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A biliary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X = 0) = 9/10, then the probability of error for an optimum receiver will be
- a)7/80
- b)63/80
- c)9/10
- d)1/10
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A biliary symmetric channel (BSC) has a transition probability of 1/8....
Py = 1)/(x = 0 P(x = 0)/Py = 1)/(x = 1P(x = 1)
P(x = 0) = 9/10
P(x = 0)+P(x = 1) = 1 So P(x = 1) = 1/10
Transition Probability is nothing but Probability of changing the value from 0 ta 1 and 1 to 0
i.e.
P(0/1) = P(1/0) = 1/8
Pe = P(x = 0)Py = 1)/(x = 0 + P(x = 1)Py = 0)/(x = 1
= (9/10) × (1/8) + (1/10) × (1/8)
= 1/8
But this nat the Pe for optimum Receiver. So,Pe = 1 − Pc
where Pc is Probability of correct detection Use MAP detection Rule:
The received symbal m = m1 if P(y/x1)P(x1)/P(y/xj)P(xj) ≥ 1
This is obtained from following:
Pc(m = mi) = P(mi/y)⋅P(y)=P(xi/y)⋅P(y)
M is the estimated output signal Using Baye's theorem
P(xi/y)⋅P(y)=P(y/xi)⋅P(xi)
when o/py = 0 then Py = 0)/(x = 0P(x = 0)/Py = 0)/(x = 1P(x = 1)
= (7/8⋅9/10)/(1/8⋅1/10) = 63 ≥ 1
So when y = 0, m = 0
Similarly when y = 1 then =(1/8⋅9/10)/(7/8⋅1/10) = 9/7 ≥ 1
So when y = 0 then alsa
So if we choose the rule such that the message signal is 0 for y =0 (or) 1 then the Probability of error is nothing but P(x = 1) = (1/10)
Most Upvoted Answer
A biliary symmetric channel (BSC) has a transition probability of 1/8....
Given:
- Biliary symmetric channel (BSC) has a transition probability of 1/8
- P(X=0) = 9/10
To find:
Probability of error for an optimum receiver
Solution:
1. Probability of error (Pe) is the probability that the received symbol is different from the transmitted symbol.
2. For the BSC, Pe can be calculated as follows:
- If X = 0 is transmitted, the probability of receiving a 1 is 1/8
- If X = 1 is transmitted, the probability of receiving a 0 is 1/8
- Therefore, Pe = P(X=0)*P(receiving 1 when X=0) + P(X=1)*P(receiving 0 when X=1)
- Substituting the given values, Pe = (9/10)*(1/8) + (1/10)*(1/8)
- Pe = 1/10
Therefore, the probability of error for an optimum receiver is 1/10, which is option D.
- Biliary symmetric channel (BSC) has a transition probability of 1/8
- P(X=0) = 9/10
To find:
Probability of error for an optimum receiver
Solution:
1. Probability of error (Pe) is the probability that the received symbol is different from the transmitted symbol.
2. For the BSC, Pe can be calculated as follows:
- If X = 0 is transmitted, the probability of receiving a 1 is 1/8
- If X = 1 is transmitted, the probability of receiving a 0 is 1/8
- Therefore, Pe = P(X=0)*P(receiving 1 when X=0) + P(X=1)*P(receiving 0 when X=1)
- Substituting the given values, Pe = (9/10)*(1/8) + (1/10)*(1/8)
- Pe = 1/10
Therefore, the probability of error for an optimum receiver is 1/10, which is option D.
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Question Description
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