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

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Mustafa Duran
Mustafa Duran 2023-1-12
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?

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采纳的回答

Torsten
Torsten 2023-1-12
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".
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Torsten
Torsten 2023-1-12
Ran in:
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
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.

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