Asked by Maddison Wagner
on 11 Nov 2018 at 5:15

I’ve been trying to write a loop where the below equations will take place for each of the 4 values in the arrays below ie x1 is first calculated with the first value of each and then calculated with the 2nd values of each, etc. I keep getting a result that moves to the end of the loop and repeats the calculation for x1 using all variables instead of just the appropriate variable of each array. Not expecting someone to write the code but an example using loops for several variables and arrays would be great .

Sm1= [2;3;1;4];

Fc1 = [5;2;3;7];

A1= [1;4;1;2];

Fs = 60;

t1 = (0:sm1-1)/Fs;

x1 = A1*cos(2*pi*Fc1*t1);

For sm1(1:end)

t1 = (0:sm1-1)/Fs;

End

For A1(1:end

For Fc1(1:end)

x1 = A1*cos(2*pi*Fc1*t1);

End

End

Answer by Stephen Cobeldick
on 11 Nov 2018 at 7:42

Edited by Stephen Cobeldick
on 11 Nov 2018 at 7:52

Accepted Answer

Use indexing and vectorize the appropriate operations:

Sm1 = [2;3;1;4];

Fc1 = [5;2;3;7];

A1 = [1;4;1;2];

Fs = 60;

N = numel(A1);

C = cell(1,N); % optional.

for k = 1:N

t1 = (0:Sm1(k)-1)/Fs

x1 = A1(k).*cos(2*pi*Fc1(k)*t1)

... your code here

C{k} = x1; % optional: store the vector x1.

end

Maddison Wagner
on 12 Nov 2018 at 0:53

Thank you! When I create the matrix from each vector of x1, I’m attempting to concatenate each x1 value into a single column , I accomplished this with

for k = 1:N

t1 = (0:Sm1(k)-1)/Fs

x1= A1(k).*cos(2*pi*Fc1(k)*t1)

M(:,k)= x1'

end

u = numel(M)

f = zeros(u,total_col)

f(:,colnum) = cat(1,M(:,1),M(:,2))

f(:,(1:6)~=colnum) = zeros %unused columns are zeros

But I’m looking to do this for any possible number of x1 arrays, so I attempted f(:,colnum) = cat(1,M(:,1(1:end))) But this did not work.

Stephen Cobeldick
on 12 Nov 2018 at 5:30

"...I’m attempting to concatenate each x1 value into a single column..."

The easiest way to get one single column as the output is to collect the intermediate vectors into the cell array, and concatenate them after the loop:

C = cell(1,N);

for k = 1:N

...

C{k} = x1(:);

end

V = vertcat(C{:});

Gives:

>> V

V =

1.00000

0.86603

4.00000

3.91259

3.65418

1.00000

2.00000

1.48629

0.20906

-1.17557

Read more about how that concatenation works:

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Answer by madhan ravi
on 11 Nov 2018 at 5:19

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