例2-4-1 3 (3)-2y+6 y=5u 解:令x1=y,x2=y,x3=y,x=y (3) 所以x 4=-6x1+2x3-3x4+5u 状态方程为: 0100 0 0010 0 J 0001 0 602-3 5例2-4-1 y 3y 2y 6y 5u (4) (3) (2) + − + = 解: (3) 1 2 3 4 令 x = y, x = y' , x = y'' , x = y 所以 = − + − + = = = x x x x u x x x x x x 4 6 1 2 3 3 4 5 3 4 2 3 1 2 状态方程为: x x − − = • 6 0 2 3 0 0 0 1 0 0 1 0 0 1 0 0 , y = 1 0 0 0x 5 0 0 0 u +