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2.1. STATE SPACE REALIZATION OF DYNAMIC SYSTEMS 11 k M x u x1 2 u1 2 m1 u k x m 2 (a) A mass-spring system (b) Double mass-spring system k1 k2 m1 m2 c c1 2 k 4 m 3 x 1 x2 x3 (c) Complicated mass-spring-damper system k 3 u c k x 1 u 1 (d) Creeping phenomenon of a steel material P x v (e) Buckling phenomenon of a rod Figure 2.1: Dynamic systems The matrices,M, C, and K, are defined as follows. • M ii is the mass value of thei th mass. • M ij,i≠j = 0. • C ii is the sum of damping coefficients of all dampers connected to thei th mass. • C ij,i≠j is the negative value of the sum of damping coefficients of all dampers connected between thei th mass and thej th mass. • K ii is the sum of spring constants of all springs connected to thei th mass. • K ij,i≠j is the negative value of the sum of spring constants of all springs connected between thei th mass and thej th mass. Digital Control Systems, Sogang University Kyoungchul Kong
2.1. STATE SPACE REALIZATION OF DYNAMIC SYSTEMS 12 [Example 2-1] Differential equations of the systems in Fig. 2.1 are as follows. (a) mẍ+kx = u [ ] x1 (b) The output vector is x = ∈ R x 2 , and the input vector is u = 2 The entries of the matrices are defined as follows. • M 11 = m 1 and M 22 = m 2 . [ u1 u 2 ] ∈ R 2 . • M 12 = M 21 = 0. • Since there is no damper in the system,C = 0. • The spring k is the only spring connected to m 1 , i.e., K 11 = k. By the same reason, K 22 = k. • The spring k is the only spring connected between m 1 and m 2 . Since K ij,i≠j is the negative value of the sum of spring constants,K 12 = K 21 = −k. Therefore, the differential equation of the two mass-spring system is [ ] [ ] [ ] m1 0 0 0 k −k ẍ+ ẋ+ x = u 0 m 2 0 0 −k k ⎡ (c) The output vector is x = ⎣ ⎤ x 1 x 2 x 3 ⎦ ∈ R 3 , and the input vector is u = ⎣ ⎡ 0 0 u ⎤ ⎦ ∈ R 3 . Note that no force is applied to m 1 and m 2 . The entries of the matrices are defined as follows. • M 11 = m 1 , M 22 = m 2 , and M 33 = m 3 . • M 12 = M 13 = M 21 = M 23 = M 31 = M 32 = 0. • The dampers connected to m 1 are c 1 and c 2 . Thus, C 11 = c 1 +c 2 . On the other hand, c 2 is the only damper connected to m 2 , i.e., C 22 = c 2 . Since no damper is connected tom 3 , C 33 = 0. • The damper(s) connected between m 1 and m 2 is c 2 . Thus, C 12 = C 21 = −c 2 . There is no damper between m 1 and m 3 , i.e., C 13 = C 31 = 0. Similarly, C 23 = C 32 = 0. • The springsk 1 ,k 2 , andk 4 are connected tom 1 , i.e.,K 11 = k 1 +k 2 +k 4 . Similarly, K 22 = k 2 +k 3 and K 33 = k 3 +k 4 . Digital Control Systems, Sogang University Kyoungchul Kong
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