10080 CHAPTER 10. CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES
The total black-body radiation intensity, Ib, is obtained by integrating over all frequencies or
wavelength [2]
The black body emissive ‡ux is given by,
where,
Integrating over all wavelengths , we obtain the total black body emissive ‡ux [2],
The temperature may be determined from Eq. (19) as,
Problem 10.22: Schematic of temperature measurement using pyrometry.
The figure on the bottom of the previous page shows the schematic of temperature measurement
References:
[1] Fraden, J., Handbook of Modern sensors: Physics, Designs, and Applications, Springer, 1996.
[2] Ozisik, M. N., Radiative Transfer and Interactions with Conduction and Convection, Wiley-
Interscience, 1973.
[3] Siegel, R. and J. R. Howell, Thermal Radiation Heat Transfer, Second Ed., Hemisphere Pub-
lishing Corp., 1981.
[4] A. Emami-Naeini, et al., “Modeling and Control of Distributed Thermal Systems,” IEEE Tran.
Contrl. Syst. Tech., pp. 668-683, September 2003.
24. Repeat the RTP case study design by summing the three sensors to form a single signal to
control the average temperature. Demonstrate the performance of the linear design, and validate the
performance on the nonlinear Simulink simulation.
Solution:
A linear model for the system was derived in the text as,
10082 CHAPTER 10. CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES
The three open-loop poles are computed from Matlab and are located at 0:0527;0:0863;and
0:1482:Since we tied the three lamps into one actuator and are only using the average temperature
for feedback, the linear model is then:
resulting in the transfer function,
We may try a simple PI controller of the form,
so as to cancel the effect of the slower pole. The linear closed-loop response is shown as well as the
10083
-5
10
30
Temperature (K)
PID (average) temperature tracking response
-5
10
30
Temperature (K)
PID (average) temperature tracking response
r
Problem 10.23: Linear closed-loop RTP response for PI controller.
10084 CHAPTER 10. CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES
020 40 60 80 100
25
Time (sec)
020 40 60 80 100
25
Problem 10.23: RTP Linear response for PI: control effort.
Next we design a state-space based controller. As in the text, we use the error space approach for
inclusion of integral control and employ the linear quadratic gaussian technique of Chapter 7. The
error system is, _e
_
=0Cavg
0 A e
+D
B; (29)
10085
and hence minimizes a measure of the temperature non-uniformity. As in the text, the factor of ten
is used as the relative weighting between the error state and the plant state. The state and control
weighting matrices, Qand R, are then,
The resulting feedback gain matrix computed from Matlab is,
K= [K1:K0];
where,
with xcdenoting the controller state and,
The resulting estimator gain matrix is,
3
10086 CHAPTER 10. CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES
The closed-loop system equations are given in the text,
where ris the reference input temperature trajectory, the closed-loop state vector is xcl = [TTxT
c^
TT]Tand
the system matrices are,
The linear closed-loop response and the associated control effort are shown. The commanded
temperature trajectory, r, is a ramp from 0C to 25C with a 1C/sec slope followed by 50 sec soak
10087
020 40 60 80 100
-5
30
Time (sec)
temperature tracking response
020 40 60 80 100
-5
30
r
10088 CHAPTER 10. CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES
14
10
-6
-2
6
lamp voltage
control effort
u
14
10
-6
-2
6
lamp voltage
control effort
u
Problem 10.23 RTP linear (average) : control effort.
10089
To Workspace2
In1 O u t1
Subsystem2
Scope3
Scope1
Saturation
K*u
M at rix
Gain5
Problem 10.23: Simulink diagram for nonlinear closed-loop RTP system to control average
temperature.
10090 CHAPTER 10. CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES
20 40 60 80 100 120 140 160 180
310
335
Time (sec)
SISO (average) temperature tracking response
r
y
20 40 60 80 100 120 140 160 180
310
335
Time (sec)
SISO (average) temperature tracking response
ry
Problem 10.23: Nonlinear closed-loop response.
10091
0.5
4
SISO (average) temperature : control effort
0.5
4
Problem 10.23: Nonlinear closed-loop response: control effort.
25. One of the steps in semiconductor wafer manufacturing during photolithography is performed
by placement of the wafer on a heated plate for a certain period of time. Laboratory experiments
have shown that the transfer function from the heater power, u, to the wafer temperature, y, is given
by
y(s)
u(s)=G(s) = 0:09
(s+ 0:19)(s+ 0:78)(s+ 0:00018)
1. (a) Sketch the 180root locus for the uncompensated system.
(b) Using the root locus design techniques, design a dynamic compensator, D(s), such that
the system meets the following time-domain specifications
i. Mp5%
ii. tr20 sec
iii. ts60 sec
iv. Steady-state error to a 1C step input command <0:1C.
Draw the 180root locus for the compensated system.
Solution:
The uncompensated root locus is shown in the following figure.
10092 CHAPTER 10. CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES
3.5 -3 2.5 -2 1.5 -1 0.5 00.5 11.5
2.5
1.5
-1
1.5
0.86
0.160.340.50.640.76
Root Locus
Real Axis
Uncompensated root locus.
(a) First, convert the time domain specifications to s-plane specifications:
Mp5% )&>0:707
(b) At this point there are several design methods one can use for this problem.
10093
RLTool: Compensator I.
10094 CHAPTER 10. CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES
Step Response
Time (sec)
Amplitude
0 5 10 15 20 25
0.5
2
0 5 10 15 20 25
0.5
2
RLTool: Step response (y) and control effort (u) for Compensator I.
Method II: We can cancel out the nearly pure integrator with our compensator zero, and
add a pure integrator. It is possible to achieve the time domain specifications with this
compensator. The compensator becomes
We can display the compensated root locus and s-plane regions in RLTool in the following
figure.
10095
RLTool: Compensator II.
10096 CHAPTER 10. CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES
Step Response
Time (sec)
010 20 30 40 50 60 70
0.2
0.6
1
1.2
010 20 30 40 50 60 70
0.2
0.6
1
1.2
RLTool: Step response (y) and control effort (u) for Compensator II.
26. Excitation-Inhibition Model from Systems Biology (Yang and Iglesias, 2005): In Dictyostelium
cells, the activation of key signaling molecules involved in chemoattractant sensing can be modeled by
the following third order linearized model. The external disturbance to the output transfer function
is: y(s)
w(s)=S(s) = (1 )s
(s+)(s+ 1)(s+)
where, wis the external disturbance signal proportional to chemoattractant concentration, and y
is the output which is the fraction of active response regulators. Show that there is an alternate
representation of the system with the “plant” transfer function
G(s) = (1 )
s2+ (1 + +)s+ (++)
and the “feedback regulator”
Dc(s) = 
(1 )s
It is known that 6= 1 for this version of the model. Draw the feedback block diagram of the
system showing the locations of the disturbance input and the output. What is the significance of
this particular representation of the system? What hidden system property does it reveal? Is the
disturbance rejection a robust property for this system? Plot the disturbance rejection response of
10097
the system for a unit step disturbance input. Assume the system parameter values are = 0:5and
= 0:2.
1. Solution: From the following figure we have
The significance of this particular representation is that it reveals the internal model, namely
the pure integrator. Hence the system is Type I with respect to disturbance rejection. It rejects
10098 CHAPTER 10. CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES
0 1 2 3 4 5 6 7 8 9 10
0.1
0.2
Step Response
Time (sec)
0 1 2 3 4 5 6 7 8 9 10
0.1
0.2
Disturbance rejection response for a unit step disturbance.