Questions
1 (a) An optical fibre system is to operate at 622 MBits/sec over a distance of 71 km without repeaters. Fibre
with a loss of 0.23 dB/km and a dispersion of 5.5 ps/km is available in maximum lengths of 1 km. The
connector loss is 0.8 dB and repair power margin is 5 dB. If the receiver sensitivity is -28 dBm and the
transmitter output power is +1 dBm determine the maximum allowable attenuation per fusion splice,
which can be tolerated during installation.
[10 Marks]
1 (b) If a system upgrade to operation at 1.2 Gbits/sec is required at a later date discuss concisely the impact
of this requirement on the fibre specification. It is known that at the higher bit rate the likely transmitter
output power will be at least +4 dBm, with a receiver sensitivity of -26 dBm.
[10 Marks]
2 (a) Define clearly what is meant by the term power penalty in an optical fibre system and list four the most
common power penalties encountered in the design of an optical fibre system.
[10 Marks]
3 (a) The transfer function of an optical fibre can be approximated by:


H ( f )  A 1  12  2 f  2  t2
where A is a constant and  t2 is the total mean square impulse spread caused by dispersion. Develop an
approximate expression for the dispersion penalty in an optical fiber in terms of the bit rate and total
mean square impulse spread. State clearly any assumptions made.
[10 Marks]
4 (a) A 560 Mb/s optical transmission system is to operate at a wavelength of 1550 nm over an unrepeatered
distance of 51 km. The transmitter available has a minimum coupled output power of +2 dBm, while the
receiver has a worst case sensitivity of -28 dBm. Two types of fibre with different specifications are
available as shown in Table 2 below. Two connectors are to be used in the system with a loss of 0.5 dB
each, while the splice loss for both fibre types is 0.05 dB maximum.
Calculate the dispersion penalty associated with the use of each fibre. Prepare a power budget for each
system and decide which fibre type should be used and why.
Table 2
Fibre type
Total dispersion
Attenuation
A
8 ps/km
0.4 dB/km
Maximum distance
between splices
700 metres
B
10 ps/km
0.35 dB/km
600 metres
[10 Marks]
5 (a) After a number of years of service an existing singlemode fibre is to upgraded to a bit rate of 622 Mb/s
at 1550 nm, using a new transmitter and receiver. The fibre span is 81 km and the fibre has a loss of 0.24
dB per km at 1550 nm and a dimensionless material dispersion coefficient Y equal to 0.007.
The fibre was originally operated at 1330 nm, where the fibre loss was 0.7 dB. The fibre was installed in
Questions
lengths 700 m long and the maximum splice loss was 0.05 dB. The original repair margin was 6 dB, but
no repairs have been required. Two connectors were used with a loss of 0.2 dB each
Two different types of transmitter are available for the upgrade. Transmitter A has an output power of
+3dBm and a RMS spectral width of 0.35 nm. Transmitter B has an output power of +4 dBm and a RMS
spectral width of 0.47 nm. Both transmitters are capable of operating at 622 Mb/s The receiver available
for the upgrade has a worst case sensitivity of -32 dBm.
Complete a power budget for the upgraded system using each of the transmitters in turn to determine
which transmitter should be chosen on the basis of the highest repair margin. State all of your
assumptions clearly.
[10 Marks]
7 (a)
[10 Marks]
8 (a)
[10 Marks]
Solutions
1 (a) 71 km system span at a bit rate of 622 Mbits/s.
To determine the maximum splice loss a system power budget is required. This is shown below:
Parameter
Transmitter output power
Receiver Sensitivity
Value
+1 dBm
-28 dBm
Power Budget
29 dB
Less 5 dB repair margin
Less connector loss 1.6 dB
Less fibre loss 16.33 dB total
Less approximate dispersion
penalty of 1.49 dB
24 dB
22.4 dB
6.07 dB
4.58 dB
Total allowable splice loss
4.58 dB
Comment
Difference between transmitter and
receiver levels.
Two connectors at 0.8 dB max. each.
71 km at 0.23 dB/km
See note 1 below.
Note 1: The dispersion penalty Pd is calculated from the formula:
Pd  10 log10 1 12( B)2 t2
a bit rate of 622 Mbits/s and a total dispersion of 390 ps (71 km x 5.5 ps/km)
Pd  10 log10 (.709)
the dispersion penalty is thus 1.49 dB.
As the fibre is available in 1 km lengths and given that the span is 71 km
the total number of splices is 70. Thus the maximum splice loss which would be tolerated during
installation would be 0.065 dB
[10 Marks]
1 (b) Upgrade plan and implications
In the planned upgrade it is assumed that the bit rate will rise to 1.2 Gbits/s and the available power
budget will rise to 30 dB. If the same fibre where used, with a dispersion of 5.5 ps/km, then the
dispersion penalty will rise to dramatically, because of the square law dependence of the dispersion
penalty on the bit rate B. Thus in order to cater for this future upgrade the fibre to be installed for 622
Mbits/s must have a lower dispersion value, even though this is unnecessary at 622 Mbits/s.
It is should also be pointed out that the fibre loss of 0.23 dB per km is close to the lowest possible using
fibre in the conventional transmission window at 1550 nm. Thus an improvement in this specification is
unlikely to be possible.
Assuming that little else changes (e.g. fibre loss, connector loss etc. ) then the extra 1 dB available in the
power budget may be included in the dispersion penalty, which becomes 2.49 dB. The total dispersion is
247 ps The allowable dispersion per km is thus 3.48 ps/km for the fibre if a future upgrade to 1.2 Gbits/s
is to be viable, while retaining the fibre attenuation specification of 0.23 dB/km.
[10 Marks]
Solutions
2 (a) A power penalty is defined as the increase in receiver power needed to eliminate the effect of some
undesirable system noise or distortion in the system. Typically power penalties can result from: (four
only required)






Dispersion.
Reflection from passive components, such as connectors.
Crosstalk in couplers.
Modal noise in the fibre.
Polarization sensitivity.
Signal distortion at the transmitter (analog systems only).
[10 Marks]
3 (a) The transfer function is given by:
H( f )  A 1  12  2 f  2  t2
A is the value of H(f) at DC, effectively the fibre attenuation.  t2 is the mean square impulse broadening
that occurs over the fibre. To determine an approximate dispersion penalty P d assume that the
transmitted pattern is very simple, e.g. the dotting pattern 10101010..... Also assume that most of the
optical power in this pattern is contained in the component at f= B/2, where B is the bit rate and NRZ
data is assumed. Finally for ease of analysis assume that A is 1.
The extra attenuation caused by dispersion can be approximated by finding H(B/2) To compensate for
this extra attenuation the transmitter output power must be increased by a factor:
1
H B2
 
The dispersion penalty is therefore:

1
Pd  10 Log 
10 
B
H 2
 



Pd  10 Log
 H B 2
Pd  10 Log
1  12  B  2  t2
10
10
This expresses the power penalty Pd in terms of the bit rate B and the mean square impulse spread.
[10 Marks]
4 (a) Fibre selection problem
Step 1: Calculate dispersion penalties for fibre type A and type B.
Find the total dispersion by multiplying the dispersion per km by the systems span of 51 km and then
using the formula above find the dispersion penalty:
Fibre Type A: total dispersion 8 ps/km by 51 km = 408 ps, so the dispersion penalty is 1.3 dB
Fibre Type B: total dispersion 10 ps/km by 51 km = 510 ps, so the dispersion penalty is 2.2 dB
Solutions
Step 2: Now prepare individual fibre power budgets to determine the power margin available after all
forms of attenuation, penalties etc. are taken into account.
Power Budget for Fibre A:
Parameter
Transmitter output power
Receiver Sensitivity
Value
+2 dBm
-28 dBm
Comment
Power Budget
30 dB
Difference between transmitter and
receiver levels.
Connector loss
Total fibre loss
1 dB
20.4 dB
Two connectors at 0.5 dB max. each.
51 km at 0.4 dB/km
Total splice loss
3.6 dB
Approximate dispersion
penalty
1.3 dB
51 km / 700 m = 72.8, so 72 splices
are needed
See calculations above
Total loss
26.3 dB
Power margin available for repair,
upgrades, rerouting etc..
3.7 dB
Power budget - Total loss
Power Budget for Fibre B:
Parameter
Transmitter output power
Receiver Sensitivity
Value
+2 dBm
-28 dBm
Comment
Power Budget
30 dB
Difference between transmitter and
receiver levels.
Connector loss
Total fibre loss
1 dB
17.85 dB
Two connectors at 0.5 dB max. each.
51 km at 0.35 dB/km
Total splice loss
4.2 dB
Approximate dispersion
penalty
2.2 dB
51 km / 600 m = 85, so 84 splices are
needed
See calculations above
Total loss
25.25 dB
Power margin available for repair,
upgrades, rerouting etc..
4.75 dB
Power budget - Total loss
Conclusion: On the basis of the higher power margin available for repair, upgrades, rerouting etc.. for
fibre type B, this is the fibre chosen for use in this system.
[10 Marks]
5 (a) Before completing the power budget we need find the dispersion parameter D using the fibre Y value.
Recall that:
D

c
d n12
d 2
Solutions


1 1

Pd  10
Log D  Y .
thus:
10 
B 
 H 2c

 
So for a Y value of 0.007 at 1550 nm the dispersion parameter is 15.06 ps/nm/km. Using this parameter,
the RMS spectral width and the fibre length it is possible to find the total RMS impulse spread, which
can be used to find the dispersion
penalty
Pd 
10using
Logthe formula:
H B2
10
  
Pd  10 Log
10
1  12  B  2  t2
This expresses the power penalty Pd in terms of the bit rate B and the RMS impulse spread t.
Power Budget System A:
+3 dBm o/p, RMS spectral width 0.35 nm.
RMS pulse broadening calculation
Source RMS spectral width (nm)
RMS pulse broadening ps/km
RMS pulse broadening s/km
RMS pulse broadening over span
0.35
5.27
5.272E-12
4.27E-10
using 15.06 ps/nm/km figure from
above
s/km
seconds
Dispersion penalty calculation
Bit rate:
RMS impulse spread
Dispersion Penalty (dB):
622.0E+6
427.1E-12
1.86E+00
Basic Budget Information
System Span in km
Assume no other penalties
Derived Information
81.00
Transmitter Output Power (dBm)
3.00
Number of Connectors
Connector Loss (dB)
2.00
0.20
Total Connector Loss (dB)
0.40
Fibre Attenuation
0.24
Total fibre attenuation (dB)
19.44
Maximum fibre length available (km)
0.70
No. of fibre lengths needed
115.71
Maximum loss per splice (dB)
0.05
Total splice loss (dB)
5.74
Dispersion Penalty
1.86
Available power margin (dB)
7.56
Receiver sensitivity (dBm)
-32.00
Solutions
Power Budget System B:
+4 dBm o/p, RMS spectral width 0.47 nm.
RMS pulse broadening calculation
Source RMS spectral width (nm)
RMS pulse broadening ps/km
RMS pulse broadening s/km
RMS pulse broadening over span
0.47
7.08
7.080E-12
5.73E-10
using 15.06 ps/nm/km figure from
above
s/km
seconds
Dispersion penalty calculator
Bit rate:
RMS impulse spread
Dispersion Penalty (dB):
622.0E+6
573.5E-12
4.30E+00
Basic Budget Information
System Span in km
Assume no other penalties
Derived Information
81.00
Transmitter Output Power (dBm)
4.00
Number of Connectors
Connector Loss (dB)
2.00
0.20
Total Connector Loss (dB)
0.40
Fibre Attenuation
0.24
Total fibre attenuation (dB)
19.44
Maximum fibre length available (km)
0.70
No. of fibre lengths needed
115.71
Maximum loss per splice (dB)
0.05
Total splice loss (dB)
5.74
Dispersion Penalty
4.30
Available power margin (dB)
6.13
Receiver sensitivity (dBm)
-32.00
Conclusion: Since transmitter A offers a higher repair margin it is the chosen transmitter. This
results from the lower dispersion found using transmitter A, resulting in a lower dispersion
penalty, which more than offsets the lower output power of transmitter A by comparison with B.
[10 Marks]
Solutions
1 (a)
[10 Marks]
1 (a)
[10 Marks]
1 (a)
[10 Marks]