Question

3.3 The differential equations of a pair of masses M1, M2 sliding on a horizontal plane without friction and connected by a spring of spring constant pare Muqi = P(92-91)+f Maga=plqı - 92) where 91.9 are the displacements from the same reference point, and is the force applied to M . The only force on M2 is the force p(91-92) exerted by the spring. Using the state vector, control vector, and output vector, respectively tu=y, y= 2 . Find the matrices A, B, C, D corresponding to the differential equations . Find the transfer function. • Find the poles of the system.

Answer 1

Solution:

Given that Mia, = P(92-9,) + f M₂ % 2 = P (91-92) X=(2,7 12 Uzf, y= V₂ Let x1 = 4, 3 x 1 = 4, = 23 ) X2 = 9 2 X 2 = 0 = x 4 ! x3 = 0, >> iz = z; Substitute all the values in above equation Mix3 = P(x2-x1) + u M2 X 4 = P(x2-x2) x y = / ₂ = X2 space Representation is State 22 + 1-PM, P/r, o a % POO 9 - [o to a 3: 157-63C47 **3 By comparing we get: 1 P/M2 - P/M2 ao C: Coi ac] ; D = [0]

(i) The transfer function is obtained by Transfer function 7F = C(SI-AJ"18+o (SI-A) & (sooo oo lo7 losooooo looso -Ple, PN, OO ooOsplo, -PHP OO 0 - 0 SI-A= S -10 1 OSO-1 Ply, Ply, so 1-P/M2 Plos P/N, -eln, s o -P/M2 P/m, os Transfer fer nction is Txy ISO- los O- 1 Pm, Plm, so 1-P/M2 P/M o s. цки чх! Thus, we get trausten function. (1) To find poles of system.- I SI-A1-o le, delenminent of ($1-A] = 0 on solning we get, s+ s (P/me + P/mi) = 0 sot ($ +P [%mi + Yme)) = 0 Then, we get poles of system as a ot/P (Ym +/m2) - Pakam, + Ymo)

(34) Considering qz-q, instead of V2 2x = x1 = Xi = = X3 2q = 92-9, =) ¿z = %, _q, =XH x3 = 2; =mg = ës *4 = 22 - 2,37 4 = üz- &, & g = V2 = 22 +21 Substitule all in given equation: нүх, = P4, +и z = l/m, x2 +ulmi å, = P/m, (92-91) + u/MI &z= PlM2 (91-92) az-a, = *4 = Plm, (91-92) - P/M, (2.2 - 2.) - Ulm, :: xy = Plma (-x2) -P/m, (*2) -4/m Stale space Representation is: tlo o flm, oo Toolkit ca o 1-ymi -[ o ) X2 () Lu)

By Corpoing we gel . A: Го о то | o o o | - " + " _ В - | 1 Ymil с: Сгоој р= [о].