# Tag Info

4

You have mistyped the formula, replace this line sum = sum + y(i).*(cos((pi.*(2.*y(i)+1).*u(j))/(2*N))); with the one below, and it works fine. sum = sum + y(i).*(cos((pi.*(2.*u(i)+1).*u(j))/(2*N)));

3

For clarity, I would write this DCT as: $$F(u) = \alpha(u)\sum_{i=0}^{N-1}f(i)\cos\left(\frac{\pi u}{2N}(2i+1)\right)$$ We note that, with this 1-indexing of Matlab: $$y[i+1] = f(i)\,.$$ Then I would modify the inner limit (from $0$ to $N-1$ instead of $1$ to $N$): for i = 0:N-1 sum = sum + y(i+1)*(cos((pi*(2*i+1)*u(j))/(2*N))); end and you can remove the ...

2

Frequencies in this context just means how fast the intensities change in the greyscale images as we move in the plane of the 2-D image. As you might be knowing the edges or boundaries or outlines of objects in the images are composed of high frequency elements. Why? Because in order to show the outline of an object in the image, the intensities at the ...

2

Detail in images require higher frequency basis functions. The frequency in this case is measuring fluctuations in intensity as a distance is traversed. With a lot of detail, a lot of fluctuations, thus higher frequency. Tamp down the higher frequency and you lose detail, i.e. the image blurs. The dampening is measured (described) best on a log scale. ...

2

I'd try a very "tinkery" approach here: Erode the image, so that the black area is shrunk by a fixed radius of pixels from its border (say, 5px). Dilate the resulting image by the same amount measure the amount of difference between original and processed image. The idea is that something that is a locally convex border doesn't suffer through erosion (it's ...

2

Unless mentioned otherwise withing the context the classic interpretation of Second Derivative Gaussian Filter is indeed (a) in your question: $$L \left( x, y, \theta \right) = \cos \left( \theta \right) {g}_{xx} \left( x, y \right) + \sin \left( \theta \right) {g}_{yy} \left( x, y \right)$$

1

In the classic framework both the Smoothing and the Difference Filter are applied using Convolution. Since it is done using convolution it implies the operation is Linear Spatially Invariant (LSI). LSI operators can be applied in any order and the result will be the same. This is also a result of the commutativity property of the convolution operator. Let's ...

1

Natural images are not very stationary, but locally, their covariance is often modeled by a first- or second-order process. The goal is to find the eigenvectors of such matrices, with less cost than matrix diagonalisation. The original DCT (DCT-II) from Discrete Cosine Transform (1974) aimed at finding basis vectors that could approximately diagonalize the ...

1

Normally, monochrome images should be stored as images with a single channel. However, I have seen digital images with three channels containing the same values. This results in a displayed image looking grayscale. Before compression multichannel data, it is common to decorrelate the channels. For RGB, one often concert them to luminance and chrominances. ...

1

The size of the tag is an important parameter. But it is a secondary parameter. The key parameter that we are looking at here (and one that is not easy to estimate) is Signal to Noise Ratio (SNR). So, very very briefly: The bigger the tag, the more data we obtain from the tag, the better the position of the estimate will be. But, the "size of the tag&...

1

The reality (as in physical reality, the phenomenon) is that a pixel's "value" is determined both by what is happening along the X dimension and the Y dimension (in k-space). If you want to reconstruct an image you have to do it from **two spatial sinusoidal waves. This is represented in the $f[m,n] \cdot e^{-j 2 \pi (u m + v n)}$ part of the DFT. ...

1

in MRI, you don't start with the image as pixels in spatial domain, but with the k-space data! So, you don't "construct k-space from the image", you "construct the image from k-space", which takes a 2D inverse Fourier transform. And that's not "do it row-by-row"; that is "do it on all rows, then all columns of the result&...

1

If you don't want to use the PyPI package for bm3d, you can use ffmpeg and run the bm3d filter as an OS command- command="ffmpeg -i "+input_image_path+" -filter_complex bm3d=sigma=30/255:block=4:bstep=2:group=1:hdthr=10000:estim=basic /path/to/output/directory/output.png" os.system(command) This takes lesser computation time.

1

Easy problem. Use silhouette image label (bwlabel), region property (regionprops) and area open (bwareaopen) functions in MATLAB after defining the fraction of the blobs to remove as a function of the area of the largest blob, which is your object. The fragment of the code is as follows: alpha=0.05;%fraction of insignificant blobs in an image [labeledImage,...

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I find that pretty well-written; just as you can have time signal that has frequency properties, you can have a spatial signal that has frequencies. That's a pretty important concept in image descriptions and processing!

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Good approach to first do a rough calculation of the bandwidths you need. Couple of remarks on that: You forgot a factor of 3, that camera has three "color pixels" per image pixel Your use case screams "I should be using a commercial off-the-shelf USB camera"; don't engineer something very complex if it doesn't have a value proposition ...

1

If you are not interested in picking out the frame that bubbles start forming and since the transition is (clearly) from order (clearly formed shapes) to disorder (mostly noise), you can use a feature based approach. That is, characterise the whole image with a few numbers. An obvious feature here is image entropy. The typical (but not the only) way to ...

1

Bilinear interpolation can be viewed as traditional sample rate conversion (insert 0s between existing samples, convolve with a triangular kernel in each dimension). Somewhat unusually, the size of the (continous-time) triangular filter kernel is not scaled relative to resampling ratio before it is sampled; each sample is always the result of 4 neighbouring ...

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