Matlab Function returns values

1 回表示 (過去 30 日間)
lilly lord
lilly lord 2021 年 2 月 8 日
コメント済み: lilly lord 2021 年 2 月 8 日
Hi. I have written a function which should return two sequences but it returns only single sequence. I have attached the code below if any one help me in this regard. Thanks in advance
function [X, Y ]= Hybrid(a,b,w0,s0,c,u);
%a=1.4, b=0.3, wo= 0.1, s0=0.1, c=0.4, u=0.9
x(1)=0;
y(1)=0;
x0=0;y0=0;
% Initialize
t=0.4-6/(1+x0^2+y0^2);
x(1,1)=1+u*(x0*cos(t)-y0*sin(t));% mu is u
y(1,1)=u*(x0*sin(t)+y0*cos(t));
% Simulate
n=10000;
X1=[];
Y1=[];
for i=2:n
t=0.4-6/(1+x(i-1,1)^2+y(i-1,1)^2);
x(i,1)=1+u*(x(i-1,1)*cos(t)-y(i-1,1)*sin(t));
x_1(i,1)=mod(ceil(abs(x(i))*10000),256);
%x2_new = unique(x_1,'stable');
y(i,1)=u*(x(i-1,1)*sin(t)+y(i-1,1)*cos(t));
y_1(i,1)=mod(ceil(abs(y(i,1))*10000),256);
%y2_new=unique(y_1,'stable');
X1=x_1;
Y1=y_1;
end
plot(x,y,'.','MarkerSize',4)
X=unique(mod(x_1,256),'stable')'; % unique pooints of x chaotic sequence
Y=unique(mod(y_1,256),'stable')'; % unique point of y choatic sequence
end
It returns only X sequence

採用された回答

the cyclist
the cyclist 2021 年 2 月 8 日
How are you calling the function? If you call it like this
Hybrid(a,b,w0,s0,c,u)
then it will only return the first output. But it should return both if you call it like this
[X,Y]= Hybrid(a,b,w0,s0,c,u)
If that does not explain it, please post more information about how you are calling the function.
  1 件のコメント
lilly lord
lilly lord 2021 年 2 月 8 日
Oh my God. U r 100 % right. Thank u very much.

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeCorrelation and Convolution 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