A frequency hopping multiple access system use 7 frequencies in the order ... 7,

Question

A frequency hopping multiple access system use 7 frequencies in the order

        ... 7, 3, 1, 4, 2, 5, 6, 7, 3, 1, 4 ...

Assume three transmitters share this set of frequencies in the same sequence, but each of them using a starting frequency selected at random. Their shift from one frequency to another is synchronized in time.

a)Compute the probability that two or more transmitters use the same frequency at the same time

b)Determine specific starting points for the three transmitters that minimize the frequency overlap.

Explanation / Answer

a) Each transistor can have frequency in 7 ways. Total possible combination = 7X7X 7 = (7)3.

P(two or more transistor having same frequency) = 1-P(no two transitors have same frequency) ----------(1)

No. of ways in which each transistor has unique frequency = 7C4 = 140

P(no two transistors have same frequency) = (No. of ways in which each transistor has unique frequency)/(Total ways)

P(no two transistors have same frequency) = 140/(7)3 = 0.408

Put this value in (1)

P(two or more transistor having same frequency) = 1-0.408=0.592

b) To minimize the frequency overlap, transistor should select 1st, 3rd and 5th frequency so that when they shift to another frequency, they will have 2nd, 4th and 6th frequency. In our case transistors should select 7, 1 and 2 as starting frequencies.

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