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A random process is a collection of random variables, one for each time instant under consideration. Typically this may be continuous time ($-\infty < t < \infty$) or discrete time (all integers $n$, or all time instants $nT$ where $T$ is the sample interval).
Stationarity refers to the distributions of the random variables. Specifically, in a ...
20
Linear Filtering
The first approach in Peter's answer (i.e. filtering white noise) is a very straightforward approach. In Spectral Audio Signal Processing, JOS gives a low-order filter that can be used to produce a decent approximation, along with an analysis of how well the resulting power spectral density matches the ideal. Linear filtering will always ...
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There are several. This site has a reasonable (but possibly old) list:
Filtering white noise to make it pink.
The Voss algorithm.
The Voss-McCartney algorithm.
10
when we observe the Random Process at a specific time $t_k$, that is the
value at $X(s_1,t_k), X(s_2,t_k),\ldots,X(s_n,t_k)$, if we denote them
by $(a_1,a_2,\ldots, a_n)$. Now the mapping between the outcomes
$(s_1,s_2,s_3,\ldots,s_n)$ and its probabilities
$(a_1,a_2,\ldots,a_n)$ are collectively called as Random Variable.
Not true. Random variable ...
7
The output of your filter is what is sometimes called band-limited white noise. In
your particular case, the autocorrelation function of the output noise is a sinc
function whose zeroes are every $100$ microseconds, that is, samples taken at
the rate of $10^4$ samples/second are uncorrelated. Your samples at $3\times 10^4$
samples per second are closer ...
7
I think this is simply an aspect of Parseval's Theorem (e.g. click me)
It simply says: sum of squares in the time time domain equals sum of squares in the frequency domain. Substitute "sum" for "integral" if using the analog domain. In other words: total energy in the time domain equals total energy in the frequency domain.
This can easily reproduce your ...
7
No. Quoting Wikipedia's article Independence (probability theory):
If $X$ and $Y$ are independent random variables,
then the expectation operator $\operatorname{E}$ has the property
$$\operatorname{E}[X Y] = \operatorname{E}[X]\operatorname{E}[Y].$$
Consider your $X(t_1)$ and $Y(t_2)$ as $X$ and $Y$ in this answer. If both $\operatorname{E}[X] \ne ...
6
Your initial algorithm creates a random variable that's uniformly distributed between 0 and 1. The variance of that is 1/12. If you sum NUM_GAUSSIAN_SUMS instances of that the variance will be NUM_GAUSSIAN_SUMS/12. In order to get to a target variance, V, you need to multiply the summed random variable with sqrt(V*12/NUM_GAUSSIAN_SUMS).
As a side note, a ...
6
Let us consider a hypothetical random process where the sample functions are DC values and are different from each other:
X1(t) = constant= mean of X1(t)
X2(t) = constant= mean of X2(t)
The temporal mean of $X_1(t)$ and $X_2(t)$ are constant but not equal.
if my process is stationary $X(t_1)$ and $X(t_2)$ are equal and RVs (refer Dilip's answer)
...
5
To elaborate on Dilip's answer (which is perfectly correct, though in practice the Ziggurat method is much more computationally efficient than Box-Mueller):
One key ingredient missing from your reasoning is whether you want your samples to be independent. It is not clear from your question, but this is the most common situation...
If you want your samples ...
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
how can I change variance of this random variable?
By multiplication, of course. The variance of $c X$, where $c$ is the multiplicative constant and $X$ is your random variable, is $c^2$ the variance of $X$.
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You can compute the power of the process from its power spectrum as well as from its PDF. Equating the two gives you a relation between the constants $A$ and $B$. More specifically you get
$$\int_{-B}^BG_x(f)df=\frac{1}{2A}\int_{-A}^Ax^2dx$$
If I'm not mistaken this should give $A=\sqrt{3B}$.
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Note that the condition
$$\int_{-\infty}^{\infty}|f(t)|^2dt<\infty\tag{1}$$
(i.e., that the signal $f(t)$ has finite energy) is very restrictive when we try to model signals, even though obviously any actually occurring signal must have finite energy. Modeling signals as random processes means that we ignore condition $(1)$. Models are always ...
4
Expanding on my comment, $\{\phi(\tau)\}$ is a non-stationary nonGaussian random process, and I doubt that there is any simple answer (or even a rather complicated one) for the probability density function of the random variable
$\phi(\tau)$ for an arbitrary value of $\tau$. But, the (time-varying) mean
function of the process is easy to calculate. We have
...
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, ...
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As pointed out in pichenettes' answer, you need to know more about the original signal $x(t)$. If the signal's power spectrum $S_x(\omega)$ is known then the power spectrum of its $n^{th}$ derivative is
$$S_y(\omega)=\omega^{2n}S_x(\omega)$$
The power of the $n^{th}$ derivative can then be computed as
$$E\{|y(t)|^2\}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\...
4
Dilip and pichenettes have already pointed out two methods for generating Gaussian random variables (the Box-Muller transform and the Ziggurat algorithm). For completeness, I will point out another: inverse transform sampling. I recently had the need to create a maximum-throughput software Gaussian random number generator, and after evaluating all of the ...
4
The output signal will still be normally distributed, but its power spectrum, i.e. its frequency content, will obviously be different from the input signal. If $S_X(\omega)$ is the power spectrum of the input signal, which is approximately flat, then the power spectrum of the output signal is
$$S_Y(\omega)=|H(\omega)|^2S_X(\omega)$$
where $H(\omega)$ is ...
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If by 'resulting discrete signal statistically independent' you mean whether the samples will be independent, i.e. whether $P(x[1],x[2],\dots,x[N]) = \prod_{i=1}^N P(x[i])$, the answer is in general no, because the filtering will induce correlations between nearby elements: the LPF operation means that values of successive elements can't change arbitrarily ...
3
This looks like a pink noise model, where the power spectral density of the noise is proportional to $\frac{1}{f}$. This model is appropriate for some real-world phenomena; I can't speak to whether it's the right choice for this particular application.
Your interpretation of the meaning of this sort of spectrum is correct, although the noise is still ...
3
As correctly pointed out in the comments, in general the process $Y(t)=x(t)M(t)$ is not wide-sense stationary (WSS), i.e. its autocorrelation function depends not only on the time difference parameter $\tau$, but also on the absolute time $t$:
$$R_Y(\tau,t)=E[Y(t+\tau)Y^*(t)]=E[x(t+\tau)M(t+\tau)x^*(t)M^*(t)]=\\
=x(t+\tau)x^*(t)E[M(t+\tau)M^*(t)]=x(t+\tau)x^...
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I will try to explain this practically using MATLAB notation.
Yet before that I must say the ergodic property sometime is limited to a level of moment, namely ergodic in the 1st , 2nd, 3rd moment, etc...
If the process is IID it is promised to be ergodic.
Now, assume we have a function myRandomProcess in MATLAB.
It returns a row vector of length n with the ...
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Though the Matched Filter is the best tool detection of a known signals under AWGN it should work well here as well.
To say something about the probabilities the question is, do you know something about the energy of the received signals?
If you do, you should easily say something about the probabilities.
Pay attention that if the assumption is a signal ...
3
The total roundoff error for the sum of $N$ numbers is:
$$
S = \sum_{i=0}^{N-1} E_i
$$
The roundoff error for the $i$-th number is represented by the random variable $E_i$. If we assume that the random number generator used by the computer yields numbers $X_i$ taken from a uniform distribution, then the difference between each $X_i$ and the nearest tenth (...
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I do not know the answer to your question, but perhaps this paper can help. I realize that you are not using single bit random waveforms, but the formulation on the distribution they calculated is fairly through.
"Probability distribution of the crosscorrelation function of finite-duration single-bit random waveforms"
Abstract
The detection errors of a ...
3
You should use the the standard formula:
s = randn(m, n) + 1i*randn(m, n);
And as pointed out by MBaz, the output should be scaled accordingly by $\frac{1}{\sqrt{2}}$
s = s/sqrt(2);
More on that topic you can find for example in a book Digital Media Processing by Hazarathaiah Malepati:
It should generate exactly what you want.
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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.
...
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The limit $\lim_{\tau\to\infty} R_x(\tau)$, if it exists, equals $E^2[X(t)]$ and so $E[X(t)]=0$ in this case.
More generally, the mean of a WSS process is nonzero only if the power spectral density has an impulse at the origin. This can be applied to
periodic autocorrelation functions such as $\cos(\omega_0t)$ pointed out
in @MattL's comment. If the ...
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If the signal and noise are uncorrelated then the variance of the sum of the two equals the sum of their variances. So, yes, you can add variances (of uncorrelated signals). However, you cannot add dB-values! First of all, a variance in dB does not make sense because dB quantifies the ratio between two values. What you probably mean is dBm or dBW. So before ...
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