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No, it will be Rician Distribution. Rayleigh distribution is a special case of Rician Distribution when the normal random variables involved are of zero mean and equal variances. I have assumed you meant :$$y=x_r + j \cdot x_i + n_r + j \cdot n_i$$ where $x_r, x_i$ are deterministic variables. Basically, you now have $y_r = (n_r + x_r),\ \sim \mathcal N(... 3 It is simply because each sample of$n(t)$has a random magnitude and phase by definition given as$n(t) = |n(t)|e^{j\phi(t)}$. With real and imaginary components as follows: $$|n(t)|e^{j\phi(t)} =|n(t)|\cos(phi(t))+j|n(t)|\sin(phi(t))$$ Real:$I(t) = |n(t)|\cos(\phi(t))$Imag:$Q(t) = |n(t)|\sin(\phi(t))$Since the phase and magnitude are independent ... 3 In general, the characteristic function of a random variable is related to the fourier transform of the distribution as follows: $$\varphi_Z(-\omega) = \mathscr F \big\{ f_Z(z) \big\}$$ Why? Because, fourier transform of a PDF$f_Z(z)$is: $$\mathscr F \big\{ f_Z(z) \big\} = \int^{\infty}_{-\infty} f_Z(z)\cdot e^{-j\omega z} \ \mathrm dz$$ And ... 3 Dirac delta function has a continuous argument, but Kronecker delta function has a discrete argument. Your example is a discrete signal so Kronecker delta is used. 2 The binomial theorem is not necessarily involved: the top waveform is simply pointwise raised to the 5-th power, as others have noted, and there are no missing peaks that should be seen. To illustrate, consider the following crude hand-drawn figure: I drew this in my simulation software, so it was automatically "digitized" as I drew it. Obviously, I am poor ... 2 In MATLAB, you can do it that way: >> nsamples = 1e9; mu = 11; sigma_squared = 18; x = mu + sqrt(sigma_squared)*randn(nsamples,1); mean(x) ans = 10.9998 var(x) ans = 17.9994 Edit: link https://www.mathworks.com/help/matlab/math/random-numbers-with-specific-mean-and-variance.html 2 So, it's generally right that a receiver adds noise due to physics (Johnson-Nyquist noise). So, yes, in any system, at least part of the noise in the received signal is caused by the receiver. There's also noise that is background noise from e.g. cosmic sources that happen neither in the receiver nor the transmitter (and also aren't other transmitters ... 2 The reason is g1 is not a complete Gaussian pdf since you limited the range to not extend past 0. The following will work for you: W=[-2500:1:2500]; g1= normpdf(W, 500, 100); g2= normpdf(W, 0, 100); ConvG1G2 = conv(g1,g2, "same"); plot(g1) hold on plot(g2) plot(ConvG1G2) Results in: 2 The covariance matrix is given by $$C_{X,Y}=\begin{bmatrix}E(XX)& E(XY) \\ E(YX )& E(YY) \end{bmatrix}$$ This can be written as below: $$C_{X,Y}=\begin{bmatrix}E(R^2cos^2(\Theta) )& E(Rcos(\Theta)Rsin(\Theta)) \\ E(Rsin(\Theta)Rcos(\Theta) )& E(R^2sin^2(\Theta)) \end{bmatrix}$$ Since$R$and$\Theta$are independent the expectation will ... 2 Important Information : Sampling at$f_{s}$will map$[-\frac {f_{s}}{2}, \frac {f_{s}}{2}]$to digital frequency$\omega=[-\pi, \pi]$, and similarly sampling at$2f_{s}$will map$[-f_{s}, f_{s}]$to digital frequency$\omega=[-\pi, \pi]$. Also, we need to look only at digital frequency$\omega = [-\pi, \pi]$as the digital spectrum is a$2\pi$-periodic ... 2 Let$X \sim B(n, p)$. At the third layer$\begin{bmatrix}1 & 3 & 3 & 1\end{bmatrix}$of the Pascal's triangle, we have$X \sim B(3, \frac{1}{2})$. Which means$P(X = k) = {3 \choose k} (\frac{1}{2})^k (\frac{1}{2})^{3-k} = \frac{{3 \choose k}}{2^3}$. The variance is given by$\sigma^2 = n p q = 3 \cdot \frac{1}{2} \frac{1}{2} = \frac{3}{4}$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: $$C_{... 2 According to your description, x = Z \cdot \cos(W),\ y = Z \cdot \sin(W). Follow this answer for the derivation of joint PDF of (Z, W) : Complex Gaussian Magnitude and Phase Joint PDF Derivation You will reach the following expression after Methid of transformation:$$f_{Z,W}(z,w) = |\mathbf J|.f_{X,Y}(z \cdot \cos(w), z \cdot \sin(w)), \ where \ \... 2 Below is a function which I wrote long back, when I needed to generate AWGN time-domain samples given Noise PSD in dBm/Hz. AWGN_NOISE() : Generates Additive White Gaussian Noise of PSD power in dBm/Hz AWGN has Gaussian PDF with 0 mean and$\sigma^{2} = N_{o}/2NoisePSD_{dBm/Hz} = 10.log_{10}(\frac{N_o}{2.BW})$, Why? Because, Output Noise Power(in dBm) :... 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 ... 2 This sounds counter intuitive to many, but as long as the difference operator and the smoothing kernel are linear and space-invariant, they can be applied in any order, and thus are often combined in a single convolution operator (for more computational efficiency), for the same. For some intuition, consider that, either for the linear smoothing and the ... 1 Adopted from the Matlab doc: N=40; mu = ones(1,2); sigma = eye(2); R = chol(sigma); z = (mu + randn(N,2)R)[1; 1j] 1 The envelope |y| follows a Rice/Rician distribution if the variances are the same. a Beckmann distribution if the variances are different. To my knowledge, there are not simplified closed-form expressions for the PDF and the moments. However, in On the Distribution Function of the Generalized Beckmann Random Variable and Its Applications in Communications ... 1 "Prior" refers to the "already known" probability (or its probability distribution function) of an event happening 1 The equation you wrote down describes the case for a narrowband (flat fading) MIMO system. Narrowband/flat fading describes the scenario when your signal bandwidth is small relative to the channel bandwidth (inverse delay spread). Whatever system you are modeling, you'd have an idea of the delay spread and signal bandwidth, and could use that as a ... 1 The equation$y = hs + n$is a discrete-time model of a certain kind of wireless communications system.$s$is a complex number, drawn from a finite set called a "constellation", that represents the information being transmitted.$n$is a complex, circularly-symmetric Gaussian random variable with zero mean and variance usually denoted by$N_0$.$h$... 1 Okay, I am not quite understanding your question. Let's start with the definition of the Gaussian, aka the Bell Curve, in its general form. $$f(t) = \frac{1}{ \sigma \sqrt{2\pi}} e^{ -\frac{(t-\mu)^2}{2\sigma^2} }$$$\mu$is the mean, and represents where the peak occurs.$\sigma$is the standard deviation, and identifies where the inflection points ... 1 I've had issue with this -- please see the errata of the reference. The eigenfunctions are not scaled in the original version of the book. This errata should solve it. 1 then according to the binomial theorem, we should get 6 peaks Where does that statement come from and why do you think it applies here? Applying the binomial theorem we get terms that consists of Gaussian raised to different powers. However, when raising the power on a Gaussian, the peak stays where it is so there is no mechanism that would move it ... 1 Tables 7.1 and 7.2 from Lattice Coding of Signals and Networks by Zamir are essentially this. Also the figures in this paper: https://arxiv.org/pdf/1103.0171.pdf 1 A weighted local histogram would mean filtering the image$I(x,y)$with a localised filter$H(x,y)$(gaussian in this example). The resultant image is the 2D convolution$Y(x,y) = I(x,y)*H(x,y)$. The normal histogram of$Y(x,y)$is the localised histogram of$I(x,y)\$. It is used to view certain specific characteristics of an image depending on the filter ...