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In general LTI System is invertible if it has no zeros in the Fourier Domain. The way to prove it is to calculate the Fourier Transform of its Impulse Response. In practice, it can be invertible if it has no zeros in the support of the input signals.


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The CWT is not, per se, superior to STFT. Due to its variations in scale, it can be better, for instance, when: you address natural signals where transients are shorter than more stationary parts, you do not know the appropriate scale of observation. which is neither the case from your signal: precise frequency on the same support. On the one hand, you ...


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If you have a one dimensional signal $s(\cdot)$, it somehow belongs to a combination of two different domains: the sampling domain: where samples are considered or taken, or how the signal is sampled. When samples are taken in some order, this is often called: the ordinal variable. This is generally related to some "physically-sound" unit, like time (you ...


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I know this is very old, and @Matt L. long since gave an excellent and informative answer. I had no idea that total variation denoising existed, so I learned something quite useful. Accordingly, I upvoted both the question and answer and want to give a little something, such as it is, back to the site. The basic idea is to use a simple digital version of the ...


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Really, there's nothing special about time and frequency domain. The math doesn't care whether you're transforming amplitude over time, gravel over mountain height, or smell intensity over fridge temperature. To the math, you map elements from a function space to elements from a function space when you do the Fourier transform. Full stop. Any scalar ...


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In the additive model $y=s+n$, when the signal is deterministic, it adds coherently over the "realizations". Hence, its variance $V(\sum s_k) = V(K s) = K^2 V( s)$. And when the noise is independent identically distributed (IID), then $V(\sum n_k) = KV( n) $. You can find this classical result detailed in Deriving the SNR for averaged signals. Warning, ...


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Very similar to what Marcus Müller suggested, this is what I did: The raw noisy waveform is compared (via comparators) with upper and lower thresholds. The upper threshold is simply a constant (0.8 here) plus the long term average (LTA). Likewise, the lower threshold is the LTA minus 0.8. The value 0.8 was chosen by inspection of the raw waveform. If a ...


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I like the answer by @Laurent Duval and will give it an upvote. I also like the comments, especially that of @robert bristow-johnson, but I thought I would try doing what I suggested in the comments, which is basically (I think) what robert bristow-johnson essentially suggested. First, I made up a fluorescence spectrum that more or less matched what the OP ...


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Second try! This question is also at the Electrical Engineering stack, so I posted a short 'starter' answer there. A mod converted it to a comment and my tiny rep there prevented me making any further comments. This is what I posted there: I suggest this link to cbeleites's answer to a question involving laser-excited Raman spectroscopy: https://stats....


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@MBaz solution is neat. Another vision is to use the idea that discrete convolutions turn into polynomial products in the $z$-transform domain: $$[1,1]\ast [1,1] = [1,2,1]$$ is equivalent to $$(1+z^{-1})(1+z^{-1})=(1+2z^{-1}+z^{-2})$$ So separating a long filter into short convolved filters is equivalent to factorizing a polynomial. To ease notations, I'll ...


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Just for the fun, and to emphasize on the importance of being precise about the domains one is talking about: if $t$ is not a real variable, but for instance $t$ belongs to an affine version of relative integers: $t=\frac{1}{5}k+b$, then the (now discrete) system rewrites $$y(k)=x(k)\sin\left(10\pi\left(\frac{1}{5}k+b\right)\right)$$ and it becomes a ...


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The term comes from the power "rails" or buses that provide the positive and negative voltage supply for an operational amplifier. (Op amps are typically used for preamplification of the raw signal before analog-to-digital conversion). If the input signal is so low or high that the amplifier's output goes as far as it can to the full positive or negative ...


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First, we define $h_3[n] = h_2[n] \star h_2[n] = [1, 2, 1]$. From this result, we know that $h_1[n]$ must have 5 elements (so that $h[n]$ ends up with 7 elements). Let's define $h_1[n] = [g_1, g_2, g_3, g_4, g_5]$. We can find these as follows, using the definition of discrete convolution. First, we know that $g_1 h_3[0] = h[0] = 1$, so $g_1 = 1$. Then, ...


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The answer given by @LaurentDuval is correct. I just wanted to also post the explanation I found in an ancient forum post right before reading his answer: [Signal processing software] shows "Railed" when microvolt magnitudes for the channels are off the top end of the scale. In other words, generally it means something wrong with the differential ...


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Just a guess. From analog electronics, amplifiers typically have a DC voltage that supplies power to the circuit. The amplified output is typically limited to that voltage. When an output is clipped, it has been said that the output is at the rails. Not really a dsp term but is a way to say that a waveform is clipped via the dynamic range of the system.


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A railed signal, or a railing signal, seems to indicate a flatline. On BIOPAC, Railing signal (flatline) says: When the amplified signal for any given channel exceeds the range -10 to +10 volts, the signal will rail. You will see a straight line at -10 or + 10 volts (more likely the reading will be close 9.99 volts). The MP system is designed to work ...


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If you're gonna use ASDF to measure periodicity and you want to window the data with something other than rectangular window, do it after subtracting and squaring: $$ Q_x[k, n_0] \triangleq \frac{2}{N} \sum\limits_{n=0}^{N-1} \left(x[n+n_0-\left\lfloor \tfrac{N+k}{2}\right\rfloor] \ - \ x[n+n_0-\left\lfloor \tfrac{N+k}{2}\right\rfloor + k] \right)^2 w\left(\...


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Stationarity is a multi-fold concept in signal processing. It can denote a wide range of behavior, encompassing deterministic or stochastic aspects. Beyond that, the main question is: do you know if your signal is stationary, and how? If you actually know how, it is probably wiser to use the generation process to build a custom, adapted model or ...


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A classical result states that a (non-trivial) time-limited signal does not have a limited bandwidth (in terms of support). The converse happens as well: a bandwidth-limited signal cannot be of finite length. And there is a similar rule-of-thumb principle: when you try to reduce the support of a signal in one domain (time or frequency), you generally ...


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All recognition tasks (doesn't even have to be speech recognition) are reductions of a very high-dimensional signal (your speech recording's dimension is the number of audio samples!) to a low-dimensional signal. As such, it is generally advisable to transform the input signal through an easy operation to a representation where the dimensionality can be ...


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Sounds like a job for dedicatedly identifying these surges and then actually subtracting them. I'd start with a low-pass filter to find the slightly time-varying mean of the signal. Use that to define lower and upper thresholds above or below you count something as surge. Identify the samples lying outside the thresholds. Find a signal model for surges, e....


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Marcus is correct, so this means something else is wrong. The left equality simply shows that, for continuous time, the unilateral PSD, $S_x(f)$, is two times the Fourier transform of the autocorrelation function, $R_x(\tau)$. See, e.g., A.B. Carlson, Introduction to Communication Systems, 2nd Ed., McGraw-Hill, NY ©1975, Chapter 2, equation 18a. Or see ...


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So, you've got the audio signal in digital on your PC. Analyze that digitally; you don't need any measurement equipment to do that. On the contrary: Measuring properties of the analog audio signal is highly undesirable, because you'd be including the sound card of your laptop in your measurement, but you probably don't care about that even remotely. If you ...


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