I am analyzing ECG data using MATLAB. The data is made up of two columns, one the time in milliseconds and the other contains the volts (mV) and is imported into MATLAB from a CSV file.
I use the built-in fft function in MATLAB (i.e fft(mV)). Now that I have the transformed data, I don't know how to plot it.
I know that I need the frequency data but I'm having trouble understanding where that comes from and what the other axis even is.
When you say "the time in milliseconds", I hope you have sampled at an even interval when performing an FFT. If you have not then you have two choices.
You can interpolate the data between the points so as to "guess" where the graph would eb at the time in the time domain.
OR
You can re-sample at a regular interval. Returning the time in milliseconds is not really necessary for this as the interval must be equal, but it could functions as a validator to prove that the data is correct.
Once you have you data with a regular sampling period then you can use this to obtain the FFT.
function [ X, f ] = ctft( x, T )
% x = sample array
% T = sampling period
% X = fft amplitude
% f = frequency
N = length(x);
X = fftshift(fft( x, N ))*( (2*pi) / N );
f = linspace( -1, 1-1/N, N)/(2*T);
Related
i've got a logical/statistical problem:
i have to find out if the fire rate according to four different stimuli given to a neuron are significantly different.
I calculated the frequencies via psth/binning method in matlab and i am not sure if it was the right way. Following i did an anova and a tukey-test via jmp. At first sight it looks good but as mentioned before i don't think the calculation was correct.
Maybe its not the right forum for my problem but perhaps some g can find my mistake or has a better solution. Thanks:D
bins is the number of bins calculated by total duration (800ms) divided by binwidth(10ms).
At the end this function should give me a histogram plotted with freq over time (ms) and the frequencies (here a 1x80 vector with the average freq per bin).
Done for four different stimuli i got 4 vectors, put in jmd and done the tukey.
function [freq] = BinFireRate(data, dur, times_snippet, binwidth)
%function that plots the firing rate of a given dataset via binning method in [hz]
%in: dataset (n x m-matrix), dur as duarion observed from trial
%time_snippet (1,n-vector) for convert data into time values [ms]
%binwidth
%out: histogram of firing rate (freq) over time and frequency [hz]
%[1x80-vector] itself
bins = dur / binwidth;
spiketimes_stim = data .* times_snippet;
spiketimes_stim = spiketimes_stim(spiketimes_stim ~= 0);
[spikes_per_bin, bincenters] = hist(spiketimes_stim, bins);
freq = ((spikes_per_bin / binwidth) / length(data(:, 1))) * 1000;
bar(bincenters, freq);
I'm studying the IFFT in Matlab by applying it to a Gaussian. According to Wikipedia tables, the Fourier transform pair would be
F(w) = sqrt(pi/a) * exp(-w^2/(4a))
in frequency, and
f(t) = exp(-at^2)
in time. I modified the code in a previous question plus Cris Luengo's answer to perform this IFFT.
a = 0.333;
ts = 1e4; % time sampling
L = 1000*ts; % no. sample points
ds = 1/ts;
f = -floor(L/2):floor((L-1)/2); % freq vector
f = f/ts;
w = 2*pi*f; % angular freq
Y = sqrt(pi/a)*exp(-w.^2/(4*a));
y = ts*ifftshift(ifft(fftshift(Y)));
t = (-L/2:L/2-1)*ts/L; % time vector
f = exp(-a*t.^2); % analytical solution
figure; subplot(1,2,1); hold on
plot(t,real(y),'.--')
plot(t,real(f),'-')
xlabel('time, t')
title('real')
legend('numerical','analytic')
xlim([-5,5])
subplot(1,2,2); hold on
plot(w,imag(y),'.--')
plot(w,imag(f),'-')
xlabel('time, t')
title('imag')
legend('numerical','analytic')
xlim([-5,5])
When I compare the result of IFFT with the analytical expression, they don't seem to agree:
I'm not sure where the mistake is. Have I scaled the IFFT properly? Is there an error in how I define the linear/angular frequency?
Edit: For some reason, when I define L=ts^2 the analytical and numerical solutions seem to agree (L = no. sampling points, ts = time sample).
Starting from the analytical solution, let's rephrase things a bit. You have a sampling of the function f(t) = exp(-a*t^2), and the way you've constructed the analytical answer you're collecting L=1000*ts=1e7 samples at a sampling rate of Ts=ts/L=1e-3. This means that your sampling frequency is Fs=1/Ts=1e3.
Since you want to compare against results obtained with fft/ifft, you should be considering digital or discrete frequencies, meaning the values you define for your transform will correspond to the digital frequencies
frd = (-L/2:L/2-1)/L;
Mapping this to angular frequencies, we have:
w = 2*pi*frd;
But when you're trying to compute the values, you also need to keep in mind that these frequencies should represent samples of the continuous time spectrum you're expecting. So you scale these values by your sampling frequency:
Y = sqrt(pi/a)*exp(-(Fs*w).^2/(4*a));
y = Fs*ifftshift(ifft(fftshift(Y)));
When you compare the analytical and computed answers, they now match.
The short answer to your question, given this, is that you are scaling y incorrectly at the end. You're scaling it by ts, which is 1e4, but you need to be scaling it by the sampling frequency which is Fs=1e3. That's why you end up off by a factor of 10.
I have two sets of data taken from experiments, and they look very similar, except there is a horizontal offset between them, which I believe is due to some bugs in the instrument setting. Suppose they have the form y1=f(x1) and y2=f(x2)= f(x1+c), what's the best way to determine c so that I can take into account the offset to superimpose two data sets to become one data set?
Edit:
let's say my data sets (index 1 and 2) have the form:
x1 = 0:0.2:10;
y1 = sin(x1)
x2 = 0:0.3:10;
y2 = sin(x2+0.5)
Of course, the real data will have some noise, but say the best fit functions have the above forms. How do I find the offset c=0.5? I have looked into the cross-correlation, but I'm not sure if they can handle two data sets with different number of data (and different step sizes). Also, what if the offset values actually fall between two data points? Cross-correlation only returns the index of the data in the array, not something in between if I understand correctly.
This Matlab script calculates the random offset from -pi/2 to +pi/2 between two sine waves:
clear;
C= pi*(rand-0.5); % should be between -pi/2 and +pi/2
N=200; % should be large enough for acceptable sampling rate
N1=20; % fraction for Ts1
N2=30; % fraction for Ts2
Ts1=abs(C*N1/N); % fraction of C for accuracy
Ts2=abs(C*N2/N); % fraction of C for accuracy
Ts=min(Ts1,Ts2); % select highest sampling rate (smaller period)
fs=1/Ts;
P=4; % number of periods should be large enough for well defined correlation plot
x1 = 0:Ts:P*2*pi;
y1 = sin(x1);
x2 = 0:Ts:P*2*pi;
y2 = sin(x2+C);
subplot(3,1,1)
plot(x1,y1);
subplot(3,1,2);
plot(x2,y2);
[cor,lag]=xcorr(y1,y2);
subplot(3,1,3);
plot(lag,cor);
[~,I] = max(abs(cor));
lagdiff = lag(I);
timediff=lagdiff/fs;
In the particular case below, C = timediff = 0.5615:
write a function which takes the time shift as an input and calculates rms between overlapping portions of the two data sets. Then find the minimum of this function using optimization (fminbnd)
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
This is the first time I'm using the fft function and I'm trying to plot the frequency spectrum of a simple cosine function:
f = cos(2*pi*300*t)
The sampling rate is 220500. I'm plotting one second of the function f.
Here is my attempt:
time = 1;
freq = 220500;
t = 0 : 1/freq : 1 - 1/freq;
N = length(t);
df = freq/(N*time);
F = fftshift(fft(cos(2*pi*300*t))/N);
faxis = -N/2 / time : df : (N/2-1) / time;
plot(faxis, real(F));
grid on;
xlim([-500, 500]);
Why do I get odd results when I increase the frequency to 900Hz? These odd results can be fixed by increasing the x-axis limits from, say, 500Hz to 1000Hz. Also, is this the correct approach? I noticed many other people didn't use fftshift(X) (but I think they only did a single sided spectrum analysis).
Thank you.
Here is my response as promised.
The first or your questions related to why you "get odd results when you increase the frequency to 900 Hz" is related to the Matlab's plot rescaling functionality as described by #Castilho. When you change the range of the x-axis, Matlab will try to be helpful and rescale the y-axis. If the peaks lie outside of your specified range, matlab will zoom in on the small numerical errors generated in the process. You can remedy this with the 'ylim' command if it bothers you.
However, your second, more open question "is this the correct approach?" requires a deeper discussion. Allow me to tell you how I would go about making a more flexible solution to achieve your goal of plotting a cosine wave.
You begin with the following:
time = 1;
freq = 220500;
This raises an alarm in my head immediately. Looking at the rest of the post, you appear to be interested in frequencies in the sub-kHz range. If that is the case, then this sampling rate is excessive as the Nyquist limit (sr/2) for this rate is above 100 kHz. I'm guessing you meant to use the common audio sampling rate of 22050 Hz (but I could be wrong here)?
Either way, your analysis works out numerically OK in the end. However, you are not helping yourself to understand how the FFT can be used most effectively for analysis in real-world situations.
Allow me to post how I would do this. The following script does almost exactly what your script does, but opens some potential on which we can build . .
%// These are the user parameters
durT = 1;
fs = 22050;
NFFT = durT*fs;
sigFreq = 300;
%//Calculate time axis
dt = 1/fs;
tAxis = 0:dt:(durT-dt);
%//Calculate frequency axis
df = fs/NFFT;
fAxis = 0:df:(fs-df);
%//Calculate time domain signal and convert to frequency domain
x = cos( 2*pi*sigFreq*tAxis );
F = abs( fft(x, NFFT) / NFFT );
subplot(2,1,1);
plot( fAxis, 2*F )
xlim([0 2*sigFreq])
title('single sided spectrum')
subplot(2,1,2);
plot( fAxis-fs/2, fftshift(F) )
xlim([-2*sigFreq 2*sigFreq])
title('whole fft-shifted spectrum')
You calculate a time axis and calculate your number of FFT points from the length of the time axis. This is very odd. The problem with this approach, is that the frequency resolution of the fft changes as you change the duration of your input signal, because N is dependent on your "time" variable. The matlab fft command will use an FFT size that matches the size of the input signal.
In my example, I calculate the frequency axis directly from the NFFT. This is somewhat irrelevant in the context of the above example, as I set the NFFT to equal the number of samples in the signal. However, using this format helps to demystify your thinking and it becomes very important in my next example.
** SIDE NOTE: You use real(F) in your example. Unless you have a very good reason to only be extracting the real part of the FFT result, then it is much more common to extract the magnitude of the FFT using abs(F). This is the equivalent of sqrt(real(F).^2 + imag(F).^2).**
Most of the time you will want to use a shorter NFFT. This might be because you are perhaps running the analysis in a real time system, or because you want to average the result of many FFTs together to get an idea of the average spectrum for a time varying signal, or because you want to compare spectra of signals that have different duration without wasting information. Just using the fft command with a value of NFFT < the number of elements in your signal will result in an fft calculated from the last NFFT points of the signal. This is a bit wasteful.
The following example is much more relevant to useful application. It shows how you would split a signal into blocks and then process each block and average the result:
%//These are the user parameters
durT = 1;
fs = 22050;
NFFT = 2048;
sigFreq = 300;
%//Calculate time axis
dt = 1/fs;
tAxis = dt:dt:(durT-dt);
%//Calculate frequency axis
df = fs/NFFT;
fAxis = 0:df:(fs-df);
%//Calculate time domain signal
x = cos( 2*pi*sigFreq*tAxis );
%//Buffer it and window
win = hamming(NFFT);%//chose window type based on your application
x = buffer(x, NFFT, NFFT/2); %// 50% overlap between frames in this instance
x = x(:, 2:end-1); %//optional step to remove zero padded frames
x = ( x' * diag(win) )'; %//efficiently window each frame using matrix algebra
%// Calculate mean FFT
F = abs( fft(x, NFFT) / sum(win) );
F = mean(F,2);
subplot(2,1,1);
plot( fAxis, 2*F )
xlim([0 2*sigFreq])
title('single sided spectrum')
subplot(2,1,2);
plot( fAxis-fs/2, fftshift(F) )
xlim([-2*sigFreq 2*sigFreq])
title('whole fft-shifted spectrum')
I use a hamming window in the above example. The window that you choose should suit the application http://en.wikipedia.org/wiki/Window_function
The overlap amount that you choose will depend somewhat on the type of window you use. In the above example, the Hamming window weights the samples in each buffer towards zero away from the centre of each frame. In order to use all of the information in the input signal, it is important to use some overlap. However, if you just use a plain rectangular window, the overlap becomes pointless as all samples are weighted equally. The more overlap you use, the more processing is required to calculate the mean spectrum.
Hope this helps your understanding.
Your result is perfectly right. Your frequency axis calculation is also right. The problem lies on the y axis scale. When you use the function xlims, matlab automatically recalculates the y scale so that you can see "meaningful" data. When the cosine peaks lie outside the limit you chose (when f>500Hz), there are no peaks to show, so the scale is calculated based on some veeeery small noise (here at my computer, with matlab 2011a, the y scale was 10-16).
Changing the limit is indeed the correct approach, because if you don't change it you can't see the peaks on the frequency spectrum.
One thing I noticed, however. Is there a reason for you to plot the real part of the transform? Usually, it is abs(F) that gets plotted, and not the real part.
edit: Actually, you're frequency axis is only right because df, in this case, is 1. The faxis line is right, but the df calculation isn't.
The FFT calculates N points from -Fs/2 to Fs/2. So N points over a range of Fs yields a df of Fs/N. As N/time = Fs => time = N/Fs. Substituting that on the expression of df you used: your_df = Fs/N*(N/Fs) = (Fs/N)^2. As Fs/N = 1, the final result was right :P