Hamming Window LUT function

Question

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, 2017 at 22:32

aksadv · Accepted Answer · 2017-01-23 19:27:22Z

0

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

answered Jan 23, 2017 at 19:21

aksadv

1266 bronze badges

Add a comment |

1 Answer 1