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First of all note that there is a certain ambiguity in the problem formulation because the approximation of $x(t)$ has to be divided into the part represented by $x^{\prime}(t)$ and the part represented by $f(y(t))$, and this division is not unique. The approach I chose is to start with a parametrization of $f(y(t))$, and approximate the given $x(t)$ as well ...


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I will divide my answer into 3 sections. The Distribution of the Derivative of Images Take a real world image, any image. Apply the derivative operator on it (Namely apply the kernel $ \left[ 1, -1 \right] $ on it. Display the histogram of the filtered image. I took this image: The histogram I got is this: This distribution is very similar to Laplace ...


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Ok, your code is very difficult to understand, but I think that we can answer your question. There is a group of processing blocks that every communication systems have. Normally, they appear in the following order in almost every simulation: Bit Stream -> Modulator -> Channel -> Equalization -> Demodulator -> Received Bit Stream Note that I suppressed ...


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What you want to do is dimension reduction. The most basic, yet very powerful and commonly used, technique to do it is principal component analysis (PCA). PCA operates on the covariance matrix. You can look it up on Wikipedia for a throughout tutorial. PCA decomposes your data $y$, having dimension $d$, into $d$ principal components. The first principal ...


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A "point spread function" denotes how a single point from the source, or a punctual object (mathematically a Dirac) would spread on the observed image through the imaging system. In blind deconvolution, both image priors and PSF priors are useful to unmix those intricate systems (even more with noise). They are different, as the PSF is somehow invariant to ...


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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} ...


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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 ...


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This is closely related to Blind Deconvolution. The only difference is we limit our self to a very specific type of blur kernels. The nice thing about the Gaussian Kernel is being defined by single parameter - The Standard Deviation of the kernel. The less nice thing is the connection isn't linear. Optimization Problem Let's define a classic non ...


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Your formula/method for computing MSE between estimated and known inputs looks good to me. For symbol error rate you could use something like a Hamming distance which simply counts the number of times the estimated symbol is different from the actual symbol. In your 10 symbol example the error rate is 4/10 i.e 40%. In Matlab you can do something like: ...


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This question is confusing, but I thought I'd try to make some sense of it. Suppose you are transmitting one of your $\mathbf{x}_i$ vectors one component at a time. That is, the transmitted signal, $s[k]$, for the $i^{\rm th}$ feature vector is: $$ s_i[k] = x_{ik} $$ If this signal is corrupted by a channel with impulse response $h[k]$ then the received ...


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You only need to find and report positions of nine specific symbols. You can do this by computing correlations (done quickly by multiplying in Fourier space) with samples of the nine digits snipped from a typical puzzle image, blurred as it is. All blurred '9' will look the same, and likewise all other digits. I'm assuming puzzle images are taken with the ...


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One way, though not directly visual, is to observe the image statistics. We have a pretty good idea about the statistics of Natural Images, more specifically, their Gradient Distribution (See Statistics of Natural Images and Models by Jinggang Huang, D. Mumford, What Makes a Good Model of Natural Images by Yair Weiss, William T. Freeman, The Statistic ...


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I'm answering the question the way I understood it - How can one find a similarity measure which isn't sensitive to scaling and shifting. An approach could be borrowed from the Computer Vision world by comparing Shift and Scale Invariant features between the two signals. I'm not sure it will work for measuring the quality of recovering signals but it ...


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As mentioned in the comments, the symbol $^H$ denotes the conjugate transpose of a matrix or vector, which means that the vector/matrix is transposed and that all of its elements are conjugated. The bar over $y_k$ means complex conjugate (note that $y_k$ is a scalar, not a vector or matrix). When computing the gradient with respect to a complex variable in ...


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If you just want to estimate the height of the square shaped component you can pass the composite signal thorough a low pass filter with a cut-off frequency close to DC or less than the minimum frequency you can expect to see in the composite signal apart from the square component. The resulting signal would have most of the non-DC components filtered out ...


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You can think of MAP as a regularization of the ML. Just like you have regularization for Least Squares Problem (They can be built, mostly, as MAP problem). The nice thing is that, as always, the best regularization is more data, namely, in most case when there is a lot of data they collide (Namely, low sensitivity fir the Posterior PDF). So they differ ...


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Maximum a Posteriori (MAP) is the same as Maximum Likelihood Estimation (MLE) except with a Bayesian prior distribution on whatever it is that you're trying to estimate. So if you have prior information on the distribution of point spread functions then MAP will work better.


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If you want to do source separation, then independent component analysis (ICA) is the correct technique, which you mention in the comments. ICA decomposes data as $x = As$ where $x$ are mixtures of some source signals $s$, and $A$ is the mixing matrix. Neither $A$ or $s$ is assumed to be known, ICA will find both of them given some $x$. The only assumptions ...


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