1991
DOI: 10.1109/50.97650
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BER performance evaluation for CPFSK phase and polarization diversity coherent optical receivers

Abstract: The following methods of CPFSK signals demodulation are compared for phase and polarization diversity receivers: single filter, dual filter, delay and multiply. BER values are obtained in each case showing that for negligible laser linewidths the delay demodulation method outperforms the dual-filter method by approximately 3 dB, and the single-filter method by 6 dB. Since the noise bandwidth for MSK and delay demodulation is approximately twice as small as for the other methods, one should add another 3 dB to … Show more

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Cited by 10 publications
(11 citation statements)
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“…The results of these simulations show that the power spectra hardly change for small values of the normalized linewidth, ΔνΤ. This latter condition has to be fulfilled in CPFSK systems in order to arrive at allowable bit error rates [1]. But since in that case the power spectra did not change significantly, as follows from the simulations, it did not make sense to convolve the analytically derived spectra with the laser spectra.…”
Section: Discussionmentioning
confidence: 91%
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“…The results of these simulations show that the power spectra hardly change for small values of the normalized linewidth, ΔνΤ. This latter condition has to be fulfilled in CPFSK systems in order to arrive at allowable bit error rates [1]. But since in that case the power spectra did not change significantly, as follows from the simulations, it did not make sense to convolve the analytically derived spectra with the laser spectra.…”
Section: Discussionmentioning
confidence: 91%
“…lasers is different from zero. In practical systems the normalized linewidth ΔνΤ is smaller than 0.01 because else the BER-floor will become too high [1]. If ΔνΤ is smaller than 0.01 the linewidth has negligible influence on the shape of the spectrum.…”
Section: Simulationsmentioning
confidence: 98%
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“…This can be seen by setting φ Ν (1), ni(t), n z (t), and n 3 (t) to zero in (10) and (11), then substituting (10) and (11) into (12) to have u d (t) = 4€ 2 8ίη(2πί α τ) if bit "1" is transmitted. One can refer to [7] for detailed discussion on bandwidth requirements. One can refer to [7] for detailed discussion on bandwidth requirements.…”
Section: Numerical Results and Discussionmentioning
confidence: 99%