Sep. 26
th
P41 2.6 Calculate the energy of a single photon at 1300 nm and at 1500 nm.
Solution: A single photon at 1300 nm:
)(10529.1
101300
10626.6103
/ 19
9
348
JchhfE p
A single photon at 1550 nm:
)(10282.1
101550
10626.6103
/ 19
9
348
JchhfE p
2.7 a. How many photons per second emanate from a laser diode radiating at 1300
nm if its power is 1 mW?
b. How many photons per second emanate at 1550 nm?
Solution: Using the result above:
a: )(101101 33 JPtE
NEE p
15
19
3
10540.6
10529.1
10
pE
E
N
b: NEE p
15
19
3
10798.7
10282.1
10
pE
E
N
P81 3.12 For a specific fiber, NA = 0.2375 and n1 = 1.4860. Find n2 (n cladding).
Solution:
2
2
2
1 nnNA hence
2
2
22 4860.12375.0 n
Hence n2 = 1.4669
3.22 An optical fiber with attenuation of 0.25 dB/km is used for 20-km
transmission. The light power launched into the fiber is 2mW. What is the output power?
Solution: Loss(dB) = -10 lg (Pout/Pin)
A(dB/km) = loss (dB) / fiber length(km)
Hence,
20
2
lg10
25.0 mW
Pout
Hence, Pout = 0.6325 mW.
3.25 Find the maximum transmission distance for a fiber link with an attenuation of
0.3 dB/km if the power launched in is 3mW and the receiver sensitivity is 100 W.
Solution: Loss(dB) = -10 lg (Pout/Pin)
A(dB/km) = loss (dB) / fiber length(km)
Hence,
hfiberlengt
3
10100
lg10
3.0
3
Hence, fiber length = 49.24 km
3.7 The core refractive index is 1.4513 and the cladding index is 1.4468. What is
(1) the critical propagation angle? (2) the acceptance angle? (3) the numerical aperture?
Solution:
(1) The critical angle is:
5131.4
4513.1
4468.1
1sin1sin
2
1
2
1
21
n
n
c
(2): The acceptance angle is:
1148.135131.4sin4513.1sin2sinsin22 11
1
ca n
(3): The NA: NA = sin a = 0.1142
3.21 What does the term “transparent windows” mean? Specify three peak
wavelengths for the transparent windows in modern optical fibers.
Solution:
The term “transparent windows” means that the regions where absorption reaches minimum
value in the typical spectral attenuation diagram (Loss (dB/km) - (nm) diagram).The three
peak wavelengths for them are: 850 nm, 1300 nm, and 1550 nm.
3.29 What is the number of modes VN . for a graded-index fiber if d is 50 m,
NA is 0.200, and the operating wavelength is 1300 nm?
Solution: The V number is:
2.24200.0
101300
1050
9
6
NA
d
V
the number of modes is:
146
4
2.24
4
22
VN
3.30 How many modes can support a step-index optical fiber whose d = 8.3 m, n-
1core = 1.4513, n2dad = 1.4468, and = 1550 nm?
Solution: 92.14468.14513.1
101550
103.8 22
9
6
2
2
2
1
nn
d
V
V < 2.405, N = 1
P71 E3.47 A graded-index fiber has n1 = 1.486 and NA = 0.200. What is the bit rate for a
1-km link?
Solution:
t
BR
cn
NAL
cN
NAL
tGI
4
1
3232 31
4
3
1
4
where N1 core group index of refraction
Hence, sGbitsbit
NAL
cn
BRGI /92.4/1092.4
200.0101
486.110388 9
43
38
4
3
1
Oct. 17
th
4.30 A graded-index fiber has the following characteristics: NA = 0.200,
dcore = 50m and = 1300 nm. What power is carried by the fiber’s cladding?
S: 17.24200.0
1300
1050 3
Z
NA
d
V
%9.3
17.243
22
3
22
VP
P
total
clad
Hence, about 3.9% of the total power is carried by the fiber’s cladding.
4.51 Calculate the pulse spreading caused by chromatic dispersion for
BF04431-02 multimode grade-index fiber from SpecTran, (See Figure 3.20) operating
at 1300 nm. Assume = 70 nm.
S: Taking 0 = 1342.5 nm, S0 = 0.097
kmnm
ps
2
3
4
00
4
S
D
kmnm
ps
329.4
1300
5.1342
1300
4
097.0
3
4
tgmat = D() = 4.329 x 70 = 303 ps/km = 0.3 ns/km
4.55 What is meant by the term dispersion power penalty? Calculate the
disperson power penalty for the pulse spread obtained in Problem 4.51. The BR is 2.5
Gbit/s.
S: The amount of transmitting power required for compensating the
increase bit error rate (BER) cause by attenuation is called dispersion power penalty.
In Problem 4.51, ttotal = tmat = 18 ns (if fiber length is 100 km)
PD(dB) = -10 log10 {exp[-(1/4 ) (ttotal)
2
(BR)2]}
= -10 log10 {exp[-1/4 x (18 x 10
-9
)
2
( x 2.5 x 109)2]}
= 2.17 x 10
4
dB,
enough to burn any type of fiber, impossible.
if the fiber length is 1 km:
P 4.34
2
4
t
2
BR
2
PD=2.17 dB ( realizable!)
4.56 What is the bit-rate length limitation caused by chromatic disperson for
the Spectran fiber referred to in Problem 4.51?
S: (BR x L)max = 1 / [4 D()]
D)2(41
1210329.427041
kmsGb /17.1
Chapter 5-6
5.13 Compute the pulse spread caused by chromatic dispersion if a fiber has a zero-
dispersion wavelength at 1312 nm, a zero-dispersion slope of
kmnm
ps
2
090.0 , a length of
100 km, and operates at 1310 nm. The LD’s spectral width is 1 nm.
S:
kmnm
ps
SD
18.0)13121310(090.000
psLDtchrom 18100118.0
5.17 Calculate pulse spread caused by polarization-mode dispersion if
kmpsDPMD /2.0 and L = 120 km.
S: psLDt PMDPMD 19.21202.0
6.2 What is the V-number of the Low-PMD SM from Plasma Optical Fibre BV,
assuming that the core diameter is equal to 8.3 m and the relative index equals 0.36%?
S: 889.1%36.0
101215
467.1103.8
2
9
6
c
dn
V
6.8 What length of DCF is needed to compensate for chromatic dispersion at a
120-km span of Low-PMD fiber from Plasma Optical Fibre BV (see Figure 5.17) if DDCF()
= -127 ps/nm*km and = 1550 nm?
S: compenschrom tt
DCFDCF LDLD 4
DCFDCF LDLS
3
4
0
0
DCFL127120
1550
1310
1550
4
092.0
3
4
L = 16.50 km
6.10 a. What means do we have to cope with PMD?
b. What bit rate can be achieved at a 120-km span with Low-PMD Fiber from
Plasma Optical Fibre BV?
S a: We use PM fiber or PM components together to make a PM Link. We also
develop PMD compensation techniques.
b: psLDt PMDPMD 19.21202.0
sGbit
t
BR
PMD
PMD /114
19.24
1
4
1
P246 7.2 A drying gas flows through a preform during the consolidation phase of the
OVD process. Why is this important?
S: It is the remove residual moisture because optical fiber’s absorption properties
are caused by some molecules of the hydroxide anion OH
-
, which is contained in water.
Without the process above, these molecules will be incorporated in silica during fabrication
and be very hard to eliminate. OH
-
molecules have major peaks of absorption at some
wavelengths. Moreover, every foreign particle found inside the core will slightly modify the
refractive index and cause scattering, including water molecules. Thus it is important to dry
the perform.
7.6 List four basic components of a fiber cable and explain their functions.
S: The four basic components of a fiber cable are fibers, buffer tube, strength
member, and outer jacket. The plastic buffer tube is the first shield protecting the fiber from
environment damage. The strength member is to release the fiber from mechanical stress
during installation and operation. Outer jacket assembles the entire construction and shields it
from an adverse environment.
8.1 a. Calculate the intrinsic connection losses for two 6.25/125 graded-index
multimode fibers manufactured by Plasma Optical Fibre (Figure 3.18) caused by (a) diameter
mismatch and (b) NA mismatch.
b. Calculate the intrinsic connection losses caused by the MFD mismatch of two
singlemode fibers made by Plasma Optical Fibre (Figure 5.17).
Solution:
a: From Figure 3.18, it is found that the tolerance of core diameter to be 62.5 25 m,
and NA 0.275 0.015.
(a) dB
a
a
Losscore 6952.0
5.25.62
5.25.62
log10log10
22
1
2
(b) 9485.0
015.0275.0
015.0275.0
log10log10
22
1
2
NA
NA
LossNA dB
b: From Figure 5.17, it is found that MFD1 = 9.3 + 0.5 m, MFD2 = 9.3 – 0.5 m
dB
W
W
W
W
LossMFD 05.0
8.9
8.8
8.8
8.9
4
log104log10
2
2
2
1
1
2
8.3 Calculate the extrinsic connection losses for two SM fibers manufactured by
Plasma Optical Fibre caused by (a) lateral misalignment for x = 0.5 m, (b) angular
misalignment for = 30”, and (c) end separation for Z = 0.5 m. (Figure 5.17).
S: From Figure 5.17, I obtained that W0 = ½ MFD = ½ x 9.3 m, = 1550 nm,
n = 1.467
(a):
2
0
2
explog10
W
xLoss lat
dB05.0
3.9
5.0
explog10
2
2
1
2
(b): 3
3
2
1
0 103793.2
101550
"30sin3.9467.1sin
ZnZW
T
dBTLossang 52 104587.2explog10
(c): dB
S
Lossend
5
2
105668.6
1
1
log10
8.22 A fiber link includes five splices at 0.02 dB/splice, four connectors at 0.2
dB/connector, transmitter power of –10 dBm, and receiver sensitivity of –25 dBm. What
length of this link will be allowed if a singlemode fiber cable with attenuation of 0.3 dB/km is
used and the required power margin is 3 dB?
S: Transmitter power, Pin = -10 dBm
Cable Loss 0.3 dB/km * L km = 0.3L dB
Splicing Loss: 0.02 dB/splice * 5 splices = 0.1 dB
Connector Loss: 0.2 dB / connector * 4 connectors = 0.8 dB
Receiver Sensitivity, PRS = -25 dBm
Power margin: 3 dB
From the above, it follows that:
Pout = Power margin + PRS = 3 + (-25) = -22 dBm
Total Loss = Pin – Pout = -10 – (-22) = 12 dB
Total Loss = Cable Loss + Splicing Loss + Connector Loss
Hence, 0.3L + 0.1 + 0.8 = 12
L = 37 km
8.23 A local data link to be installed has the following characteristics:
Maximum bit rate: 32 Mbit/s
Line code: return-to-zero (RZ) format
Fiber: 6.25/125 m
Installation length: 2000 m
Operating wavelength: 1300 nm
Rise time of lightwave equipment: 4 ns
ID spectral width: 2 nm
Will this fiber support the required bit rate? Prove your answer.
S: 222 riseltwrisesystrisefib
2
2
35.0
riseltw
BW
nsnsns
MHz
18.106289.1034
32
35.0 22
2
(Required fib-rise)
(From Figure 3.18) BWmodal = 500 MHz*km/2 km = 250 MHz
BWel-modal = 0.707 BWmodal = 0.707 x 250 = 176.75 MHz
mod-rise = 0.35 / BWel-modal = 1.9802 ns
chrom-rise = tchrom = D()L
(From Example 3.35) = 4.38 ps/(nm*km) x 2 km x 2 nm = 17.52 ps
nsrisechromriserisefib 9803.101752.09802.1
2222
mod
compared with the required rise time of 10.18ns, the chosen fiber will
support this link.
Nov 10
th
P361 9.16 Evaluate power in mW and dBm (1) coupled into a 62.5/125 m fiber and (2)
radiated by a 1.3-m (2) radiated by a 1.3-m SLED (Figure 9.8[a]) at 100 mA. Assume that
the LED radiates as a Lambertian source.
S: From Figure 3.18, it was found that NA = 0.275 for a 62.5/125 m fiber.
Pin = P0(NA)
2
= 0.275
2
P0
(1): Looking at Figure 9.8[a], when at 100 mA a 1.3 m SLED coupling power
into a 62.5 m fiber is 75 W. Thus, dBm
mW
W
PdBm 2.11
1
75
lg10
.
(2): The power radiated by a 1.3-m SLED at 100 mA is
dBm
mW
mW
dBmPmW
NA
P
P in 044.0
1
99.0
lg10,99.0
275.0
075.0
20
9.19 The 1.3 m-SLED shown in Figure 9.8(a) has a spectral width of 155 nm at
25C. What will be the spectral width if the temperature changes to 90C?
S: From Figure 9.8(a), /T = 0.38 nm/C
Hence, 38.0
2590
155
x
x = 179.7 nm
9.32 How many longitudinal modes can a Fabry-Perot laser diode generate if the
length of its resonator is 0.3 mm and the operating wavelength is 1550 nm? The width of the
gain curve is 9 mm.
S: N - N+1 =
2
/2L = 1550
2
/ (2 x 0.3 x 10
6
) = 4 nm
The width of the gain curve is 9mm, thus it can generate 2 longitudinal modes.
9.41 Calculate the slope efficiency of a DFB LD whose specifications are given in
Figure 9.19 on page 356.
S: S = (PR – Pth) / (Itot – Ith) = mAmWmAmA
mW
135.0
1587
07.9
CH10
P428 10.14 Calculate the external quantum efficiency of a laser diode if its slope
efficiency is equal to 0.08 mW/mA and the operating wavelength is 1550 nm.
S: Ep = hc/
= 6.63 x 10
-34
x 3 x 10
8
/(1550 x 10
-9
) = 1.2832 x 10
-19
J
S(W/A) = (Ep/e) ext
ext = %98.9
102832.1
106.108.0
19
19
A
W
E
eS
p
10.15 Calculate the power efficiency of the DFB laser diode whose data
sheet is given in Figure 9.19.
S: From Figure 9.19, It was found the forward current to be IF = 65 mA,
assuming that R = 1 , = 1550 nm, Pout = 2 mW.
p =
RI
e
Eg
I
Pout
2
= %5.3
4805.065
2
mWmW
mW
10.32 Explain the meaning of the term “chirp” as applied to an LD. How
does this phenomenon affect the LD’s transmission characteristics?
S:
Chirp is the deviation of laser frequency from its radiation-center frequency.
Chirp results in a broadening of a laser’s linewidth.
10.34 Calculate the laser noise power detected by a receiver for the following
data: RIN = -150 dB/Hz, Preceived= 80 W, and BW = 140 MHz.
S: RIN(dB / Hz) = 10 lg [ RIN (1/Hz)] = -150 (dB/Hz)
Hence, RIN = 10
-15
(1/Hz)
RIN(1/Hz) =
/ [
2
BW]
= HzWBWPHzRINPN
62261522 101401080101
= 3 x 10
-8
W = 0.03 W
CH11
P493 11.5 The responsivity of a PD is 0.9 A/W and its saturation power is 2 mW. What
is the photocurrent if the received power is (1) 1mW (2) 2mW (3) 3mW?
S: p = RP, R = 0.9 A/W, PC = 2 mW
(1): P = 1 mW, P = RP = 0.9 x 1 mW = 0.9 mA
(2): P = 2 mW, P = RP = 0.9 x 2 mW = 1.8 mA
(3): P = 3 mW > PC = 2mW; and as saturation means no more increase, ~ 1.8 mA.
11.23 What is the bandwidth of a germanium photodiode with w = 20 m and vsat ~
10
5
m/s?
S: pss
m
V
W
s
msat
tr 200102
10
1020 10
5
6
sGbit
ps
BW
RCtr
PD /796.0
2002
1
2
1
11.48 Calculate the RMS and bandwidth-normalized values of total noise current for
a PD whose data are given in Figure 11.24 if the average input power is 0.1 uW, = 1550 nm,
RL = 50 k, and the diode operates at room temperature.
S: Shot-noise current
Ip
*
= RPin
*
= 0.86 A/W x 0.1 W = 0.086 A
nAGHzAcBWiei PDps 42.70.2086.0106.122 19*
HznaAcei
BW
i
i p
PD
s
SN /1066.1086.0106.122
419*
Thermal-noise current
nAGHz
k
kBW
Rc
Tk
i kJPD
B
t 7.250.250
3001038.14
4 23
Hz
nA
BW
i
i
PD
t
tN
41075.5
Dark-noise current
id
*
= 1.6 nA
nAGHznAcBWei PDdd 98.00.215106.122
19*
52.19 10ddN
PD
i nAi
BW Hz
1/f noise current
Since a PD is used in high-speed application, this current is negligible.
Total noise current
The RMS value:
nAiiii dtsnoise 77.2698.07.2542.7
222222
The RMS value of bandwidth-normalized total current noise:
Hz
naiiii dNtNsNnoiseN
4252424222 1099.51019.21075.51066.1
11.52 Calculate SNR for a PD whose data are given in Figure 11.24 if the average
input power is 0.1 uW, = 1550 nm, and the diode is operating at room temperature with the
following parameters:
a. RL = 50 , RL = 500 k and RL = 50 M
b. BWPD = 50 kHz, BWPD = 50 MHz, BWPD = 50 GHz
Draw conclusions as to how load resistance and bandwidth affect SNR.
S: R = 0.86 A/W, Pin
*
= 0.1 W, hence *
* 22
2
in
R
p Pi = 7396 nA
2
From 11.48, I obtain
Hz
nAi
Hz
nAi dNsN
54 1019.2,1066.1
For RL = 50 ,
L
B
tN R
Tk
i
4
Hz
nA21082.1
For RL = 500 k,
L
B
tN R
Tk
i
4
Hz
nA41082.1
For RL = 50 M,
L
B
tN R
Tk
i
4
Hz
nA51082.1
Hence, for BWPD = 50 kHz, inoise = is
2
+ it
2
+ id
2
= (isN
2
+ itN
2
+ idN
2
) BWPD
)50(1021.5
)500(1042.2
)50(53.446
)(
**
6
6
222
2
2
2
MR
kR
R
BWiii
p
i
p
SNR
L
L
L
PDdNtNsNnoise
ii
For BWPD = 50 MHz,
)50(1021.5
)500(1042.2
)50(45.0
3
3
MR
kR
R
SNR
L
L
L
For BWPD = 5 GHz,
)50(1.52
)500(2.24
)50(105.4 3
MR
kR
R
SNR
L
L
L
From the results above, we can draw the conclusion that SNR increases with RL and
decreases with BWPD, with BWPD being the dominant factor affecting SNR.
11.54 Calculate SNRs, SNRt, and overall SNR for Si and ZnGaAs avalanche PDs if
M = 10, P = 0.1 uW, R = 0.9, RL = 50 k, BWPD = 20 GHz, and T = 300k.
S: For an Si APD, Fs = 2.49
48.56249.2106.12
1.09.0
2 19
GHzc
W
BWeF
RPSNR
PDs
s
83.12222
50
3001038.14
1.09.010
4
23
222222
GHz
k
W
BW
R
Tk
PRM
SNR
PD
L
B
t
(Overall SNR = 54)
For an InGaAs, Fs = 12.78
SNRs = 0.11
278.12106.12
1.09.0
2 19
GHzc
W
BWeF
RP
PDs
SNRt = 1222.83
(Overall SNR = 11)
11.59 a. What is BER?
b. How can you improve BER?
S a: BER stands for bit-error rate, which is the number of the erroneous bits per
total number of bits transmitted.
b: The only way to minimize BER is to make th at thopt, which introduces digital
SNR, Q. The larger the Q, the less the BER will be.
11.63 Why does BER depend on the Q parameter?
S:
01
01
ii
ii
Q
When i0 = 0, Q =
01
1
ii
i
, which is simply the ratio of signal current to noise
current. BER depends on SNR, so it depends on Q.
CH12
P582 12.12 What do we mean by the term add/drop procedure? Why is this a
problem in TDM networks?
S: When TDM or WDM is used, an add/drop node is required to extract
the desired signal from the main data stream. The problem of an add/drop procedure
is that one pushes the main data stream at a high bit rate while the individual channel
operates at a much lower bit rate. Thus, the main data stream needs to be
demultiplexed to the bit rate of the individual channel.
12.29 What is the channel spacing for a WDM with 40 channels?
S: 0.8 nm (100 GHz)
12.58 Calculate the noise figure of an optical amplifier if the input-signal
power is 250 W, the input-noise power is 25 nW in a 1-nm bandwidth, the output-
signal is 50 mW, and the output-noise power is 15 W in a 1-nm bandwidth.
S: Fn =
poweroutnoisepoweroutsignal
powerinnoisepowerinsignal
SNR
SNR
out
in
/
/
3
10151050
102510250
15
50
25
250
63
96
W
mW
nW
W
12.70 Calculate the gain of an EDFA if the input power is 150 W, the
output power is 55 mW, and the noise power is 20 W.
S: Gain (dB) = 10 lg dB
W
WmW
P
PP
in
ASEout 42.23
250
2055
lg10
12.77 Calculate the noise figure of an EDFA if the input-signal power is 300
W, the input-noise power is 30 nW in a 1-nm bandwidth, the output-signal power is
50 mW, and the output-noise power is 20 W in a 1-nm bandwidth.
S: Fn =