Question

2. The equations of motion of this system are; ma Seats 12Y"+7Y'+247-6Z'-122=0 6Z" + 6Z'+12Z- 6Y-12Y=f(t) 7W"+7W'+14W-7Y'-14Y=ft) Body Suspension Where Y,Z and W are deformations of the masses and springs. m2 Wheel Put these equations into state variable form and express the model as matrix vector equation if output of the system is Y. Road Datum level Energy Storage Element mi m2 т3 ki k2 k₃ State Variable X1 = Y' X2 = Z' X3 = W' X4 = Y XE = 2 X6 = W

Answer 1

@ the equation of motion of this system ane; 124" + 74 +244 - 62' - 122 = 0 67" + 621 + 122 - 64' - 12Y = f(t) ¥w" + 70'+ 14074-144 = ft) As we have 3 unknown, there will be 6 state variable (which are given) X4 = Y XG : Z X4 = 7 X1 = x4 X5=2 X = X8 X6 W X2= z' X6 = W Xz = wi XzX6 express the above equation of motion in state variable form, From the state variable cue can Zo say that, d²x4 dt² dt² X4=XI XS - X2 d²yct) x = dxi dt X=Xz in x2 = 2x2 d²2ct) d²x5 dt² 2 dt dt² :: xga s dxz = dx 6 d'oct) dt dt² dt²

Now represent that equation in state variable form. il 124" - 74 + 244 - 62' - 122 - 0 : 12cxi) + F(x0) + 24(X4) 6 (X2) - 12685) = 0 : 12/12 12x5 - 24 X 4 + 6x2 - 7x1 is X XG - 2 x 4 + + (-1) K2 - ( 12 ) i 62" + 62' + 122-64²-12Y = f(t) ti 6(x2) + 64 X2 + 12CX5) - 6(X1) - 120X4) = f(+) 6Cx2) = f(t) - 12 X5 + 12x4 – 6X2 +6X1 f(+) - 2x5 + 2X4 - X2 + XL 6 ili 70" + 70'+ 140-740 - 144 = f(t) 7(x3) + 7cxz) + 146X6) - 7 (xi) - 146X4) = f(+) in 7x} f(+)-14X6 + 14X5 - 7 X3 +7X1 x3 = f(+) - 2X6+ 2x4 - X3 + X1 3

AX + BY R-CX 싸 X) here f Ex x6 Y = input here input = f(t) R output here output - Y = X4. X). (-7/12) (/) o - 2 o75x12 2 O (176) + (1/7) fet) 0 . KB -- Q 2 x4 -20 0 -2 X3 0 xy 0 X5 0 0 X6 0 6 0 6 X6. O 0 0 0 * [R] = [o o 0 o osx12 x3 + [0] Хи X5 Lx61 3 cue get A, B, C & D from eqn 0, @