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

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?

Answer 1

[Torsten]

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".

[Mustafa Duran]

Thanks, I understand now. Is there any tool that do what i want correctly?

[Torsten]

Since I don't understand what you want, I cannot give advice.

Maybe you want to use "diff" ?

syms t q1(t) q1_hiz(t) q2_hiz(t)
expr = -0.4250*cos(q1(t))*q1_hiz(t);
ff = diff(expr,q2_hiz(t))

ff = 

0

[Mustafa Duran]

I just want to when i ask MATLAB for what is the coefficient of q2_hiz(t) or q3_hiz(t) in expr; it must give me 0.

[Torsten]

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]

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.