15
An image "should not be blurred using a Gaussian Kernel" in general. This can be a safe bet for a lot of basic image processing needs, and a smoothing is almost mandatory when you want to control the information lost by the downsampling.
Blurring is (often, not always) another word for low-pass filtering. When an image contains high-frequency ...
14
Noise is random, but like most random phenomena, it follows a certain pattern. Different patterns are given different names.
Consider rolling a die. This is clearly random. Roll the die 1000 times, keeping track of each result. Then, calculate the histogram of the result; you'll find that you got each of 1, 2, 3, 4, 5 and 6 approximately the same number of ...
13
Assuming a channel whose input at each time is a continuous random variable $X$ and its output is $Y=X+Z$, where $Z\sim\mathcal{N}(0,N)$ and $Z$ is independent of $X$, then $$C_{\text{CI-AWGN}}=\frac{1}{2}\log_2\left(1+\frac{P}{N}\right)$$ is the capacity of the continuous-input channel under the power constraint
$$\mathsf{E}X^2\le P$$
The mutual information ...
11
Starting at an even more basic level than the other (much smarter) answers, I'd like to pick up on this part of the question:
This seems contradictory to me as on one side it is random then on the other side their distribution is considered normally distributed.
Perhaps the issue here is what ‘random’ means?
To be clear: ‘random’ and ‘normally-...
9
The $BT$ product is the bandwidth-symbol time product where $B$ is the $-3\textrm{ dB}$(half-power) bandwidth of the pulse/filter and $T$ is the symbol duration. For different applications you will find varying recommended values. In GSM telephony for instance, a $BT=0.3$ is recommended. In satellite communications with GMSK, for near-earth missions the ...
9
Assume you have a $N\times M$ sized image.
If you know take what is classically used, a square filter kernel, of let's say size $L\times L$, you'd need to convolve that with the picture – which gives you $N\times M$ pixels, each needing $L^2$ multiply-accumulates. So you end up with $A_{2D}=L^2MN$ operations.
Now, if you can decompose that filter into an $...
8
According to (digital) sampling theorem, signals should be properly bandlimited, before they are (down) sampled.
A digital filter limits the bandwidth of the signal and makes it suitable for downsampling without aliasing.
A Gausssian kernel is very suitable as a lowpass filter, as it has a number of nice features. The Gaussian function is mathematically ...
8
normal distribution (i like to call it "gaussian") remains normal after addition of normally distributed numbers. so if gaussian goes into an LTI filter, a gaussian distribution comes out. but because of this central limit theorem, even if uniform p.d.f. random process goes into an LTI filter with a long and dense impulse response, what will come out tends ...
answered Nov 7 '18 at 2:26
robert bristow-johnson
16.2k3 gold badges29 silver badges68 bronze badges
8
The continuous Gaussian, whatever its dimension (1D, 2D), is a very important function in signal and image processing. As most data is discrete, and filtering can be costly, it has been and still is, subject of quantities of optimization and quantification/quantization schemes. In one 1D, the three most direct for a finite-length filter are illustrated below:...
7
It turns out that the rows of Pascal's Triangle approximate a Gaussian quite nicely and have the practical advantage of having integer values whose sum is a power of 2 (we can store these values exactly as integers, fixed point values, or floats). For example, say we wish to construct a 7x7 Gaussian Kernel we can do so using the 7th row of Pascal's triangle ...
7
The capacity formula
$$C = 0.5 \log (1+\frac{S}{N}) \tag{1}$$
is for discrete time channel.
Assuming you have a sequence of data $\left\lbrace a_n \right\rbrace$ to send out, you need an orthonormal waveform set $\left\lbrace \phi_n(t) \right\rbrace$ for modulation. In linear modulation, whom M-ary modution belongs to, $\phi_n(t) = \phi(t - nT)$ where $T$ ...
7
Short answer
10*log(bw/fs) to take into account the oversampling operation because the awgn() function specifies the signal-to-noise ratio per sample, in dB.
Longer answer
The discrete time AWGN model is
$$Y = X+N$$
where X is data from continuous time $X(t)$, N is noise sequence from AWGN process $N(t)$ and Y is receive symbols.
If $X(t)$ is ...
7
I'll try to clear one possible source of confusion. If picking each sample value from a single distribution feels "not random enough", then let's try to make things "more random" by adding another layer of randomness. This will be found to be futile.
Imagine that for each sample the noise is random in the sense that it comes from a distribution that is ...
7
Why Does 2D FFT of Gaussian Looks More Sharper than Gaussian Itself?
Have a look at the Fourier Transfrom of a Gaussian Signal.
$$ \mathcal{F}_{x} \left\{ {e}^{-a {x}^{2} } \right\} \left( \omega \right) = \sqrt{\frac{\pi}{a}} {e}^{- {\pi}^{2} \frac{ {\omega}^{2} }{a} } $$
First, Gaussian Signal stays Gaussian under Fourier Transform.
As you can see, the ...
6
You're correct, it has to do with the Cut Off frequency of the Gaussian Blur Filter in its Frequency Domain.
In order to see it, just apply a DFT (Using MATLAB it can be achieved by fft / fft2) and look on the absolute value.
Look for the -3dB point and you'll see.
There is also an intuitive explanation on the original article which say that blurring ...
6
Pixels outside the image borders must be extrapolated.
Now, you need to chose the model of your extrapolation.
For instance, if you're working within the Discrete Fourier model a periodic extrapolation is the native choice.
Usually the most "Eye Friendly" solution is replication based on "Nearest Neighbor".
Your choice is as valid as any choice, you just ...
5
Few notes:
First you should multiply the noise by the Standard Deviation (Root of the Variance for zero mean noise).
You can do that by multiplying the Lower Cholesky Decomposition of matrix by a column vector of Gaussian noise. Yet since you assume no correlation, you can do that by independent multiplication.
Something like:
vZ = vZ + [sigma_range; ...
5
The answer boils down to 2 issues with the practical approximations of the Gaussian Kernel:
Though the Gaussian Kernel is radially symmetric its discrete approximation has a rectangle support. Unless this support will have infinite length a rotation by any angle different from a multiplication of 90 degrees will yield a shape which has to modified to fit ...
5
Gaussian Kernel is made by using the Normal Distribution for weighing the surrounding pixel in the process of Convolution.
Since we're dealing with discrete signals and we are limited to finite length of the Gaussian Kernel usually it is created by discretization of the Normal Distribution and truncation.
I created a project in GitHub - Fast Gaussian Blur.
...
5
You should use the the standard formula:
s = randn(m, n) + 1i*randn(m, n);
And as pointed out by MBaz, the output should be scaled accordingly by $\frac{1}{\sqrt{2}}$
s = s/sqrt(2);
More on that topic you can find for example in a book Digital Media Processing by Hazarathaiah Malepati:
It should generate exactly what you want.
5
What the authors meant is to create a matrix of Weights, $ {U}^{\left( i, j \right)} $.
It is a matrix of the size of the image.
The given calculation is by the exponent of two terms (Each of them is a matrix, the calculation is element wise).
The final step is to blur it using a Gaussian Blue (2D Blur, like on an image).
In equation 10 you can see they ...
5
First of all let us assure that a Kalman filter (estimator) does not only remove Gaussian noise, but can remove (with certain success) any other type of noise as long as it's designed accordingly.
However, what lies at the heart of the standard Kalman filter is the linear estimator; and that linear estimator will be the optimum minimum mean square error ...
5
Colored Gaussian noise is by definition a wide-sense-stationary (WSS) process; that is, it has constant mean (all the random variables constituting the process have the same mean) and its autocorrelation function $R_X(t_1, t_2) = E[X(t_1)X(t_2)]$ depends only on the difference $t_2-t_1$ of the arguments. It is conventional to use $\tau$ to denote the ...
5
You can, but... you'll need to keep symmetry if your original time-domain signal is real-valued.
If a signal $x$ is real-valued, then its DFT $X$ will exhibit complex-conjugate symmetry:
$$
X[k] = X^*[N-k].
$$
So you can generate $N$ Gaussian pseudo-random noise samples, $g[n]$, and place them in the frequency domain noise vector, $\epsilon$ as:
$$
\...
5
I think Chris Luengo's answer is perfect.
The trick is that you can calculate the 2nd derivative of the image (Using Finite Differences -> Convolution) and then blur it with Gaussian Filter.
Since both are seperable kernels you can do that by 4 1D convolutions.
If the input image is given by I
Calculate Ixx (The 2nd derivative on x direction) using ...
5
With reference to $N_o$ this usually is the symbol for the power spectral density of thermal noise, where $N_o = kT$, where k is Boltzmann's Constant and T is the temperature in Kelvin. With regards to a complex baseband signal, the thermal noise signal is a complex, white Gaussian distributed noise, with half of the power in the real component and half the ...
5
In general, the characteristic function of a random variable is related to the fourier transform of the distribution as follows:
$$\varphi_Z(-\omega) = \mathscr F \big\{ f_Z(z) \big\}$$
Why?
Because, fourier transform of a PDF $f_Z(z)$ is:
$$\mathscr F \big\{ f_Z(z) \big\} = \int^{\infty}_{-\infty} f_Z(z)\cdot e^{-j\omega z} \ \mathrm dz$$
And ...
Only top voted, non community-wiki answers of a minimum length are eligible