Consider Noisy station that detects transmissions and disrupts them by beginning a competing transmission as soon as it hears the beginning of the transmitted frame, thereby causing a collision. Assume Detector machine, which is on this Ethernet with bandwidth 10 Mbps, detects collision during the transmission of its 12th bit on the wire (including any preamble). If the speed of the signal in the wire is 2 × 108 meter/second, then the distance (in meters) of the Noisy station from Detector machine is _______
@Somoshree Datta 5 @Utkarsh Joshi
please check this..correct me if i am wrong
In LAN technology
Why Tt >= 2Tp ?.. because in worst case for collision signal to come back to sender can take 2Tp time and during this time we want sender to be in sending state hence we equate Tt >= 2Tp to help Sender understand that collision happened because of him so Stop transmission..But this is for Sender only what about other how can they understand collision ..Hence JAM signal is send when collision signal reached to sender or receiver first.
If we can consider Collision detector as Simple Host in Ethernet which detects collision during its transmission
Hence Tt >= 2TP
Using the collision Detection Formula
L >= 2*Tp*B
We can derive the distance as
d <= (L/2)*(V/B )
Here d is the maximum distance up to which we can detect the collision So,
Here L = 12 , B = 107 , v = 2*108 then
distance d = 120
this same approach is also followed my Somoshree Datta 5
what's the difference ??
64.3k questions
77.9k answers
244k comments
80.0k users