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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)
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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.
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Consider a simple communication system where multiple nodes are connected by a shared broadcast medium (like Ethernet or wireless). The nodes in the system use the following carrier-sense based medium access protocol. A node that receives a packet to transmit will carrier-sense the medium for 5 units of time. If the node does not detect any other transmission in this duration, it starts transmitting its packet in the next time unit. If the node detects another transmission, it waits until this other transmission finishes, and then begins to carrier-sense for 5 time units again. Once they start to transmit, nodes do not perform any collision detection and continue transmission even if a collision occurs. All transmissions last for 20 units of time. Assume that the transmission signal travels at the speed of 10 meters per unit time in the medium.Assume that the system has two nodes P and Q, located at a distance d meters from each other. P starts transmitting a packet at time t=0 after successfully completing its carrier-sense phase. Node Q has a packet to transmit at time t=0 and begins to carrier-sense the medium. The maximum distance d (in meters, rounded to the closest integer) that allows Q to successfully avoid a collision between its proposed transmission and P’s ongoing transmission is _____. Correct answer is '50 to 50'. Can you explain this answer?

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