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412                                                               Fiber Optic Communications



            Solution:
            The bit interval for a 10-Gb/s signal is equal to
                                                  1
                                                      ps = 100 ps.                            (9.93)
                                               10 × 10 9
            The bit interval for a 40-Gb/s signal is equal to

                                                  1
                                                       ps = 25 ps.                            (9.94)
                                               40 × 10 9
            Fig. 9.23 shows a schematic of an OTDM which multiplexes four 10-Gb/s bit streams into a 40-Gb/s bit
            stream. To have a delay of 25 ps between Ch. 1 and Ch. 2, L = L + 25 ps/ .If L = 1 mm, L = 6 mm.
                                                                                          2
                                                              2
                                                                   1
                                                                           1
                                                                                1
            Similarly, L = 11 mm and L = 16 mm. Fig. 9.24 shows the pulses of 10-Gb/s channels within a bit interval
                                   4
                      3
            of the 40-Gb/s signal.
                                                  40-Gb/s signal
                                   Ch. 1      Ch. 2      Ch. 3      Ch. 4




                                         25 ps                                 t
                                                    100 ps

                            Figure 9.24  40-Gb/s signal obtained by multiplexing four 10-Gb/s signals.




            9.5.3   OTDM Experiments

            In one of the early OTDM transmission experiments [44], a 100-Gb/s OTDM signal was transmitted over
            560 km. The OTDM signal was obtained by multiplexing 16 channels at a bit rate of 6.3 Gb/s. A 40-km
            normal dispersion fiber was used in the first half of an 80-km fiber span and a 40-km anomalous dispersion
            fiber was used in the other half so that second-order dispersion was compensated. The transmission distance
            in the above experiment was mainly limited by higher-order dispersion. As the bit rate increased beyond
            40 Gb/s, the performance was degraded by the ISI caused by higher-order dispersion. This problem can be
            alleviated by using a dispersion slope compensation fiber which compensates for third-order dispersion.
            A 400-Gb/s OTDM signal was transmitted over 40 km using a dispersion slope compensation fiber [45].
            A 1-Tb/s OTDM soliton signal was transmitted over 1000 km using a DMF consisting of alternating
            sections of normal and anomalous dispersion fiber [46]. The section length in the above experiment was
            around 10 km. The normal dispersion fiber section not only compensated for second-order dispersion
            of the anomalous dispersion fiber section, but also for the dispersion slope. OTDM can be combined
            with WDM to increase the capacity. A six-channel WDM system with each channel consisting of a
            170.6-Gb/s OTDM signal was demonstrated over a 2000-km nonzero dispersion fiber using RZ-DPSK
            format [47].
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