sm8_47
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sm8_47


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Problem 8.47 In a computer-communication network, the arrival time \u3c4 between messages is modeled 
with an exponential distribution function, having the density 
 
\u23aa\u23a9
\u23aa\u23a8
\u23a7 \u2265=
\u2212
otherwise
e
fT
0
01
)(
\u3c4\u3bb\u3c4
\u3bb\u3c4
 
 
a) What is the mean time between messages with this distribution? 
b) What is the variance in this time between messages? 
 
 
Solution (Typo in problem statement, should read fT(\u3c4)=(1/\u3bb)exp(-\u3c4/\u3bb) for \u3c4>0) 
(a) The mean time between messages is 
 
[ ]
( )
( ) ( )
\u3bb
\u3bb\u3c4\u3bb
\u3c4\u3bb\u3c4\u3bb\u3c4\u3c4
\u3c4\u3bb\u3c4\u3bb
\u3c4
\u3c4\u3c4\u3c4
=
\u2212\u2212=
\u2212+\u2212\u2212=
\u2212=
=
\u221e
\u221e\u221e
\u221e
\u221e
\u222b
\u222b
\u222b
0
0
0
0
0
)/exp(0
/exp/exp
/exp
)(
d
d
dfT TE
 
 
where the third line follows by integration by parts. 
 
(b) To compute the variance, we first determine the second moment of T 
[ ]
( )
( ) ( )
[ ]
2
0
0
2
0
2
0
22
2
20
/exp2/exp
/exp
)(
\u3bb
\u3bb
\u3c4\u3bb\u3c4\u3c4\u3bb\u3c4\u3c4
\u3c4\u3bb\u3c4\u3bb
\u3c4
\u3c4\u3c4\u3c4
=
+=
\u2212+\u2212\u2212=
\u2212=
=
\u222b
\u222b
\u222b
\u221e\u221e
\u221e
\u221e
T
d
d
dfT T
E
E
 
 
 
 
Continued on next slide 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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page\u20268-57 
Problem 8.47 continued 
 
 
The variance is then given by the difference of the second moment and the first moment 
squared (see Problem 8.23) 
 [ ] [ ]( )
2
22
22
2
)(Var
\u3bb
\u3bb\u3bb
=
\u2212=
\u2212= TTT EE
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
page\u20268-58