# 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 ...

9

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 ...

7

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. $$\mathscr{F}\Big\{ r_{xx}[n] \Big\} = \mathscr{F}\Big\{ x[n] \Big\} \cdot \mathscr{F}\Big\{ x[-n] \Big\}$$ $$R_{xx}(\omega) = X(\omega)\cdot X^*(\... 7 You cant't recover the original signal through deconvolution. A Gaussian kernel is in essence a lowpass filter, i.e. it will remove information at higher frequencies from the signal. Once it's gone, it's gone and you can't recover it. This problem shows up as "divide by zero" or "divide by a very small number", which then amplifies numerically noise of ... 7 A necessary condition for invertibility is that any output has only one possible input (or injectivity, as proposed in comments). Since we are looking at counterexamples, we can look at when this condition is not satisfied. The null system, that turns every signal into a zero flat line, is not invertible, but a bit trivial. A system that computes a ... 6 The Cocktail Party Problem is a Blind Source Separation (BSS) problem. Given a linear mixture of signals:$$ \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

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 ...

5

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 ...

5

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 ...

5

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 ...

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 ...

4

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: $$X= \begin{bmatrix} x_1 & x_2 & ... & x_r \\ x_2 & x_3 & & x_{r+1} \\ x_3 & x_4 & & x_{r+2} \\ ... & &... 4 Approaches There are many methods for Deconvolution (Namely the degradation operator is linear and Time / Space Invariant) out there. All of them try to deal with the fact the problem is Ill Poised in many cases. Better methods are those which add some regularization to the model of the data to be restored. It can be statistical models (Priors) or any ... 4 The idea is to represent all operation sing Matrices. Once it is done, it is easy to solve the problems as a Least Squares problems. The way to represent Convolution Operation using a Matrix is by Toeplitz Matrix. For 1D it is pretty straight forward to do (Just pay attention to boundary). So let's take the simple model in the comment:$$ g = f \circ h + ...

4

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: $$g = {K}^{T} \left( K x - b \right) + ... 3 What you want is$$x(t) = (x(t)\otimes h(t))\otimes h'(t)$$where \otimes denotes convolution. Taking Z transform,$$X(z) = X(z) \times H(z) \times H'(z)$$or$$H'(z) = \frac{1}{H(z)} \tag{1}$$So you can deconvolve a convolution sum if you you have inverse transfer function as expressed in (1). For causal and stable system, the ROC of H'(z) must ... 3 If we can assume no noise (Or the SNR is very high) you can get the response by applying the inverse filter in frequency domain. Lets say  y [n]  are the signal samples. Given  x [n]  the samples of the ideal signal you can apply on both of them the DFT to get  Y [k]  and  X [k] . The response is given by the Inverse DFT of the division  \frac{Y[k]... 3 One way to do it is to solve a MAP problem of the up scaled video and using the High Resolution images as a prior. Try looking at the articles - Super Resolution MAP. 3 You can think of MAP as a regularization of the ML. Just like you have regularization for Least Squares Problem (They can be built, mostly, as MAP problem). The nice thing is that, as always, the best regularization is more data, namely, in most case when there is a lot of data they collide (Namely, low sensitivity fir the Posterior PDF). So they differ ... 3 The Wiener Filter can also be derived by another (Easier) way. Let's assume the following model:$$ y = h \ast x + n $$Namely the data is a result of a linear combination (Convolution) of  x  with Additive Noise. If we assume the noise model is Gaussian and our data is also formed by a Gaussian distribution then we should try to minimize (MAP ... 3 The more independent data you have, the more constrained are the possible solution sets that could produce that data, usually. If any higher frequency content in the possible solution sets is constrained to not be completely arbitrary (which data derived from sub-pixel shifted sampling might so constrain), then the solution sets could possibly becomes ... 3 In our days the Deep Neural Network methods certainly are generating best results. Due to the intense research going on in this field the best method is a moving target hence one can not pin point to one. One generation before them the best methods were based on Dictionary Learning. For example you can use the K-SVD for Single Image Super Resolution. Those ... 3 All 3 of them fall into the category of Inverse Problems in the Image Processing world. Lets assume a Linear Model and then we will be able to show all 3 of them as parameters of the same framework. Then the differences will be clear and one could generalize it into Non Linear settings as well. Have a look on the following model:$$ y \approx A x + w $$... 3 Most of the information is given in my answer to 1D Deconvolution with Gaussian Kernel (MATLAB) (Which is related to Deconvolution of 1D Signals Blurred by Gaussian Kernel). Model The least squares model is simple. The objective function as a function of the data is given by:$$ f \left( x \right) = \frac{1}{2} \left\| h \ast x - y \right\|_{2}^{2} $$... 3 In the context of image processing (and machine vision as well), blurring is an operation that reduces the sharpness of an image by some lowpass filtering applied on it. There are different causes of blurring such as lens blur, motion blur, or just LSI (linear shift invariant) lowpass filtering. Deblurring refers to any restoration performed on the image ... 3 Let me present the following Diagram: So, both Deblurring and Deconvolution are operations within the family of Image Restoration (Which is a subset of Inverse Problem set). Basically we build the Image Restoration set by different Degradation Models. The one related to the question are: Linear Degradation Model Namely, the degradation is made by a Linear ... 3 Whether LTI or not all systems are invertible if unique (distinct) inputs produce unique (distinct) outputs Causality and stability are later concerns for making sense of the obtained inverse system. For example the inverse to the delay system$$y[n] = x[n-d] $$is$$y[n] = x[n+d] $$Which is clearly noncausal for d > 0, and is not ... 3 In addition to all the answers that are correct in a mathematical sense, in a practical sense, a system whose frequency response goes below some finite but small-enough value will not be usefully invertable, even if a simple mathematical analysis would suggest that it is. In frequency-domain terms, the frequency response of a system's inverse will have gain ... 3 Your model is exactly a Convolution with Uniform Kernel where the output is what is called the Valid Part of the Convolution. In MATLAB lingo it will be using conv2(mA, mK, 'valid'). So the way to solve it will be using a matrix form of the convolution and solving the linear system of equations. Let's use the Lenna Image as input (Size was reduced for ... 3 The way I understand the problem is each sample of the output is a linear combination of the samples of the input. Hence it is modeled by:$$ \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 ...

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