Random Processes - Random Signals and Processes Primer with MATLAB - page 404

6.8 Numerical Exercises

Exercise 6.1 Find whether or not the process X ( t ) is a WS stationary process

XðtÞ ¼ A cos ðo 0 t þ yÞ; (6.158)

where A and o 0 are constants and y is a uniform variable over [0, 2 p ].
Answer The first condition for a WS stationarity says that the mean value of the
process must be constant.

ð ð

2 p 2 p

1
2 p d y ¼ 0

E XðtÞ

g ¼

A cos ðo 0 t þ yÞf y ðyÞ d y ¼ A cos ðo 0 t þ yÞ :

f (6.159)

0 0

Therefore, the first condition is satisfied.
The second condition says that the autocorrelation function has to be a function
of a time difference t .

E XðtÞXðt þ tÞ f g ¼ E A cos ðo 0 t þ yÞA cos o 0 ðt þ tÞþy f ð Þ g

A 2
2 E cos o 0 t þ cos ð 2 o 0 t þ o 0 t þ 2 yÞ

¼ f g

A 2
2

A 2
2 E cos ð 2 o 0 t þ o 0 t þ 2 yÞ

¼ cos o 0 t þ f g

A 2
2

¼ cos o 0 t: (6.160)

The second condition is also satisfied and, thus, the process is a WS stationary.

Exercise 6.2 Given the process

YðtÞ ¼ cos ðo 0 t þ XÞ; (6.161)

where o 0 is a constant and X is a random variable with the characteristic function
f X ðoÞ . Find which condition must be imposed to the characteristic function f X ðoÞ
in order for the process Y ( t ) to be a WS stationary.

Answer The mean value of the process Y ( t ) is:

E cos ðo 0 t þ XÞ f g ¼ E cos ðo 0 tÞ cos X sin ðo 0 tÞ sin X f g

¼ cos ðo 0 tÞEf cos Xg sin ðo 0 tÞEf sin Xg:

(6.162)

The mean value ( 6.162 ) is constant if the following condition is satisfied:

E cos ðXÞ f g ¼ E sin ðXÞ f g ¼ 0 : (6.163)