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Lecture 2

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McMaster University**We aren't endorsed by this school
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ENG TECH 4CT3
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Mechanical Engineering
Date
Jan 10, 2025
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30
Uploaded by BailiffDangerWalrus49
Lecture 2First & Second Order SystemsTime Domain Performance Measures14CT3 Fall 2024 Lec2 Dr. C. Tang
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Recap: Mass-Spring System•m=1.0 Kg•Spring constant k=2 N/m•Damping constant b=3 N/(m/s)•Input force f(t) = 1.0 N = u(t)•Output= mass location x(t)4CT3 Fall 2024 Lec2 Dr. C. Tang2??2𝑥(?)??2+ ??𝑥(?)??+ ?𝑥?= ?(?)?2𝑥(?)??2+ 3?𝑥(?)??+ ?𝑥?= ?(?)[?2𝑋?− ?𝑥0 − 𝑥′0 ] + 3?𝑋?− 𝑥0+ ?𝑋?= ?(?)DifferentialequationLaplaceTransform
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Solve for x(t), given f(t)=u(t), x(0)=x’(0)=04CT3 Fall 2024 Lec2 Dr. C. Tang3[?𝑋?− ?𝑥0 − 𝑥′0 ] + 3?𝑋?− 𝑥0+ 2𝑋?= ?(?)?2𝑋?+ 3?𝑋(?) + 2𝑋?= ?(?)𝑋??2+ 3? + 2 = ?(?)𝑋?=?(?)?2+ 3? + 2??=1?𝑋?=1?2+ 3? + 2×1?=1?(?2+ 3? + 2)𝑋?=1/2?−1? + 1+1/2? + 2𝑥?=12− ?−?+12?−2?(Laplace Transform)Solve for X(s)Laplace transform of the unit step function
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4CT3 Fall 2024 Lec2 Dr. C. Tang4Displacement x(t) [m]Time(sec)Final solution:
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4CT3 Fall 2024 Lec2 Dr. C. Tang5Time(sec)Displacement x(t) [m]b=3b=2b=1For different values of the damping constant ‘b’:
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Transfer Function, Input, Output, Block Diagram4CT3 Fall 2024 Lec2 Dr. C. Tang6
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System Transfer Function TF(s)In general, a system transfer function TF(s) relates output Laplace transform C(s) to input Laplace transform R(s):4CT3 Fall 2024 Lec2 Dr. C. Tang7C(s)R(s)?𝒙?+ ?𝒙 + ?C(s)/R(s) = TF(s)C(s) = TF(s) x R(s)i.e., Output = Input * System transfer function
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Poles and Zeros of Transfer FunctionsGiven a transfer function:TF?=?+2(?+1)(?+5)The zerosare the values of sthat make the transfer function to become 0 (s=-2) The polesare the values of sthat make the transfer function to become infinite. (s=-1 and s=-5)4CT3 Fall 2024 Lec2 Dr. C. Tang8poleszero
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Other Typical Input FunctionsInput Function•Unit impulse function δ(t)•Unit step function u(t)=1•Unit ramp function r(t)=tLaplace Transform•D(s)=1•U(s)=1/s•R(s)=1/s24CT3 Fall 2024 Lec2 Dr. C. Tang9
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First Order System: RL Circuit•Input = source voltage E(s)•Output = circuit current I(s)4CT3 Fall 2024 Lec2 Dr. C. Tang10Input = e(t)Output = i(t)
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Transient Response of Output i(t)4CT3 Fall 2024 Lec2 Dr. C. Tang11Time (s)Current [A]
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For different values of System Parameters4CT3 Fall 2024 Lec2 Dr. C. Tang12Increasing L
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First Order System Performance(Time Constant, Rise Time, Settling Time)The transfer function of a first order system is TF?=??+?Its step response (output=c(t)) is: 𝐶?=??+? ?in s-domain??= 1 − ?−??in time domainDefine time constant as τ= 1/aNote that after 1 τ, c(t) rises to 63% of its final value.•Rise time (Tr) is the time for the output to go from 0.1 to 0.9 of its final value. Tr=2.2/a•Settling time (Ts) is the time for the output to reach and stay within 2% of its final value, i.e., 2% settling time. •Ts=4/a4CT3 Fall 2024 Lec2 Dr. C. Tang13
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Example L2-1Given the step response of a first order system: ??= 1 − ?−??, show that:(a) Tr = 2.2/a(b) Ts = 4/a4CT3 Fall 2024 Lec2 Dr. C. Tang14
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Second Order SystemConsider a general second order system TF?=??2+??+?and its unit step response c(t)4CT3 Fall 2024 Lec2 Dr. C. Tang15R(s)=1/sC(s)The transient response c(t) is determined by the poles of the system transfer function.
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Case 1: Overdamped𝑮𝒔=𝟗𝒔?+ 𝟗𝒔 + 𝟗2 unequal real poles4CT3 Fall 2024 Lec2 Dr. C. Tang16s-planeTime domainxx
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Case 2: Critically damped𝑮𝒔=𝟗𝒔?+ 𝟔𝒔 + 𝟗2 equal real poles4CT3 Fall 2024 Lec2 Dr. C. Tang17s-planeTime domainxx
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Case 3: Underdamped𝑮𝒔=𝟗𝒔?+ ?𝒔 + 𝟗A pair of complex conjugate poles4CT3 Fall 2024 Lec2 Dr. C. Tang18xx
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Unit Step Response of 2ndOrder SystemBy partial fractions, 4CT3 Fall 2024 Lec2 Dr. C. Tang19φ = ???−1ξ1 − ξ2where
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Case 4: Undamped𝑮𝒔=𝟗𝒔?+ 𝟗Pair of complex conjugate poles on imaginary axis4CT3 Fall 2024 Lec2 Dr. C. Tang20xx
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Case 5: Negatively damped𝑮𝒔=𝟗𝒔?− ?𝒔 + 𝟗Pair of complex conjugate poles in the right half plane4CT3 Fall 2024 Lec2 Dr. C. Tang21xxpoles
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Second Order System –first 4 cases4CT3 Fall 2024 Lec2 Dr. C. Tang22undampedunderdampedoverdampedCritically damped
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Second Order System Poles•??=ω𝑛2?2+2ξω𝑛+ω𝑛2•ωn= undamped natural frequency•ξ= damping ratio•Poles are: ? = −ξω?± ω?1 − ξ2= σ + ?ωd•ωd= ωn1 − ξ2damped natural frequency•σ = damping constant4CT3 Fall 2024 Lec2 Dr. C. Tang23
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Damping Ratio ξ•ξ= 0 undamped case•0 < ξ< 1 underdamped case•ξ= 1 critically damped case•ξ> 1 overdamped case•ξ< 1 unstable case4CT3 Fall 2024 Lec2 Dr. C. Tang24
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Control System Design Specs4CT3 Fall 2024 Lec2 Dr. C. Tang25(e.g. 2%)
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Peak Time (Tp) and %Overshoot (OS)4CT3 Fall 2024 Lec2 Dr. C. Tang26
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%OS vs Damping Ratio ζ4CT3 Fall 2024 Lec2 Dr. C. Tang27The higher the damping ratio, the lower the %OS.
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Rise Time (Tr)4CT3 Fall 2024 Lec2 Dr. C. Tang28Rise time is the time it takes to go from 10% of the final value to 90% of the final value.ω𝑛𝑇?= 1.76𝜉3− 0.417𝜉2+ 1.039𝜉 + 1Approximately,
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Settling Time (Ts)4CT3 Fall 2024 Lec2 Dr. C. Tang29Ts is the time it takes for c(t) to settle within +/- 2% of the steady state value.
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Example L2-2 [Textbook Ex4.5]Given the transfer function of a 2ndorder system??=100?2+15?+100find Tp, %OS, Ts, Trof a unit step response.[Ans: Tp=0.475s, %OS=2.838, Ts=0.533s, Tr=0.23s]4CT3 Fall 2024 Lec2 Dr. C. Tang30
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