# Tag Info

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Continuous wavelet transform is suitable for a scalogram because the analysis window can be sized and placed at any position. This flexibility allows for the generation of a smooth image in both the time in scale (analogous to frequency) directions. The continuous wavelet transform is a redundant transform because the analysis window can overlap. In fact ...

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First, you need to ensure that the pixel has minimal value in its 8-neighborhood. SIFT, SURF and other keypoint detectors filter these pixels in step called "non-maximal suppression". This is basically a necessary condition for second-order approximation we will use to determine sub-pixel location. If you imagine the scale space slice $I$ as a 3D surface, ...

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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 ...

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We must produce s + 3 images in the stack of blurred images for each octave, so that ﬁnal extrema detection covers a complete octave. For $s=3$ this means you will have $s + 3 = 6$ blurred images (the Gaussian images shown in the paper in Figure 1 on the left). Having $6$ Gaussian images will result in $5$ DoG images (shown in Figure 1 on the right). This ...

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The $\sigma$ parameter is both. The Gaussian function can generate a scale-space where $\sigma$ is the scale parameter. It doesn't mean the image is scaled, instead it is the scale at which the features are being evaluated. For example, with higher $\sigma$ the image is more blurred and therefore only larger image features contribute to the gradient ...

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Actually the down sampling has no role here. It is all based on a real simple equation: $$I = A + B$$ It is always enough to keep 2 terms of the 3 to restore completely and perfectly the information. So let's look on this: $${I}_{0} = \left( \left( {I}_{0} \downarrow \right) \uparrow \right) + {R}_{0}$$ So if we keep ${R}_{0}$ and we have $\left( ... 3 The$\sigma$decides the scale of objects being simplified. This is explained here: The size of the Gaussian filter: the smoothing filter used in the first stage directly affects the results of the Canny algorithm. Smaller filters cause less blurring, and allow detection of small, sharp lines. A larger filter causes more blurring, smearing out the value ... 2 I don't know if I completely understand your question, but I will have a go at clarifying the scale space, multi-resolution ocataves and why they are important for SIFT. To understand the scale space it is helpful to consider how you recognise images at different distances (e.g far away you may be able to distinguish the shape of a person. As that person ... 2 I'm answering the question the way I understood it - How can one find a similarity measure which isn't sensitive to scaling and shifting. An approach could be borrowed from the Computer Vision world by comparing Shift and Scale Invariant features between the two signals. I'm not sure it will work for measuring the quality of recovering signals but it ... 2 For better understanding of$\sigma$I would suggest reading some about the scale spaces. Some reading recommendations may be found here: http://www.csc.kth.se/~tony/earlyvision.html 2 There's an image processing algorithm called Retinex that uses scales to perform local contrast enhancement. Here's an OK presentation on the Multi-Scale Retinex algorithm for more detail. The image size doesn't change because you are performing convolution of a Guassian kernel with the image. In practice this is generally accomplished in the frequency ... 2 The definitions behind the concepts of multiresolution or multiscale may overlap somehow, and are sometimes used interchangeably. Let me provide the following distinction: resolution encompasses spatial discretization, while scale relates to a more continuous framework. The real world can be considered as continuous, and a true image$I(x,y)$would have ... 2 In scale space the square of width of blurring kernel considered as the scale (variance of Gaussian and not its Standard deviation because if you sequentially blur an image with 2 Gaussian kernel variance of effective blurring kernel will be sum of variance of each kernel) but in multi-resolution analysis the resolution defined by the scaling parameter. 1 A very simple example on a$2\times 2$image $$I_0=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ with (very crude Gaussian) low-pass: $$g=1/4\begin{bmatrix}1&1\\1&1\end{bmatrix}$$ yields a downsampled$I_1$after filtering, with only one pixel (out of 4): $$I_1=\begin{bmatrix}(a+b+c+d/4)\end{bmatrix}$$ It can be upsampled as: $$U(I_1) = I_1^\... 1 It's hard to give a lot of useful information without knowing the details of the signals you are getting from your hardware. Typically the output from any camera is an image, so you would need to decide what is wrong with the image before trying to come up with image processing techniques to solve that problem, you don't seem to have a clear problem ... 1 Provided that the stretch is simply linear and consistent across the image, then the displacement of the point is proportional to the magnification factor. For the X-Axis: At 100px width, the point was at 30 At 150px width, where would the point be? (Let's call this our unknown x).$$x=\frac{150}{100} \times 30 = 45$$Similarly, for the Y-Axis, ... 1 I've spent a bit of time searching for an answer to this as well. I suspect since the Gaussian kernel has infinite support there will always be a little bit of aliasing, as there will always be frequencies passed that are greater than 0.25 of the sampling rate. That is, in the upper half of the spectrum. So one would choose$\sigma$to keep the amount of ... 1 Thanks for the clarification. If you're currently doing plot(y) then you should do plot(x,y) where x = linspace(0,1000,length(y)); 1 One more thing to add is that an edge detection is usually defined more or less like from Wikipedia: Identifying points in a digital image at which the image brightness changes sharply or, more formally, has discontinuities. They have a nice illustration: Here, the edge is between$4$th and$5$th pixel (or, between$3$rd and$4$th, if you count$0$-... 1 I use factor 5-6 going from the Std of the Gaussian to the Kernel -> radius = ceil(6 * STD). Though it means more computation power. 1 Without exactly seeing your code it is hard to know for certain, but I suspect this is because you are not tracking your maxima well through scale space. The position of the maxima will move in scale space. At the smallest scales you get the 4 small blobs around each circle. These are due to digitization and possibly noise/irregularities in the boundary ... 1 Generating pyramids and searching for different scales are two different things. Pyramid generation is commonly referred as a technique to speed up, where each level of the template is sought in each corresponding level of the search image. If you satisfy a sufficient match score in the higher levels of the pyramid, then you could stop searching further, ... 1 Template matching is related to finding the maximum correlation between two signals, where one of the signals (the image being explored) is usually much bigger than the reference signal (the object being searched for). Obviously, the best correlation score can only be obtained when the size of the object in the image is equal to its size in the template (... 1 The basic idea in scale space is to parametrize the image value/function space with a single parameter so that one can localize or select interesting variations of the this parameter. The various images produced for different values of the parameter can represent different things, in image analysis they are simplifications of an input image. One can perform ... 1 I can only find this is the paper: As this graph shows, the highest repeatability is obtained when sampling 3 scales per octave, and this is the number of scale samples used for all other experiments throughout this paper. It might seem surprising that the repeatability does not continue to improve as more scales are sampled. The reason is that ... 1 Yes, only scale space is sufficient, but at some point when you are scaling it down, instead of creating new Gaussian filters, it's more efficient to just resize the image and use the same/old filters (ie, don't need to keep increasing sigma, but rather decrease image size) this has the same effect as just increasing the scale (σ^2 = scale) 1 The quantity$\frac{\partial D}{\partial \textbf{x}}$is a vector, since it is the derivative of the scalar function$D(\textbf{x})$w.r.t. all the elements of$\textbf{x}$. In the formula it is assumed that all vectors are column vectors, so in order to compute the dot product of the derivative$\frac{\partial D}{\partial \textbf{x}}$and the vector$\...

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