How to get simple answer using symbolic functions?
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Muhammad Zeeshan Babar
2017년 2월 4일
댓글: Muhammad Zeeshan Babar
2017년 2월 6일
Hello, I am using syms for computing Lie derivative. My problem is that I have a robot model and when I take the 2nd lie derivative, the matrix appears like
state = [x1; x2 ;x3 ;x4];
fx = 4*1 sym function
hx = hx_new = [1 0 0 0;
0 0 0 0;
0 0 0 0;
0 0 0 0]*state;
Lf0_h_new = hx_new;
b1_new = jacobian(Lf0_h_new, state);
Lf1_h_new = simplify(b1_new*fx);
b2_new = jacobian(Lf1_h_new, state);
Lf2_h_new = simplify(b2_new*fx)
-(17932868243945039509054995298422620160000000000*x1 + 179328622760811041254165969937467169271875000*x2 - 9543444251212304856447901238125480501680537600*x1*cos(x3)^2 - 95434442512123048564479012381254805016805376*x2*cos(x3)^2 + 29067690344163397395312355854778368000000*x4^2*sin(x3) + 2340529702609818139925095463956512768000000*cos(x3)*sin(x3) - 24981799213169175398901578715708984375*x2*cos(x3))/(6773644423866147353395200*(1408908359226399653888*cos(x3)^2 - 2647447536626024609375))
0
0
0]
I dont know why it is appearing like this. Someone please help
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채택된 답변
Walter Roberson
2017년 2월 4일
Lf2_h_new = simplify(b2_new*fx, 'step', 20)
댓글 수: 2
Walter Roberson
2017년 2월 5일
That answer is different than the original Lf2_h_new.
It would help if you were to describe what you were expecting.
추가 답변 (1개)
Steven Lord
2017년 2월 5일
Use vpa to approximate the numbers in the answer to a desired number of decimal places.
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