# Tag Info

8

I don't get your downsample step when you downsampled by factor $M$. Let me go from scratch with the spectrum visualization below, with time domain, continuous frequency domain and discrete frequency domain from left to right. When we reduce the sampling frequency by a factor $k$, the signal spectrum is copied to new replicas at $f_s/k$. The discrete ...

7

I don't have a real answer but I have the feeling that this result will help you out: Bernstein's inequality says that, if the signal $x(t)$ is bandlimited to $|f|\leq B$, then $$\left| \frac{\textrm{d}x(t)}{\textrm{d}t}\right|\leq 4\pi B \,\textrm{sup}_{\tau\in\mathbb{R}}|x(\tau)| ,\,\,t\in\mathbb{R}$$ where $\textrm{sup}$ stands for "least upper bound". I ...

6

If your signal is really as simple as $$x(t)=A\sin(\omega_0t)\tag{1}$$ with known $\omega_0$, and you have observations $y(t_i)$, which are noisy samples of $x(t)$ at known time instances $t_i$, then a simple solution would be the least squares estimate $$\hat{A}=\frac{\displaystyle\sum_iy(t_i)\sin(\omega_0t_i)}{\displaystyle\sum_i\sin^2(\omega_0t_i)}\tag{2}... 5 Observations I have used +1 and -1 in the sequence instead of your 1 and 0. With \alpha=1, the band-limited continuous function f_m(T) in your first two figures (with the above mentioned modification) is:$$f_m(T) = \sum_{k=1-m}^m \operatorname{sign}\left(\operatorname{sinc}(\pi k - \pi/2)\right)\operatorname{sinc}(\pi T-\pi k),\tag{1}$$where:$$\...

4

Some recent cell phone models use something like a Cirrus Logic CS42xx series audio IO chip, which seems to use a digital polyphase interpolation filter, a sigma delta modulator, followed by a switched capacitor DAC and low-pass filter. Sinc interpolation (or, given finite hardware, a polyphase FIR kernel similar to a windowed Sinc) is one high quality ...

4

The sampling theorem requires a perfectly bandlimited signal, bandlimited to below twice the sampling frequency. The problem with this is that only an infinite length signal (e.g. exists before the big bang) can be perfectly bandlimited. This is from the Fourier theorem regarding any domain with finite support. Thus all real-world signals are ...

3

I guess this is a straightforward non-linear optimization problem (to be solved with Newton variations, such as Trust-Region methods), where you don't even need to compute the Jacobian analytically. It appears to me that the optimization problem is written over $K_i$, and thus is the input to the cost function. To compute the cost, at each call to this ...

3

The key idea is that the random sampling approach enforces more constraints on the resulting signal than the uniform sampling approach does. The POCS (projections onto convex sets) algorithm used for the reconstruction of the randomly sampled signal is the key piece: it enforces: That the signal must be from this spectrum. That the signal is real-valued. ...

3

FFT and IFFT are linear operators, and as such, the results only make a lot of sense in a linear intensity space, not if indexed into a non-linearly mapped space.

3

The general mathematical framework for interpolation is approximation theory. I guess the most important result is that for signals with bandwidth limitation, you can have perfect reconstruction via $sinc(\cdot)$ convolution; the famous sampling theorem, that has been mentioned here several times. I guess it is equally well-known that it is not really ...

3

It could be useful in the field of OCR as a pre processing step. Think about badly scanned data. You'd like to convert it into binary image. So the first step would be applying some kind of thresholding. Since no thresholding is perfect, There will be some "Holes" / "Gaps" within the text. Closing those "Holes" / "Gaps" can be done using morphological ...

3

SLAM(Simultaneous Localization and Mapping) algorithms can be used to for 3D reconstruction. They offer solutions for both monocular as well as stereo cameras. With single camera they estimate depth with few images and reconstruct the scene. You can find some of the open source solutions here. Real time 3D reconstruction can done using ORBSLAM and it is ...

3

The introduction of this paper explains the difference and gives an example. In short: Image restoration techniques presume that data are acquired in the image space; that is, the raw data represent a corrupted version of the image scene. In contrast, images are not directly observed in reconstruction problems. Instead, projections of an image are ...

3

First, a warm welcome to SE! Basically, you have a calibrated 3D reconstruction problem. The typical approach follows a 5-stage pipeline: Identify 2D features in each image along with the associated descriptors. Algorithms such as SURF, SIFT or AKAZE are heavily used and are available in many vision libraries such as OpenCV. Match the extracted keypoints ...

3

Indeed the model for the Proximal Gradient Method (Also see Proximal Gradient Methods for Learning) is in the form of: $$F \left( x \right) = f \left( x \right) + g \left( x \right)$$ Where usually $f \left( x \right)$ is convex smooth function and $g \left( x \right)$ is convex non smooth function. Yet the model is quite flexible and you may define ...

3

Our goal is to obtain proximal operator of the following function $$g \left( x \right) = {\left\| x \right\|}_{1} + \operatorname{TV}(x).$$ The involved optimization problem for any $z \in \mathbb{R}^d$ is the following $$\text{argmin}_{x}\left\{g(x) + \frac{1}{2}\|x-z\|^2_2\right\}$$ Denote the following $$g_1(x) := {\left\| x \right\|}_{1} + \frac{1}... 3 Zero-order hold will result in a piecewise-constant waveform. Linear interpolation will result in a piecewise-linear waveform. If you want a piecewise-quadratic or piecewise-cubic or higher order polynomial interpolation, it will not appear much different from the original bandlimited waveform. 3 Build a basis set with your frequency and match your signal. It is straightforward linear algebra: C is portion of cosine S is portion of the sine U is a vector of ones (DC)$$ X = a C + b S + c U  X \cdot C = a (C \cdot C) + b (S \cdot C) + c (U \cdot C)  X \cdot S = a (C \cdot S) + b (S \cdot S) + c (U \cdot S)  X \cdot U = a (C \cdot ...

2

Alright, I'll try to take a shot at giving you an explanation of what's going on with the DWT. So in the CWT, basically what you are doing is you are generating the wavelet coefficients by convolving the signal with each scale and shift of the mother wavelet function. However, the problem with doing this is that when you analyze the low frequency components ...

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

There are some mistakes in the question and your description. When we sample a signal of frequency $f_m$ with a sampling rate $f_s$, the sampled signal contains the frequencies $f=f_m \pm nf_s$, where $n \in \mathbb{Z}$. Here the frequencies available in the sampled signal are calculated as, $f_m=14100 Hz$ and $f_s=400 Hz$. we know that, $f=f_m \pm ... 2 Yet another possibility is to zero-pad the signal sample vector, FFT it, rotate the phase of each FFT result bin linearly with the bin index, and IFFT a time shifted result. 2 Because the sampling frequency is above the Nyquist rate, the original signal can be written in terms of its samples$y(n)$: $$x(t)=\sum_{m=-\infty}^{\infty}y(m)\frac{\sin[\pi(t-mT)/T]}{\pi(t-mT)/T}\tag{1}$$ By setting$t=nT+T/2$you get from (1) $$z(n)=\sum_{m=-\infty}^{\infty}y(m)\frac{\sin[\pi(n-m)+\pi/2]}{\pi(n-m)+\pi/2}$$ and since$$\sin[\pi(n-m)+\... 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 I agree with Jim Clay's answer, but I think it is important to point out two things. First of all, there are no phase distortions due to the hold operation, just a simple delay of half a sampling interval. So nothing needs (and can) be done about the phase. Second, it is important to realize that the gain roll-off due to the sinc shape is relatively mild. ... 2 What you are describing is the distortion introduced by an ideal digital-to-analog converter (DAC) in the analog domain. Two things are typically done to reduce this distortion: Analog filtering Oversampling As you note in your question, the distortion is modeled, in the frequency domain, by a sinc rolloff. Increasing the sample rate before converting ... 2$x(t)$must be a band pass signal. Under certain conditions on the sampling frequency and its relation to the lower and upper band edges of the signal,$x(t)$can be sampled at a frequency that is lower than twice its maximum frequency, without introducing aliasing. Have a look at this article and at this answer to a related question. For the numbers given ... 2 First, since$t>0$in your case, you can write your function as \begin{equation} C(t)=1.6925\left(\exp^{-0.136t}- \exp^{-1.192t} \right) u(t) \end{equation} where$u(t)$is a unit-step function. Then, denote the FT of$C(t)$as$C(F)$, which is given by \begin{equation} C(f)=1.6925\left(\frac{1}{0.136+2\pi jf}- \frac{1}{1.192+ 2\pi jf} \right) \end{... 2 The Poisson equation may suffer from ill-posedness if the boundary condition is not suﬃcient to yield a unique solution to the problem. The most often used boundary condition is Dirichlet boundary condition, in which the values of$I(x,y)$are specified for those points$(x,y)$on the boundary of$I\$. In your case, after you stretch your solution to another ...

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