8

One word for that technique is superresolution. Robert Gawron has a blog post here and the Python implementation here. Usually, this technique relies on each image being slightly offset from the others. The only gain you'd get from not moving between shots would be to reduce the noise level.


6

The objective is to derive a higher resolution image from a low resolution image. When performing this upsampling, inevitably, you have to fill in information that you do not have. The differences between the two approaches are in how you fill that information in. In the standard techniques an underlying model is assumed. So, for example, in nearest ...


4

The term super-resolution is used very loosely nowadays. But I think the following problem was the original idea, or at least, the one that made the term famous. Suppose you have a scene and several observations of that scene, i.e. several frames of a video with slight camera motion between them. Super-resolution algorithms combine those several observation ...


4

Intuitively, if You move the sensor $ N $ steps each at the size of $ \frac{1}{N} $ of its resolution you can get $ \times N $ more resolution. It is like a polyphase representation of the signal. Using estimation methods, any movement which is not an (Event with zero probability) integer multiplication of the resolution of the sensor, namely, fractional ...


3

Images are technically limited to the original pixel resolution. Objects in an image are spatially quantized at this resolution. However, an edge for instance may be localized "between" native pixels. Traditional tools like basic gradients often stick to that initial resolution. However, using additional information (multiple data, motion vectors, prior), ...


2

Sub-pixel means that instead of getting a location in the image in terms of $(x,y)$ coordinates at the pixel level, which means integer values for $x$ and $y$, the location is calculated to possibly give fractional pixel locations. For example, instead of getting $(10,12)$ as a location, the algorithm may give $(9.934, 12.1234)$. Suppose you have two ...


2

I have recently discussed image restoration with a team of one of the French forensics police departments. They have issues related to dirty/spider web nested lens, wrong illumation, occlusions related to poor camera location, very low frame rates in MPEG compression, shaking with low-res smartphone lens, proprietary video file formats. Problems are ...


2

Entertainment industry greatly exaggerates what video enhancement can do. Spatial and temporal noise reduction Noise reduction by Neat Video and MSU Denoiser Filter, which are somewhere at (according to a comparison from 2007) or near state-of-the-art for hard noise, takes advantage of that adjacent frames usually are of the same scene but with independent ...


2

One simple approach would be taking the mean square error (MSE) by using fitness_1 = mean((inputimage(:) - reconstructedimage_1(:)).^2) though, as your image size won't change, you can ommit the mean and use sum instead. fitness_1 = sum((inputimage(:) - reconstructedimage_1(:)).^2) In general you want to have some kind of distance measure. Look at ...


2

The more independent data you have, the more constrained are the possible solution sets that could produce that data, usually. If any higher frequency content in the possible solution sets is constrained to not be completely arbitrary (which data derived from sub-pixel shifted sampling might so constrain), then the solution sets could possibly becomes ...


2

It sounds like $W_k$ is just a matrix of 1's and 0's. When $W_k$ is a 1, then the corresponding range value in $D_k$ is used. If it's a 0, then that range value is not used. Similarly, $T_k$ should be mostly 1, and 0 when the corresponding $D_k$ values are unreliable. Does that tally with your understanding? If not, can you elaborate on your question ...


2

Broadly similar techniques are used in a variety of applications when it's necessary to capture more samples of an event than is possible with available equipment. The idea is really quite simple: an interpolation is just "stretching" but it does not provide any new data, which is why interpolation requires a low pass filter. By contrast, if you add ...


2

([a b c] .+ [d e f]).^2 = [a+d b+e c+f].^2 = [(a+d)^2 (b+e)^2 (c+f)^2] = [a^2+2ad+d^2 b^2+2be+e^2 c^2+2cf+f^2] I suspect that: The problem may be in your assumptions: "We should see three peaks in this case" sounds like a hypothesis that may be false. You may have a bug in your code, which might be why "We ...


2

The binomial theorem is not necessarily involved: the top waveform is simply pointwise raised to the 5-th power, as others have noted, and there are no missing peaks that should be seen. To illustrate, consider the following crude hand-drawn figure: I drew this in my simulation software, so it was automatically "digitized" as I drew it. Obviously, I am poor ...


2

All 3 of them fall into the category of Inverse Problems in the Image Processing world. Lets assume a Linear Model and then we will be able to show all 3 of them as parameters of the same framework. Then the differences will be clear and one could generalize it into Non Linear settings as well. Have a look on the following model: $$ y \approx A x + w $$ ...


1

In super resolution problem, the goal is finding sub pixel values based on multiple observations or a prediction.If we know values of black, red, blue and yellow pixels, there is absolutely no way to know even guess the middle pixel, if we do not have any priory knowledge. They are just four random values on a grid that may or may not have any relation to ...


1

then according to the binomial theorem, we should get 6 peaks Where does that statement come from and why do you think it applies here? Applying the binomial theorem we get terms that consists of Gaussian raised to different powers. However, when raising the power on a Gaussian, the peak stays where it is so there is no mechanism that would move it ...


1

If your image is modeled as an image which is noisy, blurry and heavily decimated the optimal thing to do is estimate the image given that model. The model is well defined in @Laurent Duval's answer. I'd remark that in most real world cases the blurring is spatially variant hence it can't be modeled by convolution (Well, it is a generalized convolution). ...


1

If H is representation of shift invariant operator then its a convolution operator. If you are implementing your algorithm in MATLAB, there is a intrinsic function with name convmtx2 which exactly do what you want. Algorithms which warp image don't use a linear operator (a matrix) to do the job (considering large size of this operator it's not practical, ...


1

In our days the Deep Neural Network methods certainly are generating best results. Due to the intense research going on in this field the best method is a moving target hence one can not pin point to one. One generation before them the best methods were based on Dictionary Learning. For example you can use the K-SVD for Single Image Super Resolution. Those ...


1

Another word is "stacking". It is used to reduce CCD noise, to increase focal depth (by stacking images that are focused slightly differently), to improve very low-light astronomical photos, and to obtain high dynamic range (HDR) from a series of normal range images. See http://en.wikipedia.org/wiki/Focus_stacking http://www.instructables.com/id/...


1

There are many algorithms for "edge-aware" upsampling. Not sure what Photoshop itself uses, but for example Alien Skin Blow Up plugin gives similar results and they use vectorization using triangulation. The triangles are carefully smoothed while keeping the upsampled image sharp enough. You can also take a look on NEDI (New Edge-Directed Interpolation), or ...


1

It's really nothing beyond Bi Cubic Interpolation with "Sharpening". http://www.lynda.com/Photoshop-tutorials/interpolation-settings/124096/140573-4.html They haven't updated their interpolation algorithms for ages.


1

You will get good frequency information from a periodogram if, as you say, you have enough data to resolve the frequencies to the desired resolution. It is important that your signal be band limited to 1/2 the sample rate of the data for your periodogram.


1

A periodogram created from a finite number of samples using an DFT is not really a "correct" frequency estimator, as much as it is a (leaky) frequency energy binning sieve, where the width of each bin is set by the DFT length and the window function applied. The width and spacing of each bin may or may not be narrow enough for you to consider an error of ...


1

The window shape chosen for your wavelet basis will generally only have a slight effect on frequency resolution. To increase frequency resolution more, increase a scale factor that is related or proportional to the window length for a given frequency. This will also decrease temporal resolution.


1

As I've said in the (copious) comments, the signal modeled using $$ y[t]=A_1(\sin(\omega_1 t+\phi_1)+A_2 \sin(\omega_2 t+\phi_2)+....+A_p \sin(\omega_p t+\phi_p) + z(t) $$ is stationary. Your statement i am not interested estimation of parameters,just i am concerned about spectral resolution seems like an oxymoron. I think what you mean is that you're ...


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