Optical Communication Components and Subsystems
ECE 565
Spring 2005
Homework 4 Solutions
Problem 2.16.
Received power in dm = 10log
10
0.3x10
3
= 35.2288 dBm
Losses:
Fiber loss = (50)(0.5)=25 dBm
Connector loss = (2)(1)=2 dBm
Number of sections =50/5;
Number of tips = 11;
Number of spliced ends = 11 –2=9;
Splice loss = (9)(0.2)=1.8 dBm
Total loss = 28.8 dBm
Power budget equation (in dBm):
Launched power
 total loss
≥
Receive power
Or minimum launched power =
28.8 dBm
35.2288 dBm = 6.4288 dBm = 0.2276 mW.
Problem 2.17.
Assuming that fiber loss is negligible, Eq. (2.6.14) becomes:
φ
NL
=
γ
P
in
L.
We can calculate
γ
using
γ
=
λ
π
eff
A
n
/
2
2
. Thus, with the given parameters,
γ
= 2.1 W
1
/km.
With P
in
= 6dBm and by setting
φ
NL
= 2
π
, we obtain L=745 km.
We need to minimize (T
1
/T
0
) in Eq.(2.4.17) with respect to z. For, simplicity, we equivalently
minimize (T
1
/T
0
)
2
by differentiating and setting the derivative to zero. (Note that the
coefficient of z
2
in Eq.(2.4.17) is always positive and the critical point of the above quadratic
equation is therefore a minimum.)
d(T
1
/T
0
)
2
/dz = (d/dz)[(1+C
β
2
z/ T
0
2
)
2
+ (
β
2
z/ T
0
2
)
2
] = 2(C
β
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 Spring '11
 Rahmir
 Critical Point, Quadratic equation, Elementary algebra, dBm Connector loss

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