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What's the relationship between sigma and radius? I've read that sigma is equivalent to radius, I don't see how sigma is expressed in pixels. Or is "radius" just a name for sigma, not related to pixels?
There are three things at play here. The variance, ($\sigma^2$), the radius, and the number of pixels. Since this is a 2-dimensional gaussian function, it ...
19
You would generate bandlimited Gaussian noise by first generating white noise, then filtering it to the bandwidth that you desire. As an example:
% design FIR filter to filter noise to half of Nyquist rate
b = fir1(64, 0.5);
% generate Gaussian (normally-distributed) white noise
n = randn(1e4, 1);
% apply to filter to yield bandlimited noise
nb = filter(b,1,...
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 ...
11
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-...
10
I understand that a Laplacian-of-Gaussian filter can be approximated by a Difference-of-Gaussians filter, and that the ratio of the two sigmas for the latter should be 1:1.6 for the best approximation
In theory, the smaller the ratio between two sigmas, the better the approximation. In practice, you'll get numerical errors at some point, but as long as you'...
10
It should not, the need really depends on your application. However, this is a safe bet for most needs, and almost mandatory when you want to control the information lost by the downsampling.
Blurring is often another word for low-pass filtering. When an image contains high-frequency content (fast variations), downsampling can produce visually weird or ...
9
Just as a small addition to Jason's answer: usually you need to generate bandlimited noise with a given variance $\sigma^2$. You can add this code to the code given in Jason's answer:
var = 3.0; % just an example
scale = sqrt(var)/std(nb);
nb = scale*nb; % nb has variance 'var'
Note that you have to do the scaling after filtering, because in general ...
8
The rightmost one (where you divide by the sum) ensures that the output of the filter wil have the same dynamic range as the input.
Actually, the output blurred images are not the same because their pixel values are different. You have the feeling that they are similar because your software (I assume Matlab) does remap the intensity values of the pixels to ...
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
12.8k3 gold badges20 silver badges53 bronze badges
7
For you questions:
1. After applying gaussian filter on a histogram, the pixel value of new histogram will be changed.
2. The sum of pixels in new histogram is almost impossible to remain unchanged.
Visually speaking, after your applying the gaussian filter (low pass), the histogram shall become more smooth than before. Thus, the new histogram is ...
7
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 ...
7
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 $...
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
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 ...
6
There are many. In general, steerable filters are not separable. Few are, but many are not.
Here is a short list:
Gabor filters in certain directions. Of course, one can use the bases that are separable due to the fact that Gabor filters are steerable. Gabor filters have the advantage that they are optimally localized in time and frequency. As a ...
6
The sum of a gaussian kernel cannot be zero, because all the elements are going to be positive. The first kernel you have shown, is most likely an edge detection kernel, (which is a type of high pass filter), so the elements add up to zero because you want to completely null out any DC/constant component.
The second kernel you have shown however, is a low ...
6
Since the DFT is representable by multiplication with the Fourier matrix, your question is equivalent to asking what the eigenvectors of the Fourier matrix are.
Actually, Wikipedia provides the answer (http://en.wikipedia.org/wiki/Discrete_Fourier_transform#Eigenvalues_and_eigenvectors).
However, since the eigenvalues ($1, -1, i, -i$) are not simple, the ...
6
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 ...
5
The filter you referenced is known as a Binomial filter. It is an approximation to a Gaussian, but for smaller filters it's a very crude one. The design is oriented more toward efficiency than accuracy. Rather than sampling a Gaussian directly, the idea behind the approximation is based on the Central Limit Theorem. In this case, it means that a small moving ...
5
The Deriche and van Vliet filters are heuristics. In both cases they choose the locations of poles and zeros to minimize either the RMS difference or the maximum difference of the filter's impulse response from a Gaussian.
Both filters are causal-anti-causal pairs. So I think they have no phase error or group delay, but you need to be able to run them ...
5
To elaborate on Dilip's answer (which is perfectly correct, though in practice the Ziggurat method is much more computationally efficient than Box-Mueller):
One key ingredient missing from your reasoning is whether you want your samples to be independent. It is not clear from your question, but this is the most common situation...
If you want your samples ...
5
The parameter sigma is enough to define the Gaussian blur from a continuous point of view. In practice however, images and convolution kernels are discrete. How to choose an optimal discrete approximation of the continuous Gaussian kernel?
The discrete approximation will be closer to the continuous Gaussian kernel when using a larger radius. But this may ...
5
Well, let's look at the two issues: 1) linearity and 2) Gaussianity.
Linearity
If you're imaging moving 3D objects (people) with a single camera, then you're working with a 2D projection of those 3D objects. That dimensionality reduction can cause non-linearities to appear.
Take a 2D to 1D example: an object moving in a circle in 2D. The object is ...
5
In addition to Peter's answer, if you have a nonlinear system that is well-behaved in a sense of being only mildly nonlinear or at least exhibiting no discontinuities, special variants of the Kalman filter can still be applied.
Extended Kalman Filter
This filter linearizes the system at the current state of the system using a first order Taylor Series ...
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
According to (digital) sampling theory, 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 filter is very suitable as a filter, as it has a number of nice features. The Gaussian function is mathematically tractable. ...
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