6
Let's look at the case $x[n] \in \mathbb{R}$, where $x[n]$ is real.
Autocorrelation is basically convolution of the signal with it's time inverse. This can be easily expressed in the frequency domain.
$$ \mathscr{F}\Big\{ r_{xx}[n] \Big\} = \mathscr{F}\Big\{ x[n] \Big\} \cdot \mathscr{F}\Big\{ x[-n] \Big\} $$
$$R_{xx}(\omega) = X(\omega)\cdot X^*(\...
5
There is a book by Basseville and Nikiforov called "Detection of Abrupt Changes : Theory and Application" that they released to the public as a PDF several years ago (it's out of print, now, I believe).
That book looks at the basic CUSUM (cumulative sum) algorithm and how to choose appropriate thresholds for it.
5
Since the bulk of R’s DSP capability comes from the signal package which was ported over from the open source project Octave (itself influenced by MATLAB), there's no intrinsic limitation of R.
What you have picked up on, are ecosystem preferences. We learned MATLAB in university, picked up numpy/scipy/sklearn at work, so R isn't the first weapon of choice.
...
4
I would try a median filter.
Let your original signal be $f[n]$.
Median filter $f[n]$ using $N$ pixels, where $N > 2 \times S + 1$, where $S$ is the maximum number of samples in the spike. The resulting signal, lets call it $g[n]$ should have all the spikes removed.
Find the absolute of the difference between the two signals, $h[n] = |f[n] - g[n]|$. This ...
4
Types of signals:
According to their range set (values): Real Valued, Complex valued ;
According to their dimensions: Scalar, Vector ;
According to their values: Continuous Amplitude, Quantized ;
According to their domain set (arguments): Continuous-time, Discrete-time :
According to their mappings: Deterministic, Stochastic (Random) ;
According to their ...
4
The default parameters of signal.spectrogram are:
nperseg = 256
noverlap = nperseg/8 = 32
This means that:
The length of analysis window is $256$ samples ($256/250 = 1.024$ second)
The overlap between consecutive windows is $32$ samples ($32/250 = 0.128$ second)
The timestamps returned by signal.spectrogram correspond to the centres of a window. So in ...
4
This a very complicated question, and I would say a still open topic. The concept of stationarity is manifold, from pure statistics to applied DSP (strict, strong, wide-sense, quasi-stationarity, cyclo-stationarity, to refer to a recently closed question). The lack of access to sound models and faithful realizations renders the quest quite difficult.
Non-...
4
There is in general, as @Hilmar's answer points out, no unique solution to the question of a sequence that has the given perodic autocorrelation function. In the simplest case, that a shifted version $y$ of any sequence $x$ (e.g. $y[n] = x[n-3]$ for all $n$) has the same autocorrelation function as $x$. Similarly, $y[n] = x[-n]$ for all $n$ has the same ...
3
Q1: should the model generate a time series of length 'N=16` i.e, would the output of the above model $\mathbf{y} = [y_1,y_2,\ldots,y_N]$ contain 16 elements where $n = 1,2,\ldots,16$?
If one thinks about your question in terms of FIR filtering, then an input signal, $x$ of length $N$ filtered with an FIR filter of length $p$ ($p$ coefficients) will have an ...
3
A centered moving average filter is a finite impulse response (FIR) filter that affects the same weight to all the samples in the window. If you only care about time domain properties, and do not care about its relatively poor performance in the spectral domain, for a signal $s$ that is quite stationary across the window, you can use it. It has extremely ...
3
Suppose you are given a system with transfer function
$$H(z)=\frac{(1-3z^{-1})(1-7z^{-1})}{(1-4z^{-1})(1-6z^{-1})} $$
Poles
Poles are the values of $z$ for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right?
Look here, what values of $z$ will turn the transfer function tend to infinity? ...
3
Mathematically, shifting the frequency of a signal is pretty easy:
Following @OlliNiemitalo's answer, the 0.003 frequency shift can either be done in
time domain, or
frequency domain.
I recommend doing it in time domain, by multiplying the signal with a complex sinusoid,
$$e^{j 2 \pi f_\text{shift} n},\, n \in\{1,2,\ldots\}$$
that way you can get ...
3
In addition to what others have already said, I'll try to answer it from a purely practical point of view (this is also a variant of the overlap-add technique).
If your FFT length is 2048, then an overlap of 1024 (50%) means that you have twice as many analysis (FFT) frames (as compared to the number of frames without any overlapping). A 512 samples overlap ...
3
LPC reduces to AR modelling only if the stochastic time process is stationary (does not change distribution parameters over time) and ergodic (average over time is equivalent to mean of ensemble average).
This connection between auto-regressive coefficients and the autocovariance of the process is described by the Yule-Walker-Equations (mentioned in the ...
3
From further research I've discovered that the frequency is given by the index of the FFT multiplied by the sampling rate and divided by the size of the array. And the amplitude is the magnitude of the complex number.
So the full code for such a plot would be as follows
# Load ggplot library
library(ggplot)
# X is some set of Wait times between spikes, ...
3
...best results come from a weighted ensemble of techniques...
Maybe they do, depending on the application. But each one of the similarities mentioned, is equivalent to the other at least when we are talking about signals originating from linear systems.
Cross correlation provides very good results especially if you are trying to figure out if a signal is ...
3
The product $x(t)y(t)$ of two periodic signals with fundamental periods $T_x$ and $T_y$ is not a periodic signal unless $T_x$ and $T_y$ are rational multiples of one another; that is, $T_x = aT_y$ where $a$ is a rational number. Thus, except when such a relationship holds, $x(t)y(t)$ does not have a Fourier series.
When $T_x$ is a rational multiple of $T_y$,...
3
To be honest, I don't think CNNs, RNNs and LSTM are useful for this kind of problem – a bandpass filter followed by a threshold would be.
Now, that would have three parameters:
Lower cutoff frequency
Upper cutoff frequency
threshold value
and what is usually called "Machine Learning" is nothing but finding local minima over some (loss) function with real ...
3
I will introduce some terminology and intuition that will be helpful when reading other references. It will be neither complete nor completely rigorous.
The measures that we first encounter in real analysis assign sizes (non-negative real numbers) to measurable subsets of $\mathbb{R}$; Lebesgue measure is the measure that agrees with the intuition we build ...
3
If I understand this problem correctly you have access to 2 signals:
Noise Signal - $ w \left[ n \right] $. It is composed of a linear combination of harmonic signals. Something like $ w \left[ n \right] = \sum_{i}^{m} {a}_{i} \sin \left[ 2 \pi \frac{ {f}_{i} }{ {f}_{s} } n + {\phi}_{i} \right] $.
Input Signal - $ y \left[ n \right] $ which is composed of ...
3
It's almost a matter of philosophy, i.e., difficult to argue hard facts.
On the one hand all the features you mention can be extracted from the raw signals. So in theory the network should be able to learn how to do that if they provide meaningful information for the task at hand. This is what part of the ML community is claiming: feature engineering is dead,...
2
For discarding events where the signal is not very different from the threshold (special case: oscillation), have you considered using a hysteresis?
If the signal rises above the threshold ($t_{on}$-event), temporarily decrease the threshold (by either a factor (a few percent) or an absolute value, the best value will depend on the your system/model). This ...
2
If the signals are as you've drawn them (flat, with abrupt changes), then perhaps this approach might work:
Take the absolute value of the differences between successive time samples.
The non-zero entries are the transition times.
Find these times for both signals.
Subtract the two.
I've made an attempt to do this in the R code below.
Example output:
&...
2
Yes indeed, there is a huge body of works on the topic of variational methods for signal restoration (more general than denoising), one example being total variation denoising, or proximal methods.
If you can be more specific, you could get more accurate and precise answers.
2
Well, any input-output representation obviously admits a state-sapce form. for your equation in $y[k]$ you can easily construct one as follows. Create a "shift" system (an integrator chain) as
$$
\begin{aligned}
x_1[k+1] &= x_2[k],\\
x_2[k+1] &= x_3[k],\\
&\vdots\\
x_n[k+1] &= y[k]
\end{aligned}
$$
In this way indeed you have $x_n[k] = y[k-1]$...
2
Given the strong correlation of the downward trend in each cycle, I offer the following that would offer an improved prediction over one linear regression, but not be significantly more complicated (but still would like to see what a precise "best estimator" would be):
Do a linear regression to determine the general upward trend. This would be the Income - ...
2
Since you have the x/y position for each object it should be fairly straightforward. Call the position of each $p_1(t)$ through $p_4(t)$. Subtract the mean from all of them since it sounds like you only care about their movement, not their position.
Once you have done that it should be a fairly simple matter of cross-correlating the position functions. ...
2
First remove the DC offset (C2) from each signal by subtracting the average level from each sample, then removing the arbitrary scaling (C1) of each signal by dividing each sample by the average sample magnitude.
2
Pay attention that for Scalar Random Process the Power Spectrum Density is non negative.
Namely, let $ y \left[ n \right] \in \mathbb{R} $ a WSS Random process with its Auto Correlation function given by:
$$ {R}_{y, y} \left[ m \right] = \mathbb{E} \left[ y \left[ n \right] y \left[ n - m \right] \right] $$
Then the Power Spectrum Denisty is:
$$ {S}_{y, ...
2
Assuming your time series are the same length, take your data and produce spectrograms, $S_k$, for each row of the $4 \times 3000$ data matrix $D$ that you have. Since these individual time series should be similar, this means you can try to extract features by stacking these spectrograms next to each other horizontally into a large matrix. So if you have $...
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