Exercise 4.09: Cyclo-Ergodicity

To illustrate the "cycloergodicity" property

We consider two different random processes whose pattern functions are harmonic oscillations, each with the same frequency $f_0 = 1/T_0$ $T_0$ denotes the period duration.

In the random process shown above $\{x_i(t)\}$ the stochastic component is the amplitude, where the random parameter $C_i$ can take all values between $1\hspace{0.05cm}\rm V$ and $2\hspace{0.05cm}\rm V$ with equal probability:

$$\{ x_i(t) \} = \{ C_i \cdot \cos (2 \pi f_{\rm 0} t)\}. $$

In the process $\{y_i(t)\}$ all pattern functions have the same amplitude: $x_0 = 2\hspace{0.05cm}\rm V$. Here the phase $\varphi_i$ varies, which averaged over all pattern functions is equally distributed between $0$ and $2\pi$ :

$$\{ y_i(t) \} = \{ x_{\rm 0} \cdot \cos (2 \pi f_{\rm 0} t - \varphi_i)\}. $$

The properties "cyclostationary" and "cyclergodic" state,

that although the processes cannot be described as stationary and ergodic in the strict sense,
but all statistical characteristics are the same for multiples of the perion duration $T_0$ in each case.

In these cases, most of the calculation rules, which actually apply only to ergodic processes, are also applicable.

Hint:

This exercise belongs to the chapter Auto-Correlation Function.

Questions

	The process $\{x_i(t)\}$ is stationary.
	The process $\{x_i(t)\}$ is ergodic.
	The process $\{y_i(t)\}$ is stationary.
	The process $\{y_i(t)\}$ is ergodic.

$\varphi_y(\tau=0)\ = \ $

$\ \rm V^2$

$\varphi_y(\tau=0.25 \cdot T_0)\ = \ $

$\ \rm V^2$

$\varphi_y(\tau=1.50 \cdot T_0)\ = \ $

$\ \rm V^2$

	All pattern signals are free of equal signals.
	All pattern signals have rms value $2\hspace{0.05cm}\rm V$.
	The ACF has twice the period $(2T_0)$ as the pattern signals $(T_0)$.

Solution

(1) Correct are proposed solutions 3 and 4:

At time $t = 0$ $($and all multiples of the period $T_0)$ each pattern signal $x_i(t)$ has a value between $1\hspace{0. 05cm}\rm V$ and $2\hspace{0.05cm}\rm V$. The mean value is $1.5\hspace{0.05cm}\rm V$.
In contrast, for $t = T_0/4$ the signal value of the entire ensemble is identically zero. That is,t:
Even the linear mean does not satisfy the stationarity condition: The process $\{x_i(t)\}$ is not stationary and therefore cannot be ergodic.
In contrast, for the process $\{y_i(t)\}$ the same moments are expected at all times due to the uniformly distributed phase ⇒ the process is stationary.
Since the phase relations are lost in the ACF calculation, each individual pattern function is representative of the entire process Therefore, ergodicity can be hypothetically assumed here.
At the end of the exercise, check whether this assumption is justified.

(2) Due to ergodicity, any pattern function can be used for ACF–calculation. We arbitrarily use here the phase $\varphi_i = 0$.

Because of the periodicity, it is sufficient to report only one period $T_0$. Then holds:

$$\varphi_y (\tau) = \frac{1}{T_0} \cdot \int_0^{T_0} y(t) \cdot y (t+\tau) \hspace{0.1cm}{\rm d} t = \frac{{ x}_0^2}{{ T}_0} \cdot \int_0^{{\it T}_0} \cos (2 \pi {f_{\rm 0} t}) \cdot \cos (2 \pi {f_{\rm 0} {t+\tau}) \hspace{0.1cm}\rm d \it t.$$

Using the trigonometric relation $\cos (\alpha) \cdot \cos (\beta)= {1}/{2} \cdot \cos (\alpha + \beta) + {1}/{2} \cdot \cos (\alpha - \beta)$ it further follows:

$$\varphi_y (\tau) = \rm \frac{{\it x}_0^2}{{2 \it T}_0} \cdot \int_0^{{\it T}_0} \rm cos (4 \pi \it{f_{\rm 0} t} + {\rm 2} \pi \it{f_{\rm 0} \tau}{\rm )} \hspace{0.1cm}\rm d \it t \ {\rm +} \ \rm \frac{{\it x}_0^2}{{2 \it T}_0} \cdot \int_0^{{\it T}_0} \rm cos (-2 \pi \it{f_{\rm 0} \tau}{\rm )} \hspace{0.1cm}\rm d \it t. $$

The first integral is zero (integration over two periods of the cosine function).
The second integrand is independent of the integration variable $t$. It follows: $\varphi_y (\tau) ={ x}_0^2}/{\rm 2} \cdot \cos (2 \pi {f_{\rm 0} \tau}). $
For the given time points, with $x_0 = 2\hspace{0.05cm}\rm V$:

$$\varphi_y (0)\hspace{0.15cm}\underline{ = 2\hspace{0.05cm}{\rm V}^2}, \hspace{0.5cm} \varphi_y (0.25 \cdot { T}_{\rm 0}{\rm )} \hspace{0.15cm}\underline{ = 0}, \hspace{0.5cm} \varphi_y (\rm 1.5 \cdot {\it T}_{\rm 0} {\rm )} \hspace{0.15cm}\underline{= \rm -2\hspace{0.05cm}{\rm V}^2}.$$

(3) Correct are both first proposed solutions:

The mean $m_y$ can be obtained from the limit of the ACF for $\tau \to \infty$ if one excludes the periodic parts. It follows $m_y= 0$.
The variance (power) is equal to the ACF–value at the point $\tau = 0$, so $2\hspace{0.05cm}\rm V^2$. The rms value is the square root of it: $\sigma_y \approx 1.414\hspace{0.05cm}\rm V$.
The period of a periodic random process is preserved in the ACF, that is, the period of the ACF is also $T_0$.