10

Most realistic signals are both random and periodic. For example, you can modulate a harmonic oscillator with a slow enough random signal that moves its frequency around a $\mu_{f}, \sigma_f$. This looks like: $$y= \sin \left( \frac{2 \pi \mathcal{N_s}(\mu_f, \sigma_f) n}{Fs} \right )$$ Where $\mathcal{N_s}(\cdot)$, denotes a normally distributed random ...


6

If you are talking about a given signal as "a deterministic realization of a phenomenon", it can be periodic, but not really random. However, some physical systems are prone to produce randomness and periodicity, like rotating machines, gears, cyclic engines, that produce signals similar to: Naturally rotating bodies (stars, planets) also produce random ...


4

You got some definitions wrong. It's correct that orthogonality means that $E[XY]=0$. Uncorrelated means that $X-\mu_X$ and $Y-\mu_Y$ are orthogonal, i.e., $E[(X-\mu_X)(Y-\mu_Y)]=0$. If you work that out you should arrive at the equivalent condition $E[XY]=\mu_X\mu_Y$ for uncorrelatedness (not for independence!). Consequently, if at least one of the two RVs $...


3

Answer and hint: Just add the mean to expression of zero mean random variable. Along with this scale the expression for unit variance gaussian with the standard deviation of the required variance. You would now have non zero mean gaussian RV with given variance


3

There are several questions on this forum dealing with various aspects of strict-sense and wide-sense stationarity and ergodicity, and some of the answers give counterexamples to the Venn diagram that you have constructed, such as A strictly stationary process that is not wide-sense-stationary because its mean is undefined A wide-sense-stationary process ...


2

It is not the expectation operator that makes sure that the limit exists. The expectation just results in an ensemble average, which we need to obtain a deterministic function $S(f)$ for the power spectrum. Assume we're given a deterministic power signal $x(t)$, i.e., a signal with finite non-zero power, and, consequently, infinite energy. Its Fourier ...


2

There are two mistakes in your code/method. The first is the term $\sqrt{\Delta t}$ in your second formula; it should be replaced by $\Delta t$. The second is in the computation of the power spectrum from the estimated auto-correlation. What you do is square the result of the FFT Y to obtain mY, but that's not correct. First of all, Y is complex-valued, and ...


2

Histograms of images can differ, widely. However, when features are inspected, one often uses derivative filters at different scales, or morphological decompositions, or independent component analysis. A traditional and heuristic model for the resulting coefficients of a component is that of the Generalized Gaussian-Laplacian Distribution, or GGD: $$ C_{...


2

The sample mean can be a choice for the estimator with the following pros and cons Pros: it is unbiased Cons: 1. The variance of the estimator increases with increase in absolute value of $x$ The estimator will converge to true value of $x$ only when number of observations over which the mean is taken is large. A different approach is illustrated ...


2

In most of the engineering literature I'm familiar with, white noise is introduced as an idealized random process $n(t)$ with a flat power spectrum $$S_N(f)=\frac{N_0}{2}\tag{1}$$ and the corresponding autocorrelation function $$R_N(\tau)=\frac{N_0}{2}\delta(\tau)\tag{2}$$ The reason for defining white noise in this way is because it closely approximates the ...


2

White noise is not "WSS by nature" whatever you mean by that phrase but it can be treated as a (zero-mean) WSS process insofar as its effects in linear systems are concerned. For example, standard linear system theory ways when the input to an LTI system is an ordinary WSS process $\{X(t)\}$ with autocorrelation function $R_X(\tau)$, then the ...


2

Disclaimer: this might very well be wrong. Still pondering it, but Dilip Sarwate has convincing points. When you say "white" you assume it's WSS to begin with. For non-WSS processes, "white" isn't defined, since no only lag-dependent autocorrelation can be found. (And a process is white, exactly if its autocorrelation takes the form of a ...


1

Once you talk about the Spectrum of noise / process you implicitly says it is stationary in the wide sense. What does it mean to have a signal with uncorrelated samples? Do you understand it means you can't using linear predictor, to have any information from all past samples ion the current one? Usually this is not how signals behave. This is exactly why ...


1

The definition of auto-correlation depends on the type of signal. For random processes, the auto-correlation function is defined by the expectation given in Eq. $(1)$ of your question. For deterministic signals, there are two definitions, depending on whether the signal is an energy signal (i.e., has finite energy), or a power signal (i.e., has finite power ...


1

first order stationary The CDF of any sample is time invariant. $F_X(x;~~ t) = F_X(x;~~ t+\tau)~~~\forall \tau$ Thus, the PDF is time invariant: $f_X(x;~~ t) = f_X(x;~~ t+\tau)~~~\forall \tau$ And Thus, all first order statistics are constant: $\mu_X(t)=\mu=\text{constant}$ $\sigma^2(t)=\sigma^2=\text{constant}$ $E\Big[X^2(t)\Big] = \text{constant}~~~~\text{(...


1

I would say that $R$ is an estimate of the instantaneous power in $x(t)$, as opposed to the average power. But this assumes that you implement the expectation by averaging over multiple realizations of $x(t)$. In this case, the average is still a function of time, $$ R(t) = E \left\{ x^2(t) \right\},$$ and is an estimate of the instantaneous power in $x(t)$...


1

The answer is yes. Even if we could generate several realizations of the random process by conducting the random experiment a number of times. We could only have a limited number of realizations and for a limited period of time practically. Now to take expectation of the random process we will have to take freeze time and look at the distribution of the ...


1

Given the definition of the correlation matrix $\mathbf{R}_{\mathbf{x}}$ here, I am assuming that $\mathsf{E}[\mathbf{x}] = \mathbf{0}$. I do this because the correlation matrix is usually defined as $\mathsf{E}[(\mathbf{x} - \mathsf{E}[\mathbf{x}])(\mathbf{x} - \mathsf{E}[\mathbf{x}])^{\dagger}]$, where $\dagger$ indicates complex conjugate tranpose. Note ...


1

Taking the average values of these measurements should work. The estimator is $$ \hat{X}=\frac{1}{L}\sum_0^{L-1}K=\frac{1}{L}\sum_0^{L-1}(X+X*N) $$ .To show that it is the unbiased estimator, $$ E(\hat{X}) = \frac{1}{L}LX + \frac{1}{L}XL\times 0 = X $$ assuming the uniform is distributed between +0.5 and -0.5 as you mentioned.


1

This is my understanding: the statistics of the source described in the paper depend on which character is produced first. If the first character is $a$, then one of the source properties is that letters in odd positions are always $a$. However, if the first letter is $b$ (in other words, a shift of $1$ in the circuit), then letters in odd positions can be ...


1

I see what the upsampling has done but I fail to understand how the amplitude distribution is periodic? The reason they can say this is because the symbols are assumed to be equally likely. Think about an upsampled BPSK stream, you'll have approximately the same amount of $+1$ pulses as $-1$ pulses so if you take a large enough block of samples it is ...


1

Since all transitions are equally likely, there is no information in the channel output about the source. Hence, the mutual information, and, consequently, the channel capacity, are zero in this case. You might as well look at the output of a random number generator and from that guess the source symbols. Expressed in formulas you have $$H(Y)=H(Y|X)\tag{1}$...


1

You need the joint distribution of $X_{t_1}$ and $X_{t_2}$ in order to calculate $E[X_{t_1}X_{t_2}]$ and nobody has claimed that the joint distribution of $X_{t_1}$ and $X_{t_2}$ is the same as the joint distribution of $X_{t_1}$ and $X_{t_3}$. Stationarity requires that the joint distribution of $X_{t_1}$ and $X_{t_2}$ be the same as the joint distribution ...


1

Summation or integration of the spectral components yields the total power (for a physical process) yes, that's how "power density" is defined ... or variance (in a statistical process) For any bound density, that's true. But would the total power correspond to the uncentralized second moment, so variance + meanĀ²? Yes, but if you have a process ...


1

As suggested I am adding my comments as an answer. The sample autocorrelation function (ACF) for $n$ observations is given by $\hat{p}_x(m) = \frac{\sum_{n=0}^{N-m-1}(x_n - \bar{x})(x_{n+m} - \bar{x})}{\sum_{n=0}^{N-m-1}(x_n - \bar{x})^2} $ To understand why you have a difference between figure 1 and figure 2, lets assume we are looking at observations $...


1

There are a lot of misconceptions in the way that the problem has been posed (in particular, $X(\omega)$ as defined by the OP in the first version of his question -- he has since then deleted the definition -- ) has nothing to do with the matter), but when $\{x(t)\}$ is a wide-sense-stationary (WSS) process, then the processes $\{y_1(t)\}$ and $\{y_2(t)\}$ ...


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