exp4-result & discussion

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

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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,

()

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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

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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

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Figure 2: Kp=40, =10

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Figure 4: Kp=40, =5

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Figure 7: Kp=10, A=40, B=14

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Figure 8: Kp=10, A=18, B=2

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Figure 9: Kp=10, A=42.25, B=13

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Figure 11: Kp=10, A=40, B=20

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Figure 12: System identification problem 1

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Figure 13: System identification problem 2

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