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Hot answers tagged inverse-problem

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The Cocktail Party Problem is a Blind Source Separation (BSS) problem. Given a linear mixture of signals: $$\boldsymbol{y} \left[ n \right] = A \boldsymbol{x} \left[ n \right]$$ We're trying to estimate the signal $\boldsymbol{x} \left[ n \right]$. The model can get even more complex with $A$ being time varying: $$\boldsymbol{y} \left[ n \right] = A \... 5 Basically your problem is called Blind Deconvolution. It means we want to estimate both the operator and the input given the output. You model is Linear Time Invariant Operator so we have LTI Blind Deconvolution. In general blind deconvolution is ill poised problem. So we need to make assumptions about the model. The more assumptions the better the chance ... 3 Similar to The Concepts Behind SVD Based Image Processing the horizontal axis are the samples index of the SVD basis. The idea in the chapter you linked is generalizing the Wiener Filter. While the Wiener Filter uses the Fourier Transform as a basis the SVD uses the data adaptive basis. 3 The way I understand the problem is each sample of the output is a linear combination of the samples of the input. Hence it is modeled by:$$ \boldsymbol{y} = H \boldsymbol{x} $$Where the  i  -th row of  H  is basically the instantaneous kernel of the  i  -th sample of  \boldsymbol{y} . The problem above is highly ill poised. In the classic ... 3 Solving a deconvolution isn't easy even in simulated environment not to mention in practice. The main trick to solve it is using the proper model / prior for the problem and very good measurements (High SNR). So basically, for deconvolution we're after:$$ \hat{\boldsymbol{x}} = \arg \min_{\boldsymbol{x}} \frac{1}{2} {\left\| H \boldsymbol{x} - \boldsymbol{y}...

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You can approach this problem as a special case of the "$k$-simple bounded signal" class described in (Donoho & Tanner, 2010 - Precise Undersampling Theorems ), see page 2, Example 3. Particularly, your signal is a "0-simple" signal, i.e. your values are either 0 or some constant. The problem can easily be scaled to 0 or "some constant" instead of 0 or 1....

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Your $h(t)$ is an ideal low pass filter with cutoff frequency of $75$ Hertz. At the output, everything above $75$ Hertz will be gone and equal to zero. If you're thinking about the inverse system, you need to think about getting back the original input. What can you say about the original input signal at frequencies greater than $75$ Hertz? Isn't it anybody'...

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Any Deconvolution method which takes into account the noise in the image is basically a stochastic approach. Usually, the model for Deconvolution is: So having the noise in there makes it a problem with stochastic properties. Remark If by stochastic you meant sampling from the Posterior Distribution then you may have a look at Stochastic Image Denoising by ...

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The operation aimed to restore a source signal from the result of the signal's convolution with a transfer function is called deconvolution. Your question concerns the rather innocuous kind of deconvolution problems, because the transfer function is given as an input. The transfer function being one of the unknowns, the problem is called a blind ...

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Speech Source Separation (SSS) or Audio Source Separation (ASS) can be seen as a specialized version of source separation. I mention these expressions under which one can find additional works. One acceptation of the "Cocktail Party Problem" is the task of hearing/recovering one specific sound of interest in a complex environment (one-source ...

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For time-based systems, I understand that it is difficult to imagine a memory of the future. But for general systems, $-t$ and $t$ are just left and right (think of a spatial system). Other discussions are in LTI system $y(t)=x(t−T)$ with or without memory, What is a memory less system?, or A question about the concept of the time. By definition of ...

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Note that the original mathematical back-projection method assumes that the value of the Radon transform is known for all lines. This is an infinite amount of information, so it is pure mathematics at that level. Since practical tomography works with a finite number of values of the Radon transform, the inversion is ill-posed. This cannot be ignored due to ...

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As an answer probably require to have more details on the look-up table (smoothed and regularity of the kernels), here is a couple of recent papers, including a review: Satellite image restoration in the context of a spatially varying point spread function, 2010 Efficient shift-variant image restoration using deformable filtering, 2012 Fast Approximations ...

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If the signal is oversampled and the PSF variation corresponds (approximately) to a smooth local compression/expansion, perhaps you can resample y so as to make the PSF approximately LTI, then apply conventional methods (somewhat akin to homomorphic processing) If the input signal is convolved with a small discrete set of PSFs, perhaps you can devonvolve ...

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Yes, your understanding is correct. The differentiation is not an invertible system. And, the reason is precisely, because of the possible unknown constant which gets nulled by differentiation. Basically, the differentiation of $f(t)$ and $(f(t) + c)$ is indistinguishable $f^{'}(t)$, which makes the system, many-to-one, and hence non-invertible.

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