Experimental Demonstrations of Electronic Dispersion Compensation for Long-Haul Transmission Using Direct-Detection Optical OFDM
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Citations
OFDM for Optical Communications
Comparison of Asymmetrically Clipped Optical OFDM and DC-Biased Optical OFDM in AWGN
OFDM for Next-Generation Optical Access Networks
Kramers–Kronig coherent receiver
Terabit Optical Access Networks Based on WDM-OFDMA-PON
References
Power efficient optical OFDM
Analysis of new and existing methods of reducing intercarrier interference due to carrier frequency offset in OFDM
Orthogonal-frequency-division multiplexing for dispersion compensation of long-haul optical systems
Performance of Optical OFDM in Ultralong-Haul WDM Lightwave Systems
Orthogonal Frequency Division Multiplexing for Adaptive Dispersion Compensation in Long Haul WDM Systems
Related Papers (5)
Frequently Asked Questions (16)
Q2. How many samples were used for the training overhead?
For both experimental systems, a CP length of 32 samples equivalent to 3.2 ns for the 12 Gbit/s system and 1.6 ns for the 20 Gbit/s system was used for an overhead of 6.25%.
Q3. Why is the AWG used in the digital system?
The available 123 subcarriers are modulated with 32 QAM, giving a data rate of 12 Gbit/s. A cyclic prefix of 32 samples is used in the two digital systems and a cyclic prefix of 64 samples is used in the system using an RF carrier.
Q4. Why was the data rate limited to 10 Gbit/s?
Because of the imperfections of the RF electronics, only 4-QAM modulation could be supported and so the data rate was limited to 10 Gbit/s.
Q5. What is the frequency response of the AWG?
The frequency response ripples are due to the AWG in return-to-zero mode and the bit errors can be seen to line up with the frequency response dips.
Q6. How many bits of data were recovered?
Three hundred OFDM symbols (184 500 bits) were transmitted over a distance of 400 km and recovered with a BER of 3.03 at a measured OSNR of 25 dB.
Q7. How did the third transmitter design work?
The third transmitter design used an analog mixer to up-convert the signal so that the DAC requirements were reduced and 10 Gbit/s using 4-QAM was achieved.
Q8. What is the meaning of a frequency guard band?
The use of a frequency guard band means that all of the results of mixing between OFDM subcarriers fall out of band and do not degrade performance.
Q9. What was the main limitation to the data rates for the two digital systems?
The main limitation to the achievable data rates for the two digital systems was the sampling rate of the AWG which can generate either two independent 10-GS/s outputs each with a 5-GHz bandwidth or a single 20-GS/s output with a 5.8-GHz bandwidth.
Q10. What is the way to bias the optical modulator?
To provide the optimum noise performance, the transmitter optical modulator should be biased for equal carrier and sideband powers [22] as this provides the peak electrical SNR and lowest bit error rate (BER) for a given optical SNR (OSNR).
Q11. How many bits per second were used for the colorless transmitter?
The bit rate of the digital systems were limited by the DAC speeds to 12 Gbit/s with 32-QAM modulation for the colorless transmitter and 20 Gbit/s with 16-QAM for the system which used an optical filter.
Q12. What is the BER of a 7.5-GHz RF system?
A direct, back to back, electrical connection between the AWG and the DSO resulted in a BER of 3 and an OSNR of 26 dB was required for a system BER to 2.5 .C. 10-Gbit/s Up-Converted SystemFig.
Q13. How many symbols did the equalizer capture?
To determine the training overheads required for the equalizer, the long term stability of the channel must be known, however the memory depth of the DSO limited the data capture lengths to 300 symbols.
Q14. How long would the CP length be needed for the higher bandwidth system?
For distances significantly longer than 1000 km, the CP length for the higher bandwidth 20-Gbit/s system would need to be increased.
Q15. Why is the high QAM system used?
The high QAM system also illustrates the low levels of IMD that can be achieved using a modulator biased in the linear optical field region.
Q16. What is the amplitude response for all subcarriers?
While the falling amplitude response shown in Fig. 4 is apparent on the frequency response axis, the amplitude response for all subcarriers is constant with time except for the random fluctuations caused by the ASE noise.