# Tag Info

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The SVD Decompose the image into the (One way to look at it) many matrices. For instance, given an Image $I$ its SVD is given by: $$I = U S {V}^{T} = \sum_{i=1}^{\textrm{rank}(I)} {s}_{i} {u}_{i} {v}_{i}^{T}$$ As you can see, the image is decomposed into linear combination of matrices (Given by ${u}_{i} {v}_{i}^{T}$) where the weight of the $i$ -th ...

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Nilesh Padhi, Welcome to the DSP Community. The classic definition of separable means the data (2D) given by $X \in \mathbb{R}^{m \times n}$ can be written as: $$X = \sigma u {v}^{T}$$ Where $\sigma \in \mathbb{R}$, $u \in \mathbb{R}^{m}$ and $v \in \mathbb{R}^{n}$. This is called Rank 1 Matrix. How can you get those parameters and vectors ...

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Indeed you can do that. You may look on my answer to How to Prove a 2D Filter Is Separable? By the SVD for any filter $A$: $$A = \sum_{i = 1}^{n} {\sigma}_{i} {u}_{i} {v}_{i}^{T}$$ Since we're talking about separable filter then: $$A = {\sigma}_{1} {u}_{1} {v}_{1}^{T}$$ So the columns filter is $\sqrt{\sigma}_{1} {u}_{1}$ and the rows filter ...

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One way to interpret the Tikhonov Regularization is using the Maximum A Posteriori (MAP) framework. Lets' say we have a model of the form: $$\boldsymbol{y} = H \boldsymbol{x} + \boldsymbol{n}$$ Where $\boldsymbol{n} \sim N \left( 0, {\sigma}_{n}^{2} \right)$, namely Additive White Gaussian Noise, and the prior knowledge about $\boldsymbol{x}$ is $\... 5 Similar to The Concepts Behind SVD Based Image Processing the horizontal axis are the samples index of the SVD basis. The idea in the chapter you linked is generalizing the Wiener Filter. While the Wiener Filter uses the Fourier Transform as a basis the SVD uses the data adaptive basis. 4 [EDIT: some code made available] A common framework for (multivariate) image processing is to suppose that its useful features (edges, textures, spectral correlation) contain redundancy, while the noise is pretty uncorrelated. A major preprocessing task consists in decorrelating or compacting the image features prior to processing, hoping to separate them ... 4 That's not true, it's not better. The thing is: the matched filter just implements the projection in the signal vector space, onto the signal vector itself (or a multiple thereof). (You'll find correlation is just an inner product in that space.) The line through that vector is the signal subspace, the plane to which that vector is normal is the noise space. ... 4 If the data is stationary and the noise is white then you should use the Wiener Filter. If data isn't stationary you should look into the family of adaptive filters (LMS Filter and RLS Filter for starter). 4 This is an example of the Fidelity Term and Prior Term model. In many Inverse Problems we assume some model on the additive noise. This part is modeled by the Fidelity Term ($ \mathcal{D} \left(A \boldsymbol{f}, \tilde{\boldsymbol{g}} \right) $in your example). For Gaussian Noise it is given by Least Squares Term: $$\frac{1}{2} {\left\| A \boldsymbol{f} - \... 4 You may think on the SVD as a generalization of the Discrete Fourier Transform. Namely, it is generates an orthogonal basis to represent the data. The nice thing about it, it generates the basis according to data (Where the Discrete Fourier Transform basis is the same for any data). Just like the Fourier Spectrum, you have the "Energy" - The ... 3 A first difficulty with your question is that "usual instances" may not be fully publicized nor documented. A second one is related to the fact that "compression standard" are indeed "decompression procedures". In other words, they offer several options to compress a file, and make sure anybody could decompress, given the ... 3 The two methods differ, above all, by their applicability to matrix classes. col (cholesky) decomposes Hermitian, positive-definite rectangular matrices into the product of a lower triangular matrix and its conjugate transpose; svd (singular value decomposition) factorizes any m×n matrix into the form UΣV*, where U and V are square real or compex unitary ... 3 I will try to illustrate why it is important to remove the Mean from data when doing the PCA (SVD is the tool to basically do the PCA approach to dimensionality reduction -> Compression). If the data isn't centered you will have one (Most probably the most dominant) component (Direction) which will point to the center of the data. Since all directions must ... 3 I found an answer which is good-enough for me. As @Stanley Pawlukiewicz has pointed out in the comments, this is hard to do for a general case when there is little correlation between the images. I, however, want to work with real images of actual things in the real world. This means there will be a lot of low frequency components (that's why jpeg ... 3 Don't have enough reputation to comment but you need to use the conjugate transpose in your formula for the result to be correct. So try stftb=U*S*V'; in the last line of code. Note that I removed the . which makes a difference since the matrices you are working with are complex. 3 The typical thing to do is the low-rank approximation on separate channels. Assume that C is a channel of the RGB image I: rank = 10; [U,S,V] = svd(C); L = U(:,1:rank) * S(1:rank, 1:rank) * V(:, 1:rank)'; Now, L should be the compressed image. If you do this operation and compose the channels back, you should get a compressed RGB image. However, such ... 2 I'm not sure I understand the question, but if you have the exact waveform you want to recover, you can basically employ a matched filter to detect the existence of the signal in the acquired data. This does not really constitute denoising, but if you have full knowledge of the signal and the noise functions, I'm not sure how denoising would be useful, as ... 2 Assuming the simplest case with a square image x[n,m] of size N \times N and a square filter kernel h[n,m] of size M \times M, the raw 2D convolution to produce the, cropped, output image y[n,m] of size N \times N requires about N^2 \times M^2 MACs. The raw 2D convolution between x[n,m] and the filter h[n,m] is implemented by using two ... 2 It has been said that the DCT reasonably closely matches the KLT for a representative set of images. KLT is essentially the same as PCA, I believe and SVD is only a different way to compute the same? An image coder that computes the DCT, sorts the result, then truncate could perhaps be said to be an approximate «truncated SVD»? Usually, you dont sort ... 2 You can rewrite the equation in a block matrix form: \begin{eqnarray}X &=& U_s \Lambda_s V_s^H + U_n \Lambda_n V_n^H \\ &=& \left[ \begin{matrix} U_s & U_n\end{matrix} \right] \left[ \begin{matrix} \Lambda_s & 0 \\ 0 & \Lambda_n\end{matrix} \right] \left[ \begin{matrix} V_s^H \\ V_n^H\end{matrix} \right] \\ &=& U_t \... 1 A simple approach is to just take the last value of your time series and keep repeating it. If I repeat the last value 100 times, then I still get the large drop at the end, but the repetition means the end data is not as affected. Zooming in on the part that is not repeated: Python code import numpy as np import pandas_datareader as pdr from datetime ... 1 The reason x[n] must be white is because the solution will effectively spectrally weight the channel response based on the amount of energy present in each spectral frequency location. A white noise source provides equal weight to all frequencies. If energy is not present in any particular frequency bin, a proper solution cannot be found for that frequency. ... 1 If noise is available and it is sure that noise will remain same/almost same for entire duration then one can go for Spectral Subtraction or Wiener filter techniques for noise reduction, which will take advantage of prior knowledge of noise and clean signal. 1 What you're describing is a spectrum estimation based on signal representation in a vector space spanned by eigenvectors of a particular matrix generated from your signal observation. That reminds me extremely much of MUSIC, and ESPRIT. I think the best way to go here is to read up on spectral estimation using MUSIC, which is based on the idea that there's ... 1 You can write$$ R=YY^H $$where Y is a matrix of size N\times N_f and N is the dimension of y_k. Y contains all the measured y_k as its columns. Then, the rank of R is upper bounded by N_f. In particular, if N_f<N, R will always be a singular matrix. So, if you have too few measurements, you will likely run into the problem you ... 1 The piece I was missing was the distribution of the initial phase values \varphi_1 and \varphi_2. It is standard to assume that these are uniformly distributed [^]. This leads to:$$ \mathbf{R_x} = \mathbf{E[xx^H]} = \mathbf{S \Lambda S^H}$$where$\mathbf{\Lambda} = \begin{bmatrix} A_1^2 & 0 \\ 0 & A_2^2 \end{bmatrix}$and$\mathbf{S} = \begin{...

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