Random Processes - Random Signals and Processes Primer with MATLAB - page 395

Taking the expectation of ( 6.121 ), we arrive at:

R XX ð 0 ÞþR YY ð 0 Þ 2 R XY ðtÞ 0 : (6.122)

From here, we easily obtain the desired result ( 6.120 ).
Example 6.6.2 Verify the properties of the cross-correlation function R XY ( t ) ¼
s 2 sin o 0 t from Example 6.6.1 for the variables X and Y .
Solution

P.1 For t ¼ 0 the cross-correlation function becomes zero and thus, it does not
take its maximum value for t ¼ 0.

P.2 Next we investigate the property ( 6.116 ). To this end, we find R XY ( t ) from
the cross-correlation function ( 6.112 ):

R XY ðtÞ¼s 2 sin o 0 t:

(6.123)

We also find R YX ( t ),

R YX ðtÞ¼EYðtÞXðt þ tf g
¼ E r cos o 0 t x sin o 0 t

f ½ x cos o 0 ðt þ tÞþr sin o 0 ðt þ tÞ ½ g

¼ Efr 2 g cos o 0 t sin o 0 ðt þ tÞEfx 2 g sin o 0 t cos o 0 ðt þ tÞ

þ Efxrg cos o 0 t cos o 0 ðt þ tÞEfxrg sin o 0 t sin o 0 ðt þ tÞ
¼ s 2 cos o 0 t sin o 0 ðt þ tÞ sin o 0 t cos o 0 ðt þ tÞ

:

¼ s 2 sin ðo 0 t þ o 0 t o 0 tÞ ½

¼ s 2 sin ðo 0 tÞ

(6.124)

Comparing ( 6.123 ) and ( 6.124 ), we see that the following is satisfied:

R XY ðtÞ¼R YX ðtÞ: (6.125)

P.3 In order to verify the property ( 6.117 ) we use the results ( 6.112 ) and ( 6.108 ):

R XX ð 0 Þ¼R YY ð 0 Þ¼s 2

(6.126)

resulting in the following:

p

R XX ð 0 ÞR YY ð 0 Þ j ¼ s 2 sin o 0 t ¼ s 2

j R XY ðtÞ j j ; (6.127)

As

j sin o 0 tj 1 ; (6.128)

the inequality ( 6.117 ) is satisfied.