# Tag Info

14

Noise is random, but like most random phenomena, it follows a certain pattern. Different patterns are given different names. Consider rolling a die. This is clearly random. Roll the die 1000 times, keeping track of each result. Then, calculate the histogram of the result; you'll find that you got each of 1, 2, 3, 4, 5 and 6 approximately the same number of ...

10

Yeah, it can mess you up pretty badly if you don't get the fundamentals right off the get-go. This is how I interpret correlation, and it has worked for me for what I do for a living. Let's start off with a relatively simple example. Take a look at the following figure (pulled from dspguide... this is actually a great online book for knowing the basics of ...

10

I currently work in the design of atomic clocks and precision frequency sources and pleased to report that the Allan Variance is still quite relevant and useful. In fact it's utility extends to convenient characterization of many non-stationary processes, well beyond its primary tool as a frequency stability assessment. (And as mentioned in its comments, it’...

9

Because each step in the processing chain is linear we consider a case with only noise and no coherent signal. Denote the noise $\xi(t)$. The $I$ and $Q$ signals are \begin{align}\ I(t) &= \xi(t) \cos(\Omega t) \\ Q(t) &= - \xi(t) \sin(\Omega t) \, . \end{align} We express the effect of the filter as a convolution with the time response function $h$, ...

6

A binary symmetric channel (BSC) can be characterized by its complemented probability $p$. Its well-known capacity is $$C = 1 - H(p) = 1 - (-p\log(p) - (1-p)\log(1-p))$$ where $H(p)$ is binary entropy function: A $L-$concatenated BSC, which is also a BSC characterized by $p_L$, can be visualized as in the figure below The complemented probability $p_L$ ...

6

Your formulation: $$\arg \min_{\boldsymbol{x}} \frac{1}{2} {\left\| A \boldsymbol{x} - \boldsymbol{y} \right\|}_{2}^{2} + \lambda {\left\| \boldsymbol{x} \right\|}_{1}$$ Has 2 elements: The Fidelity Term This is basically measurements term with the model of AWGN with IID noise. The Regularization Term This is a sparse promoting model by using the Laplace ...

5

Given data ${ \left\{ {x}_{i} \right\} }_{i = 1}^{N}$ the Empirical STD of the data is well defined: $$STD = \sqrt{ \frac{1}{N - 1} \sum_{i = 1}^{N} { \left( {x}_{i} - \bar{x} \right) }^{2} }$$ Where $\bar{x}$ is the empirical mean of the data given by: $$\bar{x} = \frac{1}{N} \sum_{i = 1}^{N} {x}_{i}$$ Now, if there's a model on the data (Such ...

5

Even though the signals are sampled you can get accuracy which is well above the accuracy offered by the samples as long as you sample using Nyquist. Actually, Using the Matched Filter you can achieve the CRLB (Cramer Rao Lower Bound) for Delay Estimation (Easy to derive for white noise). If you calculate the CRLB for Time Delay Estimation you'll see it ...

5

I suppose you mean the cross-correlation at lag zero. Well take an Hilbert space $H$ (i.e. a metric space in which you can define a scalar product $\langle\cdot ,\cdot\rangle$). Then $x,y\in H$ are orthogonal if $\langle x,y\rangle=0$, by definition. If your Hilbert Space is $L_2(\mathbb{R})$ (the space of real square integrable functions) then the scalar ...

5

What are reasons to choose for cross-correlation or cross-covariance when comparing signals with non-zero mean? Well, part of the issue is that cross-correlation as defined in your equation: $$(f \star g)[n]\ \stackrel{\mathrm{def}}{=} \sum_{m=-\infty}^{\infty} f^*[m]\ g[m+n].$$ will not exist (or be infinite) if $f$ and $g$ have non-zero mean. So, in ...

5

Even though your calculation yields the correct result (in this case), the steps are not completely correct. First of all, the covariance matrix of a random variable $n$ is given by $$C = E[nn^H].$$ The nomenclature of variance is more suitable for scalar random variables, in my opinion (or it refers to the diagonal of the covariance matrix, i.e. the ...

5

The answer to the question (a counterexample) Properties of random processes will in general be time-dependent. They are not only when talking about stationary processes. Another related concept (not relevant here but that you might find interesting) is ergodicity. Most of us find it difficult to understand the difference between both concepts, but then you ...

5

If sensor A has a defect, the clear answer is to only use sensor B. A preferred solution to minimize noise would be to do a weighted average based on the quality of each sensor, when that can be actively characterized. This can be easily done for the case the OP has presented of taking a reading from a single rotating axis. The optimum combining for the ...

5

In the case of uniform quantization, and under some light hypothesis for the signal, the error can be modeled as an additive IID signal, independent of the signal, and with uniform distribution between +/- half LSB. The power of such error is then $\Delta^2/12$, where $\Delta$ is the amplitude of one LSB. Taking the square root and calling it standard ...

4

Note that in general the Fourier transform of a stationary process $x(t)$ does not exist. The Wiener-Khinchin theorem only states that under certain conditions the power spectral density of $x(t)$ exists, and it can be computed as the Fourier transform of the autocorrelation function of $x(t)$. Having said that, if for some reason one assumes that the ...

4

Since $M \in \mathbb{S}^{N}_{++}$ (In other convention $M \succ 0$) by Cholesky Decomposition there is a Triangular Matrix $R \in \mathbb{R}^{N \times N}$ such that $M = {R}^{T} R$. Using this fact one could prove $1 \iff 2$ as following: \begin{align*} \arg \min_{\hat{x}} \mathbb{E} \left[ {\left( \hat{x} - x \right)}^{T} M \left( \hat{x} - x \... 4 The output signal will still be normally distributed, but its power spectrum, i.e. its frequency content, will obviously be different from the input signal. If S_X(\omega) is the power spectrum of the input signal, which is approximately flat, then the power spectrum of the output signal isS_Y(\omega)=|H(\omega)|^2S_X(\omega)$$where H(\omega) is ... 4 Correlation between 2 signals means you can say something about one of them by observing the other. If you mean the standard correlation,  E \left[ x y \right] , it means you knowledge second moment statistics. Which implies one can, using only linear functions, estimate one from the other to some level. 4 If you define the SNR as the ratio of the signal power and the noise power in dB, you have$$SNR_{dB}=10\log \left(\frac{P_s}{P_w}\right)\tag{1}$$where P_s is the power of the desired signal and P_w is the noise poiwer. If the noise w has a mean of zero, then P_w=\sigma^2_w=1. From (1) (with P_w=1) you get the desired value of P_s for a given ... 4 You need to do it numerically, as a related toy problem 2^a + 3^a = 4 appears to have no symbolic solution. Bisection method (binary search) is probably the easiest solver to implement: minexponent = 0; maxexponent = 2; targetsum = 10; precision = 32; loop precision times { exponent = (minexponent + maxexponent)*0.5; if (sum(s.^exponent) < ... 4 OK, let's have a look at one of the problematic terms:$$ \frac{\delta}{\delta x} \bigg[ \bigg(\frac{\tilde{d}[n]}{x}-\mu_A\bigg)^2 \bigg ] = - \frac{2 \tilde{d}[n] \bigg (\tilde{d}[n] - \mu_A x\bigg) }{x^3} $$which can be verified by Wolfram Alpha. The full derivative of the summation term is then just this summed over n. 4 You can regulate its second derivative which is the curvature. Something like:$$ \hat{x} = \arg \min_{x} \frac{1}{2} \left\| x - y \right\|_{2}^{2} + \frac{\lambda}{2} \left\| D D x \right\|_{2}^{2} $$Where  y  is samples you have and  D  is the Derivative Operator (In Matrix Form). By applying it twice we're regulating the Second Derivative. You ... 4 edit: to be clear this answer describes why Lab can be described as a decorrelated color space. This does not imply that decorrelation is the main benefit of using Lab (see many answers on why Lab is useful) If you plot all the RGB colors of a standard RGB image of the natural world you will notice something (see below), the values tend to fall on a ... 4 For the first case, as you wrote, it means the elements are not correlated. Since this is a Gaussian Random Vector it means the elements are independent. It means that at most only one element of  \boldsymbol{\mu}  is not zero. Since if there were more than 1, the matrix  \boldsymbol{R}  wasn't diagonal. Update Let's define  \hat{\boldsymbol{x}} = \... 4 If you filter a Gaussian random process with an LTI system, the output will also be Gaussian. You can make intuitive sense of this by considering that a linear combination (which is what filtering does) of jointly Gaussian random variables is a Gaussian random variable. You can find an in-depth treatment of filtering random processes in this MIT ... 4 After you equalize the histogram you can think of your data as a stream of variables  {X}_{i}  where  X \sim U \left[ 0, 1 \right] . Now all you need is to transform samples of Uniform Random Variable into Gaussian Variable. You should do that by applying the Inverse CDF of Gaussian Distribution. Basically applying the Inverse Transform Sampling ... 4 What you did is the reasonable solution. From here you can do 2 things to mitigate your issues: Computation Efficiency You can use online calculation of the Mean and the STD. Remember when you move one sample to the right 24 samples are shared. Hence if you take advantage of that you'll get a performance boost. Threshold Use adaptive threshold per signal if ... 4 The author is modeling quantization noise as being white (i.e., each sample is independent of previous or following samples) with each sample being a zero-mean, uniformly distributed random number with a span of one:$$p(x) = \begin{cases}1 & -\frac{1}{2} < x < \frac{1}{2} \\ 0 & \mathrm{otherwise}\end{cases}. \tag 1 Do the math (or look it ...

3

It's the key point of array signal processing, I suppose. Say $x$ is the input vector of $[N,1]$ dimension collected from $N$ array sensors. $x(k)$ is its realization at the $k$ moment of time. By its definition covariance matrix (sometimes it's called autocorrelation matrix): $R = E[x\cdot x^H]$ , where $E[]$ is expectation operator and $x^H$ is Hermitian ...

3

After many weeks I give the answer to my own question. There is a limit in which we can solve this problem in a reasonably simple way. Suppose we sum enough points in our DFT that the central limit theorem guarantees that the distribution of the sum's real and imaginary parts are Gaussian distributed. Then we only need to compute the variance. If we ...

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