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FIGURE 5.5TDMA: bit rate r b for 1 user.The throughput of TDMA depends on the number of bursts (number ofaccesses or traffic stations), n p , to the frame. If we allow x to be the number ofbits in the header and y the number of bits in the guard time, and assume thatthe frame contains m reference bursts, then the throughput can be written asZ ¼ 1ðn p þ mÞðx þ yÞR r T sð5:3ÞExample 5.2: Consider the frame format of the INTELSAT=EUTELSATgiven by Fig. 5.4. If an offset-quadrature phase shift keying (OQPSK)modulation scheme is used for transmission, express and plot the TDMAthroughput for 200 accesses.SolutionWe can write the following:R r ¼ 0:1208 Gbit=sec, since 1 symbol ¼ 2 bits and T s ¼ 2T b .T s ¼ 2 msec, m ¼ 2; x ¼ 560 and y ¼ 128.Substituting these values in (5.3), we can write the throughput as a linearfunction of the number of traffic stations or accesses n p :Z ¼ 1 0:00285 n p þ 2 ð5:4ÞCopyright © 2002 by Marcel Dekker, Inc. All Rights Reserved.