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Help with loops _ multiple variables

Asked by Maddison Wagner on 11 Nov 2018 at 5:15
Latest activity Edited by Stephen Cobeldick on 12 Nov 2018 at 6:01
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

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2 Answers

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

  2 Comments

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.
"...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

"Not expecting someone to write the code but an example using loops for several variables and arrays would be great"

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