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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 ...


8

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

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 ...


5

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.


5

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^...


5

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 $A=\sqrt{3B}$.


5

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 ...


5

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

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

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

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 ...


4

Your approach will not model the actual channel but can be used to create the received waveform as having passed through a Rayleigh fading channel. I describe considerations to actually create the received waveform using your approach in either flat and frequency selective conditions below, which should give insight into why it would not be a channel ...


3

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. ...


3

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 ...


3

Since the signal is discrete and the operation is linear it be formed using a Filter. Assuming the signal is given by $ x \left[ n \right] $. Then its derivative is given by: $$ y \left[ n \right] = \frac{ x \left[ n \right] - x \left[ n - 1 \right] }{ \frac{1}{5000} } $$ Since $ x \left[ n \right] $ samples are independent the STD is given by: $$ \...


3

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

I hope this video (from the Florida Institute of Technology. Titled "what is wide sense staionary, strict sense, ergodic signals" by Dr. Ivica Kostanic in his Communications Theory class) from 16:55 could clear your doubts


3

Let me explain it in another way. Consider you have 6 different function of time. You only throw your dice once and regarding the outcome you choose on of six functions and the chose one is one outcome of your random process. Now expand this concept to all possible functions of time instead of only 6 functions and instead of dice throwing consider another ...


3

I think simple. We want to model a random physical phenomenon for analysis purpose. One way is to model it by a stochastic process $X(t)$, i.e. a time series of random variables $\left\lbrace X(t_k) = X(t=t_k), t_k \in \mathbb{R} \right\rbrace$. The random variable $X(t_k)$ is associated with a probability distribution function (PDF) with some finite ...


3

If the imaginary and real components each has a variance of 0.5, then what would be the total variance? The variance of the complex RV in that case would be $0.5 + 0.5 = 1$. If on the other hand, if $w∼CN(0,2σ^2_w)$ then does it mean that the real and imaginary components each has variance 1, so the total variance is 2. Is my understanding correct? Each ...


3

As you correctly say, if $\{X(t)\colon t \in \mathbb T\}$ is a Gaussian process, then for every possible choice of $n$ random variables $X(t_1), X(t_2), \ldots, X(t_n),$ where $t_k \in \mathbb T$ for $ k = 1,2,\ldots, n$, the $n$ variables are jointly Gaussian. Indeed, that is the definition of a Gaussian random process: a collection of random variables such ...


2

I have been using Corsini and Saletti's algorithm since 1990: G. Corsini, R. Saletti, "A 1/f^gamma Power Spectrum Noise Sequence Generator", IEEE Transactions on Instrumentation and Measurement, 37(4), December, 1988, 615-619. The gamma exponent is between -2 and +2. It works well for my purposes. Ed If this attempt to add a screenshot works, the figure ...


2

It depends on the frequency content of the signal. Consider sinusoidal signals with an amplitude of 1 and frequencies of 1Hz and 1kHz. Both have the same standard deviation, but the standard deviation of their derivatives are different - 1000 times as large for the latter.


2

If you are generating sample of white noise (Normal distribution value) with frequency of 48 kHz, then you already have this in frequency range of 0 - 24 kHz (at least you should if generator is truly random). If you use same samples with different sampling frequency, then they will also be a white noise. Now if you want to make it a band-limited white ...


2

So you have a complex impulse response, and the real and imaginary parts of each tap are modeled as i.i.d. zero mean Gaussian variables with variance $\sigma^2/2$ (correct me if I'm wrong). The DFT of the impulse response is a weighted sum of the filter taps: $$H[k]=\sum_{n=0}^{N-1}h[n]e^{-j2\pi nk/N}\tag{1}$$ where I assume $N\ge L$ (where $L$ is the ...


2

randn gives you Gaussian noise with std. dev. of 1. (x/255)*(randn(size(image))) gives you a noisy image of std. dev. (x/255)


2

The problem with quantizing Gaussian distributed signals (like the real/imaginary part of an OFDM signal) is that they can take any value in theory. It is thus necessary to clip such signals at threshold $C$ prior to quantization. Low $C$ increases the distortion noise in this process, while large $C$ will lead to strong quantization noise. It can be shown, ...


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