Review: step response of 1st order systemsReview: poles, zeros, and the forced/natural responsesGoals for todayDC motor system with non-negligible inductanceStep response of 2nd order system – large R/LStep response of 2nd order system – large R/LStep response of 2nd order system – large R/LComparison of 1st order and 2nd order overdampedStep response of 2nd order system – small R/LComparison of 1st order and 2nd order underdampedOverdamped DC motor: derivation of the step responseOverdamped DC motor in the s-domainUndamped DC motor: no dissipationUndamped DC motor in the s-domainUnderdamped DC motor: small dissipationUnderdamped DC motor: small dissipationUnderdamped DC motor: small dissipationWhat the real and imaginary parts of the poles doUnderdamped DC motor in the s-domainThe general 2nd order systemThe general 2nd order systemThe underdamped 2nd order systemThe underdamped 2nd order systemThe underdamped 2nd order systemTransients in the underdamped 2nd order systemTransient qualities from pole location in the s-planeLecture 07 – Wednesday, Sept. 192.004 Fall ’07 Review: step response of 1storder systemsFigure 4.3Step responsein the s—domainas(s + a);in the time domain¡1 − e−at¢u(t);time constantτ =1a;rise time (10%→90%)Tr=2.2a;settling time (98%)Ts=4a.1.00.90.80.70.60.50.40.30.20.101a2a3a4atc(t)5aTsTr63% of final valueat t = one time constantInitial slope = time constant1= asteady state(final value)Figure by MIT OpenCourseWare.Lecture 07 – Wednesday, Sept. 192.004 Fall ’07 Review: poles, zeros, and the forced/natural responsesσjωσjω0−50−5−2input pole –forced responsesystem pole –natural responsesystem zero –derivative & amplificationLecture 07 – Wednesday, Sept. 192.004 Fall ’07 Goals for today• Second-order systems response– types of 2nd-order systems• overdamped• underdamped• undamped• critically damped– transient behavior of overdamped 2nd-order systems– transient behavior of underdamped 2nd-order systems– DC motor with non-negligible impedance• Next lecture (Friday):– examples of modeling & transient calculations for electro-mechanical 2ndorder systemsLecture 07 – Wednesday, Sept. 192.004 Fall ’07 DC motor system with non-negligible inductanceRecall 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)Including the DC motor’s inductance, we find⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩Ω(s)Vs(s)=KmLJ1s2+µbJ+RL¶s +µbR + KmKvLJ¶I(s)Vs(s)=1Rµs +bJ¶s2+µbJ+RL¶s +µbR + KmKvLJ¶Quadratic polynomial denominatorSecond—order systemLecture 07 – Wednesday, Sept. 192.004 Fall ’07 Step response of 2ndorder system – large R/LL =0.1H, Kv=6V· sec, Km=6N· m/A, J =2kg· m2,R =6Ω, b =4kg· m2· Hz; vs(t)=30u(t)V.Overdampedresponse↔dissipation >energy storageLecture 07 – Wednesday, Sept. 192.004 Fall ’07 Step response of 2ndorder system – large R/LL =0.1H, Kv=6V· sec, Km=6N· m/A, J =2kg· m2,R =6Ω, b =4kg· m2· Hz; vs(t)=30u(t)V.OverdampedresponseLecture 07 – Wednesday, Sept. 192.004 Fall ’07 Step response of 2ndorder system – large R/LL =0.1H, Kv=6V· sec, Km=6N· m/A, J =2kg· m2,R =6Ω, b =4kg· m2· Hz; vs(t)=30u(t)V.OverdampedresponseLecture 07 – Wednesday, Sept. 192.004 Fall ’07 Comparison of 1storder and 2ndorder overdamped1storder(L≈0)2ndorder(L=0.1H)Lecture 07 – Wednesday, Sept. 192.004 Fall ’07 Step response of 2ndorder system – small R/LL =1.0H, Kv=6V· sec, Km=6N· m/A, J =2kg· m2,R =6Ω, b =4kg· m2· Hz; vs(t)=30u(t)V.Underdampedresponse↔dissipation <energy storageovershootLecture 07 – Wednesday, Sept. 192.004 Fall ’07 Comparison of 1storder and 2ndorder underdamped1storder(L≈0)2ndorder(L=1.0H)overshootNO overshootLecture 07 – Wednesday, Sept. 192.004 Fall ’07 Overdamped DC motor: derivation of the step responseUsing the numerical values L =0.1H, Kv=6V· sec, Km=6N· m/A, J =2kg · m2, R =6Ω, b =4kg· m2· Hz we findKmLJ=30radsec · V;bJ+RL=62rad/sec;bR + KmKvLJ=300(rad/sec)2.Therefore, the transfer function for the angular velocity isΩ(s)Vs(s)=30s2+62s +300.We find that the denominator has two real roots,s1= −5.290Hz,s2= −56.71Hz ⇒Ω(s)Vs(s)=30(s +5.290)(s +56.71).To compute the step response w e substitute the Laplace transform of t he voltagesource Vs(s)=30/s and carry out the partial fraction expansion:Ω(s)=900s(s +5.290)(s +56.71)=3s−3.3s +5.290+0.3s +56.71⇒ω(t)=£3 − 3.3e−5.29t+0.3e−56.71t¤u(t).This is the function whose plot we analyzed in slides #5—8.A2nd—order system is overdampedif the transfer function denominatorhas two real roots.Lecture 07 – Wednesday, Sept. 192.004 Fall ’07 Overdamped DC motor in the s-domain−5.29−56.710inputpolesystem poles(2ndorder overdamped)−51stordersystem poleσjωjωLecture 07 – Wednesday, Sept. 192.004 Fall ’07 Undamped DC motor: no dissipationConsider the o pposite extreme where the dissipation due to both the resistor andbearings friction is negligible, i.e. R =0andb = 0. Using the same remainingnumerical values L =0.1H, Kv=6V· sec, Km=6N· m/A, J =2kg· m2,wefindKmLJ=30radsec · V;bJ+RL=0;bR + KmKvLJ=180(rad/sec)2.Therefore, the transfer function for the angular velocity isΩ(s)Vs(s)=30s2+180.The denominator has a conjugate pair of two imaginary roots,s1,2= ±j13.42Hz ⇒Ω(s)Vs(s)=30(s + j13.42)(s − j13.42).Again, the step r esponse is found by partial fraction expansion:Ω(s)=900s(s + j13.42)(s − j13.42)=5s−5ss2+(13.42)2⇒ω(t)=[5− 5cos(13.42t)] u(t).A2nd—order system is undampedif the transfer function denominator has aconjugate pair of two imaginary roots.Lecture 07 – Wednesday, Sept. 192.004 Fall ’07 Undamped DC motor in the s-domainj13.420inputpolesystem poles(2ndorder undamped)−51stordersystem poleσjω−j13.42Natural frequency ωn=13.42rad/sec.Period T =2π/ωn=4.24sec.Lecture 07 – Wednesday, Sept. 192.004 Fall ’07 Underdamped DC motor: small dissipationFinally, let us return to what we previously labelled as “underdamped” case,i.e. L =1.0H, Kv=6V· sec, Km=6N· m/A, J =2kg· m2, R =6Ω,b =4kg· m2· Hz. The values of L, R are such t hat the dissipation in the systemis negligible compared to the energy storage capacity. We then findKmLJ=3radsec · V;bJ+RL= 8rad/sec;bR + KmKvLJ=30(rad/sec)2.Therefore, the transfer function for the angular velocity
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