Question

5. A binary Z-channel is show in the figure. Assume the input is "0" with probability p and "1" with probability 1-p. (a) What can you say about the input bit if " is received? (b) Find the the probability that the input was "1" given that the output is "0" 0 Input Output 6. A transmitter randomly sends one of the messages in fa1, a2,.. ,an). The receiver either receives the transmitted message with probability p, or mistakenly receives one of the other messages with equal probabilities (a) What is the probability of receiving ai at the receiver? (b) If is received at the receiver, what is the probability that the transmitted message was a?

Answer 1

5)  solution :

Given data :

A binary z- channel has the input is '" 0 " with probability = p

A binary z- channel has the input is '" 1 " with probability = 1-p

Now we have to find out the;

a) What can you say about the input bit if " 1 " is received :-

From the diagram, we say that

• We can see that 1 is gotten as yield just when the information bit is 1.
• On the off chance that the information bit is 0, it will dependably give 0 as yield .

Hence, the input bit is " 1" .

b) If $\alpha _\ {1}$is received at the receiver, what is the probability that the transmitted message was $\alpha _\ {1}$ :-

Now we have to find :

probability that the transmitted message was $\alpha _\ {1}$ is p( Input = 1 / output = 0)

  p( Input = 1 $\mid$ output = 0) = $\frac{p( input =1 \cap output =0 )}{ p( output = 0)}$

we know ,

p( Input = 1  $\cap$ output = 0 ) =   p( output = 0 $\mid$ input = 1 ) p( Input = 1)

   = $\frac{p(output = 0\mid input = 1) p ( input = 1 )}{ p( output = 0)}$

From the diagram,

p( output = 0 $\mid$ input = 1 ) = $\epsilon$

And p( Input = 1) = 1-p ( From the given data )

p( output = 0 ) =   (input  = 0 with probability ) + (input = 1 with probability ) * $\epsilon$

= p + ( 1-p) $\epsilon$

p( output = 0 ) = p + ( 1-p) $\epsilon$

Now,    p( Input = 1 $\mid$ output = 0) = $\frac{p(output = 0\mid input = 1) p ( input = 1 )}{ p( output = 0)}$

= $\frac{ \epsilon ( 1-P) }{P + ( 1-P)\epsilon }$   

  $\therefore$ p( Input = 1 $\mid$ output = 0) = $\frac{ \epsilon ( 1-P) }{P + ( 1-P)\epsilon }$

Probability that the transmitted message was $\alpha _\ {1}$ is     $\frac{ \epsilon ( 1-P) }{P + ( 1-P)\epsilon }$

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