# Experiment 4 Study on Dynamics of First Order and Second Order

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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

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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

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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

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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

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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 - - - - -

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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 − 015 − 0 𝑲𝒑 = 𝟎. 𝟐𝟏𝟒𝟒 Final output value minus initial output value

𝜏𝑃 = 𝑓𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑎𝑙𝑢𝑒 − 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑎𝑙𝑢𝑒 0 20 40 60 80 100 120 140 160 180 200 0 2 4 6 8 10 12 14 16 Input Profile Time (sec) M agni tude ( -) 0 20 40 60 80 100 120 140 160 180 200 0 1 2 3 4 5 Output Profile Time (sec) M agni tude ( -) FROM RESULT: Step time = 10 Initial value = 0 Final value = 15

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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) = 𝐾𝑝

𝜏𝑠+1

𝑌(𝑠) = 𝜏𝑠 + 1𝐾𝑝 𝒀(𝒔) = 𝟑. 𝟐𝟏𝟓𝟔𝒔 + 𝟏𝟎. 𝟐𝟏𝟒𝟒 • 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 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) = ∆𝒙(𝒕𝒊𝒎𝒆) = ∆𝒚 𝜏𝑠 + 1𝐾𝑝

Then Laplace it to be,

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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

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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 − 14.96814.968 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 90 100 0 5 10 15 Input Profile Time (sec) M agni tude ( -) 0 10 20 30 40 50 60 70 80 90 100 0 500 1000 1500 2000 2500 Output Profile Time (sec) M agni tude ( -) FROM RESULT: Initial value = 0 Final value = 15

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9 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 = �π2[ln(OS)] + [ln(OS)]2 2 , OS = Overshoot

ζ = �π2 + [ln 0.4691][ln 0.4691]2 2 ζ = 𝟎. 𝟐𝟑𝟒𝟐 𝜏 = respond time 𝜏 = �1 − ζ2π P 2 𝜏 = �1 − 0.2342 2(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) = 𝜏2𝑠2 + 2ζτs + 1𝐾

𝐺(𝑠) = 𝜏2𝑠2 + 2ζτs + 1 𝐾

𝐺(𝑠) = (6.7275)2𝑠2 + 2(0.2342)(6.7275)s + 1 15

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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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11 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

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