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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;
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^2), -1/tf^2 | |1| |-15/tf^4|
|C|=
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