ENC3250 Solutions pg 47

ENC3250 Solutions pg 47 - The total data transmission rate...

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Unformatted text preview: The total data transmission rate is 5 × 11.25 = 56.25 Mbps Efficiency = 56.25/60 = 0.9375 9.8 The transponder must carry a total data bit rate of 15 + 10 + 5 = 30 Mbps. Thus, each frame carries 30 Mbps × 0.001 s = 30 kb The three preamble and guard times take up 3 × (10 + 2) = 36 µs in each frame, leaving 1000 – 36 = 964 µs for transmission of the data. Therefore, the burst bit rate is Rbit = 30 kb/964 µs = 31.12 Mbps For QPSK, the symbol rate is half the bit rate = 15.56 Mbaud -47- ...
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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.

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