# Tag Info

7

The point is that circular convolution of two 1-D discrete signals can be expressed as the product of a circulant matrix and the vector representation of the other signal. The circulant matrix is a toeplitz matrix which is constructed by different circular shifts of a vector in different rows. For example, consider two signls $h[n]$ and $g[n]$, each of ...

4

If you want to solve for single value of $\lambda$ in the model: $$\arg \min_{x} \frac{1}{2} {\left\| A x - b \right\|}_{2}^{2} + \lambda {\left\| x \right\|}_{1}$$ Then you can use Coordinate Descent method which is the fastest and simplest and doesn't require any matrix inversion. I have a MATLAB code for in my ${L}_{1}$ Regularized Least Squares ...

4

Lets take the given matrix as $$A=\left[\begin{array}{cccccc}&0 &0 &0 &0 &0 &0 &\\ &0 &\color{red}{1} &0 &0 &0 &0 \\ &\color{red}{1} &\color{red}{1} &0 &0 &0 &0 \\ &\color{red}{1} &\color{red}{1} &0 &0 &0 &0 \\ &0 &0 &0 &0 &0 &0 \\ &... 4 An orthogonal matrix has orthogal columns, i.e. the scalar product of two different columns is zero (the case i\neq j). For the case i=j you have a_i^Ta_i=\|a_i\|^2>0. So, all you can say about an orthogonal matrix with colums a_i is:$$ a_i^Ta_j=\begin{cases}c_i>0 & i=j\\ 0 & i\neq j\end{cases}$$where c_i>0 is some constant, ... 4 Least Squares solution is always well defined for Linear System of Equations. In your case, which is under determined it means there are many solutions to the Linear Equations. The Least Squares solution has nice property, it also minimizes the  {L}_{2}  norm of the solution (Least Norm Solution) hence it is well defined. In practice, in order to solve ... 4 It seems to me that we can directly calculate ... which can be done in a few second in matlab. Who says that Matlab is calculating it directly, or that it isn't using Gershgorin's circle theorem in its algorithm? Can you look at the code to see what they're doing? Finding the eigenvalues of an n \times n matrix is tantamount to factoring an n^{th}-... 4 We do use eigenvalues, because they behave like invariants in linear systems (invariant inputs are well connected to outputs) and generally finding invariants of unknown or model systems provides a great deal of information on how they act on other inputs. Most problems solved using methods based on eigenvalues rely on some technical hypotheses, related to ... 3 In general, a kernel is a function that acts as a parameter to some algorithm. Filtering: For example, it's possible to call the impulse response of a filter h[n] a kernel, so that it is the parameter that defines the filter operation:$$ y[n] = h[n] * x[n]. $$The use of the term kernel in the filtering context is much more common in 2D filtering or ... 3 HINT If we have the diagonal matrix:$$ D = \left[\begin{array}{cccc} d_1&0&0&0\\ 0&d_2&0&0\\ 0&0&\ddots&0\\ 0&0&0&d_n \end{array}\right]$$Multiplying another matrix$$M_r = \left[\begin{array}{c} r_1\\ r_2\\ \vdots\\ r_n \end{array}\right]$$to the left with it multiplies each row like this:$$DM_r = \left[\...

3

Variance is defined as $V(x)=\frac{\sum_{i=1}^n(x_i-\mu)^2}{n}$. Just in case for you, mean $\mu$ is defined as $\mu=\frac{\sum_{i=1}^nx}{n}$. Covariance between two random variables $x$ and $y$ (or columns of a matrix) is defined as $Cov(x,y)=\frac{\sum_{i=1}^n[(x_i-\mu_x)(y_i-\mu_y)]}{n}$ and $Cov(x,x)=V(x)$. The term covariance matrix may be misleading ...

3

I would build a tensor of those matrices and use low rank or some thresholding methods on it. You may have a look for Tensor SVD.

3

The method used was a parametric representation of the matrix $H$. For intuition, think of rotation matrix in $\mathbb{R}^{2}$, namely $H \in \mathbb{R}^{2 \times 2}$. Since we have series of those (In case of video, a series in time) we can define $H \left( t \right)$. As a rotation matrix it can be parameterized by a single parameter $\theta$ ...

3

Let's say you have an $M \times N$ image. If you turn it into a vector in ${R}^{\left( M N \right)}$ and create a matrix which is $\left(M N \right) \times \left( M N \right)$ by multiplication and rearranging the vector from the multiplication result you can reposition any pixels where ever you want. Being practical, by padding the original image ...

3

It's the key point of array signal processing, I suppose. Say $x$ is the input vector of $[N,1]$ dimension collected from $N$ array sensors. $x(k)$ is its realization at the $k$ moment of time. By its definition covariance matrix (sometimes it's called autocorrelation matrix): $R = E[x\cdot x^H]$ , where $E[]$ is expectation operator and $x^H$ is Hermitian ...

3

MVDR is a narrowband beamformer. For broadband signals it is usually applied for each frequency bin. That means that $\mathbf{R}_{xx}$ is frequency dependent. In other words, for each time you should have $M$ matrices, each one is $3\times 3$. Now, since you usually cannot compute $\mathbf{R}_{xx}$ exactly, you perform covariance estimation $\tilde{\mathbf{... 3 You don't chose transforms by whether they are involutions or not. If invertibility is of interest, any simple form of inverse is sufficient. Useful transforms reveal structure of some sort or separate wanted from unwanted information. That said, there are plenty involutions in signal processing. Time inversion is one, polarity inversion is another one, as ... 3 Note that$\mathbf{P} _{k\mid k-1}$, just like$\mathbf{R}_k$, is also a covariance matrix, and for this reason it is (at least) positve semi-definite, i.e.,$\mathbf{y}^T\mathbf{P}_{k\mid k-1}\mathbf{y}\ge 0$for$\mathbf{y}\neq\mathbf{0}$. Now set$\mathbf{y}=\mathbf{H}_k^T\mathbf{x}$to see that also$\mathbf {H} _{k}\mathbf {P} _{k\mid k-1}\mathbf {H} _{...

3

It is not clear what are you asking but I will try answer both things. Deriving the Matrix Inversion Lemma The Matrix Inversion Lemma goes as: $${\left( A + U C V \right)}^{-1} = {A}^{-1} - {A}^{-1} U {\left( {C}^{-1} + V {A}^{-1} U \right)}^{-1} V {A}^{-1}$$ Deriving it is by utilizing these useful identities: \begin{align} U + U C V {A}^{-1} U &... 3 A sensing matrix maps input vector to measurement vector through linear wighted summation of input. What makes a specific matrix good, is application dependent. Now, both distributions more or less satisfy RIP. However hardware implementation of the Bernoulli matrix (binary or bipolar) is much much easier especially in analog domain. A Bernoulli wight is ... 3 The generalized eigenvalue problem is given byBw=\lambda Cw\tag{1}$$where \lambda is the generalized eigenvalue of the matrices B and C. Multiplying (1) from the left with w^H (with ^H denoting the Hermitian conjugate) and dividing both sides by w^HCw (assuming that this term is non-zero), we obtain$$\frac{w^HBw}{w^HCw}=J(w)=\lambda\tag{...

3

If $H$ is a matrix form of Circular Convolution then it is a Circulant Matrix. Being a Circulant Matrix means it can be diagonalized by the Fourier Matrix ${F}$: $$H = {F}^{H} D F$$ Where the matrix $D$ id a Diagonal Matrix with the Fourier Coefficients of $\mathcal{F} \left( h \right)$ on it main diagonal. Also pay attention that we use the ...

3

If you use the Euler's formula, you can simplify like this: $$[d]_{k,n} = \frac{\sqrt{2}}{N}\left( \cos{\left[ \frac{(k-1)(2n-1)\pi}{2N} \right]} e^{j\frac{2 \pi nk}{N}} \right)$$ I think we can't simplify more. PS: If you use your expressions of $A$ and $B$ and again the Euler's formula, you will get the same result.

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

Could it be that you are indeed looking for the closest orthogonal matrix $Y$? Then, there is a solution which involves computing the square root of $D^TD$ . If $E=(D^TD)^{1/2}$ were invertible, the solution would be its inverse. Yet, it is not invertible here. Then, there is a trick. If I remember well, you have to perform an eigenvalue/eigenvector ...

2

Well I will try to explain. Let us first discission in time domain: 1) Let us say you have two signals, x and y. By Convolution in time domain, you mean that you flip(invert) one of the signals(lets say ,y) and then slide it over the other(x). By corerlation, you just don't flip them, rest is the same. 2) Now when we talk about correlation, we first ...

2

For an overlapped STFT (overlap > 0), the output vector will be larger than the signal input vector, so the matrix form will not be square.

2

Try using bsxfun if your version has it. A = bsxfun(@max, B, C.') As per their documentation, bsxfun expands the dimensions of the argument matrices on-the-fly, so does not use as much memory as repmat.

2

If the matrices are not too big, repmat could work: B = [2 3]; C = [0 1 2 3]; % Vectorize the vectors for a simplication B = B(:); C = C(:); A = max(repmat(B,1,length(C)),repmat(C',length(B),1)); For those interested, Comparing BSXFUN and REPMAT and Matlab - bsxfun no longer faster than repmat? address its relative efficiency with respect to repmat. ...

2

The function audioread doesn't generate any values, it just reads audio samples stored in a file. If you want to generate the sound of a guitar, you need to look into sound synthesis, such as the Karplus-Strong method.

2

The commonly called 5.1 format uses only surround channels, which are defined as rear/side channels in ITU-R BS 775. The case you want to deal with (turning rear surround channels to side surround channels) is therefore not explicitly defined. Notice that in the referenced ITU document, the case of changing the number of rear/sides loudspeakers reproducing ...

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