Values of vectors in matrix (changes in time)

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martin martin
martin martin 2019-3-11
评论： martin martin 2019-3-11
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?

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采纳的回答

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.
  1 个评论
martin martin
martin martin 2019-3-11
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

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更多回答（2 个）

Andrei Bobrov
Andrei Bobrov 2019-3-11
编辑：Andrei Bobrov 2019-3-11
t =0:pi/20:4*pi;
x =reshape(cos((1:4)'*t),2,2,[]);
KSSV
KSSV 2019-3-11
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.

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