# Tag Info

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You need to define what you mean by "invertible". Do you mean invertible by a causal and stable system? If yes, then any system that is not minimum-phase is not invertible (because the inverse system can't be causal and stable). Example of a system that cannot be inverted by a causal and stable system: a simple delay $y(t)=x(t-T)$, $T>0$, could only be ...

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Since convolution describes the operation of a linear time-invariant (LTI) system, the question is if the effect of an LTI system can be compensated by another LTI system. In the discrete-time domain you can use the $\mathcal{Z}$-transform to analyze LTI systems. If a signal $x(n)$ (with $\mathcal{Z}$-transform $X(z)$) is filtered by a system with impulse ...

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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. $$\mathscr{F}\Big\{ r_{xx}[n] \Big\} = \mathscr{F}\Big\{ x[n] \Big\} \cdot \mathscr{F}\Big\{ x[-n] \Big\}$$ $$R_{xx}(\omega) = X(\omega)\cdot X^*(\... 7 You cant't recover the original signal through deconvolution. A Gaussian kernel is in essence a lowpass filter, i.e. it will remove information at higher frequencies from the signal. Once it's gone, it's gone and you can't recover it. This problem shows up as "divide by zero" or "divide by a very small number", which then amplifies numerically noise of ... 7 A necessary condition for invertibility is that any output has only one possible input (or injectivity, as proposed in comments). Since we are looking at counterexamples, we can look at when this condition is not satisfied. The null system, that turns every signal into a zero flat line, is not invertible, but a bit trivial. A system that computes a ... 7 Solving a deconvolution isn't easy even in simulated environment not to mention in practice. The main trick to solve it is using the proper model / prior for the problem and very good measurements (High SNR). So basically, for deconvolution we're after:$$ \hat{\boldsymbol{x}} = \arg \min_{\boldsymbol{x}} \frac{1}{2} {\left\| H \boldsymbol{x} - \boldsymbol{y}...

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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: $$\boldsymbol{y} \left[ n \right] = A \... 6 The Wiener Filter can also be derived by another (Easier) way. Let's assume the following model:$$ y = h \ast x + n $$Namely the data is a result of a linear combination (Convolution) of  x  with Additive Noise. If we assume the noise model is Gaussian and our data is also formed by a Gaussian distribution then we should try to minimize (MAP ... 6 Intuitively, if You move the sensor  N  steps each at the size of  \frac{1}{N}  of its resolution you can get  \times N  more resolution. It is like a polyphase representation of the signal. Using estimation methods, any movement which is not an (Event with zero probability) integer multiplication of the resolution of the sensor, namely, fractional ... 6 I will divide my answer into 3 sections. The Distribution of the Derivative of Images Take a real world image, any image. Apply the derivative operator on it (Namely apply the kernel  \left[ 1, -1 \right]  on it. Display the histogram of the filtered image. I took this image: The histogram I got is this: This distribution is very similar to Laplace ... 6 One general form of Inverse Problem in Imaging which assumes Linear Operator is given by:$$ \arg \min_{x} \frac{1}{2} {\left\| A x - b \right\|}_{2}^{2} + \lambda R \left( x \right) $$Where  R \left( x \right)  is the regularization function. One way, which is pretty intuitive in my opinion, is to treat the above as a solution to Maximum a Posteriori ... 6 I would take approach based on Blind Deconvolution. Since we're dealing with ill posed problem some assumptions should be made. The intuitive approach would be using the information as a prior for the signal. Another idea is to add LPF assumption of the Filter by setting the sum of its coefficients to be 1 and non negative. Yet since we have Discrete ... 6 Basically your problem is called Blind Deconvolution. It means we want to estimate both the operator and the input given the output. You model is Linear Time Invariant Operator so we have LTI Blind Deconvolution. In general blind deconvolution is ill poised problem. So we need to make assumptions about the model. The more assumptions the better the chance ... 5 The idea is to represent all operation sing Matrices. Once it is done, it is easy to solve the problems as a Least Squares problems. The way to represent Convolution Operation using a Matrix is by Toeplitz Matrix. For 1D it is pretty straight forward to do (Just pay attention to boundary). So let's take the simple model in the comment:$$ g = f \circ h + ...

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Approaches There are many methods for Deconvolution (Namely the degradation operator is linear and Time / Space Invariant) out there. All of them try to deal with the fact the problem is Ill Poised in many cases. Better methods are those which add some regularization to the model of the data to be restored. It can be statistical models (Priors) or any ...

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Here's the way I think about a discrete Wiener Filter Consider a sequence of observations $\mathbf{y} \in \Re^n$ Form a matrix from the input $\mathbf{x} \in \Re^{n+r-1}$ by shifting columns one sample each: $$X= \begin{bmatrix} x_1 & x_2 & ... & x_r \\ x_2 & x_3 & & x_{r+1} \\ x_3 & x_4 & & x_{r+2} \\ ... & &... 5 Some Remarks: 2nd Order Knowledge It seems you only have a knowledge about the Auto Correlation of your data (2nd Order) and not on its distribution. Hence methods you can apply are ones which minimizes only 2nd order functions of the noise (Such as MMSE). Wiener Filter Minimizes the MMSE As can be seen in the derivation of the Wiener Deconvolution Filter ... 5 The question really depends on  f \left( \cdot \right) . Yet in order to show how to use FFT we can even use 1D signals. Let's rewrite the problem:$$ \hat{x} = \arg \min_{x} \frac{1}{2} \left\| K x - b \right\|_{2}^{2} + \frac{\lambda}{2} \left\| f \left( x \right) \right\|_{2}^{2} $$The derivative is given by:$$ g = {K}^{T} \left( K x - b \right) + ...

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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 ...

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All 3 of them fall into the category of Inverse Problems in the Image Processing world. Lets assume a Linear Model and then we will be able to show all 3 of them as parameters of the same framework. Then the differences will be clear and one could generalize it into Non Linear settings as well. Have a look on the following model: $$y \approx A x + w$$ ...

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Most of the information is given in my answer to 1D Deconvolution with Gaussian Kernel (MATLAB) (Which is related to Deconvolution of 1D Signals Blurred by Gaussian Kernel). Model The least squares model is simple. The objective function as a function of the data is given by: $$f \left( x \right) = \frac{1}{2} \left\| h \ast x - y \right\|_{2}^{2}$$ ...

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Let me present the following Diagram: So, both Deblurring and Deconvolution are operations within the family of Image Restoration (Which is a subset of Inverse Problem set). Basically we build the Image Restoration set by different Degradation Models. The one related to the question are: Linear Degradation Model Namely, the degradation is made by a Linear ...

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Usually Tikhonov Regularization is applied in the following form: $$\arg \min_{\boldsymbol{x}} \frac{1}{2} {\left\| A \boldsymbol{x} - \boldsymbol{y} \right\|}_{2}^{2} + \frac{\lambda}{2} {\left\| \boldsymbol{x} \right\|}_{2}^{2}$$ This formulation can be seen as: MAP Estimator with the prior of $\boldsymbol{x} \sim N \left( 0, {\sigma}_{x}^{2} \right)$....

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This is a nice question. The math is actually pretty simple once you embrace the method I derived the Wiener Filter in - How Is the Formula for the Wiener Deconvolution Derived? So, here is the model: $$\boldsymbol{y}_{i} = \boldsymbol{h}_{i} \ast \boldsymbol{x} + \boldsymbol{w}, \; i = 1, 2, \ldots, n$$ Where $\boldsymbol{w}$ is an additive white ...

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Think of it in the frequency domain. Some of the data is highly damped. Namely energy in High Frequency is damped with low values. Yet, unless it is zero identically "There is hope...". Basically, this is the idea behind deconvolution algorithms. Restore the energy in frequencies before it was damped. Practically, we are limited in DR. Namely if it is ...

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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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You can always use the Least Squares Solution: $$\arg \min_{x} \frac{1}{2} \left\| A x - y \right\|_{2}^{2}$$ Now, in cases $A$ isn't full rank you won't be able to solve it using the Normal Equations. What you should do (And always works for Least Squares Solution) is use the Pseudo Inverse of $A$. In MATLAB it will go like this: vX = pinv(mA) * ...

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Another approach (Though the same). Let's assume we're in a Finite Dimension Space and the convolution operator is the Circular Convolution (Namely, the same we would do using the DFT). Then $y = h \ast x$ can be written as: $$y = H x$$ Where $H$ is a circulant matrix. Now, the Deconvolution Operator is basically the Inverse of $H$. Yet, since $H$ ...

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If we can assume no noise (Or the SNR is very high) you can get the response by applying the inverse filter in frequency domain. Lets say $y [n]$ are the signal samples. Given $x [n]$ the samples of the ideal signal you can apply on both of them the DFT to get $Y [k]$ and $X [k]$. The response is given by the Inverse DFT of the division \$ \frac{Y[k]...

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One way to do it is to solve a MAP problem of the up scaled video and using the High Resolution images as a prior. Try looking at the articles - Super Resolution MAP.

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