plotting an exponential will a matrix over time

8 ビュー (過去 30 日間)
IP KWAN SO
IP KWAN SO 2019 年 4 月 13 日
編集済み: David Wilson 2019 年 4 月 18 日
Hi guys, i want to plot Y respect to time in 40 seconds. AF is a 4x4 matrix,
t= [0:2:40];
iden= eye(4);
AF= [ 0 1 0 0; 0.00010714 0.00275 10.049 0.98385; 0 0 0 1; -7.6531e-05 -0.0019643 -0.17761 -0.70275];
meme=iden*AF;
figure(1)
Y = exp(AF.*meme);
plot(t,Y);
It results in vector must be in the same length, i do not know how to fix it.
  3 件のコメント
IP KWAN SO
IP KWAN SO 2019 年 4 月 13 日
Sorry i have typing error,
i want to plot Y= exp(AF.*t) overtime in 40 seconds.
Vineeth Nair
Vineeth Nair 2019 年 4 月 18 日
The matrix dimensions must agree before multiplication can happen. Hence the above expression Y= exp(AF.*t) is incorrect.
To read more about elementwise multiplication please visty this link https://in.mathworks.com/help/matlab/ref/times.html

サインインしてコメントする。

回答 (1 件)

David Wilson
David Wilson 2019 年 4 月 18 日
編集済み: David Wilson 2019 年 4 月 18 日
OK, your question is a bit vauge, and there are missing bits, i.e. the start point for y.
I'm assuming you are trying to solve a control problem. As mentioned above, you need to carefully look at dimensions, and I think you want the matrix exponental function, expm, (don't forget the "m").
I started my simulation below at some random point.
t= [0:2:40]; % time vector
AF= [ 0 1 0 0;
0.00010714 0.00275 10.049 0.98385;
0 0 0 1;
-7.6531e-05 -0.0019643 -0.17761 -0.70275];
Y = randn(1,4); % start position (who knows??)
for i=2:length(t)
Y(i,:) = expm(AF*t(i))*Y(i-1,:)';
end
plot(t,Y);
Of course if you have the control toolbox, there are many better ways to do this, e.g.
>> help lti/initial
or,
G = ss(AF, zeros(4,1), eye(4,4), 0)
Y0 = randn(1,4); % another random start poit
[Y,t] = initial(G,Y0,t) % simulate the free response

カテゴリ

Help Center および File ExchangeCreating and Concatenating Matrices についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by