Faculty of Information and
Engineering Technology
Prof. Dr. Abdelmegid Allam
Bar Code
OFCS (NETW 702)
TWO hours
Jan 2014
Final Examination
Please read carefully before proceeding.
1. The duration of this exam is TWO hours
2. Point value of this exam is 40%
3. Only calculators are permitted for this exam
4. This exam booklet contains 17 pages including this one.
Good Luck!
Problem Number
1
2
Total
Possible Marks
30
27
57
Final Marks
1
NETW702-OFCS Formula Sheet:
OPTICAL FIBER PROPERTIES:
ωo /a = 0.65 + 1.619V -3/2 + 2.879V -6
for SI fiber
2
1/2
ωo /a = (1/n1k)(2/Δ)
for GI near parabolic profile
L=1.7(0.85/λ)4 dB/km , NA = (n12- n22)1/ 2 , ∆ = [(n1-n2) /n1]
P2 / P1 = 1- ((α+2)/2α∆)[2a/R +(3λ/4πn2R)2/3]
ηSI = (NA)2 min[1,(a /rs )2 ]
ηGI =2n12 ∆[1-(2/(α+2))(rs /a)α]
Lt = 0.75y/a
Losscd= -10log(dR/dT )2
for dR< dT otherwise =0dB
Lossα = -10log (αR(αT+2))/ (αT (αR+2))] for dR< dT
otherwise =0 dB
2
LossNA= -10log(NAR/NAT )
for NAR< NAT otherwise =0 dB
Losssz= -10log[4(ωoR /ωoT+ ωoT /ωoR ) -2 ] dB
σnGI =
M=(λ/c)|d2n1/dλ2
|
σnGI
=
20 √3 c
σ = σ = σ LM
m c λ
2
RECEIVER PERFORMANCE:
ρ = ηeλ/hc [A/W] h= 6.626x10-34 J.sec e=1.602x10-19C
eP / hf  R 
MeP / hf  R 
2
2
SNR 
k= 1.38 x 10-23 J /K
SNR 
L
2eI D  eP / hf f .RL  4kTf
MeP / hf  R 
(M
L
2
2
SNR 
(M
L
n
SNR 

).2eI D  eP / hf f .RL  4kTf
m / 2.MeP / hf  R 
).2eI  eP / hf f .R  4kTf
2
2
(M

).2eI D  eP / hf f .RL  4kTf
L
n
D
L
2
SNR 
PES
PNS  PNT
 e 
2 RL PS PL  
 hf 


ePL  PS 
1    4kT (f )
2eRL (f )  I D 
hf  PL 

PNA  4kTA f
NEP 
2
2
iNSD
 iNT
 f
PNout  GPN  G 4k (Te )f  G 4k ( FT )f
3
Problem1
[30 Marks]
(A) Derive the formula for the acceptance angle and the numerical aperture of an optical
fiber line, considering the surrounding medium is water of index of refraction nw
(A) LECTURE
(5 Marks)
(B) LECTURE
(9 Marks)
(6 Marks)
(C)
1
2
3
4
5
6
B C D A C A
(10 Marks)
(D)
Given V = 5 , from the graph we can get the modes :
LP01 …………………………………………………….. (1)
LP11 …………………………………………………….. (1)
LP21 …………………………………………………….. (1)
LP02 …………………………………………………….. (1)
Propagation coefficient β for each mode in terms of k :
=
+ 2.1
Using this expression for β and substituting b for each mode , we get for modes :
LP01 : β = 1.474k …………………………………………………….. (1)
LP11 : β = 1.467k …………………………………………………….. (1)
LP21 : β = 1.457k …………………………………………………….. (1)
LP02 : β = 1.455k …………………………………………………….. (1)
Fraction of power that can travel in the cladding :
…………………………………………………….. (2)
4
Problem2
[27 Marks]
(14Marks)
(A) LECTURE
1
D
2
D
3
A
4
B
5
A
6
A
7
B
8
B
9
C
10
C
11
D
E
S
5
6
(B)
7
(C)
8