# Tag Info

3

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.

3

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 ...

2

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

2

If I understand you correctly, what you are asking for is called camera response function (CRF) and in general it is nonlinear and depends on camera device. For fixed device denote its CRF by $f$. Then $f$ maps the set of possible scene irradiances (or illuminances) $\mathcal{I}$ at a given spatial location to the set of possible pixel intensity values $\... 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 ... 2 Brightness. You can find more detailed information in here: Bi-Histogram Equalization with Brightness Preservation Using Contras Enhancement 2 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 ... 1 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 ... 1 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 ... 1 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,... 1 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  | 1  | 1 | 1 | +-------+-------+---+---+ | 1  | 1  | 1 | 1 | +-------+-------+---+---+ | 0 | 0 | 1 | 1 | +-------+-------+---+---+ | 1 | 1 | 1 | 1 | +----... 1 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. ... 1 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 ... 1 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 ... 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 ...

1

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.

1

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 ...

1

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 ...

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- ...

1

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)

1

turn to float first!!!!!!!! turn to float first!!!!!!!! turn to float first!!!!!!!! def compute_psnr(img1, img2): img1 = img1.astype(np.float64) / 255. img2 = img2.astype(np.float64) / 255. mse = np.mean((img1 - img2) ** 2) if mse == 0: return "Same Image" return 10 * math.log10(1. / mse)

Only top voted, non community-wiki answers of a minimum length are eligible