13.2 Common Network Topologies
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13.2.2 Performance of Star Networks
To see how a star coupler can be applied to a given network, consider the various
optical power losses associated with the coupler. Section 10.2.4 gives the details of
how an individual star coupler works. As a quick review, the excess loss is defined
as the ratio of the input power to the total output power. That is, it is the fraction of
power lost in the process of coupling light from the input port to all the output ports.
From Eq. (10.25), for a single input power P in and N output powers, the excess loss
in decibels is given by
Fiber star excess loss = L excess = 10 log
P in
N
i=1 P out,i
(13.2)
Here P out,i is the optical power output from branch i. In an ideal star coupler the
optical power from any input is evenly divided among the output ports so that all
values of P out,i are equal. The total loss of the device consists of its splitting loss plus
the excess loss in each path through the star. The splitting loss is given in decibels
by
Splitting loss = L split = −10 log
1
N
= 10 log N
(13.3)
Drill Problem 13.1 Assuming that there is no excess loss, show that the splitting losses of ideal 8 × 8, 16 × 16, and 32 × 32 star couplers are 9.03, 12.04,
and 15.05 dB, respectively.
To find the power-balance equation, the following parameters are used:
• P S is the fiber-coupled output power from a source in dBm
• P R is the minimum optical power in dBm required at the receiver to achieve a
specific bit-error rate
• α is the fiber attenuation
• All stations are located at the same distance
L from the star coupler
• L c is the connector loss in decibels.
Then, the power-balance equation for a particular link between two stations in a
star network is
P S − P R = Link Loss + Margin
= L excess + α(2L) + 2L c + L split + Margin
517
13.2.2 Performance of Star Networks
To see how a star coupler can be applied to a given network, consider the various
optical power losses associated with the coupler. Section 10.2.4 gives the details of
how an individual star coupler works. As a quick review, the excess loss is defined
as the ratio of the input power to the total output power. That is, it is the fraction of
power lost in the process of coupling light from the input port to all the output ports.
From Eq. (10.25), for a single input power P in and N output powers, the excess loss
in decibels is given by
Fiber star excess loss = L excess = 10 log
P in
N
i=1 P out,i
(13.2)
Here P out,i is the optical power output from branch i. In an ideal star coupler the
optical power from any input is evenly divided among the output ports so that all
values of P out,i are equal. The total loss of the device consists of its splitting loss plus
the excess loss in each path through the star. The splitting loss is given in decibels
by
Splitting loss = L split = −10 log
1
N
= 10 log N
(13.3)
Drill Problem 13.1 Assuming that there is no excess loss, show that the splitting losses of ideal 8 × 8, 16 × 16, and 32 × 32 star couplers are 9.03, 12.04,
and 15.05 dB, respectively.
To find the power-balance equation, the following parameters are used:
• P S is the fiber-coupled output power from a source in dBm
• P R is the minimum optical power in dBm required at the receiver to achieve a
specific bit-error rate
• α is the fiber attenuation
• All stations are located at the same distance
L from the star coupler
• L c is the connector loss in decibels.
Then, the power-balance equation for a particular link between two stations in a
star network is
P S − P R = Link Loss + Margin
= L excess + α(2L) + 2L c + L split + Margin