Control ENG HW_Part_31

Control ENG HW_Part_31 - 8.6 verify that for a second order...

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Unformatted text preview: 8.6 verify that for a second order system subjected to a step response, 1 Y (t ) = 1 − 1− ξ 2 e − ξt τ [ sin 1 − ξ 2 τt + tan −1 1− ξ 2 ξ With ξ <1 1 1 22 s (τ s + 2ξτs + 1) Y (s) = τ 2 s 2 + 2ξτs + 1 = ( s − s1 )(s − s 2 ) where s1 = s2 = −ξ τ + −ξ τ + ξ 2 −1 =− a + b τ ξ 2 −1 =− a −b τ 1 2 τ Y (s) = s ( s − s1 )( s − s 2 ) Y (s) = 1 A B C + + 2 τ s ( s − s1 ) ( s − s 2 ) Y (s) = 1 A B C + + 2 τ s ( s − s1 ) ( s − s 2 ) A( s 2 − ( s1 + s s ) s + s1 s 2 ) + Bs( s − s 2 ) + Cs ( s − s1 ) = 1 A+ B +C = 0 − A( s1 + s s ) − Bs 2 − Cs1 = 1 As1 s s = 1; A = 1 1 = B+C = − s1 s 2 s1 s 2 1 s1 = Bs 2 + Cs 2 = − = Cs1 + Cs 2 = s1 + s 2 1 1 −= s1 s 2 s1 s 2 =C= 1 1 1 1 =B=− − =− s 2 ( s 2 − s1 ) s 2 ( s 2 − s1 ) s1 s 2 s1 ( s 2 − s1 ) Y (s) = 1 1 τ 2 s1 s 2 1 1 1 1 1 . − s s (s = s ) . (s − s ) + s ( s − s ) . (s − s ) 12 1 1 2 2 1 2 11 1 1 − e S1t + e s2t 2 s2 ( s 2 − s1 ) τ s1 s2 s1 ( s2 − s1 ) 8.6 Y (t ) = 1 =1 τ s1 s2 2 Y (t ) = 1 − 1 S1t 1 S2t s e − s e τ ( s2 − s1 ) 1 2 Y (t ) = 1 − 1 2 [ 1 s2 e S1t − s1e S2t ( s2 − s1 ) Y (t ) = 1 + τ 2 ξ −1 2 [s e 2 S1t − s1e S2t ...
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