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If your system is powered by a switching power supply that is not synchronized to the PRF (or multiples of), then you may get reliable spurs over time as seen in your histogram. In your case, it's been found that was indeed the problem! The 500 KHz frequency fell nicely in the middle of your 1 MHz histogram. Hopefully this will help hunt down similar issues ...


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To downsample an FFT result for both magnitude and phase, it may work best to do an FFTShift before the FFT, then downsample the real component vector and then the imaginary component vector separately, before taking the atan2() of them to estimate downsampled phase.


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It looks to me as if you are trying to do a FIR filtering with a long impulse response. The correct way of doing this using FFT is either the "Overlap add" or the "Overlap save" method. Both methods are well described on Wikipedia even with programming examples.


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I believe, there are two aspects of your question. For the FFT calculation, the power density spectrum is the square of the FFT modulus multiplied by the sampling period (in your notation: T) and divided by the number of samples (in your notation:L): T/L * abs(FFT(X))^2 or T/L *abs(Y(f))^2 Here is now where I believe you have an error, you have to add ...


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Try using the complement operator on the color bytes. For example, a lot of software used #define's for their colors: #define MAGENTA 0xF81F Try adding the complete operator before the numeric value: #define MAGENTA ~0xF81F Notice the tilde before the number.


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Here I am, reading your original PROBLEM in 2019 (7 yrs later). You start out with TIMEDOMAIN data TD[], a set of equally spaced sampled amplitudes of a realtime laser signal. Let's verify what your data looks like. Does it have dead space followed by pulses (0 to 2V ?). Does it resemble a clock of off-time then on-time pulses ? And, was the data ...


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The FFT is just a fast method to implement the Discrete Fourier Transform (DFT). It's based on the idea that you can break a large DFT down into a bunch of smaller ones and the combine the results for the final DFT. That requires the DFT length to be broken down into it's prime factors. The more and the smaller the prime factors are, the more efficient the ...


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Yes. Simply sum all the polyphase outputs and the sum result will have higher resolution. Consider that each polyphase output is a delayed version of the same signal, so that if you commutated through all the outputs, you would get a higher sampled version of your same signal and the quantization noise of this signal would be approximately white across this ...


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To increase the frequency resolution of your signal, if it can be done by increasing the sampling rate at measurement, this would be the ideal way to improve frequency resolution.


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Zero padding does not increase the frequency resolution— it just interpolates more samples in frequency but does not give you any more information. The frequency resolution in Hz of an non-windowed (rectangular window) Fourier Transform is 1/T where T is the length of your time domain waveform in seconds. You must have a longer data sample in order to get ...


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Periodogram is an DTFT / FFT based nonparametric PSD estimator which is computed in Matlab/Octave by the following: $$ \hat{S}[k] = \frac{1}{N} |X[k]|^2 $$ Where $L$ is the length of the signal $x[n]$. It has got a number of options, including different windows, instead of a rectangular, to use. Welch method is a variation of periodogram method in that it ...


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This is an untested time domain solution, but the math looks solid. This will be impossible to implement unless you solve the reciever synchronization problem first. That is either a hardware fix or a calibration operation. Assume it is solved and your two signals are coming in as time aligned sequences. Assume also your sampling rates (I don't like "...


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If your signal is at least somewhat oversampled, you could try the following time domain approach; 1) Apply the output of each A/D to a Hilbert Transform filter to generate a complex signal. 2) Derive the sample-by-sample angle for each complex signal by using ATAN2. 3) Designate 1 channel as the reference channel. For every reference clock and subsequent ...


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Is there any single shot solution for comparing the relative phase difference between the signals received with different sampling frequencies as mentioned? Yes it is, as long as you have an exact knowledge of the timing relationship between the samples for each receiver. It's complicated, but if you understand the properties of the Fourier transform it's ...


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Assuming that the relevant portion of a continuous-time signal $x(t)$ is inside (or has been shifted to) the interval $[0,T]$, the DFT of a sampled version of the signal approximates the continuous-time Fourier transform (CTFT) in the following way: $$\begin{align}X(f)&=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt\\&\stackrel{\textrm{truncation}}{\...


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I would not rule out hardware effects. The input stage of the AC measurement path in a DMM will include a variable-gain amplifier to auto-range the input so that the ADC noise and non-linearity are minimized. It's possible that the bandwidth (and therefore phase) is not constant as a function of gain (although it sounded as if you are inputting a pretty low ...


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Attached Matlab sample code to apply an matched filter to a sine sweep for finding the playback position (the playback position estimate can be derived as the position of the peak in the matched filter plot): % params sampling_rate = 44000; f_low = 10; f_high = 100; shift_percent = 0.0; shift_samples = 30000; % init t = 0:1/sampling_rate:1; N = length(t); ...


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You could also do this (in numpy notation): np.arctan( (signal*cos).sum() / (signal*sin).sum() )) where signal is your phase-shifted signal, cos and sin are the reference signals, and you generate an approximation of an integral over a certain time via summing over the two products.


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must maintain a specific FFT length and small FFT bin width, down at baseband. Well, the FFT bin width is defined as $\frac{f_\text{sample}}{N}$, with $N$ being the FFT length and $f_\text{sample}$ the sampling rate of the signal that undergoes the FFT. So, if both these parameters are fixed, you have no freedom in choosing the bin width whatsoever. Note ...


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As mentioned, a FT or DFT can be used to analyse a transient signal but, as a really hand-wavy rule of thumb, the closer the basis vectors resemble the signal itself the less additional terms may be needed to accurately characterise it. This can help with the analysis phase if it leads to simplified data. This is typically where e.g. wavelets and lifting ...


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It depends on your application. The term “transient” means different things to different people. Analyzing power line hiccups is different than looking at dolphin chirps. The start up of a steady state signal can be considered of interest. You also have complex sound scapes where there are transients in the presence of steady state signals where the ...


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You are right. Fourier transforms are not useful for analysing transient signals compared to time-domain analysis or even wavelet analysis. Transient signal analsysis is a complex endeavour. However, keep in mind that all the information in the signal is retianed in the Fourier transform, albeit in a quite hard to interpret form. A better tool should be ...


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Fourier transforms are multi-purpose tools. While they are well-suited to "stationary" signals, there are several ways to use them in different contexts. Think about a $2$-point DFT. Up to a scaling factor (for orthonormality), it consists in the matrix $F_2=\begin{bmatrix}1&1\\1&-1\end{bmatrix}$. It computes something that is essentially (thanks ...


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If the signal is zero-valued outside the time window being analyzed ("analysis window"), then the discrete Fourier transform (DFT) exactly samples the discrete-time Fourier transform (DTFT) of the signal at harmonic frequencies. DTFT implies no time-domain periodicity. The sampling of DTFT by DFT is evident from definitions of the two transforms, if we apply ...


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