Random Processes - Random Signals and Processes Primer with MATLAB - page 407

The process X ( t ) is WS stationary because both conditions ( 6.32 ) and ( 6.33 ) are
satisfied.

Exercise 6.6 The autocorrelation function of the process X ( t ) is given as:

R XX ðtÞ¼p 2 þ pð 1 pÞ e cjtj ;

(6.181)

where p and c are constants. Find the variance of the process.

Answer The constant term in ( 6.181 ) is equal to a squared mean value, resulting in:

2

¼ p 2 EXðtfg : (6.182)

Similarly, from ( 6.181 ), we have:

ðtÞ ¼ p 2

R XX ð 0 Þ¼EX 2 þ pð 1 pÞ: (6.183)

From ( 6.182 ) and ( 6.183 ), we get:

ðtÞ EXðtfg

2

s 2 ¼ EX 2 ¼ p 2 þ pð 1 pÞp 2 ¼ pð 1 pÞ: (6.184)

Exercise 6.7 Determine whether or not a process ( 6.185 ) is a WS stationary
process .

YðtÞ¼X cos ðo 0 t þ yÞ; (6.185)

where o 0 and y are constants and X is a random variable.
Answer Let us verify that the first condition ( 6.32 ) is satisfied:

EYðtfg¼ EX cos ðo 0 t þ yÞ f g ¼ cos ðo 0 t þ yÞEfXg: (6.186)

This result indicates that the mean value E { Y ( t )} is not constant but dependent
on time, unless E { X } ¼ 0. In this case, E { Y ( t )} ¼ 0 and the first condition ( 6.32 )is
satisfied.
To verify the second condition ( 6.33 ), we must find the autocorrelation function:

R XX ðt; t þ tÞ¼EYðtÞYðt þ tÞ f g ¼ EX cos ðo 0 t þ yÞX cos o 0 ðt þ tÞþy f ð Þ g

;

¼ EfX 2 g cos ðo 0 t þ yÞ cos o 0 ðt þ tÞþy ð Þ

(6.187)