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3 Optical Signal Attenuation and Dispersion
Drill Problem 3.7 A 10 km transmission link consists of a step-index multimode fiber that has a core index n 1 = 1.480 and a core-cladding refractive
index difference = 0.01.
(a) Using the approximation on the right-hand side of Eq. (3.18), show that
the delay difference between the fastest and slowest modes is 493 ns.
(b) Using Eq. (3.20), show that the rms pulse broadening resulting from
intermodal delay is 142 ns.
(c) Using Eq. (3.18) and the condition that the maximum bit rate B
should satisfy the condition B < 0.1/T, show that the maximum bit
rate-distance product is BL = (2.03 Mb/s) km.
A successful technique for reducing modal delay in multimode fibers is through
the use of a graded refractive index in the fiber core, as shown in Fig. 2.15. In any
multimode fiber the ray paths associated with higher-order modes are concentrated
near the edge of the core and thus follow a longer path through the fiber than lowerorder modes (which are concentrated near the fiber axis). However, if the core has
a graded index profile, then the higher-order modes encounter a lower refractive
index near the core edge. Because the speed of light in a material depends on the
refractive index value, the higher-order modes travel faster in the outer core region
than those modes that propagate through a higher refractive index along the fiber
center. Consequently this reduces the delay difference between the fastest and slowest
modes. A detailed analysis using electromagnetic mode theory gives the following
absolute modal delay at the output of a graded-index fiber that has a parabolic (α =
2) core index profile:
σ s ≈
Ln 1
2
20
√
3 c
(3.21)
Thus for an index difference of = 0.01, the theoretical improvement factor for
intermodal rms pulse broadening in a graded-index fiber is 1000.
Example 3.9 Consider the following two multimode fibers: (a) a step-index fiber
with a core index n 1 = 1.458 and a core-cladding index difference = 0.01; (b)
a parabolic-profile (α = 2) graded-index fiber with the same values of n 1 and .
Compare the rms pulse broadening per kilometer for these two fibers.
Solution
(a) From Eq. (3.20)
σ s
L
≈
n 1
2
√
3 c
=
1.458(0.01)
2
√
3 × 3 × 10 8 m/s
= 14.0 ns/km
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