construct two dimensional array in matlab

Question

let us consider following equation

x(t)=sum(a(i)*sin(2*pi*f(i)*t+b(i)*cos(2*pi*f(i)*t))

where i=1,2,......m and frequencies f=[f1,f2,.....fm] and t=[t1,t2,....tn]

i want to create matrix by sin(2*pi*f(i)*t) and cos(2*pi*f(i)*t),clearly it would be matrix with dimension NX2*m,i have tried following code

function [amplitudes]=determine_amplitudes(y,f,t,n,m);

X=zeros(n,2*m);
for i=1:n
     for k=1:m
      if mod(k,2)==1 
          X(i,k)=sin(2*pi*f(k)*t(i));
      else
           X(i,k)=cos(2*pi*f(k)*t(i);
      end

     end
end
end

i used mod operator to determine that if k is odd index,then there should be written sin value,else cosine value,but problem is that i am not sure that given matrix would be with dimension NX2*m,so how to create such matrix so that not exceed index of bounds of frequency array ,recall that frequency array is following f=[f1,f2,..fm],so my problem simple is how to apply m frequency at 2*m position,thanks for help

UPDATE:

let say m=3, and frequencies f=[12.5 13.6 21.7]

then we have following matrix ,also assume

 n=4 t=[0.01 0.02 0.03 0.04]

 sin(2*pi*f(1)*t(1))  cos(2*pi*f(1)* t(1)) sin(2*pi*f(2)*t(1)) cos(2*pi*f(2)*t(1)) sin(2*pi*f(3)*t(1)) cos(2*pi*f(3)*t(1))

I think the final matix would be of size nXm because with the if conditional it's just selecting one of sin or cos terms, isn't that right? — Divakar
– Divakar, Commented Mar 21, 2014 at 5:51
no no,select sin or cosine yes,but ok i will update my question to clarify it — user466534
– user466534, Commented Mar 21, 2014 at 5:53
sorry there must be 2*m components,there is no space,it should sin(f(1) cos(f(1)) sin(f(2) cos(f(2)) and so on — user466534
– user466534, Commented Mar 21, 2014 at 5:59

Divakar · Accepted Answer · 2014-03-21 06:36:20Z

1

No loop version

[f1,t1] = meshgrid(f',t');

p1 = sin(bsxfun(@times,f1,2*pi*t')); %%// Create sin copy '
p2 = cos(bsxfun(@times,f1,2*pi*t')); %%// Create cos copy '

d1 = [p1;p2];
final_matrix = reshape(d1,size(p1,1),[]);

Naive loop version

final_matrx2 = zeros(n,2*m);
for k1=1:n
    for k2=1:2*m
        if rem(k2,2)==1 
            final_matrx2(k1,k2) = sin(2*pi*f(ceil(k2/2))*t(k1));
        else
            final_matrx2(k1,k2) = cos(2*pi*f(ceil(k2/2))*t(k1));
        end
    end
end

edited Mar 21, 2014 at 6:36

answered Mar 21, 2014 at 6:10

Divakar

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2 Comments

user466534 Over a year ago

please if it possible using loops?it would be more readily for me

user466534 Over a year ago

no i have not tested this code,just need to have code like i have posted

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