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Unformatted text preview: ECE320 Solution Notes to Homework 3 Spring 2006 Cornell University T.L.Fine 1. The average current emitted by a cathode is I amps (coulombs/sec.). The charge emitted over time T seconds is Q coulombs, where a coulomb is approximately (but assume it to be exactly) 6 × 10 18 electrons. (a) If I is one nanoampere and T is one nanosecond then what is the average number of electrons emitted in this time? The average charge is 10 9 × 10 9 × 6 × 10 18 = 6 electrons. (b) What is the probability that no electrons will be seen in this time? Use Poisson with mean of 6 to determine P ( K = 0) = e λ = e 6 . (c) What is the probability that 2 electrons will be seen in this time? P ( K = 2) = e 6 6 2 2! = 18 e 6 . (d) By examining the ratio of probabilities, is seeing 7 electrons more probable than seeing 6 electrons? P ( K = 7) P ( K = 6) = 6 7 7! ÷ 6 6 6! = 6 7 < 1 . Hence, seeing 7 electrons is less probable than seeing 6. 2. Let 10 2 be the probability of error per bit transmitted in a digital commu nication system, and let there be 1000 bits in a packet. (a) What is the average number of errors made in transmitting a packet? Bi nomial B (1000 ,. 01) and EK = np = 1000 × . 01 = 10. (a) Are 100 errors more probable than 101 errors? P ( K = 100) P ( K = 101) = 1000 100 p 100 (1 p ) 900 ÷ 1000 101 p 101 (1 p ) 899 = 101!899! 100!900! 1 p p = 101 900 1 p p = 101(1 . 01) 9 > 1 . Therefore, seeing 100 errors is much more probable than seeing 101 errors....
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This homework help was uploaded on 09/25/2007 for the course ECE 3200 taught by Professor Fine during the Spring '06 term at Cornell.
 Spring '06
 FINE
 Probability theory, Code word, Kraft's inequality, uniquely decodable code

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