how can i rewrite an equation base on variables

Question

shahin hashemi 2017-12-9

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/372009-how-can-i-rewrite-an-equation-base-on-variables

评论： Walter Roberson 2017-12-22

采纳的回答： Birdman

在 MATLAB Online 中打开

dear all

i have 3*1 matrix like matrix E below

syms qd td ed q teta
E =[ ed*cos(q)*sin(teta) - cos(teta)*(qd*cos(q)*sin(teta) + td*cos(teta)*sin(q)) - td*sin(q)*sin(teta)^2 ; cos(teta)*(td*cos(teta)*sin(q) - qd*sin(q)*sin(teta)) + ed*sin(q)*sin(teta) + td*cos(q)*sin(teta)^2  ;  ed*cos(teta) + cos(q)*sin(teta)*(qd*cos(q)*sin(teta) + td*cos(teta)*sin(q)) - sin(q)*sin(teta)*(td*cos(teta)*sin(q) - qd*sin(q)*sin(teta))]

that i want to rewrite it in [3*3]*[qd;td;ed]=[E] this form

is there any Command that can help me to calculate this [3*3] matrix by matlab

answer should be like this:

[                 -cos(q)*cos(teta)*sin(teta),                        - cos(teta)^2*sin(q) - sin(q)*sin(teta)^2, cos(q)*sin(teta)]
[                 -cos(teta)*sin(q)*sin(teta),                          sin(q)*cos(teta)^2 + cos(q)*sin(teta)^2, sin(q)*sin(teta)]
[ cos(q)^2*sin(teta)^2 + sin(q)^2*sin(teta)^2, cos(q)*cos(teta)*sin(q)*sin(teta) - cos(teta)*sin(q)^2*sin(teta),        cos(teta)]

i really apreciated if you could help me

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

Birdman 2017-12-9

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/372009-how-can-i-rewrite-an-equation-base-on-variables#answer_295446

在 MATLAB Online 中打开

Try the following code:

syms qd td ed q teta
E =[ ed*cos(q)*sin(teta) - cos(teta)*(qd*cos(q)*sin(teta) + td*cos(teta)*sin(q)) - td*sin(q)*sin(teta)^2 ; cos(teta)*(td*cos(teta)*sin(q) - qd*sin(q)*sin(teta)) + ed*sin(q)*sin(teta) + td*cos(q)*sin(teta)^2  ;  ed*cos(teta) + cos(q)*sin(teta)*(qd*cos(q)*sin(teta) + td*cos(teta)*sin(q)) - sin(q)*sin(teta)*(td*cos(teta)*sin(q) - qd*sin(q)*sin(teta))]
Ecollected=collect(E,[qd,td,ed])
for i=1:size(E,1)
    E3by3form(i,:)=coeffs(Ecollected(i),[qd,td,ed])
end

Let me know the results. The answer you want is stored in E3by3form.

5 个评论
显示 3更早的评论隐藏 3更早的评论

shahin hashemi 2017-12-22

在 MATLAB Online 中打开

dear mr walter raberson thank you for your attention

function wk = consnt7(x)
L=3e-3;
N=1;
for j=2:N+1
   for i=j-1
    Rl(:,:,j)=[cos(x(3*i-2))*cos(x(3*i-1))*cos(x(3*i))-sin(x(3*i-2))*sin(x(3*i)) -cos(x(3*i-2))*cos(x(3*i-1))*sin(x(3*i))-sin(x(3*i-2))*cos(x(3*i)) cos(x(3*i-2))*sin(x(3*i-1));sin(x(3*i-2))*cos(x(3*i-1))*cos(x(3*i))+cos(x(3*i-2))*sin(x(3*i)) -sin(x(3*i-2))*cos(x(3*i-1))*sin(x(3*i))+cos(x(3*i-2))*cos(x(3*i)) sin(x(3*i-2))*sin(x(3*i-1));-sin(x(3*i-1))*cos(x(3*i)) sin(x(3*i-1))*sin(x(3*i)) cos(x(3*i-1))];   end
end
    R(:,:,2)=Rl(:,:,2);
    R(:,:,1)=[1 0 0;0 1 0;0 0 1];
for i=3:N+1
    R(:,:,i)=R(:,:,i-1)*Rl(:,:,i);
end
for j=2:N+1
    for i=j-1
    wl(:,:,j)=[-sin(x(3*i-2))*x(6*i-1)+cos(x(3*i-2))*sin(x(3*i-1))*x(6*i);cos(x(3*i-2))*x(6*i-1)+sin(x(3*i-2))*sin(x(3*i-1))*x(6*i);x(6*i-2)+cos(x(3*i-1))*x(6*i)];
    w(:,:,2)=wl(:,:,2);
    end
end
for i=3:N+1
    w(:,:,i)= w(:,:,i-1)+R(:,:,i-1)*wl(:,:,i);
end
for i=2:N+1
       wc(:,:,i)=collect(w(:,:,i),[x(4),x(5),x(6)]);%,xd(7),xd(8),xd(9),xd(10),xd(11),xd(12) ,xd(13),xd(14),xd(15),xd(16),xd(17),xd(18),xd(19),xd(20),xd(21),xd(22),xd(23),xd(24)]);
end
for i=2:N+1
    for j=1:size(w(:,:,i),1)
        [wtemp,vtemp]=coeffs(wc(j,:,i),[x(4),x(5),x(6)]);%,xd(7),xd(8)]);%,xd(9),xd(10),xd(11),xd(12),xd(13),xd(14),xd(15),xd(16),xd(17),xd(18),xd(19),xd(20),xd(21),xd(22),xd(23),xd(24)]);
        [~,idx]=ismember(vtemp,[x(4),x(5),x(6)]);%,xd(7),xd(8)]);%,xd(9),xd(10),xd(11),xd(12),xd(13),xd(14),xd(15),xd(16),xd(17),xd(18),xd(19),xd(20),xd(21),xd(22),xd(23),xd(24)]);
        wk(j,idx,i)=wtemp
    end
end

this is a part of my cod and as you can see i use fucntions and then i run this :

x0=[0;0.1;0;0.01;0.001;0.0001];
options = optimoptions('fsolve','Display','iter');
[x,fval] = fsolve(@consnt7,x0,options)

but i face this error :

Undefined function 'collect' for input arguments of type 'double'.

Walter Roberson 2017-12-22

在 MATLAB Online 中打开

That cannot be solved.

x = sym('x',[1 6]);
consnt7(x)
ans(:,:,1) =
[ 0, 0, 0]
[ 0, 0, 0]
[ 0, 0, 0]
ans(:,:,2) =
[ 0, -sin(x1), cos(x1)*sin(x2)]
[ 0,  cos(x1), sin(x1)*sin(x2)]
[ 1,        0,         cos(x2)]

The first pane of the 3D answers can be solved, as it is already all 0. However, in the second pane, ans(:,:,2), there is no x1 such that -sin(x1) == 0 at the same time that cos(x1) == 0 -- and even if there were, the 1 in the bottom left corner, position ans(3,1,2), can never be equal to 0, so there is no possible input vector such that the result of the function is all zeros.

请先登录，再进行评论。