Aufgaben:Exercise 4.09: Cyclo-Ergodicity

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"Cyclo–ergodicity property"

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

• In the random process  $\{x_i(t)\}$  shown above,  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 uniformly distributed between  $0$  and  $2\pi$

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

The properties  "cyclo-stationary"  and  "cyclo-ergodic"  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 period 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

1

Which of the following statements are true?

	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.

2

Calculate the auto-correlation function  $\varphi_y(\tau)$  for different  $\tau$  values.

$\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)\ = \ $ $\ \rm V^2$

3

Which of the following statements are true regarding  $\{y_i(t)\}$ ?

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

Solution

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  $T_0$.