6 - DynamicResponse6 filled

9 n ormalized such that as t y n 1 y n t y

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Unformatted text preview: o ICs) zero 1 & τy + y = Ku ⇒ y ( t ) = K (1 − e −t Normalized Response 0.9 τ) N ormalized (such that as t → ∞ , y n → 1) : ⇒ y n (t ) = y (t ) −t = (1 − e τ ) K 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1τ 0 τ Time t ( 1− e − t/ τ 2τ 3τ 2τ 3τ Time [ t ] 4τ 4τ 5τ 6τ 5τ ) School of Mechanical Engineering Purdue University ME575 Session 6 – Dynamic Response Slide 5 Step Response of 1st Order System Effect of Time Constant τ : 1 & τy + y = Ku Slope at t = 0: 0: d ⇒ y (t ) = dt d y (0) = ⇒ dt −t τ) 0.8 Normalized Response ⇒ y (t ) = K (1 − e 0.9 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 ME575 Session 6 – Dynamic Response 2 School of Mechanical Engineering Purdue University 4 6 Time [sec] 8 10 Slide 6 3 ME575 Handouts Example – Vehicle Acceleration v m 1 & v + v = F = vmax b b F m b Standing-Start Acceleration; Lincoln Aviator SUV Standing-Start Acceleration; Dodge Viper SRT-10 140 120 120 100 100 Speed (MPH) 160 140 Speed (MPH) 160 80 80 60 60 40 40 20 20 0 0 5 10 15 20 Time (sec) 25 30 35 0 0 40 5 10 15 20 Time (sec) 25 30 35 40 School of Mechanical Engineering Purdue University ME575 Session 6 – Dynamic Response Slide 7 Example – Automotive Suspension y m && & & my + by + ky = br + kr g k b r Response to Initial Conditions 0.02 b k && + y + y = 0 & y m m && + 28 y + 400 y = 0 & y 0 -0.02 Amplitude for free response: && & my + by + ky = 0 -0.04 -0.06 -0.08 -0.1 ME575 Session 6 – Dynamic Response 0 0.05 0.1 School of Mechanical Engineering Purdue University 0.15 0.2 Time (sec) 0.25 0.3 0.35 0.4 Slide 8 4 ME575 Handouts Step Response of 2nd Order System • Standard Form of Stable 2nd Order System (no zeros) && + a1 y + a 0 y = b u & y ⇒ && + 2 ζ ω n y + ω n 2 y = K ω n 2 u & y where ωn > 0 : Natural Frequency [rad/s] ζ > 0 : Damping Ratio K : Static (Steady State, DC) Gain Static Pole locations s = −ζωn ± ωn Img. ωn ( ζ 2 − 1) Overdamped (ζ > 1): Critically damped (ζ = 1): Real Underdamped (ζ < 1): −ωn School of Mechanical Engineering Purdue University ME575 Session 6 – Dynamic Response Slide 9 Step Response of 2nd Order System Unit Step Response of Underdamped 2nd Order Systems ( u = 1 and zero ICs ) && + 2 ζ ω n y + ω n 2 y = K ω n 2 u & y Y ( s ) = G ( s )U ( s ) = Y (s) = K ωn2 s + 2ζ ω n s + ω n 2 2 • K ωn2 1 = s s ⎡ ( s + ζ ω )2 + ω 2 ⎤ n d ...
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This note was uploaded on 12/09/2011 for the course ME 575 taught by Professor Meckl during the Fall '10 term at Purdue University-West Lafayette.

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