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Symbolic seems to have an error

1 回表示 (過去 30 日間)
Stephen Wilkerson
Stephen Wilkerson 2020 年 9 月 11 日
閉鎖済み: MATLAB Answer Bot 2021 年 8 月 20 日
Two typos in my original post corrected
  2 件のコメント
John D'Errico
John D'Errico 2020 年 9 月 11 日
編集済み: John D'Errico 2020 年 9 月 11 日
Then why did you feel the need to post it again? Just edit your original post. Anyway, it seems like everytime you don't understand a piece of code, you declare it to be a bug.
Stephen Wilkerson
Stephen Wilkerson 2020 年 9 月 11 日
We'll do you have an answer?

回答 (3 件)

Stephen Wilkerson
Stephen Wilkerson 2020 年 9 月 11 日
Yes that's what I ended up doing.

Stephen Wilkerson
Stephen Wilkerson 2020 年 9 月 11 日
Now I'm unsure how to delete the second post?

Paul
Paul 2020 年 9 月 11 日
編集済み: Paul 2020 年 9 月 12 日
The RHS of your differential equation should be 3*u.
In the frequency domain, Y(s) = H(s)*U(s). It looks like you're trying to get the step repsonse of your second order system, so U(s) = 1/s. You forgot to multiply by U(s) in your symbolic approach.
num = 3;
den = [1 2 5];
syms s
nums = poly2sym(num,s);
dens = poly2sym(conv(den,[1 0]),s); % here!
ilaplace(nums/dens)
ans =
3/5 - (3*exp(-t)*(cos(2*t) + sin(2*t)/2))/5
  1 件のコメント
Stephen Wilkerson
Stephen Wilkerson 2020 年 9 月 11 日
Thank you

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