MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.004 Dynamics and Control II Fall 2007 Problem Set #3 Solution Posted: Friday, Sept. 28, ’07 1. A second–order system has the step response shown below.1 Determine its trans­ fer function. 2 f(t) [a.u.] 1.5 1 0.5 0 0 0.5 1 t [sec] 1.5 2 Answer: This is under–damped 2nd order system. Starting from the transfer function of the second order system A ωn2 , s2 + 2ζωn s + ω 2 we have to decide the parameters of A(constant),ζ(damping ratio) and ωn (natural frequency). From the final value theorem, 1 Aωn2 =A lim s 2 s→0 s s2 + 2ζωn s + ωn 1 a.u. denotes arbitrary units; its use appropriate when we consider a function that does not correspond to any particular physical quantity. 1 and the steady state value is 1 (from the given figure). Therefore, A = 1. The step response of the under–damped second order system is � � 1 − ae−σd t cos (ωd t − φ) u(t), where σd = ζωn and ωd = ωn � 1 − ζ 2. � � ζπ From the lecture note 7 (pp. 26), %OS = exp − 1−ζ : 72%. 2 Thus the damping ratio ζ ≈ 0.1. To get the natural frequency, we choose two peak points at t1 = 0.35 sec and t2 = 0.95 sec. The cosine term will be 1 at the peaks, so that we can consider exponential decay term only. f (t1 ) = 1 − ae−σd t1 = 1.72 f (t2 ) = 1 − ae−σd t2 = 1.4 Dividing the two equations, we obtain ae−σd t1 1 − 1.72 = . −σ t 1 d ae 1 − 1.4 � � �� From that σd = ln 0.72 / {t2 − t1 } = 0.9796. Therefore ωn ≈ 9.8. (The reason 0.4 why I picked two points instead of one point is to cancel the constant a). The transfer function is 64 , + 1.96s + 96 and its step response by MATLAB is s2 Step Response 1.8 1.6 1.4 Amplitude 1.2 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 Time (sec) Note that the estimated parameters might be slightly different than the original because our reading of the plot can never be completely accurate. 2 2. Consider again the system of a DC motor with a parallel current source connected via a gear pair to an inertia that we saw in Problem 5 of PS02. Substituting numerical values is = 1.0u(t) A, R = 5 Ω, Km = 0.5 N · m/A, Kv = 0.5 V · sec, Jm = 0.1 kg · m2 , (N2 /N1 ) = 10, J = 6 kg · m2 , K = 1 N · m/rad, derive and plot the step response for the following two cases: a) b = 9.4 N · m · sec/rad; b) b = 0.76 N · m · sec/rad. Answer: The transfer function is Θ(s) (N2 /N1 )Km � � � =� . 2 N N 2 Is (s) J + N 22 Jm s 2 + b + N 22 KvRKm s + K 1 1 Rearranging it, we can re–write it as (N2 /N1 )Km Θ(s) = Is (s) K s 2 + K/(J + (N2 /N1 )2 Jm ) bR+(N22 /N12 )Kv Km s R(J+(N22 /N12 )Jm ) + J +(N 2K/N 2 )Jm 2 . 1 The general form of the transfer function is ωn2 Θ(s) = A 2 , Is (s) s + 2ζωn s + ωn2 � where the natural frequency ωn = J+(N 2K/N 2 )Jm and the damping ratio ζ = 2 bR+(N22 /N12 )Kv Km 1 . R(J+(N22 /N12 )Jm ) 2ωn 1 a) b = 9.4 N · m · sec/rad; ωn = 0.25 and ζ = 1.8 > 1. (Over–damped system) Transfer function is 5/16 Θ(s) = 2 Is (s) s + (14.4/16)s + (1/4)2 whose poles are p1 = −0.8242 and p2 = −0.0758. To obtain its step re­ sponse, we do partial fraction expansion. K2 K3 K1 ωn2 = + + , A 2 2 s + 2ζωn s + ωn s s + 0.8242 s + 0.0758 where K1 = 5, K2 = 0.5067, K3 = −5.5067. The step response is f (t) = (K1 + K2 e−0.8242t + K3 e−0.0758t )u(t). 3 Step Response 5 4.5 4 3.5 θ(t) [rad] 3 2.5 2 1.5 1 0.5 0 0 10 20 30 40 50 60 70 80 t [sec] (sec) b) b = 0.76 N · m · sec/rad. ωn = 0.25 and ζ = 0.72 < 1. (Under–damped system) The transfer function is 5/16 Θ(s) = 2 Is (s) s + (5.76/16)s + (1/4)2 whose poles are p1 = −0.18 + j0.1735 and p1 = −0.18 − j0.1735. Doing partial fraction expansion, we obtain A K2 s + K3 K1 ωn2 = + 2 , 2 2 s + 2ζωn s + ωn s s + 2ζωn s + ωn2 where K1 = 5, K2 = −5, K3 = −1.8. The step response is f (t) = (K1 + K2 e−σd t cos(ωd t) + K3 e−σd t sin(ωd t))u(t). Step Response 6 5 θ(t) [rad] 4 3 2 1 0 0 5 10 15 20 t [sec] (sec) 4 25 30 35 3. Problem 8 from Nise textbook, Chapter 4 (page 234). Answer: Plotting the step response was not required; we did it here for completeness. a) T (s) = 2 s+2 Answer: Pole: p=-2, Zero: none Pole−Zero Map Step Response 2 1.5 1 0.8 0.5 Amplitude Imaginary Axis 1 0 −0.5 0.6 0.4 −1 0.2 −1.5 −2 −4 −3 −2 −1 0 0 0 0.5 1 Real Axis 1.5 2 2.5 3 2 2.5 3 Time (sec) Step response (1 − e−2t )u(t). 1st order system. b) T (s) = 5 (s+2)(s+6) Answer: Poles: p1 = −2, p2 = −6, Zero: none Pole−Zero Map Step Response 2 0.45 0.4 1.5 0.35 0.3 0.5 Amplitude Imaginary Axis 1 0 −0.5 0.2 0.15 −1 0.1 −1.5 −2 −8 0.25 0.05 −6 −4 −2 0 0 0 Real Axis 0.5 1 1.5 Time (sec) √ 2 ωn = 12 and ζ = 4/ 12 = 1.15 > 1. 2nd order overdamped system. Step response [1 + K1 e−p1 t + K2 e−p2 t ] u(t). c) T (s) = 10(s+7) (s+10)(s+20) Answer: Poles: p1 = −10, p2 = −20, Zeros: z1 = −7 5 Step Response 0.4 1.5 0.35 1 0.3 0.5 0.25 Amplitude Imaginary Axis Pole−Zero Map 2 0 0.2 −0.5 0.15 −1 0.1 −1.5 0.05 −2 −35 −30 −25 −20 −15 −10 −5 0 0 0 0.1 Real Axis 0.2 0.3 0.4 Time (sec) √ ωn2 = 200 and ζ = 30/2/ 200 = 1.06 > 1. 2nd order overdamped system. Step response [1 + K1 e−p1 t + K2 e−p2 t ] u(t). d) T (s) = 20 s2 +6s+144 Answer: Poles: p1 = −3 + j11.619, p2 = −3 − j11.619, Zeros: none Pole−Zero Map Step Response 15 0.25 10 0.2 Amplitude Imaginary Axis 5 0 0.15 0.1 −5 0.05 −10 −15 −4 −3 −2 −1 0 0 0 0.5 Real Axis 1 1.5 2 Time (sec) √ ωn2 = 144 and ζ = 6/2/ 144 = 0.25 < 1. 2nd order underdamped system. Step response [1 − Ae−σd t cos (ωd t − φ)] u(t). e) T (s) = s+2 s2 +9 Answer: Poles: p1 = 3j, p2 = −3j, Zero: z = −2 6 Pole−Zero Map Step Response 4 0.7 3 0.6 0.5 0.4 1 Amplitude Imaginary Axis 2 0 −1 0.2 0.1 −2 0 −3 −4 −3 0.3 −0.1 −2.5 −2 −1.5 −1 −0.5 0 −0.2 0 0.5 5 10 Real Axis 15 20 Time (sec) ωn2 = 9 and ζ = 0. 2nd order undamped system. Step response [1 − K1 sin (3t) + K2 cos (3t)] u(t). f ) T (s) = (s+5) (s+10)2 Answer: Poles: p = −10(double), Zeors: z = −5 Pole−Zero Map Step Response 2 0.06 1.5 0.05 0.04 0.5 Amplitude Imaginary Axis 1 0 −0.5 0.03 0.02 −1 0.01 −1.5 −2 −12 −10 −8 −6 −4 −2 0 0 0 0.1 Real Axis 0.2 0.3 0.4 Time (sec) ωn2 = 100 and ζ = 1. 2nd order critically damped system. Step response [K0 + K1 e−10t − K2 te−10t ] u(t). 7 0.5 0.6 0.7