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That looks a lot like an exponentially decaying sinusoid. What you are primarily interested in is the decay rate. Where it ends would then have to be defined as when it reached some threshold level. Create a subsequence consisting of the peak values and the negative values of the troughs. This should give you a nice exponential decay function. Then ...

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The answer is yes but one has to specify $B_n$ properly to avoide possible confusions. In case if one uses a pulse compression, the bandwidth through which the receiver collects the noise will normally be $B_n = \beta_c$. Then, the "new" signal-to-noise ratio should be written as: $SNR = \dfrac{P_TG_TG_R\lambda^2\sigma{P_g}}{(4\pi)^3R^4(kT_{sys}\beta_c)} = \... 3 Assuming that the relevant portion of a continuous-time signal$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}}{\... 3 Well, by definition of the \delta distribution, you have: \int_{-\infty}^{\infty} f(t) \delta(t-T)\, \textrm{d}t = f(T) The autocorrelation of a function g(t) can be computed via: \int_{-\infty}^{\infty} g^{*}(t)g(t + \tau)\, \textrm{d}t, with g^* as the complex conjugate of g. Since \delta(t) is real-valued, this is conjugation can be ... 2 My first comment would be why the heck are you using R if you are concerned with processing speed, or are you just prototyping algorithms? Anyway, Without getting into how I derived it, here is a formula that is much much faster: Take the log of your signal (-1 if 0): g[x] = \ln(y[x]) $$Calculate the following value:$$ B = \frac{ \begin{array}{c} ... 2 Ha just figured out a faster and better method just using BIC-optimized selection of optimal peak width, using a banded covariate matrix with shifted Gaussian peak shapes of given width & using nonnegative least squares fits (which is solved using an active set method and regularizes the problem a bit, though less of course than with LASSO or L0 norm ... 2 Yes. Pulse compression is really just running the returned signal through a pulse matched filter, which is equivalent to cross correlation. If you view it as a pulse matched filter, matched filters are optimal for detection of a signal in AWGN. If you view it as a cross correlation, the output of the correlation will peak when the signal best matches with ... 2 It's a combination of two processes: first, you create a discrete-time signal that is pulse-shaped. Then, the DAC converts it to an anlog signal. (And, in a radio, the analog front-end following the DAC converts it to an RF signal). The discrete-time part is usually done as follows (using Matlab here for convenience). Assume a BPSK data vector data. First, ... 1 You could define a weight sequencew(n)$that weighs the error differently for different time indices$n$: $$\epsilon^2=\sum_{n=0}^{N+K-2} w(n)\big|d(n)-z(n)\big|^2\tag{1}$$ Samples that are too large get a higher weight, whereas samples that are smaller than necessary can get a lower weight. Of course, you cannot know in advance which samples should get ... 1 If the number of samples per symbol (sps) is 4, then you should upsample by 4. The input to the pulse shaping filter is 1 sample/symbol, and the output is 4 samples per symbol, so the upsampling rate is 4. The span is the number of symbol periods in the filter's impulse response. Intuitively, this means that a particular symbol influences the pulse-shaped ... 1 The answer is simple. I will give 3 points to solve it: The Fourier transform is linear. Hence$ \mathcal{ F } \left\{ \alpha f \left( x \right) + \beta g \left( x \right) \right\} = \alpha \mathcal{ F } \left\{ f \left( x \right) \right\} + \beta \mathcal{ F } \left\{ g \left( x \right) \right\} $. Shift in time$ f \left( x - {x}_{0} \right) \$ equals ...

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Judging only from one bad example, it is hard to give a good answer. If this is a representative example of your data, then I would see it more important to identify the repetition of the peaks, instead of the peaks itself. Two thoughts on that: Use a peak detection algorithm. You already mentioned that you experimented with one, which certainly can be used ...

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"Sharp peaks" in the time domain translates to high frequency content. Your simple algorithm of min and max values would be improved by first passing your data through a high pass filter - otherwise slow (low frequency) rumbles could degrade your detection methods. To take your solution to the next level, you should use an "onset detection" algorithm. These ...

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