Method to Resolve the Adjacent-Channel Interference in 100G Transmission.

In a non-coherent WDM system, each optical channel on the line side uses only one binary channel to carry service information. The service transmission rate on each optical channel is called bit rate while the binary channel rate is called baud rate. In this sense, the baud rate was equal to the bit rate. The spectral width of an optical signal is determined by the baud rate. Specifically, the spectral width is linearly proportional to the baud rate, which means a higher baud rate generates a larger spectral width.

• Baud (pronounced as /bɔ:d/ and abbreviated as “Bd”) is the unit for representing the data communication speed. It indicates the signal changes occurring in every second on a device, for example, a modulator-demodulator (modem). During encoding, one baud (namely, the signal change) actually represents two or more bits. In the current high-speed modulation techniques, each change in a carrier can transmit multiple bits, which makes the baud rate different from the transmission speed.

In practice, the spectral width of the optical signal cannot be larger than the frequency spacing between WDM channels; otherwise, the optical spectrums of the neighboring WDM channels will overlap, causing interference among data streams on different WDM channels and thus generating bit errors and a system penalty.

For example, the spectral width of a 100G BPSK/DPSK signal is about 50 GHz, which means a common 40G BPSK/DPSK modulator is not suitable for a 50 GHz channel spaced 100G system because it will cause a high crosstalk penalty. When the baud rate reaches 100 Gbaud/s, the spectral width of the BPSK/DPSK signal is greater than 50 GHz. Thus, it is impossible to achieve 50 GHz channel spacing in a 100G BPSK/DPSK transmission system.