# Wide sense cyclostationary process

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

## 1)

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.

## 2)

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.

• Do you think I should have asked this question in Maths Stack Exchange? Jan 22, 2022 at 17:34

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 $$t=0$$.
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.