Error using sym/coeffs. First argument must be a scalar.

Question

Mustafa Duran 2023-1-12

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1893020-error-using-sym-coeffs-first-argument-must-be-a-scalar

编辑： Mustafa Duran 2023-1-12

syms q1(t) q2(t) q3(t) q4(t) q5(t) q6(t) q1_hiz(t) q2_hiz(t) q3_hiz(t) q4_hiz(t) q5_hiz(t) q6_hiz(t) q1_ivme(t) q2_ivme(t) q3_ivme(t) q4_ivme(t) q5_ivme(t) q6_ivme(t) t

After various transactions, i get for example that equation:

VG1_O =

-0.4250*cos(q1(t))*q1_hiz(t)

-0.4250*sin(q1(t))*q1_hiz(t)

0

When i take the coefficients of that matrix as q2_hiz(t); it is expected that solution will be 0 however the solution is like this:

ff=coeffs(VG1_O(1,1),[q2_hiz(t)]);

ff=

-0.4250*cos(q1(t))*q1_hiz(t)

How can i solve this problem?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Torsten 2023-1-12

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1893020-error-using-sym-coeffs-first-argument-must-be-a-scalar#answer_1147365

编辑：Torsten 2023-1-12

My guess is that VG1_0(1,1) is interpreted as a polynomial in the variable q2_hiz(t).

Since your expression does not contain q2_hiz(t), -0.4250*cos(q1(t))*q1_hiz(t) is the constant term of the polynomial

-0.4250*cos(q1(t))*q1_hiz(t) + 0*q2_hiz(t) + 0*q2_hiz(t)^2 + ...

and is thus returned by "coeffs".

5 个评论
显示 3更早的评论隐藏 3更早的评论

Torsten 2023-1-12

在 MATLAB Online 中打开

That's what the diff command does - if a variable does not appear in an expression, the derivative with respect to this variable is 0. And if you want the coefficient of q1_hiz(t), the command will give you -0.4250*cos(q1(t)).

syms t q1(t) q1_hiz(t) q2_hiz(t) q3_hiz(t)

expr = -0.4250*cos(q1(t))*q1_hiz(t);

ff1 = diff(expr,q1_hiz(t))

ff1 =

ff2 = diff(expr,q2_hiz(t))

ff2 =

0

ff3 = diff(expr,q3_hiz(t))

ff3 =

0

Mustafa Duran 2023-1-12

编辑：Mustafa Duran 2023-1-12

You're right. Solution is so basic however i didn't think it is that much simple even if i know derivative. Sometimes it is better to take opinions of others. Thanks so much for responding my questions.

请先登录，再进行评论。