Random Processes - Random Signals and Processes Primer with MATLAB - page 406

Exercise 6.4 Find the mean value and variance of the process X ( t ) if its autocorrela-
tion function is given as

2
1 þ t 2 :

R XX ðtÞ¼ (6.172)

Answer The autocorrelation function does not have a constant term. This indicates
that the mean value of the process is zero:

EXðtÞ f ¼ 0 : (6.173)

For t ¼ 0, we have:

ðtÞ ¼

2
1 þ 0 ¼ 2 :

g ¼ EX 2 R XX ð 0 Þ¼EXðtÞXðt þ 0 Þ f (6.174)

Using ( 6.172 ) and ( 6.174 ), we get the variance of the process:

ðtÞ EXðtfg ðtÞ ¼ 2

2

s 2 ¼ EX 2 ¼ EX 2 (6.175)

Exercise 6.5 Determine whether or not the process X ( t ) is a WS stationary and find
its variance if it is known that the autocorrelation function is:

4
1 þ t 2 þ 16

R XX ðtÞ¼ : (6.176)

Answer The autocorrelation function has a constant term which is equal to a
squared mean value:

2

16 ¼ EXðtfg : (6.177)

From here, it follows that the mean value of the process X ( t ) is a constant:

EXðtfg¼ 4 : (6.178)

From ( 6.176 ), we can calculate the mean squared value:

ðtÞ ¼

4
1 þ 0 þ 16 ¼ 20 :

R XX ð 0 Þ¼EX 2

(6.179)

Using ( 6.178 ) and ( 6.179 ), we get:

ðtÞ EXðtfg

2

s 2 ¼ EX 2

¼ 20 16 ¼ 4 : (6.180)