您正在看的电子信息论文是:柔性机械臂逆动力学问题的分析和求解。 f coordinates in x and y directions for the end movement

　　由图2中可以看出，由式(7)得到的弹性坐标(不含振动)与机械臂的动力学仿真得到的弹性坐标(含振动)的慢变部分十分相似，所以在式(2)中去掉弹性坐标的一阶、二阶时间导数项相当于滤掉了弹性坐标中的振动部分，说明这种方法是合理的。

　　由图3与图4给出的仿真结果可以看出，轨迹跟踪很好，由此可见，得到的力矩精度很高.

　　4 结束语

　　由图2可以看到，机械臂在运动过程中，其弹性坐标由两方面组成，一方面是振动部分(快变部分)，另一方面是与载荷、惯性力有关的慢变部分。而弹性坐标速度、加速度的慢变部分很小，在逆动力学求解中将其略去是合理的，由式(7)得到了比较准确的弹性坐标慢变部分并非偶然。

　　由以上分析可以看出，对于柔性机械臂系统，振动部分的精确逆动力学解是发散的，进行逆动力学求解时，应滤掉振动部分，在慢变的意义上进行，才能得到比较好的前馈力矩。

　　参 考 文 献

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