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SISTEMA COM 3 GRAUS DE LIBERDADE

(1) x = k1x1 + c1ẋ1 + k2(x1-x2) + c2(ẋ1- ẋ2) + M1ẍ1
(2) k2(x1-x2) + c2(ẋ1- ẋ2) = k3(x2-x3) + c3(ẋ2- ẋ3) + M2ẍ2
(3) k3(x2-x3) + c3(ẋ2- ẋ3) = k4x3 + c4ẋ3 + M3ẍ3
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(1) → (4) x(t) = x1(k1+k2) + ẋ1(c1+c2) - x2k2 + ẋ2c2 + ẍ1M1
(2) → (5) 0 = x1k2 + ẋ1c2 - x2(k2+k3) - ẋ2(c2+c3) - ẍ2M2 + x3k3 + ẋ3c3
(3) → (6) 0 = x2k3 + ẋ2c3 - x3(k3+k4) - ẋ3(c3+c4) - ẍ3M3

VARIAVEIS DE ESTADO y1 = x1 ẏ1 = ẋ1 = y2 y2 = ẋ1 ẏ2 = ẍ1 = (4) y3 = x2 ẏ3 = ẋ2 = y4 y4 = ẋ2 ẏ4 = ẍ2 = (5) y5 = x3 ẏ5 = ẋ3 = y6 y6 = ẋ3 ẏ6 = ẍ3 = (6) ẏ1 = 0y1 + 1y2 + 0y3 + 0y4 + 0y5 + 0y6 + 0x(t) ẏ2 = - (k1+k2)y1/M1 - (c1+c2)y2/M1 + k2y3/M1 + c2y4/M1 + 0y5 + 0y6 + x(t)/M1 ẏ3 = 0y1 + 0y2 + 0y3 + 1y4 + 0y5 + 0y6 + 0x(t) ẏ4 = k2y1/M2 + c2y2/M2 - (k2 + k3)y3/M2 – (c2 + c3) y4/M2 + k3y5/M2 + c3y6/M2 + 0x(t) ẏ5 = 0y1 + 0y2 + 0y3 + 0y4 + 0y5 + 1y6 + 0x(t) ẏ6 = 0y1 + 0y2 + k3y3/M3 + c3y4/M3 - (k3 + k1)y5/M3 - (c3 + c4)y6/M3 + 0x(t)

ẏ1ẏ2ẏ3ẏ4ẏ5ẏ6=010000-(k1+k2)M1-(c1+c2)M1 k2M1c2M100000100k2M2c2M2-(k2 + k3)M2–(c2 + c3) M2k3M2c3M200000100k3M3c3M3-(k3 + k1)M3-(c3 + c4)M3*y1y2y3y4y5y6+01/M10000*x(t)

x1 = 1y1 + 0y2 + 0y3 + 0y4 + 0y5 + 0y6 x2 = 0y1 + 0y2 + 1y3 + 0y4 + 0y5 + 0y6 x3 = 0y1 + 0y2 + 0y3 + 0y4 + 1y5 + 0y6 x1x2x3=100000001000000010*y1y2y3y4y5y6 Atribuir valores: k1 = 10, c1 = 5, M1 = 9, k2 = 10, c2 = 1, M2 = 2, k3 = 9, c3 = 2, M3 = 4, k4 = 4, c4 = 8 e x(t) = U. ẏ1ẏ2ẏ3ẏ4ẏ5ẏ6=010000-2,22-0,661,110,110000010050,5-9,5-1,54,51000001002,250,5-4,75-2,5*y1y2y3y4y5y6+00,110000*U RESPOSTA AO IMPULSO UNITÁRIO

RESPOSTA AO DEGRAU