ECEN4533    Exam 1    4 March 2005   

1) Many data formats add framing bits to a sequence, which are specific bit patterns that the receiver can use to synchronize in order to help determine where a key component, such as a packet header or data field, commences.  Suppose the lead byte in the packet header of each ATM cell is set = 10101010 for synchronization purposes.  Call this byte the cell sync byte.  Since ATM cell slots always have some sort of cell within, either a filler or a cell carrying traffic, this cell sync byte will always occur every 53 bytes.  The same pattern might also show up at random in the other 52 bytes that make up each ATM cell.
[10] If you have a completely random bit stream with equally likely logic 1's and 0's, compute the probability the sequence 10101010 occurs in a randomly chosen sequence of 8 bits.  [Answer = 0.58 = 0.003906]
[15] If synchronization is lost, a phase-locked loop on the receiver will first reacquire bit sync. Then the system will begin to search for a sequence of 10101010 in eight consecutive bits.  Once found, this sequence might be the correct cell sync byte, or it might, by chance, be a sequence elsewhere with this same pattern.  Once the pattern is found, the receiver will then go 53 bytes away and check to see if the pattern at this second location is also a sequence of 10101010.  To further reduce the probability of locking to the wrong location, the receiver might also check the bit sequence at other locations N*53 bytes (N = 2, 3, 4, ...) from the original match.  If the probability of a false lock is desired to be less than 1 in 3 million, how many consecutive locations should be checked?  [Answer:  If you're looking in the wrong byte location, there is a 1 in 256 chance that you'll get a 10101010 by chance, a 1 in 65,536 chance you'll get that pattern in two consecutive locations, and a 1 in 16.78 million chance you'll get the sync pattern in three consecutive locations.  Hence the syncronizer should check three consecutive locations.]

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2) The Federal Communications Commission has allocated your company 1 MHz of RF bandwidth that will be used to develop a low cost proprietary wireless LAN protocol.  Frequency Division Multiplexing will be used to split this bandwidth up among multiple users.
[10] How many users can the 1 MHz bandwidth support if each user is to be allocated 200 KHz, with a 20 KHz guard band required to separate users on any adjacent frequencies.  [4 users]
[15] In theory, what must be the minimum SNR if it is desired to support a 400 Kbps bit rate in each 200 KHz channel?  [SNR = 3]

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3) Suppose that 400 32.2Kbps modems are located at America Off-Line's Tulsa modem bank.  At 8:42 pm, assume that every modem is being used by a customer, and that the P(a modem is active and moving packet traffic) = 0.39.  Hence the P(a modem is idle and waiting for packets to be generated by its end user) = 0.61.
[10] Compute the average amount of traffic being generated by this modem bank at 8:42 pm. [5.023 Mbps]
[15] Compute the probability that > 350 modems are active and moving traffic at 8:42 pm.  [Q(19.84) = 0.0]

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4) A PC and two servers are connected to a small 10BaseT Ethernet switched hub.  The PC has been downloading a one-way video stream from server #1.  Server #1 has been outputting a full-sized Ethernet packet every 2 msec, commencing at time t = 0 seconds.  At time t = 10 seconds, the PC also commences downloading a 1.4 MB application file from server #2.  Server #2 injects these packets back-to-back onto its Ethernet connection.  The Ethernet switch will move these packets on a first-in, first-out basis.
[25] Estimate the amount of time it will take to download this 1.4 MB application file.  Ignore the time required to open and close any logical links. [1.886 seconds]

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