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...
Related
I created a sinusoid with frequency 550Hz that goes for 1 second
fs=44100;
Duration=1; %second
Len=Duration * fs; %length of sinusoid
t=(0:Len-1)/fs;
x=sin(2*pi*550*t);
for the purpose of exploring and learning, I have decided to take the short time Fourier transform of this signal. I did it as below:
window_len=0.02*fs; %length of the window
hop=window_len/3; %hop size
nfft=2^nextpow2(window_len);
window=hamming(window_len,'periodic');
[S,f,t]=spectrogram(x,window,hop,nfft,fs);
Now I want to plot the real versus imaginary value of S for the frequency equal to 550 and see what happens. First of all, in the frequency vector I didn’t have the exact 550. There was one 516.5 and 559.6. So, I just looked at the spectrogram and chose whichever that was close to it and picked that. When I tried to plot real vs imaginary of S for the frequency I chose (over all time frames), the values all fall in 3 points as it shows in the attached plot. Why three points?
Each STFT window can have a different complex phase depending on how the start (or middle) of the window is synchronized (or not) with the sinusoids period. So the real-complex IQ plot for the peak magnitude DFT result bin can be a circular scatter plot, depending on the number of DFT windows and the ratio between the stepping distance (or length - overlap) and the period of the sinusoid.
The phase of the STFT coefficients for the different windows depends on which data exactly the window "sees". So for your particular choice of window length and hop, it so happens that as you slide through your single-frequency sinusoid, there only three different data chunks that you window "sees". To see what I mean, just plot:
plot(x(1:window_len),'x')
plot(x(1+hop:window_len+hop),'x')
plot(x(1+2*hop:window_len+2*hop),'x')
plot(x(1+3*hop:window_len+3*hop),'x')
.. and if you continue you will see that the pattern repeats itself, i.e., the first plot for instance is the same as the fourth, the second as the fifth etc. Therefore you only have three different real-imaginary part combinations.
Of course, this will change if you change the window length and the hopsize, and you will get more points. For instance, try
window_len =nfft;
hop=ceil(window_len/4)
I hope that helps.
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 try to find the strongest frequency component with Matlab. It works, but if the datapoints and periods are not nicely aligned, I need to zero-pad my data to increase the FFT resolution. So far so good.
The problem is that, when I zero-pad too much, the frequency with the maximal power changes, even if everything is aligned nicely and I would expect a clear result.
This is my MWE:
Tmax = 1024;
resolution = 1024;
period = 512;
X = linspace(0,Tmax,resolution);
Y = sin(2*pi*X/period);
% N_fft = 2^12; % still fine, max_period is 512
N_fft = 2^13; % now max_period is 546.1333
F = fft(Y,N_fft);
power = abs(F(1:N_fft/2)).^2;
dt = Tmax/resolution;
freq = (0:N_fft/2-1)/N_fft/dt;
[~, ind] = max(power);
max_period = 1/freq(ind)
With zero-padding up to 2^12 everything works fine, but when I zero-pad to 2^13, I get a wrong result. It seems like too much zero-padding shifts the spectrum, but I doubt it. I rather expect a bug in my code, but I cannot find it. What am I doing wrong?
EDIT: It seems like the spectrum is skewed towards the low frequencies. Zero-padding just makes this visible:
Why is my spectrum skewed? Shouldn't it be symmetric?
Here is a graphic explanation of what you're doing wrong (which is mostly a resolution problem).
EDIT: this shows the power for each fft data point, mapped to the indices of the 2^14 dataset. That is, the indices for the 2^13 data numbered 1,2,3 map to 1,3,5 on this graph; the indices for 2^12 data numbered 1,2,3 map to 1,5,9; and so on.
.
You can see that the "true" value should in fact not be 512 -- your indexing is off by 1 or a fraction of 1.
Its not a bug in your code. It has to do with the properties of the DFT (and thus the FFT, which is merely a fast version of the DFT).
When you zero-pad, you add frequency resolution, particularly on the lower end.
Here you use a sine wave as test, so you are basically convolving a finite length sine with finite sines and cosines (see here https://en.wikipedia.org/wiki/Fast_Fourier_transform details), which have almost the same or lower frequency.
If you were doing a "proper" fft, i.e. doing integrals from -inf to +inf, even those low frequency components would give you zero coefficients for the FFT, but since you are doing finite sums, the result of those convolutions is not zero and hence the actual computed fourier transform is inaccurate.
TL;DR: Use a better window function!
The long version:
After searching further, I finally found the explanation. Neither is indexing the problem, nor the additional low frequency components added by the zero-padding. The frequency response of the rectangular window, combined with the negative frequency components is the culprit. I found out on this website explaining window functions.
I made more plots to explain:
Top: The frequency response without windowing: two delta peaks, one at the positive and one at the negative frequency. I always plotted the positive part, since I didn't expect to need the negative frequency components. Middle: The frequency response of the rectangular window function. It is relatively broad, but I didn't care, because I thought I'd have only a single peak. Bottom: The frequency response of the zero-padded signal. In time domain, this is the multiplication of window function and sine-wave. In frequency domain, this amounts to the convolution of the frequency response of the window function with the frequency response of the perfect sine. Since there are two peaks, the relatively broad frequency responses of the window overlap significantly, leading to a skewed spectrum and therefore a shifted peak.
The solution: A way to circumvent this is to use a proper window function, like a Hamming window, to have a much smaller frequency response of the window, leading to less overlap.
I am trying to use the ifft function in MATLAB on some experimental data, but I don't get the expected results.
I have frequency data of a logarithmic sine sweep excitation, therefore I know the amplitude [g's], the frequency [Hz] and the phase (which is 0 since the point is a piloting point).
I tried to feed it directly to the ifft function, but I get a complex number as a result (and I expected a real result since it is a time signal). I thought the problem could be that the signal is not symmetric, therefore I computed the symmetric part in this way (in a 'for' loop)
x(i) = conj(x(mod(N-i+1,N)+1))
and I added it at the end of the amplitude vector.
new_amp = [amplitude x];
In this way the new amplitude vector is symmetric, but now I also doubled the dimension of that vector and this means I have to double the dimension of the frequency vector also.
Anyway, I fed the new amplitude vector to the ifft but still I don't get the logarithmic sine sweep, although this time the output is real as expected.
To compute the time [s] for the plot I used the following formula:
t = 60*3.33*log10(f/f(1))/(sweep rate)
What am I doing wrong?
Thank you in advance
If you want to create identical time domain signal from specified frequency values you should take into account lots of details. It seems to me very complicated problem and I think it need very strength background on the mathematics behind it.
But I think you may work on some details to get more acceptable result:
1- Time vector should be equally spaced based on sampling from frequency steps and maximum.
t = 0:1/fs:N/fs;
where: *N* is the length of signal in frequency domain, and *fs* is twice the
highest frequency in frequency domain.
2- You should have some sort of logarithmic phases on the frequency bins I think.
3- Your signal in frequency domain must be even to have real signal in time domain.
I hope this could help, even for someone to improve it.
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.