Download Optical Comms 2004 (Summer)

1. (a) Explain with the aid of a diagram what is meant by the terms acceptance angle and numerical
aperture for an optical fibre. Hence derive an expression for the numerical aperture of a step
index fibre with a core refractive index of n1 and a cladding refractive index of n2.
[6 marks]
(b) Assuming an impulse-like input pulse to a step index fibre, derive from first principles an
expression for the worst case pulse spreading caused by modal dispersion only, in terms of the
numerical aperture, the fibre length and the fibre core refractive index. State all of your
assumptions clearly.
[10 marks]
(c) Hence, based on the results in part (a) and (b) discuss the trade-off between numerical aperture
and fibre bandwidth and comment on the impact of this trade-off in an actual optical fibre system.
[4 marks]
2. (a) Explain clearly how each of the following physical mechanisms causes attenuation in a silica
optical fibre, stating in each case how the attenuation depends on wavelength:
(i) Intrinsic absorption;
(ii) Extrinsic absorption;
(iii) Rayleigh scattering.
[6 marks]
(b) For a singlemode optical fibre explain concisely what meant by the terms mode field diameter
and normalised spot size.
[4 marks]
(c) For a singlemode fibre the mode field radius w is related to the fibre core radius a by the
expression:
- 3/2
-6
w
= 0.65 + 1.619V
+ 2.879V
a
For a given optical fibre at a wavelength of 1550 nm the mode field diameter is 9.8 µm, when the
normalised frequency is 2.1. Using this information find the core radius of the fibre and the fibre
numerical aperture. Hence determine if singlemode operation is still possible at 1320 nm for this
fibre. Comment briefly on the decision with reference to the two possible definitions of cut-off
wavelength in common use.
[10 marks]
3. (a) What is meant by the term parametric mismatch used in reference to fibre-to-fibre joints?
Illustrate your answer by concisely defining the three most common forms of parametric
mismatch for a multimode fibre.
[4 marks]
(b) The claddings of two multimode step index fibres are perfectly aligned together in an optical
fibre joint. Each fibre has a nominal core radius of a µm. The core of the first fibre is perfectly
concentric within its cladding, but the core of the second fibre is not concentric in its cladding.
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S220/110
The value of the small concentricity error for the second fibre is d µm. No other forms of
parametric mismatch are present.
Assuming a uniform power distribution in the fibre core derive from first principles an expression
for the coupling efficiency between the fibres in terms of the concentricity error d and the core
radius a.
[12 marks]
(c) Hence determine the worst-case joint attenuation in dB for a pair of 50/125 µm multimode
fibres if the maximum core concentricity error per fibre is known to be 2 µm, for the case where
both fibres have a concentricity error. All other joint losses can be considered negligible.
[4 marks]
S220/110
3
Q1(a)
Cladding


Fibre Axis

Core
If  must be greater than c, the critical angle, for TIR and thus propagation to take place,
then the maximum value of 1 under these circumstances is the fibre acceptance angle.
Visualised in space the acceptance angle is defined is a conical half angle, for the fibre. The
numerical aperture for a fibre is the sine of the acceptance angle.
The analysis below applies only to rays entering along the axis, so called meridional rays.
Assuming  is equal to c, the critical angle, what is the value of 
But
and by Snells Law
so
now
 2  90   c
n0 sin  1  n1 sin  2
 1  sin 1(n1 sin(90  c ))
 1  sin 1(n1 sin(90  c ))
 1  sin 1 n 1
1 
hence
n0 is the refractive
index of air =1

 1
sin n 1

cos 2  c
n 22
1 n 2
1
Simple trigonometry



Simple trigonometry
 1  sin 1  n 12  n 22 
This last value is the maximum value that can take on if TIR is to take
place, it is therefore called the fibre acceptance angle.
Numerical aperture (NA) is defined as the sine of the acceptance angle for a fibre. From the
analysis above the NA can be written as:
NA  (n12 - n22 )
(Assumes refractive index of air is approx 1.)
[6 marks]
S220/110
4
Q1(b) In this simple analysis the difference in path length delay is analysed for two light
rays, axial and extreme meridional. The resultant time delay difference effectively tells us the
maximum possible pulse spread or “modal dispersion”
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S220/110
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numerical aperture and other parameters as:
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[10 marks]
Q1(c)
Based on results in part (a) and (b) we can state that
1. As the acceptance angle for a fibre increases the numerical aperture also
increases.
2. As the numerical aperture increases the time delay spread t for a pulse input
also increases, in effect the modal dispersion increases. It is known that an
increase in dispersion reduces the bandwidth of a fibre.
6
S220/110
Overall then it is apparent that there is significant trade-off between a high acceptance angle
and a large fibre bandwidth. The impact of this trade-off in an actual optical fibre system is
important, for good coupling efficiency for optical sources such as lasers or particularly
LEDs a high fibre acceptance angle is very useful. But in effect this means that by selecting a
fibre with a high NA value (eg. plastic optical fibre) and thus improving the coupling
efficiency the result will be a reduction the fibre bandwidth. In simple terms the trade-off
could be defined as one between transmitted (launched) power and bandwidth. As the
transmitted power level influences the received power levels this means that in pure
communications terms a difficult trade-off between received power and transmission
bandwidth can arise for system designers.
[4 marks]
Q2(a)
Loss mechanisms in a Silica Optical Fibre
Intrinsic absorption loss:
Intrinsic absorption is caused by the interaction of the light with one or more of the
components of the glass itself. For silica glass there is a low loss window between 800 and
1600 nm where intrinsic absorption is negligible, by comparison with other loss
mechanisms, such as scattering loss (see below). Intrinsic absorption in this window falls
between 700 nm and 1500 nm, then rises again toward 1700 nm
Extrinsic absorption loss:
Absorption of light caused by impurities in the fibre, such as water and metals ions. One of
the most common impurities is dissolved water in the glass, present as the hydroxyl or OH
ion. In this case the fundamental processes takes place between 2700 nm and 4200 nm, but
gives rise to so called absorption overtones at 1380, 950 and 720 nm. Extrinsic absorption
depends only on the absorption wavelength of a particular impurity and on the level of the
impurity. Very recently newly developed fibre manufacturing techniques have virtually
eliminated absorption loss peaks giving rise to silica fibres which show no absorption peaks,
This in turn opens up transmission at wavelengths circa 1400 nm and 1000 nm, which have
not been utilised to date.
Scattering Loss:
Scattering is a process whereby all or some of the optical power in a mode is transferred into
another mode. This frequently causes attenuation, since the transfer is often to a mode which
does not propagate well. (also called a leaky or radiation mode). One of the most common
forms of scattering is Rayleigh, a the dominant loss mechanism in the low loss silica window
between 800 nm and 1600 nm. The attenuation caused by Rayleigh scattering falls off with
wavelength as a function of the 4th power of wavelength.
[6 marks]
7
S220/110
Q2(b)
Mode field diameter (MFD) is an important property of SM fibres. The amplitude
distribution of the HE11 mode in the transverse plane in the fibre is not uniform, but is
approximately gaussian in shape. The MFD is defined as the width of this amplitude
distribution at a level 1/e (37%) from the peak or for power 13.5% from the peak
The spot size is the mode field radius w. Its value relative to core radius “a” is given by the
expression:
3
/
2
6
w

=
0
.
6
5
+
1
.
6
1
9
V
+
2
.
8
7
9
V
a
where V is the fibre “V-value”.
[4 marks]
Q2(c)
Problem:
Firstly it is noted that the two unknown parameters are the fibre core radius and the
numerical aperture.
Firstly calculate the actual fibre core radius “a”. Using the formula relating w/a to the
normalised frequency V then since V is 2.1 then w/a is 1.22.
The MFD is 9.8 µm, so the spot size is 4.9 µm. Thus the core radius is 4 µm. Using this
value of core radius and the normalised frequency V of 2.1 it is possible to also find the
numerical aperture for the fibre:
2

V
=
a
.
N
A

For a wavelength of 1550 nm from this expression the NA is found to be 0.128
To determine if singlemode operation is possible at 1320 nm we must find the fibre cutoff
wavelength. Using the all of the parameter values available the fibre cutoff wavelength can
be calculated by rearranging the expression above and using the cutoff value of the
normalised frequency V (= 2.405).
2

a
N
A
c
=

V
c
Using the above expression the cutoff wavelength is found to be 1353 nm. Below this
wavelength V > 2.405, so singlemode operation is not possible at 1320 nm according to the
strict criteria above.
In practice the theoretical cutoff wavelength above is difficult to measure. An alternative is
EIA (Electronics Industry Association of America) cutoff wavelength, which states that the
cutoff wavelength is:
S220/110
8
“The wavelength at which the power in the HE21
mode is 10% of the power in the HE11
(fundamental mode)”
Since the EIA cutoff wavelength can be 100 nm less than the theoretical cutoff wavelength it
is possible that singlemode operation defined as above could still take place at 1320 nm. To
determine this the power in both the HE11 and the HE21 mode would need to be calculated.
[10 marks]
Q3(a)
Parametric mismatch: In a fibre-to-fibre joint even where there is perfect “mechanical”
alignment of the fibre cores (e.g. no lateral misalignment etc.) there still exists the possibility
of loss because of slight parameter differences between the two fibres, this is referred to as
parametric loss resulting from parametric mismatch.
The three most common forms of parametric mismatch are:

Core diameter mismatch: If the exit fibre core in a joint is smaller any mismatch will
cause loss

Numerical aperture mismatch: If the exit fibre has a lower NA then loss will occur

Core concentricity. If the core of either is not centred within the cladding loss will
occur in joint types where the claddings are aligned (eg. connectors, mechanical
splices) rather than the actual core (eg. a fusion splice from a full three axis fusion
splice machine).
[4 marks]
Q3(b)
Model assumes optical power is uniformly distributed over the fibre core, so it is best suited
to step index multimode fibres. Diagram shows two fibre cores out of alignment by a
distance "d", caused by a concentricity error in one or both of the fibres. The overlap of cores
is shown by crosshatched area. Fibre radius is "a"
Coupling efficiency is defined in this simple case as the ratio of overlap to core area
d
+
+
a
In general for a circle with a radius "a" the area of a segment defined by the angle is given
by:
S220/110
9
Area = 1/2 a2 ( sin  )
To find the coupling efficiency we need to find the total overlap area. The overlap area is
twice the area of the segment defined by the vertical line x-y in the diagram below. From the
formula above we can now find the area of each segment and thus the overlap area
x
+
+
a
y
d
To proceed we need to express the area of a segment in terms of the core radius and the
misalignment "d" Consider the right angle triangle shown below as a detail of the segment.
From this we can express  as a function of d/a:
a
22
2
d

a
/


a
+
+
 a
+
/
d
2
2
a

(
d
/
)
1
2
c
o
s
A
n
g
l
e
=
d
A
l
l
a
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n
r
a
d
i
a
n
s
10
S220/110
Thus:
[12 marks]
11
S220/110
Q3(c)
Exercise:
To find the attenuation at the joint we need to find the coupling efficiency, assuming that the
only source of attenuation is this parametric mismatch.
Where both fibres have concentricity error the worst case loss occurs where the fibre cores
are aligned in such a way that the concentricity errors add. For a 50 µm fibre the core radius
“a” is 25 µm. The worst case occurs when both concentricity errors add to give an effective
core separation value d of 4 µm. The ratio d/2a is 0.08. Substituting d/2a into the coupling
efficiency formula derived above we get a coupling efficiency of approx. 89.8%. In dB this
represents an attenuation of 0.467 dB.
[4 marks]