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2

That is because correlation (and convolution) are not meant to "match" exactly a given pattern. They are multiplicative operators in their nature so they are strongly related to signal amplitude if you multiply the reference signal by N, the output gets twice bigger. for instance your operator will return a peak twice higher when encountering this piece of ...


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As it was already posted multiple times: The problem comes from an inaccurate definition of correlation in your application. The Pearson correlation coefficient does require the data to be centered, ie the mean must be subtracted normalized, ie the data must be divided by the standard deviation This centering and normalization must be done for the mask ...


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You picked a tough example. Short answer: Change your "0"s to another value, e.g., 2, and it should work much better. What's really happening: your signals are not zero mean, correlation requires to center the signals (i.e., subtract means). Example, since it's easer to understand in 1-D: say you want to find the pattern p=[0,2,2,0] in the sequence s=[2,...


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This might be hard with trying to illustrate this way but here I go. Take your original image and put the filter in the upper left corner like so: +-------+-------+---+---+ | 1 [0] | 1 [0] | 1 | 1 | +-------+-------+---+---+ | 1 [1] | 1 [1] | 1 | 1 | +-------+-------+---+---+ | 0 | 0 | 1 | 1 | +-------+-------+---+---+ | 1 | 1 | 1 | 1 | +----...


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Taking a step back, a standard image $I$ is composed of pixels. They are defined by a location (spatial coordinates, here denoted by $p$), and a "value", denoted by $I_p$. Smoothing or enhancing an image, in a large sense, consists in replacing each $I_p$ by $\hat{I}_p$, a value that would be: more probable, more consistent, more visually pleasant (choose ...


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I would start with the many resources on this site: Is the Bilateral Filter a Solution of Some Variational Method? How to Validate Bilateral Filter Implementation? Comparison Between Guided Filter (Edge Preserving Filter) and Gaussian Filter. What Is the Bilateral Filter Category: LPF, HPF, BPF or BSF? Understanding the Parameters of the Bilateral Filter. ...


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I would rephrase your terms as: the "directional derivatives" are not so directional (although sometimes called similarly in lecture, they are only horizontal and vertical. Truer "directional derivatives" would allow angular refinement, cf. non-separable filters (Deformable Kernels for Early Vision, Perona) the "directional derivatives" are not so ...


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A simple way to calculate contrast is by computing the standard deviation of the greyed image pixel intensities.


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Well, look at your original picture: it's constant for all points but the edges, which means your derivative is zero for all points but these edges. By applying a "rounding, smoothing" filter to it, you "smear" the edges enough to make the derivative be non-zero for multiple pixels, in every direction.


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My guess would be that making an image sharper translates into a spectrum that have more high frequency and less low frequency: i.e. the weighted median frequency would increase. Now you can define a relative measure of image sharpness. To get an absolute measure of image sharpness, you need to define what is maximum and minimum sharpness, and find a way to ...


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Some references on image sharpness metrics: Encoding Visual Sensitivity by MaxPol Convolution Filters for Image Sharpness Assessment, IEEE Transactions on Image Processing, 2019 A Fast Approach for No-Reference Image Sharpness Assessment Based on Maximum Local Variation, IEEE Signal Processing Letters, 2014 Image Sharpness Assessment Based on Local Phase ...


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I suspect you are aiming at what is known as eigenimage or eigenface analysis. A key step resides in computing the eigenvectors, often performing with preprocessing (e.g. mean subtraction) followed by vectorizing 2D images (aligning pixels in a 1D vector), before concatenating vectors and computing the eignevectors, with fast alternatives.


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It's been said so many times before that multiplying the Fourier transform coefficients of an image with a mask of ones and zeros in the frequency domain in order to achieve a short-cut to the ideal brickwall filtering in the time-domain is not a recommended method; and almost never preferred due to unavoidable, and unforeseen, errors it would introduce. ...


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@SOFUser, Welcome to our community. Since the High Pass Filter, such as the Laplacian Kernel, output has zero mean (As the DC Component is removed) it indeed can (Should) have negative values as the output oscillates around zero. What to do with it depends on the objective: Display Purpose In case on wants to display the output of the Sharpening / HPF ...


1

Richardson Lucy does not need to necessarily work in linear space. It works by minimizing a log-likelihood function, so as far as it is concerned it does not matter whether the data is an array of photoelectrons e, xe, (xe)^y, or similar, with x and y being constants: minimizing the log of any of those will result in the 'same' solution in e-, ADU (ADU = e- ...


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For most fonts, just digitally smear the image horizontally, then look at the height of the resulting gray and white bars. Assuming typical text the frequency of capitals, descenders and ascenders should be small enough to just contribute a bit of gray. A more sophisticated method would try to pick out the lines, then find the tips of the ascenders vs. the ...


1

There are different approaches to histogram equalization. Most typical approach essentially maps intensity levels $I_k$ into new ones $I_m$ based on a premise that new image would look better in contrast. This mapping is reversible assuming you keep simple a record of the associated intensity mapping; an array of 256 bytes for 8-bit images. Note that if ...


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The answer rely depends on the histogram equalization you are using. If in the process there is either differentiation, quantization, rebinning or clipping, some information will be lost. Its extends depends on data and data range.


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Brightness. You can find more detailed information in here: Bi-Histogram Equalization with Brightness Preservation Using Contras Enhancement


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The expectation of information is called entropy. The loss of information can hence be understood as difference in entropy between source and processed image, assuming no random effect was added. Together with my answer on why contrast is not an appropriate measure of entropy, this gives us the simple answer: The loss in information is simply the number ...


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Simple Description Imagine you're in a car that is traveling at 70MPH with cruise control. Because the cruise control isn't perfect, your actual speed might vary slightly. This imperfection is called "process noise". Now lets also imagine the car is being tracked using GPS. Because GPS isn't perfect, there will be some noise in the sensor reading. This ...


2

Perhaps an analogy might be constructive. Consider a submarine commander with a fat tanker in the cross hairs of his periscope. He needs to shoot his torpedoes, not at where the target is now, but at some place where the torpedo will intersect with the target. A skilled commander will have knowledge about how fast or slow the tanker can go. knowledge ...


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See: What is the relationship between a Kalman filter and polynomial regression? In over-simplified form, eyeball a line though a cloud of data samples, look where that line might point one sample into the future; and, when you get a new sample check, how good that estimate might have been; then redo, but optimize for a lot less arithmetic per step. A ...


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You can use cv2.PSNR like this example: import cv2 img1 = cv2.imread('img1.bmp') img2 = cv2.imread('img2.bmp') psnr = cv2.PSNR(img1, img2)


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KF is actually a mixture of a deterministic state propagator and a statistical estimator. Despite it's name including the term filter, Kalman filter is not a simple frequency selective one. It's indeed a statistical recursive estimator of a state of a (linear) dynamic system. Yet on a broader sense it's called as a filter as it will separate a desired ...


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Please read this document here. If I understood your question clearly, then you need to take out the mean from the image. After that probably you would find cross-correlation for the matching patch.


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