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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 ...


4

This answer complements @CedronDawg's answer which introduced this family of eigenvectors. More specifically, this answer presents three algorithms and a hybrid algorithm for generating for a given length $N$ the unique (except for normalization) non-negative eigenvector of discrete Fourier transform (DFT) with the maximal number of consecutive zeros. This ...


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I have made a tremendous amount of progress on this issue in the last few weeks. The Zeroing Sine Family of Window Functions I have discovered a previously unrecognized class of window functions. What is particularly amazing about these is that the Eigenvectors of the DFT seem to be regularly placed members in this family. This provides not only a ...


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 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 ...


3

Let's analyze it in 1D as the intuition is the same. First, let's have a look on a few different Gaussian Kernels: As expected, they are wider as the Standard Deviation (STD) increase. It means that when the kernel is applied using the convolution, more information is aggregates from farther samples. On the other side it means data is spread. Now, in your ...


3

Since $W(t)$ is assumed to be zero-mean, also the RV $Y$ is zero-mean. Hence, the variance of $Y$ is given by $$\begin{align}\sigma_Y^2&=E\left\{Y^2\right\}\\&=E\left\{\int_{-\infty}^{\infty}W(t_1)\phi(t_1)dt_1\int_{-\infty}^{\infty}W(t_2)\phi(t_2)dt_2\right\}\\&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi(t_1)\phi(t_2)E\big\{W(t_1)W(t_2)\...


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

I might be a tad late to this, but I'll only reply to the part about the "similarities" between the (analog) Bessel and Gaussian. They are not the same. The Bessel filter is meant to approximate an ideal delay: $$B(s)=\mathrm{e}^{-s\tau}\tag{1}$$ while the Gaussian filter tries to approximate a Gaussian bell: $$|G(j\omega)|^2=\mathrm{e}^{-\alpha\...


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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 ...


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Actually, the purpose of all this is to approximate a Laplacian of Gaussian! This computation is part of the corner detection of SIFT. You can find corners by examining extrema of the Laplacian of Gaussians (2nd order derivative). You use Gaussians for denoising, and a Laplacian to find inflection points. However, it is classical to not deal directly with ...


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In the past I derived it as following: It is a little different approach. If it answers your question I will rewrite it in a proper LaTeX. Regarding your question about the steps in the derivation you presented, it is using the Woodbury Matrix Identity (Both 12.26 to 12.27 and 12.28 to 12.29). Related Answers: Deriving the Matrix Inversion Lemma for RLS ...


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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

The OP's updated working is incorrect. Following up what Hilmar suggested gives \begin{align} Y(t) &= a\left(X(t)\right)^2\\ &= a\left(S(t) + N(t)\right)^2\\ &= a\left(S(t)\right)^2 + 2aS(t)N(t) + a\left(N(t)\right)^2\\ &{\large\Downarrow}\\ E[Y(t)]&= aE\left[\left(S(t)\right)^2 \right] + 2aE\left[S(t)N(t)\right] + aE\left[\left(S(t)\...


2

Put your second equations into your first equation, express $Y(t)$ as a function of $S(t)$ and $N(t)$ Apply the definition for mean and autocorrelation. Simplify and solve


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Using $$\int\limits_{-\infty}^\infty|s(t)*h(t)|^2 \,\mathrm dt$$ requires that this integral exists – and that requires, unless an infinite amount of the energy of $s$ is in $\ker((*h))$ (i.e. gets mapped to 0) that $s$ is an energy signal, not only a power signal (as you're probably used to dealing with). (An energy signal is exactly a signal that is ...


2

The choice of window is a tradeoff between various different properties: main lobe width, time domain extension, side lobe locations and amplitudes, pass band flatness, stop band attenuation, transition width, re-constructability, time domain discontinuities, etc. Each window type represents a certain set of tradeoffs and you should choose the one that's ...


2

First, a Gaussian window will have some parameter controlling the ‘width’ of the Gaussian pulse. So there is no singular Gaussian window like there would be with many other window types. Second, because it is a window, it has a defined beginning and end such that the signal function is zero outside the window. The is mathematically equivalent to multiplying ...


2

You can add random variables with different means and variances. For example, look at two continuous uncorrelated normal-distributed random variables $x_1$ and $x_2$ with means and variances $\mu_1, \sigma_1^2$ and $\mu_2, \sigma_2^3$ Then $x_3 = x_1 + x_2 $ is also a normal-distributed random variable with $\mu_3 = \mu_1 + \mu_2$ and $ \sigma_3^2 = \sigma_1^...


2

You may be talking about a sine wave (carrier) of frequency $f_c$ modulated by an Hermite polynomial shape function. If a parameter $1/r$ of your UWB pulse waveform is commensurate with a carrier frequency $f_c$, this setup can also be considered an ultra wide band scenario. But if this is the case, the $\cos(2πf_ct)$ wave and the waves, which higher order ...


2

Well the expression for the $\frac {d^n}{dt^n} p(t) $ is $V_n(t) p(t)$, where $V_n(t)$ is a polynomial of degree $n$, and $V_n(t)$ may be determined recursivelyy $$ V_{n+1}(t)e^{-t^2/(2\tau^2)} = \frac{d}{dt}\left(V_{n}(t)e^{-t^2/(2\tau^2)}\right) $$ Solving for $V_{n+2}(t)$ we have $$ V_{n+1} = \frac{d}{dt} V_{n}(t) - \frac{t}{2\tau^2} V_n(t)$$ In order to ...


2

Let P(0) ( P(1) ) is a probability of transmitting bit zero (one); P(e|0) ( P(e|1) ) is a probability of error when detecting bit zero (one). The probability of erroneous detection is $$ P(e) = P(e|0)P(0) + P(e|1)P(1) \tag {1} $$ Let $V_0$ ( $V_1$ ) be a nominal signal voltage of bit zero ( one ) signal at the transmitter. $$ P(e|0) = \int_T^{\infty}{{\frac ...


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(Assuming the context is FSK signals, where the pulse shape is applied to the phase, instead of the amplitude as in linear modulation). The main reason is that the sideband and out-of-band emissions of GMSK are lower than those produced with a raised cosine filter. See for example this plot from Wikipedia: https://en.wikipedia.org/wiki/File:GMSK_PSD.png This ...


2

GMSK doesn't just use a Gaussian filter instead of a raised-cosine filter. It uses a Gaussian filter on the phase, before applying it to the modulator. This makes it a nonlinear operation. When a raised-cosine filter is applied to some modulated signal, it is applied after modulation, as a linear operation. So there's a whole lot of convenient rules about ...


1

I think we need to be a fair bit more specific: No matter what random variable has its distribution, … If that distribution has a finite variance and mean (counterexample: Cauchy-distributed variables, e.g. $\operatorname*{Im}(z)/\operatorname*{Re}(z)$ of complex normal $z$), if we pick n samples and mean (or just sum) them and do this many time, those ...


1

First of all, you're understating your problem. The error probability depends not only on your noise power (density), but also on the actual distribution of noise. Also, you need to realize that bit error probability is not the same as symbol error probability, not even proportional, so at best, the formula you pasted (from an uncited source) is only an ...


1

I'll give this a shot. The Fourier Transform of a Gaussian is also a Gaussian. The standard deviations in each domain are related as $\sigma_t \cdot \sigma_F = \frac{1}{2\pi}$ The time standard deviation, $\sigma_t$ has units of time and the frequency domain standard deviation $\sigma_F$ has units of Hz. We can define the "bandwidth" of a gaussion ...


1

The fact that it's modulated with a sinusoid doesn't change the FWHM bandwidth of your pulse – the $e^{jx}$ function has $\left\lvert e^{jx}\right\rvert\equiv 1$ at every point. That doesn't change the amplitude, so the FWHM of a sinusoid-modulated gaussian is just the same as of the unmodulated gaussian.


1

No such thing as a single frequency of the noise. That's exactly why it's called white; it has power in all frequency ranges, but not at a single frequency. Finally, is there a frequency-domain representation of Gaussian white noise? Yes, a constant power spectral density for all frequencies. That's like white light (which contains also a continuum of all ...


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