ENE 623/EIE 696
Optical Communication
Lecture 3
Example 1

A common optical component is the equal-power splitter which splits the
incoming optical power evenly among M outputs. By reversing this
component, we can make a combiner, which can be made to deliver to a
single output the sum of the input powers if multimode fiber is used, but
which splits the power incoming to each port by a factor of M if singlemode fiber is used.
(a) Compare the loss in dB between the worst-case pair of nodes for the
3 topologies if the number of nodes is N = 128 and multimode is used.
Assume that, for the tree, there are 32 nodes in each of the top two
clusters and 64 nodes in the bottom one.
(b) What would these numbers become if single-mode fiber were used?
(c) How would you go about reducing the very large accumulated splitting
loss for the bus?
Example 1

Soln
Example 2

A light wave communication link, operating at a wavelength of 1500 nm and
a bit rate of 1 Gbps, has a receiver consisting of a cascaded optical amplifier,
narrow optical filter, and a photodetector. It ideally takes at least 130
photons/bit to achieve 10-15 bit error rate.
(a) How many photons/bit would it take to achieve the same error rate
at 10 Gbps?
(b) At this wavelength, 1 mW of power is carried by 7.5 x 1015
photons/s, what is the received power level for 10-15 bit error rate at 1
Gbps?
(c) Same as (b) but at 10 Gbps?
Example 2

Soln
Four-port optical couplers

By definition:
Pout1  Pout 2  LPin
Pout1
r
LPin
rLPin


1 1 r
1
r
 rPout 2
Pout 2 
Pout1
Pout1
Pout 2
LPin

1 r
Pout1
r
Pout 2
Four-port optical couplers

Power division matrix
 Pout1   C11 C12   Pin1 
 P   C
 P 
C
22   in 2 
 out 2   21
rL
C11  C22 
1 r
L
C12  C21 
1 r
C11  C12  L
where Cij is called “incoherent additional input amplitude.
Four-port optical couplers

r = 0  C11 = 0, C12 = L
C 
ij


0 1 

L

1 0 
All input power crossover to output 2.
r = ∞  C11 = L, C12 = 0
C 
ij

1 0 

L

0 1 
All input power goes straight through.
Four-port optical couplers

r = 1  C11 = C12 = L/2
C 
ij

 0.5 0.5

L

 0.5 0.5
3-dB coupler or ‘50-50’ coupler.
Example 3

For the 4-port fiber optic directional coupler, the network below uses 8 of these
couplers in a unidirectional bus. Assume that the excess loss of each coupler is 1
dB.
(a) If the splitting ratio is 1 for all of the couplers, what is the worst case loss
between any Tx and Rx combination in dB?
(b) What is the least loss between any Tx and Rx?
Example 3

Soln
Multimode fiber (SI fiber)


Rays incident at an angle to axis travel further than rays
incident parallel to an axis.
Low-length bandwidth product (<100 MHz-km)  not
widely used in telecommunications.
Multimode fiber (GRIN fiber)


Rays incident at angle to the axis travel farther but also
low average index (
)
n  n1
Length bandwidth product is lot greater than one of SI
multimode fiber.  Useful for telecommunications.
Single-mode fiber



Only one mode propagates: neglecting dispulsion all
incident light arrives at fiber end at the same time.
Length bandwidth product > 100 GHz-km.
Much greater bandwidth than any multimode fiber. 
suitable for long live intercity applications.
Modes in fibers
Modes in fibers

It begins with Maxwell’s equations to define a wave
equation.

In an isotropic medium:
2
2
n

E
2 E  2 2
c t
2
2
2
2
  2 2 2
x
y
z
n2
2


,


n
0
2
c
Modes in fibers


We have 3 equations with solution of Ei for each axis
which is not generally independent.
Assume that wave travels in z-direction:
E ( x, y , z , t )  E ( x , y ) e i (  z   t )
  propagation constant
  2
Substitute these into a wave equation, it yields
Modes in fibers
 2 E ( x, y )  2 E ( x, y )


2
2
x
y
n 2 n

c
c
2
We know k0 
.


 2 E ( x, y )  2 E ( x, y )


2
2
x
y
0
Modes in fibers

For guided mode: n2 < neff < n1

For radiation mode: neff < n2
Modes in Fibers

If we rewrite a wave equation in scalar, we get
 2u  2u
2 2
2


n
k


u0
0
2
2
x
y
u ( x, y, z )  Ex or E y (propagation in z-direction)


n ( x, y )  n ( r )
2
2
2 1 
1 2
From 2  2  2 
 2
r r r  2
x
y
r
x  r cos  , y  r sin  , r  x 2  y 2 ,  
Modes in Fibers

Solutions for the last equation are
u (r )eil
for l  0, 1, 2,...
Solutions to scalar wave equation for step index fiber:
u (r ,  ) 
AJ lm ( pr )eil ; r  a
J m ( x)  Bessel Function of 1st kind of order m.
K n ( x)  Modified Bessel Function of 1st kind of order m.
Modes in Fibers
Modes in Fibers

It is convenient to define a useful parameter called ‘Vnumber’ as
V  a p2  s2
V


V is dimensionless.
V determines


Number of modes.
Strength of guiding of guided modes.
Modes in Fibers
Modes in Fibers

Mode designation LPlm


l = angular dependence of field amplitude eil (l = 0,1,..)
m = number of zeroes in radial function u(r)
 Fundamental LP01 mode: no cutoff. It can guide no
matter how small r is.
 u(r,) = u01(r) ….circular symmetric maximum at r = 0.
Modes in Fibers
 Two mode fiber guide LP01 and LP11 modes:
Modes in Fibers
Mode
Cutoff condition
LP0m
l=0
J-1(r)=0
LP1m
l=1
J0(r)=0
LP2m
l=2
J1(r)=0
V at cutoff @ m =1,2,3
0
LP01  HE11
LP11  TE01, TM01, HE21
For large V, number of guide modes = V2/2
3.832
7.016
Example 4

Find a core diameter for a single-mode fiber with λ=1330
nm.