《Robotics, Vision and Control — Fundamental Algorithms in MATLAB》第三章课后习题

最新推荐文章于 2025-05-10 14:51:43 发布

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最新推荐文章于 2025-05-10 14:51:43 发布 · 3.4k 阅读

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本博客主要探讨《Robotics, Vision and Control》一书第三章的课后习题，包括：1)研究lspb轨迹中恒定速度段的设定对效果的影响；2)对比tpoly和lspb轨迹在最大速度限制下的时间步数需求；3)使用tranimate比较四元数、欧拉角和航向-俯仰-偏航角的旋转插值；4)分析通过奇异点时不同旋转插值方法的行为；5)建立定量评估不同方向插值方法性能的方法；6)重新运行mstraj例子，调整初始和最终速度以及加速时间，研究总时间与加速时间的关系。" 90717949,5054199,Vue.js 实战项目大全,"['前端开发', 'Vue', 'Vue全家桶', 'webapp开发', '单页应用']

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前言

本博客主要是对《Robotics, Vision and Control》第三章课后习题答案进行总结，与大家交流学习。

2.For a lspb trajectory from 0 to 1 in 50 steps explore the effects of specifying the velocity for the constant velocity segment. What are the minimum and maximum bounds possible?

%Code:
[s1,sd1,sdd1]=lspb(0,1,50,0.025);
[s2,sd2,sdd2]=lspb(0,1,50,0.03);
[s3,sd3,sdd3]=lspb(0,1,50,0.035);
[s4,sd4,sdd4]=lspb(0,1,50,0.04);
plot(sd1,'r'),hold on,plot(sd2,'g'),plot(sd3,'b'),plot(sd4,'y'),grid on;

这里写图片描述

Answer:
1. [s,sd,sdd] = lspb(q0, q1, t, V), where q0=0, q1=1,t=50, tf=max(0:t-1) and abs(q1-q0)/tf <= abs(V) <= 2*abs(q1-q0)/tf
   in this question, tf=49, so we have 0.020408 <= abs(V) <= 0.040816.

2. The constant velocity segment is descrease along with the increasing of the specific velocity. 
   The minimum and maximum value of the specific velocity are 0.020408 and 0.040816 respectively.

3.For a trajectory from 0 to 1 and given a maximum possible velocity of 0.025 compare how many time steps are required for each of the tpoly and lspb trajectories?

fprintf('\nFor lspb() function, 41<= time_steps <=81; for tpoly() function, time_steps=76\n')

Answer:
For lspb function, from question2, we know that abs(q1-q0)/tf <= abs(V) <= 2*abs(q1-q0)/tf. 
In this function, abs(V)=0.025, so we get 40 <= tf<= 80, and 41<= time_steps <= 81.

For tpoly function, the solution is as following:
1.   |q0 |  |0,          0,        0,      0,     0,   1| |A|
     |qf |  |tf^5,     tf^4,     tf^3,   tf^2,    tf,  1| |B|
     |qd0|= |0,          0,        0,      0,     1,   0|*|C|
     |qdf|  |5*tf^4,   4*tf^3,   3*tf^2, 2*tf,    1,   0| |D|
     | 0 |  |0,          0,        0,      2,     0,   0| |E| 
     | 0 |  |20*tf^3,  12*tf^2,  6*tf,     2,     0,   0| |F|,  where qf=1, q0=qd0=qdf=0

2.   |A|   |-6/tf^5,   6/tf^5, -3/tf^4, -3/tf^4, -1/(2*tf^3), 1/(2*tf^3) | |0| | 6/tf^5 |
     |B|   |15/tf^4, -15/tf^4,  8/tf^3,  7/tf^3,  3/(2*tf