Random Processes - Random Signals and Processes Primer with MATLAB - page 390

For t ¼ 0, from ( 6.84 ), we have:

2

¼ s XX 2 C XX ð 0 Þ¼R XX ð 0 ÞXðtÞ ¼ const : (6.85)

Example 6.5.3 Find autocovariance and variance of a process with the following
autocorrelation function,

1
1 þ t 2 :

R XX ðtÞ¼ 9 þ (6.86)

Solution According to ( 6.60 ), the constant term in ( 6.86 ) corresponds to the
squared mean value:

2

9 ¼ XðtÞ (6.87)

resulting in:

EXðtfg¼ 3 : (6.88)

Additionally, the autocorrelation function ( 6.86 ) depends on the time distance t ,
thus indicating that a process is a WS stationary process.
According to ( 6.55 ) and ( 6.86 ), we get:

1
1 þ 0 ¼ 10

R XX ð 0 Þ¼X 2 ðtÞ¼ 9 þ : (6.89)

From ( 6.85 ), we have:

2

s 2

XX ¼ R XX ð 0 ÞXðtÞ

¼ 10 9 ¼ 1 : (6.90)

The obtained result confirms that for a WS stationary process, variance is
constant.
From ( 6.82 ), the autocovariance is equal to:

1
1 þ t 2 9 ¼

1
1 þ t 2 :

2

C XX ðtÞ¼R XX ðtÞXðtÞ ¼ 9 þ (6.91)

6.6 Cross-Correlation Function

6.6.1 Definition

When we are concerned with two random processes X ( t ) and Y ( t ) we define a
cross-correlation function as:

R XY ðt 1 ; t 2 Þ¼EXðt 1 ÞYðt 2 Þ f g; (6.92)