MATLAB Answers

root of nonlinear equation

2 ビュー (過去 30 日間)
Lidaou Ali
Lidaou Ali 2020 年 3 月 28 日
コメント済み: Lidaou Ali 2020 年 3 月 28 日
i am working on an electical concepts design problem for a emergency exit door release mechanism. I solved the circuit and found that i(t)=0.25*e^(-t/8c)*cos(t/8c)
i want i(t)=0.1 at t=3 sec
i got the code but it doesnt work
%ft=0.25*exp(-3/c)*cos(3/(8*c))-0.1;
%c= fzero(ft,3);
but i cant get result for c
can someone help me?

  0 件のコメント

Sign in to comment.

採用された回答

Torsten
Torsten 2020 年 3 月 28 日
ft = @( c ) 0.25* ...
instead of
ft = 0.25* ...

  5 件のコメント

表示 2 件の古いコメント
Lidaou Ali
Lidaou Ali 2020 年 3 月 28 日
thank you. I am getting different answers for different c values
Torsten
Torsten 2020 年 3 月 28 日
If you plot the function, you'll see that it has two roots:
c=-0.238
c=3.297
Lidaou Ali
Lidaou Ali 2020 年 3 月 28 日
that is correct. i believe i missed an "8"
the right equation is ft=@(c)0.25*exp(-3/(8*c))*cos(3/(8*c))

Sign in to comment.

その他の回答 (0 件)


Translated by