I am trying to compare two data sets in MATLAB. To do this I need to filter the data sets by Fourier transforming the data, filtering it and then inverse Fourier transforming it.
When I inverse Fourier transform the data however I get a spike at either end of the red data set (picture shows the first spike), it should be close to zero at the start, like the blue line. I am comparing many data sets and this only happens occasionally.
I have three questions about this phenomenon. First, what may be causing it, secondly, how can I remedy it, and third, will it affect the data further along the time series or just at the beginning and end of the time series as it appears to from the picture.
Any help would be great thanks.
When using DFT you must remember the DFT assumes a Periodic Signal (As a Superposition of Harmonic Functions).
As you can see, the start point is exact continuation of the last point in harmonic function manner.
Did you perform any Zero Padding in the Spectrum Domain?
Anyhow, Windowing might reduce the Overshooting.
Knowing more about the filter and the Original data might be helpful.
If you say spike near zero frequencies, I answer check the DC component.
You seem interested by the shape, so doing
x = x - mean(x)
or
x -= mean(x)
or
x -= x.mean()
(I love numpy!)
will just constrain the dataset to begin with null amplitude at zero-frequency and to go ahead with comapring the spectra's amplitude.
(as a side-note: did you check that you approprately use fftshift and ifftshift? this has always been the source of trouble for me)
Could be the numerical equivalent of Gibbs' phenomenon. If that's correct, there's no way to remedy it except for filtering.
Related
Is the Matlab code available anywhere?
I'm trying to understand what exactly does it do. As I understood, it divides the data into segments with length tau (when tau increases each time) and then averages the data within each segment. After that, it compares the value attained from successive segments.
Am I correct in my understanding?
Thanks in advance!
You can view the code by typing edit allanvar into the command window.
Of course, this assumes you've already downloaded the associated toolbox.
I tried the same approach, but was also UNable to open the allanvar.p file. Despite the documentation I've found, if I use simple white noise and run allanvar on it, when I plot sqrt(this output), I get a line with a slope of -1/2 on a log-log scale (one order of magnitude drop in ADEV for two orders of magnitude increase in averaging interval value). A simple program to compute the OVERLAPPING Allan variance yields a -1/1 slope. Thus I am suspicious that the allanvar function does not compute the overlapping version!
I want to calculate the Fourier series of a signal using FFT on matlab. And I came across the following unexpected issue. Mere example
If I define a grid and then compute the fft as:
M=59;
x= deal(1*(0:M-1)/M);
y=3*cos(2*pi*x);
Yk=fftshift(fft2(y)/(M));
which gives me the exact analytic values expected: Yk(29)=1.5; Yk(31)=1.5; zeros anything else
but if I define the grid as, and repeat the fft calculation:
x=0:1/(M-1):1;
y=3*cos(2*pi*x);
Yk=fftshift(fft2(y)/(M));
got the Yk's values completely screwed up
This is an annoying issue since I have to analyse many signals data that was sampled as in the second method so the Yk's values will be wrong. Is there a way to workaround this? an option to tell something to the fft function about the way the signal was sampled. Have no way to resample the data in the correct way.
The main reason to avoid have spectral leaking, is that I do further operations with these Fourier terms individually Real and Imag parts. And the spectral leaking is messing the final results.
The second form of sampling includes one sample too many in the period of the cosine. This causes some spectral leaking, and adds a small shift to your signal (which leads to non-zero imaginary values). If you drop the last point, you'll cosine will again be sampled correctly, and you'll get rid of both of these effects. Your FFT will have one value less, I don't know if this will affect your analyses in any way.
x = 0:1/(M-1):1;
y = 3*cos(2*pi*x);
Yk = fftshift(fft2(y(1:end-1))/(M-1));
>> max(abs(imag(Yk)))
ans =
1.837610523517500e-16
I am trying to find power spectrum of the signal. The length of the signal is 100000, sample frequency is 1000Hz,and the number of points is 100000. I found the power spectrum using two approaches. The first one is by taking all the length as one part and found power spectrum for it while the second approach is by dividing the signal into 100*1000and find spectrum for each row then get the mean for all rows. My problem is that I must get the same answer in both approaches but I got different answers. I do not know what is the error in my code.
N=100000;
SF=1000;
a=0.1;
b=0.3;
amplitude1=1;
amplitude2=0.5;
t=0:1/SF:100;
f1=SF*a;
f2=SF*b;
A=amplitude1*sin(2*pi*f1*t)+amplitude2*sin(2*pi*f2*t);
Y=2*randn(1,length(A))+A;
bin=[0 :N/2];
fax_Hz=(bin*SF)/N;
FFT=fft(Y);
spectra=2/(SF*length(Y))*(FFT.*conj(FFT));
plot(fax_Hz,spectra(1,1:50001));
D=reshape(Y(1,1:100000),[100,1000]);
M=length(D(1,:));
for i=1:100
FFT_1(i,:)=fft(D(i,:));
S(i,:)=(2/(SF*M))*(FFT_1(i,:).*conj(FFT_1(i,:)));
end
S_f=mean(S);
figure
plot (S_f);
I just update the code. I do not know but when I added noise to signal the two plots looks shifted.
The main problem is with reshape you are working with each row being a separate sequence. Reshape however fills the first column before moving to the second one.
You can use the following instead.
D=reshape(A(1,1:100000),[1000,100]).';
Normalization is another problem. You can either use ifft instead of fft as it is normalized by default (not sure why). Or alternatively keep your normalization and instead of using mean you should can use sum, maybe that is due to a mistake you might have made. There still seems to be a small discrepancy in the amplitudes, not sure where that is coming from.
At the end to plot use the following:
bin=[0 :N];
fax_Hz=(bin*SF)/N;
FFT=ifft(A);
spectra=FFT.*conj(FFT);
plot(fax_Hz,spectra); hold on
D=reshape(A(1,1:100000),[1000,100]).';
M=length(D(1,:));
for i=1:100
FFT_1(i,:)=ifft(D(i,:));
S(i,:)=FFT_1(i,:).*conj(FFT_1(i,:));
end
S_f=mean(S);
plot(fax_Hz(1:100:end-1), S_f);
Note: the fax_Hz(1:100:end-1) is a hacky way of getting the length of the vectors to be the same.
I am having a samll problem while converting a spectrum to a time series. I have read many article sand I htink I am applying the right procedure but I do not get the right results. Could you help to find the error?
I have a time series like:
When I compute the spectrum I do:
%number of points
nPoints=length(timeSeries);
%time interval
dt=time(2)-time(1);
%Fast Fourier transform
p=abs(fft(timeSeries))./(nPoints/2);
%power of positive frequencies
spectrum=p(1:(nPoints/2)).^2;
%frequency
dfFFT=1/tDur;
frequency=(1:nPoints)*dfFFT;
frequency=frequency(1:(nPoints)/2);
%plot spectrum
semilogy(frequency,spectrum); grid on;
xlabel('Frequency [Hz]');
ylabel('Power Spectrum [N*m]^2/[Hz]');
title('SPD load signal');
And I obtain:
I think the spectrum is well computed. However now I need to go back and obtain a time series from this spectrum and I do:
df=frequency(2)-frequency(1);
ap = sqrt(2.*spectrum*df)';
%random number form -pi to pi
epsilon=-pi + 2*pi*rand(1,length(ap));
%transform to time series
randomSeries=length(time).*real(ifft(pad(ap.*exp(epsilon.*i.*2.*pi),length(time))));
%Add the mean value
randomSeries=randomSeries+mean(timeSeries);
However, the plot looks like:
Where it is one order of magnitude lower than the original serie.
Any recommendation?
There are (at least) two things going on here. The first is that you are throwing away information, and then substituting random numbers for that information.
The FFT of a real sequence is a sequence of complex numbers consisting of a real and imaginary part. Converting those numbers to polar form gives you magnitude and phase angle. You are capturing the magnitude part with p=aps(fft(...)), but you are not capturing the phase angle (which would involve atan2(...)). You are then making up random numbers (epsilon=...) and using those to replace the original numbers when you reconstruct your time-series. Also, as the FFT of a real sequence has a particular symmetry, substituting random numbers for the phase angle destroys that symmetry, which means that the IFFT will in general no longer be a real sequence, but a sequence of complex numbers - and again, you're only looking at the real portion of the IFFT, so you're throwing away information again. If this is an audio signal, the results may sound somewhat like the original (or they may be completely different), but the waveform definitely won't match...
The second issue is that in many implementations, ifft(fft(...)) will scale the result by the number of points in the signal. There are several different ways to avoid that, with differing results, but sometimes more attractive in different scenarios, depending on what you are trying to do. You can either scale the fft() result before you do the ifft(), or scale the ifft() result at the end, or in some cases, I've even seen both being scaled by a factor of sqrt(N) - doing it twice has the end result of scaling the final result by N, but it is a bit less efficient since you do the scaling twice...
I am trying to measure the PSD of a stochastic process in matlab, but I am not sure how to do it. I have posted the exact same question here, but I thought I might have more luck here.
The stochastic process describes wind speed, and is represented by a vector of real numbers. Each entry corresponds to the wind speed in a point in space, measured in m/s. The points are 0.0005 m apart. How do I measure and plot the PSD? Let's call the vector V. My first idea was to use
[p, w] = pwelch(V);
loglog(w,p);
But is this correct? The thing is, that I'm given an analytical expression, which the PSD should (in theory) match. By plotting it with these two lines of code, it looks all wrong. Specifically it looks as though it could need a translation and a scaling. Other than that, the shapes of the two are similar.
UPDATE:
The image above actually doesn't depict the PSD obtained by using pwelch on a single vector, but rather the mean of the PSD of 200 vectors, since these vectors stems from numerical simulations. As suggested, I have tried scaling by 2*pi/0.0005. I saw that you can actually give this information to pwelch. So I tried using the code
[p, w] = pwelch(V,[],[],[],2*pi/0.0005);
loglog(w,p);
instead. As seen below, it looks much nicer. It is, however, still not perfect. Is that just something I should expect? Taking the squareroot is not the answer, by the way. But thanks for the suggestion. For one thing, it should follow Kolmogorov's -5/3 law, which it does now (it follows the green line, which has slope -5/3). The function I'm trying to match it with is the Shkarofsky spectral density function, which is the one-dimensional Fourier transform of the Shkarofsky correlation function. Is it not possible to mark up math, here on the site?
UPDATE 2:
I have tried using [p, w] = pwelch(V,[],[],[],1/0.0005); as I was suggested. But as you can seem it still doesn't quite match up. It's hard for me to explain exactly what I'm looking for. But what I would like (or, what I expected) is that the dip, of the computed and the analytical PSD happens at the same time, and falls off with the same speed. The data comes from simulations of turbulence. The analytical expression has been fitted to actual measurements of turbulence, wherein this dip is present as well. I'm no expert at all, but as far as I know the dip happens at the small length scales, since the energy is dissipated, due to viscosity of the air.
What about using the standard equation for obtaining a PSD? I'd would do this way:
Sxx(f) = (fft(x(t)).*conj(fft(x(t))))*(dt^2);
or
Sxx = fftshift(abs(fft(x(t))))*(dt^2);
Then, if you really need, you may think of applying a windowing criterium, such as
Hanning
Hamming
Welch
which will only somehow filter your PSD.
Presumably you need to rescale the frequency (wavenumber) to units of 1/m.
The frequency units from pwelch should be rescaled, since as the documentation explains:
W is the vector of normalized frequencies at which the PSD is
estimated. W has units of rad/sample.
Off the cuff my guess is that the scaling factor is
scale = 1/0.0005/(2*pi);
or 318.3 (m^-1).
As for the intensity, it looks like taking a square root might help. Perhaps your equation reports an intensity, not PSD?
Edit
As you point out, since the analytical and computed PSD have nearly identical slopes they appear to obey similar power laws up to 800 m^-1. I am not sure to what degree you require exponents or offsets to match to be satisfied with a specific model, and I am not familiar with this particular theory.
As for the apparent inconsistency at high wavenumbers, I would point out that you are entering the domain of very small numbers and therefore (1) floating point issues and (2) noise are probably lurking. The very nice looking dip in the computed PSD on the other hand appears very real but I have no explanation for it (maybe your noise is not white?).
You may want to look at this submission at matlab central as it may be useful.
Edit #2
After inspecting documentation of pwelch, it appears that you should pass 1/0.0005 (the sampling rate) and not 2*pi/0.0005. This should not affect the slope but will affect the intercept.
The dip in PSD in your simulation results looks similar to aliasing artifacts
that I have seen in my data when the original data were interpolated with a
low-order method. To make this clearer - say my original data was spaced at
0.002m, but in the course of cleaning up missing data, trying to save space, whatever,
I linearly interpolated those data onto a 0.005m spacing. The frequency response
of linear interpolation is not well-behaved, and will introduce peaks and valleys
at the high wavenumber end of your spectrum.
There are different conventions for spectral estimates - whether the wavenumber
units are 1/m, or radians/m. Single-sided spectra or double-sided spectra.
help pwelch
shows that the default settings return a one-sided spectrum, i.e. the bin for some
frequency ω will include the power density for both +ω and -ω.
You should double check that the idealized spectrum to which you are comparing
is also a one-sided spectrum. Otherwise, you'll need to half the values of your
one-sided spectrum to get values representative of the +ω side of a
two-sided spectrum.
I agree with Try Hard that it is the cyclic frequency (generally Hz, or in this case 1/m)
which should be specified to pwelch. That said, the returned frequency vector
from pwelch is also in those units. Analytical
spectral formulae are usually written in terms of angular frequency, so you'll
want to be sure that you evaluate it in terms of radians/m, but scale back to 1/m
for plotting.