Random Processes - Random Signals and Processes Primer with MATLAB - page 403

If ( 6.152 ) is satisfied, we can write

T= 2

ð 1 1

1
T

lim
T!1

xðtÞxðt þ tÞ d t ¼ x 1 x 2 f X 1 X 2 ðx 1 ; x 2 ; tÞ d x 1 d x 2 ; (6.153)

1 1

T= 2

where X 1 and X 2 are random variables of the process X ( t ) in time instants t and t + t ,
respectively.

Example 6.7.2 Determine whether the process

XðtÞ¼Y cos ðo 0 t þ yÞ (6.154)

is a mean and autocorrelation ergodic, if Y is uniform over [0, 1], and if o 0 and y are
constants.

Solution The mean value is:

ð

1

EXðtfg¼ y cos ðo 0 t þ yÞf Y ðyÞ d y ¼ cos ðo 0 t þ yÞ= 2 : (6.155)

0

The mean value is not a constant and thus the process is not a first-order
stationary and consequently, it is not a mean ergodic.
The autocorrelation function is:

EXðtÞXðt þ tÞ f g ¼ EY cos ðo 0 t þ yÞY cos ðo 0 ðt þ tÞþyÞ f g

¼ EfY 2

g cos o 0 t þ cos ð 2 o 0 t þ o 0 t þ 2 yÞ ½ = 2

1
6 cos o 0 t þ cos ð 2 o 0 t þ o 0 t þ 2 yÞ

¼ ½ : (6.156)

From ( 6.156 ), the autocorrelation function is a function of time t and thus cannot
be equal to a time-averaged autocorrelation function which does not depend on
time t . Therefore, the process is not autocorrelation ergodic.
Similarly, for mutually ergodic processes X ( t ) and Y ( t ), we can make a time and a
statistical cross-correlation function equal as:

t

XðtÞYðt þ tÞ¼XðtÞYðt þ tÞ

¼ Ð 1
1

Ð 1

T= 2

Ð

xðtÞyðt þ tÞ d t :

1
T

x 1 y 2 f X 1 Y 2 ðx 1 ; y 2 ; tÞ d x 1 d y 2 ¼ lim
T!1

1

T= 2

(6.157)