tailieunhanh - Optical Networks: A Practical Perspective - Part 34

Optical Networks: A Practical Perspective - Part 34. This book describes a revolution within a revolution, the opening up of the capacity of the now-familiar optical fiber to carry more messages, handle a wider variety of transmission types, and provide improved reliabilities and ease of use. In many places where fiber has been installed simply as a better form of copper, even the gigabit capacities that result have not proved adequate to keep up with the demand. The inborn human voracity for more and more bandwidth, plus the growing realization that there are other flexibilities to be had by imaginative use of the fiber, have led people. | 300 Transmission System Engineering a b Figure Sources of intrachannel crosstalk a A cascaded wavelength demultiplexer and a multiplexer and b an optical switch. The crosstalk penalty is highest when the state of polarization SOP of the crosstalk signal is the same as the SOP of the desired signal. In practice the SOPs vary slowly with time in a system using standard single-mode fiber nonpolarization preserving . Similarly the crosstalk penalty is highest when the crosstalk signal is exactly out of phase with the desired signal. The phase relationship between the two signals can vary over time due to several factors including temperature variations. We must however design the system to work even if the two SOPs happen to match and the signals are exactly out of phase. Thus for the calculations in this section we will assume that the SOPs are the same and compute the penalty when the signals are out of phase which is the worst-case scenario. The power penalty due to intrachannel crosstalk can be determined as follows. Let P denote the average received signal power and eP the average received crosstalk power from a single other crosstalk channel. Assume that the signal and crosstalk are at the same optical wavelength. The electric field at the receiver can be written as Eft V2Pds t cos 27r ft 5 t V2ePdx t cos 2ttfct t . Here ds t 0 1 depending on whether a 0 or 1 is being sent in the desired channel dx t 0 1 depending on whether a 0 or 1 is being sent in the crosstalk channel fc is the frequency of the optical carrier and .s t and t are the random phases of the signal and crosstalk channels respectively. It is assumed that all channels have an ideal extinction ratio of oo. Crosstalk 301 The photodetector produces a current that is proportional to the received power within its receiver bandwidth. This received power is given by Pr Pds t Pdx t 2 Pds .f dAf cos l s .t Assuming e 1 we can neglect the e term compared to the y e term. Also the worst case above

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