sR )(sE
0 0.5 1 1.5 2 2.5 30
0.2
0.4
0.6
0.8
1
1.2
1.4Step Response
Time (sec)
Am
plit
ude
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Time domain design طراحی حوزه زمانی
Example 6: Design a controller such that the velocity constant be 50 and
percent overshoot be less than 16%.
% 16و درصد فراجهش کمتر از 50در سیستم مقابل کنترلری طراحی کنید که ثابت خطای شیب: 6مثال .باشد
Lag controller ….. )10)(5(
1
sss
)(sC
ps
zsk
+
-
)(sR )(sE
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Lag controller design
)10)(5(
1
sss
)(sC
01.0
2976.084
s
s+
-
)(sR )(sE
P.O. is not ok
Tune the controller to
derive the performance
0 2 4 6 8 10 120
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Am
plit
ude
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Lag controller design
طراحی کنترلر پس فاز
When the design of lag controller is
not possible?
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Lead controller design procedure
رویه طراحی کنترلر پیش فاز
)(sG)(sC
ps
zsk
+
-
)(sR )(sE
pz
1- From the time-domain specifications
obtain the desired location of the closed-
loop dominant poles.
2- Select the zero of controller. Place the
zero on the real value of desired location of
the closed-loop dominant poles or on the
pole for pole-zero cancellation.
3- Locate the compensator pole so that the angle criterion is satisfied.
4- Determine the compensator gain k, such that the magnitude criterion is satisfied.
7 - If the overall response rise time, overshoot and settling time is not satisfactory, choose
another location of the closed-loop dominant poles.
5 - Choose the controller as:ps
zsksGc
)(
6 - Check the controller.
Lecture 7
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Time domain design طراحی حوزه زمانی
Example 7: Design a controller such that the steady state error to unit
ramp be less than 0.01 and damping ratio of complex poles be 0.707?
و نسبت 0/01در حالت دائم کمتر از در سیستم مقابل کنترلری طراحی کنید که خطای شیب واحد: 7مثال .گردد0.707میرائی قطبهای مختلط سیستم
Lead controller …..
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Exercises
1 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 )(sE2 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.
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Exercises
3 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
4 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.
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Exercises
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 in series with a
capacitor Cf .
a) Show that this choices lead to form a PID controller with high
frequency gain limit as;
5 Consider following structure:
b) Derive the parameters in the controller with respect to resistors and
capacitors.
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Exercises
6 In the following system design
a lag controller such that the
damping ratio of complex poles
be 0.6 and ramp error constant be 80.
)5.48(
40
ss
)(sC)(sGlag
+
-
)(sR )(sE
)10)(5(
4
sss
)(sC)(sGlag
+
-
)(sR )(sE7 In the following system design a
lag controller such that the damping
ratio of complex poles be 0.6 and
ramp error constant be 80.
8 Design a lead controller for exercise 6.
9 Design a lead controller for exercise 7.
10- In the following system design a controller such that leads to zero
steady state error to step and 4.3% P.O.
(Final)
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11 In the following system design
a lag controller such that that the
settling time be less than 3 sec.
2
1
s
)(sC)(sGlag
+
-
)(sR )(sE
12 In the following system design
a lead controller such that that the
settling time be less than 3 sec.
2
1
s
)(sC)(sGlead
+
-
)(sR )(sE
Exercises
13 In the following system design
a Lead controller such that the
damping ratio of complex poles
be 0.4.(Final 1390)
2
1
s
)(sC)(sGc
+
-
)(sR )(sE
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14- In the following system (Final 1395)
a) Draw root loci for K>0.)10(
81
ss
s )(sC
K+
-
)(sR )(sE
Exercises
b) Derive K for Ts=0.1s (Ts is Settling time)
c) Derive P.O. and velocity error constant for derived K in part “b”.
d) Derive a controller such that the velocity error constant multiplied
by 10 but P.O. and Ts remain the same as part “c”.
e) Derive a controller such that the velocity error be zero but P.O. and
Ts remain the same as part “c”.
f) Derive a controller such that the closed loop dominant poles are on
the -15±15j.
g) Derive step response of system for part “c” , “d” , “e” and “f”.
Lecture 7
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Supplementary Exercises
15- In the following system design a
PID controller with
Ziegler-Nichols Oscillation Method
)5.48(
40
ss
)(sC
)(sGPID
+
-
)(sR )(sE
16- In the following system design a PID controller with Ziegler-Nichols
Oscillation Method)10)(5(
4
sss
)(sC
)(sGPID
+
-
)(sR )(sE
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Appendix: 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.
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Appendix: 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 7
Dr. Ali Karimpour Dec 2017
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PIDKT
T19.0
Kp Ki Kd
2
127.0
DKT
T
PIDDKT
T12.1
P
طراحی زیگلر نیکولز حالت حلقه باز
DKT
T1
2
16.0
DKT
T
K
T16.0
Appendix: Ziegler-Nichols Reaction Curve Method(Open-Loop Case)
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Example 8: Following figure shows the step response of an
open-loop transfer function system. Design a PID controller.
s
esGTTCK
DsT
D201
40)(sec20sec,5,40 :So 1
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PI 09.09.0 1
DKT
T
Kp Ki Kd
0054.027.0
2
1 DKT
T
PID 12.02.1 1
DKT
T
P 1.01 DKT
T
012.06.0
2
1 DKT
T3.0
6.0 1 K
T
s
esGTTCK
DsT
D201
40)(sec20sec,5,40 :So 1
1.0)( sKP
ssKPI
0054.009.0)(
ss
sKPID 3.0012.0
12.0)(
Example 8: Following figure shows the step response of an
open-loop transfer function system. Design a PID controller.
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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
(حلقه بسته)طراحی زیگلر نیکولز بروش نوسانی
Lecture 7
Dr. Ali Karimpour Dec 2017
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PI cK45.0
Kp Ki Kd
T
Kc54.0
PIDT
Kc2.1 TKc075.0cK6.0
P cK5.0
Appendix: Ziegler-Nichols Oscillation Method(Closed-loop)
(حلقه بسته)طراحی زیگلر نیکولز بروش نوسانی
Lecture 7
Dr. Ali Karimpour Dec 2017
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.
33
Example 9: Design a PID controller for following system.
Lecture 7
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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 7
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Response to step reference
0 5 10 150
0.5
1
1.5Step response for PID control
Time (sec)
Am
plit
ude
پاسخ سیستم به پله
ss
sCPID 17.265.2
8.4)(
117.201.0
17.265.28.4)(
s
s
ssCPID