Random Processes - Random Signals and Processes Primer with MATLAB - page 405

The autocorrelation function is:

EYðtÞYðt þ tÞ f g ¼ E cos ðo 0 t þ XÞ cos o 0 ðt þ tÞþX f ð Þ g

1
2 cos ðo 0 tÞþ

1
2 E cos

¼ f ð ð 2 o 0 t þ o 0 tÞþ 2 X Þ g

1
2 cos ðo 0 tÞþ

1
2 E cos ð 2 o 0 t þ o 0 tÞ cos ð 2 XÞ sin ð 2 o 0 t þ o 0 tÞ sin ð 2 XÞ

:

¼ f g

1
2 cos ðo 0 tÞþ

1
2 cos ð 2 o 0 t þ o 0 tÞE cos ð 2 XÞ

1
2 sin ð 2 o 0 t þ o 0 tÞE sin ð 2 XÞ

¼ f g f g

(6.164)

The expression ( 6.164 ) will be dependent only on t if

E cos ð 2 XÞ f g ¼ E sin ð 2 XÞ f g ¼ 0 : (6.165)

Next we relate the conditions ( 6.163 ) and ( 6.165 ) with the characteristic function
f X ðoÞ .

f X ðoÞ¼Efe joX g¼E cos ðoXÞ f þ jE sin ðoXÞ f g: (6.166)

From here, we have:

f X ð 1 Þ¼E cos ðXÞ f þ jE sin ðXÞ f g; (6.167)

f X ð 2 Þ¼E cos ð 2 XÞ f þ jE sin ð 2 XÞ f g: (6.168)

Placing ( 6.167 ) and ( 6.165 ), we get the following conditions for WS stationarity
of the process Y ( t ):

f X ð 1 Þ¼ 0 ;

(6.169)

f X ð 2 Þ¼ 0 :

Exercise 6.3 Express the autocorrelation function of the process

YðtÞ¼Xðt þ bÞXðtÞ (6.170)

in terms of the autocorrelation function of the process X ( t ), R XX ( t 1 , t 2 )
Answer

R YY ðt 1 ; t 2 Þ¼EXðt 1 þ bÞXðt 1 Þ f ½ Xðt 2 þ bÞXðt 2 Þ ½ g

¼ EXðt 1 þ bÞXðt 2 þ bÞ f g EXðt 1 ÞXðt 2 þ bÞ f g

f þEXðt 1 ÞXðt 2 f g
¼ R XX ðt 1 þ b; t 2 þ bÞR XX ðt 1 ; t 2 þ bÞR XX ðt 1 þ b; t 2 ÞþR XX ðt 1 ; t 2 Þ:
(6.171)

EXðt 1 þ bÞXðt 2 Þ