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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 ...


5

Simplest reason for downsampling is simply to reduce the amount of data you have. Say you have a audio raw stream sampled at 44.1 kHz. You may simply want to reduce your data rate to 22.05 kHz because your algorithm can work with that sampling rate. Apply the same principle anywhere else. Just because you have more data does not mean you always need it. It ...


4

The name of this effect is Spectral Leakage. Remember the relation $$\frac{k}{N} = \frac{F}{F_S}$$ where $k$ is the bin number, $N$ is the FFT size, $F$ is the continuous frequency in Hz and $F_S$ is the sample rate in Hz. It can be seen that $k$ varies from $-N/2$ to $N/2-1$ (or from $0$ to $N-1$). So there are only $N$ continuous frequencies $F$ for which ...


4

The up-sampling process will always change the signal in some measurable way. However, if it's done properly the changes are negligible and don't result it any audible difference. Most commercially sample rate converters (hardware or software implementations), do a really good job at this. Off course, if done badly, upsampling can result in clearly audible ...


4

The problem with downsampling is that it can be lossy -- since you're reducing the sampling rate, you can introduce aliasing. So, you can reverse the order whenever downsampling does not result in aliasing. For example, say your discrete-time signal $x[n]$ contains energy in frequencies up to $f_N/3$, where $f_N$ is the Nyquist frequency. Then, downsampling ...


4

Sampling at a higher rate will distribute the quantization noise over a wider frequency, thus reducing the noise spectral density due to that quantization noise, with a lot of caveats. For more details on that see What are advantages of having higher sampling rate of a signal? In your last paragraph, if you are referring to running the same filter at a ...


4

When discussing DFT, you have to remember two things: you're windowing your true signal $x[n]$ (which is periodic, and then infinitely long) with a window $w[n]$ (here a rectangular window) to obtain a truncated version $x'[n]$ of it $$x'[n] = x[n]w[n].$$ you're sampling the Fourier Transform $X'(e^{j\Omega})$ of your signal at normalized pulsation $\...


3

Think of it this way: You have a 10Hz signal. You try to make a 10Hz signals with some building blocks. 1) Your building blocks are this signals: 1Hz, 2Hz,.. 10Hz, .. 20Hz. Quite easy right? You just make your 10Hz signals using the 10Hz "building block". 2) Your building blocks are: 0.6Hz, 1.2Hz, .. 9.6Hz, 10.2Hz, ... Now what can we do? Well.. the ...


3

A common reason for upsampling is rate matching, for instance mixing two signals with different sample rates, or sending different lesser rates of audio to a player that only plays at 44.1kHz. And you are correct that downsampling does potentially destroy information in a signal (assuming there was spectral energy in the portion that has to be low-pass ...


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

You should go to see wavelet transform, which uses the filter-bank structure to process signals. In a forward wavelet transform, it decompose a signal using down-sampling techniques; and in a inverse wavelet transform, it reconstructs a signal using up-sampling techniques. For a quick explanation, here is an example. Suppose you have a signal $S$ with a ...


2

HINT: Note that if $$Y(z)=X(z^2)$$ then obviously $$Y(z^{1/2})=X(z)$$ but also $$Y(-z^{1/2})=X((-z^{1/2})^2)=X(z)$$ from which the desired result follows in a straightforward way.


2

You can use a multirate structure where the stages have increasingly wider transition bands. This is very close to the situation answered in another question


2

Upsampling by 100 can be factored into upsampling by 2X, 2X, 5X and 5X for each output block (output blocks sized to be data cache resident), thus requiring a smaller phase table for each FIR interpolation kernel. Some output points may either need to be overlapped across blocks or re-calculated, depending on computation speed vs cache load latency.


2

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

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

Let's say you've sampled an analog signal $x(t)$ with spectrum $X(\omega)$ at rate $1/T$ that is high enough to satisfy the sampling theorem. The spectrum (i.e. the discrete time Fourier transform (DTFT)) of the sampled signal $x_{1,k}$ will be a periodic repetition of $X(\omega)$. The repetition period is $1/T$. Now you sample the same signal with a higher ...


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.


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