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Understanding First and Second Order Control Systems

School
McMaster University**We aren't endorsed by this school
Course
SFWR ENG 4AA4
Subject
Electrical Engineering
Date
Dec 11, 2024
Pages
24
Uploaded by ChancellorPorcupinePerson766
Real Time Systems and Control ApplicationsContentsFirst Order SystemsSecond Order SystemsFall, 2024Prof. Wenbo He@CAS, McMaster1
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Time Constant of First Order Systemsβ€’First order systemβ€’The output of a general first order system to a step input: π‘Œπ‘Œπ‘ π‘ =𝑋𝑋𝑠𝑠𝐺𝐺(𝑠𝑠) =π‘Žπ‘Žπ‘ π‘ π‘ π‘ +π‘Žπ‘Žβ€’This results in a time domain output given by: y𝑑𝑑= 1βˆ’ π‘’π‘’βˆ’π‘Žπ‘Žπ‘‘π‘‘, where parameter π‘Žπ‘Žis the only parameter that affects the output.𝑑𝑑=1π‘Žπ‘Ž, y𝑑𝑑= 0.63.1π‘Žπ‘Žis the time constant of the response, and is the time it takes for the step response to rise to 63% of its final value.Fall, 2024Prof. Wenbo He@CAS, McMaster2π‘Žπ‘Žπ‘ π‘ + aX(s)=1𝑠𝑠Y(s)G(s)
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Response in Time Domainβ€’Rise Time (π‘‡π‘‡π‘Ÿπ‘Ÿ): time for the waveform to go from 0.1 to 0.9 of its final value. For first order systems: π‘‡π‘‡π‘Ÿπ‘Ÿ=2.2π‘Žπ‘Žβ€’Settling Time (𝑇𝑇𝑠𝑠): time for the response to reach and stay within 2% of its final value. For first order systems: 𝑇𝑇𝑠𝑠=4π‘Žπ‘ŽFall, 2024Prof. Wenbo He@CAS, McMaster3
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Another First Order Response Exampleβ€’Give 𝐺𝐺𝑠𝑠=1𝑠𝑠+π‘Žπ‘Ž, what is the time constant, rise time, and settling time?β€’Y𝑠𝑠=1π‘Žπ‘Ž1sβˆ’1𝑠𝑠+π‘Žπ‘Žβ‡’ 𝑦𝑦𝑑𝑑=1π‘Žπ‘Ž(1βˆ’ π‘’π‘’βˆ’π‘Žπ‘Žπ‘‘π‘‘)Solution:β€’Timeconstant𝜏𝜏makes π‘¦π‘¦πœπœ=1π‘Žπ‘Žβˆ—0.63, so 𝜏𝜏=1π‘Žπ‘Žβ€’Rise time π‘‘π‘‘π‘Ÿπ‘Ÿ=𝑑𝑑2βˆ’ 𝑑𝑑1, where 𝑑𝑑2makes 𝑦𝑦𝑑𝑑2=1π‘Žπ‘Žβˆ—0.9and 𝑑𝑑1makes 𝑦𝑦𝑑𝑑1=1π‘Žπ‘Žβˆ—0.1, hence π‘‘π‘‘π‘Ÿπ‘Ÿ=2.3βˆ’0.1π‘Žπ‘Ž= 2.2βˆ—1π‘Žπ‘Žβ€’Settling time 𝑑𝑑𝑠𝑠makes 𝑦𝑦𝑑𝑑𝑠𝑠=0.981π‘Žπ‘Ž, hence 𝑑𝑑𝑠𝑠≅4π‘Žπ‘ŽFall, 2024Prof. Wenbo He@CAS, McMaster4
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More ExampleThe figure shows the response of three first order systems having transfer function 𝐾𝐾𝑠𝑠+π‘Žπ‘Ž, where the values of K are different for the three systems. Answer the following questions:β€’1. Which of the three curves (1, 2, 3) represents a system with the lowest time constant?β€’2. The big dots on the three graphs represent the time when the response settles within 2% of the final value. Find the transfer function for each of the three systems. Fall, 2024Prof. Wenbo He@CAS, McMaster5
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SolutionFall, 2024β€’The settling time for first order systems is given by Ts=4π‘Žπ‘Ž.β€’From the figure, the values of Tsare 7.8, 3.9 and 0.8 respectively, so the value π‘Žπ‘Žfor 3 systems are roughly 0.5, 1.0 and 5 respectively.Since the steady state value of each system is 1, so K=a. Therefore the transfer function of the systems are 0.5𝑠𝑠+0.5, 1𝑠𝑠+1, and 5𝑠𝑠+5.Prof. Wenbo He@CAS, McMaster6
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Second Order Systemsβ€’Most real-world systems are not first order systems. A general second order system defined by the transfer function:𝐺𝐺(𝑠𝑠) =𝑏𝑏𝑠𝑠2+π‘Žπ‘Žπ‘ π‘ +𝑏𝑏‒Find the poles of this transfer function to examine the behaviour of the output response. Using quadratic formula:𝑠𝑠1,𝑠𝑠2=βˆ’π‘Žπ‘ŽΒ±π‘Žπ‘Ž2βˆ’4b2Fall, 2024Prof. Wenbo He@CAS, McMaster7
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A Special Case (a=0)β€’If a = 0, the transfer function is Gs=𝑏𝑏𝑠𝑠2+𝑏𝑏, and the poles will have only imaginary part ±𝑗𝑗𝑗𝑗and by definition the natural frequency𝑗𝑗𝑛𝑛=𝑏𝑏is the frequency of oscillation of this system.Fall, 2024Prof. Wenbo He@CAS, McMaster8Undamped Case
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Damping Coefficient ΞΆ, whenπ‘Žπ‘Žis not zeroβ€’The complex poles have a real part Οƒ=βˆ’π‘Žπ‘Ž2.β€’The magnitude of Οƒis called the exponential decay frequency, and 𝑗𝑗𝑛𝑛the natural frequency. We define the Damping Ratio or Damping Coefficient, 𝜁𝜁asΞΆ=πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπ‘’π‘’πΈπΈπ‘‘π‘‘πΈπΈπ‘Žπ‘ŽπΈπΈ π‘‘π‘‘π‘’π‘’π‘‘π‘‘π‘Žπ‘Žπ‘¦π‘¦ π‘“π‘“π‘“π‘“π‘’π‘’π‘“π‘“π‘“π‘“π‘’π‘’πΈπΈπ‘‘π‘‘π‘¦π‘¦π‘π‘π‘Žπ‘Žπ‘‘π‘‘π‘“π‘“π‘“π‘“π‘Žπ‘ŽπΈπΈ π‘“π‘“π‘“π‘“π‘’π‘’π‘“π‘“π‘“π‘“π‘’π‘’πΈπΈπ‘‘π‘‘π‘¦π‘¦βˆ΄ΞΆ=|Οƒ|𝑗𝑗𝑛𝑛=π‘Žπ‘Ž2𝑗𝑗𝑛𝑛𝑠𝑠𝐸𝐸 π‘‘π‘‘π‘‘π‘Žπ‘Žπ‘‘π‘‘ π‘Žπ‘Ž= 2΢𝑗𝑗𝑛𝑛Fall, 2024Prof. Wenbo He@CAS, McMaster9
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General Second Order Transfer Function β€’The general second order transfer function can now be written as:Gs=𝑗𝑗𝑛𝑛2𝑠𝑠2+ 2΢𝑗𝑗𝑛𝑛𝑠𝑠+𝑗𝑗𝑛𝑛2𝑠𝑠1,𝑠𝑠2=βˆ’ΞΆπ‘—π‘—π‘›π‘›Β±π‘—π‘—π‘›π‘›ΞΆ2βˆ’1β€’From above we can examine the effect of parameter ΞΆon the output of a second order system.Fall, 2024Prof. Wenbo He@CAS, McMaster10
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Effect Of Parameter ΞΆFall, 2024Prof. Wenbo He@CAS, McMaster11
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Summary of Observationsβ€’Two imaginary poles at Β±π‘—π‘—πœ”πœ”π‘›π‘›: ΞΆ= 0 (undamped)Natural response: undamped sinusoid of frequency πœ”πœ”π‘›π‘›equal to the imaginary part of the poles. Or 𝑑𝑑𝑑𝑑=𝐴𝐴cos(πœ”πœ”π‘›π‘›π‘‘π‘‘ βˆ’ πœ‘πœ‘)β€’Two complex poles at Οƒπ‘‘π‘‘Β±π‘—π‘—πœ”πœ”π‘‘π‘‘:0 <ΞΆ< 1 (underdamped)Natural response: Underdamped response in the form of sinusoid with an exponential envelope whose time constant is equal to the reciprocal of the pole's real part. Or 𝑑𝑑𝑑𝑑=𝐴𝐴 𝑒𝑒(βˆ’πœŽπœŽπ‘‘π‘‘)𝑑𝑑cos(πœ”πœ”π‘‘π‘‘π‘‘π‘‘ βˆ’ πœ‘πœ‘), where 𝑗𝑗𝑑𝑑=𝑗𝑗𝑛𝑛1βˆ’ΞΆ2.Fall, 2024Prof. Wenbo He@CAS, McMaster12
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Continued…‒Two real poles at Οƒ1: ΞΆ= 1 (critically damped)Natural response: critically damped system has the time domain response as:𝑑𝑑𝑑𝑑=𝐾𝐾𝑑𝑑𝑒𝑒σ1𝑑𝑑‒Two real poles at Οƒ1and Οƒ2: ΞΆ> 1 (overdamped)Natural response: overdamped with two exponentials having timeconstants equal to the reciprocal of the pole locations. Or𝑑𝑑𝑑𝑑=𝐾𝐾(𝑒𝑒σ1𝑑𝑑+𝑒𝑒σ2𝑑𝑑)Fall, 2024Prof. Wenbo He@CAS, McMaster13
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Second Order Impulse ResponseFall, 2024Prof. Wenbo He@CAS, McMaster14
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Underdamped Second Order Step Responseβ€’The general Transfer Function of a second order system is:Gs=𝑗𝑗𝑛𝑛2𝑠𝑠2+ 2΢𝑗𝑗𝑛𝑛𝑠𝑠+𝑗𝑗𝑛𝑛2β€’Consider response for a step input. The transfer function of response C(s) is given by:Cs=𝑗𝑗𝑛𝑛2𝑠𝑠(𝑠𝑠2+ 2΢𝑗𝑗𝑛𝑛𝑠𝑠+𝑗𝑗𝑛𝑛2)β€’Taking the inverse LT to get response in time domain results in:𝑑𝑑𝑑𝑑= 1βˆ’11βˆ’ΞΆ2π‘’π‘’βˆ’ΞΆπ‘€π‘€π‘›π‘›π‘‘π‘‘cos(1βˆ’ΞΆ2πœ”πœ”π‘›π‘›π‘‘π‘‘+πœ‘πœ‘)where πœ‘πœ‘=π‘‘π‘‘π‘Žπ‘ŽπΈπΈβˆ’1(ΞΆ1βˆ’ΞΆ2).Fall, 2024Prof. Wenbo He@CAS, McMaster15
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Fall, 2024Prof. Wenbo He@CAS, McMaster16
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Peak Time, TpThe time required to reach the first or maximum peak. This can be found by differentiating c(t), and equating to zero which gives:𝑇𝑇𝑝𝑝=πœ‹πœ‹πœ”πœ”π‘›π‘›1βˆ’ΞΆ2Because c’(t)= βˆ’11βˆ’ΞΆ2π‘’π‘’βˆ’ΞΆπ‘€π‘€π‘›π‘›π‘‘π‘‘sin1βˆ’ΞΆ2πœ”πœ”π‘›π‘›π‘‘π‘‘+πœ‘πœ‘1βˆ’ΞΆ2πœ”πœ”π‘›π‘›βˆ’11βˆ’ΞΆ2βˆ’ΞΆπ‘—π‘—π‘›π‘›π‘’π‘’βˆ’ΞΆπ‘€π‘€π‘›π‘›π‘‘π‘‘cos1βˆ’ΞΆ2πœ”πœ”π‘›π‘›π‘‘π‘‘+πœ‘πœ‘= 0βˆ΄π‘‘π‘‘π‘Žπ‘ŽπΈπΈ1βˆ’ΞΆ2πœ”πœ”π‘›π‘›π‘‘π‘‘+πœ‘πœ‘=ΞΆ1βˆ’ΞΆ2Therefore, 1βˆ’ΞΆ2πœ”πœ”π‘›π‘›π‘‘π‘‘=πœ‹πœ‹οƒ¨π‘‡π‘‡π‘π‘=πœ‹πœ‹πœ”πœ”π‘›π‘›1βˆ’ΞΆ2Fall, 2024Prof. Wenbo He@CAS, McMaster17𝑑𝑑𝑑𝑑= 1βˆ’11βˆ’ΞΆ2π‘’π‘’βˆ’ΞΆπ‘€π‘€π‘›π‘›π‘‘π‘‘cos(1βˆ’ΞΆ2πœ”πœ”π‘›π‘›π‘‘π‘‘+πœ‘πœ‘)πœ‘πœ‘=π‘‘π‘‘π‘Žπ‘ŽπΈπΈβˆ’1(ΞΆ1βˆ’ΞΆ2)
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Percent Overshoot, % OSThe amount that the waveform overshoots the steady state of final value at peak time, expressed as percentage of steady state value: %OS =π‘π‘π‘šπ‘šπ‘šπ‘šπ‘šπ‘šβˆ’π‘π‘π‘“π‘“π‘“π‘“π‘›π‘›π‘šπ‘šπ‘“π‘“π‘π‘π‘“π‘“π‘“π‘“π‘›π‘›π‘šπ‘šπ‘“π‘“Γ— 100, where π‘‘π‘‘π‘“π‘“π‘“π‘“π‘›π‘›π‘Žπ‘Žπ‘“π‘“= 1and π‘‘π‘‘π‘šπ‘šπ‘Žπ‘Žπ‘šπ‘š=𝑑𝑑𝑇𝑇𝑝𝑝.Substituting the expression for 𝑑𝑑𝑇𝑇𝑝𝑝= 1 +π‘’π‘’βˆ’πœ‹πœ‹ΞΆ1βˆ’ΞΆ2in previous subsection and some manipulation results in: %OS =π‘’π‘’βˆ’πœ‹πœ‹ΞΆ1βˆ’ΞΆ2Γ— 100.Note that %OS is a function of ΞΆ, the damping ratio only. The above expression gives an expression for ΞΆin terms of %OS.ΞΆ=βˆ’ln(%𝑂𝑂𝑂𝑂100)πœ‹πœ‹2+𝐸𝐸𝐸𝐸2(%𝑂𝑂𝑂𝑂100)Fall, 2024Prof. Wenbo He@CAS, McMaster18
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Finding 𝑇𝑇𝑝𝑝and %OS From Transfer Functionβ€’Given a transfer function 𝐺𝐺𝑠𝑠=100𝑠𝑠2+15𝑠𝑠+100, find 𝑇𝑇𝑝𝑝and %OS.Solution: πœ”πœ”π‘›π‘›=100 = 10and ΞΆ=π‘šπ‘š2πœ”πœ”π‘›π‘›=15/20 = 0.75.∴ 𝑇𝑇𝑝𝑝=πœ‹πœ‹πœ”πœ”π‘›π‘›1βˆ’ΞΆ2= 0.475𝑠𝑠And finally, %OS =π‘’π‘’βˆ’πœ‹πœ‹ΞΆ1βˆ’ΞΆ2Γ— 100 =π‘’π‘’βˆ’0.75πœ‹πœ‹1βˆ’0.752Γ— 100 = 2.838%.Fall, 2024Prof. Wenbo He@CAS, McMaster19
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How are 𝑇𝑇𝑝𝑝And 𝑇𝑇𝑠𝑠Related To Location of Poles𝑇𝑇𝑝𝑝=πœ‹πœ‹πœ”πœ”π‘›π‘›1βˆ’ΞΆ2𝑇𝑇𝑠𝑠≅4ΞΆπœ”πœ”π‘›π‘›Fall, 2024Prof. Wenbo He@CAS, McMaster20Poles of a second order underdamped systemGs=𝑗𝑗𝑛𝑛2𝑠𝑠2+ 2΢𝑗𝑗𝑛𝑛𝑠𝑠+𝑗𝑗𝑛𝑛2𝑠𝑠1,𝑠𝑠2=βˆ’ΞΆπ‘—π‘—π‘›π‘›Β±π‘—π‘—π‘—π‘—π‘›π‘›ΞΆ2βˆ’1
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Lines of Constants 𝑇𝑇𝑝𝑝, 𝑇𝑇𝑠𝑠and %OSon s-Plane β€’Note that horizontal lines on s-plane are lines of constant πœ”πœ”π‘‘π‘‘consequently they represent lines of constant Tp. Also vertical lines represent constant values of and are therefore lines of constant Ts.β€’Finally, since ΞΆ=π‘‘π‘‘πΈπΈπ‘ π‘ πœƒπœƒ, radial lines represent lines of constant damping ratio. But %Overshoot depends only on ΞΆ.Fall, 2024Prof. Wenbo He@CAS, McMaster21%OS =π‘’π‘’βˆ’πœ‹πœ‹ΞΆ1βˆ’ΞΆ2Γ— 100
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Location Of Poles vs Response (1)Fall, 2024Prof. Wenbo He@CAS, McMaster22
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Location Of Poles vs Response (2)Fall, 2024Prof. Wenbo He@CAS, McMaster23
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Location Of Poles vs Response (3)Fall, 2024Prof. Wenbo He@CAS, McMaster24
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