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 ...


6

No. It's only LTI (Linear and Time-Invariant) systems that can be modeled with convolution through a unique single impulse response. For example the systems $$ y(t) = g(t) x(t) $$ or $$ y[n] = \sum_{k=0}^{k < n} x[n-k] $$ are both linear but not time-invariant and their output $y[n]$ cannot be computed with the convolution operation ( $\star$ denoting ...


3

Essentially, your code does not respect the inherent Hermitian symmetry of the output of the FFT. Here, your signal is odd-sized $2K+1$. Hence, this FFT yields a complex vector of coefficients $d$ (real) and $a_k$ (generally complex), arranged as: $$ \left[d,a_1,a_2,\ldots,a_K,\overline{a_K},\ldots,,\overline{a_2},\overline{a_1} \right]$$ If you want to ...


3

Any LTI system can be completely characterized (among other things) by it's transfer function or it's impulse response. If your filter represents an LTI system, that you can calculate it's output by either convolving the input with the impulse response or multiplying the transfer function with the spectrum of the input signal. In theory these things are ...


2

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 ...


2

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 ...


2

I'm not sure why the problem occurs for you with the x22 example, but I believe the problem is because the deconvolution is not being performed over the right length. Deconvolution of an FIR (inverting an FIR) requires, in general, convolution using an infinite duration impulse response. So just using the default lengths in performing the deconvolution ...


2

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: $$ f \left( x \right) = \frac{1}{2} \left\| h \ast x - y \right\|_{2}^{2} $$ ...


2

In general, one method to handle the issue that generalizes substantially to a problem of extracting two or more components is to take the spectra G¹, G² ⋯, Gⁿ of signals #1, #2, ..., #n, tabulate the total square Γ(ν) = |G¹(ν)|² + |G²(ν)|² + ⋯ + |Gⁿ(ν)|² at each frequency ν, and normalize G₁(ν) ≡ G¹(ν)* / Γ(ν), G₂(ν) ≡ G²(ν)* / Γ(ν), ..., G_n(ν) ≡ Gⁿ(ν)* / ...


2

The easiest approach would be writing each case using Matrix Form of the convolution. In this answer we assume the discrete convolution is applied only on valid support (Matching MATLAB's valid parameter for the convolution). Namely, given $ x \in \mathbb{R}^{m \times n} $ and $ h \in \mathbb{R}^{k \times l} $ then $ h \ast x \in \mathbb{R}^{ \left( m - k + ...


2

The convolution integral is a special case of the Fredholm equation of the first kind. https://en.wikipedia.org/wiki/Fredholm_integral_equation I believe that it covers linear time varying systems, as do linear time varying state space equations, so it’s a no but... kind of answer.


1

How can I justify the expression of G′(ν)? That's easy enough. Denominator is the sum of the signal energy $|X(\omega)|^2$ and the noise energy $\lambda ^2$. If the signal energy is significantly larger, then the whole expression simplifies to $G(\omega)$. If the noise is larger, we can't do a anything useful with the information and the $1/ \lambda ^2$ ...


1

To unveil part of the mystery, let us recall how the convolution operation and the properties of linearity and time-invariance are related. In other words, if a discrete system $\mathcal{S}$ is linear and time-invariant, what would be the output for a discrete signal $x[n]$? To do that, let us rewrite the signal on the basis of Kronecker symbols $\delta_n$,...


1

The problem is that the formula $$H(f) = \frac{Y(f)}{X(f)},$$ where $H(f)$ is the frequency response, and $X(f)$, $Y(f)$ are the input and output, is valid only for frquencies $f$ where $|X(f)| \neq 0$. Furthermore, if $|X(f_0)| \approx 0$, you're going to run into numerical errors, since $|H(f_0)| \rightarrow \infty$. Your input signal has a bunch of ...


1

A few comments: An overdetermined system (with more equations than unknowns) can have exact solutions. An overdetermined system can have approximate solutions; for example, in the least-squares sense, where $\mathbf{x}=(H^TH)^{-1}H^T\mathbf{y}$ minimizes $||H\mathbf{x}-\mathbf{y}||$. Deconvolution is usually performed in the frequency domain, where $X(f) = ...


1

Richardson Lucy does not need to necessarily work in linear space. It works by minimizing a log-likelihood function, so as far as it is concerned it does not matter whether the data is an array of photoelectrons e, xe, (xe)^y, or similar, with x and y being constants: minimizing the log of any of those will result in the 'same' solution in e-, ADU (ADU = e- ...


1

I think the answer is in Jim Kasson's Post (The Zone System and Digital Cameras) you linked to: If you are only moderately familiar with the Zone System, you may be asking yourself about the utility of recording light brighter than 100% reflectance. The 100% reflectance calibration assumes a matte subject with perfectly Lambertian reflectance. The higher ...


1

A memoryless time-invariant non-linearity, a waveshaper, has an input $W$ output $X$ relationship $X = f(W).$ To arbitrarily small error for a given range of input values, the function $f(x)$ can be thought to be a polynomial of sufficient degree. A polynomial $f(x)$ is a weighted sum of powers $x^n$ with nonnegative exponents $n$. The input $W$ to the ...


1

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 ...


1

Actually, the question is not clear. But the answers carified what you've asked for. You can build a system of linear algebraic equations as some people advice, that is correct, but the matrix built on known signal is so-called poorly conditioned. That means when you try to invert it, the truncation errors kill solution and you receive random numbers in ...


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