I'm trying to find the peak frequency for two signals 'CA1' and 'PFC', within a specified range (25-140Hz).
In Matlab, so far I have plotted an FFT for each of these signals (see pictures below). These FFTs suggest that the peak frequency between 25-140Hz is different for each signal, but I would like to quantify this (e.g. CA1 peaks at 80Hz, whereas PFC peaks at 55Hz). However, I think the FFT is not smooth enough, so when I try and extract the peak frequencies it doesn't make sense as my code pulls out loads of values. I was only expecting a few values - one each time the FFT peaks (around 2Hz, 5Hz and ~60Hz).
I want to know, between 25-140Hz, what is the peak frequency in 'CA1' compared with 'PFC'. 'CA1' and 'PFC' are both 152401 x 7 matrices of EEG data, recorded
from 7 separate individuals. I want the MEAN peak frequency for each data set (i.e. averaged across the 7 test subjects for CA1 and PFC).
My code so far (based on Matlab help files and code I've scrabbled together online):
Fs = 508;
%notch filter
[b50,a50] = iirnotch(50/(Fs/2), (50/(Fs/2))/70);
CA1 = filtfilt(b50,a50,CA1);
PFC = filtfilt(b50,a50,PFC);
%FFT
L = length(CA1);
NFFT = 2^nextpow2(L);
%FFT for each of the 7 subjects
for i = 1:size(CA1,2);
CA1_FFT(:,i) = fft(CA1(:,i),NFFT)/L;
PFC_FFT(:,i) = fft(PFC(:,i),NFFT)/L;
end
%Average FFT across all 7 subjects - CA1
Mean_CA1_FFT = mean(CA1_FFT,2);
% Mean_CA1_FFT_abs = 2*abs(Mean_CA1_FFT(1:NFFT/2+1));
%Average FFT across all 7 subjects - PFC
Mean_PFC_FFT = mean(PFC_FFT,2);
% Mean_PFC_FFT_abs = 2*abs(Mean_PFC_FFT(1:NFFT/2+1));
f = Fs/2*linspace(0,1,NFFT/2+1);
%LEFT HAND SIDE FIGURE
plot(f,2*abs(Mean_CA1_FFT(1:NFFT/2+1)),'r');
set(gca,'ylim', [0 2]);
set(gca,'xlim', [0 200]);
[C,cInd] = sort(2*abs(Mean_CA1_FFT(1:NFFT/2+1)));
CFloor = 0.1; %CFloor is the minimum amplitude value (ignore small values)
Amplitudes_CA1 = C(C>=CFloor); %find all amplitudes above the CFloor
Frequencies_CA1 = f(cInd(1+end-numel(Amplitudes_CA1):end)); %frequency of the peaks
%RIGHT HAND SIDE FIGURE
figure;plot(f,2*abs(Mean_PFC_FFT(1:NFFT/2+1)),'r');
set(gca,'ylim', [0 2]);
set(gca,'xlim', [0 200]);
[P,pInd] = sort(2*abs(Mean_PFC_FFT(1:NFFT/2+1)));
PFloor = 0.1; %PFloor is the minimum amplitude value (ignore small values)
Amplitudes_PFC = P(P>=PFloor); %find all amplitudes above the PFloor
Frequencies_PFC = f(pInd(1+end-numel(Amplitudes_PFC):end)); %frequency of the peaks
Please help!! How do I calculate the 'major' peak frequencies from an FFT, and ignore all the 'minor' peaks (because the FFT is not smoothed).
FFTs assume that the signal has no trend (this is called a stationary signal), if it does then this will give a dominant frequency component at 0Hz as you have here. Try using the MATLAB function detrend, you may find this solves your problem.
Something along the lines of:
x = x - mean(x)
y = detrend(x, 'constant')
Related
I would like to create 500 ms of band-limited (100-640 Hz) white Gaussian noise with a (relatively) flat frequency spectrum. The noise should be normally distributed with mean = ~0 and 99.7% of values between ± 2 (i.e. standard deviation = 2/3). My sample rate is 1280 Hz; thus, a new amplitude is generated for each frame.
duration = 500e-3;
rate = 1280;
amplitude = 2;
npoints = duration * rate;
noise = (amplitude/3)* randn( 1, npoints );
% Gaus distributed white noise; mean = ~0; 99.7% of amplitudes between ± 2.
time = (0:npoints-1) / rate
Could somebody please show me how to filter the signal for the desired result (i.e. 100-640 Hz)? In addition, I was hoping somebody could also show me how to generate a graph to illustrate that the frequency spectrum is indeed flat.
I intend on importing the waveform to Signal (CED) to output as a form of transcranial electrical stimulation.
The following is Matlab implementation of the method alluded to by "Some Guy" in a comment to your question.
% In frequency domain, white noise has constant amplitude but uniformly
% distributed random phase. We generate this here. Only half of the
% samples are generated here, the rest are computed later using the complex
% conjugate symmetry property of the FFT (of real signals).
X = [1; exp(i*2*pi*rand(npoints/2-1,1)); 1]; % X(1) and X(NFFT/2) must be real
% Identify the locations of frequency bins. These will be used to zero out
% the elements of X that are not in the desired band
freqbins = (0:npoints/2)'/npoints*rate;
% Zero out the frequency components outside the desired band
X(find((freqbins < 100) | (freqbins > 640))) = 0;
% Use the complex conjugate symmetry property of the FFT (for real signals) to
% generate the other half of the frequency-domain signal
X = [X; conj(flipud(X(2:end-1)))];
% IFFT to convert to time-domain
noise = real(ifft(X));
% Normalize such that 99.7% of the times signal lies between ±2
noise = 2*noise/prctile(noise, 99.7);
Statistical analysis of around a million samples generated using this method results in the following spectrum and distribution:
Firstly, the spectrum (using Welch method) is, as expected, flat in the band of interest:
Also, the distribution, estimated using histogram of the signal, matches the Gaussian PDF quite well.
I have a series of 2D measurements (time on x-axis) that plot to a non-smooth (but pretty good) sawtooth wave. In an ideal world the data points would form a perfect sawtooth wave (with partial amplitude data points at either end). Is there a way of calculating the (average) period of the wave, using OCTAVE/MATLAB? I tried using the formula for a sawtooth from Wikipedia (Sawtooth_wave):
P = mean(time.*pi./acot(tan(y./4))), -pi < y < +pi
also tried:
P = mean(abs(time.*pi./acot(tan(y./4))))
but it didn't work, or at least it gave me an answer I know is out.
An example of the plotted data:
I've also tried the following method - should work - but it's NOT giving me what I know is close to the right answer. Probably something simple and wrong with my code. What?
slopes = diff(y)./diff(x); % form vector of slopes for each two adjacent points
for n = 1:length(diff(y)) % delete slope of any two points that form the 'cliff'
if abs(diff(y(n,1))) > pi
slopes(n,:) = [];
end
end
P = median((2*pi)./slopes); % Amplitude is 2*pi
Old post, but thought I'd offer my two-cent's worth. I think there are two reasonable ways to do this:
Perform a Fourier transform and calculate the fundamental
Do a curve-fitting of the phase, period, amplitude, and offset to an ideal square-wave.
Given curve-fitting will likely be difficult because of discontinuities in saw-wave, so I'd recommend Fourier transform. Self-contained example below:
f_s = 10; # Sampling freq. in Hz
record_length = 1000; # length of recording in sec.
% Create noisy saw-tooth wave, with known period and phase
saw_period = 50;
saw_phase = 10;
t = (1/f_s):(1/f_s):record_length;
saw_function = #(t) mod((t-saw_phase)*(2*pi/saw_period), 2*pi) - pi;
noise_lvl = 2.0;
saw_wave = saw_function(t) + noise_lvl*randn(size(t));
num_tsteps = length(t);
% Plot time-series data
figure();
plot(t, saw_wave, '*r', t, saw_function(t));
xlabel('Time [s]');
ylabel('Measurement');
legend('measurements', 'ideal');
% Perform fast-Fourier transform (and plot it)
dft = fft(saw_wave);
freq = 0:(f_s/length(saw_wave)):(f_s/2);
dft = dft(1:(length(saw_wave)/2+1));
figure();
plot(freq, abs(dft));
xlabel('Freqency [Hz]');
ylabel('FFT of Measurement');
% Estimate fundamental frequency:
[~, idx] = max(abs(dft));
peak_f = abs(freq(idx));
peak_period = 1/peak_f;
disp(strcat('Estimated period [s]: ', num2str(peak_period)))
Which outputs a couple of graphs, and also the estimated period of the saw-tooth wave. You can play around with the amount of noise and see that it correctly gets a period of 50 seconds till very high levels of noise.
Estimated period [s]: 50
I'm trying to understand how the FFT in matlab works, particularly, how to define the frequency range to plot it. It happens that I have read from matlab help links and from other discussions here and I think (guess) that I'm confused about it.
In the matlab link:
http://es.mathworks.com/help/matlab/math/fast-fourier-transform-fft.html
they define such a frequency range as:
f = (0:n-1)*(fs/n)
with n and fs as:
n = 2^nextpow2(L); % Next power of 2 from length of signal x
fs = N/T; % N number of samples in x and T the total time of the recorded signal
But, on the other hand, in the previous post Understanding Matlab FFT example
(based on previous version of matlab), the resulting frequency range is defined as:
f = fs/2*linspace(0,1,NFFT/2+1);
with NFFT as the aforementioned n (Next power of 2 from length of signal x).
So, based on that, how these different vectors (equation 1 and final equation) could be the same?
If you can see, the vectors are different since the former has n points and the later has NFFT/2 points! In fact, the factor (fs/n) is different from fs/2.
So, based on that, how these different vectors (equation 1 and final equation) could be the same?
The example in the documentation from Mathworks plots the entire n-point output of the FFT. This covers the frequencies from 0 to nearly fs (exactly (n-1)/n * fs). They then make the following observation (valid for real inputs to the FFT):
The first half of the frequency range (from 0 to the Nyquist frequency fs/2) is sufficient to identify the component frequencies in the data, since the second half is just a reflection of the first half.
The other post you refer to just chooses to not show that redundant second half. It then uses half the number of points which also cover half the frequency range.
In fact, the factor (fs/n) is different from fs/2.
Perhaps the easiest way to make sense of it is to compare the output of the two expressions for some small value of n, says n=8 and setting fs=1 (since fs multiplies both expressions). On the one hand the output of the first expression [0:n-1]*(fs/n) would be:
0.000 0.125 0.250 0.500 0.625 0.750 0.875
whereas the output of fs/2*linspace(0,1,n/2+1) would be:
0.000 0.125 0.250 0.500
As you can see the set of frequencies are exactly the same up to Nyquist frequency fs/2.
The confusion is perhaps arising from the fact that the two examples which you have referenced are plotting results of the fft differently. Please refer to the code below for the references made in this explanation.
In the first example, the plot is of the power spectrum (periodogram) over the frequency range. Note, in the first plot, that the periodogram is not centered at 0, meaning that the frequency range appears to be twice that of the Nyquist sampling frequency. As mentioned in the mathworks link, it is common practice to center the periodogram at 0 to avoid this confusion (figure 2).
For the second example, taking the same parameters, the original plot is of amplitude of the fourier spectrum with a different normalization than in the first example (figure 3). Using the syntax of Matlab's full frequency ordering (as commented in the code), it is trivial to convert this seemingly different fft result to that of example 1; the identical result of the 0-centered periodogram is replicated in figure 4.
Thus, to answer your question specifically, the frequency ranges in both cases are the same, with the maximum frequency equal to the Nyquist sampling frequency as in:
f = fs/2*linspace(0,1,NFFT/2+1);
The key to understanding how the dfft works (also in Matlab) is to understand that you are simply performing a projection of your discrete data set into fourier space where what is returned by the fft() function in matlab are the coefficients of the expansion for each frequency component and the order of the coefficients is given (in Matlab as in example 2) by:
f = [f(1:end-1) -fliplr(f(1,2:end))];
See the Wikipedia page on the DFT for additional details:
https://en.wikipedia.org/wiki/Discrete_Fourier_transform
It might also be helpful for you to take the fft omitting the length as a power of 2 parameter as
y = fft(x).
In this case, you would see only a few non-zero components in y corresponding to the exact coefficients of your input signal. The mathworks page claims the following as a motivation for using or not using this length:
"Using a power of two for the transform length optimizes the FFT algorithm, though in practice there is usually little difference in execution time from using n = m."
%% First example:
% http://www.mathworks.com/help/matlab/math/fast-fourier-transform-fft.html
fs = 10; % Sample frequency (Hz)
t = 0:1/fs:10-1/fs; % 10 sec sample
x = (1.3)*sin(2*pi*15*t) ... % 15 Hz component
+ (1.7)*sin(2*pi*40*(t-2)); % 40 Hz component
% Removed the noise
m = length(x); % Window length
n = pow2(nextpow2(m)); % Transform length
y = fft(x,n); % DFT
f = (0:n-1)*(fs/n); % Frequency range
power = y.*conj(y)/n; % Power of the DFT
subplot(2,2,1)
plot(f,power,'-o')
xlabel('Frequency (Hz)')
ylabel('Power')
title('{\bf Periodogram}')
y0 = fftshift(y); % Rearrange y values
f0 = (-n/2:n/2-1)*(fs/n); % 0-centered frequency range
power0 = y0.*conj(y0)/n; % 0-centered power
subplot(2,2,2)
plot(f0,power0,'-o')
% plot(f0,sqrt_power0,'-o')
xlabel('Frequency (Hz)')
ylabel('Power')
title('{\bf 0-Centered Periodogram} Ex. 1')
%% Second example:
% http://stackoverflow.com/questions/10758315/understanding-matlab-fft-example
% Let's redefine the parameters for consistency between the two examples
Fs = fs; % Sampling frequency
% T = 1/Fs; % Sample time (not required)
L = m; % Length of signal
% t = (0:L-1)*T; % Time vector (as above)
% % Sum of a 3 Hz sinusoid and a 2 Hz sinusoid
% x = 0.7*sin(2*pi*3*t) + sin(2*pi*2*t); %(as above)
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
% NFFT == n (from above)
Y = fft(x,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);
% Plot single-sided amplitude spectrum.
subplot(2,2,3)
plot(f,2*abs(Y(1:NFFT/2+1)),'-o')
title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')
% Get the 0-Centered Periodogram using the parameters of the second example
f = [f(1:end-1) -fliplr(f(1,2:end))]; % This is the frequency ordering used
% by the full fft in Matlab
power = (Y*L).*conj(Y*L)/NFFT;
% Rearrange for nicer plot
ToPlot = [f; power]; [~,ind] = sort(f);
ToPlot = ToPlot(:,ind);
subplot(2,2,4)
plot(ToPlot(1,:),ToPlot(2,:),'-o')
xlabel('Frequency (Hz)')
ylabel('Power')
title('{\bf 0-Centered Periodogram} Ex. 2')
I am trying to compare the FFT of exp(-t^2) to the function's analytical fourier transform, exp(-(w^2)/4)/sqrt(2), over the frequency range -3 to 3.
I have written the following matlab code and have iterated on it MANY times now with no success.
fs = 100; %sampling frequency
dt = 1/fs;
t = 0:dt:10-dt; %time vector
L = length(t); %number of sample points
%N = 2^nextpow2(L); %necessary?
y = exp(-(t.^2));
Y=dt*ifftshift(abs(fft(y)));
freq = (-L/2:L/2-1)*fs/L; %freq vector
F = (exp(-(freq.^2)/4))/sqrt(2); %analytical solution
%Y_valid_pts = Y(W>=-3 & W<=3); %compare for freq = -3 to 3
%npts = length(Y_valid_pts);
% w = linspace(-3,3,npts);
% Fe = (exp(-(w.^2)/4))/sqrt(2);
error = norm(Y - F) %L2 Norm for error
hold on;
plot(freq,Y,'r');
plot(freq,F,'b');
xlabel('Frequency, w');
legend('numerical','analytic');
hold off;
You can see that right now, I am simply trying to get the two plots to look similar. Eventually, I would like to find a way to do two things:
1) find the minimum sampling rate,
2) find the minimum number of samples,
to reach an error (defined as the L2 norm of the difference between the two solutions) of 10^-4.
I feel that this is pretty simple, but I can't seem to even get the two graphs visually agree.
If someone could let me know where I'm going wrong and how I can tackle the two points above (minimum sampling frequency and minimum number of samples) I would be very appreciative.
Thanks
A first thing to note is that the Fourier transform pair for the function exp(-t^2) over the +/- infinity range, as can be derived from tables of Fourier transforms is actually:
Finally, as you are generating the function exp(-t^2), you are limiting the range of t to positive values (instead of taking the whole +/- infinity range).
For the relationship to hold, you would thus have to generate exp(-t^2) with something such as:
t = 0:dt:10-dt; %time vector
t = t - 0.5*max(t); %center around t=0
y = exp(-(t.^2));
Then, the variable w represents angular frequency in radians which is related to the normalized frequency freq through:
w = 2*pi*freq;
Thus,
F = (exp(-((2*pi*freq).^2)/4))*sqrt(pi); %analytical solution
I am very interested to know how the top right figure in :http://en.wikipedia.org/wiki/Spectrogram
is generated (the script) and how to analyse it i.e what information does it convey?I would appreciate a simplified answer with minimum mathematical jargons. Thank you.
The plot shows time along the horizontal axis, and frequency along the vertical axis. With pixel color showing the intensity of each frequency at each time.
A spectrogram is generated by taking a signal and chopping it into small time segments, doing a Fourier series on each segment.
here is some matlab code to generate one.
Notice how plotting the signal directly, it looks like garbage, but plotting the spectrogram, we can clearly see the frequencies of the component signals.
%%%%%%%%
%% setup
%%%%%%%%
%signal length in seconds
signalLength = 60+10*randn();
%100Hz sampling rate
sampleRate = 100;
dt = 1/sampleRate;
%total number of samples, and all time tags
Nsamples = round(sampleRate*signalLength);
time = linspace(0,signalLength,Nsamples);
%%%%%%%%%%%%%%%%%%%%%
%create a test signal
%%%%%%%%%%%%%%%%%%%%%
%function for converting from time to frequency in this test signal
F1 = #(T)0+40*T/signalLength; #frequency increasing with time
M1 = #(T)1-T/signalLength; #amplitude decreasing with time
F2 = #(T)20+10*sin(2*pi()*T/signalLength); #oscilating frequenct over time
M2 = #(T)1/2; #constant low amplitude
%Signal frequency as a function of time
signal1Frequency = F1(time);
signal1Mag = M1(time);
signal2Frequency = F2(time);
signal2Mag = M2(time);
%integrate frequency to get angle
signal1Angle = 2*pi()*dt*cumsum(signal1Frequency);
signal2Angle = 2*pi()*dt*cumsum(signal2Frequency);
%sin of the angle to get the signal value
signal = signal1Mag.*sin(signal1Angle+randn()) + signal2Mag.*sin(signal2Angle+randn());
figure();
plot(time,signal)
%%%%%%%%%%%%%%%%%%%%%%%
%processing starts here
%%%%%%%%%%%%%%%%%%%%%%%
frequencyResolution = 1
%time resolution, binWidth, is inversly proportional to frequency resolution
binWidth = 1/frequencyResolution;
%number of resulting samples per bin
binSize = sampleRate*binWidth;
%number of bins
Nbins = ceil(Nsamples/binSize);
%pad the data with zeros so that it fills Nbins
signal(Nbins*binSize+1)=0;
signal(end) = [];
%reshape the data to binSize by Nbins
signal = reshape(signal,[binSize,Nbins]);
%calculate the fourier transform
fourierResult = fft(signal);
%convert the cos+j*sin, encoded in the complex numbers into magnitude.^2
mags= fourierResult.*conj(fourierResult);
binTimes = linspace(0,signalLength,Nbins);
frequencies = (0:frequencyResolution:binSize*frequencyResolution);
frequencies = frequencies(1:end-1);
%the upper frequencies are just aliasing, you can ignore them in this example.
slice = frequencies<max(frequencies)/2;
%plot the spectrogram
figure();
pcolor(binTimes,frequencies(slice),mags(slice,:));
The inverse Fourier transform of the fourierResult matrix, will return the original signal.
Just to add to Suki's answer, here is a great tutorial that walks you through, step by step, reading Matlab spectrograms, touching on only enough math and physics to explain the main concepts intuitively:
http://www.caam.rice.edu/~yad1/data/EEG_Rice/Literature/Spectrograms.pdf