how to vectorise this for loop?

 for i = 2:numHarm
            h1HatN = h1HatN + alpha(i-1)*sin(i*2*pi*Jint*n/M) + beta(i-1)*cos(i*2*pi*Jint*n/M);
            h1Hat2N = h1Hat2N + alpha(i-1)*sin(i*2*pi*Jnew2*n/M) + beta(i-1)*cos(i*2*pi*Jnew2*n/M);
    end

[Stephen23]

What are n, M, Jint, alpha and beta? What are the initial values of h1HatN and h1Hat2N? What value does numHarm have?

[AKHILESH KESAVANUNNITHAN]

n = 262144 x 1 array M = 262144 x 1 array Jint = 263 alpha = 1 x 400 array beta = 1 x 400 array

Initial values of h1HatN and h1Hat2N will be an array of size 262144 x 1

[AKHILESH KESAVANUNNITHAN]

numHarm = 310

[Jan]

@AKHILESH KESAVANUNNITHAN: Please post a piece of code, which is running by copy&paste.

[AKHILESH KESAVANUNNITHAN]

alpha = rand(1, numHarm ); beta = rand(1, numHarm ); % h1HatN = rand(z, 1 ); % h1Hat2N = rand(z, 1 );

    alpha = zeros(1,numHarm);
    beta = zeros(1,numHarm);
    for i = 2:numHarm
        k = round(i*Jnew2);
            cons1 = sin(pi*(i*Jnew2-k))/sin(pi*(i*Jnew2-k)/M);
            cons2 = sin(pi*(i*Jnew2+k))/sin(pi*(i*Jnew2+k)/M);
            theta1 = pi*(M-1)*(i*Jnew2-k)/M;
            theta2 = pi*(M-1)*(i*Jnew2+k)/M;

            cons3 = ((sin(pi*(Jnew2-k))/sin(pi*(Jnew2-k)/M))*(exp((1i)*(a*(Jnew2-k)+phi))) + (sin(pi*(Jnew2+k))/sin(pi*(Jnew2+k)/M))*(exp(-(1i)*(a*(Jnew2+k)+phi))))*(A/(2*M));

            Y = [cons1*sin(theta1)+cons2*sin(theta2) cons1*cos(theta1)+cons2*cos(theta2); cons2*cos(theta2)-cons1*cos(theta1) cons1*sin(theta1)-cons2*sin(theta2)];
            Z = [2*M*real((yfft(mod(round(i*Jnew2),M)+1)-cons3)) 2*M*imag((yfft(mod(round(i*Jnew2),M)+1)-cons3))]';
            % Y inversion start
            detY = (Y(1,1)*Y(2,2)-Y(1,2)*Y(2,1));
            invY = [Y(2,2) -Y(1,2); -Y(2,1) Y(1,1)]/detY;
            % Y inversion end
            SecParams = (invY)*Z;
            alpha(i-1) = SecParams(1);
            beta(i-1) = SecParams(2);
    end

    % ----- Construct the non-coherent and coherent fundamental -----
    n=ones(size(dataRec)); 
    n=cumsum(n)-1;
    h1HatN = Amp*cos((2*pi()*Jint*n/M)+phi);        % Coherent fundamental
    h1Hat2N = Amp*cos((2*pi()*Jnew2*n/M)+phi);      % Non-coherent fundamental
    dataProc2Ne = dataRec - h1Hat2N;                % Error containing the information of harmonics and noise.

    % ----- Add the non-coherent harmonics to the non-coherent fundamental
    % and coherent harmonics to the coherent fundamental -------------
 for i = 2:numHarm
          h1HatN = h1HatN + alpha(i-1)*sin(i*2*pi*Jint*n/M) + beta(i-1)*cos(i*2*pi*Jint*n/M);
          h1Hat2N = h1Hat2N + alpha(i-1)*sin(i*2*pi*Jnew2*n/M) + beta(i-1)*cos(i*2*pi*Jnew2*n/M);
end

Answer 1

[OCDER]

Jint = 263;
Jnew2 = 360;
z = 262144;
numHarm = 400;
n = rand(z, 1);
M = rand(z, 1);
alpha = rand(1, numHarm);
beta = rand(1, numHarm);
h1HatN = rand(z, 1);
h1Hat2N = rand(z, 1);

i = [2:numHarm];
h1HatN  = h1HatN  + sum(alpha(i-1).*sin(i*2*pi*Jint.*n./M)  + beta(i-1).*cos(i*2*pi*Jint.*n./M), 2);
h1Hat2N = h1Hat2N + sum(alpha(i-1).*sin(i*2*pi*Jnew2.*n./M) + beta(i-1).*cos(i*2*pi*Jnew2.*n./M), 2);

%check this   n/M vs n./M   . n/M for 262144x1 / 262144x1 matrix will generate a 512 GB matrix!

[AKHILESH KESAVANUNNITHAN]

This change improved execution from 7s to 5.6s