Control ENG HW_Part_32

Control ENG HW_Part_32 - Y (t ) = 1 − τje − ξt τ 2...

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Unformatted text preview: Y (t ) = 1 − τje − ξt τ 2 ξ −1 2 [(−a − jb)(cos bt + j sin bt) − (−a + jb)(cos bt − j sin bt] Y (t ) = 1 − τje − ξt τ 2 ξ 2 −1 Y (t ) = 1 − e ξ − 2 [− 2 jb(b cos bt + a sin bt )] ξt τ [ 1−ξ −1 2 cos(αt ) + ξ sin(αt ) [ 1−ξ ] 2 α= ξ [ 1−ξ 2 ξ φ = tan −1 verified 8.14 From the figure in your text Y(4) for the system response is expressed b) verify that for ξ = 1, and a step input t Y (t ) = 1 − 1 + τ t −τ e Y ( s) = Y ( s) 1 1 s τ 2 s 2 + τs + 1 A Bs + C 1 =+ 2 s (τs + 1) B (τ + 1) 2 2 A(τ 2 s 2 + 2τs + 1) + Bs Cs = 1 Aτ 2 + B = 0 2 Aτ + C = 0 A=1; B = τ 2 ; C = 2τ Y ( s) = 1 τ (τs + 1) + τ − s τ (s + 1)2 Y ( s) = τ τ 1 − − s (τs + 1) (τs + 1) 2 Y (t ) = 1 − e − t τ t 1− − te τ τ t Y (t ) = 1 − (1 + )te τ − t τ proved c) for ξ > 1, prove that the step response is Y (t ) = 1 − e − ξt τ [cosh(αt ) + β sinh(αt )] ...
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