eee3420 lecture01 rev2011
TRANSCRIPT
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│ Lecture 1 │
Introduction to Control Systems
EEC3420 Industrial ControlDepartment of Electrical Engineering
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EEE3420 Industrial Control
Learning objectives
Understand the basic concept in control systems.
Know what is a Transfer Function.
Appreciate the PID control process.
Know what criteria leading to a stable control system.
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EEE3420 Industrial Control
Open-loop control system
An open-loop control system is one in which the control signal of the process is independent of the process output
control accuracy is determined by the calibration of the plant
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Advantages & disadvantage of open-loop control system
Advantages
simple and inexpensive
no stability problem
Disadvantages
cannot compensate for any disturbances that add to the controller’s driving signal
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Closed-loop control system
A closed-loop control system depends on the output of the process to adjust the signal controlling the closed loop
process output is compared to the user command and an output from the plant
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Advantages & disadvantages of closed-loop control system
Advantages
less sensitive to noise, disturbances and changes in the environment
transient response and steady-state error can be controlled more conveniently and with greater flexibility
Disadvantages
relatively expansive
may be unstable if not properly designed
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Architecture of a closed-loop control system
Controlled Variable (CV).
Set point.
Error = set point – current value of CV.
Manipulated Variable.
Feedback Loop.
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Feedback control real-time scheduling
Choices for control variables, manipulated variables, set points
Choice of appropriate control functions
Stability problem of feedback control in the context of real-time scheduling?
How to tune control parameters?
How significant is the overhead and how to minimize it?
How to integrate a runtime analysis of time constraints with scheduling algorithms?
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Using negative feedback control system
typically more stable
less sensitive to variation in component values
more immune to noise
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Transfer function of a control system
The transfer function of a control system is defined as the ratio of the output to the input
predict how the system will perform if the transfer function is known
output depends on both the present input and the past history of the input, so the output c(t) is a convolution product of the input r(t) and the system g(t)
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Transfer function of a control system
c(t) = r(t) * g(t)
∫∞
⋅−=0
)()( τττ dgtr
with the use of Laplace transform, we get
∫ ∫∫∞ −∞∞ − ⋅
⋅−=⋅==
0 00)()()()]([)( dtedgtrdtetctcLsC stst τττ
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Transfer function of a control system
Let t’ = t – τ then t = t’ + τ
the convolution product of r(t)*g(t) is now transformed into an algebraic product of R(s).G(s)
)()()(')'(
')()'()()()(
0 0
'
0
)'(
00 0
sGsRdegdtetr
dtedgtrdtedgtrsC
sst
tsst
⋅=
⋅⋅⋅=
⋅
⋅=⋅
⋅−=
∫ ∫
∫ ∫∫ ∫∞ ∞ −−
∞ +−∞∞ −∞
ττ
τττττ
τ
τ
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EEE3420 Industrial Control
Transfer function of a control system
time-domain output c(t) may be obtained by the inverse Laplace operation
open-loop control system can be represented in frequency domain as shown above
the output C(s) is given by G(s) x R(s).
∫∞+
∞−
− ⋅==jk
jk
st dsesCj
sCLtc )(2
1)]([)( 1
π
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PID Control System PID – Controller is the most widely used control strategy
in industry used for various control problems such as automated
systems or plants consists of three different elements
P Proportional control I Integral control D Derivative control
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PID Control System
for control loop to work properly, the PID loop must be properly tuned
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The PID transfer function
)(teKP P ×=
∫ ⋅×= dtteKI I )(
dt
tedKD D
))((×=
P Proportional control,
D Derivative control,
I Integral control,
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The PID transfer function
The total controller output,
We get,
Use the Laplace transform,
∫ ×+⋅×+=dt
tedKdtteKteKtu DIP
))(()()()(
∫ ∫ →→→→s
xdtx
sdtsx
dt
dxs
dt
d;1
;;
)1()()(
)(sK
s
KKsG
sE
sUD
IPC ⋅++==
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The PID transfer function
Re-arrange to get, sT
sTsTTKsG
I
IDIPC ⋅
+⋅+⋅⋅= )1()(
2
Where
KP is the proportional gain
TI is the integral time constant
TD is the derivative time constant
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The PID transfer function
the three different adjustments (KP, TI, TD) interact with each other
it can be very difficult and time consuming to tune these three values in order to get the best performance according to the design specifications of the system.
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A Thermal Control System
• electrical heater of heat capacity Ch & thermal resistance Rho
• oven of heat capacity Co & thermal resistance Ro
• environment temperature Te, set-point temperature Ts
• temperature controller adjusts the power W by comparing To with Ts
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On-Off Control of the Thermal System• the simplest form of control
• when the oven is cooler than the set-point temperature the heater is turned on at maximum power M
• once the oven is hotter than the set-point temperature the heater is switched off completely
Red line: set-point temperature
Green line: actual temperature
Blue line: Delivered Power
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Proportional Control of the Thermal System
• proportional controller attempts to perform better than the On-Off type by applying power W to the heater in proportion to the difference in temperature between the oven and the set-point
• KP is known as the proportional gain of the controller
( )OSP TTKW −×=
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PD Control of the Thermal System
• add D-Control (proportional to the time-derivative of the error signal) to mitigate the stability and overshoot problems that arise from a high gain proportional controller
• adjust KD (the damping constant) to achieve a critically damped response
( ) ( )dt
TTdKTTKW OSDOSP
−×+−×=
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PID Control of the Thermal System• add I-Cotnrol (proportional to
the time-integral of the error signal) to change the heater power until the time-averaged value of the temperature error is zero
• KI is the integral gain parameter
( ) ( ) ( )dt
TTdKdtTTKTTKW OSDDSIOSP
−×+−×+−×= ∫
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Analysis of a Control System using Transfer Function
given that,6116
10)(:
23 +++=
ssssGprocessThe P
and the feedback path: H(s) = 1
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Analysis of a Control System using Transfer Function
for P-control only, if KP = 3, then GC(s) = KP =3, and
36116
30
)()(1
)()(23 +++
=⋅+
⋅=ssssGsG
sGsG
PC
PC
i
o
θθ
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Analysis of a Control System using Transfer Function
for PI-control, if KP = 2.7 and TI = 1.5, then
and the transfer function is:
275.495.1695.1
275.40
)()(1
)()(234 ++++
+=⋅+
⋅=ssss
s
sGsG
sGsG
PC
PC
i
o
θθ
s
s
sT
sTK
sTKsG
I
IP
IPC 5.1
)15.1(7.2)1(11)(
+=⋅
+⋅=
⋅
+=
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Analysis of a Control System using Transfer Function
for PID-control, if KP = 2,
TI = 0.9 and TD = 0.6, then
and the transfer function is:
204.237.204.59.0
20188.10
)()(1
)()(234
2
++++++=
⋅+⋅=
ssss
ss
sGsG
sGsG
PC
PC
i
o
θθ
s
ss
sT
sTsTTKsG
I
IDIPC 9.0
)19.054.0(2)1()(
22 ++=⋅
+⋅+⋅⋅=
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Analysis of a Control System using Transfer Function
The transfer functions give the step responses as shown on the right
• for P-control (the red curve) – a steady state error occurs
• for PI-control (the blue curve) – the response becomes more oscillatory and needs longer to settle, the error disappears
• for PID-control (the green curve) – the overshoot and the number of oscillatory cycles are much reduced
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Concluding remarks of the PID Effect
• In general, we may observe that
• P term is used to adjust the speed of response.
• I term provides zero error.
• D term introduces damping.
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3. Stability of Control System• a control system responds to an input by undergoing a
transient response before reaching a steady-state
• the total response of a system consists of two parts, namely, the natural response and the forced response
• natural response describes the way the system dissipates or acquires energy, the nature of this response is dependent only on the system
• the nature of the forced response is dependent on the input
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3. Stability of Control System• for a linear system, we can write
Total response = Natural response + Forced response
• for a control system to be useful, the natural response must eventually approach zero, thus leaving only the forced response
• a stable control system will always return to a stable operating state
• in an unstable system, any disturbance will result in oscillations building up until some parts fails
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3. Stability of Control System
• Oscillatory System
• between the stable state and the unstable state lies the conditionally stable system in which oscillations neither increase nor decrease
• each cycle being identical to the previous one and results in sustainable oscillation.
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Stability of First Order SystemConsider the response of a control
system with transfer function (s+2)/(s+5) under a step input with R(s) = 1/s
• a pole on the real axis generates an exponential response of the form e-αt, where –α is the pole location on the real axis.
• if α is positive, the transient response will decay to zero
• if α is negative, then the transient response will grow and the system will be unstable.
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Stability of Second Order System
As long as the poles of the output function lies on the left hand side of the complex plane
• the system output will not grow without bound
• it will be stable
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Stability of Second Order System
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Stability of Higher Order System
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Stability of Higher Order System - Routh table
The Routh table method will yield the stability information of a system without the need to solve for the system poles.
The method requires two steps:
(1) generate a Routh table
For a fourth order system given on the right,
Construct the table for the denominator by using the formulae shown on the right
Similar formulae are used for system with order higher than 4.
s4 a4 a2 a0
s3 a3 a1 0
s2 b1=(a2*a3-a1*a4)/a3 b2=(a0*a3- 0*a4)/a3 0
s1 c1=(a1*b1-a3*b2)/b1 0 0
s0 d1=(b2*c1-0*b1)/c1 0 0
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Stability of Higher Order System - Routh table
(2) interpret the Routh table The number of poles that are
in the right half plane is equal to the number of sign change in the first column of the Routh table.
Note: For the special cases such as the element in the first column is equal to zero or the elements in the entire row are equal to zero will not be treated here, please refer to descriptions in books dealing with control theory.
s4 a4 a2 a0
s3 a3 a1 0
s2 b1=(a2*a3-a1*a4)/a3 b2=(a0*a3- 0*a4)/a3 0
s1 c1=(a1*b1-a3*b2)/b1 0 0
s0 d1=(b2*c1-0*b1)/c1 0 0
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Stability of Higher Order System - Routh table
Example
Determine the stability of the system on the right by Routh table method.
Solution
As there are two sign changes in the first column of the Routh table, so there are two poles lying in the right half plane and hence the system is not stable.
s3 a3 = 1 a1 = 31 0
s2 a2 = 10 a0 = 1030 0
s1 b1=(31*10-1*1030)/10
= -72
0 0
s0 c1 = (1030*(-72)-0*10)/(-72)
= 1030
0 0
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Summary of Introduction to Control System
• Closed-loop control system is less sensitive to noise, disturbances and changes in the environment.
• The transfer function of a control system is the ratio of the output to the input.
• Proportional control is used to adjust the speed of response.
• Integral control provides zero error.
• Differential control introduces damping.
• If the poles of the output function lies on the left hand side of the complex plane, the system will be stable.
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Introduction to Control System
End of Lecture 1
RevisionNorman S. Nise, Control Systems Engineering, Fourth Edition, Johne Wiley & Sons, Inc., page 177 to page 183.