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Design of double-pass discrete Raman amplifier and the impairments induced by Rayleigh backscattering

M. Tang, P. Shum, and Y.D Gong

Open Access Open Access

Abstract

We report on the investigation of discrete Raman fiber amplifier in double-pass configuration based on the dispersion-compensated fiber and high reflection FBG. We proved in simulation and experiments that the double-pass configuration requires nearly 50% less pump power and the same fiber length to provide the same Raman gain and double-dispersion-compensation performance compared to the typical counter-pumped Raman amplifier. We also analyzed the equivalent noise figure (NF) and the Rayleigh backscattering impairments. The theoretical results shown that the impact of multipath interference (MPI) noise is the dominating limitation factor of this system operated at very high Raman gain region.

©2003 Optical Society of America

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Figures (7)
Fig. 1.
Fig. 1. Double-pass discrete Raman amplifier configuration.

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Fig. 2.
Fig. 2. Signal and noise light (ASE and MPI) flows.

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Fig. 3.
Fig. 3. Signal and pump power along the 3-km DCF for double- and single-pass schemes.

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Fig. 4.
Fig. 4. Raman gain versus pump power at double- and single-pass configuration.

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Fig. 5.
Fig. 5. ASE and RB/DRB noise light power versus Raman gain.

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Fig. 6.
Fig. 6. Equivalent NF with/without considering the MPI noise.

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Fig. 7.
Fig. 7. Overall equivalent NF for different reflection ratio R.

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Equations (11)

Equations on this page are rendered with MathJax. Learn more.

± d P SK dz = − α s P SK + g · P SK P Pf
d P Pf dz = − α P P Pf − g · ∑ P SK · P Pf
G Rf ( z 1 , z 2 ) = exp ⁡ { g P Pf ( 0 ) [ exp ⁡ ( − α P z 1 ) − exp ⁡ ( − α P z 2 ) ] α P }
G Rb ( z 1 , z 2 ) = exp ⁡ { g P Pf ( 0 ) α P [ exp ⁡ ( α P ( z 2 − L ) ) − exp ⁡ ( α P ( z 1 − L ) ) ] }
N Sb ASE ( z ) = N Sf ASE ( L ) T ( 0 , L ) G Rb ( 0 , L ) + h υ ∫ 0 L − z P Pf ( x ) T ( 0 , x ) G Rb ( L − x , L ) dx
N Sf ASE ( L ) = h υ ∫ 0 L P Pf ( x ) T ( x , L ) G Rf ( x , L ) dx
P Sf RB ( 0 ) = r P Sf ( 0 ) G b ( 0 , L ) ∫ 0 L G f ( 0 , z ) G b ( 0 , z ) dz
P Sf DRB ( L ) = r 2 P Sf ( 0 ) G f ( 0 , L ) ∫ 0 L 1 G f 2 ( 0 , z ) ∫ z L G f 2 ( 0 , x ) dx · dz
P Sb RB ( L ) = r P Sb ( L ) G f ( 0 , L ) ∫ 0 L G b ( z , L ) G f ( 0 , z ) dz
P Sb DRB ( 0 ) = r 2 P Sb ( L ) G b ( 0 , L ) ∫ 0 L 1 G b 2 ( z , L ) ∫ z L G b 2 ( x , L ) dx · dz
NF = 1 G ON − OFF [ 2 N Sb ASE ( 0 ) h υ + ( 5 9 ) P RB ( 0 ) h υ ( B e 2 + B s 2 2 ) 1 2 + 1 ]
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