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# lectr15 - 15. Poisson Processes In Lecture 4, we introduced...

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1 15. Poisson Processes In Lecture 4, we introduced Poisson arrivals as the limiting behavior of Binomial random variables. (Refer to Poisson approximation of Binomial random variables.) From the discussion there (see (4-6)-(4-8) Lecture 4) where " , 2 , 1 , 0 , ! " duration of interval an in occur arrivals " = = k k e k P k λ λ (15-1) = = µ λ T T np (15-2) Fig. 15.1 PILLAI 0 T ±² ±³ ´ arrivals k 2 0 T ´ arrivals k

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2 PILLAI It follows that (refer to Fig. 15.1) since in that case From (15-1)-(15-4), Poisson arrivals over an interval form a Poisson random variable whose parameter depends on the duration of that interval. Moreover because of the Bernoulli nature of the underlying basic random arrivals, events over nonoverlapping intervals are independent. We shall use these two key observations to define a Poisson process formally. (Refer to Example 9-5, Text) Definition: X ( t ) = n (0, t ) represents a Poisson process if (i) the number of arrivals n ( t 1 , t 2 ) in an interval ( t 1 , t 2 ) of length t = t 2 t 1 is a Poisson random variable with parameter Thus (15-3) . 2 2 2 1 λ µ µ= = = T T np (15-4) . t λ 2 " arrivals occur in an (2 ) , 0, 1, 2, , interval of duration 2 " ! k k Pe k k λ λ  ==   "
3 PILLAI and (ii) If the intervals ( t 1 , t 2 ) and ( t 3 , t 4 ) are nonoverlapping, then the random variables n ( t 1 , t 2 ) and n ( t 3 , t 4 ) are independent. Since n (0, t ) ~ we have and To determine the autocorrelation function let t 2 > t 1 , then from (ii) above n (0, t 1 ) and n ( t 1 , t 2 ) are independent Poisson random variables with parameters and respectively. Thus 1 2 2 1 , , 2 , 1 , 0 , ! ) ( } ) , ( { t t t k k t e k t t n P k t = = = = " λ λ (15-5) ), , ( 2 1 t t R XX ), ( t P λ t t n E t X E λ = = )] , 0 ( [ )] ( [ (15-6) . )] , 0 ( [ )] ( [ 2 2 2 2 t t t n E t X E λ λ+ = = (15-7) 1 t λ ) ( 1 2 t t λ ). ( )] , ( [ )] , 0 ( [ )] , ( ) , 0 ( [ 1 2 1 2 2 1 1 2 1 1 t t t t t n E t n E t t n t n E = (15-8)

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4 PILLAI But and hence the left side if (15-8) can be rewritten as Using (15-7) in (15-9) together with (15-8), we obtain Similarly Thus ) ( ) ( ) , 0 ( ) , 0 ( ) , ( 1 2 1 2 2 1 t X t X t n t n t t n = = )]. ( [ ) , ( )}] ( ) ( ){ ( [ 1 2 2 1 1 2 1 t X E t t R t X t X t X E XX = (15-9) . , )] ( [ ) ( ) , ( 1 2 2 1 2 1 1 2 1 2 1 2 2 1 t t t t t t X E t t t t t R XX + = + = λ λ λ (15-10) (15-12) (15-11) . , ) , ( 1 2 2 1 2 2 2 1 t t t t t t t R XX < + λ ). , min( ) , ( 2 1 2 1 2 2 1 t t t t t t R XX λ λ+ =
5 PILLAI From (15-12), notice that the Poisson process X ( t ) does not represent a wide sense stationary process. Define a binary level process that represents a telegraph signal (Fig. 15.2). Notice that the transition instants { t i } are random. (see Example 9-6, Text for the mean and autocorrelation function of a telegraph signal). Although X ( t ) does not represent a wide sense stationary process, ) ( ) 1 ( ) ( t X t Y = (15-13) Fig. 15.2 0 1 t i t t ) ( t X t ) ( t Y t 1 + Poisson arrivals 1 1 t

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6 PILLAI its derivative does represent a wide sense stationary process. To see this, we can make use of Fig. 14.7 and (14-34)-(14-37).
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## lectr15 - 15. Poisson Processes In Lecture 4, we introduced...

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