32
The meaning of that formula is really quite simple. Imagine you take two same-sized small areas of an image, the blue one and the red one:
The window function equals 0 outside the red rectangle (for simplicity, we can assume the window is simply constant within the red rectangle). So the window function selects which pixels you want to look at and assigns ...
15
Many years ago I wrote this tutorial on the Kalman filter. It derives the filter using both the conventional matrix approach as well as showing it's statistical assumptions as an 'optimal' least squares filter.
11
The "projection" that is referred to is a vector projection. To calculate the projection of vector $\mathbf{a}$ onto vector $\mathbf{b}$, you use the inner product of the two vectors:
$$
\mathbf{a_{proj}} = \langle \mathbf{a}, \mathbf{b} \rangle \mathbf{b}
$$
$\mathbf{a_{proj}}$ in this case is the vector component of $\mathbf{a}$ that lies in the same ...
8
I think you are constructing the convolution matrix incorrectly. Let's do an easy example with some numbers to see how it is constructed
Let's imagine that youhave the following vectors of dimensions $M$ and $N$, And let's for the sake of it, say that $M = 5$ and $N = 4$
$\mathbf{a} = [a_{1},a_{2},a_{3},a_{4}]^{T}$
$\mathbf{b} = [b_{1},b_{2},b_{3},b_{4},...
8
The book's formula is right.
Let $$H(w) = 1 - r e^{j(\theta - w)} = [1-r \cos(\theta - w)] + j [-r \sin(\theta - w)]$$ Since the group delay $\tau$ is the negative of the derivative of the phase of $H(w)$, we first define the phase as:
$$\phi(w) = \tan^{-1}\left( \frac{-r \sin(\theta - w)}{1-r \cos(\theta - w)} \right)$$
Using the derivative rule for the ...
7
Slope from all samples obtained
To summarize the question's problem, you want to calculate the slope based on all samples obtained thus far, and as new samples are obtained, update the slope without going through all the samples again.
On the page you cite is the equation for calculation of the slope $m_n$ that together with $b_n$ minimizes the sum of ...
6
I would like to recommend a great book to you (see below), but before that I would like to point out that there is some confusion in your question. If you see a convolution integral in a textbook, then this refers to the convolution of two continuous time signals, which can't be represented by a matrix/vector multiplication. You must be referring to the ...
6
Hint:
According to Euler's formula we have
$$e^{-j2\pi k}=\cos(2\pi k)-j\sin(2\pi k)=\ldots$$
6
This is a slightly tedious but nevertheless straightforward exercise in computing the derivative of a function:
$$\begin{align}\tau(\omega)&=-\frac{d\phi(\omega)}{d\omega}=-\frac{d}{d\omega}\arctan(f(\omega))\tag{1}\end{align}$$
with
$$f(\omega)=\frac{r\sin(\omega-\theta)}{1-r\cos(\omega-\theta)}\tag{2}$$
From $(1)$ we have
$$\tau(\omega)=-\frac{f'(\...
5
Basic linear algebra. M is a symmetric matrix, so there's a decomposition:
$M=Q^{\mathsf{T}} \Lambda Q$
where Q is a rotation and $\Lambda$ is a diagonal matrix.
We can rotate $[\text{$\Delta $x},\text{$\Delta $y}]$ by $Q^{\mathsf{T}}$
$Q^{\mathsf{T}} [\text{$\Delta $x},\text{$\Delta $y}]^{\mathsf{T}}=\left[\text{$\Delta $x}',\text{$\Delta $y}'\right]^{\...
5
Remember that $e^z$ has a very different meaning than $e^x$ (taking $z\in\mathbb{C}$ and $x\in\mathbb{R}$).
If the exponent was real, then, as you state in your question:
$$e^x = 1 \iff x=0$$
However, when the exponent is complex, this function acquires a very different meaning. Let $z=x+jy$, where $x,y\in\mathbb{R}$ and $j$ is the imaginary unit. Then
$$...
5
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 ...
4
I'll give you a hint: You need the probability distribution of the alphabet (English and Turkish). For English see the Wikipedia "Letter Frequency" page. Now it's possible to compute the Entropy of this distribution.
Note that this is a rough approximation (and an upper bound) since in practice, the letters of a word (and a sentence) are correlated. So, for ...
4
First of all, the minimum norm least square solution is $A^+b$, where $A^+$ is the pseudoinverse. Only when the left inverse $A_L^{-1}$ (or right inverse $A_R^{-1}$) exists, you have $A^+=A_L^{-1}=(A^TA)^{-1}A^T$ (or $A^+=A_R^{-1}=A^T(AA^T)^{-1}$).
Likewise, the projection matrix onto the column space is $P=AA^+$, or $P=AA_R^{-1}=AA^T(AA^T)^{-1}=I$ if $A_R^{...
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:
$$
X= \begin{bmatrix}
x_1 & x_2 & ... & x_r \\
x_2 & x_3 & & x_{r+1} \\
x_3 & x_4 & & x_{r+2} \\
... & &...
4
It's not valid.
If any matrix $W$ is invertible (such as the DFT matrix is), then there's the inverse $W^{-1}$ with
$$\begin{align}
W^{-1}W &=I\\
&\implies\\
X &= Wx \\
&\iff\\
W^{-1} X &= W^{-1}W x\\
&= Ix\\
&=x\text{ .}
\end{align}$$
Now, the Discrete Fourier transform can be defined to be unitary, so that its inverse
$$W^{-1}...
3
I created a function to create a Matrix for Image Filtering (Similar ideas to MATLAB's imfilter()):
function [ mK ] = CreateImageFilterMtx( mH, numRows, numCols, operationMode, boundaryMode )
% ----------------------------------------------------------------------------------------------- %
% [ mK ] = CreateImageFilterMtx( mH, numRows, numCols, ...
3
Since $ \epsilon $ is a parameter you need to set why not trade it with another parameter you need to set to create an easily solvable problem (Relaxation of the Problem)?
You can transform the problem into the following form ($ {L}_{1} $ Regularized Least Squares):
$$ \arg \min_{x} \frac{1}{2} \left\| A x - z \right\|^{2} + \lambda \left\| x \right\|_{1} $...
3
This seems to be a nice write-up of the Kalman filter.
3
From here:
$$
(AB)(B^{-1}A^{-1}) = A (BB^{-1}) A^{-1} = A A^{-1} = I
$$
So $(AB)^{-1} = B^{-1} A^{-1}$ provided $A$ and $B$ are invertible.
3
When talking about image deconvolution, it is referring to the process of correcting an image for distortion caused by the optical properties of the camera or imaging system. This distortion is characterized by the point spread function of the optical system. The point spread function describes the response of an imaging system to a point source. For a non ...
3
How to make the $\ell_2$ norm of all columns and rows of an $n \times n$ matrix equal to $\sqrt{n}$?
HINT If we have the diagonal matrix: $$ 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
There's a very simple way to check controllability, indeed if you define the reachability matrix
$$
R = \begin{pmatrix}B & AB & \dots & A^{n-1}B\end{pmatrix}
$$
then the reachable subspace is the image of R. Hence to check complete controllability you just have to check that $R$ is full rank.
First, I think there's an error in the question, $B$ ...
3
I admit I did not really thought about it before. I hope my notations won't be too sloppy.
I assume that given an operator matrix $A(u,v)$, you can apply this operator as a transform on an image $I$, to obtain an image in a novel domain $J(u,v)$. For instance a Fourier kernel would give $$a_{m,n}(u,v) = \exp^{-2\pi \imath \left(um/M+vn/N\right)}\,.$$
For ...
3
The main advantage of OMP is that the residual is orthogonal to the current solution.
Let's say you select all $k$ columns from $A$ (also called atoms) at once and let us also presume that $A$ is an overcomplete basis (this is more or less the standard in OMP literature).
Now, with your method, if the atom that correlates the most with your measurements $...
3
Did you ever think about where $\pi $ came from? Watch out...
Let us first draw this weird function $e^{-2j\pi t}$ for several $t\in[0,10]$ (the little blue circles joined by line segments):
On one axis, the variable $t$, on the others the real and imaginary parts, respectively. It looks like an everlasting spring. Now draw the function at integers $k\in [...
3
Using the logarithmic derivative of the transfer function, as detailed in Julius O. Smith's Numerical Computation of Group Delay, the following computations seem to involve a little less of derivatives (and less risks of mistakes), which could be useful for more complicated frequency responses and related group delays (like rational fractions). And you can (...
3
To write what Laurent says in the comments a bit more fully, what you want to show is that if
$$f_1(x, y) = 56g_1(x,y)+93g_1(x−1,y)+92g_1(x+1, y)−57g_1(x, y−1)+555g_1(x, y+1) $$
and
$$f_2(x, y) = 56g_2(x,y)+93g_2(x−1,y)+92g_2(x+1, y)−57g_2(x, y−1)+555g_2(x, y+1) $$
and
$$f_3(x, y) = 56(g_3(x,y)+93g_3(x−1,y)+92g_3(x+1, y)−57g_3(x, y−1)+555g_3(x, y+1) $$
...
3
There are really great answers.
I will try to give the Sequential Least Squares approach which generalizes to any Linear Model.
Sequential Least Squares Model
We're after solving the Linear Least Squares model:
$$ \arg \min_{\boldsymbol{\theta}} {\left\| H \boldsymbol{\theta} - \boldsymbol{x} \right\|}_{2}^{2} $$
Now imagine that we have new measurement ...
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