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If you could leave some thoughts on the following, it would be really helpful.

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In Wikipedia one can find the definition of a wide sense cyclostationary process as the stohastic process in which $$E[X(t)] = E[X(t+T_0)] ,\forall t$$ $$R_x(t,\tau) = R_x(t+T_0,\tau) ,\forall t,\tau$$

Then, because $R_x$ is periodic it can be expanded in Fourier Series as $$R_x(t,\tau)= \sum_{n = -\infty}^{\infty} R_x^{n/T_0}(\tau)\cdot e^{i2\pi \cdot n/T_0 \cdot t}$$ In the last sentence (in wikipedia) it says: "Wide-sense stationary processes are a special case of cyclostationary processes with only $ R_{x}^{0}(\tau )\neq 0$."

But if that is true, I can conclude that $$R_x(t,\tau)= R_x^{0}(\tau) = 1/T_0 \int_{-T_0/2}^{T_0/2} R_x(t,\tau) dt$$

I suspect an error here. If anyone could say anything useful what maybe the writer wanted to say about it, I would be grateful. Thanks in advance.

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In university the professor told us that a cyclostationary process $\{X(t)\}$ becomes a wss process when we add a rv $\theta$ with uniform distribution like this $\{X(t+\theta)\}$ And then we can calculate the spectral density: $$S_{xx}(f) = \int_{-\infty}^{\infty}\overline{R}_{x}(\tau) \cdot e^{-i2\pi f\tau}d\tau$$ where $$\overline{R}_{x}(\tau) = 1/T_0 \int_{-T_0/2}^{T_0/2}R_x(t,\tau)dt$$ (which is equal to $R_x^{0}(\tau)$)

I cannot see how the uniformly distributed variable is connected to the above.

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  • $\begingroup$ Do you think I should have asked this question in Maths Stack Exchange? $\endgroup$
    – Anonymous
    Jan 22, 2022 at 17:34

1 Answer 1

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1) WSS processes

Here is the definition of cyclic autocorrelation function from Wikipedia: $$ R_x^{n/T_0}(\tau) = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} R_x(t,\tau)e^{-j2\pi\frac{n}{T_0}t} \mathrm{d}t. $$ We can evaluate this quantity for a WSS process ($R_x^0(\tau)\ne 0$) for the special case of $n/T_0 = 0$ and get $$ R_x^{0}(\tau) = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} R_x(t,\tau)\mathrm{d}t = R_x(0, \tau) $$ in accordance with the definition, because the function being integrated is 0 everywhere w.r.t. the integrating variable, except in $t=0$.

2) Randomized time-index

Notice that with $\{X(t+\theta)\}$ and a random $\theta$ you are completely randomizing the time-structure of the process, thus every cyclic property is lost. Think of the example of temperature measurements in a city. There will be a cyclic trend that depends on the seasons. But if you shuffle the measurements over time, you will be left with a completely random process.

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