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8/10/2019 Exp4-Result & Discussion
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3.0 RESULTS AND DISCUSSIONS
3.1 RESULTS
3.1.1 PART A: FIRST ORDER SYSTEM
Table 3.1: Results for first order system
No Kp p Time Output
1a 10 10 50.3327 9.9464
2a 40 10 50.6351 39.6836
3a 10 20 99.3246 9.9464
4a 40 5 25.5343 39.6836
5a 20 20 100.8367 19.8429
6a 20 10 50.9375 19.8650
aThe first order behaviour is attached in the Appendices.
3.1.2 PART B: SECOND ORDER SYSTEM
Table 3.2: Results for second order system
No. Kp A B Type OvershootDecay
Ratio
Rise
Time
Settling
TimePeriod
1a 10 40 14 Overdamped - - - - -
2a 10 18 2 Underdamped 0.4668 0.2163 7.6915 67.2681 32.9637
3a 10 42.25 13 CriticallyDamped - - - - -
4a 20 42.25 13Critically
Damped- - - - -
5a 10 40 20 Overdamped - - - - -aThe first order behaviour is attached in the Appendices.
The type of response (i.e. overdamped, critically damped, underdamped) can be determined
theoretically (Seborg et al., 2011). The second order transfer function follows the following
equation:
() Table 3.3: Theoretical results for second order system
No. Kp Damping coefficient Type1 10 40 14 1.1068 Overdamped2 10 18 2 0.2357 Underdamped3 10 42.25 13
1
Critically Damped
4 20 42.25 13 1 Critically Damped5 10 40 20 1.5811 Overdamped
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2. Calculate the final output value minus the initial output value.
Determining the final output value minus initial output value:
3. Fill in the following table with the parameter values you calculated and give the first
order transfer function of this unknown system.
Calculating the parameter values:
Given: A=15
From the behavior of system identification problem 1, at steady state,
() At steady state,
For first order,
() ()()()( ) Table 3.4: Calculated parameter values for system identification problem 1
Kp 0.2133
p 27.3244
Derivation of first order transfer function for this system
()
()
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EXERCISE
1. What effect does increasing the gain have on the system output?
When the value of gain (Kp) is increase, the value of system output also increase and the
value of system output is approaching to the value of Kp.
2. What is meant physically by a system with a large gain?
As the changes of the gain in the input is small, it will lead to a large changes on the output
gain.
3. What effect does decreasing the time constant have on the system output?
Decrease in the time constant does not affect or change the value of the system output.
4. What is meant physically by a system with a small time constant?
With a small time constant, the system will result in fast response, thus it will reach steadystate faster.
5. Is it possible for a system to have a negative gain? What is the expected behaviour?
Yes, it is possible since the system tends to reach steady state over the time. The behaviour of
the system with negative gain can be seen in Figure 3.2 below.
Figure 3.2: Behaviour of the system with negative gain
0 50 100 1500
0.5
1
1.5
2Input Profile
Time (sec)
Magnitude(-)
0 50 100 150
-10
-5
0
5
10Output Profile
Time (sec)
Magnitude(-)
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6. Is it possible for a system to have a negative time constant? What is the expected
behaviour?
No, it is not possible since the system does not reach steady state over the time. The
behaviour of the system with negative time constant can be seen in Figure 3.3 below.
Figure 3.3: Behaviour of system with negative time constant
7. What is the expected response from a first order system driven by a sinusoidal input?
A pure sinusoidal response.
3.2.2 PART B: SECOND ORDER SYSTEM
QUESTIONS
1. What is the overshoot in the response?
Based on Figure 13 in the Appendices,
()
0 50 100 1500
0.5
1
1.5
2Input Profile
Time (sec)
Magnitude(-)
0 50 100 150-12
-10
-8
-6
-4
-2
0x 10
6 Output Profile
Time (sec)
Magnitude(-)
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2. What is the period of the oscillatory response?
Based on Figure 13 in the Appendices,
3. Calculate the final output value minus the initial output value.
4. Fill in the following table with the parameter values you calculated and derive the
second order transfer function for this unknown system.
Determining the parameter values:
()
()
() Table 3.4: Calculated parameter values for system identification problem 1
Kp 15
11.7668 0.2477
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Derivation of second order transfer function for this system:
()
() () () ()() ()
EXERCISE
Consider the following values for the damping coefficient for a second order dynamic
system.
Region I Region II Region III
1
1. What types of poles does this system have? What types of response would be expected
for a system with a damping coefficient in Region I, II and III?
Region Region I Region II Region III
Damping
coefficient1
Types of polesComplex conjugate
poles
Real and multiple
poles
Real and distinct
poles
Types of response Underdamped Critically damped Overdamped
2. Sketch the corresponding response of the output variable to a step input in Region I,
II and III.
Region I Region II Region III
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3. How does a decrease in the damping coefficient affect the speed of response?
Decrease in the damping coefficient will increase the speed of response since the system
tends to achieve steady state faster.
4. Which of the three responses would be expected to have a shorter response time and
sluggish?
Among the three regions, region I has the shortest response time with overshoot that has a
damping coefficient which is 1. Region II with =1 has the fastest response without
overshoot.
5. What is the trade-offs from a control perspective of the different responses?
Different responses help engineers to identify the type control that need to fit into the system
and performance of the system. The responses fall into Region II is the ideal control where
the control reaches steady-state without overshoot. The responses fall into Region I is the
most common responses face by engineer. The control overshoots a few times to be able to
reach steady state. The responses fall into Region III does not overshoot but has the slowest
responses among the three. The control reaches steady state in a long period of time. With
different responses, engineers can determine the best control for a process because some
process cannot have overshoot in their process such as, dosing of bleaching agent into food
products. Overdosing can bring health hazards to consumers, therefore responses fall in
Region II and Region I are more preferable.
3.3 OVERALL DISCUSSION
In this experiment, MATLAB was used to conduct this experiment. Dealing with the first
order system, we determined the system output by manipulating the inputs such as system
gain, KP and the time constant, . By manipulating the system inputs, we are required todetermine the changes in the system output by observing the graphs generated. The new
system output is when the system reaches the steady-state. The value taken tends to be
slightly deviate due to human error, in addition the software is unable to give an actual value.
To avoid the mentioned problem, an average value is taken from the five persons in the
group. For the System Identification Problem 1, the slope at the most oblique gradient was
taken as it showed the most exact value of the system gain, KP.
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On the other hand, dealing with second order system, the system is totally different from the
first order although we are manipulating the same inputs which are system gain, KPand the
time constant, . This is because the second order system is influence by the dampingcoefficient . A large value of yield a sluggish response and a small value of yield a fast
response. The characterisation or types of response and roots of equation or types of poles
vary with different values of damping coefficient, .
For only under-damped of response, the damping coefficient, can be easily determine as the
overshoot can be easily obtain or detect from the graph. By observing the trend of the graph
we can easily determine the characterisation of the response.
While for the overdamped and critically damped responses, we could hardly determine the
characterisation of their response since they were almost the same. Therefore, it is necessary
to calculate the damping coefficient, by using the overshoot obtained from the graph and
then compare the value calculated with the value of damping coefficient,
shown
in the table
below to determine the characterisation of the response.
Table 3.5: Characteristic of first order system response
Damping Coefficient Characterisation/ Types of
Response
Roots of Characteristic
Equation/ Types of Poles
1 Over-damped Complex conjugates
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REFERENCES
Seborg, D., Edgar, T., Mellichamp, D., & Doyle, F. (2011).Process dynamics and control
(1st ed.). New York: Wiley.
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APPENDICES
Figure 1: Kp=10,
=10
0 50 100 1500
0.5
1
1.5
2Input Profile
Time (sec)
Magnitude(-)
0 50 100 1500
2
4
6
8
10Output Profile
Time (sec)
Magnitu
de(-)
Behaviours for first order system
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Figure 2: Kp=40, =10
0 50 100 1500
0.5
1
1.5
2Input Profile
Time (sec)
Magnitude(-)
0 50 100 150
0
10
20
30
40
50
Output Profile
Time (sec)
Magnitude(-)
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Figure 3: Kp=10, =20
0 50 100 1500
0.5
1
1.5
2Input Profile
Time (sec)
Magnitude(-)
0 50 100 150
0
2
4
6
8
10
Output Profile
Time (sec)
Magnitude(-)
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Figure 4: Kp=40, =5
0 50 100 1500
0.5
1
1.5
2Input Profile
Time (sec)
Magnitude(-)
0 50 100 150
0
10
20
30
40
50
Output Profile
Time (sec)
Magnitude(-)
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Figure 5: Kp=20, =20
0 50 100 1500
0.5
1
1.5
2Input Profile
Time (sec)
Magnitude(-)
0 50 100 150
0
5
10
15
20
Output Profile
Time (sec)
Magnitude(-)
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Figure 6: Kp=20, =10
0 50 100 1500
0.5
1
1.5
2Input Profile
Time (sec)
Magnitude(-)
0 50 100 150
0
5
10
15
20
Output Profile
Time (sec)
Magnitude(-)
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Figure 7: Kp=10, A=40, B=14
0 50 100 1500
0.5
1
1.5
2 Input Profile
Time (sec)
Magnitude(-)
0 50 100 1500
2
4
6
8
10
Output Profile
Time (sec)
Magnitude(-)
Behaviours for second order system
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Figure 8: Kp=10, A=18, B=2
0 50 100 1500
0.5
1
1.5
2Input Profile
Time (sec)
Magnitude(-)
0 50 100 150
0
5
10
15
Output Profile
Time (sec)
Magnitude(-)
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Figure 9: Kp=10, A=42.25, B=13
0 50 100 1500
0.5
1
1.5
2Input Profile
Time (sec)
Magnitude(-)
0 50 100 150
0
2
4
6
8
10
Output Profile
Time (sec)
Magnitude(-)
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Figure 11: Kp=10, A=40, B=20
0 50 100 1500
0.5
1
1.5
2Input Profile
Time (sec)
Magnitude(-)
0 50 100 150
0
2
4
6
8
10
Output Profile
Time (sec)
Magnitude(-)
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Figure 12: System identification problem 1
0 50 100 150 200 2500
5
10
15Input Profile
Time (sec)
Magnitude(-)
0 50 100 150 200 2500
1
2
3
4
5Output Profile
Time (sec)
Magnitude(-)
Behaviours for system identification problem 1
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Figure 13: System identification problem 2
0 50 100 1500
0.05
0.1
0.15
0.2 Input Profile
Time (sec)
Magnitude(-)
0 50 100 1500
5
10
15
20
25
30Output Profile
Time (sec)
Magnitude(-)
Behaviours for system identification problem 2