61_adv_control_eng

56 when 0 1 the frequency of transient oscillation is

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Unformatted text preview: 8.2001 2:26PM 54 Advanced Control Engineering Inverse transform 4 @ xo (t) ˆ K 1 À 2  p !n p sin !n 1 À  2 t e !n 1 À  2 @ A 5 n   p o 2 À!n t À p e sin !n 1 À  2 t 1 À 2 À!n t  p cos !n 1 À  2 t À 3A (3:55) Equation (3.55) can be simplified to give 2 3 4 @ A5  p  p  (3:56) xo (t) ˆ K 1 À eÀ!n t cos !n 1 À  2 t ‡ p sin !n 1 À  2 t 1 À 2 When  ˆ 0 xo (t) ˆ K [1 À e0 fcos !n t ‡ 0g] ˆ K [1 À cos !n t] (3:57) From equation (3.57) it can be seen that when there is no damping, a step input will cause the system to oscillate continuously at !n (rad/s). Damped natural frequency w d From equation (3.56), when 0 <  > 1, the frequency of transient oscillation is given by p (3:58) ! d ˆ !n 1 À  2 where !d is called the damped natural frequency. Hence equation (3.56) can be written as 4 @ A5 2 3  À!n t xo (t) ˆ K 1 À e cos !d t ‡ p sin !d t (3:59) 1 À 2 4 5 eÀ!n t ˆ K 1 À p sin (!d t ‡ ) (3:60) 1 À 2 where p 1 À 2 tan  ˆ  (3:61) When  ˆ 1, the unit step response is xo (t) ˆ K [1 À eÀ!n t (1 ‡ !n t)] and when  > 1, the unit step response from equation (3.46) is given by 4 @2 3À pÁ 1  2 ‡ p e À ‡  À1 !n t xo (t) ˆ K 1 À 2À1 22 A5 2 3À pÁ 1  À À  2 À1 ! n t À p e ‡ 2 2 2 À 1 (3:62) (3:63) //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC03.3D ± 55 ± [35±62/28] 9.8.2001 2:27PM Time domain analysis 55 1.6 ζ = 0.2 1.4 ζ = 0.4 1.2 ζ = 0.6 ζ = 0.8 xo(t ) 1 0.8 0.6 ζ = 2.0 0.4 ζ = 1.0 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5...
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This note was uploaded on 01/18/2014 for the course MECH 107 taught by Professor Sali mon during the Winter '12 term at Bingham University.

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