Control ENG HW_Part_33

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

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.

Ask a homework question - tutors are online