61_adv_control_eng

61_adv_control_eng

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Unformatted text preview: : s2 ‡ and the roots are V sW  2 1 `C C Ka Æ s1 , s2 ˆ À4 2 Xm m mY (3:50) For critical damping, the discriminant is zero, hence the roots become s1 ˆ s2 ˆ À Cc 2m Also, for critical damping 2 C c 4K ˆ m m2 2 Cc ˆ giving 3.6.4 4Km2 m p Cc ˆ 2 Km (3:51) Generalized second-order system response to a unit step input Consider a second-order system whose steady-state gain is K, undamped natural frequency is !n and whose damping ratio is  , where  < 1. For a unit step input, the block diagram is as shown in Figure 3.18. From Figure 3.18 Xo (s) ˆ s( s2 K !2 n ‡ 2!n s ‡ !2 ) n (3:52) //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC03.3D ± 53 ± [35±62/28] 9.8.2001 2:26PM Time domain analysis 53 Xo(s ) Kωn2 Xi (s) = 1/s 2 s + 2ζωns + s ωn2 Fig. 3.18 Step response of a generalized second-order system for  < 1. Expanding equation (3.52) using partial fractions A Bs ‡ C ‡2 s …s ‡ 2!n s ‡ !2 † n À2 Á Equating (3.52) and (3.53) and multiply by s s ‡ 2!n s ‡ !2 n Xo (s) ˆ (3:53) À Á K !2 ˆ A s2 ‡ 2!n s ‡ !2 ‡ Bs2 ‡ Cs n n Equating coefficients (s2 ) X 0ˆA‡B 1 0 ˆ 2!n A ‡ C 0 K !2 ˆ !2 A n n (s ) X (s ) X giving A ˆ K, B ˆ ÀK and C ˆ À2!n K Substituting back into equation (3.53) Xo (s) ˆ K & '! 1 s ‡ 2!n À2 s s ‡ 2!n s ‡ !2 n Completing the square 4 @ A5 1 s ‡ 2!n Xo (s) ˆ K À s (s ‡ !n )2 ‡ !2 À  2 !2 n n WQ b a s ‡ 2!n T1 U ˆ KR À  p2 S b sb X(s ‡ !n )2 ‡ !n 1 À  2 Y P V b ` (3:54) The terms in the brackets { } can be written in the standard forms 10 and 9 in Table 3.1. Às Term (1) ˆ  p2 (s ‡ !n )2 ‡ !n 1 À  2 W p b a 2!n !n 1 À  2 p Term (2) ˆ À  p !n 1 À  2 b…s2 ‡ ! †2 ‡ ! 1 À  2 2 b Y X n n @ V Ab ` //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC03.3D ± 54 ± [35±62/28] 9....
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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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