Selection of sigma variable for gaussian function - filtering

Question

I am new to signal processing and would like to apply low pass filter using fft.
I found this post which answers my question. While using it, I have one question:
given that my cutoff frequency is 3500Hz and sampling rate is 25600Hz,
what will be the value of sigma to use while generating Gaussian curve, as given by code from eigenchris' answer below?
gauss = zeros(size(Y));
sigma = 8; % just a guess for a range of 20
gauss(1:r+1) = exp(-(1:r+1).^ 2 / (2 * sigma ^ 2)); % +ve frequencies
gauss(end-r+1:end) = fliplr(gauss(2:r+1)); % -ve frequencies
y_gauss = ifft(Y.*gauss,1024);
Below is function code I am using:
clf; clc;
Fs = 25600;
file = '01cKhaitan181015M4_Opp_LeftS1H8_a.dat';
signal = dlmread(file); % read file from specified location
signal = signal - mean(signal);
N = size(signal, 1);
time = 1000*(0 : N-1)/Fs; % in msec
freq = (-Fs/2 : Fs/N : Fs/2-Fs/N)';
Y = fft(signal, 1024);
r = 141; % range of frequencies we want to preserve
gauss = zeros(size(Y));
sigma = 119.75;
gauss(1:r+1) = exp(-(1:r+1).^ 2 / (2 * sigma ^ 2)); % +ve frequencies
gauss(end-r+1:end) = fliplr(gauss(2:r+1)); % -ve frequencies
y_gauss = ifft(Y.*gauss,1024);
hold on;
plot(time, signal, 'k'); plot(time, abs(y_gauss), 'c');
legend('signal', 'gaussian', 'Location', 'southwest')
hold off;
Below is link of data file
https://www.dropbox.com/s/edb8g43j4a54jvq/01cKhaitan181015M4_Opp_LeftS1H8_a.dat?dl=0

Answer 1

The cutoff frequency is defined as the frequency where the attenuation is 0.5 or ~6dB. You already know the desired cutoff frequency to be 3500Hz. The next step would be to get the corresponding index with:
cutoff_frequency = 3500;
sampling_rate = 25600;
cutoff_index = 1 + cutoff_frequency/sampling_rate * length(Y);
At that cutoff_index you would want to have the desired 0.5 attenuation. Solving for sigma which yields an attenuation of 0.5 at the cutoff_index in your Gaussian formula:
%% Derivation of sigma value
% 0.5 = exp(-cutoff_index^2 / (2 * sigma ^ 2));
% log(0.5) = -cutoff_index^2 / (2 * sigma ^ 2));
% sigma^2 = -cutoff_index^2 / (2 * log(0.5));
% sigma^2 = cutoff_index^2 / (2 * log(2));
yielding:
sigma = cutoff_index / sqrt(2 * log(2));
So, if for example length(Y) is 1024, you would get
sigma = (3500/25600 * 1024) / sqrt(2 * log(2)); % approx. 119.75