Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 1.8
A classical example of communication over an unreliable channel can be set up using a probabilistic model
of the channel.
Let us assume that two symbols A and B are sent over the channel characterized by four transition
probabilities as shown in the table below. The rows of the table represent the sending signals and the
columns represent the received symbols or ra and rb. The entries in the table are the channel transitional
probabilities or P[ra was received | A was sent] = P raA. The unreliable channel is modeled by probabilistic
transitions in the symbols {A, B} into {ra, rb}. Clearly, an ideal channel will have P[ra|A] = P[rb|B] =1
and P[ra|B] = P[rb|A] = 0.
The problem we now set up uses this model: Given the model described, we would like to construct a table
showing the aposteriori probabilities or P[an A was sent | received a ra] = P Ara. In addition we are given
that the probability that symbol A or B is sent = P A and PB, respectively.
PA=0.6;
PB=0.4;
Channel probabilities PraA
A
B
PraA=0.9;
PrbA=0.1;
ra
0.9
0.2
PraB=0.2;
rb
0.1
0.8
PrbB=0.8;
SOLUTION: We must first compute the total probability of a ra and rb being received
Pra= PraA*PA+PraB*PB
Pra =
0.6200
Prb= PrbA*PA+PrbB*PB
Prb =
0.3800
Compute the aposteriori probabilities. Aposteriori means we desire the probability that the symbol A was
actually sent after we have knowledge that symbol was received. Apriori is used to mean we want to
compute the probability of symbol A being sent when we do not have any receiver information. The
aposteriori Probabilities are now tabulated below:
PAra=PraA*PA/Pra;
PArb=PrbA*PA/Prb;
PBrb=PrbB*PB/Prb;
[PAra PArb; PBra PBrb]
ans =
0.8710
0.1290
0.1579
0.8421
PBra=PraB*PB/Pra;
The sum of certain transition probabilities must sum to unity because of the way the channel model was set
up. On the sending side, if an A is sent then P raA and PrbA must sum to unity because when an A is sent it
must either be received as ra or rb and these conditional probabilities are mutually exclusive. Therefore,
[PraA+PrbA PraB+PrbB]
ans =
1
1
On the receiver side, if a ra is received either A or B must have been sent. Again these conditions are
mutually exclusive. We may verify that
[ PAra+PBra
ans =
1.0000
PArb+PBrb]
1.0000