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Review: Step Response of 1St Order Systems We've Seen Review: step response of 1st order systems we’ve seen • Inertia with bearings (viscous friction) Step input Ts(t)=T0u(t) ⇒ Step response∗ T0 t/τ J ω(t)= 1 − e− , where τ = . b b ³ ´ • RC circuit (charging of a capacitor) R + + Step input vi(t)=V0u(t) ⇒ Step response vi C vC t/τ − − vC (t)=V0 1 − e− , where τ = RC. ³ ´ • DC motor with inertia load, bearings and negligible inductance Step input vs(t)=V0u(t) ⇒ Step response Km t/τ ω(t)= V 1 − e− , R 0 ³ ´ J where τ = . KmKv ∗When writing the step response, b + we may omit the step function u(t) à R ! implying the result holds for t>0only. 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 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 (1st order: 1 pole) – current (1st order: 1 pole, 1 zero) • Two new properties of the Laplace transform: – final value theorem – initial value theorem • Next two lectures: –2nd order systems 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 Current dynamics in DC motor system 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) ⎩⎪ Neglecting the DC motor’s inductance (i.e.,assumingL/R ≈ 0), we find Km Transfer function for the angular velocity Ω(s) RJ is of the form ⎧ = Vs(s) ⎪ 1 KmKv A 1 KmKv ⎪ s + b + ,p= b + ⎪ J R pole ⎪ µ ¶ s + p J R ⎪ µ ¶ ⎪ ⎪ 1 b Transfer function for the current ⎨ s + zero I(s) R J is of the form ⎪ = µ ¶ ⎪ Vs(s) ⎪ 1 KmKv B (s + z) 1 KmKv b ⎪ s + b + ,p= b + ,z= ⎪ J R s + p J R J ⎪ µ ¶ µ ¶ ⎪ ⎩⎪ 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 Poles and zeros in the s-plane jω left half—plane right half—plane s—plane imaginary axis real axis −p −z σ pole zero 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 DC motor step response: numerical example L ≈ 0, R =6Ω, Kv =6V· sec, 2 2 Km =6N· m/A, J =2kg· m , b =4kg· m · 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, Ω(s) 1 1 = V (s) 2 s +5[Hz] ⎧ s whereas the Laplace transform of the input is we find ⎪ ⎪ I(s) 1 (s +2[Hz]) 30 ⎨ = Vs(s) ≡ L vs(t) = L 30u(t) = . V (s) 6 s +5[Hz] s ⎪ s h i h i ⎪ I. Angular velocity ⎩⎪ 15 K K 1 1 1 1 Ω(s)= =15 1 + 2 where K = = ,K = = − ⇒ s (s +5) s s +5 1 s +5 5 2 s 5 µ ¶ ¯s=0 ¯s= 5 ¯ ¯ − ¯ ¯ 1 1 5t¯ rad ¯ Ω(s)=3 − ⇒ ω(t)=3 1 − e− u(t) . s s +5 sec µ ¶ Forced response ¡ ¢ Homogeneous response (Natural response) 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 DC motor step response: numerical example L ≈ 0, R =6Ω, Kv =6V· sec, 2 2 Km =6N· m/A, J =2kg· m , b =4kg· m · 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, Ω(s) 1 1 = V (s) 2 s +5[Hz] ⎧ s whereas the Laplace transform of the input is we find ⎪ ⎪ I(s) 1 (s +2[Hz]) 30 ⎨ = Vs(s) ≡ L vs(t) = L 30u(t) = . V (s) 6 s +5[Hz] s ⎪ s h i h i ⎪ II. Current ⎩⎪ 5(s +2) K10 K20 s +2 2 s +2 3 I(s)= =5 + where K 0 = = ,K0 = = ⇒ s (s +5) s s +5 1 s +5 5 2 s 5 µ ¶ ¯s=0 ¯s= 5 ¯ ¯ − ¯ ¯ 2 3 5t ¯ ¯ I(s)= + ⇒ i(t)= 2+3e− u(t)A. s s +5 µ ¶ ¡ ¢ Forced response Homogeneous response (Natural response) 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 DC motor system in the s-plane Input: Voltage source system Output: Angular velocity representation 30 Gω(s) V (s)= s s 1 1 Ω(s) 2 s +5 Input: Voltage source Output: Current system representation 30 Gi(s) V (s)= s s 1 s +2 I(s) 6 s +5 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 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. 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 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. 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 1st order system response from s-plane representation • 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 responses • Pole at –α generates response e–αt (exponentially decreasing if • Pole at zero generates step pole on the right half-plane; function increasing if on the left half-plane) jω jω αt e− u(t) σ σ u(t) −α 0 Time constant τ=1/α 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 1st order system: transient response properties Step response steady state in the s—domain c(t) Initial slope = 1 = a time constant (final value) a ; 1.0 s(s + a) 0.9 in the time domain 0.8 at 0.7 1 − e− u(t); 63% of final value 0.6 at t = one time constant ¡ ¢ time constant 0.5 1 0.4 τ = ; a 0.3 rise time (10%→90%) 0.2 0.1 2.2 Tr = ; 0 t a 1 2 3 4 5 a a a a a settling time (98%) Tr 4 Ts T = . s a Figure by MIT OpenCourseWare. Figure 4.3 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 DC motor step responses 5t 5t ω(t)=3 1 − e− u(t)rad/sec i(t)= 2+3e− u(t)A ω(¡∞)=3¢ rad/sec ¡i(∞)=2¢ A dω di (0+) = 3 × 5=15 rad/sec2 (0+) = ∞ dt dt 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 The Final Value theorem: steady-state We will now learn two additional properties of the Laplace transform, which we will quote without proof. Let F (s) denote the Laplace transform of the function f(t). The first property is the Final Value theorem: f(∞)=lims 0 sF (s); → Let us see how this applies to the step response of a general 1st—order system with a pole at −a and without a zero (e.g., the angular velocity response of the DC motor.) We select the system gain such that the steady—state will equal 1. The step response in the s—domain then is a 1 1 a F (s)= = − ;also,sF (s)= . 1 s(s + a) s s + a 1 s + a Using the partial fraction expansion above, we find the time domain step re- sponse as at f1(t)= 1 − e− u(t) ⇒ f1(∞)=1 (asadvertised.) Applying the final value¡ theorem,¢ we find, consistently, a f1(∞) = lims 0 sF1(s)=lims 0 =1. → → s + a 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 The Initial Value theorem: initial slope The second property of the Laplace transform is the Initial Value theorem f(0+) = lims sF (s). →∞ Let us use this property to compute the initial slope of the step response, i.e. the value of the derivative of the step response at t = 0+ for the same general 1st—order system with steady state equal to unity, a pole at −a and without a zero. Since we are interested in the derivative of f(t), the Laplace transform of interest is df (t) as H (s)=L 1 = sF (s) ⇒ sH (s)=s2F (s)= . 1 dt 1 1 1 s + a · ¸ Applying the final value theorem, df1 as a (0+) = lims = lims = a. dt →∞ s + a →∞ 1+a/s Again, this is consistent with the result we get directly from the time domain: at df1 at df1 f (t)= 1 − e− ⇒ (t)=− (−a)e− ⇒ (0+) = a. 1 dt dt ¡ ¢ 2.004 Fall ’07 Lecture 06 – Monday, Sept. 17 Initial and final value of 1st-order system with a zero We now consider the same 1st—order system with unity steady state, a pole at −a, but we also add a zero at −z. In that case, the Laplace transforms of the step response and its derivative are s + z z a − z s + z s(s + z) F (s)=a = + ; sF (s)=a ; s2F (s)=a . 2 s(s + a) s s + a 2 s + a 2 s + a We can readily see that df2 2 f2(∞)=lims 0sF2(s)=z; f2(0+) = lims sF2(s)=a; (0+) = lims s F2(s)=∞. → →∞ dt →∞ Youshouldverifythattheseresultsareconsistent with the time—domain solution for this system. We can see that the effects of the zero −z on the 1st—order system are (in comparison to a system with the same pole at −a but without the zero) • amplify the steady—state response by z; • raise the initial value from zero to A; • raise the initial slope to infinity.
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