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Review: Step Response of 1St Order Systems

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Review: Step Response of 1St Order Systems Review: step response of 1st order systems Step response in the s—domain c(t) 1 steady state Initial slope = = a (final value) a time constant ; s(s + a) 1.0 0.9 in the time domain 0.8 at 1 − e− u(t); 0.7 63% of final value ¡time constant¢ 0.6 at t = one time constant 0.5 1 τ = ; 0.4 a 0.3 rise time (10%→90%) 0.2 2.2 0.1 Tr = ; a 0 t 1 2 3 4 5 settling time (98%) a a a a a Tr 4 T = . Ts s a Figure 4.3 Figure by MIT OpenCourseWare. 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Review: poles, zeros, and the forced/natural responses jω jω system pole – input pole – system zero – natural response forced response derivative & amplification σ σ −5 0 −5 −2 0 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 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 2nd order systems 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 DC motor system with non-negligible inductance Recall combined equations of motion LsI(s)+RI(s)+KvΩ(s)=Vs(s) ⇒ JsΩ(s)+bΩ(s)=KmI(s) ) LJ Lb KmKv Km s2 + + J s + b + Ω(s)= V (s) R R R R s ⎧ · µ ¶ µ ¶¸ ⎨⎪ (Js + b) Ω(s)=KmI(s) ⎩⎪ Including the DC motor’s inductance, we find Ω(s) Km 1 = Vs(s) LJ b R bR + KmKv ⎧ s2 + + s + Quadratic polynomial denominator ⎪ J L LJ Second—order system ⎪ µ ¶ µ ¶ ⎪ ⎪ ⎪ b ⎪ s + ⎨ I(s) 1 J = µ ¶ ⎪ Vs(s) R b R bR + K K ⎪ 2 m v ⎪ s + + s + ⎪ J L LJ ⎪ µ ¶ µ ¶ ⎪ ⎩⎪ 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Step response of 2nd order system – large R/L 2 L =0.1H, Kv =6V· sec, Km =6N· m/A, J =2kg· m , Overdamped 2 R =6Ω, b =4kg· m · Hz; vs(t)=30u(t)V. response ↔ dissipation > energy storage 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Step response of 2nd order system – large R/L 2 L =0.1H, Kv =6V· sec, Km =6N· m/A, J =2kg· m , Overdamped 2 R =6Ω, b =4kg· m · Hz; vs(t)=30u(t)V. response 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Step response of 2nd order system – large R/L 2 L =0.1H, Kv =6V· sec, Km =6N· m/A, J =2kg· m , Overdamped 2 R =6Ω, b =4kg· m · Hz; vs(t)=30u(t)V. response 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Comparison of 1st order and 2nd order overdamped 1st order 2nd order (L≈0) (L=0.1H) 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Step response of 2nd order system – small R/L 2 L =1.0H, Kv =6V· sec, Km =6N· m/A, J =2kg· m , Underdamped 2 R =6Ω, b =4kg· m · Hz; vs(t)=30u(t)V. response ↔ dissipation < energy storage overshoot 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Comparison of 1st order and 2nd order underdamped NO overshoot overshoot 1st order 2nd order (L≈0) (L=1.0H) 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Overdamped DC motor: derivation of the step response Using the numerical values L =0.1H, Kv =6V· sec, Km =6N· m/A, J = 2kg · m2, R =6Ω, b =4kg· m2 · Hz we find K rad b R bR + K K m =30 ; + =62rad/sec; m v =300(rad/sec)2 . LJ sec · V J L LJ Therefore, the transfer function for the angular velocity is nd Ω(s) 30 A2 —order system is overdamped = 2 . if the transfer function denominator Vs(s) s +62s +300 has two real roots. We find that the denominator has two real roots, Ω(s) 30 s1 = −5.290Hz,s2 = −56.71Hz ⇒ = . Vs(s) (s +5.290)(s +56.71) To compute the step response we substitute the Laplace transform of the voltage source Vs(s)=30/s and carry out the partial fraction expansion: 900 3 3.3 0.3 Ω(s)= = − + ⇒ s(s +5.290)(s +56.71) s s +5.290 s +56.71 5.29t 56.71t ω(t)= 3 − 3.3e− +0.3e− u(t). This is the function whose£ plot we analyzed in slides #5—8.¤ 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Overdamped DC motor in the s-domain jω system poles (2nd order overdamped) input pole σ −56.71 −5.29 −5 0 1st order system pole jω 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Undamped DC motor: no dissipation Consider the opposite extreme where the dissipation due to both the resistor and bearings friction is negligible, i.e. R =0andb = 0. Using the same remaining 2 numerical values L =0.1H, Kv =6V· sec, Km =6N· m/A, J =2kg· m ,we find K rad b R bR + K K m =30 ; + =0; m v =180(rad/sec)2 . LJ sec · V J L LJ Therefore, the transfer function for the angular velocity is A2nd—order system is undamped Ω(s) 30 if the transfer function denominator has a = 2 . Vs(s) s +180 conjugate pair of two imaginary roots. The denominator has a conjugate pair of two imaginary roots, Ω(s) 30 s1,2 = ±j13.42Hz ⇒ = . Vs(s) (s + j13.42)(s − j13.42) Again, the step response is found by partial fraction expansion: 900 5 5s Ω(s)= = − ⇒ s(s + j13.42)(s − j13.42) s s2 +(13.42)2 ω(t)=[5− 5cos(13.42t)] u(t). 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Undamped DC motor in the s-domain Natural frequency ωn =13.42rad/sec. Period T =2π/ωn =4.24sec. system poles (2nd order undamped) jω j13.42 input pole σ −5 0 1st order −j13.42 system pole 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Underdamped DC motor: small dissipation Finally, let us return to what we previously labelled as “underdamped” case, 2 i.e. L =1.0H, Kv =6V· sec, Km =6N· m/A, J =2kg· m , R =6Ω, b =4kg· m2 · Hz. The values of L, R are such that the dissipation in the system is negligible compared to the energy storage capacity. We then find K rad b R bR + K K m =3 ; + = 8rad/sec; m v =30(rad/sec)2 . LJ sec · V J L LJ Therefore, the transfer function for the angular velocity is Ω(s) 30 A2nd—order system is underdamped = 2 . Vs(s) s +8s +30 if the transfer function denominator has a conjugate pair of two complex roots. This denominator has a conjugate pair of two complex roots, Ω(s) 30 s1,2 = −4 ± j3.74 rad/sec ⇒ = . Vs(s) (s +4+j3.74)(s +4− j3.74) We will now develop the partial fraction expansion method for this case, aiming to find the step response: 90 90 K K s + K Ω(s)= = =90 1 + 2 3 . s(s +4+j3.74)(s +4− j3.74) s(s2 +8s +30) s s2 +8s +30 µ ¶ 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Underdamped DC motor: small dissipation Using the familiar partial fraction method, we can find K1 =1/30, K2 = −1/30, K3 = −8/30, therefore 1 s +8 Ω(s)=3 − . s s2 +8s +30 µ ¶ To findtheinverseLaplacetransform,werewrite the denominator as a complete square plus a constant, and break down the numerator into the sum of the same factor that appeared in the denominator’s complete square plus another constant: 1 (s +4)+4 Ω(s)=3 − 2 . à s (s +4) +14! Ifthecompletesquareinsteadof(s+4)2 were of the form s2, the inverse Laplace transform would have followed easily from Nise Table 2.1: 1 s +4 1 s +4 L− = L− =cos(3.74t)+4sin(3.74t) . s2 +14 2 2 · ¸ "s +(3.74) # To take the extra factor of 4 into account, we must use yet another property of Laplace transforms, which we have not seen until now: at L e− f(t) = F (s + a). (Nise Table 2.2, #4). £ ¤ 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Underdamped DC motor: small dissipation We apply this “frequency shift” property as follows: 1 s +4 L− 2 =cos(3.74t)+4sin(3.74t) ⇒ "s2 +(3.74) # 1 (s +4)+4 4t ⇒ L− 2 2 =e− [cos (3.74t)+4sin(3.74t)] . "(s +4) +(3.74) # Combining all of the above results, we can finally compute the step response for the angular velocity of the DC motor as 4t ω(t)= 3 − 3 e− [cos (3.74t)+4sin(3.74t)] u(t). ½ ³ ´¾ With a little bit of trigonometry, which we leave to you to do as exercise, we can rewrite the step response as 4t ω(t)= 3 − 4.39e− cos (3.74t − 0.82) u(t). ½ ¾ So the step response of the 2nd—order underdamped system is characterized by a phase—shifted sinusoid enveloped by an exponential decay. This step response was analyzed in slides #9—10 of today’s notes. 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 What the real and imaginary parts of the poles do c(t) Exponential decay generated by real part of complex pole pair Sinusoidal oscillation generated by imaginary part of complex pole pair t Damping ratio Figure by MIT OpenCourseWare.
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