MATLAB: Calculating SEF (spectral edge frequency) of a EEG

eeg

I currently have a file with a 4 channel EEG, and I want to calculate the SEF (95%) using that raw data. I wanted to know if there is as automated way of doing so using Matlab. I also have computed data Hz x time x dB of the given raw EEG file.

ADICIONAL INFO: The SEF is a Hz by time chart where the Hz shown is the threshold frequency where 95% of the EEG power lies beneath it at a given time (I don't know which algorithm to use) though.

Thank you in advance!

Best Answer

I would use cumtrapz to do the integration, and interp1 using the cumtrapz result and the frequency vector to find the SEF.

Example —

% First, Create Data
t = 0 : 0.01 : 1;
for f = 1 : 2 : 10
    y(f,:) = sin (2 * pi .* f .* t);
end
L = length(t);                                          % Signal Length
Ts = t(2) - t(1);                                       % Sampling Time Interval
Fs = 1/Ts;                                              % Sampling Frequency
Fn = Fs/2;                                              % Nyquist Frequency
Fv = linspace(0, 1, fix(L/2)+1)*Fn;                     % Frequency Vector
Iv = 1:length(Fv);                                      % Index Vector
Fourier = fft(y, [], 2)/L;                              % Fourier Transform (Along Rows Of Matrix)
Fouriers = sum(abs(Fourier));                           % Spectrum
IntSpectrum = cumtrapz(Fv, Fouriers(Iv));               % Numeric Integration
SEF = interp1(IntSpectrum, Fv, 0.95*IntSpectrum(end), 'linear');    % Interploate To Find ‘SEF’
figure(1)
plot(Fv, Fouriers(Iv)*2)
hold on
plot([1 1]*SEF, ylim, '--r')
hold off

Related Solutions

MATLAB: I get the peaks at the wrong frequency and with the wrong amplitude.

hello Anna

look at my fft function at the end of my code

clc
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%





% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% data
data = importdata('SFC5S_nov25_ST_1_1.txt');
dt = 1e-6; % 1 micro seconds
signal = data(:,1);
samples = length(signal);
Fs = 1/dt; % sampling frequency (Hz)
%% decimate (if needed)
% NB : decim = 1 will do nothing (output = input)
decim = 1;
if decim>1
    signal = decimate(signal,decim);
    Fs = Fs/decim;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 512;    % 
OVERLAP = 0.95;
% spectrogram dB scale
spectrogram_dB_scale = 80;  % dB range scale (means , the lowest displayed level is XX dB below the max level)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% options 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% if you are dealing with acoustics, you may wish to have A weighted
% spectrums 
% option_w = 0 : linear spectrum (no weighting dB (L) )
% option_w = 1 : A weighted spectrum (dB (A) )
option_w = 0;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% display 1 : averaged FFT spectrum
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[freq, sensor_spectrum] = myfft_peak(signal,Fs,NFFT,OVERLAP);
% convert to dB scale (ref = 1)
sensor_spectrum_dB = 20*log10(sensor_spectrum);
% apply A weigthing if needed

if option_w == 1
    pondA_dB = pondA_function(freq);
    sensor_spectrum_dB = sensor_spectrum_dB+pondA_dB;
    my_ylabel = ('Amplitude (dB (A))');
else
    my_ylabel = ('Amplitude (dB (L))');
end
figure(1),plot(freq,sensor_spectrum_dB,'b');grid
title(['Averaged FFT Spectrum  / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(freq(2)-freq(1)) ' Hz ']);
xlabel('Frequency (Hz)');ylabel(my_ylabel);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% display 2 : time / frequency analysis : spectrogram demo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[sg,fsg,tsg] = specgram(signal,NFFT,Fs,hanning(NFFT),floor(NFFT*OVERLAP));  
% FFT normalisation and conversion amplitude from linear to dB (peak)
sg_dBpeak = 20*log10(abs(sg))+20*log10(2/length(fsg));     % NB : X=fft(x.*hanning(N))*4/N; % hanning only
% apply A weigthing if needed
if option_w == 1
    pondA_dB = pondA_function(fsg);
    sg_dBpeak = sg_dBpeak+(pondA_dB*ones(1,size(sg_dBpeak,2)));
    my_title = ('Spectrogram (dB (A))');
else
    my_title = ('Spectrogram (dB (L))');
end
% saturation of the dB range : 
% saturation_dB = 60;  % dB range scale (means , the lowest displayed level is XX dB below the max level)
min_disp_dB = round(max(max(sg_dBpeak))) - spectrogram_dB_scale;
sg_dBpeak(sg_dBpeak<min_disp_dB) = min_disp_dB;
% plots spectrogram
figure(2);
imagesc(tsg,fsg,sg_dBpeak);colormap('jet');
axis('xy');colorbar('vert');grid
title([my_title ' / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(fsg(2)-fsg(1)) ' Hz ']);
xlabel('Time (s)');ylabel('Frequency (Hz)');
function pondA_dB = pondA_function(f)
	% dB (A) weighting curve
	n = ((12200^2*f.^4)./((f.^2+20.6^2).*(f.^2+12200^2).*sqrt(f.^2+107.7^2).*sqrt(f.^2+737.9^2)));
	r = ((12200^2*1000.^4)./((1000.^2+20.6^2).*(1000.^2+12200^2).*sqrt(1000.^2+107.7^2).*sqrt(1000.^2+737.9^2))) * ones(size(f));
	pondA = n./r;
	pondA_dB = 20*log10(pondA(:));
end
function  [freq_vector,fft_spectrum] = myfft_peak(signal, Fs, nfft, Overlap)
% FFT peak spectrum of signal  (example sinus amplitude 1   = 0 dB after fft).
% Linear averaging
%   signal - input signal, 
%   Fs - Sampling frequency (Hz).
%   nfft - FFT window size
%   Overlap - buffer overlap % (between 0 and 0.95)
signal = signal(:);
samples = length(signal);
% fill signal with zeros if its length is lower than nfft
if samples<nfft
    s_tmp = zeros(nfft,1);
    s_tmp((1:samples)) = signal;
    signal = s_tmp;
    samples = nfft;
end
% window : hanning
window = hanning(nfft);
window = window(:);
%    compute fft with overlap 
 offset = fix((1-Overlap)*nfft);
 spectnum = 1+ fix((samples-nfft)/offset); % Number of windows
%     % for info is equivalent to : 
%     noverlap = Overlap*nfft;
%     spectnum = fix((samples-noverlap)/(nfft-noverlap));	% Number of windows
    % main loop
    fft_spectrum = 0;
    for i=1:spectnum
        start = (i-1)*offset;
        sw = signal((1+start):(start+nfft)).*window;
        fft_spectrum = fft_spectrum + (abs(fft(sw))*4/nfft);     % X=fft(x.*hanning(N))*4/N; % hanning only 
    end
    fft_spectrum = fft_spectrum/spectnum; % to do linear averaging scaling
% one sidded fft spectrum  % Select first half 
    if rem(nfft,2)    % nfft odd
        select = (1:(nfft+1)/2)';
    else
        select = (1:nfft/2+1)';
    end
fft_spectrum = fft_spectrum(select);
freq_vector = (select - 1)*Fs/nfft;
end

MATLAB: Return X coordinate of a fft plot where 80% of the total area under the curve is achieved

Try this:

t = linspace(0, 10, 150);                                   % Create Data

y = sin(2*pi*t*3) .* cos(2*pi*t*2) + 0.01*randn(size(t));   % Create Data
L = numel(t);
Ts = mean(diff(t));                                         % Sampling Interval
Fs = 1/Ts;                                                  % Sampling Frequency
Fn = Fs/2;                                                  % Nyquist Frequency
FTy = fft(y)/L;                                             % Fourier Transform
Fv = linspace(0, 1, fix(L/2)+1)*Fn;                         % Frequency Vector
Iv = 1:numel(Fv);                                           % Index Vector
areaFrac = 0.8;                                             % Desired Fraction Of Total Area
areaTot = trapz(Fv, abs(FTy(Iv)));                          % Total Area
areaCum = cumtrapz(Fv, abs(FTy(Iv)));                       % Cumulative Integral
Fv80 = interp1(areaCum, Fv, areaTot*areaFrac, 'spline')     % Frequency Corresponding To 80% Total Area
FTy80 = interp1(Fv, abs(FTy(Iv))*2, Fv80, 'spline')         % Value of Fourier Transform At ‘Fv80’
figure
plot(Fv, abs(FTy(Iv))*2, '-b')
hold on
plot(Fv, areaCum, '-k')
plot([0 max(Fv)], [1 1]*areaTot, '--k')
plot([0 max(Fv)], [1 1]*areaFrac*areaTot, ':k')
plot([1 1]*Fv80, [0 FTy80], '-r')
plot(Fv80, areaTot*areaFrac, 'r+')
hold off
legend('Fourier Transform', 'Cumulative Area', 'Maximum Area', '80% Maximum Area', 'Fourier Transform at 80% Maximum Area')

Experiment to get the result you want.

Best Answer

Related Solutions

MATLAB: I get the peaks at the wrong frequency and with the wrong amplitude.

MATLAB: Return X coordinate of a fft plot where 80% of the total area under the curve is achieved

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