Experiment 4 Study on Dynamics of First Order and Second Order
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Transcript of Experiment 4 Study on Dynamics of First Order and Second Order
0
PROCESS CONTROL & INSTRUMENTATION LAB
(BKF4791)
2014/2015 Semester I
Title of Experiment : 4 - STUDY ON DYNAMICS OF FIRST ORDER AND SECOND ORDER SYSTEMS
Date of Experiment : 25 OCTOBER 2014
Instructor’s Name : Dr Noorlisa Binti Harun
Group of Member :
NAME ID
QASTALANI BT MOHD GHAZALI KE 11004
TEOH TZE SIANG KA11103
ASHWINDER A/P CHELLIAH KA11138
NURUL ATIKAH BINTI KAMARUDIN KA11032
NUR SABRINA BINTI RAHMAT
KA11050
Group No. : 04
Section : 05
Marks :
FACULTY OFCHEMICAL AND NATURAL RESOURCES ENGINEERING
UNIVERSITY MALAYSIA PAHANG
Tear here
Please keep for student reference.
Received by;
( )
Submitted by;
( )
1
ABSTRACT
The first purpose of this experiment is to demonstrate the properties of first and second order
systems for different input values while the next purpose is to illustrate the dynamic response
of first and second order systems to different input signals. For the first order system, the
system gain Kp (numerator coefficient) was set to 10, 40, 10, 10, 20 and 30 pairing with the
system time constant τP (denominator coefficient) which was set to 10, 10, 20, 5, 10 and 20
accordingly. Next, the step time was set to 10.0 and the step function from 0.0 to 1.0
constantly throughout the experiment. Then, the simulation was started to see the response
curve. While for the second order system, the procedure is almost same but the value for Kp
was different and also the value for A and B must be change before starting each simulations.
If the system is under damped, the overshoot, decay ratio, rise time, settling time and the
period of oscillation will be calculated. For first order, the increase in Kp values when τP was
kept constant resulting the decrease in time but the output values increased. As for when the
τP was set decreasing and Kp was set constant, the time and output values also decreased. As
for second order, underdamped system was obtained only when the value of Kp , A and B
were 10, 18 and 2 respectively.
2
2.0 METHODOLOGY
2.1 First Order System
To start the first order system, on the First and Second Order Systems button from the Main Menu was click and then the First Order System button was selected. The two
windows will be display; the first is the system window and the second is the input/output window.
First, the system gain Kp (numerator coefficient) and the system time constant τP (denominator coefficient) both to 10.0 was set by clicking once on the first order
system block. The step time to 10.0 was set, initial value of the step function to 0.0 and the final value of the step function to 1.0 by clicking once on the step function block.
Click OK after you have done.
To start the simulation, Start button was selected from the Simulation menu. The new steady state value and the length of time it takes for the output to reach the new steady state (sec) were recorded. The Pointer button to take several points along the response
curve was used in our analysis.
Now the value of Kp was increased to 40.0 and step 3 was repeated. The response differ from the response in step 3 was recorded.
Kp value was set back to 10.0 and the value of τP was increased to 20.0. The simulation was repeated. This response differ from the response in step 3 was recorded.
Now the value of the Kp was maintain at 10.0 and the value of τP was decreased to 5.0. The simulation was repeated. The new steady state value and the length of time it takes
for the output to reach the new steady state (sec) were recorded.
For syestem identification problem, From the Main Menu, the System Identification Problem 1 button was selected. A step input was used; the simulation was run to
generate output data that can be used to determine the system gain (Kp) and the system time constant (τP). Remember to use the Pointer button and to take several points along
the response curve in your analysis of the system output.
3
2.2 Second Order System
To start the second order system, the First and Second Order Systems button was click from the Main Menu then the Second Order Systems button was selected. The two
windows will be display; the first is the system window and the second is the input/output window.
The system gain (Kp) was set to 10.0 (numerator coefficient), the value of A to 40.0 and the value of B to 14.0 (denominator coefficient) by clicking Second Order System
block and the initial value of the Step Function was set to 0.0 and the final value of the Step Function to 1.0 by clicking once on the Step Function block.
To start the simulation, Start from the Simulation menu was selected. The system was recorded and analyze whether the system is overdamped, underdamped or critically
damped. The overshoot, decay ratio, rise time, settling time and the period of oscillation was recorded if the systems are underdamped.
The value of A was changed to 18 and the value of B also changed to 2. The simulation was repeated. The system was recorded and analyze whether the system is
overdamped, underdamped or critically damped. The overshoot, decay ratio, rise time, settling time and the period of oscillation was recorded if the systems are
underdamped.
The value of A was changed to 42.25 and the value of B also changed to 13. The simulation was repeated. The system was recorded and analyze whether the system is overdamped, underdamped or critically damped. The overshoot, decay ratio, rise time,
settling time and the period of oscillation was recorded if the systems are underdamped.
Now the value of the Kp was increased to 50.0 and the value of A and B was increased to 60.0 and 15.0. The simulation was repeated. The new steady state value and the length
of time it takes for the output to reach the new steady state (sec) were recorded.
For syestem identification problem, close the two windows by clicking the left mouse button on the upper left hand box of both windows and selecting Close. From the First
and Second Order Systems Menu, the System Identification Problem 2 button was selected. After that, by using a step input, run the simulation to generate data that
can be used to determine the system gain (Kp), the system time constant (τP) and the damping coefficient (ζ).
4
RESULTS
Part A
No. Kp 𝜏p Time Output 1 10 10 54.6169 9.9831 2 40 10 51.996 39.653 3 10 20 114.093 9.9831 4 10 5 27.5 9.939 5 20 10 62.715 19.9178 6 30 20 118.125 30.0982
Part B
No Kp A B Type Overshoot Decay Ratio
Rise Time
Settling time Period
1 10 40 14 Over
damped - - - - -
2 10 18 2 Under
damped 0.4711 0.2131 17.608 s 70.296 s 27.419s
3 10 42.25 13 Critically damped - - - - -
4 20 42.25 13 Critically damped - - - - -
5 50 60 15 Under
damped - - - - -
5
DISCUSSION
• Problem identification problem for first order system
Figure 1: Graph generated for first order identification problem
Slope of initial response
𝐾𝑝 =∆𝑂𝑢𝑡𝑝𝑢𝑡∆𝐼𝑛𝑝𝑢𝑡
𝐾𝑝 =3.2156 − 0
15 − 0
𝑲𝒑 = 𝟎.𝟐𝟏𝟒𝟒
Final output value minus initial output value
𝜏𝑃 = 𝑓𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑎𝑙𝑢𝑒 − 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑎𝑙𝑢𝑒
0 20 40 60 80 100 120 140 160 180 2000
2
4
6
8
10
12
14
16Input Profile
Time (sec)
Magnitude (
-)
0 20 40 60 80 100 120 140 160 180 2000
1
2
3
4
5Output Profile
Time (sec)
Magnitude (
-)
FROM RESULT:
Step time = 10
Initial value = 0
Final value = 15
6
𝜏𝑃 = 3.2156 − 0
𝝉𝑷 = 𝟑.𝟐𝟐𝟓𝟔
Parameter values calculated
Table 1: Parameter values calculated for unknown first order system
𝐾𝑝 0.2144
𝜏𝑃 3.2156
Derivation of first order transfer function for this system
Standard form: 𝑌(𝑠) = 𝐾𝑝𝜏𝑠+1
𝑌(𝑠) = 𝐾𝑝
𝜏𝑠 + 1
𝒀(𝒔) = 𝟎.𝟐𝟏𝟒𝟒
𝟑.𝟐𝟏𝟓𝟔𝒔 + 𝟏
• Discussion on first order system (Exercise)
MATLAB software is used to conduct this experiment. When the First order system
in the MATLAB is used, the system output is determined by manipulating the inputs which
are system gain, KP and the time constant, 𝜏𝑃. The changing in the output by observing the
graph is detected by manipulating the system inputs. When the system reaches the steady-
state is the new system output. When taking the value at steady state, the value taken is
ensure at initial value where the system become steady and not the value behind it as the time
where the system start to reach steady state need to be detect. As the value is taken by human,
the value will be slightly deviate and not shown automatically by the software and hence it
may deviate from the actual data.
To avoid this problem average value, in order to obtain the exact value as possible,
the pointer is point several times at the graph shown to obtain the average value. The changes
can be calculated manually by using first order transfer function standard form.
𝑌(𝑠) = ∆𝒚
∆𝒙(𝒕𝒊𝒎𝒆)=
𝐾𝑝𝜏𝑠 + 1
Then Laplace it to be,
y(t) = A 𝐾𝑝 (1 - 𝑒𝜏𝑡)
7
Compare the result for output from the MATLAB and changes from manually
calculated, if the value is nearly ± 0.1, the value is confirm correct.
Other than that, For the System Identification Problem 1, the slope is needed to be
calculated from the value that been used in the system input and system output.
On the other hand, the system of Second order system, is totally different from the
first order although we are manipulating the same inputs which are system gain, 𝐾𝑝 and the
time constant, 𝜏𝑃. This is because the second order system is influence by the damping
coefficient ξ. The large value of ξ yield a sluggish response and small value of ξ yield a fast
response. The characterisation or types of response and roots of characteristic equation or
types of poles are different with different values of damping coefficient, ξ.
For only under-damped of response, the damping coefficient, ξ can be easily
determine as the overshoot can be easily obtain or detect from the graph. The characterisation
of the response can be easily determined by observing the trend of the graph.
While the over-damped and critically-damped of responses, the characterisation of the
response cannot easily determine as the trend of graph of both responses are almost the same.
Hence the damping coefficient, ξ have to be calculated by using overshoot from the graph
and then the value calculated is compared with the value of damping coefficient, ξ as shown
in the table below to determine the characterisation of the response.
Table 4: Characteristic of first order system response
Damping
Coefficient
Characterisation/ Types of
Response
Roots of Characteristic
Equation/ Types of Poles
ξ<1 Under-damped Real and unequal
ξ =1 Critically damped Real and equal
ξ>1 Over-damped Complex conjugates
8
• Problem identification problem for second order system
Figure 2.Graph generated for second order identification problem
𝑎 = 21.99-14.968 𝑏 = 14.968
tpeak1 = 25.36 sec tpeak2 =55.846 sec
Initial output value = 0 Final output value = 15
Overshoot in the response
𝑂𝑣𝑒𝑟𝑠ℎ𝑜𝑜𝑡 = 𝑎𝑏
𝑂𝑣𝑒𝑟𝑠ℎ𝑜𝑜𝑡 = 21.99 − 14.968
14.968
𝑶𝒗𝒆𝒓𝒔𝒉𝒐𝒐𝒕,𝑶𝑺 = 𝟎.𝟒𝟔𝟗𝟏
Period of the oscillatory response
Period = time between two successive peaks
Period = tpeak2 − tpeak1
Period = 55.846 − 25.36
𝐏𝐞𝐫𝐢𝐨𝐝,𝐏 = 𝟑𝟎.𝟒𝟖𝟔 𝐬𝐞𝐜
0 10 20 30 40 50 60 70 80 90 1000
5
10
15Input Profile
Time (sec)
Mag
nitu
de (-
)
0 10 20 30 40 50 60 70 80 90 1000
500
1000
1500
2000
2500
Output Profile
Time (sec)
Mag
nitu
de (-
)
FROM RESULT:
Initial value = 0
Final value = 15
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Different between initial and final output value (system gain), Kp
𝐾𝑝 = 𝑓𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑎𝑙𝑢𝑒 − 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑎𝑙𝑢𝑒
𝐾𝑝 = 15 − 0
𝑲𝒑 = 𝟏𝟓
Parameter values calculated
𝐾𝑝 = 𝑓𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑎𝑙𝑢𝑒 − 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑎𝑙𝑢𝑒
𝐾𝑝 = 15 − 0
𝑲𝒑 = 𝟏𝟓
𝐷𝑎𝑚𝑝𝑖𝑛𝑔𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡, ζ = � [ln(OS)]2
π2 + [ln(OS)]2 , OS = Overshoot
ζ = �[ln 0.4691]2
π2 + [ln 0.4691]2
ζ = 𝟎.𝟐𝟑𝟒𝟐
𝜏 = respond time
𝜏 = �1 − ζ2
2πP
𝜏 = �1 − 0.23422
2π(30.486)
𝝉 = 𝟔.𝟕𝟐𝟕𝟓
Kp 15 τ 6.7275 ζ 0.2342
Derivation of second order transfer function for this system
Standard form: 𝐺(𝑠) = 𝐾𝜏2𝑠2 + 2ζτs + 1
𝐺(𝑠) = 𝐾
𝜏2𝑠2 + 2ζτs + 1
𝐺(𝑠) = 15
(6.7275)2𝑠2 + 2(0.2342)(6.7275)s + 1
𝑮(𝒔) = 𝟏𝟓
𝟒𝟓.𝟑𝟓𝟗𝟑𝒔𝟐 + 𝟏.𝟓𝟕𝟓𝟔𝐬 + 𝟏
10
• Discussion on second order system (Exercise)
1.
Region Region 1 Region 2 Region 3 Damping coefficient
ζ<1 ζ =1 ζ>1
Types of poles Complex conjugate poles
Real and multiple poles
Real and distinct poles
Types of response Underdamped Critically damped Overdamped
2.
Region 1 Region 2 Region 3
3. As the damping coefficient is increase, it will decrease the speed of response and vice
versa.
4. Based on those three region, the shortest response time with overshoot is the region that
has the smaller damping coefficient which is ζ<1. The sluggish response time is the one that
has the larger damping coefficient which is ζ>1. The region with ζ =1 has the fastest response
without overshoot.
5. The 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 reach steady-state without overshoot. The responses fall into Region I is
the most common responses face by engineer. The control overshoot 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 is more preferable.
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CONCLUSION AND RECOMMENDATIONS
First order systems are, by definition, systems whose input-output relationship is a
first order differential equation. Most important parameters in first order system is the system
gain, Kp and the time constant, p. From the result, it was found that the increasing in Kp will
increase the system output. Kp with the highest value which is 30 and 40 give the highest
output result which is 30.0982 and 39.653. The larger the time constant p is, the slower the
system response is. From the result it showed that the largest p which is 20 gives the longest
time response which is 118.125.
Second order system is different than first order system where second order system is
more complex and has characteristics like decaying oscillations which are not to be seen in
first order systems (Mastascusa, 2002). From the result, it was found that for the second order
system, with the Kp value of 10, A=18 & B=2 the system is underdamped with the decay
ratio of 0.2131, overshoot of 0.4711, rise time of 17.608, settling time of 70.296 and period
of 27.419. While for other values of Kp, A and B the system is critically damped.
In this experiment, the graphs displayed were small. To get the accurate result, we
should zoom in the graphs until we can get a more accurate data display which consists of
four decimal points. Besides, we can prepare a programme coding in the MATLAB which
able to calculate the result data after key in the number of system gain KP to get a more
precious result data.
REFERENCES
1) Mastascusa E.J (2003). Retrieved on April 13rd 2014 from
http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/SysDyn/SysDyn1.html
2) RDM, (2000). Retrieved on April 13rd 2014 from
http://www.engin.umich.edu/group/ctm/working/mac/first_order/systemid/index.htm
3) RDM, (2000). Retrieved on April 13rd 2014 from
http://www.engin.umich.edu/group/ctm/working/mac/second_order/system_id.htm
4) Seborg, D. E. (2011). Process dynamics and control. Hoboken, NJ: Wiley.