I have retrieved some signal in my Abaqus simulation for verification purpose. The true signal shall be a perfect sinusoid at 300kHz and I performed fft on the sampled signal using scipy.fftpack.fft.
But I got a strange spectrum as shown below (sorry that I am too lazy to scale the x-axis of the spectrum to the correct frequency). In the same figure, I sliced the signal into pieces and plotted in the time domain. I also repeated the same process for a pure sine wave.
This totally surprises me. As indicated below in the code, sampling frequency is 16.66x of the frequency of the signal. At the moment, I think it is due to the very little error in the sampling period. In theory, Abaqus shall sample it in a regular time interval. As you can see, there is some little error so that the dots in my signal appear to be thicker than the perfect signal. But does such a small error give a striking difference in the frequency spectrum? Otherwise, why is the frequency spectrum like that?
FYI1: This is the magnified fft spectrum of my signal:
FYI2: This is the python code that was used to produce the above figures
def myfft(x, k, label):
plt.plot(np.abs(fft(x))[0:k], label = label)
plt.legend()
plt.subplot(4,1,1)
for i in range(149800//200):
plt.plot(mysignal[200*i:200*(i+1)], 'bo')
plt.subplot(4,1,2)
myfft(mysignal,150000//2, 'fft of my signal')
plt.subplot(4,1,3)
[Fs,f, sample] = [5e6,300000, 150000]
x = np.arange(sample)
y = np.sin(2 * np.pi * f * x / Fs)
for i in range(149800//200):
plt.plot(y[200*i:200*(i+1)], 'bo')
plt.subplot(4,1,4)
myfft(y,150000//2, 'fft of a perfect signal')
plt.subplots_adjust(top = 2, right = 2)
FYI3: Here is my signal in .npy and .txt format. The signal is pretty long. It has 150001 points. The .txt one is the raw file from Abaqus. The .npy format is what I used to produce the above plot - (1) the time vector is removed and (2) the data is in half precision and normalized.
Any standard FFT algorithm you use operates on the assumption that the signal you provide is uniformly sampled. Uniform in this context means equally spaced in time. Your signal is clearly not uniformly sampled, therefore the FFT does not "see" a perfect sine but a distorted version. As a consequence you see all these additional spectral components the FFT computes to map your distorted signal to the frequency domain. You have two options now. Resample your signal i.e. it is uniformly sampled and use your off the shelf FFT or take a non-uniform FFT to get your spectrum. Here is one library you could use to calculate your non-uniform FFT.
Related
I'm currently trying to calculate 'minimum-phase band-with limited step functions' (BLEP, see (1) and (2)) for synthesizing alias-free saw/square/triangle waves.
I think I understand most of the theory, and are exploring/prototyping in Jupyter notebooks and in MATLAB. I'm getting some strange (or at least, strange to me) results in the minimum-phase reconstruction of the windowed sinc function I want to use to calculate the BLEP.
The idea is as follows:
Calculate sinc function with specified number of zero-crossings and oversampling rate
Apply a Blackman window to the sinc function to remove periodic discontinuities
Calculate minimum-phase reconstruction of the windowed sinc function to move the energy of the impulse response to the left (to remove the need for delay lines)
Integrate the minimum-phase reconstruction to get the BLEP
Step 3 is what confuses me. The minimum phase reconstruction is defined as the ifft of the weighted cepstrum of the signal, which I can calculate step-by-step in a Jupyter notebook, but I can also directly calculate in MATLAB using the rceps function. Both are giving me the same confusing results, which I like to understand better before moving forward.
This is the MATLAB code I am using for experimentation:
clear;
L = 16;
OMEGA = 64;
N = 2*L*OMEGA+1;
x = linspace(-L, L, N);
h = blackman(N);
%h = [zeros(floor(N/4), 1); blackman(ceil(N/2)); zeros(floor(N/4), 1)];
s = sinc(x);
sw = s .* h';
sw_f = fft(sw);
[y,ym] = rceps(sw);
figure;
plot(y);
figure;
plot(ym);
What is happening is that the 'right half' of the minimum-phase reconstruction of the windowed sinc (the result from step 3) contains signal energy that I don't understand.
For example, the windowed sinc looks like this:
windowed-sinc
Which gives me the following minimum-phase reconstruction and cepstrum:
min-phase reconstruction
cepstrum
As you can see, stuff is going on in the right-half of the min-phase reconstruction, where I would have expected the graph to converge to 0 and stay there.
The effect is even more pronounced when I narrow the window function by 2x, filtering out all but the first few zero crossings:
min-phase reconstruction, narrow window
It looks like a lot of energy 'leaks' into bins that affect the right half of the reconstructed signal, but I have no idea why. The input windowed sinc function is periodic, the sample rate is high enough, and the left half of the reconstructed signal looks correct.
The cepstrum for the narrow-window sinc function also shows a spike right in the middle, which I don't understand:
min-phase reconstruction, narrow window
Now, I'm aware that to use the min-phase reconstructed signal to generate a BLEP that will be used for at most up to L samples, I don't really care what the signal looks like after the first L zero-crossings. I can just ignore it and calculate the BLEP from the left part of the graph. But I still like to understand what is going on, if only to be sure I'm not missing something.
One thing I already found out is that if I reduce the ration between zero-crossings and oversampling factor too much, I can introduce FFT leakage that even affects the left-half of the min-phase reconstructed signal, to make it unusable. I think I understand this phenomenon and just need to make sure the sample rate is high enough. But bumping the oversampling to some very high number does not remove any energy in the right-hand part of the reconstructed signal, so it does not appear it is a result of FFT leakage. What explains it though?
(1) https://www.experimentalscene.com/articles/minbleps.php
(2) https://www.kvraudio.com/forum/viewtopic.php?t=364256
I am implementing weiner filtering in python which is applied on an image blurred using disk shape point spread function, i am including code of making disk shape psf and weiner filter
def weinerFiltering(kernel,K_const,image):
#F(u,v)
copy_img= np.copy(image)
image_fft =np.fft.fft2(copy_img)
#H(u,v)
kernel_fft = np.fft.fft2(kernel,s=copy_img.shape)
#H_mag(u,v)
kernel_fft_mag = np.abs(kernel_fft)
#H*(u,v)
kernel_conj = np.conj(kernel_fft)
f = (kernel_conj)/(kernel_fft_mag**2 + K_const)
return np.abs(np.fft.ifft2(image_fft*f))
def makeDiskShape(arr,radius,centrX,centrY):
for i in range(centrX-radius,centrX+radius):
for j in range(centrY-radius,centrY+radius):
if(l2dist(centrX,centrY,i,j)<=radius):
arr[i][j]=1
return arr/np.sum(arr)
this is blurred and gaussian noised image
this is what i am getting result after weiner filtering for K value of 50
result does not seem very good, can someone help
seems noise is reduced but amount of blurred is not, shape of disk shaped psf matrix is 20,20 and radius is 9 which seems like this
Update
using power spectrum of ground truth image and noise to calculate K constant value, still i am getting strong artifacts
this is noised and blurred image
this is result after using power specturm in place of a constant K value
Reduce your value of K. You need to play around with it until you get good results. If it's too large it doesn't filter, if it's too small you get strong artifacts.
If you have knowledge of the noise variance, you can use that to estimate the regularization parameter. In the Wiener filter, the constant K is a simplification of N/S, where N is the noise power and S is the signal power. Both these values are frequency-dependent. The signal power S can be estimated by the Fourier transform of the autocorrelation function of the image to be filtered. The noise power is hard to estimate, but if you have such an estimate (or know it because you created the noisy image synthetically), then you can plug that value into the equation. Note that this is the noise power, not the variance of the noise.
The following code uses DIPlib (the Python interface we call PyDIP) to demonstrate Wiener deconvolution (disclaimer: I'm an author). I don't think it is hard to convert this code to use other libraries.
import PyDIP as dip
image = dip.ImageRead('trui.ics');
kernel = dip.CreateGauss([3,3]).Pad(image.Sizes())
smooth = dip.ConvolveFT(image, kernel)
smooth = dip.GaussianNoise(smooth, 5.0) # variance = 5.0
H = dip.FourierTransform(kernel)
F = dip.FourierTransform(smooth)
S = dip.SquareModulus(F) # signal power estimate
N = dip.Image(5.0 * smooth.NumberOfPixels()) # noise power (has same value at all frequencies)
Hinv = dip.Conjugate(H) / ( dip.SquareModulus(H) + N / S )
out = dip.FourierTransform(F * Hinv, {"inverse", "real"})
The smooth image looks like this:
The out image that comes from deconvolving the image above looks like this:
Don't expect a perfect result. The regularization term impedes a perfect inverse filtering because such filtering would enhance the noise so strongly that it would swamp the signal and produce a totally useless output. The Wiener filter finds a middle ground between undoing the convolution and suppressing the noise.
The DIPlib documentation for WienerDeconvolution explains some of the equations involved.
My general aim is to calculate the power spectral density of an input signal exactly as what is seen in the QT GUI Frequency Sink block. I'll need to process the PSD values later on. Here is my current setup.
These graphs are produced when a signal (no transmission) is inputted.
Blue = signal, green = max of signal, pink = min of signal.
My implementation of PSD produces a graph that has the same height and shape as the actual PSD shown by the QT GUI Frequency Sink block, but wrong offset - the actual PSD is about 66 dB lower than mine.
I eyeballed the rough max/min/signal y-axis values:
Actual PSD has max = -76dB, min = -115dB, signal = -86dB.
My PSD has max = -10dB, min = -50dB, signal = -20dB.
I'm not really sure what I'm doing wrong; the offset of -66dB seems pretty arbitrary and I think I'm on the right track generally.
You need to scale result of FFT by FFT length. 20Log10(2048) = 66 dB. You are estimating the PSD by calculating the Periodogram which is magnitude squared of the FFT divided by length of FFT.
I've been experimenting with a few different techniques that I can find for a freq shifting (specifically I want to shift high freq signals to a lower freq). At the moment I'm trying to use this technique -
take the original signal, x(t), multiply it by: cos(2 PI dF t), sin(2
PI dF t)
R(t) = x(t) cos(2 PI dF t)
I(t) = x(t) sin(2 PI dF t)
where dF is the delta frequency to be shifted.
Now you have two time series signals: R(t) and I(t).
Conduct complex Fourier transform using R(t) as real and I(t) as
imaginary parts. The results will be frequency shifted spectrum.
I have interpreted this into the following code -
for(j=0;j<(BUFFERSIZE/2);j++)
{
Partfunc = (((double)j)/2048);
PreFFTShift[j+x] = PingData[j]*(cos(2*M_PI*Shift*(Partfunc)));
PreFFTShift[j+1+x] = PingData[j]*(sin(2*M_PI*Shift*(Partfunc)));
x++;
}
//INITIALIZE FFT
status = arm_cfft_radix4_init_f32(&S, fftSize, ifftFlag, doBitReverse);
//FFT on FFTData
arm_cfft_radix4_f32(&S, PreFFTShift);
This builds me an array with interleaved real and imag data and then FFT. I then inverse the FFT, but the output im getting is pretty garbled. Results seem huge in comparison to what I think they should be, and although there are a few traces of a freq shifted signal, its hard to tell as the result seems mostly pretty noisy.
I've also attempted simply revolving the array values of a standard FFT of my original signal to get a freq shift, but to no avail. Is there a better method for doing this?
have you tried something like:
Use a Hanning window for each framed data
Once you have your windowed frame of audio data, you do an FFT on it
Do some kind of transformation in the frequency domain (you can use
Flanagan - phase vocoder)
Now you need to go back to the time domain with an IFFT
Apply Hanning window in the IFFT data
Use overlap-add at each new frame of time-domain data into the output
stream
My results:
I created two concatenated sinusoids (250Hz and 400Hz) and move one octave UP!
Blue waveform is the original and red was changed, you can see one fadeIN-fadeOut caused by overlap add and hann window !
If you want the frequency shift to sound more "natural", you will have to maintain the ratios between all the initial frequency bins, where the amount of shift will depend on the FFT bin, thus requiring lots of interpolation. The Phase Vocoder algorithm will use multiple FFTs to reduce phase distortion in the result.
I'm just doing a power spectral density analysis of a signal in time domain. I'm following the fft method described in :
http://www.mathworks.com/support/tech-notes/1700/1702.html
It gives the real physical unit for the PSD. However, the unit is "power", is that mean "V^2/Hz"?
If I take 10*log10(power) or 10*log10(V^2/Hz), do I get the unit of "dB/Hz"?
Then how can I convert it to dBm/MHz?
It depends on the unit of your timeseries. Often we think of this as just "amplitude", but if your timeseries is a series of voltage amplitude vs. time, then your PSD estimate will be Volts^2/Hz. This is because the PSD is the Fourier Transform of the autocorrelation of your original signal: The autocorrelation has units of Volts^2, and running it through the Fourier Transform decomposes these units over frequency, instead of time, resulting in units of Volts^2/Hz. This is commonly referred to as Watts/Hz, but the conversion from Volts^2 to Watts is not very physically meaningful, as W = V^2/R.
10*log10(power) will result in a unit of dB/Hz, but remember that decibels are always a comparison between two power levels; you are quantifying a ratio of powers. A better definition of decibels is 10*log10(P1/P0), as explained here. If you simply plug a PSD bin estimate into this equation, you are setting your PSD bin to P1 and implicitly comparing it to a P0 value of 1. This may be what you want, and it may not be. For visualization purposes, this is fairly typical, but if you have a standard reference power you should be comparing to, you should use that for P0 instead.
Assuming that you are attempting to plot a dB Power Spectral Density estimate, to convert from Hz to MHz, you simple rescale the x-axis of your frequency graph. Remember that a MHz is just 1 million Hz, so the only difference is that 240000Hz = 0.24MHz
EDIT
The point brought up by mtrw is a very valid one; if you are dealing with large amounts of data and are averaging FFT vectors, I highly suggest the Multitaper method; it's a much more statistically sound method of sacrificing frequency resolution for greater confidence on your PSD estimate.
If you have a PSD in W/Hz i.e. 100 W/Hz then you have 50 dBm/Hz. dB/Hz or is often vaguely and generically used instead of dBm/Hz. Audacity uses dB as shorthand for dBFS (not dBFS/Hz, because it is computing a DFT, and discrete frequencies use a power spectrum and not a density) . A digital signal that reaches 50% of the maximum level has an amplitude of −6 dBFS, which is 6 dB below full scale – the removal of the MSB, hence the 6dB/bit figure (because 50% of maximum level is 25% of maximum power; 1/4 = - 6dB)
dBm is the logarithmic ratio of the power with respect to 1mW, you divide the power by 1mW to get a unitless ratio, and then take the logarithm to get dB units, which in this case makes more sense to be clarified as dBm.
dBc/Hz is the ratio with respect to the carrier power, which is a ratio of two dBm/Hz values, meaning you subtract them and you get dBc/Hz; you get the same result if you divide the two linear power levels in W and then convert the ratio to dB (or more appropriately dBc).
dB-Hz is a logarithmic measure of bandwidth with respect to 1Hz and
dBJ is a measure of spectral density as a logarithmic ratio to 1 joule, seeing as W/Hz is indeed J.
Power spectral density is a density function, so you need to integrate it to get the actual quantity, like a line Integral of a V/m electric field, or a probability density of probability per x. This does not make sense for discrete quantities and instead the power spectrum is used akin to a probability mass function. If you see dB (which should be used for the discrete frequency domain) instead of dBm/Hz then it's wrong, but if you see it instead of dBm then it's right, as long as it's made clear what the reference is.