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Reti in fibra ottica
Prima Esercitazione – Lunghezze d’onda e potenze
Exercise 1
A WDM optical transmission system is characterized by an available bandwidth, set by the EDFA gain
curve, equal to 30 nm centered around 1550 nm. The channel spacing is equal to 200 GHz. Evaluate the
maximum number of channels that the system can carry.
Exercise 2
A WDM optical transmission system is characterized by an available bandwidth equal to 30 nm centered
around 1550 nm. For every channel it is used a directly modulated laser, so the emitted optical spectrum
extends over 500 GHz. Consider also a 1 nm guard-band left vacant between two channels to provide a
margin of safety against mutual interference. Evaluate the maximum number of channels that the system
can carry.
Exercise 3
Let’s consider a WDM optical transmission system based on 32 channels regularly spaced around 1550
nm with channel spacing equal to 200 GHz. The transmitted power is 0 dBm for every channel. The
received power must be greater than –20 dBm for each channel. The fiber attenuation is wavelength
dependent and the relation between attenuation and wavelength is approximated by the following
formula:
  0.2  A  (  1550nm) 2
dB / km
where: A  30 10 6
.
nm 2
Evaluate the maximum distance of the fiber link.
Exercise 4
Let’s consider a WDM optical transmission system based on channels regularly spaced around 1550 nm
with channel spacing equal to 200 GHz. The fiber link length is 50 km. The transmitted power is 0 dBm
for every channel. The received power must be greater than –20 dBm for each channel. The fiber
attenuation is wavelength dependent and the relation between attenuation and wavelength is
approximated by the following formula:
  0.2  A  (  1550nm) 2
dB / km
where A  30 10 6
.
nm 2
Evaluate the maximum number of channels that the system can carry.
Exercise 5
Let’s consider the optical transmission system used for the distribution of the signal generated by the transmitter TX presented
in the following figure.
20 km
5 km
RX1
RX3
1x2
15 km
2 km
Splitter
RX2
RX4
1 km
1x4
TX
Splitter
1 km
10 km
RX5
5 km
1 km
RX6
1x4
Splitter
10 km
RX7
2 km
RX8
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Each fiber link is characterized by a connector every kilometer. The fiber attenuation is equal to 0.22 dB/km and every splitter
has 1 dB of excess loss. The input power of every receiver must be greater than –20 dBm.
Evaluate the minimum output power that the transmitter must emit.
Exercise 6
Let’s consider the WDM optical transmission system in the following figure.
RX1
TX1
TX2
4
channel
TX3
AWG1
10 km
4
channel
RX2
AWG2
RX3
RX4
TX4
The 4 channels are regularly spaced around 1550 nm and the channel spacing is 200 GHz. The transfer
function for each port of the two AWG is given by the following formula:
T e
 f  f0 


 f 
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where f 0 is the central frequency of the single channel and  f =100 GHz. Every port of AWG1 is
perfectly aligned with the corresponding port of AWG2. Evaluate the power penalty due to an error of
0.5 nm in the wavelength of the signal emitted by the laser.
Exercise 7
Let’s consider the WDM optical transmission system in the following figure.
RX1
TX1
TX2
4
channel
TX3
AWG1
10 km
4
channel
RX2
AWG2
RX3
RX4
TX4
The 4 channels are regularly spaced around 1550 nm and the channel spacing is 200 GHz. Let’s be
each port of AWG1 and
T1 the transfer function for
T2 the transfer function for each port of AWG2. They are given by the following formula:
T1, 2  e
 f  f1, 2 


 f 
2
f1 and f 2 are the central frequencies of the considered port of respectively AWG1 and AWG2, f =100 GHz. Every port
of AWG1 is affected by an error of f / 2 in the central frequency respect with the central frequency of the channel. The
central frequency of each port of AWG2 is affected by an error of  f / 2 respect with the central frequency of the channel.
Evaluate the power penalty due to the total frequency deviation  f .
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