# Hamming Window LUT function

I'm working on a project where I need to generate a LUT for Hamming Window to perform a FFT.

I'm using this function (doing the table with MATLAB):

n = 0:4095;
LUT_length = length(n);
WIN_HAM_v = round(4095*(0.54-(0.46*cos(2*pi*n/(LUT_length-1)))));


I have some doubts if the function is correct

• Can you ask a more specific question? For instance, what do you expect your code to do, and what is it actually doing? – MBaz Jan 23 '17 at 22:32

## 1 Answer

Your code is correct.

This can also be implemented by calling the built-in MATLAB function hamming:

WIN_HAM_v = round(4095*hamming(LUT_length));


You can (should) compare the two methods yourself:

win1 = 4095*(0.54-(0.46*cos(2*pi*n/(LUT_length-1))));
win2 = 4095*hamming(LUT_length);
maxError = max(abs(win1(:)-win2(:)));
% result: 2.2737e-12