Hi
To determine the damping ratio (psij and psik ), i have to solve the attached system where all other name
are constant.
How can i solve this system ?

3 件のコメント

Jan
Jan 2017 年 1 月 25 日
編集済み: Jan 2017 年 1 月 25 日
The attached image does not contain any "psi"s. Do you mean zeta or omega? Do you want to solve this symbolically or numerically?
What have you tried so far? Which problems occur? Is this a homework question? We should know the last detail, because this demands for another kind of answering.
Mallouli Marwa
Mallouli Marwa 2017 年 1 月 26 日
I mean zeta
Mallouli Marwa
Mallouli Marwa 2017 年 1 月 26 日
I want to solve it numerically.

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 採用された回答

Torsten
Torsten 2017 年 1 月 26 日

1 投票

A = 2*omegaj*omegak/(omegaj^2-omegak^2)*[Y*I/omegak -Y*I/omegaj ; -m*omegak m*omegaj];
b = [cs*I ; ca];
zeta = A\b;
zetaj = zeta(1);
zetak = zeta(2);
Best wishes
Torsten.

5 件のコメント

Mallouli Marwa
Mallouli Marwa 2017 年 1 月 26 日
Hi How can i stock zetaj and zetak in two vectors ?
ca_m = 19.295;
cs_I = 2.93e-6;
for i=1:8
A = 2 * wr(i)* wr(i+1) / (wr(i)^2 - wr(i+1)^2) * [1/ wr(i+1) -1/ wr(i) ; -wr(i+1) wr(i)];
b = [cs_I ; ca_m];
zeta = A\b;
zetaj = zeta(1);
zetak = zeta(2);
end
Torsten
Torsten 2017 年 1 月 26 日
zetaj(i)=zeta(1);
zetak(i)=zeta(2);
Best wishes
Torsten.
Mallouli Marwa
Mallouli Marwa 2017 年 1 月 26 日
For this loop i obtain this error :
Warning: Matrix is singular, close to singular or badly scaled.
Results may be inaccurate. RCOND = NaN.
In program at 5 (zeta)
Please help me.
for i=1:10
for k=1:10
A = 2* wr(i)*wr(k)/(wr(i)^2-wr(k)^2)*[1/wr(k) -1/wr(i) ; -wr(k) wr(i)];
b = [cs_I ; ca_m];
zeta = A\b;
zetaj(k)=zeta(2);
zetak (i)=zeta(1);
end
end
Torsten
Torsten 2017 年 1 月 27 日
編集済み: Torsten 2017 年 1 月 27 日
If abs(wr(i))=abs(wr(k)), your matrix A becomes singular and you won't get a solution.
This will at least be the case if i=k in your nested-loop construction.
Best wishes
Torsten.
Mallouli Marwa
Mallouli Marwa 2017 年 1 月 27 日
Please how can i correct this error ?

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その他の回答 (1 件)

Torsten
Torsten 2017 年 1 月 26 日

0 投票

Hint:
The inverse of
[a b ; c d]
is
1/(a*d-b*c)*[d -b; -c a]
Best wishes
Torsten.

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