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Control ENG HW_Part_53

# Control ENG HW_Part_53 - As k → ∞ ξ = 4π(kτ D τ 1...

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Unformatted text preview: As k → ∞, ξ = 4π (kτ D + τ 1 ) T= τ 4k k D τ I τ1 + τ I − (k + 1) 2 kτ I τ D + τ 1τ I k τ= τ = τ Iτ D + τ 1 τI = 0.3535 2 τ1 K →∞ τ Iτ 1 k = τ I τ D =1 As k → ∞, T = 4πτ D τ 4 D −1 τI = 7.2552 1 τ 1s + 1 C (d) = U 1 k 1 + τ D s + 1+ τ 1s + 1 τIs τIs C = U (k + 1)τ I s + (τ 1 + kτ D )τ I s 2 + k U= C= 1 s τI (k + 1)τ I s + (τ 1 + kτ D )τ I s 2 + k Lt sf ( s ) = S →0 0 =0 k K=2 For a unit step change in U C (∞) = 0 Offset = 0 (e) k = 0.5, ξ =0.671 & τ =2.236 T= 2πτ 1− ξ 2 = 18.95 If k = 0.5 C 2s s ;C= 2 =2 U 5s + 3s + 1 5s + 3s + 1 If k = 2 C 0.5s 0.5 ;C = 2 =2 U 2 s + 1. 5 s + 1 2s + 1.5s + 1 In general C (t ) = 1 1 τ 1− ξ 2 e The maximum occurs at t = − ξt τ sin 1 − ξ 2 τ 1− ξ 2 If k = 0.5 tmax = 2.52 Cmax=0.42 If k = 2 tmax = 1.69 Cmax=0.19 tan −1 t τ 1− ξ 2 ξ ...
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Control ENG HW_Part_53 - As k → ∞ ξ = 4π(kτ D τ 1...

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