Control ENG HW_Part_33

Control ENG HW_Part_33 - ξ 2 −1 ξ β= 2 α= τ ξ −1...

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Unformatted text preview: ξ 2 −1 ξ β= 2 α= τ ξ −1 Now Y ( s ) = 1/τ 2 s ( s − B1 )( s − s2 ) Where s1 = − ξ 2 −1 ξ + τ τ ξ 2 −1 ξ s2 = − − τ τ from 8.6(a) Y (t ) = 11 1 1 1 1 1 − + 2 τ s1 s2 s s1 ( s2 − s1 ) ( s − s1 ) s2 ( s2 − s1 ) ( s − s2 ) Y (t ) = 1 − [ 1 s2 e S1t − s1e S2t ( s2 − s1 ) − ξ − 1 − ξ 2 τ Y (t ) = 1 + τ 2 ξ 2 − 1 e Y (t ) = 1 + − ξ e 2 ξ 2 −1 − ξt τ ξ 2 −1 t τ −ξt e τ e 1 ξ 2 −1 τ t + ξe − ξt τ −ξ + 1−ξ 2 − τ − ξ 2 − 1e − ξ 2 −1 t τ − ξt − e τ e − ξ 2 − 1e e − ξ 2 −1 τ ξ 2 −1 t τ t Y (t ) = 1 + e − ξ eαt − e −αt − 2 ξ 2 −1 ξt τ Y (t ) = 1 − e − ξt τ eαt + e −αt − 2 [cosh(αt ) + β sinh(αt )] 8.7 Verify that for a unit step-input − πξ (1) overshoot = exp 1−ξ 2 − 2πξ (2) Decay ratio = exp 1−ξ 2 For a unit step input the response (ξ<1): ξt − τ 1 − ξ 2 t Sin 1 − ξ 2 + tan −1 2 ξ τ 1−ξ (1) we have to find time t where the maxima occurs e Y (t ) = 1 − = dY/dt = 0 ξt − τ 1−ξ 2 dY ) ξe t Sin 1 − ξ 2 + tan −1 = ξ dt τ 1 − ξ 2 τ − e ξt − τ τ = tan 1 − ξ 2 t = 0 Cos 1 − ξ 2 + tan −1 ξ τ 2 1−ξ 2 2t −1 1 − ξ = + tan 1−ξ ξ τ ξ 1−ξ 2t ξ = nπ for maxima = 1−ξ 2t ξ = 2 nπ ...
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This note was uploaded on 11/13/2011 for the course COP 4355 taught by Professor Koslov during the Spring '10 term at University of Florida.

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