# Tag Info

15

Many years ago I wrote this tutorial on the Kalman filter. It derives the filter using both the conventional matrix approach as well as showing it's statistical assumptions as an 'optimal' least squares filter.

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 ...

11

A signal having a mean-value or DC component of zero is commonly referred to as mean-free or as having no DC component. It does not mean that it cannot be averaged, just that the average comes out as zero. Might be a little inexact but it is very common.

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 $h$, ...

8

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 ...

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 $p_L$ ...

5

For any single chunk (window) of data the coherence will, as you observed, be 1. In order to properly estimate coherence you must average the spectra and cross-spectra for multiple windows, and THEN calculate coherence. The auto-spectra XX and YY can be averaged the conventional way. For the cross-spectrum XY you must average the real and imaginary ...

5

Try something like this: $\begin{eqnarray} \mu(n) &=& (1 - \alpha_1) \mu(n) + \alpha_1 x(n) \\ \bar{x}(n) &=& x(n) - \mu(n) \\ s(n) &=& (1 - \alpha_2) s(n) + \alpha_2 \bar{x}(n)^2 \\ \sigma(n) &=& \sqrt{s(n)} \\ \end{eqnarray}$ This is equivalent to sending your input signal to a 1-pole DC blocking high-pass filter with a ...

5

It depends on what you mean by a "randomly moving object". If you are trying to track something that truly moves around in a totally uncorrelated manner from sample to sample (like, say, a laser pointer that flickers on and off and randomly changes position in your camera images) then a linear tracker will not give you insight into the object's state. ...

5

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) < ...

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

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 ...

4

You should define a linear $F$ to use Kalman filtering recursions. However, I think, random movements can not be described very well with linear models (an object which is tied to certain and linear physical rules can be tracked by the Kalman filter). Therefore, your state space model will not be linear. Then, your question about $F$ is critical. Because, ...

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

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

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 is $$S_Y(\omega)=|H(\omega)|^2S_X(\omega)$$ where $H(\omega)$ is ...

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

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 ...

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 ...

3

A complex random variable with dimension being 2 $Z=X+jY$ can be defined as a Random vector with components $X$ and $Y$ which take values along 2 different dimensions. The Probability Density Function of $Z$ is defined using the PDFs of $X$ and $Y$. The Expectation is defined as $\mathrm{E}(Z)=\mathrm{E}(X)+j\mathrm{E}(Y)$ and the variance is defined the ...

3

This seems to be a nice write-up of the Kalman filter.

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 ...

3

An individual inner product does produce a scalar, but often when a cross correlation is calculated multiple individual cross correlations (i.e. dot products) are calculated at different time offsets. These individual scalar results form a vector that is indexed by the relative time offset.

3

It comes from the Nyquist rate. The Nyquist sample rate for a signal with bandwidth $B$ is $2B$. In other words, that is the lowest sample rate that will contain all of the information in the signal without distorting it (i.e. without losing information). The question implicitly assumes that the sample rate is $2B$, and thus that in time $T$ you will get $... 3 As mentioned above, both signals 3 and 5 appear to be quite correlated and have a similar period. We can think of two signals being correlated if we can shift one of the sources left or right and increase or decrease its amplitude so that it fits on top of the other source. Note we are not changing the frequency of the source, we are just performing a phase ... 3 Similar to the preferred answer above (Jason S.), and also derived from the formula taken from Knut (Vol.2, p 232), one can also derive a formula to replace a value, i.e. remove and add a value in one step. According to my tests, replace delivers better precision than the two-step remove/add version. The code below is in Java, mean and s get updated ("... 3 As a fellow astrophysicist, I suggest you use the standard 3 sigma above noise for detection, and 5 sigma above noise if you want to do analysis on what you find. If your noise does not have a flat background, you need to subtract a "Flat", an estimation of that the background might be. If you know what the noise should look like, such as exponential, ... 3 There is no best pdf for a periodic signal. There is also no way to find the 'exact' pdf of a measured signal. What you have to do is to measure the pdf from the data. Use a histogram to approximate the pdf of your data. Define a number of intervals within the amplitude range of your signal and simply count the numbers of data samples per interval. This ... 3 If$h(t)$is continuous at$a$, then$h(t)\delta(t-a) = h(a)\delta(t-a)$. Since the$\operatorname{sinc}$function is continuous everywhere,$\$\begin{align} Y(t) &= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)h(t)\\ &= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)h(n\tau)\\ &= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)\operatorname{sinc}(n)\\ &...

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