Influence of polarization-mode coupling on the transmission bandwidth of single-mode fibers

C. J. Nielsen

doi:10.1364/JOSA.72.001142

Journal of the Optical Society of America
Vol. 72,
Issue 9,
pp. 1142-1146
(1982)
•https://doi.org/10.1364/JOSA.72.001142

Influence of polarization-mode coupling on the transmission bandwidth of single-mode fibers

C. J. Nielsen

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C. J. Nielsen, "Influence of polarization-mode coupling on the transmission bandwidth of single-mode fibers," J. Opt. Soc. Am. 72, 1142-1146 (1982)

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Abstract

A model describing transmission of intensity- and frequency-modulated light through a birefringent single-mode fiber with polarization-mode coupling is presented. It is shown that the distortion caused by the presence of uniform mode coupling is negligible. Polarization-mode-dispersion figures of the order of 40 psec and source wavelength shift of 0.1 nm will, in the worst-case situation, limit the bandwidth to 12 GHz with total harmonic distortion of −80 dB.

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Equations (36)

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i	ξ_i	C_xi	C_yi
1	ψ − βz	½ cos θ	½ sin θ
2	− ψ − βz	½ cos θ	½ sin θ
3	μ + ψ − βz	$(cos θ + \frac{α}{∣ α ∣} sin θ) \frac{1}{4 j}$	$- (\frac{α *}{∣ α ∣} cos θ + sin θ) \frac{1}{4 j}$
4	μ − ψ − βz	$- (cos θ + \frac{α}{∣ α ∣} sin θ) \frac{1}{4 j}$	$(\frac{α *}{∣ α ∣} cos θ + sin θ) \frac{1}{4 j}$
5	−μ + ψ − βz	$- (cos θ - \frac{α}{∣ α ∣} sin θ) \frac{1}{4 j}$	$(- \frac{α *}{∣ α ∣} cos θ + sin θ) \frac{1}{4 j}$
6	−μ − ψ − βz	$(cos θ - \frac{α}{∣ α ∣} sin θ) \frac{1}{4 j}$	$- (- \frac{α *}{∣ α ∣} cos θ + sin θ) \frac{1}{4 j}$

f	f′₀	f″₀
βL = ½(β_x + β_y)L	½(τ_x + τ_y)L	½(β″_x₀ + β″_y₀)L
ψ = \|α\|CL	$\frac{η}{2 C} Δ τ L$	$\frac{η}{2 C} Δ {β^{″}}_{0} L + \frac{{(Δ τ)}^{2}}{4 ∣ α ∣ C^{3}} L$
$μ = arcsin (\frac{η}{C})$	$\frac{Δ τ}{2 ∣ α ∣ C^{2}}$	$\frac{Δ {β^{″}}_{0}}{2 ∣ α ∣ C^{2}} - \frac{{(Δ τ)}^{2}}{2 ∣ α ∣^{2}} \frac{η}{C^{4}}$

i	ξ_i	C_xi	C_yi
1	ψ − βz	½ cos θ	½ sin θ
2	− ψ − βz	½ cos θ	½ sin θ
3	μ + ψ − βz	$(cos θ + \frac{α}{∣ α ∣} sin θ) \frac{1}{4 j}$	$- (\frac{α *}{∣ α ∣} cos θ + sin θ) \frac{1}{4 j}$
4	μ − ψ − βz	$- (cos θ + \frac{α}{∣ α ∣} sin θ) \frac{1}{4 j}$	$(\frac{α *}{∣ α ∣} cos θ + sin θ) \frac{1}{4 j}$
5	−μ + ψ − βz	$- (cos θ - \frac{α}{∣ α ∣} sin θ) \frac{1}{4 j}$	$(- \frac{α *}{∣ α ∣} cos θ + sin θ) \frac{1}{4 j}$
6	−μ − ψ − βz	$(cos θ - \frac{α}{∣ α ∣} sin θ) \frac{1}{4 j}$	$- (- \frac{α *}{∣ α ∣} cos θ + sin θ) \frac{1}{4 j}$

f	f′₀	f″₀
βL = ½(β_x + β_y)L	½(τ_x + τ_y)L	½(β″_x₀ + β″_y₀)L
ψ = \|α\|CL	$\frac{η}{2 C} Δ τ L$	$\frac{η}{2 C} Δ {β^{″}}_{0} L + \frac{{(Δ τ)}^{2}}{4 ∣ α ∣ C^{3}} L$
$μ = arcsin (\frac{η}{C})$	$\frac{Δ τ}{2 ∣ α ∣ C^{2}}$	$\frac{Δ {β^{″}}_{0}}{2 ∣ α ∣ C^{2}} - \frac{{(Δ τ)}^{2}}{2 ∣ α ∣^{2}} \frac{η}{C^{4}}$

Abstract

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Equations (36)

Journal of the Optical Society of America