Other Important Random Variables - Random Signals and Processes Primer with MATLAB - page 342

Answer The exponential random variable X describes the time between two suc-
cessive Poisson events. The random variable Y is then:

Y ¼ X= 2 ; (5.241)

f Y ðyÞ¼ 2 f X ð 2 yÞ¼ 2 l e 2 ly

; for y> 0 ; (5.242)

f Y ðyÞ¼ 0 for y< 0 : (5.243)

Exercise 5.11 Find the probability of error in a detecting binary signals “0” and
“1” in an impulse noise X which is described using a Laplace PDF:

5e jxj f X ðxÞ¼ 0 : ; 1<x<1: (5.244)

The probability that signals “0” and “1” are sent is equal to ½.

Answer The received signal is:

v ¼ s þ x; (5.245)

where the signal s is

V; when ''1'' is sent ;

s ¼

: :

(5.246)

;

0 when ''0'' is sent

Therefore, the received signal is:

V þ x; when ''1'' is sent ;

v ¼ (5.247)

x; when ''0'' is sent :

If “1” was sent, the corresponding PDF is:

5e jvVj

f vj 1 ðvÞ¼ 0 : : (5.248)

Similarly, if “0” was sent, the received signal is only impulse noise and the PDF is:

5e jvj

f vj 0 ðvÞ¼ 0 : : (5.249)

The intersection of the PDFs corresponds to the value of V /2 for the received
signal.
The following rule is applied:
When a received signal is higher than V /2, the decision is made that “1” was sent.
When a received signal is less than V /2, the decision is made that “0” was sent.