Control System Engineering Chapter 1 - Introduction
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Transcript of Control System Engineering Chapter 1 - Introduction
Control System Engineering - BETR 3393
Chapter 1 : Introduction of Control Design
Mr. Mohd Hanif Che [email protected]
Learning Outcome
At the end of this lecture, you should be able to:
• Briefly know how control system engineering involves in daily life application.
• Recap fundamental control theory• Explain control system design step by step
procedure
2
What do these two have in Common?
o Highly nonlinear, complicated dynamics!o Both are capable of transporting goods and people over long distances
BUT
o One is controlled, and the other is not.o Control is “the hidden technology that you meet every day”o It heavily relies on the notion of “feedback”
CONTROL consists ofSensing , Computation , Actuation
5
SenseVehicle Speed
ComputeControl “Law”
ActuateGas Pedal
In Feedback “Loop”
Example of Control System Block Diagram in Vehicle Auto Cruise Control
Goals in Vehicle Speed Control
Stability: system maintains and hold steadily on desired speed operating point.
Performance: system have fast responds and rapidly changes once desired speed is set.
Robustness: system tolerates various disturbances dynamics (mass, drag, etc)
1.2 REVIEW ON CONTROL FUNDAMENTAL
1. Various cases of poles location for second order/approx. system
2. Four (4) cases of damping ratio and relation to poles position for second order/approx. system
3. Step responses for second-order system damping cases
4. Underdamped second order system characteristics
5. Steady state error6. 2nd order approximation
Various cases of poles location for 2nd order system
Damping ratio, ζ Poles location Step Response
Overdamped, ζ>1Matlab Command:>>y=zpk([],[-1 -3],3)>> step(y)
Critically damped, ζ=1Matlab Command:>>y=zpk([],[-2 -2],4)>> step(y)
Under damped, 1>ζ>0Matlab Command:>>y=zpk([],[-2+3i -2-3i],13)>> step(y)
Undamped, ζ=0Matlab Command:>>y=zpk([],[5i -5i],25)>> step(y)
Re
Im
Re
Im
Re
Im
Re
Im
0 1 2 3 4 5 6 7 80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Step Response
Time (seconds)
Am
plit
ude
0 1 2 3 4 5 6 7 80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Step Response
Time (seconds)
Am
plit
ude
0 1 2 3 4 5 6 7 80
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Step Response
Time (seconds)
Am
plit
ude
0 1 2 3 4 5 6 7 8-0.01
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
Step Response
Time (seconds)
Am
plit
ude
Four (4) cases of damping ratio and relation to poles position for 2nd order system
Damping ratio, ζ Case Location of poles
ζ>1 Overdamped Different locations on the real axis
ζ=1 Critically damped Overlapping on the real axis
1>ζ>0 Under damped Conjugated and complex
ζ=0 Undamped On the imaginary axis, with same magnitude
2
21 1, nn jss
1, 2
21 nnss
nss 21 ,
njss 21 ,
Step responses for second-order system damping cases
0 1 2 3 4 5 6 7 80
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Step Response
Time (seconds)
Am
plit
ude
Undamped, ζ=0
Under damped, 1>ζ>0
Critically damped, ζ=1
Overdamped, ζ>1
General Formula Constants
With Position Constant,
With Velocity Constant,
With Acceleration Constant,
Steady state errorR(s) C(s)
+
-)(sG
E(s)
Closed loop negative feedback system
P
stepK
ee
1
1)(lim
0sGK
sp
v
rampK
ee1
)(lim0
ssGKs
v
a
parabolaK
ee1
)(lim 2
0sGsK
sa
2nd order approx. system
• System Response with Additional Poles
0 0.5 1 1.5 2 2.5 3 3.50
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (seconds)
Am
plit
ude
c1
c2
c3
Step response of system T1(s), system T2(s),and system T3(s)
)32)(32)(5(
65
)134)(5(
65
)32)(32)(10(
130
)134)(10(
130
)32)(32(
13
134
13
23
22
21
jsjsssssT
jsjsssssT
jsjsssT
0 0.05 0.1 0.15 0.2 0.25 0.3 0.350
5
10
15
20
25
Step Response
Time (seconds)
Am
plit
ude
0 0.5 1 1.5 2 2.5 3
x 104
0
1
2
3
4
5
6x 10
5 Step Response
Time (seconds)
Am
plit
ude
DC MOTOR
Motor Speed
Motor Position
25.179687.112
54.364212
ss
G
ss
sG
8.24
0546455.162191.4632
Example of Matlab command :ms_sys= tf(36421.54,[1 112.87 1796.25])step(ms_sys)
HEAT TRANSFER
Transfer function157
5.1
sG
0 50 100 150 200 250 300 3500
0.5
1
1.5
Step Response
Time (seconds)
Am
plit
ude
WATER TANK
Transfer functions
G39315.2491
004323.10
0 500 1000 15000
1
2
3
4
5
6
7
8
9
10
Step Response
Time (seconds)
Am
plit
ude
1.4 DESIGN METHOD
18
1- Design in the s-plane. (Classical method)
2- Design in the frequency domain. (Classical method)
3- Design in the State Variable Feedback Systems
Design in
the s-plane (Classical)
CONTROLLER DESIGN
19
Design in
the
frequency
domain
(Classical)
Design in
state
variables
Classical design: mainly use transfer function as a system model, design of simple controller using intuitive technique
Modern design: based on state space description, for complex system (MIMO, nonlinear), analytical design
• Root locus displayed both transient response and stability information • Changes in gain resulted changes in performance• Adding poles and/or zeros change the root locus and also the performance
Design of Control Systems
1. Modeling of System - model and behavior
2. Determine control goals (requirements)
3. Selection on control architecture (cascaded, feedback, etc.)
4. Set control gains (have various methods)
5. Are goals (requirements) are fulfilled?
6. Investigate sensitivity of performance to changes in system parameters
7. modify gains or architecture if needed