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Review: step response of 1st order systems we’ve seenGoals for todayCurrent dynamics in DC motor systemPoles and zeros in the s-planeDC motor step response: numerical exampleDC motor step response: numerical exampleDC motor system in the s-planeDC motor step response (angular velocity)DC motor step response (current)1st order system response from s-plane representation1st order system: transient response propertiesDC motor step responsesThe Final Value theorem: steady-stateThe Initial Value theorem: initial slopeInitial and final value of 1st-order system with a zeroHow the zero actsLecture 06 – Monday, Sept. 172.004 Fall ’07 Review: step response of 1storder systems we’ve seen• Inertia with bearings (viscous friction)• RC circuit (charging of a capacitor)• DC motor with inertia load, bearings and negligible inductance+−vi+−CvCRStep input Ts(t)=T0u(t) ⇒ Step response∗ω(t)=T0b³1 − e−t/τ´, where τ =Jb.Step input vi(t)=V0u(t) ⇒ Step responsevC(t)=V0³1 − e−t/τ´, where τ = RC.Step input vs(t)=V0u(t) ⇒ Step responseω(t)=KmRV0³1 − e−t/τ´,where τ =JÃb +KmKvR!.∗When writing the step response,we may omit t he step function u(t)implying the result holds for t>0only.Lecture 06 – Monday, Sept. 172.004 Fall ’07 Goals for today• First-order systems response– pole, zero definitions– the significance of poles and zeros:from s-domain representation to transient characteristics• DC motor dynamics:– angular velocity (1storder: 1 pole)– current (1storder: 1 pole, 1 zero)• Two new properties of the Laplace transform:– final value theorem– initial value theorem• Next two lectures:–2ndorder systemsLecture 06 – Monday, Sept. 172.004 Fall ’07 Current dynamics in DC motor systemRecall combined equations of motionLsI(s)+RI(s)+KvΩ(s)=Vs(s)JsΩ(s)+bΩ(s)=KmI(s))⇒⎧⎪⎨⎪⎩·LJRs2+µLbR+ J¶s +µb +KmKvR¶¸Ω(s)=KmRVs(s)(Js + b) Ω(s)=KmI(s)Neglecting the DC motor’s inductance (i.e.,assumingL/R ≈ 0), we find⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩Ω(s)Vs(s)=KmRJs +1Jµb +KmKvR¶I(s)Vs(s)=1Rµs +bJ¶s +1Jµb +KmKvR¶Transfer function for t he angular velocityis of the formAs + p,p=1Jµb +KmKvR¶Transfer function for the currentis of the formB (s + z)s + p,p=1Jµb +KmKvR¶,z=bJpolezeroLecture 06 – Monday, Sept. 172.004 Fall ’07 Poles and zeros in the s-planeσjωs—plane−ppole−zzeroleft half—planeright half—planeimaginary axisreal axisLecture 06 – Monday, Sept. 172.004 Fall ’07 DC motor step response: numerical exampleL ≈ 0, R =6Ω, Kv=6V· sec,Km=6N· m/A, J =2kg· m2, b =4kg· m2· Hz.We will compute the system’s response(both angular velocity and current)to the step input vs(t)=30u(t)V.Substituting the numerical values into the system TF,we find⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩Ω(s)Vs(s)=121s +5[Hz]I(s)Vs(s)=16(s +2[Hz])s +5[Hz]whereas the Laplace transform of the input isVs(s) ≡ Lhvs(t)i= Lh30u(t)i=30s.I. Angular velocityΩ(s)=15s (s +5)=15µK1s+K2s +5¶where K1=1s +5¯¯¯¯s=0=15,K2=1s¯¯¯¯s=−5= −15⇒Ω(s)=3µ1s−1s +5¶⇒ ω(t)=3¡1 − e−5t¢u(t)radsec.Forced responseHomogeneous response(Natural response)Lecture 06 – Monday, Sept. 172.004 Fall ’07 DC motor step response: numerical exampleL ≈ 0, R =6Ω, Kv=6V· sec,Km=6N· m/A, J =2kg· m2, b =4kg· m2· Hz.We will compute the system’s response(both angular velocity and current)to the step input vs(t)=30u(t)V.Substituting the numerical values into the system TF,we find⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩Ω(s)Vs(s)=121s+5[Hz]I(s)Vs(s)=16(s+2[Hz])s+5[Hz]whereas the Laplace transform of the input isVs(s) ≡ Lhvs(t)i= Lh30u(t)i=30s.II. CurrentI(s)=5(s +2)s (s +5)=5µK10s+K20s +5¶where K10=s +2s +5¯¯¯¯s=0=25,K20=s +2s¯¯¯¯s=−5=35⇒I(s)=µ2s+3s +5¶⇒ i(t)=¡2+3e−5t¢u(t)A.Forced responseHomogeneous response(Natural response)Lecture 06 – Monday, Sept. 172.004 Fall ’07 DC motor system in the s-planeVs(s)=30s121s +5Ω(s)Gω(s)Vs(s)=30s16s +2s +5I(s)Gi(s)Input: Voltage sourceOutput: Angular velocityInput: Voltage sourceOutput: CurrentsystemrepresentationsystemrepresentationLecture 06 – Monday, Sept. 172.004 Fall ’07 DC motor step response (angular velocity)Image removed due to copyright restrictions.Please see: Fig. 4.1 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.Lecture 06 – Monday, Sept. 172.004 Fall ’07 DC motor step response (current)Image removed due to copyright restrictions.Please see: Fig. 4.1 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.Lecture 06 – Monday, Sept. 172.004 Fall ’07 1storder system response from s-plane representation• Pole at –α generates responsee–αt(exponentially decreasing if pole on the right half-plane;increasing if on the left half-plane)• Pole at zero generates step function• Pole in the input function generates forced response• Pole in the transfer function generates natural response• Zero in the transfer function does not alter the speed of settling to steady state (i.e. the time constant) but it does alter the relative amplitudes of the forced and natural responsesσjω−αe−αtu(t)σjω0u(t)Time constant τ=1/αLecture 06 – Monday, Sept. 172.004 Fall ’07 1storder system: transient response propertiesFigure 4.3Step responsein the s—domaina;s(s + a)in the time domain¡1 − e−at¢u(t);time constant1τ =;arise time (10%→90%)2.2Tr=;asettling time (98%)4Ts=.a1.00.90.80.70.60.50.40.30.20.101a2a3a4atc(t)5aTsTr63% of final valueat t = one time constantInitial slope = time constant1= asteady state(final value)Figure by MIT OpenCourseWare.Lecture 06 – Monday, Sept. 172.004 Fall ’07 DC motor step responsesω( t)=3¡1 − e−5t¢u(t)rad/secdωdt(0+) = 3 × 5=15 rad/sec2ω(∞)=3 rad/seci(t)=¡2+3e−5t¢u(t)Adidt(0+) = ∞i(∞)=2 ALecture 06 – Monday, Sept. 172.004 Fall ’07 The Final Value theorem: steady-stateWe will now learn two additional properties of the Laplace transform, which wewill quote without proof. L et F (s) denote the Laplace transform of the functionf(t). The first property is theFinal Value theorem:f(∞)=lims→0sF (s);Let us see how this applies to the step response of a general 1st—order systemwith a pole at −a and without a zero (e.g., the angular velocity response of theDC motor.) We select the system gain such that the steady—state will equal 1.The step


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