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Unformatted text preview: = 20 log [(4π × 4 × 10 7 )/(2.727 x 10–2) = –205.3 dB
received_power = 10 + 17 + 52.3 – 205.3 = –126 dBW
9.5 The total bandwidth is 500 MHz. The channel bandwidth is 12 × 36 = 432 MHz. So
the overhead is ((500  432)/500) × 100% = 13.6%
9.6 a. The data rate R = 2 × QPSK baud rate = 120.272 Mbps
The frame duration T = 0.002 s
The number of frame bits bF = R × T = 1.20272 × 10 8 × 2 × 10–3 = 240544 bits
Overhead calculation: Overhead bits bo = NRb R + NT b p + (NR + NT )b G
where
NR = number of participating reference stations = 2
b R = number of bits in reference burst = 576
NT = number of participating traffic stations
b p = number of preamble bits =560
b G = number guard bits = 24
b o = (2)(576) + 560 N T + (2)(24) + 24 N T = 1200 + 584 N T
b F – b o = 240544 – (1200 + 584 N T ) = 239344 – 584 NT
NT = (bF – b o)/16512 = (239344 – 584 N T )/16512
(16512 + 584) NT = 239344
Therefore, NT = 14
b. (b F – b o)/bF = (239344 – 584 × 14)/240544 = 0.96
Source: [GLOV98]
9.7 a. The time, Td, available in each station burst for transmission of data bits is
Td = [T frame – N(tg + tpre)]/N
That is, take the frame time, subtract out all the guard and preamble times,
and divide by the number of stations N.
Td = [2000 – 5(5 + 20)]/5 = 375 µs
A burst transmission rate of 30 Mbaud is 30 million signal elements per
second and QPSK signal elements carry 2 bits, so the transmitted bit rate in each
burst is Rb = 60 Mbps
The capacity of each earth station, Cb, is the number of data bits
transmitted in one burst divided by the frame time:
C b = (375 × 60)/2000 = 11.25 Mbps
The number of 64kbps channels that can be carried is:
(11.25 × 10 6 )/(64,000) = 175
b. 11.25 Mbps
c. The total available capacity is 60 Mbps.
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This note was uploaded on 02/20/2012 for the course ENC 3250 taught by Professor Dr.timothycerner during the Fall '11 term at UNF.
 Fall '11
 Dr.TimothyCerner

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