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# Tag Info

12

An example you run typically across in a text book (Papoulis as an example) is the sine with random phase $$x(t)=\sin(2\pi f + \phi)$$ where $\phi$ is a random variable, distributed uniformly, over $0$ to $2\pi$. Any realization will have $\phi$ take on a particular value, but it’s random, just like a 6 on a dice after a throw. You could not predict it ...

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 ...

3

Since your signal isn't sampled uniformly some strange things might happen when you apply FFT and look at the results. What you should do is estimate the Uniform DFT of the Non Uniform Time Series. One easy way to do it is use the reference code and analysis I posted on the question - Frequency Analysis of a Signal Without a Constant Sampling Frequency (...

3

As $T\rightarrow \infty$, the integral becomes the convolution integral. You can use the fact that convolution in the time domain is multiplication in the frequency domain and then you have a single carrier being passed through a low pass filter. The term where $\omega=\omega_c$ just has to do with how you define the ideal low pass filter and in this case ...

2

The frequency resolution of your FFT is determined by its length and the sampling rate. In your second plot you show Fs = 10e6 and 3200 samples. Assuming you don't zero-pad these samples when you pass them to the FFT this gives you a frequency resolution of 3125Hz: $$3215 = \frac{10 \times 10^6}{3200}$$ So at lower frequencies you have frequency bins as ...

2

Your FFT method of finding the phase of a single tone is only valid if your 's' signal contains an exact integer number of cycles. I.E., no spectral leakage. And that is NOT the case for your 's' signal.

2

The DFT Matrix for Non Uniform Time Samples Series Problem Statement We have a signal $x \left( t \right)$ defined on the interval $\left[ {T}_{1}, {T}_{2} \right]$. Assume we have $N$ samples of it given by $\left\{ x \left( {t}_{i} \right) \right\}_{i = 0}^{N - 1}$. The samples time ${t}_{i}$ is arbitrary and not necessarily uniform. We're ...

2

My guess would be that the DC peak is part of the transient response, thus it decreases to zero over time T as the oscillating parts of your signal continue. But I don't know what your signal is so that's a guess

2

Well, by definition of the $\delta$ distribution, you have: $\int_{-\infty}^{\infty} f(t) \delta(t-T)\, \textrm{d}t = f(T)$ The autocorrelation of a function $g(t)$ can be computed via: $\int_{-\infty}^{\infty} g^{*}(t)g(t + \tau)\, \textrm{d}t$, with $g^*$ as the complex conjugate of $g$. Since $\delta(t)$ is real-valued, this is conjugation can be ...

1

Without more details, when a signal is properly recorded at $F_n$ Hz, one is entitled to expect that all information below the half of it, namely $F_n/2$ Hz, can be recovered, somehow. This is an application of the classical Shannon Sampling Theorem. Twice the maximum frequency is sufficient to sample a signal... in theory. The question says: max frequency ...

1

A finite power periodic signal will have the following property: $$P_x = \frac{1}{T_0} \int_{0}^{T_0} |x(t)|^2 dt < \infty \tag{1}$$ where $P_x$ is the power averaged over a period of the signal. Then from Parseval's theorem, the total power can also be shown to be (for a real signal I'm assuming): P_x = |a_0|^2+2 \sum_{n=1}^{\infty} |a_n|^2 < \...

1

Here is a pretty clear paper explaining secrecy rate and uses over fading channels. http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.105.2844&rep=rep1&type=pdf See equation 3 for the formal definition. One way to think about it is like a tug-of-war between the main channel and wiretapping channel. If the main channel has a higher capacity ...

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