I have a hard time wrapping my head around this since there is non-linearity involved. How do I convert a spectrum I have in units "1/m" (or cycles per m) to "cycles/deg"?
Example: For simplicity let's work in one dimension and the screen of a laptop is 768 pixels on which I show the highest pixel frequency. I use MATLAB/Octave style code but I hope it's clear enough to act as pseudo code:
X = 768
x = 0:X-1
y = cos(2*pi/X*(X/2)*x)
The frequency in this signal is now pi radiants or 0.5 cycles/pixel (sampling rate is 1 pixel). I can create the frequency vectors:
w = linspace(0, 2*pi, length(x)) % in radiants
f = linspace(0, 1, length(x)) % in 1/pixel or cycles/pixel
I can easily convert this to spatial frequency in units [m] and frequency [cycles/m] when I know the pixel size. Let's assume one pixel is 0.5mm:
xm = x*0.5e-3 % for image in m
fm = f/0.5e-3 % frequency in 1/m
Plotting y versus xm shows now a waveform with 1 cycle per mm. Plotting the FFT of y versus fm shows a peak at 1000 cycles/m (or 1 cycle per mm), as expected.
How do I need to modify fm to arrive at fdeg, in other words, to show the FFT of y in cycles per degree, assuming distance from the screen is 60cm? I would assume
fdeg = 1/(2*atan(1/fm)/(2*0.6)*180/pi)
That would give me 20.97 cycles/deg. However, using the small angle approximation, 1mm should be roughly 1mm/0.6*180/pi=0.9549e-3 degree, so we have 1cycle per 0.9549e-3 degrees or 1/0.9549e-3=10.47 cycles per degree. This is off by a factor of 2.
Related
This is a question for a hand in in one of my courses, just want to state that.
What I am trying to do is sampling a square wave, take the fourier transform (fft) and plot the answer to graph. This is how I have achieved this:
Fs = 100;
Ts = 1/Fs;
N = 8192;
Tmax = (N - 1)*Ts;
t = 0:Ts:Tmax;
x = square(t);
X = fft(x,N);
plot(t, abs(X))
What it return is a graph that looks like this
This looks almost as inspected, but since I do not know what to expect with the square wave I also try to do it with a $\sin(2*t)$ wave. If I take the fourier transform on this, I should get 2 spiks, each at 2 and -2 (right side). But what I get is something like this
(Note! I have zoomed in on the left hand side of the graph to show that the spike is not at 2) As you can see the spike is not where it is supposed to be. I can than conclude that probably the 1 graph is not eater how it should be.
Is it something wrong with my x axis representation? And if so, how do I convert the x axis into the frequency plane?
The frequencies resulting from the FFT range from 0 to the sampling frequency. Specifically, the horizontal axis of the FFT corresponds to frequencies 0, fs/N, 2*fs/N, ... ,(N-1)*fs/N, where fs is the sample frequency and N is the FFT size.
So, you should modify the horizontal axis in the plot to the following, where N is numel(t) and fs is computed as 1/(t(2)-t(1)):
freq_axis = (0:numel(t)-1)/numel(t)/(t(2)-t(1));
plot(freq_axis, abs(X))
You may also want to apply fftshift to observe frequencies from -fs/2 to fs/2, instead of from 0 to fs. In that case:
freq_axis = (-numel(t)/2:numel(t)/2-1)/numel(t)/(t(2)-t(1));
plot(freq_axis, fftshift(abs(X)))
As a check, with your example x = sin(2*t) the second plot gives:
Comparing your sin(2*t) with the generic expression sin(2*pi*f*t), the frequency f of that sinusoid is seen to be 1/pi = 0.3183, in agreement with the figure.
I have a time series of air temperatures, that are measured in degrees C and the frequency of the signal is monthly. I compute the power spectra of the time series as:
L = length(temp);
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(temp,NFFT)/L;
Fs = 1; % one sample per month
f = Fs/2*linspace(0,1,NFFT/2+1); % frequency
Pxx = 2*abs(Y(1:NFFT/2+1)); % power
% Plot single-sided amplitude spectrum.
plot(f,Pxx,'-k');
xlabel('Frequency (c/month)'); % cycles per unit time
which results in the following plot.
What should the units be for the yaxis? Should it be ^{o}C?
The range of the time series is approx 20, so I guess 10 would work as the amplitude...
Let's say the original axes were T (vertical) and t (horizontal); after the transform, they are Y and X.
You correctly found that the units of X are reciprocals of the units of t. This is necessary for formulas like exp(2*pi*i*t*X) to make sense: the product t*X must be unitless.
Parseval's identity can be used to infer the units on the vertical axis. According to it,
(T units)2 (t units) = (Y units)2 (X units)
hence Y units = (T units) (t units). In your case this is degree-times-second, but this is better expressed as degree-per-frequency unit.
Indeed, the function produced by Fourier transform is amplitude density, expressing how much oscillation is there in every particular frequency range.
I'm working on sound signals of a walking pattern, which has obvious regular patterns:
Then I thought I can get the frequency of walking (approximately 1.7Hz from the image) using FFT function:
x = walk_5; % Walking sound with a size of 711680x2 double
Fs = 48000; % sound frquency
L=length(x);
t=(1:L)/Fs; %time base
plot(t,x);
figure;
NFFT=2^nextpow2(L);
X=fft(x,NFFT);
Px=X.*conj(X)/(NFFT*L); %Power of each freq components
fVals=Fs*(0:NFFT/2-1)/NFFT;
plot(fVals,Px(1:NFFT/2),'b','LineSmoothing','on','LineWidth',1);
title('One Sided Power Spectral Density');
xlabel('Frequency (Hz)')
ylabel('PSD');
But then it doesn't give me what I expected:
FFT result:
zoom image has lots of noises:
and there is no information near 1.7Hz
Here is the graph from log domain using
semilogy(fVals,Px(1:NFFT));
It's pretty symmetric though:
I couldn't find anything wrong with my code. Do you have any solutions to easily extract the 1.7Hz from the walking pattern?
here is the link for the audio file in mat
https://www.dropbox.com/s/craof8qkz9n5dr1/walk_sound.mat?dl=0
Thank you very much!
Kai
I suggest you to forget about DFT approach since your signal is not appropriate for this type of analysis due to many reasons. Even by looking on the spectrum in range of frequencies that you are interested in, there is no easy way to estimate the peak:
Of course you could try with PSD/STFT and other funky methods, but this is an overkill. I can think of two, rather simple methods, for this task.
First one is based simply on the Auto Correlation Function.
Calculate the ACF
Define the minimum distance between them. Since you know that expected frequency is around 1.7Hz, then it corresponds to 0.58s. Let's make it 0.5s as the minimum distance.
Calculate the average distance between peaks found.
This gave me an approximate frequency of 1.72 Hz .
Second approach is based on the observation to your signal already has some peaks which are periodic. Therefore we can simply search for them using findpeaks function.
Define the minimum peak distance in a same way as before.
Define the minimum peak height. For example 10% of maximum peak.
Get the average difference.
This gave me an average frequency of 1.7 Hz.
Easy and fast method. There are obviously some things that can be improved, such as:
Refining thresholds
Finding both positive and negative peaks
Taking care of some missing peaks, i.e. due to low amplitude
Anyway that should get you started, instead of being stuck with crappy FFT and lazy semilogx.
Code snippet:
load walk_sound
fs = 48000;
dt = 1/fs;
x = walk_5(:,1);
x = x - mean(x);
N = length(x);
t = 0:dt:(N-1)*dt;
% FFT based
win = hamming(N);
X = abs(fft(x.*win));
X = 2*X(1:N/2+1)/sum(win);
X = 20*log10(X/max(abs(X)));
f = 0:fs/N:fs/2;
subplot(2,1,1)
plot(t, x)
grid on
xlabel('t [s]')
ylabel('A')
title('Time domain signal')
subplot(2,1,2)
plot(f, X)
grid on
xlabel('f [Hz]')
ylabel('A [dB]')
title('Signal Spectrum')
% Autocorrelation
[ac, lag] = xcorr(x);
min_dist = ceil(0.5*fs);
[pks, loc] = findpeaks(ac, 'MinPeakDistance', min_dist);
% Average distance/frequency
avg_dt = mean(gradient(loc))*dt;
avg_f = 1/avg_dt;
figure
plot(lag*dt, ac);
hold on
grid on
plot(lag(loc)*dt, pks, 'xr')
title(sprintf('ACF - Average frequency: %.2f Hz', avg_f))
% Simple peak finding in time domain
[pkst, loct] = findpeaks(x, 'MinPeakDistance', min_dist, ...
'MinPeakHeight', 0.1*max(x));
avg_dt2 = mean(gradient(loct))*dt;
avg_f2 = 1/avg_dt2;
figure
plot(t, x)
grid on
hold on
plot(loct*dt, pkst, 'xr')
xlabel('t [s]')
ylabel('A')
title(sprintf('Peak search in time domain - Average frequency: %.2f Hz', avg_f2))
Here's a nifty solution:
Take the absolute value of your raw data before taking the FFT. The data has a ton of high frequency noise that is drowning out whatever low frequency periodicity is present in the signal. The amplitude of the high frequency noise gets bigger every 1.7 seconds, and the increase in amplitude is visible to the eye, and periodic, but when you multiply the signal by a low frequency sine wave and sum everything you still end up with something close to zero. Taking the absolute value changes this, making those amplitude modulations periodic at low frequencies.
Try the following code comparing the FFT of the regular data with the FFT of abs(data). Note that I took a few liberties with your code, such as combining what I assume were the two stereo channels into a single mono channel.
x = (walk_5(:,1)+walk_5(:,2))/2; % Convert from sterio to mono
Fs = 48000; % sampling frquency
L=length(x); % length of sample
fVals=(0:L-1)*(Fs/L); % frequency range for FFT
walk5abs=abs(x); % Take the absolute value of the raw data
Xold=abs(fft(x)); % FFT of the data (abs in Matlab takes complex magnitude)
Xnew=abs(fft(walk5abs-mean(walk5abs))); % FFT of the absolute value of the data, with average value subtracted
figure;
plot(fVals,Xold/max(Xold),'r',fVals,Xnew/max(Xnew),'b')
axis([0 10 0 1])
legend('old method','new method')
[~,maxInd]=max(Xnew); % Index of maximum value of FFT
walkingFrequency=fVals(maxInd) % print max value
And plotting the FFT for both the old method and the new, from 0 to 10 Hz gives:
As you can see it detects a peak at about 1.686 Hz, and for this data, that's the highest peak in the FFT spectrum.
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')
(Disclaimer: I thought about posting this on math.statsexchange, but found similar questions there that were moved to SO, so here I am)
The context:
I'm using fft/ifft to determine probability distributions for sums of random variables.
So e.g. I'm having two uniform probability distributions - in the simplest case two uniform distributions on the interval [0,1].
So to get the probability distribution for the sum of two random variables sampled from these two distributions, one can calculate the product of the fourier-transformed of each probabilty density.
Doing the inverse fft on this product, you get back the probability density for the sum.
An example:
function usumdist_example()
x = linspace(-1, 2, 1e5);
dx = diff(x(1:2));
NFFT = 2^nextpow2(numel(x));
% take two uniform distributions on [0,0.5]
intervals = [0, 0.5;
0, 0.5];
figure();
hold all;
for i=1:size(intervals,1)
% construct the prob. dens. function
P_x = x >= intervals(i,1) & x <= intervals(i,2);
plot(x, P_x);
% for each pdf, get the characteristic function fft(pdf,NFFT)
% and form the product of all char. functions in Y
if i==1
Y = fft(P_x,NFFT) / NFFT;
else
Y = Y .* fft(P_x,NFFT) / NFFT;
end
end
y = ifft(Y, NFFT);
x_plot = x(1) + (0:dx:(NFFT-1)*dx);
plot(x_plot, y / max(y), '.');
end
My issue is, the shape of the resulting prob. dens. function is perfect.
However, the x-axis does not fit to the x I create in the beginning, but is shifted.
In the example, the peak is at 1.5, while it should be 0.5.
The shift changes if I e.g. add a third random variable or if I modify the range of x.
But I can't get figure how.
I'm afraid it might have to do with the fact that I'm having negative x values, while fourier transforms usually work in a time/frequency domain, where frequencies < 0 don't make sense.
I'm aware I could find e.g. the peak and shift it to its proper place, but seems nasty and error prone...
Glad about any ideas!
The problem is that your x origin is -1, not 0. You expect the center of the triangular pdf to be at .5, because that's twice the value of the center of the uniform pdf. However, the correct reasoning is: the center of the uniform pdf is 1.25 above your minimum x, and you get the center of the triangle at 2*1.25 = 2.5 above the minimum x (that is, at 1.5).
In other words: although your original x axis is (-1, 2), the convolution (or the FFT) behave as if it were (0, 3). In fact, the FFT knows nothing about your x axis; it only uses the y samples. Since your uniform is zero for the first samples, that zero interval of width 1 is amplified to twice its width when you do the convolution (or the FFT). I suggest drawing the convolution on paper to see this (draw original signal, reflected signal about y axis, displace the latter and see when both begin to overlap). So you need a correction in the x_plot line to compensate for this increased width of the zero interval: use
x_plot = 2*x(1) + (0:dx:(NFFT-1)*dx);
and then plot(x_plot, y / max(y), '.') will give the correct graph: