Ec2402 Optical QB 2bsub
Ec2402 Optical QB 2bsub
Ec2402 Optical QB 2bsub
UNIT I INTRODUCTION
Introduction, Ray theory transmission- Total internal reflection-Acceptance angle –
Numerical aperture – Skew rays – Electromagnetic mode theory of optical propagation –
EM waves – modes in Planar guide – phase and group velocity – cylindrical fibers – SM
fibers.
PART-A
1) A multimode step index fiber with a core diameter of 80 µm and a relative index difference
of 1.5% is operating at a wavelength of 0.85 µm. If the core refractive index is 1.48
determine (AU April/May 2010)
(a) Normalized frequency of fiber.
(b) The number of guide modes.
V= (2πa/λ) n1 (2Δ)1/2 =75.7954
M= V2/2 = 2872
2) Define the numerical aperture of a step index fiber. (AU April/May 2005 & 2010)
It is the relationship between the acceptance angle and the RI of the three media involved
namely core, cladding and air. The NA is a dimensionless quantity, which is less than unity
with values ranging from 0.14 t0 0.50.
NA = (n1 2 – n2 2) ½
3) Mention the major advantages of optical fiber communication system over microwave
communication system. (AU Nov/Dec 2010)
Small size and weight
Electrical isolation
Immunity to interference and cross talk
Signal security and low transmission loss
Ruggedness and flexibility
System reliability and ease of maintenance
4) Mention the advantages and disadvantages of monomode fiber over multimode fiber.
(AU Nov/Dec 2010 AU Apr/May’2008 R-2004)
Single mode fiber sustains only one mode of propagation. Multimode fiber contains
hundreds of modes.
Advantages of monomode fiber:
1. Monomode fiber is free from inter modal dispersion
2. Higher bandwidth is possible in monomode fiber.
Disadvantages of Monomode fiber:
1. Only LASER optical source has to be used which is costly than LED source.
2. Smaller core radii prevents easier launch of optical power in to the fiber.
5) Define acceptance angle and critical angle for fiber. (AU Apr/May2009)
Acceptance Angle: Light acceptance is how much light can get into the fiber for
transmission.
Critical angle: At which angle the refracted ray lies in the core cladding interface, that
angle is called as critical angle. Sinθc = n2 / n1
6) What is tunnel effect? (AU Nov/Dec 2009)
The leaky modes are continuously radiating their power out of the core as they propagate along
the fiber. This power radiation out of the waveguide in quantum mechanical phenomenon is
referred to as ‘tunnel effect’.
7) What is skew ray? (AU Nov/Dec 2009)
Skew rays are not transmitted through the fiber axis. The skew rays follow a helical path in
the optical fiber. It is very difficult to track the skew rays as they do not lie in a single plane.
8) A Silica optical fiber with a core diameter large enough to be considered by ray theory
analysis has a core refractive index of 1.5 and a cladding refractive index of 1.47. Determine
the numerical aperture and acceptance angle in air of the fiber.
(AU Apr/May’2008,Nov/Dec 2007 R-2004)
NA=√ (n12-n22)
n1=1.5, n2=1.47
NA=√ (1.52-1.472) =0.3
Acceptance angle θ=sin-1(NA) = sin-1(0.3) =17.46o.
9) A typical relative refractive index difference for an optical fiber designed long distance
transmission is 1%.Estimate the numerical aperture for the fiber when the core index is
1.47. (AU Nov/Dec 2008)
1/2
N.A=n1 (2Δ) = 0.20788
10) What are the advantages and disadvantages of the ray optic theory?(AU Nov/Dec 2008)
Advantage:
It is easy to track as it travels along the fiber because it lies in a single plane.
The bounded rays are trapped in the core and propagate along the fiber axis according to the laws
of geometrical optics. (Meridional ray).
Disadvantage:
Skew rays are not confined to a single plane. These rays follows helical path along the fiber.
These rays are very difficult to track as they travel along the fiber, since they do not lie on the
single plane.
11) A step index fiber has a normalized frequency V=26.6 at 1300 nm wavelength .If the core
radius is 25µm, Find the numerical aperture. (AU Apr/May 2007)
Use the formula V= (2πa/λ)*NA
V=26.6, a=25µm, λ=1300 nm.
NA= (λ/2πa)*V= (1300nm/2*π*25 µm) * 26.6=0.169
12) Define Mode-Field Diameter? (or) What is the fundamental parameter of single mode
fiber? (AU April/May 2005, April/May 2007)
The fundamental parameter of single mode fiber is the mode field diameter (MFD). This
parameter can be determined from the mode field distribution of the fundamental LP 01 modes.
The MFD is the cross sectional dimension 2wo, Where the beams intensity drops to 1/e 2 =
0.135 of its peak value.
13) Give the expression for the effective number of modes guided by a curved multimode fiber
“a”. (AU Nov/Dec 2004, April/May 2005, Nov/Dec 2005)
M = V2 /2
Where V = (2πa/) (n1 2 – n2 2) ½
14) Consider a parabolic index wave-guide with n1 = 1.75, n2 = 1.677 and core radius 25
micrometer. Calculate the numerical aperture at the axis and at a point 20 micrometer
from the axis. (AU Nov/Dec 2005)
Given n1 = 1.75, n2 = 1.677, a= 25µm,
Numerical aperture at the axis NA (0) = n1 (2*Δ) 1/2
Δ= (n1-n2)/ n1= (1.75-1.677)/1.75=0.04171
NA (0) =1.75 * (2*0.04171)1/2=0.5054
NA (20 µm) = NA (0)* √(1-(r/a)2=0.5054*√(1-(20/25)2=0.3324.
15) It is desired to make a single –mode fiber at an operating wavelength = 1300 nm with n core
=1.505 and nclad = 1.502. Find the numerical aperture and core radius.
(AU Nov/Dec 2006)
NA=√ n21−n22 =√1 . 5052−1. 5022=0. 095
λ V 1300 nm 2. 405
a= = =5 .2 μm
2 π NA 2 π 0. 095
16) Give the refractive index expression for graded index fiber. (AU Nov/Dec 2006)
{ [ ( )] }
α 1/2
r
n(r )=¿ n1 1 −2 Δ for 0≤r≤a ¿ ¿ {}
a
17) What is meant by mode coupling? What causes it? (AU Nov/Dec 2006)
Mode coupling refers to interaction of higher order modes in the core medium with
the radiation modes in the cladding. Harmonic variation of higher order mode results when the
mode travels through the core and it undergoes exponential decay in the boundary which interacts
with the radiation mode and results in mode coupling.
Boundary condition n2 k< β <n1 k.
18) A point source of light is 12 cm below the surface of a large body water (n=1.33). What is
the radius of the largest circle on the water surface through which the light can emerge?
(AU Nov/Dec 2004 ,Nov/Dec 2005)
n1 sinθ1= n2 sinθ2
n1=1.33,
sinθ1=12*10-2/x
1.33*(12*10-2/x) =1
x=16 cm.
19) Write the expression for refractive index in graded index fiber?
(AU Nov/Dec 2004)
In GI fiber the RI decreases gradually with increasing radial distance r from the core axis.
But it is constant in the cladding. The most commonly used construction for the RI
variation in the core is the power law relationship.
PART-A
1) The optical power launched into the fiber is 10 µm. The transportation distance is 10 km.
The optical power at the output of fiber is 2 µm. (AU Apr/May2010)
(a) Calculate the signal attenuation/ unit km length.
(b) Calculate the overall signal attenuation.
α= (10/Z) log(p(0)/p(Z)) => α=0.6989dB/Km
Overall attenuation (For 10 Km) = 6.989dB/Km
5) A 100 Km fiber is used in a communication system. The fiber has 3.0dB/km loss.
What will be the output power, when the input power fed at the input of fiber is
500 µW? (AU Apr/May2009)
α=3dB/Km; α = 4.343 αp; αp = 0.6907
p(Z)=p(0) e-αpz = 5.037*10-34Watts
6) What is the need for mode coupling in optical fiber? (AU Apr/May2009)
An optical fiber mode coupling device, capable of being readily connected to a
conventional optical fiber with a high degree of ruggedness, is provided. The mode coupling
device only allows transmission of at least one supported fiber mode there through, and is
preferably configured to maximize the coupling, of at least one desired fiber mode, to the at least
one supported fiber mode.
7) What are micro bends? How they are formed? (AU Nov/Dec 2009)
Micro bends are repetitive small scale fluctuations in the radius of curvature of the fiber axis. They
are formed either by non-uniformities in the manufacturing of the fiber or by non-uniform lateral
pressures created during cabling of fibers.
8) A multimode GI fiber exhibits a total pulse broadening of 0.1µsec over a distance of 15 km.
Estimate maximum possible bandwidth on the line. (AU Nov/Dec 2009)
-6
Bandwidth Bopt = BT = (1/2ζ) = (1/2x0.1x10 ) = 5MHz.
9) Distinguish intrinsic and extrinsic absorption: (AU Apr/May’2008,R-2004)
Intrinsic absorption is due to the absorption of basic constituent atoms of the fiber material
extrinsic absorption is by the absorption of impurity atoms in the glass material.
10) What is inter modal dispersion? What causes it? (AU Apr/May’2008 R-2004)
Modes in a given optical pulse arrive at the fiber end at slightly different times thus
causing the pulse to spread out in time as it travels along the fiber. This is known as intermodal
dispersion.
11) A multimode graded index fiber exhibits total pulse broadening of 0.1µs over a distance of
15 km. Estimate the maximum possible bandwidth on the link assuming RZ coding without
inter symbol interference. (AU Nov/Dec 2008)
15) Find the coupling loss for two fibers having core refractive index profiles αE=2.0 and αR=1.5
(AU May/June 2006)
{ ( )
L F ( α )=¿ −10 log
α R ( α E +2 )
α E ( α R +2 ) }
for α R ≤α E ¿ ¿{}
{
L F ( α )= −10 log
((
1 . 5 ( 2+2 )
2 1 .5+2 ) )}
=0 .6694 dB
16) List the basic attenuation mechanisms in an optical fiber. (AU Nov/Dec 2006)
The basic attenuation mechanisms are
1. absorption
2. scattering
3. radiation losses
17) What are the causes of absorption? (AU April/May 2005)
Absorption occurs when the incoming photon has such a frequency that its energy is equal
to the energy gap of the material.
Classification of absorption:
Atomic defects.
extrinsic absorption
intrinsic absorption
18) Define normalized propagation constant. (AU Nov/Dec 2005)
The number of modes that can exist in a waveguide as a function of V may be
conveniently represented in terms of a normalized propagation constant defined by
b = a2 w2 / v 2 = (β /k) – n2 2/ n1 2 – n2 2
19) An optical signal has lost 55% of its power after traversing 3.5 Km of fiber. What is the loss
in Db/Km of this fiber? (AU Nov/Dec 2004)
1 P 1 P
log i log i
Attenuation α= L P 0 = 3. 5 Po = 0.55
Loss=0.55*3.5=1.9 Db/km
20) What is fiber birefringence and fiber beat length? (AU Apr/May2009)
In the multi mode fiber modes propagate with different velocity. So the difference between their
effective refractive indices as called as fiber birefringence.
Bf = ny - nx
The length over which the beating occurs is called as fiber beat length. Lp = 2π / β
21) Commonly available single mode fibers have beat lengths in the range 10 cm<L P<2 cm.
What range of refractive index differences does this correspond to for λ=1300nm?
(AU Apr/May 2006)
Fiber birefingence Bf= ny-nx = (λ/Lp)
When Lp= 10 cm, Bf = (1300 *10-9/ 10*10-2) = 130*10-7
When Lp= 2 cm, Bf = (1300 *10-9/ 2*10-2) = 650 * 10-7
22) List the different types of mechanical misalignments that can occur between two joined
fibers (AU Nov/Dec’2007 R-2004)
Lateral misalignment
Longitudinal misalignment
Angular misalignment
23) Calculate the ratio of stimulated emission rate to the spontaneous emission rate for lamp
operating at a temperature of 1000 K. Assume average operating wavelength 0.5
micrometer. (AU May/June 2006)
(Refer page number 288 of the book “optical fiber communication” by John. M. Senior, 2 nd
Edition)
24) Name few splicing methods in fiber optics. (AU Nov/Dec 2006)
The different fiber Splicing techniques are
Fusion splicing
V groove and tube mechanical splicing
Elastic tube splicing
Rotary splicing
25) What are splices? What are the requirements of splices?
The splices are generally permanent fiber joints, whereas connectors are temporary fiber joints.
Splicing is a sort of soldering. The requirements of splices are:
Should cause low attenuation
Should be strong & light in weight
should have minimum power loss
Should be easy to install
26) Define Raleigh scattering loss.
It’s the dominant loss mechanism in the ultraviolet region. Its tail extends upto infrared
region. Its inversely proportional to the fourth power of wavelength. It arises due to the
microscopic inhomogeneties caused by density fluctuations, refractive index fluctuations and
compositional variations.
27) Define Mie scattering loss
It’s a linear scattering which arises from the inhomogenities, which are comparable in size
to the guided wavelength, in the forward direction. Its also due to the imperfect cylindrical
structure of the waveguide, irregularities in the core-cladding interface, core-cladding refractive
index difference along the fiber and diameter fluctuations. It can be reduced by defect free fiber
fabrication and increasing the relative refractive index difference.
28) How are micro-bending losses reduced?
These are the losses due to the bends in the fiber axis, during cabling and stress acting on
the fiber. These produce mode coupling and radiation losses. These can be reduced by extruding a
compressible jacket over the fiber. When external forces are applied, the jacket will be deformed
but the fiber will tend to stay relatively straight.
PART-A
1) A lens coupled surface emitting LED launches 190µm of optical power into a multimode
step index fiber when a forward current of 25mA is flowing through the device. Determine
the overall power conversion efficiency when the corresponding forward voltage across the
diode is 1.5v. (AU April/May 2010)
(Refer page number 397 of the book “optical fiber communication” by John. M. Senior, 2 nd
Edition)
2) Draw the three key transition processes involved in laser action. (AU April/May 2010) (AU
Nov/Dec 2005) (AU April/May 2009)
The three key transition processes involved in laser action
Absorption
Spontaneous emission
Stimulated emission
3) Write short notes on Direct and Indirect band gap materials.
(AU April/May2005, 2006, Nov/Dec 2005, 2006 &2008 & 2010)
The simplest and most probable recombination process ill be where the electron and hole have the
same momentum value. This is called direct band gap material.
Indirect band gap materials are materials whose conduction band minimum and the valence band
maximum energy levels occur at the different values of momentum. Here the band-to-band
recombination must involve a third particle to conserve momentum, since the photon momentum
is very small.
4) A double hetero junction InGaAsP LED emitting at a peak wavelength of 1310 nm has
radioactive and nonradioactive recombination times of 30 and 100 ns respectively.
Calculate the bulk recombination time. (AU Nov/Dec 2010)
1/τ=(1/τr)+(1/τnr) = 0.23ns
5) When a LED has 2V applied to its terminals, it draws 100mA and produces 2 mW of
optical power.Determine the LED conversion efficiency from electrical to optical power?
(AU April/May 2009)
Pin = V*I = 2*100m = 200 mW
Pout = 2 mW (given)
η = pout/pin = 0.01
6) Define three modes of cavity. (AU Nov/Dec 2009)
Longitudinal modes are related to the length L of the cavity.
Lateral modes lie in the plane of the PN junction. These modes depend upon the side wall
preparation and width of the cavity.
Traverse modes are associated with the electromagnetic field and beam profile in the direction
perpendicular to the plane of the PN junction. These modes determine the radiation pattern of the
laser.
7) What does population Inversion mean?
(AU Nov/Dec 2004, April/May 2007, Apr/May’2008 R-2004)
To achieve optical amplification, it is necessary to create a non equilibrium distribution of
atoms such that the population of the upper energy level is greater than that of the lower energy
level. This is known as population Inversion.
8) Compare LED and LASER diodes: (AU Apr/May 2007 Apr/May’2008 R-2004)
LED can be used in optical communication systems where bit rate is less than 100
Mb/s to 200Mb/s together with multimode fiber coupled optical power. Laser diodes can be used
for systems with bit rate greater than 200 Mb/s with single mode fiber. LED s require less
complex drive circuit than laser diodes since no thermal or optical stabilization are
required .LED s can be fabricated less expensively than LASER diodes with higher yields.
Ip = (ηq/(hν))p0 = (0.65*1.6*10-19*9*10-7*5*10-7)/(6.625*10-34*3*108)
0.235µA
M=IM/IP = 10µA/0.235µA = 43
18) Define quantum efficiency and responsivity of photo detector. (AU April/May 2009& 2008)
Quantum efficiency is defined as the ratio of number of electron hole pair generated to
number of incident photons
η = (IP/q)/(PO/ hν).
Responsivity specifies the photo current generated per unit optical power
R= IP/ PO
19) List out the values of Operating wavelength and Responsivities of Si,Ge and InGaAs
Photodiodes. (AU Nov/Dec 2009)
Si: Operating wavelength = 400-1100nm & Responsivity = 0.4-0.6
Ge: 800-1650nm & 0.4-0.5
InGaAs: 1100-1700nm & 0.75-0.95
20) What is meant by (1/f) noise corner frequency? (AU Nov/Dec 2009)
The (1/f) noise corner frequency fc is defined as the frequency at which (1/f) noise,which
dominates the FET noise at low frequencies and has (1/f) power spectrum becomes equal to the
high frequency channel noise given by Г.
21) Ga As has a band gap energy of 1.43 eV at 300K. Determine the wavelength above which
an intrinsic photo detector fabricated from this material will cease to operate.
(AU Apr/May’2008 R-2004,Apr/May 2007)
Given Eg=1.43 eV=1.43*1.602*10-19 J =2.29086*10-19 J.
Eg=hν=hc/λ.
h=6.625*10-34 J-s.
c=3*108 m/s.
λ= hc/ Eg=6.625*10-34*3*108/2.29086*10-19=8.6757*10-7.
λ=8.6757*10-7 m.
22) Compare the performance of APD and PIN diode. (AU Nov/Dec 2008)
Refer table 6-1. Page No: 267
23) Define quantum limit. (AU Nov/Dec’2007 R-2004)
The minimum received optical power required for specific bit error rate performance in a
digital system is known as quantum limit.
24) What are the desired features of a photodetector? (AU Nov/Dec’2007 R-2004)
High response
Minimum addition of noise
Fast response speed
Sufficient bandwidth
Insensitive to variation in temperature.
27) A given APD has a quantum efficiency of 65% at a wavelength of 900 nm. If 0.5
microwatts of optical power produces a multiplied photocurrent of 10 microamperes, find
the multiplication M. (AU April/May 2005)
ηq ηqλ (0 .65 )∗(1. 6∗10−19 c )∗(9∗10−7 m)
P 0= P0 = −34 8
∗5∗10−7 W=0. 235 μA
Ip=RP0= hν hc (6 . 625∗10 J −s )∗(3∗10 m/s )
I M 10 μA
= =43
Multiplication M= I P 0. 235 μA
28) Define long wavelength cut off related to photo diode. (AU Nov/Dec 2004)
The upper wavelength cutoff λC is determined by the band gap energy Eg of the material. If
Eg is expressed in units of electron volts (eV) then c is given in units of micrometers (μ m) by
hc 1.24
λC(μm) = ---- = ----------
Eg Eg (e V)
29) What does bulk dark current mean? (AU Nov/Dec 2004)
The bulk dark current iDB arises from electrons and/or holes which are thermally generated in
the pn junction of the photodiode.
30) Define radiance. (AU Nov/Dec 2004)
Radiance is the optical power radiated into a unit solid angle per unit emitting surface area
and is generally specified in terms of watts per square centimeter per steradian. Since the optical
power that can be coupled into a fiber depends on the radiance (i-e, on the spatial distribution of
the optical power), the radiance of an optical source rather than the total output power is the
important parameter when considering source to fiber coupling efficiencies.
31) Define modal noise and mode partition noise. (AU April/May 2009 & 2010)
Modal Noise: Interference of a multimode optical communications fiber with a laser light when a
speckle pattern in the light intensity in the fiber alters because of motion of the fiber or changes in
the laser spectrum. Also known as modal distortion.
Mode partition noise: In an optical communications link, phase jitter of the signal caused by the
combined effects of mode hopping in the optical source and intramodal distortion in the fiber.
32) What is meant by modal noise? (AU April/May 2005, April/May 2007)
Modal noise arises when light from a coherent laser is coupled into a multimode fiber. This
is generally not a problem for links operating below 100 Mbps but becomes disastrous at speeds
around 400 Mbps and higher. The following factors can produce modal noise in an optical fiber
link
Mechanical disturbances along the line
Fluctuations in the frequency of an optical source
PART-A
1) List the important requirements of an optical receiver. (AU Nov/Dec 2006)
It has the task of first converting the optical energy emerging from the end of a fiber into
an electrical signal, and then amplifying this signal to a large enough level. The BER of a optical
receiver should be less.
2) What are the benefits of transimpedance amplifier? (AU April/May 2007)
It has a wide dynamic range compared to the high –impedance amplifier
Usually little or no equalization is required because the combination of Rin and the feedback
resistor Rf is very small, which means the time constant of the detector is also small.
The output resistance is small, so that the amplifier is less susceptible to pickup noise ,cross
talk ,electromagnetic interference etc
The transfer characteristic of the amplifier is actually its transimpedance, which is the
feedback resistor. Therefore, the transimpedance amplifier is very easily controlled and stable.
Although the transimpedance amplifier is less sensitive than the high impedance amplifier, this
difference is usually only about 2 to 3 dB for most practical wide band designs.
3) What is meant by pre-amplifier? What are the advantages of pre-amplifier?
Pre-amplifiers are the circuits that are designed to maximize the receiver sensitivity while maintaining
a suitable bandwidth since the receiver’s sensitivity and B.W are dominated by noise sources. Its
advantages are:
(a) Has low noise level and high gain
(b) Has high B.W
(c) Has high dynamic range
(d) Has high sensitivity to avoid non-linearities
4) What are the drawbacks of high impedance amplifier?
It produces a large input RC time constant, the front-end B.W is less than the signal B.W. Thus,
the input signal is integrated and equalization techniques must be employed to compensate for
this.
5) Define extinction ratio
Its defined as the ratio of the optical power in a 0 pulse to the power in a 1 pulse. Biasing the light
source slightly on during a 0 time slot results in a non-zero extinction ratio ,є. Its effect is a power
penalty in receiver sensitivity.
6) A 2km length of multimode fiber is attached to apparatus for spectral loss measurement the
measured output voltage from the photo receiver using the full 2km fiber length is 2.1v at a
wavelength of 0.85 micrometer. When the fiber is then cut back to leave a 2km length the
output voltage increases to 10.7V.Determine the attenuation per km for the fiber at a
wavelength of 0.85 micrometer and estimate the accuracy of the result.
Solution:
αdb=10/(L1-L2)log10(V2/V1) =10/(1.998)log10(10.7/2.1) =3.5 db km-1
7) A He-Ne laser operating at a wavelength of 0.63 µm was used with a solar cell cube to
measure the scattering loss in a multimode fiber sample. With a constant optical output
power the reading from the solar cell cube was 6.14 nV.The optical power measurement at
the cube without scattering was 153.38 µV .The length of the fiber in the cube was
2.92cm.Determine the loss due to scattering in dB km -1 for the fiber at a wavelength of 0.63
µm.
Solution:
αsc= 4.343 X 105 /L(cm) (Vsc/Vopt)
=4.343 X 105/2.92 (6.14 x 10-9/153.38 x 10-6)
= 6.0 db km-1
8) A trigonometrical measurement is performed in order to determine the numerical aperture
of a step index fiber. The screen is positioned 10.0 cm from the fiber end face. When
illuminated from a wide angled visible source the measured output pattern size is 6.2 cm.
Calculate the approximate numerical aperture of the fiber.
Solution:
NA= A/(A2+4D2) (1/2)
=6.2/(38.44+400) (1/2)
=0.30
9) The shadow method is used for the on-line measurement of the outer diameter of an optical
fiber. The apparatus employs a rotating mirror with an angular velocity of 4 rads -1 which is
located 10cm from the photo detector. At a particular instant in time a shadow pulse of
width 300µs is registered by the photo detector. Determine the outer diameter of the optical
fiber in µm at this instant in time.
Solution:
ds/dt = L . Dф/dt = 0.1 x 4 = 0.4 ms-1 = 0.4 µm µs-1
hence the fiber outer diameter do in µm is ,
do = We.ds/dt = 300µs x 0.4 µm µs-1 = 120 µm.
UNIT V OPTICAL NETWORKS
Basic Networks – SONET / SDH – Broadcast – and –select WDM Networks – Wavelength
Routed Networks – Non linear effects on Network performance – Performance of WDM +
EDFA system – Solitons – Optical CDMA – Ultra High Capacity Networks.
PART-A
1) What are the main parameters used for characterizing the performance of optical amplifiers
in a communication system? (AU April/May 2010)
External pumping and amplifier gain are the main parameters of optical amplifier.
2) What is meant by ‘soliton’? (AU April/May 2005 & Nov/Dec 2010)
A soliton is a non dispersive pulse that makes use of nonlinear dispersion properties in a fiber
to cancel out chromatic dispersion effects. It refers to a special kind of wave that can propagate
undisturbed over long distances and remain unaffected after collisions with each other.
3) What are the advantages of using soliton signals through fiber? (AU April/May 2009)
(i) Solitons are very narrow, high intensity optical pulses that retain their shape through the
interaction of balancing pulse dispersion with the nonlinear properties of an optical fiber.
ii) It can propagate undistorted over long distances and remain unaffected after collisions
with each other.
4) What is chirping? (AU Nov/Dec 2009)
A laser which oscillates in a single longitudinal mode under CW operation may experience
dynamic line broadening when the injection current is directly modulated. This line broadening is a
frequency chirp associated with modulation induced changes in the carrier density.
5) What is EDFA? (AU Apr/ May’2008,R-2004)
An important class of fiber amplifiers makes use of rare earth elements as again medium by
doping the fiber core during the manufacturing process. Amplifier properties such as the operating
wavelength and the gain bandwidth are determined by the dopants rather than by silica fiber which
plays the role of host medium. Erbium is one of the rare earth elements which is doped along with the
fiber core. and they operate near the wavelength region near 1.55µm
6) List down the system requirements needed in analyzing a point-to-point link
(AU Nov/Dec 2005, Apr/May 2007)
Desired (or) possible transmission distance
Data rate or channel BW
Bit error rate (BER).
7) What are the advantages of WDM? (AU Nov/Dec’2007 R-2004)
WDM corresponds to the scheme in which multiple optical carriers at different
wavelengths are modulated using independent electrical bit streams and are then transmitted over
the same fiber. WDM has the potential of exploiting the large bandwidth offered by optical fibers
WDM was used to transmit two channels in different transmission windows of optical fiber.
8) Distinguish fundamental and higher order solitons. (AU Nov/Dec’2007 R-2004)
When an input pulse having an initial amplitude U (0, τ) =N sech (τ) is launched in to the
fiber its shape remains unchanged during propagation when N=1 is called fundamental soliton ,but
follows a periodic pattern for integer values of N>1 are called higher order solitons.
9) A soliton communication is operating at 1.55 micro meter by using fibers
β2 = -1 ps2 and r=2/ (w-km). What is the peak power required to maintain a fundamental
soliton of width (TFWHM) 10 ps? (AU Nov/Dec 2006)
The FWHM pulse width and T0 are related by TFWHM=1.763T0.
T0= TFWHM/1.763= 10ps/1.763= 5.67 ps
Ppeak=1/(2*5.672)=15.5 mW.
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UNIT I INTRODUCTION
Introduction, Ray theory transmission- Total internal reflection-Acceptance angle –
Numerical aperture – Skew rays – Electromagnetic mode theory of optical propagation –
EM waves – modes in Planar guide – phase and group velocity – cylindrical fibers – SM
fibers.
13) A multimode step index fiber with a core diameter of 80 µm and a relative refractive index
difference of 1.5% is operating at a wavelength of 0.85 µm. If the core refractive index is
1.48, estimate the normalized frequency for the fiber and number of guided modes.
(AU Nov/Dec’2007R-2004)
(Refer page number 46 of the book “optical fiber communication” by John. M. Senior, 2nd Edition)
14) Discuss the electromagnetic mode theory of optical propagation (AU April/May 2004)
**
(Refer page number 23-25 of the book “optical fiber communication” by John. M. Senior, 2 nd
Edition)
Note: For problems please refer notes which I gave for all units.
For pass mark, First concentrate unit 3, 4 & 5