Schneider/Freude Schneider/Freude
WS 2011/12
FPC Problem Set 1 – Proposed Solution Wednesday, Wednesday, October 26, 2011. Problem 1: Fiber attenuation (1)
When the mean optical power launched into an 8 km length of fiber is 120 µW, the mean optical power at the fiber output is 3 µW. Determine: a) The overall signal attenuation or loss in decibels through the fiber.
120µW = −9.2 dBm 1mW 3µ W Pout = 3 µ W ≙ 10 lg = −25.2 dBm 1µW Pin
= 120 µ W ≙ 10 lg
a dB
=
Pin
−
dBm
Pout
(1.1)
= 16,0dB
dBm
b) The signal attenuation per kilometer for the fiber, assuming that there are no connectors or splices that would introduce i ntroduce additional losses. a 16,0dB = = 2dB/km (1.2) α = L 8 km c) The overall signal attenuation for a 10 km optical link using the same fiber with splices at 1 km intervals, each giving an attenuation of 1 dB. dB aF = 2 ⋅ 10 km = 20 dB km dB aSpl = 1 ⋅ 9 Spl = 9 dB (1.3) Spl a = aF + aSpl
=
29dB
d) The value of the input/output power ratio in c). a lin
a dB
29
10
10
= 10
= 10
=
794
(1.4)
Problem 2: Fiber attenuation (2)
The mean optical power launched into an optical fiber link is 1.5 mW and the fiber has an attenuation of 0.5 dB / km. km. Determine the maximum possible link length (assuming lossless connectors) when the optical power level required at the detector is 2 µW. amax
=
Lmax
=
Pin
dBm
− Pout
28.7dB
0.5dB 5dB km Problem 3: Phase velocity and group velocity
Bm d
=
28.7 28.7 dB
= 57.5km
(2.1)
A light pulse at wavelength λ = 1500 nm propagates over a length of 6 km in a medium with refractive index n = 1.5 . At the same time, a second light pulse propagates in parallel in free space. The dispersion of the medium at λ = 1500 nm is given by dn dλ
= −1× 10
−5
nm −1 .
a) Which time delay would result after 6 km due to the different phase velocities?
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Schneider/Freude
WS 2011/12
∆t =
t m − tv
∆t =
6 km ⋅
c
=
Medium: vm
=
m
c = 3 ⋅108
=
Vacuum: vv
L
3 ⋅10
=
n
n
1.5 − 1 3 ⋅108
1.5 1
−
c
s 8 m s
m s
=
c
L
=
2 ⋅ 108
m s
n −1
(2.2)
c
= 10µs
b) Which time delay would result after 6 km due to the different group velocities? Same calculation but with group-velocities. vg
=
ng
=
d ω dk
=
c ng
dk
c
d ω dk dn ng = cn0 0 + ck0 dω d ω dn d λ 1 ng = cn0 + ck0 c d λ d ω 2 2π λ 0
dn
ng
=
n0 − c
ng
=
n0 − λ 0
ng
= 1.5 + 1500 nm ⋅10
⋅
(2.3)
λ0 2π c d λ
dn d λ
∆t g =
L
ng
−1
c
−5
nm −1
= 1.515
= 10.3µs
(2.4)
c) After which time does the pulse arrive? A pulse, as a package of waves, travels with its group velocity. t g
tv
=
=
L ⋅ ng c L c
=
= 30.3µs
in medium (2.5)
20µs in vacuum
For questions and suggestions on the FPC tutorial please contact:
Simon Schneider, Building: 30.10, Room 2.23, Phone: 0721 608 41935, E-mail:
[email protected]
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