Values of vectors in matrix (changes in time)

Hello guys, how may I do this..

I have 4 vectors (signals in time)

t =0:pi/20:4*pi;
x1 = cos(t);
x2 = cos(2*t);
x3 = cos(3*t);
x4 = cos(4*t);

And I want to put current value of signal to matrix:

x = [x1 x3]
    [x2 x4]
    

But for t = 0 values of signals in t = 0; for t = t0 + t_step ... etc, Just changes values in matrix in time, I hope you understard :)

Any idea?

Answer 1

[Enthusiastic Student]

Since all the x variables are functions of the same t variable you should be able to create a matrix by:

for m = 1:length(t)
    x(m,:,:) = [x1(m) x2(m);x3(m) x4(m)];
end

This should create a multidimensional array with the first dimension having the same length as t and the two other dimension having a length of 2.

x(10,:,:)

will access the 2x2 matrix for t = t0+9*t_step.

[martin martin]

Guys, very thanks for your help. This is it .. ouput matrix x changes values in time

t = 0:1:10;
x1 = t;
x2 = t.^2;
x3 = t.^3;
x4 = t.^4;

for m = 1:length(t)
    x = [x1(m) x2(m);x3(m) x4(m)];
end

Answer 2

[Andrei Bobrov]

t =0:pi/20:4*pi;
x =reshape(cos((1:4)'*t),2,2,[]);

Answer 3

[KSSV]

t =0:pi/20:4*pi;
x1 = cos(t);
x2 = cos(2*t);
x3 = cos(3*t);
x4 = cos(4*t);

A = zeros(2,2,length(t)) ;
for i = 1:length(t)
    A(:,:,i) = [x1(i) x3(i) ; x2(i) x4(i)] ;
end

It can eb achieved without loop also. Read about reshape.