New answers tagged fft
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I think that's because of the DC offset of the signal. A time domain or frequency domain plot provided by you could have been helpful here for the confirmation of your doubt. Following are suggestions
1.try to filter the signals before FFT depending upon the signal band you are interested in
2.Subtract the signal by the mean of the entire signal (which is ...
1
I will assume:
The probe is like a line, infinitely narrow.
We move the probe along the $x$ dimension, parallel to the large-scale surface, and from each position, probe the surface:
The probe is extended to the direction it is pointing to until it meets the surface. (An alternative assumption would be that probing is done to the direction $y$, that of the ...
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You can use expansion analysis to deduce what happens when a sequence is zero staffed in between its samples.
Let $x[n]$ be your original length $N$ sequence. Then its expansion by $2$ yields the new sequence $y[n]$ of length $2N$ :
$$ y[n] = \begin{cases} { x[n/2] ~~~,~~~n = 2m \\ ~~~~~ 0 ~~~,~~~ \text{otherwise} } \end{cases} $$
Then the DTFT relation ...
2
the lomb scargle in matlab, (plomb) returns a frequency vector as the second output. i would be surprised if the routine you are using doesn’t do the same. there is a common tendency for python signal processing libraries to be functionally equivalent to matlab conventions.
your plots show symmetry around a “center” frequency which is the same as what a ...
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has some compensation in the signal chain to compensate for the DC spike which would appear at 370MHz the center frequency.
Yes, a typical problem of IQ demodulators is the LO leakage/DC spike. You really can't do much about that. You don't mention the SDR you're using, but many can be used with offset tuning, completely moving the LO out of your band of ...
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Depends on what kind of “sense” you want.
A 1 cycle sine wavelet might give you good time resolution for locating a mostly positive to negative zero-crossing waveform slice, but almost no frequency resolution, as the nearest orthogonal sinewave is a whole octave higher.
Add more cycles and you get more frequency resolution, but less time resolution.
A ...
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With all due caution, no in both cases (title and body question). I'll start with the second one.
Continuous wavelets use all dilations of the mother wavelet, which are not accessed with the STFT
The STFT is complex in general, and the windowed sine is not.
For the first one: I never tried it, and do not remember having seen it in use, and one should ...
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No, since the STFT gives you complex values by not only correlating to sines of different periods, but also cosines.
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Not really.
Unless you fully understand the math behind the DFT, you are likely to run into problems like circular vs linear convolution, time domain aliasing, excessive time domain ringing, etc.
For "light weight" frequency selectivity application, time domain filter is typically a lot easier.
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That's aliasing in its most classic form.
The anti-alias filter in your receiver isn't perfect, so that frequencies slightly outside of what you want to sample appear on the other end.
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I think there are a couple things going on here.
First, the flatness of your baseline plot is a little deceiving because the tones that you have added (1 - 500 Hz) are all almost exact multiples of the fundamental frequency of the fft (fs/nfft = 1.024 Hz/bin).
If you increased nfft by a factor of 10, for example, the finer frequency resolution of the zero-...
3
Assuming:
That you limit yourself to LTI filters.
That you can characterize both the noise and the signal of interest.
Then:
(a) If you want to detect a signal of interest (e.g. detect footsteps), use a matched filter.
(b) If you want to estimate the value of such signal, use a Wiener filter.
These are "the best" you can do (under a bunch of assumptions)....
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Overlapping FFTs, and averaging them, provide an excellent way to more accurately render the noise floor of the signal being observed. That said, you cannot actually start the FFT until you have enough samples, in part because the "last" sample will be used even on the very first butterfly computation.
That said, the time to compute the FFT can be quite ...
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1) No, if you have a particularly low level in one of the FFT points it will indicate that you have low energy in that particular frequency, but you can have several other frequencies at the same time.
2) In theory yes, but in practice the numerical precision will make the resulting signal slightly different than the original one.
Note: before applying the ...
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The most precise way to deal with them is to compare them using the algorythm used in the Wigner timer frequency distribution.
It's a processing intensive way of comparing the real and imaginary results of the FFT, I am completely ignorant of the other types of FFT implementations.
The wigner time frequency distribution uses both products to construct a ...
2
It depends on your spectrum analyzer.
If you want to analyze a one channel real baseband signal, the mirrored frequencies are redundant. you should keep the first N/2+1 points.
For communications signals, there can be an I and Q (two reals as a single complex), you want to keep them all. the frequencies will not be (in general) symmetric.
If you want to ...
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