ECE 455 1

ECE 445 – Optical Fiber Communications

Lecture 01 - Introduction
Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus
iezekiel@ucy.ac.cy

• ECE 445
• Lecture 01
• Fall Semester 2016
ECE 455 2

ECE 455
Why is this course useful to you?
ECE 455 3

• There are many ways we can answer this question.

• Perhaps the most obvious answer is to consider the social impact of optical
communications.

• In 1966, two engineers at Standard Telecommunication Laboratories in England wrote a

paper which in essence started the race to develop optical fibre for communications:

• One year later, the Philco-Ford company tried to predict the future; they were in many
ways very close to what we have.

https://www.youtube.com/watch?v=rpq5ZmANp0k
ECE 455 4
ECE 455 5

All of this is made possible by a global infrastructure of optical cables

http://www.submarinecablemap.com/#/
ECE 455 6

The aim of this course is to study optical fibre technology and its application to optical
communication links and systems

The basic questions we will seek to answer in the first few lectures include:

1. What are the advantages of using photonics for communications?

2. What is optical fibre?

3. What is an optical communications link? (Basic architecture)

4. How does light propagate in an optical fibre?

ECE 455 7

http://www.arthitectural.com/wp-content/uploads/2013/04/02-TABLE-OF-OPTICKS-SIR-ISAAC-NEWTON-1704.jpg

"Light is waves on Mondays, Wednesdays, and

Fridays; it's particles on Tuesdays, Thursdays,
and Saturdays, and on Sundays, we think about
it!"

W.H. Bragg, 1930.

WHAT IS LIGHT?
ECE 455 8

Optics is one of the oldest branches of science.

It is concerned with the generation, propagation, manipulation and detection of light. For
many centuries, the development of optical sources and optical detectors was very slow,
hence progress was strongest in studies of light propagation and light manipulation, e.g.:

Refraction (ray optics) Interference

(wave optics)

Polarisation (electromagnetic optics)

ECE 455 9

By the late 19th century, the theoretical work of Maxwell and the experiments of Hertz had
resulted in the electromagnetic view of light, in which it holds that light consists of
coupled time-varying electric and magnetic fields that satisfy a wave equation (which itself
can be derived from Maxwell’s equations):

c = speed of light = 2.998 × 10-8 ms-1 in vacuo

c = fλ
Solution is a travelling-wave:

2π 2π
k= ω=
λ T
ECE 455 10

However, the development of modern physics (and especially the work of Planck and
Einstein) led to the photon view of light.

Energy of a photon: hc
E = hf =
λ

h = Planck’s constant = 6.626 × 10-34 J·s

ECE 455 11

Light as photons: Photoelectric effect

• Increasing the intensity of the light increases the number of photoelectrons, but not their
maximum kinetic energy.

• Red light will not cause the ejection of electrons from potassium, no matter what the
intensity.

• Weak violet light will eject only a few electrons, but their maximum kinetic energies are
greater than those for intense light of longer wavelengths!

hc
Explained by Planck relationship: E = hf =
λ
ECE 455 12

• Nowadays, most electronic communication (e.g. wireless) is in the microwave region.

• Typically there is a three orders of magnitude difference between microwaves and photonics

1 mm

1 nm
1 km

1 µm
1 cm
Wavelength (m)
104 103 102 10 1 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9

Ultrashortwave

ultrashortwave

Sub-Mm-wave
Mediumwave

Mid-infrared
Far-infrared
Microwave
Shortwave

Ultraviolet
Mm-wave
Longwave

Extremely

Visible

X-ray
Electronic techniques Optics
THz
Microwaves Photonics
Gap

30 kHz 30 MHz 3 GHz 30 GHz 3 THz 3 PHz

Frequency
ECE 455 13

OPTICAL COMMUNICATIONS
ECE 455 14

The most basic optical communication link:

Optical Optical
Modulation Channel
source detector
ECE 455 15

Optical communications has a long history, having been used by many civilizations. One
example is the friktories of ancient Greece:

This was a very early

example of digital optical
communications.

Το σύστημα με τις Φρυκτωρίες παρουσιάζεται στον ειδικό

χάρτη όπου εμφανίζει με φωτεινές ενδείξεις τους πυρσούς
να ανάβουν και να μεταδίδουν το μήνυμα από την Τροία
στις Μυκήνες....

/
ECE 455 16

Digital optics, 1793-1852:

http://www.ec-lyon.fr/tourisme/Chappe/

• Claude Chappe’s Optical Telegraph (France)

• Based on a semaphore system
• Repeater spacing ≈ 6 miles
• Message could cover 100 miles in 30 minutes
• Bit rate < 1 bit/s
ECE 455 17

Bell’s photophone 1880

- Analogue optical link

Transmitter Receiver

• Light modulated by vibrating mirror • Light is photodetected using selenium

(i.e. opto-mechanical) (resistance decreases with increasing light
intensity)
• First example of optoelectronic receiver

https://www.youtube.com/watch?v=Kc9Mjzfowcs
© Alexander Graham Bell Foundation
ECE 455 18

One of the problems with these early systems was the fact that there was no guided
channel between the transmitter and receiver, in other words the channel was free-space
optics.
For some applications, such as satellite-
to-satellite free space optical links, this is
not a problem.

But for terrestrial free space

optical communications,
weather conditions have to be
considered:
ECE 455 19

Each of these factors can attenuate the signal. Sunlight

However, there are ways to mitigate each environmental factor.

Window
Attenuation
Fog

Building
motion

Scintillation Alignment
Σπινθηροβολία

Obstructions

Low Cloud
ECE 455 20

An efficient way of guiding light is useful for modern long distance optical
communications….
ECE 455 21

Kao and Hockham proposed the use of optical fibres for communications - 1966

More specifically, they showed that a target figure of 20 dB/km for fibre
attenuation would make optical communications viable. At this stage, losses were
way too high (1000 dB/km for glass, as opposed to tens of dB/km at most for
coaxial cable), but they showed this was mainly due to impurities.
ECE 455 22

Work at Corning in the early 1970’s eventually led to fiber losses of 20 dB/km, and
over time these have been reduced to as low as 0.2 dB/km (at 1550 nm).

Egyptian
107

106
Venetian
105
Optical
104
Loss Optical fibre
(dB/km) 103 Optical glass

102

10
1

0.1
3000 BC 1000 AD 1900 1966 1979 1983
ECE 455 23

Fibre offers a lot of bandwidth!

ECE 455 24

Optical Fibres: Basic Structure

• dielectric waveguides that operate at optical wavelengths; mostly made from silica
glass, but plastic versions (for multimode) also available

• confine electromagnetic energy in the form of light within core and guide the light
parallel to the longitudinal axis:

CORE CLADDING BUFFER COATING

Not to scale!

•A circular core of refractive index n1 is surrounded by cladding with a slightly lower

value of refractive index (n2 < n1). The fibre is encapsulated by the buffer and
additional layers as appropriate.

• Light is confined to the core of the fibre by total internal reflection – TIR at the
core-cladding interface.
ECE 455 25

Advantages of optical fibre

• Very wide bandwidth compared to metallic transmission lines, i.e. potentially

thousands of GHz
• Very low loss (as low as 0.2 dB/km)
• Can achieve low dispersion (depends on wavelength of source and fibre type)
• Small size and weight
• Electrical isolation (glass and plastic)

A fiber-optic cable (right) containing 144 tiny glass

fibers is compared with a cross section of a
conventional copper cable.
ECE 455 26

However, even though fibre itself is small in cross-section, in some applications the overall
cable is not so small or light:

A lot of optical fibre is installed in

undersea (submarine) systems,
and must be well protected.
ECE 455 27

The basic ingredients of a classical communications link include:

• coherent oscillator (i.e. laser)

• mixer (e.g. directly modulated laser or modulator)
• envelope detector (e.g. photodiode)

However, there are other components (analogous to electronic

components) that are also used in optical communications:

• amplifiers (Erbium-doped fibre)

• couplers, combiners and splitters
• wavelength selective components – filters, multiplexers
• isolators and circulators
ECE 455 28

Basic architecture of an optical fibre link

ECE 455 29

Most fibre links are digital, and consequently we worry about bit rate – distance
products and bit error rates:

© G.D. Keiser
ECE 455 30

Figures of merit

• The designers of a long distance high-bit rate fibre link have a number of
objectives.
• One is to achieve as high a bit rate as possible.
• However, it is also important to maximise the distance between optical amplifiers
or repeaters (i.e. the repeater spacing).
• The two figures are multiplied to give a key figure of merit used to assess link
performance:

Bit-rate - repeater spacing product (bits/s - km)

ECE 455 31

Bit rate – distance product improvements

ECE 455: Lecture 02 1

Introduction to Optical Links

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 445
• Lecture 02
• Fall Semester 2014
ECE 455: Lecture 02 2

BASIC CONSIDERATIONS IN
COMMUNICATION LINKS
ECE 455: Lecture 02 3

• The aim of any communication link is the faithful (i.e. accurate) transmission of
information from the transmitter (sender) to the receiver, over a channel.

• The channel is far from perfect, and can introduce noise and distortion.

• Ideally, the demodulated message at the receiver should match that from the source. Most
systems today are digital, so for these we require error-free transmission. However, there
are still some analogue systems, and here we require a high signal-to-noise ratio.

• Many modern communication systems use complicated modulation schemes, especially

since the advent of digital signal processing.
ECE 455: Lecture 02 4

The Impact of Noise

• Detection in the presence of noise is

one of the major challenges in a
communications link, whether it is
analogue or digital.

Analogue

Digital
ECE 455: Lecture 02 5

Transmitted
Received
signal Receiver
signal
Channel Analogue Communication
Noise
t
Distorted
and noisy
Noise
Nonlinearity
Dispersion
Attenuation

Transmitted
Regenerative Received Digital Communication
Channel receiver signal
signal
(Fibre)

t
• Regenerated
pulse
Noise • Effects of channel
Nonlinearity can be mitigated
Dispersion
Attenuation
ECE 455: Lecture 02 6

Digital and Analogue Communications: Comparison

• Advantages of digital communications
– 1. Signal regeneration is possible
– 2. Error detection and error correction is possible
– 3. Greater dynamic range
– 4. Digital hardware is relatively cheap and advanced

• Disadvantages of digital communications:

– 1. Generally requires more bandwidth than analogue
– 2. Digital detection requires synchronisation
(i.e. clock must be recovered at the receiver from the incoming data
stream).
- 3. Bit errors are possible (e.g. if there is insufficient SNR), as is
quantisation noise
ECE 455: Lecture 02 7

Digital vs. Analogue Performance Criteria

• Analogue systems deal with continuous waveforms. Evaluate performance

by fidelity criterion, e.g. signal-to-noise ratio (SNR).
– i.e. SNR is a figure of merit largely used for analogue communications

• Digital communication systems transmit signals from a finite set

(“alphabet”) that is known by the receiver. (e.g. Morse code).
• Hence a figure of merit for digital communications is the probability of
incorrectly detecting a digit
– also called the probability of error
– often specify the bit error rate (BER)
– a BER of 10-9 or less is typically specified for many modern systems

• Note: Although SNR and BER are two different measures, for optical
systems they can be related to each other as we shall see later on.
ECE 455: Lecture 02

Probability of Error (Digital performance criterion)

• The bit error rate (BER) is obtained by dividing the number of

errors (Ne) occurring over a time interval t by the number of
pulses (ones and zeros) transmitted during this interval (Nt):

Ne Ne
BER  
N t BT t

• BT is the bit rate (bits/sec) and is equivalent to 1/TB where TB is

the bit duration. We assume that a “1” is “high” and “0” is low
for the duration of TB, but other line coding schemes are
possible.
ECE 455: Lecture 02

Common data (line coding) formats

Many optical systems

use NRZ or RZ codes.
ECE 455: Lecture 02 10

So in a digital optical link, the decision points are made at the mid-point of the bit:
ECE 455: Lecture 02 11

This is shown in more detail for the photoreceiver:

Bit errors can be made here;

the number depends on the SNR
of the received signal

Recovered
pulse train
(output voltage)
ECE 455: Lecture 02 12

BASIC PRINCIPLES OF OPTICAL

LINKS
ECE 455: Lecture 02 13

Basic architecture of the simplest type of optical fibre link

Information Optical Optical Information

source transmitter receiver recipient
Optical
fibre

Electrical signal, e.g. Electrical-to-optical Optical-to-electrical The recovered

current drive to a (E/O) conversion. (O/E) conversion. An electrical signal
laser or voltage drive The input electrical input optical signal is
to an electro-optic signal modulates converted to current.
modulator the light.

The modulated optical signal

propagates along the optical
fibre; as it does so it suffers
loss and dispersion

Alternative names for optical receiver: photoreceiver and photodetector

ECE 455: Lecture 02 14

In reality, an optical link will be more complex.

However, for the moment we will focus on the simpler version from the previous slide.
ECE 455: Lecture 02 15

There are two main reasons why we are interested in using photonics for digital (and
sometimes analogue) communications:

1. Bandwidth
2. Low loss optical fibres

3 108
c
Typical DFB laser f    200 THz
spectrum  1.5 10 6

Lasers offer a very high carrier frequency

c1  2  c
Typical single-mode f  f 2  f1  
fibre attenuation 12 20
plus optical
amplifier gain, both e.g. A wavelength span of 0.8 nm centred on
versus wavelength. 1550 nm gives 100 GHz bandwidth

Optical fibre offers large bandwidths

ECE 455: Lecture 02 16

Lasers are of interest because they offer an (almost) monochromatic source of very
high quality (coherent) light.

We can describe the light emitted from a laser by an electric field:

E (t )  E0 exp jot  o  kz 

which implies that we can use amplitude modulation (E0), frequency modulation (ω0)
or phase modulation (φ0). Note that the electric field as described above is complex,
i.e. E0 is a complex number. In the following analysis we will ignore position z.

Note: in some cases, when we modulate the amplitude, we also end up modulating
the frequency (wavelength) – when this happens, we say we have chirp.
ECE 455: Lecture 02 17

The optical intensity (unit Wm-2) is directly proportional to the square of the electric
field magnitude:
cn 0
I (t ) 
2
E (t )
2

where c is the speed of light, n is the refractive index of the medium, and ε0 is the
permittivity of free space. Hence:

I (t )  E02

For a fixed cross-sectional area, the optical power is directly proportional to the
optical intensity, and so:

POptical (t )  E02

We mostly work with optical power, not intensity for two reasons. First, it can be
measured directly. Second, as electrical engineers we prefer to use the symbol I for
current, so if we used intensity instead of optical power it would lead to confusion
in our equations!
ECE 455: Lecture 02 18

So we have two of the basic ingredients – an optical source (laser) and an optical
transmission medium (optical fibre). The third basic ingredient is an optical receiver to
convert the light coming from the output of a fibre into an electronic signal (usually
current).

The simplest photo-receiver is a photodiode connected to a load resistor:

IP

Incoming signal (from laser):

ER  ER cos  R t   R 
~ Load
resistor
RL
Photodiode

optical
phase

optical frequency (of order 200 THz) = c / R

electric field amplitude; optical power is  ER2

Electric field of incident optical signal

ECE 455: Lecture 02

The frequency response of a photodiode is limited to about 100 GHz in state-of-the-art

devices. Consider an optical OOK (on-off keying) waveform:

Incoming electric field

Envelope
Optical carrier

The photodiode cannot detect the fast variations of the optical carrier, it can only
respond to the modulation envelope, i.e. it acts as an envelope detector. (We also
use the term direct detection.)

Photodiodes respond to the magnitude of the incoming electric field, and convert
this to current.

In other words, photocurrent is proportional to the optical power incident on the

photodiode.
ECE 455: Lecture 02 20

Slope is given by
Photocurrent

responsivity

IP
ty 
Responsivi (A/W)
PO

This is with reference to the static

characteristic

Input optical power

• In an ideal photodiode (no noise, no nonlinearity), there is a linear correspondence

between input optical power and photocurrent.

• One consequence of this is that optical loss in dB is double the corresponding electrical loss
in dB (1 dBo = 2 dBe). More about this later.

• Note that the photodiode is actually classified as a square law device, since optical power
varies directly with the square of the electric field magnitude.
ECE 455: Lecture 02 21

IP
Incoming electric field at the input to the photodiode
(originating from an intensity modulated laser):

Load
ER  ER cos  Rt   R 
~ resistor
RL
Photodiode

The incident optical power is proportional to the square of the E-field, i.e.:

Pincident  ER2  ER2 cos 2  R t   R 

~

 Pincident  12 ER2  1  cos 2 R t  2 R 

the photodiode cannot detect the term 2R. Hence for intensity modulation/direct
detection schemes (IM/DD),
I P , DD  ER2
ECE 455: Lecture 02 22

• The majority of optical links are digital and are based on direct detection of intensity
modulation

• This means that the optical power emitted by the source is modulated, and the
modulated power is then detected by a simple photoreceiver (like the one shown on
the previous slide) after passing through a length of optical fibre.

• The optical power can be modulated either directly (the current into a laser diode is
modulated) or externally, as shown below:

Transimpedance
Amplifier

• The advantage of external modulators is that they can be modulated to many tens of
GHz, and they can also be used to implement optical phase modulation.
ECE 455: Lecture 02 23

• In digital optical communications, an important figure of merit for

photoreceivers is the receiver sensitivity. The receiver sensitivity PR sets the
lower limit on the optical power needed to achieve binary transmission at BT
bits per second with a specified BER.

• Sensitivity gets worse as we go up in bit rate:

ECE 455: Lecture 02 24

• One way of improving sensitivity is by using coherent detection of phase modulated

light:

Note: PDs can only

respond to intensity
variations. Phase
modulation is
converted to intensity
variation through
mixing with LO laser.

but the receiver is more complex (and therefore more expensive also).

• However, we can improve the performance of IM/DD (intensity modulation/direct

detection) by using optical amplifiers:
ECE 455: Lecture 02 25

As a result of optical amplifiers, in long distance networks, WDM is pervasive:

1 Multiplexer 1
TX1 RX1
2 EDFA 2
TX2 RX2
3 3
TX3 DMUX RX3
MUX

1, 2, 3 ..... N

N N
TXN Demultiplexer RXN

1000

2008
which means we can relax bandwidth

Number of wavelength channels

2001
1998 2003
100 2006
requirements by increasing the number of 2001

Improving photonics
2003
1998
wavelength channels and the channel 1996
10
density. Even so, the trend is to higher bit 1995
1993
rates for optoelectronics. 1
1977 1995

1986 1991
Improving electronics

0.1
0.01 0.1 1 10 100 1000
Data rate per channel (Gb/s) Total
capacity
ECE 455: Lecture 02 26

Optical Optical RF
Photodiode
input coupler out
Source and detector RF
Optical out
options input CW
+ Square-law
Detection
Laser Photo-
& LPF
(LO) diode

Direct detection Coherent detection

Direct intensity modulation
Intensity
RF input / Direct detection (IM/DD)
modulated
optical
Directly • Simple technique, cheap
signal
Modulated • Problems can include:
(“AM”)
Laser Diode • Chirp
• Nonlinearity
Direct modulation

Modulated External Intensity Coherent detection of:

optical modulation / Direct • Amplitude
External Signal detection (IM/DD) • Phase
modulator • or Frequency
Intensity, • No chirp problems
Phase, • Larger bandwidth Offers better sensitivity,
CW or Frequency compared to direct but increased receiver
Laser RF input
modulation complexity compared to
External modulation • Relatively expensive direct detection
ECE 455 – Lecture 03 1

Signal Degradation in Optical Fibres

- Attenuation, Dispersion and their System Impact
Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 445
• Lecture 03
• Fall Semester 2016
ECE 455 – Lecture 03 2

SIGNAL DEGRADATION
ECE 455 – Lecture 03
3

Signal Degradation in Optical Fibres

• The simplest optical fibre communications system is a point-to-
point link in which an optical transmitter and receiver are
connected to one another via an optical fibre. This simple
architecture is typical of those used in trans-oceanic links for
example.

Information Optical Optical Information

source transmitter receiver recipient

Optical
fibre
• Without looking into the detail of the optical fibre itself, in this
lecture we will look at how two important parameters –
attenuation and dispersion – can affect the above system.
ECE 455 – Lecture 03 4

• In an ideal fibre, “what goes in, is what comes out”:

f (t) f (t - )

L
 = Ln1/c

In reality, the signal going through the fibre is degraded by:

• attenuation (i.e. optical signal loss)
• dispersion (i.e. optical signal distortion)

Pin Pout
Attenuation on its own
reduces the power of the
pulse. We will see the
impact of dispersion later. input pulse output pulse
ECE 455 – Lecture 03
5

• In digital communications, the key aim is to minimise the number of bit

errors. A typical bit error rate (BER) for many systems is 10-9.

• The other aim is to maximise the repeater spacing L for a given bit rate BT.
These two are lumped together to give the bit-rate - repeater spacing
product.

• For a given BT, the minimum allowable power at the photoreceiver is called
the receiver sensitivity PR.

• If the optical power emitted by the laser diode is given by PS, then the total
allowable link loss is given by:

PS
loss link 
PR
ECE 455 – Lecture 03 6

• In other words, losslink represents the loss allowed between

the output of the optical source and the input to the
photoreceiver:

Optical Optical
transmitter receiver

Optical fibre loss

PS (mW) = losslink PR (mW)

PS
loss link 
PR
ECE 455 – Lecture 03 7

• Now, for an optical fibre: attenuation is per unit length, i.e. the longer the fibre,
the more the attenuation.

• As light travels down an optical fibre, its power (in mW) decreases exponentially
according to Beer’s law:
P z   P0  e  Az

P(z): power at a distance z along the fibre

P(0): power at input to fibre
A: attenuation constant (per unit length): nepers per m

z0 zL
Pz 
P0

z
0
ECE 455 – Lecture 03 8

• Hence for a fibre of length L, the attenuation in dB is:

• Noting that log10 x = ln x / ln 10, we get:

F is the fibre attenuation per unit length, in units of dB/km

Note that attenuation in dB is a positive number. Gain in dB is also a positive number

by convention. We have to remember this in link budget calculations.
ECE 455 – Lecture 03
9

• If we return to the equation for optical powers along a

fibre expressed in mW,

P(L) = P(0) e -AL

and then take logarithms:

10 log10 P(L) = 10 log10 {P(0) e -AL}

= 10 log10 {P(0)} + 10 log10 {e -AL}

= 10 log10 {P(0)} - AL 10 log10 {e}

P(L) in units of
dBm
P(0) in units of dBm FL in units of dB
ECE 455 – Lecture 03 10

DECIBELS ETC.
ECE 455 – Lecture 03 11

The dB and dBm units

The dB unit is widely used in optical link design because:

• It allows the various loss and gain contributions to be included via

addition/subtraction, rather than by multiplication/division which is what
would be required if linear gain/loss units were used.

• The logarithmic nature of the dB also allows large ratios to be expressed

with more manageable numbers and allows power levels differing by
many orders of magnitude to be easily compared.

• Consider, for example, an optical amplifier with a gain (dimensionless) of

G:
POUT mW   G PIN mW 
• We define gain G in dB as follows:
 POUT 
G(dB)  10 log 10  
 PIN 
ECE 455 – Lecture 03 12

• Examples of decibel measures are listed below:

Ratio dB Ratio dB
10N 10 N 2N 3.01 N

1000 30 8 9.03
100 20 4 6.02
10 10 2 3.01
1 0 1 0
0.1 -10 0.5 -3.01
0.01 -20 0.25 -6.02

0.001 -30 0.125 -9.03

10 -N
- 10 N 2-N - 3.01 N
ECE 455 – Lecture 03 13

• In optical communications, it is useful to have a logarithmic

measure of the absolute power at any point in the system. This
can be achieved with the dBm unit, which is the decibel level
referenced to 1 mW:

 P(mW ) 
P(dBm )  10 log 10  
 1 mW 

• For example:
mW 0.01 0.1 0.5 1 2 10 100

dBm -20 -10 -3.01 0 3.01 10 20

ECE 455 – Lecture 03 14

• The usefulness of the dBm comes with it being compatible

with the dB unit for gain/loss.

• For example, we know that if we take a power of 1 mW and

multiply by a gain of 2, we get 2 mW.

• Now consider the same situation with dB and dBm. The

corresponding values of power are 0 dBm and 3 dBm; and
the factor 2 corresponds to 3 dB.

• So we could also say that if we take a power level of 0 dBm

and pass it through an optical amplifier of 3 dB gain, the
output power will be 0 + 3 = 3 dBm.
ECE 455 – Lecture 03 15

• To summarise:

POUT (mW )  G PIN (mW )

log 10 POUT (mW )  log 10 G PIN (mW )

 log 10 G  log 10 PIN (mW )

 POUT (mW )   PIN (mW ) 

10 log 10    10 log 10 G  10 log 10  
 1 mW   1 mW 

POUT (dBm)  G(dB)  PIN (dBm)

ECE 455 – Lecture 03 16

LINK POWER BUDGET &

ATTENUATION-LIMITED DISTANCE
ECE 455 – Lecture 03 17

Link Power Budget

• The link power budget determines how much optical power can be lost
between the transmitter and the receiver for a given receiver sensitivity
(which depends on the bit rate) and transmitter power output.

• The dB ratio and dBm units are used in the link power budget.

FL (dB) PHOTODIODE

LASER
PS (dBm)
FIBRE PR (dBm)
Source power
Receiver sensitivity

PS  PR   F L
ECE 455 – Lecture 03 18

In practice, we also have to include optical fibre connector losses and also a
system margin:
Total fibre losses

PS  PR   F L   TC  M
System margin
Total connector losses

Hence this allows us to calculate the maximum allowed length of link (without
using intermediate optical amplifiers):

PS  PR   TC  M
Lmax 
F
ECE 455 – Lecture 03 19

However, we emphasised in the first two lectures that an important figure of

merit is:
Bit-rate - repeater spacing product (bits/s - km)

So how do we include bit-rate in the previous equation?

PS  PR  TC  M
Lmax 
F

We saw in Lecture 02 that the receiver sensitivity is a function of bit rate. Hence
knowing how the sensitivity varies with bit rate will allow us to see how
attenuation will affect the maximum distance for a particular bit rate.
ECE 455 – Lecture 03 20

Receiver sensitivity versus bit rate

In a later lecture on quantum limited receivers, we will show that for a given BER,
there must be a minimum average bit energy at the photoreceiver, corresponding
to a minimum average number of photons per bit N:

hc Pb
Eb  N

Eb  PbTb
Tb
Bit period
The receiver sensitivity is given by:

Eb
PR   Eb BT
Tb

This is the minimum optical power needed to maintain the specified BER, and is a
function of bit rate BT (and also wavelength), i.e.

PR BT   Eb BT
ECE 455 – Lecture 03 21

PR BTO   Eb BTO

PR BT   Eb BTO   PR BTO  

BO BO (Normalisation)
BTO BTO
 BO 
In dBm, we have: PR BT dBm  PR BTO dBm  10 log10  
 BTO 
PR BT  dBm
PR BTO   20
Sensitivity worsens by 10 dB/dec

PR BTO   10

BO
PR BTO 
1 10 100
BTO
ECE 455 – Lecture 03 22

Hence for a given wavelength and BER, the maximum fibre length (due to
attenuation limits) will depend on bit rate, and it decreases with increasing bit
rate:
ECE 455 – Lecture 03 23

At “high” bit rates, we notice that the curves of maximum fibre length versus bit-
rate change shape:

Attenuation-
limited

Limited by???

This is because a different type of signal degradation dominates over attenuation

at higher bit rates – this type of degradation is dispersion.
ECE 455 – Lecture 03 24

DISPERSION – IMPACT ON PULSES

IN THE TIME-DOMAIN
ECE 455 – Lecture 03 25

z=0 z=L
Attenuation

z=0 z=L
Dispersion

Attenuation leads to a reduction of power (which becomes worse with increasing

length, i.e. attenuation is specified in dB/km)

Dispersion leads to temporal pulse broadening (this too becomes worse with
increasing length, so we might expect it to be specified in ns/km for example).
ECE 455 – Lecture 03 26

• “What comes out, is not what goes in”

pIN (t) pOUT (t)
Fibre
p (t) p(t - )
No change in
pulse shape
t
t

Attenuation only
• Reduction in pulse energy

 t

Attenuation & dispersion

• Reduction in pulse energy
• Pulse spreading
 t
ECE 455 – Lecture 03 27

• We can (usually, but not always!) consider the fibre to be a linear

system, with an impulse response as shown:

pin(t)
pout(t)

 (t) pin(t) pout(t)

h(t)
t t
t
pin(t) =  (t), hence pout(t) = h(t)

t = mean arrival time
 = rms pulse spread
ECE 455 – Lecture 03 28

Pulse shape definitions • Energy content

• E = area under pulse
pout(t) 
E p

out (t ) dt

FWHM = 
•  is root mean square spread of
t pulse around mean arrival time
• It gives a measure of the dispersion
t
 

 1
   t t
2 2
pout (t ) dt
E 
• Mean time of pulse arrival 
1 2
  t pout (t ) dt  t
2

1 E 
t   t pout (t ) dt
E  • An alternative measure is the
full width at half maximum (FWHM)
ECE 455 – Lecture 03 29

pin(t) pout(t)

h(t)

t t
m1 m2

1 2
• If a pulse with an rms pulse width of 1 is applied to a fibre,
then the output pulse spread will be given by:

  
2
2
2
1
2
ECE 455 – Lecture 03 30

In a digital system, inter-symbol interference (ISI) will occur, leading to bit errors:
ECE 455 – Lecture 03 31

Consider an extreme example: a pulse sequence 101 for which   Tb

1 0 1 1 1 1
Bit stream Bit stream
at fibre at fibre
input output
S R

t t
Tb
Pulses overlap to such an extent
   R   S as to cause a bit error

  FWHM 100%

50% 
Full width at half maximum
t
ECE 455 – Lecture 03 32

For a given bit rate therefore, there will be some upper limit to the possible fibre length
before inter-symbol interference starts to have an impact.

In other words, we must try to limit the pulse spread relative to the bit duration.

A general rule of thumb is that for a bit period Tb, the rms pulse spread should be confined
to:
Tb

4

Hence the maximum bit rate will be:

1
BT 
4

Because the pulse spread will be proportional to fibre length, we see that there will be a
dispersion-limited value for the bit rate – distance product BTL.
ECE 455 – Lecture 03 Continued 1

LOSS AND BANDWIDTH

DEFINITIONS: OPTICAL & ELECTRICAL
ECE 455 – Lecture 03 Continued 2

Optical Loss / Electrical Loss

• Why should optical loss be “different” to electrical loss?
– After all, power is power, whether it be electrical or optical,
and SI units are the same, i.e. watts (W). And the ratio of
power will be dimensionless for both.

• The clue is in the conversion from electronic signals to optical

signals by, for example, laser diodes, and also the
corresponding optical to electronic conversion by photodiodes.

• In light emitting diodes (LEDs) and laser diodes, photons are

generated by electron-hole recombination. Electrons are
provided by an injection current. In photodiodes, electron-holes
are generated by incident photons.
ECE 455 – Lecture 03 Continued 3

• Consider a photodiode:
...over the same period,
Ne electrons (and holes)
In a time period T, are generated and flow
Np photons of through the load resistor
wavelength λ are
detected... I
Neq
PO I
N p hc Load T
Po  resistor
RL
T

Ne I hc
N.B. quantum efficiency is:  PD  
N p Po q

q = electron charge, h = Planck’s constant

ECE 455 – Lecture 03 Continued 4

• In other words, the photocurrent is directly proportional to the input optical

power. Consider an ideal photodiode characteristic:

I
q
slope =  PD 
hc

Note: the slope is called the

Po
responsivity , and has units A/W

There is a linear relationship between the current and optical power. However,
the power dissipated in the load resistor is given by:

Pe  I 2 RL
ECE 455 – Lecture 03 Continued 5

• Hence optical power emitted is directly proportional to

current: Po  I

• In contrast, electrical power dissipated is directly

proportional to the square of the current:
Pe  I 2

• What does this mean in practice?

• The electrical power dissipated in the load resistance of the

photodetector is proportional to the square of the incident
optical power. (Remember, photodetectors are often
referred to as square law detectors.)
ECE 455 – Lecture 03 Continued 6

Consider the following link, in which power is in mW and current in mA:

Fibre length L
Pe1  Por1  RL
2
Pos Por1

We now keep everything the same, but increase the fibre length (and hence
increase the optical loss) such that the optical power at the receiver is halved:

Pe 2  Por 2  RL
2
Pos Por 2


Por1

Por1 
2
R
2 4
L

The loss of the optical part has increased, and so has the loss of the overall link
(which is going to be electrical, since the input to the source is current). But in
what way?
ECE 455 – Lecture 03 Continued 7

Optical loss Because the input power to the fibre is kept constant, we simply
compare the output power from the fibre for both cases:

Por1
 2  3 dB
Por 2

Electrical loss Because the input current to the source is kept constant, we
simply compare the power dissipated in the load resistor for both cases:

Per1
 4  6 dB
Per 2
To avoid confusion, we often use dBo for optical losses and dBe for the
corresponding electrical losses, and the relationship between them is:

1 dBo  2 dBe
ECE 455 – Lecture 03 Continued 8

Optical BW / Electrical BW
current ratio iout(j)/ iin(j)
electrical 3 dB point

1.000
optical 3 dB point
0.707

0.500

electrical bandwidth
frequency

optical bandwidth
(Note: the default BW definition is electrical BW)
ECE 455 – Lecture 04 1

Optical Fibres
- Introduction
Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 445
• Lecture 04
• Fall Semester 2016
ECE 455 – Lecture 04 2

OPTICAL FIBRES – THE BASICS

ECE 455 – Lecture 04 3

Types of Optical Fibre

We can subdivide the different types of optical fibre in many ways, for example:

• Material: Plastic, glass or glass and plastic.

• Mode of Propagation: Single-mode or multimode. Light can only propagate through

certain modes – hence in a multimode fibre there are different modes (ways) the light
can propagate.

• Refractive index profile: Step-index, graded index or double-clad (including triangular)

There are also other more “specialised” fibres, such as photonic crystal fibres, holey
fibres and multicore fibres.

So we see that even though we started to discuss optical fibres in lectures 01 – 03 from a
general system perspective (e.g. looking at attenuation and link power budget), the
subject of optical fibres is large and covers many aspects.
ECE 455 – Lecture 04 4
ECE 455 – Lecture 04 5

We begin our treatment of the subject of optical fibres by considering their main
function – waveguiding (i.e. guiding light from an optical source).
https://www.youtube.com/watch?v=uRgdJYPh9G8

• In 1841, Daniel Colladon demonstrated that light could

be guided inside a flowing jet of water

• A similar experiment was demonstrated by John

Tyndall to the Royal Institution in London, and he
explained it as:

“the total reflexion of light at the common surface of

two media of different refractive indices.

• The total internal reflection is made possible due to

the fact that water has a higher refractive index than
air.

• It is this mechanism that is used in optical waveguides.

ECE 455 – Lecture 04 6

A waveguide is a structure that is used to guide electromagnetic waves by confining

them so as to allow propagation, e.g.:

Rectangular metallic waveguide, used at microwave

frequencies.

The dimensions (a, b) of the waveguide cross-section

are important – they determine the frequencies at
which the waveguide can be used.

The electromagnetic waves propagate through

various modes that must “fit” the geometry:

Electric field Ex component of the TE31 mode

inside an X-band hollow metal waveguide.
ECE 455 – Lecture 04 7

For the example of a parallel plate waveguide, we can see how the wave is guided by
total internal reflection from either plate, which sets up a pair of waves that then add up
to produce a wave travelling along the axis of propagation.

http://demonstrations.wolfram.com/ElectromagneticWavesInAParallelPlateWaveg
uide/
ECE 455 – Lecture 04 8

Total internal reflection is also the mechanism that is used to guide electromagnetic
waves (i.e. light) in an optical waveguide.

Here, the materials used are dielectrics (or semiconductors).

For example, we can fabricate planar optical waveguides (semiconductor-based):

Light y Light
x
n2
Light Light
n2
n1 > n2 z Propagation direction

These however are only suitable for

chip-scale applications such as diode
lasers and integrated photonics
ECE 455 – Lecture 04 9

• However, for long distance applications

we use flexible optical waveguides with
a cylindrical structure constructed from
dielectric materials – these are known as
optical fibres.

• The core of an optical fibre has a higher

refractive index than the cladding that
surrounds it, as in the water jet
experiment of Colladon.

• Hence light propagates down the optical

fibre by repeated total internal
reflection at the core-cladding interface:
ECE 455 – Lecture 04 10

• The two main dielectric materials that are used to fabricate optical fibres are plastic
and glass (specifically silica glass – SiO2).

• Fibre cross-sections come in various sizes according to the application:

Relative sizes of different

fibre types Core
POF = plastic optical fibre
Cladding
HCS = hard clad fibre (silica fibre
core, plastic cladding), also
known as plastic clad fibre (PCF)

Note the relatively small core

size of single mode fibre
compared to the cladding. Silica
ECE 455 – Lecture 04 11

• Although the structure of an optical fibre looks very simple, we should be aware that
this is a precision piece of engineering, especially for single mode fibre which has a
typical core diameter of between 8 m and 10 m:

Comparison between cross sectional size

of a human hair and the core of a single
mode fibre

• This precision is also reflected in the design of an optical fibre and the fabrication
process.

• The relative size of components is also important in terms of how we analyse them;
specifically we are interested in how the component size compares with the
wavelength.
ECE 455 – Lecture 04 12

• We can (and do) analyse optical fibres using electromagnetic theory, i.e. solution of the
wave equation (derived from Maxwell’s equations) for a cylindrical system.

• However, the simplest model of light is based on the assumption that light propagates
as a ray:

• Whenever an optical component has dimensions that are significantly larger than the
wavelength of the light, we can use geometrical optics (also known as ray optics) to
analyse that component.

• In multimode optical fibres, the core diameter is typically 50 μm or 62.5 μm, which is
significantly bigger than the wavelength of light used (of the order of 1 μm).

• So we will begin by using ray optics to look at multimode fibres, although some of the
results (e.g. numerical aperture) will also be used for single mode fibres.
ECE 455 – Lecture 04 13

RAY OPTICS ANALYSIS OF OPTICAL

FIBRES
ECE 455 – Lecture 04 14

• We begin by defining the refractive index (n)

of a material by:

n
c
(1)
t
v n2
• Here v is the speed of light in the material,
which is less than the speed of light in vacuo c. n1
• We then examine what happens at the
interface between two different materials i r
with:
n1  n2 (2)

• From an approach based on electromagnetics,

we can obtain these laws:

i   r Law of reflection

n1 sin i  n2 sin t Law of refraction (Snell’s law) (3)

ECE 455 – Lecture 04 15

Snell’s law for Refraction

n1 sin 1  n2 sin  2 (4)

If we choose n1  n2 , we have:
Total internal reflection
Refraction Critical angle (TIR)

2=900
n2 2 n2 n2
n1 n1 n1
1 1 1

1 < C 1 = C 1 > C

Total internal reflection (TIR) is the mechanism with which light propagates in
multimode optical fibres.
ECE 455 – Lecture 04 16

Example: Reflection coefficients for S and P

polarisation (more about this in ECE333)

 Here the incident ray goes from air

(refractive index 1) to glass and total
internal reflection does not occur.

 Here the incident ray goes from glass

towards air and total internal reflection will
occur for angles above the critical angle.

Note that the reflection coefficient is 100% for

TIR. This is important – the very large number
of reflections in optical fibre would create a lot
of loss if this was not the case.
ECE 455 – Lecture 04 17

By forming a “sandwich” of a central core with refractive index n1 higher than the
surrounding material refractive index n2, we can produce waveguiding via multiple
total internal reflections:

n2
n1 1 1 1
1 1 1 1
1   C

n2

This “sandwich” can be in the form of

the planar waveguide:
ECE 455 – Lecture 04 18

Or it can be “rolled up” to form fibre:

You will notice that the light rays travel in straight lines. This is because the refractive
index of the core region is uniform throughout. (The refractive index of the cladding
is also uniform.)

We term this a step index refractive index profile:

ECE 455 – Lecture 04 19

Standard single mode fibre is step-index, but for multimode fibres we can have both
step-index and a parabolic profile called graded-index:

When we return to the subject of dispersion, we will see that special dispersion-
compensated fibres have more complex refractive index profiles, such as:
ECE 455 – Lecture 04 20

We usually use laser diodes for high

performance optical links. The light
from a laser diode actually diverges:

Provided the light from the laser

can be coupled into the entrance
of the fibre within the correct
range of angles, the light will
propagate within the fibre.

We say that the light must be

within the acceptance angle of
the fibre:
ECE 455 – Lecture 04 21

Note: previous pictures and also the analysis to follow assumes meridional rays:

http://www.fiberoptic.institute/fiber-optic-guide/light-propagation-through-optical-fiber/
ECE 455 – Lecture 04 22

Propagation in an ideal step-index fibre

Refracted
ray n2 Cladding
Air
n0
 Reflected ray
n1 > n2 > n0 

0 Core n1

n2 Cladding

Does not propagate;  < critical angle at core-cladding interface

Propagates through repeated TIR at core-cladding interface;

0 denotes the acceptance angle.
ECE 455 – Lecture 04 23

Numerical aperture (NA) in a step-index fibre

What is the maximum acceptance angle 0?

Apply Snell’s law:

Cladding n2
Air n0

C n1 sin C  n2
n0 sin  0  n1 sin 

0   C   2
Core n1
ECE 455 – Lecture 04 24

n0 sin  0  n1 sin  Cladding n2

 n1 sin ( 2  C )
 n1 cos C C n1 sin C  n2

 n1 1  sin C2

2
0   C   2
 n2 
 n1 1    Core n1
 n1 
 n12  n22 (5)

This equation defines the numerical aperture (NA) of a step-index fibre:

NA  n0 sin  0  n12  n22 (6)

ECE 455 – Lecture 04 25

Normalised Frequency – The V-number

Using an electromagnetic analysis, we can determine the so-called dispersion relations for
an optical fibre:
Normalised propagation constant

V  2.405
Cut-off frequency – below this only one mode
ECE 455 – Lecture 04 26

For a step-index fibre, the normalised frequency or V-number is related to the

numerical aperture by:
2 2
V a n12  n22  a . NA (7)
 

Core radius

The fibre is single mode if V < 2.405. Above this number, the number of modes rises
quickly:
V2
M (8)
 / 22

Number of modes
ECE 455 – Lecture 04 27

From equation (7), we have multimode operation if:

2 2
a n12  n22  a . NA  2.405
 
If we have either a large numerical aperture or a large core radius (a) or both, then
we will end up with multimode operation. In terms of ray optics, this leads to a
picture as follows:

Animation 01
ECE 455 – Lecture 04 28

The problem with step-index multimode fibres is that the different modes (i.e. ray paths)
all have the same speed (uniform refractive index) but they travel different distances for
a fibre of length L (because of different angles of TIR). So if a pulse is launched into a fibre
and excites multiple modes, we end up with pulse broadening:

t +
input
t
This is called intermodal dispersion (or multimode =
dispersion):
t
output

Animation 02
ECE 455 – Lecture 04 29

We can try to minimise intermodal dispersion by using an optical fibre with a graded index
profile:

n2
n1
Replace this (step index 3
multimode): 2
O
1 n

with this (graded index

multimode): n2

3
2
O 1 n
O' O'' 2 n1
3

n2
ECE 455 – Lecture 04 30

n2

3
2
O 1 n
O' O'' 2 n1
3

n2
Light on path 3 (blue) covers a greater overall distance than path 2 (red) or path 1
(green), but because of the varying refractive index, all three paths have the same
average speed and exit the fibre at the same time.

Animation 03
ECE 455 – Lecture 04 31

Alternatively we can return to a step-index profile and to equation (9):

2 2
a n12  n22  a . NA  2.405
 

This shows that for a fixed NA, we can reduce the fibre diameter a until we have only
single mode operation:

Animation 04
ECE 455 – Lecture 04 32

Hence we have now seen three types of optical fibre:

Refractive index Fibre cross sections & Typical

profile ray paths dimensions
n2 n1

Step-index single mode 125 μm cladding

n 8 - 12 μm
core

n2 n1

Step-index multimode 125 - 400 μm cladding

50 - 200 μm
n
core

n2 n1

Graded index multimode 125 – 140 μm cladding

50 - 100 μm
n
core
ECE 455 – Lecture 05 1

Optical Fibres
- Dispersion Part 1
Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 445
• Lecture 05
• Fall Semester 2016
ECE 455 – Lecture 05 2

Optical pulse
This is distorted as it propagates along a fibre

z
Dispersive medium,
Direction of propagation
e.g. optical fibre

DISPERSION – WHAT WE KNOW SO

FAR
ECE 455 – Lecture 05 3

From Lecture 02: The signal travelling through an optical fibre can be degraded due to:

• Attenuation (leads to loss of power)

• Nonlinearity (we have not discussed this yet) and

• Dispersion – leads to pulse spreading. Pulses become wider as they cover more distance in
the fibre, for example:
Pulses overlap to cause
a bit error
1 0 1 1 1 1
Bit stream
at fibre Bit stream
input at fibre
output
t t
Tb

L σ • rms pulse spread per

Rule of thumb BT L ≤ L
distance
4σ • Units: ns/km
ECE 455 – Lecture 05 4

From Lecture 04: Main types of fibre and refractive index profile

Note: Standard single mode

fibre is step-index
ECE 455 – Lecture 05 5
Silica fibres

Single-mode (Step index) Multimode (Step index)

• Do not have modal dispersion • Exhibit modal dispersion (also

(because they are single mode) called intermodal dispersion)

• This does not mean that they

have no dispersion at all.

Multimode (Graded index)

From Lecture 04: Main types of fibre
and modal dispersion

• Minimises modal dispersion

ECE 455 – Lecture 05 6

http://fobasics.blogspot.com.cy/?view=classic
Step-index multimode fibres suffer from modal dispersion because the different modes
(rays of light) travel different distances in covering the length of the fibre.

Note: the different colours are used simply to show different ray paths. A purely
monochromatic source will still lead to modal dispersion.

From Question Sheet 3:

δτ n1  n1 − n2  (NA )
2
=   ≈
L c  n2  2c n2 Modal dispersion (ns/km) considering simple
ray analysis and looking at difference between
shortest and longest (meridional) ray paths.
ECE 455 – Lecture 05 7

In this lecture we will start to consider other types of dispersion:

Dispersion

Modal Dispersion
Chromatic Dispersion Polarization-mode
Also called: Dispersion (PMD)
Intermodal Dispersion Also called:
Multimode Dispersion Mostly a problem for
Intramodal Dispersion single-mode fibres that are
Occurs in: compensated for chromatic
Occurs in: single-mode and dispersion,
Multimode fibres multimode fibres
NOT in single-mode and at long distances

Material Waveguide
Dispersion Dispersion
Due to nonlinear wavelength Due to light propagating in cladding
dependence of refractive index of fibre, can be engineered with
with wavelength different refractive index profiles
ECE 455 – Lecture 05 8

PHYSICAL CAUSES OF DISPERSION

ECE 455 – Lecture 05 9

Modal dispersion in a few words

• There must be more than one mode for this type of dispersion

• Input light is launched into a multimode fibre, exciting multiple modes of

propagation. Because the core diameter is sufficiently large, we can use ray
optics to model the propagation.

• Different modes have the same speed (strictly speaking the same group
velocity), but they travel different distances relative to fibre length.

• This leads to different arrival times at the output and thus pulse spreading.

• A graded index profile helps to reduce modal dispersion.

ECE 455 – Lecture 05 10

Chromatic dispersion in a few words

• Also called intramodal – within a mode. Essentially occurs due to the light
being launched into the fibre being having a spread of wavelengths (i.e.
“colours”, hence the name chromatic dispersion).

• Has two components:

• Material dispersion (due to refractive index dependency on wavelength)
• Waveguide dispersion

• Occurs in both multimode and single-mode fibres, but we neglect it in

multimode fibres because their intermodal dispersion is larger. But we
cannot neglect it in single-mode fibres.
ECE 455 – Lecture 05 11

Chromatic dispersion in a few words – material dispersion

Optical sources (even lasers) are not purely

monochromatic, but instead they have a spectral width if
we look at their optical spectrum

Hence different wavelengths (or different “colours”,

hence the name chromatic) will be present in the input
pulse.

t t
IN OUT
v g (λ2 )
Leading to
Silica has a nonlinear variation v g (λ1 ) < vg (λ2 ) pulse spreading
of refractive index with
wavelength Hence different wavelengths
travel at different group velocities
ECE 455 – Lecture 05 12

Chromatic dispersion in a few words – waveguide dispersion

• Because of the small core diameter (typically 8 μm) relative to the wavelength of the
light, light launched into a single-mode fibre actually travels with a mode field diameter
that is larger than the core diameter.

• Hence part of the optical power is coupled into the cladding, which has a lower
refractive index than the core, leading to a higher velocity.

• The mode field diameter increases with wavelength, leading to more power being
coupled into the cladding and thus a higher overall group velocity.

Material and waveguide dispersion

combine to give overall chromatic
dispersion
ECE 455 – Lecture 05 13

Polarisation mode dispersion (PMD) in a few words

Light can travel in two
orthogonal states of polarisation;
if these see different values of
refractive index (due to
birefringence), then pulse
spreading occurs:

Birefringence can be caused either by material or waveguide effects.

Waveguide PMD is essentially due to manufacturing variations leading to non-

concentric cores, elliptical cores or elliptical fibre:
ECE 455 – Lecture 05 14

Comparing chromatic & modal dispersion, single and multimode fibres

• Chromatic (intramodal) dispersion also exists in multimode fibres, but for this type of fibre
the biggest cause of dispersion is modal dispersion.
- For multimode fibres, we usually ignore chromatic dispersion when we calculate the total dispersion.

• For chromatic dispersion

- Material dispersion makes a larger contribution compared to waveguide dispersion.

Multimode fibres

Caused
by: Material dispersion
Modal dispersion Chromatic (intramodal)
Waveguide dispersion

Single-mode fibres

Multimode fibres have worse dispersion performance compared to single mode.

ECE 455 – Lecture 05 15

Optical fibre
© U. of Washington

If we have a pulse of light which is not monochromatic (it contains a group of

wavelengths), then we will have dispersion:

CHROMATIC DISPERSION
ECE 455 – Lecture 05 16

n2

n1
1 n
O

n
In single-mode fibres, there is no intermodal dispersion (because there is only one mode
of propagation).

However, we have chromatic (intramodal) dispersion

Material dispersion Waveguide dispersion

Chromatic dispersion Due to refractive

Due to nonlinear index profile of the
relationship between fibre. Changes with
• Dispersion due to fact that group
n and λ velocity changes with wavelength λ.
ECE 455 – Lecture 05 17

Dispersion for a standard single mode fibre (silica)

• Dispersion is minimised at 1310 nm. Note: unit is ps/(nm.km)

• But dispersion is non-zero at 1550 nm, which is the wavelength of minimum

attenuation.
ECE 455 – Lecture 05 18

We begin by looking at the contribution of material dispersion. This is caused because of

the finite spectral width of optical sources and the nonlinear dependence of refractive
index with wavelength. This results in group velocity dispersion, so we need to know
something about group velocity.
ECE 455 – Lecture 05 19

PHASE VELOCITY/GROUP VELOCITY

ECE 455 – Lecture 05 20

Group and phase velocity

• All optical sources (including lasers) have a finite spectral width:

Intensity (arbitrary units)

∆λ: spectral width, FWHM

λ
λ0
n
Peak wavelength The refractive index
varies nonlinearly with
wavelength
• Each wavelength will see a different value of
refractive index, and so travel at different speeds:

λ
ECE 455 – Lecture 05 21

• Although we mainly specify wavelength rather than frequency, it will be more

convenient to use frequency for this discussion. We will also consider just two, very
closely spaced frequencies within the group:

Intensity
(arbitrary units)

ω
ω0

δω = ω2 - ω1
ω1 ω2
ECE 455 – Lecture 05 22

• At any given wavelength, we can consider the light to be an electromagnetic wave

whose electric field is a sinusoidal travelling wave (in the + z direction):

E ( z , t ) = E0 cos (kz − ωt ) (1)

2π 2π ω
k= (2) ω= (3) v= ( = fλ ) (4)
λ T k

phase constant angular frequency phase velocity

2π
2π 2π ω ω
k= = =n = nk 0 k= = = nk0 k0 = free space
λ λ0 / n λ0 vp c/n
phase constant
ECE 455 – Lecture 05 23

Transverse electromagnetic wave

x
z
ECE 455 – Lecture 05 24

• Hence if we take the simplified picture of assuming that our optical source emits two
closely spaced frequencies ω1 and ω2, the corresponding waves are:

E1 = E0 cos (k1 z − ω1t )

E2 = E0 cos (k 2 z − ω2t )

• The superposition (addition) of these two waves gives the total waveform as:

ET = E0 [cos (k1 z − ω1t ) + cos (k 2 z − ω2t )] (5)

Superposition of two waves

Total waveform
ECE 455 – Lecture 05 25

Destructive Constructive
Interference interference

E0 cos (k1 z − ω1t )

E0 cos (k1 z − ω1t )
ECE 455 – Lecture 05 26

cos α + cos β = 2 cos 12 (α − β ) cos 12 (α + β )

 (k1 − k 2 ) z − (ω1 − ω2 )t   (k1 + k 2 ) z − (ω1 + ω2 )t 

∴ ET = 2 E0 cos   cos   (6)
 2   2

E ~ kg ωg
0

ET = 2 E0 cos [12 (k1 − k 2 ) z − (ω1 − ω2 )t ]

1
2

× cos [12 (k1 + k 2 ) z − 12 (ω1 + ω2 )t ]

kp ωp
ECE 455 – Lecture 05 27

[ ] [
ET = E0~ cos k g z − ω g t cos k p z − ω p t ] (7)

• If the frequencies are closely spaced, then:

ω1 ≈ ω2 ≈ ω ω p = 12 (ω1 + ω 2 ) ≈ ω

ω g = 12 (ω1 − ω 2 ) << ω

ω p >> ω g
ECE 455 – Lecture 05 28

• We can think of the resultant electric field ET as an amplitude-modulated wave:

ET = E~
0 [
cos k g z − ω g t ] [
cos k p z − ω p t ]
ENVELOPE CARRIER
Modulation frequency = ωg Carrier frequency = ωp

ET1
Superposition of the
two waves is equivalent
Normalised field

to amplitude modulation
0
0 100
t (DSB-Suppressed
carrier)

-1
Time
ECE 455 – Lecture 05 29

[
ET = E0~ cos k g z − ω g t ] [
cos k p z − ω p t ]
ENVELOPE CARRIER

• Velocity of “carrier” is:

ωp ω1 + ω 2 ω
vp = = ≈
kp k1 + k 2 k Phase velocity (8)
ECE 455 – Lecture 05 30

[
ET = E0~ cos k g z − ω g t ] [
cos k p z − ω p t ]
ENVELOPE CARRIER

• Velocity of “envelope” is:

ωg ω1 − ω 2
dω
vg = = ≈
kg k1 − k 2 dk Group velocity (9)
ECE 455 – Lecture 05 31

• The signal propagates at the group velocity vg.

• N.B. The envelope does not exist as a physical artefact; it represents the maximum
excursion of the wave amplitude.
Normalised field
1
vg

0
0 100 vp

-1
Time
ECE 455 – Lecture 05 32

• From (8): ω = kvp and substituting into (9):

dω dv p
vg = = vp + k (10)
dk dk
dλ dv p
∴ vg = v p + k
dk dλ

• Now, k = 2π/λ, hence: dk − 2π − k

= 2 =
dλ λ λ

dv p
∴ vg = v p − λ (11)
dλ
ECE 455 – Lecture 05 33

• If the phase and group velocities are equal, then the envelope will travel at
the same speed as the carrier wave, and there will be no dispersion.

• From equation (11), this implies that the phase velocity should not depend
on wavelength if we are to achieve dispersion-less transmission.

∴ vg = v p ⇒ no dispersion

∴ vg ≠ v p ⇒ dispersion
ECE 455 – Lecture 05 34

• The plot between ω and k is known as the dispersion relation.

• From (9), the gradient of this curve will yield the group velocity:

dω
ωx vg =
vp = dk k =kx
kx
ωx

k
kx
ECE 455 – Lecture 05 35

ω vg = v p

• In normal dispersion, the vg < v p

group velocity is less than the ⇒ normal dispersion
phase velocity.

v g > v p ⇒ anomalous dispersion

• In anomalous dispersion, ω
the group velocity exceeds vg = v p
the phase velocity.

http://www.csupomona.edu/~ajm/materials/animations/packets.html k
ECE 455 – Lecture 05 36

Animation 01
ECE 455 – Lecture 06 1

Optical Fibres
- Dispersion Part 2
[Group Velocity Dispersion]
Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 445
• Lecture 06
• Fall Semester 2016
ECE 455 – Lecture 06 2

CHROMATIC DISPERSION – WHAT

WE KNOW SO FAR
ECE 455 – Lecture 06 3

If we have a pulse of light which is not monochromatic (it contains a group of

wavelengths), then we will have chromatic dispersion:
© U. of Washington

Optical fibre

Material dispersion Waveguide dispersion

Intramodal (chromatic) dispersion

Due to refractive
Due to nonlinear
index profile of the
relationship between • Dispersion due to fact that group fibre. Changes with
n and λ velocity changes with wavelength λ.
ECE 455 – Lecture 06 4

ω
vp = Phase velocity
k
dω
vg = Group velocity
dk
vg ≠ v p ⇒ dispersion

vg = v p ⇒ no dispersion

ω
vg = v p
vg > v p
⇒ anomalousdispersion

vg < v p
⇒ normal dispersion

k
ECE 455 – Lecture 06 5

GROUP VELOCITY DISPERSION

ECE 455 – Lecture 06 6

Wavepackets

• So far we have considered just two, very closely spaced frequencies within the
group emitted by an optical source such as a laser:

Intensity
(arbitrary units) We now look at the whole
spectrum.

Wave packet –
A short pulse composed of

ω0 ω the sum of waves over a

finite bandwidth.
2 δω
• If we consider the entire spectrum emitted by the source, we still obtain a
modulated waveform, with a group velocity and phase velocity as before.
ECE 455 – Lecture 06 7

© UCSD
ECE 455 – Lecture 06 8

• We can prove the properties of the wavepacket by using the Fourier transform:

∞ Fourier ∞
1
∫ F (ω )e ∫
− jω t
f (t ) = dω ↔ F (ω ) = f (t )e jω t dt (12)
−∞
2π −∞

Time domain Frequency domain

F(ω)
This represents optical source
spectrum; has a gaussian
profile

ω
ω0 - δω ω0 ω0 + δω
peak frequency
ECE 455 – Lecture 06 9

• We can think of F(ω) as being equal to some spectrum G(ω) which is identical in
shape, but centred at ω = 0 instead of ω0:

G(ω) F(ω)

- δω 0 δω ω0 - δω ω0 ω0 + δω ω

• By inspection, F (ω ) = G (ω − ω 0 ) (13)

∞
F (ω ) = 1 j (ω − ω 0 ) t
G (ω − ω 0 ) =
2π −∞
∫ g (t ) e dt
∞
1
∫ f (t )e
jω t ∞
dt 1 − jω 0 t jω t
2π −∞
=
2π −∞
∫ g (t ) e e dt
(14) (15)
ECE 455 – Lecture 06 10

− jω 0 t
• Hence: f (t ) = g (t ) e (16)

Corresponds to sinusoid
at optical frequency ω0
1,5

Impulse response of: 0,5

G(ω)
0
0 0,5 1 1,5 2

-0,5

-1

-1,5

0
gives:
g(t)

N.B. Fourier transform of a gaussian

0 pulse is also gaussian in shape
ECE 455 – Lecture 06 11

• In other words, the impulse response associated with the optical source takes on the
form of a wavepacket:

g(t)

f (t)

• This wavepacket represents a pulse of light emitted by the optical source, and it contains
a range of frequencies (i.e. wavelengths).

• We now need to examine what will happen to the group velocity of this pulse as it
propagates along a fibre.
ECE 455 – Lecture 06 12

• Consider an optical pulse launched into a single mode fibre. Due to the spectral width
of the source, this pulse consists of a group of wavelengths which travel at the group
velocity:

optical power dω
vg =
dk
wavelength λ

λ0 distance
ECE 455 – Lecture 06 13

• So the time taken for the wavegroup to travel a distance L along the fibre is given by the
group delay τg:
L dk
τg = = L (17)
vg dω

• The phase velocity of the peak wavelength λ0 is given by:

ω c ωn
vp = = ⇒ k= (18)
k n c
• Substituting (18) into (17):

τg dk 1  dn 
= = n + ω (19)
L dω c  dω 
ECE 455 – Lecture 06 14

• Eqn. (3) shows that the group delay per unit length depends on both n and dn/dω. It is
also dependent on the frequency ω. However, we prefer to work with wavelength λ
instead:

n n

instead of....

λ ω

• From the inverse relationship between frequency and wavelength (c = fλ = ωλ/2π), we

find that:

τg 1 dn  1  dn 
= n + ω  = n − λ  (20)
L c dω  c  dλ 
ECE 455 – Lecture 06 15

Derivation of equation (20):

c 2πc dλ 2πc λ
c = fλ ⇒ λ = ⇒λ = ⇒ =− 2 =−
f ω dω ω ω

τg 1 dn 
∴ = n + ω
L c  dω 
1 dn dλ 
= n + ω
c dλ dω 
1 dn 
= n − λ
c dλ 
ECE 455 – Lecture 06 16

Group Refractive Index

• Imagine we have a fibre with core refractive index n. In this case,

cω
vp = = (21)
k n

• If we transmit a spread of wavelengths, then we can regard the resulting group as

encountering a group refractive index, and this is defined via:

dω c
vg = = (22)
dk ng

c
ng = (23)
vg
ECE 455 – Lecture 06 17

τg dn
∴ ng = c = n − λ
L dλ (24)

• n varies with wavelength:

dn
≠ 0 ⇒ ng ≠ n ⇒ v g ≠ v ⇒ dispersion
dλ

• In fact, ng will also be wavelength dependent:

dng d 2n
=−λ 2 (25)
dλ dλ
ECE 455 – Lecture 06 18

dng d 2n
ng =−λ 2
dλ dλ

dng
Minimum =0
n dλ

d 2n
Point of inflection
=0
dλ 2

For silica glass: At 1.31 µm, n has

a point of inflection, ng is
λ minimum, and the group velocity
1.31 µm
is therefore maximum.
ECE 455 – Lecture 06 19

Group velocity dispersion (GVD)

• We know that:

• An optical source emits a spread of wavelengths centred on λ0.

• This can be represented by a wavepacket which travels at the group velocity
and therefore “sees” a group index ng.

• However, ng and thus the group velocity vg and delay τg are all wavelength
dependent.

• Each different spectral component emitted by the source will travel at

different group velocities, and this GVD is the cause of material dispersion.
ECE 455 – Lecture 06 20

• Consider the delay difference (per unit length) for a wavelength δλ away from the
central wavelength λ0:

τg 1
τ g (λ0 )
L
L δτg
 ng  1
 =  τ g (λ0 + δλ)
δλ L
 c 

λ
λ0 λ0 + δλ

• If the wavelength difference is sufficiently small, we can neglect second-order

terms in a Taylor series expansion to get:

τ g (λ 0 + δλ) − τ g (λ 0 ) δλ dτ g
= (26)
L L dλ λ0
ECE 455 – Lecture 06 21

• Consider the delay difference (per unit length) for a wavelength δλ away from the
central wavelength λ0:

1 δτ g 1 dτ g
=
L δλ L dλ λ0

• From (20):
τg 1 dn 
=  n−λ 
L c dλ 

Material
dispersion
1 δτ g − λ d 2n
∴ = (27)
Dmat
L δλ c dλ2 Units: ps/(nm.km)
ECE 455 – Lecture 06 22

Group index, refractive index

and material dispersion for
silica glass (SiO2)

material dispersion
(ps/nm-km)
length (km)
(11)

σ mat = Dmat σ λ L
− λ d 2n
spread Dmat =
in time spread in c dλ2
(ps) wavelength (nm)
ECE 455 – Lecture 06 23

DISPERSION MANAGEMENT
ECE 455 – Lecture 06 24

Dispersion modified fibres

• For conventional single-mode optical fibre:

– minimum attenuation occurs at 1.55 µm

– minimum dispersion occurs at 1.3 µm

• Furthermore, optical amplifiers operate in the 1.55 µm

region

• In response to this, dispersion modified fibres have been

developed to provide minimal dispersion at 1.55 µm
ECE 455 – Lecture 06 25

• Structure dependent losses (waveguide losses) have little

effect on overall attenuation, so changing the refractive
index profile in single-mode fibre will have negligible impact
on attenuation.
– However, changing the refractive index will modify the
waveguide dispersion term, and this can be used to our
advantage.

• In fact, the refractive index profile can be tailored to shift

the dispersion zero to 1.55 µm or to flatten the dispersion
vs. wavelength profile so that dispersion is almost zero
between 1.3 µm and 1.55 µm
ECE 455 – Lecture 06 26

Changing the refractive index profile changes the waveguide

dispersion:

Dispersion
flattened

Dispersion
shifted
ECE 455 – Lecture 06 27

Dispersion Dispersion
shifted flattened
ECE 455 – Lecture 07 1

Optical Fibres
- Attenuation

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 445
• Lecture 07
• Fall Semester 2016
ECE 455 – Lecture 07 2

LOOKING AT REFRACTIVE INDEX IN A

NEW WAY
ECE 455 – Lecture 07 3

Silica optical fibre attenuation varies with wavelength, and over time it has
been reduced through improved manufacturing methods. This has influenced
the evolution of the first, second and third generations of optical fibre
communications.
ECE 455 – Lecture 07 4

When we looked at group velocity dispersion in Lecture 06, we saw that

refractive index is an optical material parameter that also varies with
wavelength. Is there a connection?

In electromagnetism, you will have seen that Maxwell’s equations can be used
to derive the wave equation, e.g. in free space for one dimension:

∂ 2Ex ∂ 2Ex 1 ∂ 2Ex

= µ 0ε 0 = 2 (1)
∂z 2
∂t 2
c ∂t 2

1
c= (2)
µ 0ε 0

µ 0 = Permeability
of free space
ε 0 = Permittivity
of free space
ECE 455 – Lecture 07 5
1 c
In a material with refractive index n, we have the phase velocity: v= = (3)
µε n

c µε
∴n = = (4)
v µ 0ε 0

µ = µr µ0 µr = Relative permeability = dimensionless number

ε = ε rε 0 εr = Relative permittivity = dimensionless number

n= µ rε r (5)

We consider that optical materials used for optical fibres are non-magnetic, such that:

µr ≈ 1 (6)

This then gives Maxwell’s relation for the refractive index of a dielectric: n ≈ εr (7)

You might have seen in ECE 331 that relative permittivity can be a complex number……
ECE 455 – Lecture 07 6

From Lecture 05, we saw that a monochromatic wave of light can be written as a
travelling wave:
E ( z, t ) = E cos (ωt − kz)
0
(8)

This can also be written as a complex number:

E ( z, t ) = E0 exp[ j (ωt − kz)] (9)

ω c ωn
Phase velocity v= = (10) k= (11) Phase constant
k n c

Substituting (11) into (9):

  n 
E ( z, t ) = E0 exp jω  t − z  (12)
  c 

So far in ECE 455 we have considered the refractive index to be real.

ECE 455 – Lecture 07 7

What would happen if we assumed a complex refractive index?

n = n′ − jn′′ (13) n′′ ≥ 0

Our monochromatic wave now becomes:

  (n′ − jn′′) 
E ( z, t ) = E0 exp jω  t − z 
  c 
 ω    n′ 
′
′
= E0 exp− n z  ⋅ exp jω  t − z  (14)
 c    c 

The imaginary part of the refractive The real part of the refractive index
index corresponds to attenuation as corresponds to a phase change term
described by Beer’s law in Lecture 03. that can also be seen as a delay.
ECE 455 – Lecture 07 8

INTERACTION OF LIGHT WITH

DIELECTRIC MATERIALS
ECE 455 – Lecture 07 9

Returning to slide 3, the question is, why does

the attenuation vs wavelength profile have this
general shape?

To find the answer, we have to consider what

happens to light as it travels through a
dielectric material (like silica glass).

r r
A dielectric is a material in which charges do
E p = qx
not flow like in a conductor. Instead, we have a
collection of dipoles which under an applied
+
electric field will be slightly distorted.

We define the dipole moment as the vector x

with the same direction as the E-field and
magnitude:
p = qx (15) -
ECE 455 – Lecture 07 10

So the electric field causes a slight displacement of the electron compared to the much
heavier nucleus. This separation between the electron and the positively charged
nucleus will create an electric field that wants to restore things to their original state
when the external field is removed.

This situation is very similar to that of a spring:

d 2x dx
F = m 2 + 2ζm + mωo2 x (16)
dt dt

Newton’s Damping Restoring force

second law (Hooke’s law)

And so we can use a spring-mass-damper system

to model a dielectric dipole:
ECE 455 – Lecture 07 11

We can use this spring model to analyse how light, in the form of an incident travelling
electric wave will displace a dipole. The sinusoidal displacement takes the form of simple
harmonic motion.

The oscillating dipole will then act like a mini-antenna, radiating its own electric field:

https://phet.colorado.edu/sims/radiating-charge/radiating-charge_el.html
ECE 455 – Lecture 07 12

So as the wave propagates through the dielectric material (e.g. glass):

it excites dipole oscillations, which then emit their own light:

We will return to this concept of dipole oscillations in later exercises, in which we obtain
more detail on the nature of the refractive index and its link with attenuation.
ECE 455 – Lecture 07 13

PHYSICAL CAUSES OF OPTICAL FIBRE

ATTENUATION
ECE 455 – Lecture 07 14

Light can interact with matter in different ways:

Once light is coupled into a standard optical fibre, the attenuation is mainly
caused by absorption and scattering.

We will not consider amplification (e.g. erbium-doped fibre) yet, and we will not
look at nonlinear effects like Raman scattering.
ECE 455 – Lecture 07 15

Various sources of fibre attenuation

Source: C-L Chen, Elements of Optoelectronics & Fiber Optics

ECE 455 – Lecture 07
16

• Attenuation is caused by:

– Absorption: depends on material and impurities

• Intrinsic absorption by atoms of fibre material
• Extrinsic absorption by impurity atoms
• Absorption by atomic defects in glass

– Scattering: due to inhomogeneous material

• Rayleigh scattering
• Mie scattering

– Radiation: due to discontinuities, e.g. bending of fibre

• Macrobends and microbends
ECE 455 – Lecture 07 17

Absorption losses: extrinsic

• Extrinsic absorption is caused by metal (iron,cobalt, copper
and chromium) and hydroxyl (OH) ions. In the early years,
fibres had a high water impurity content, and hence high
overtones of the water absorption peak.

• In high purity modern fibres (low OH), loss due to extrinsic

absorption has been significantly reduced. (This is achieved by
drying the glass in chlorine gas to leach out the water vapour).

• For both OH and metal ions, ion concentrations of one part

per billion or less are needed to minimise losses to acceptable
levels.
ECE 455 – Lecture 07 18

Absorption spectrum for OH in silica

ECE 455 – Lecture 07 19

Absorption losses: intrinsic

• Intrinsic absorption results from electronic absorption bands
in the UV region and atomic vibration bands in the near
infrared region. It is the loss associated with the pure fibre
material, and therefore sets the lower limit on absorption.
– In other words, loss due to absorption cannot be reduced
below this limit.

• Attenuation caused by intrinsic absorption in the UV and IR

regions is wavelength dependent as follows:

αUV = AUV exp (λUV / λ)

αIR = AIR exp (-λIR / λ)

ECE 455 – Lecture 07 20

• Lattice absorption through a crystal structure

– The EM wave (near infrared light) forces ions to vibrate at
the frequency of the wave; some energy is then lost by
being coupled into lattice vibrations (heat).

- + - + -
Solid material
+ - + - + Ions form a lattice

- + - + - αIR = AIR exp (-λIR / λ)

Ex

z
http://en.wikipedia.org/wiki/Transparency_and_translucency
ECE 455 – Lecture 07 21

• Absorption of ultraviolet light leading to electronic

transitions:

αUV = AUV exp (λUV / λ)

ECE 455 – Lecture 07 22

Scattering losses
• Scattering mechanisms cause the transfer of some or all of the optical
power contained in one propagating mode to be transferred linearly into
a different mode. There are two major types of scattering:

– Rayleigh scattering: caused by inhomogeneities of a random nature

occuring on a small scale compared with the wavelength of the light.

– Mie scattering: occurs at inhomogeneities where the discontinuity is

comparable to the wavelength.
ECE 455 – Lecture 07 23

• Rayleigh scattering:
– The EM wave forces dipole oscillations in the dielectric
particle that it encounters. The particle then acts like a dipole
antenna, radiating waves in many directions.
Scattered wave

Incident wave Through wave

+
-

Scattered wave
Scattered wave
Dielectric particle
smaller than wavelength
ECE 455 – Lecture 07 24

• Rayleigh scattering has a λ-4 dependence, i.e.

αRayleigh = AR λ-4
Attenuation (dB/km) 1.0

0.1
1200 1400 1600
Wavelength (nm)
ECE 455 – Lecture 07 25

Intrinsic attenuation for a pure silica fibre

ECE 455 – Lecture 07 26

Measured attenuation for ultra-low loss single-

mode silica fibre
ECE 455 – Lecture 07 27

Radiation losses usually occur at bends in the optical fibre:

Critical bend radius

usually 3 - 4 cm for standard
SM fibre
ECE 455 – Lecture 07 28

Radiation losses also occur at microbends

introduced due to uneven pressures in cabling of fibre:

Keiser
ECE 455 – Lecture 07 29
αB (m-1) for 10 cm of bend © 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
102

10

10−1

10−2 λ = 633 nm
V ≈ 2.08 λ = 790 nm
V ≈ 1.67
10−3
0 2 4 6 8 10 12 14 16 18
Radius of curvature (mm)

Measured microbending loss for a 10 cm fiber bent by different amounts of radius of

curvature R. Single mode fiber with a core diameter of 3.9 µm, cladding radius 48 µm,
∆ = 0.004, NA = 0.11, V ≈ 1.67 and 2.08 (Data extracted and replotted with ∆ correction
from, A.J. Harris and P.F. Castle, IEEE J. Light Wave Technology , Vol. LT14, pp. 34-
40, 1986; see original article for discussion of peaks in αB vs. R at 790 nm).
ECE 455 – Lecture 07 30

SUMMARY OF ATTENUATION

Attenuation

Scattering Radiative
Absorption Losses losses/ Bending
losses

Extrinsic
Intrinsic Atomic
(Impurity
Absorption Defects
atoms, e.g. OH)

Absorption Absorption
in in Rayleigh Mie Microscopic Macroscopic
Infrared Ultraviolet Scattering Scattering bends bends
region region
ECE 455 Lecture 08 1

Optical Fibre Amplifiers

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• ECE 445
• Lecture 08
• Fall Semester 2016
ECE 455 Lecture 08 2

SYSTEM CONSIDERATIONS
ECE 455 Lecture 08 3

pIN (t) pOUT (t)  pIN(t - )

Fibre
p (t)
L = c/ng

t
t
to 
Attenuation & dispersion
• Reduction in pulse energy
• Pulse spreading
ECE 455 Lecture 08 4

At low bit rates, the maximum transmission distance is limited by attenuation, while
at high bit rates the distance is limited by dispersion:

Attenuation-limited

Dispersion-limited

FP = Fabry-Perot laser diode

DFB = distributed feedback laser diode
ECE 455 Lecture 08 5

REGENERATION
ECE 455 Lecture 08
Optical Signal Regeneration
• It is necessary to re-amplify and reshape the pulses at regular
intervals using regeneration:

Optical FIBRE Photo-

3R
source receiver

Regenerator 3R =
Fibre Photoreceiver • retiming
input • reshaping
• re-amplification
Electronics:
Clock recovery,
pulse reshaping

Laser transmitter Fibre

output
ECE 455 Lecture 08 7
Optical regenerators are classified into three categories by the 3-R's scheme.

1R : re-amplification of the data pulse alone is carried out.

2R : in addition to re-amplification, pulse reshaping is carried out.
3R : in addition to re-amplification and reshaping, retiming of data pulse is
done.

• Disadvantages:
• Advantages:
 O/E & E/O conversion
– Clock recovery needed
– Pulse reshaping  Bit rate is “locked in” –
no upgrades
 Single wavelength only
ECE 455 Lecture 08 8

OPTICAL AMPLIFICATION
ECE 455 Lecture 08
Optical Amplifiers

• All-optical components (i.e. optical input/output). Fibre-based amplifiers

also contain lasers, but this is to create a population inversion in the gain
medium.

• Have replaced electronics-based regenerators, in which optical signals had

to be hotodetected, amplified electronically and then applied to optical
source.

• Have revolutionised optical communications

– used in wavelength division multiplexed (WDM) systems
– allow the use of soliton transmission at ultra high bit rates (1000s of
Gb/s) over thousands of km
– Have removed the speed and wavelength bottleneck associated with all-
electronic regeneration.
ECE 455 Lecture 08
• An optical amplifier provides gain over a useful
spectral range, as shown here for an erbium-doped fibre amplifier:

Fibre
Attenuation
(dB/km)


1550 nm
Optical
amplifier
gain
(dB)
40 nm


ECE 455 Lecture 08
• This broad spectral range enables a number of wavelengths to be
multiplexed onto a fibre, thus increasing the bit rate that can be
transmitted.

Spectrum of 16 amplified WDM channels (using EDFA)

ECE 455 Lecture 08

Optical
Fibre
gain medium Fibre
input
output

Optical Pump
amplifier

• Advantages: • Disadvantages:
– Optical input & output • No pulse reshaping
– Photons in – more photons out •  Needs dispersion
– Transparent to both bit rate & compensation
modulation format • Adds noise to output
signal
– Supports many wavelengths
• WDM: Wavelength division
multiplexing
ECE 455 Lecture 08
Ideal amplifier: POUT
Output
Input

GAIN PIN
Gain
Gain
Phase

• Flat gain response PIN

• Linear phase response
ECE 455 Lecture 08
Real amplifier: Output POUT

Input

PIN
GAIN + Gain
NOISE

Gain
Phase

PIN
f • Gain saturation
• Nonlinearity
ECE 455 Lecture 08 15

TYPES OF OPTICAL AMPLIFIER

ECE 455 Lecture 08
Types of optical amplifier
• Semiconductor optical amplifiers (SOAs)
• Fibre amplifiers
1. Making use of nonlinear effects, such as stimulated Raman
scattering (these are also known as distributed fibre
amplifiers).
2. Rare earth doped fibres: most common type is erbium-
doped (1.55μm central wavelength), but praseodymium-
doped also available (1.3μm).

In this course, we will only consider erbium-doped fibre

amplifiers (EDFA).
ECE 455 Lecture 08
The most commonly used amplifiers are EDFAs
17

EDFAs have replaced the

approach taken with early
generation links that used all-
electronic 3R regenerators.

EDFAs are used in all modern

long-distance optical links,
but usually the regeneration
is 1R.
ECE 455 Lecture 08
Packaged erbium-doped fibre amplifier (EDFA)
Coupler Pump laser diode

Erbium-doped fibre
Input and output fibres
ECE 455 Lecture 08 19

Other doped fibre amplifiers

Band name Meaning Wavelength (nm) Technology

O Original 1260-1360 Praseodymium
E Extended 1360-1460
S Short 1460-1530 Thulium
C Conventional 1530-1565 Erbium
L Long 1565-1625 Erbium
U Ultra-long 1625-1675
ECE 455 Lecture 08 20

APPLICATIONS OF OPTICAL
AMPLIFIERS
ECE 455 Lecture 08
• Application 1: As in-line amplifiers in long-haul links to compensate for
attenuation in the 1550 nm window. Mostly EDFAs and Raman.

Optical Optical
Source Receiver

Optical Optical amplifier compensates

fibre for fibre loss at periodic intervals

Optical amplifiers boost the

signal at regular intervals (e.g.
between 30 km to 80 km) to
make sure power level in link
does not drop below the
required receiver sensitivity
ECE 455 Lecture 08
Optical amplifiers compensate for loss, but they also introduce noise:

Amplifier
80-100km

Power Input After Loss After Amplifier

Added Noise

Wavelength

Hence a low noise figure is important, as well as saturation power (being able to handle
medium power levels)
ECE 455 Lecture 08
30

The other problem is that for

20 1300nm

Fibre Loss (dB/km)

Di sper sion (ps/nm)
standard single-mode fibre, the 10 1550nm
1550 nm window offers low window
loss, but minimal chromatic 0
dispersion is at 1310 nm.
-10

-20

-30
1250 1350 1450 1550 1650
Wavelength (nm)

Hence some kind of “dispersion

management” or dispersion compensation is
required, e.g. by using dispersion-shifted fibre
(DSF)
ECE 455 Lecture 08
• Application 2: As power amplifiers to increase source power (post-amplifiers):

PS (dBm) G(dB)

Optical
Transmitter Output power (dBm)
= PS + G

• Most laser diodes used in optical transmitters have powers of a few mW, but
fibre can handle of the order of 100 mW before optical nonlinear effects occur.
So a power amplifier can be used to boost signal immediately after the source.

• SOAs are useful because they can be integrated with lasers, but EDFA power
amplifiers are also available with output powers around 100 mW.

• Amplifier adds noise, but this is attenuated by the fibre

• Important that the amplifier is not saturated by the transmitter

ECE 455 Lecture 08
• Application 3: As pre-amplifiers to improve receiver sensitivity:

Optical input
Optical
Receiver

• Optical amplifier is placed immediately before the optical receiver in order

to improve sensitivity.

• At this point the signal is weak, so good gain is required, but even more
important is the fact that the amplifier must not add a lot of noise, so a low
noise figure is required (typically less than 5 dB).
ECE 455 Lecture 08
• Application 4: As booster amplifiers in distribution networks (e.g. local access)
to compensate for losses in a fibre splitter:

Star coupler: splits into N fibres; has insertion and splitting loss
ECE 455 Lecture 08 27

FIGURES OF MERIT FOR OPTICAL

AMPLIFIERS
ECE 455 Lecture 08

Important figures of merit & considerations for an amplifier

• Include:
– Gain
– Bandwidth
– Gain saturation
– Noise
ECE 455 Lecture 08
Properties of Ideal Optical Amplifiers
• Provide high gain
– (30 dB or more)
• Have a wide spectral bandwidth
– to allow several wavelengths to be transmitted
• Provide uniform (i.e. flat) gain vs. 
– to maintain relative strength of spectral components
• Allow bi-directional operation
– i.e. gain in both directions
• Have low insertion loss
– to maximise benefits of amplifier gain
• Have no crosstalk
– i.e. no interference between different spectral components
• Have wide dynamic range
– gain should not saturate with high input powers
• Have a good conversion efficiency
– pump power converted to amplifier gain
ECE 455 Lecture 08
Gain profile of erbium-doped silica fibre

High gain over a wide spectral bandwidth, but the gain profile is not flat.
ECE 455 Lecture 08
Spectrum of EDFA with1480 nm pump

ASE: Amplified spontaneous emission noise

ECE 455 Lecture 08

Typical gain versus power profile for optical amplifier:

ECE 455 Lecture 09 1

Optical Fibre Amplifiers – Continued

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• ECE 445
• Lecture 09
• Fall Semester 2016
ECE 455 Lecture 09 2

ERBIUM-DOPED FIBRE AMPLIFIERS

– BASIC PHYSICS
ECE 455 Lecture 09
Energy Transitions in Er3+ - Doped Silica Fibre
ECE 455 Lecture 09

EDFA Basic Structure

Wavelength Narrowband
Isolator multiplexer optical filter

Weak input
signal at • Amplified
1.55μm signal at
Amplification 1.55m
section with • Gain
erbium doped 20 to 30 dB.
silica fibre,
Laser diode a few tens of metres 30 dB gain means
1000 photons out
pump at 980 nm (Er3+ ions, 100 – 100 ppm) for 1 photon in
(or 1480 nm,
up to 50 mW power)
ECE 455 Lecture 09
Power exchange

Power level
Power level

980 nm 1550 nm 980 nm 1550 nm

signal data signal signal data signal

Wavelength
Isolator multiplexer Narrowband
optical filter

Input
Output

Pump
ECE 455 Lecture 09 6

Gain as a function of length of erbium-doped fibre

If the fibre is too long, there will be more absorption than gain, but if the
fibre is too short we will not have as much gain as we could. Optimum
length depends on the pump power.
ECE 455 Lecture 09 7

NOISE IN ERBIUM-DOPED FIBRE

AMPLIFIERS
ECE 455 Lecture 09
Random spontaneous
emission (SE)
Amplified
Amplification along fibre spontaneous
emission (ASE)

Erbium randomly emits photons between 1520 and 1570 nm

• Spontaneous emission (SE) is not polarized or coherent

• Like any photon, SE stimulates emission of other photons
• With no input signal, eventually all optical energy is consumed into
amplified spontaneous emission
ECE 455 Lecture 09

Optical Amplifier Chains

 Optical amplifiers allow one to extend linkFibre

distance
Link between a transmitter and
receiver
 Amplifier can compensate for attenuation

 Cannot compensate for dispersion (and crosstalk in DWDM systems)

 Amplifiers also introduce noise, as each amplifier reduces the Optical SNR by a small
amount (noise figure)

Optical
Transmitter
Receiver
1 2 N

Optical Amplifiers Fibre Section

ECE 455 Lecture 09

Amplifier Chains and Signal Level

 Example: system uses fibre with 0.25 dB/km attenuation, 80 km fibre sections,
Fibre Link
amplifiers with 19 dB gain a noise figure of 5 dB

10

0
Signal level (dBm)

-10

-20

-30
0 100 200 300 400 500 600 700 800
Location (km )

 Each amplifier restores the signal level to a value almost equivalent to the level
at the start of the section - in principle reach is extended to 700 km +
ECE 455 Lecture 09

Amplifier Chains and Optical SNR

Fibre
 Same system: Transmitter SNR is 50 dB, Link noise figure of 5 dB,
amplifier

60

50
Optical SNR (dB)

40

30

20

10

0
0 100 200 300 400 500 600 700 800

Location (km)

 Optical SNR drops with distance, so that if we take 30 dB as a reasonable limit, the
max distance between T/X and R/X is only 300 km
ECE 455 Lecture 09 12

GAIN PROFILE OF ERBIUM-DOPED

FIBRE AMPLIFIERS
ECE 455 Lecture 09

EDFA Output Spectra

+10 dBm

Amplified signal spectrum

(input signal saturates the
optical amplifier)

ASE spectrum when no

input signal is present

-40 dBm
1575 nm
1525 nm
ECE 455 Lecture 09

Gain Characteristics of EDFA

Gain (amplifier) - is the ratio in decibels of
input power to output power.
Gain at 1560 nm is some 3 dB higher than
gain at 1540 nm (this is twice as much).
In most applications (if there is only a
single channel or if there are only a few
amplifiers in the circuit) this is not too
much of a limitation.

WDM systems use many

wavelengths within the amplified
band. If we have a very long WDM
link with many amplifiers the
difference in response in various
channels adds up.
ECE 455 Lecture 09

Gain Flattening Concept

ECE 455 Lecture 09 16

gain G, noise figure F

input signal  gain + noise
analogous to DC bias

SYSTEM PERFORMANCE OF
OPTICAL AMPLIFIERS
ECE 455 Lecture 09

EDFA output versus wavelength

ASE = amplified spontaneous emission: noise

ECE 455 Lecture 09

Gain versus EDFA length

ECE 455 Lecture 09

EDFA gain versus pump level

ECE 455 Lecture 09

Typical gain versus power profile for optical amplifier:

ECE 455 Lecture 09

SNR degradation for a chain of EDFAs

ECE 455 Lecture 09 22

EDFA CHAINS
ECE 455 Lecture 09
Optical Amplifier Gain Control
• Consider in-line amplifier application, as in long haul links:

L L L
G G G

• Set amplifier gain to compensate for loss of inter-

connecting fibres of length L, i.e.:

G = L

• So if the link consists of equal number of amplifiers and

interconnecting fibres, overall link loss should be zero.
ECE 455 Lecture 09

L L L
G G G

Px G + Px G + Px - L = Px {If G = L}

Note! All powers expressed in dBm, all gains and losses

expressed in dB.

• Consider next example, with three in-line amplifiers, and

length L chosen to be maximum for given source power and
receiver sensitivity.
ECE 455 Lecture 09
Optical Photo-
source receiver

L L L L
PS G G G PR

Ps

PR = receiver sensitivity
PR = PS - L

G + PS - L = PS {If G = L}
: output power from first amplifier

PS - L : power entering first amplifier

Ps : source power (dBm)
ECE•455
NowLecture 09
consider
situation where power at some point in link
drops suddenly (e.g. due to fault at laser):

L L L L
PS G G G PR

Ps - Px

G + PS - Px - L PS - Px - L < PR

PS - Px - L
Ps - Px

Bad news: drop in power means that the power incident on the
photoreceiver is now less than the receiver sensitivity, which in
a digital system means the BER specification is not met.
ECE 455 Lecture 09
• One solution is passive gain control: relies on using the amplifier
in its saturation region:

• If input power drops (rises), gain increases (decreases) to

compensate for this. Similar effect to feedback (but it is not f/b!).
ECE 455 Lecture 09
• For example, consider an amplifier with a gain/input power
slope of - 1 dB/dBm in the saturation region:

POUT = Pnom -  + Gnom +  = Pnom + Gnom

G(dB)
POUT = Pnom + Gnom

Gnom + 
POUT = Pnom +  + Gnom - 
= Pnom + Gnom
Gnom
slope = -1 dB/dBm
Gnom - 

PIN(dBm)
Pnom -  Pnom Pnom + 
ECE 455 Lecture 09

• This leads to a self-healing effect in systems where

cascades of amplifiers are used (such as in-line).

• The disadvantage is that the gain is low, because the

amplifiers operate in the saturation region.

• The slope in general is not -1 dB/dBm, but even when it

is not, self-healing will occur, but not immediately after
the first amplifier.

We will see this in the next example.

ECE 455 Lecture 09
Example
Consider a long-distance transmission system containing a cascaded
chain of erbium-doped fibre amplifiers (EDFAs). Assume each EDFA
is operated in saturation and that the slope of the gain-versus-input
power curve is –0.5; for example, the gain changes by
±
± 2 dB for a 4 dB variation in input power. The EDFAs in the link
have the following operational parameters:

Nominal gain: Gnom = 7.3 dB

Nominal optical output power: POUT = 3 dBm
Nominal optical input power: PIN = -4.3 dBm

Suppose there is a sudden 4 dB drop in signal level at some point

in the link. Find the output power levels after the degraded signal
has passed through 1,2, and 3 succeeding EDFAs.
ECE 455 Lecture 09
• Before power drop:
L
G = 7.3 dB = 7.3 dB
L
G G G

- 4.3 dBm 3 dBm - 4.3 dBm ...................

• After power drop:

G(PIN1) G(PIN2) G(PIN3)
L L L

- 4.3 dBm - 4 dB
= - 8.3 dBm
ECE 455 Lecture 09

• After power drop:

G(PIN1) G(PIN2) G(PIN3)
L = 7.3 dB L L

- 4.3 dBm - 4 dB
= - 8.3 dBm

- 4 dB drop for PIN1 (relative to the

nominal value of -4.3 dBm) means that
G for amp 1 goes up by 2 dB from Gnom,
hence

G(PIN1) = 7.3 + 2 = 9.3 dB

ECE 455 Lecture 09
G(PIN1) =
9.3 dB
L =
L 7.3 dB L

- 8.3 dBm

- 8.3 dBm + 9.3 dB PIN2 is 2 dB below the

= 1 dBm nominal value of - 4.3 dBm

So G for amp 2 will be 1 dB

1 dBm - 7.3 dB
above the nominal value of
= -6.3 dBm
7.3 dB, i.e. G(PIN2) = 8.3 dB
ECE 455 Lecture 09
G(PIN2) =
8.3 dB L =
L L 7.3 dB

- 6.3 dBm 2 dBm - 7.3 dB

= -5.3 dBm

- 6.3 dBm + 8.3 dB

= 2 dBm
PIN3 is 1 dB below the
nominal value of - 4.3 dBm

So G for amp 2 will be 0.5 dB

above the nominal value of
7.3 dB, i.e. G(PIN3) = 7.8 dB
ECE 455 Lecture 09
G(PIN3) =
7.8 dB L =
L L 7.3 dB

- 5.3 dBm
- 5.3 dBm + 7.8 dB
= 2.5 dBm
2.5 dBm - 7.3 dB
= -4.8 dBm

PIN4 is 0.5 dB below the

nominal value of - 4.3 dBm
ECE 455 Lecture 09

G(dB)

1 self-healing
G1 = 9.3

G2 = 8.3 2 nominal
3 point
G3 = 7.8

Gnom = 7.3

PIN(dBm)
PIN1 PIN2 PIN2 Pnom = -4.3
-8.3 -6.3 -5.3
ECE 455 Lecture 10 1

Photodiodes

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• ECE 445
• Lecture 10
• Fall Semester 2016
ECE 455 Lecture 10 2

OPTOELECTRONIC (O/E) CONVERSION

ECE 455 Lecture 10 3

• In optical-to-electrical (O/E) conversion, our aim is to convert an incoming time-

varying optical power into corresponding variations of electrical signal (current,
perhaps followed by a transimpedance amplifier).

In addition to having sufficient bandwidth and

adequate conversion efficiency, we require that the output is an
exact copy of the input
– i.e. noise is bad news.

• Definitions used here: a photodiode is a one-port electrical device with an

optical input which is based on a semiconductor diode. A photodetector is a
circuit containing a photodiode as the front end followed by electronic
amplification.
ECE 455 Lecture 10 4

IP
Slope is given by ty 
Responsivi (A/W)
Photocurrent

PO
responsivity
This is with reference to the static
characteristic

numberof e  hpa i rs
Qua ntum effi ci ency
no.of i nci dentphotons
I q
 P
PO / hf
Input optical power
hc
R
q

• In an ideal photodiode (no noise, no nonlinearity), there is a linear correspondence

between input optical power and photocurrent.

• One consequence of this is that optical loss in dB is double the corresponding electrical
loss in dB (1 dBo = 2 dBe).

• Note that the photodiode is actually classified as a square law device, since optical
power varies directly with the square of the electric field magnitude.
ECE 455 Lecture 10 5

Photodiode Requirements
• High sensitivity at operating wavelengths
• Minimum noise
• High e/o conversion efficiency
• Fast response times
• High linearity
• Small size
• Low bias voltages
• High reliability
• Efficient coupling of light (anti-reflection coating)
ECE 455 Lecture 10 6

TAXONOMY
ECE 455 Lecture 10 7

Classification according to electrical (microwave) properties

Vertical illumination Travelling wave – fully distributed

Lumped p p
i Distributed i
n n

Travelling wave – periodically distributed

p
Edge i
illumination
p p p p
n i i i i
n n n n

Tapered p
waveguide i
n Electrode configuration
ECE 455 Lecture 10 8

Classification according to optical collection/propagation

Vertical illumination

Edge illumination
ECE 455 Lecture 10 9

BASIC PRINCIPLES OF
PHOTODETECTION – PIN STRUCTURES
ECE 455 Lecture 10 10

Vertically-illuminated PIN photodiode

Bias voltage usually

needed to fully deplete
the intrinsic “i” region
Device Layer Structure for high speed
operation

Band diagram Incident photons must

showing carrier have sufficient energy
movement in E-field to meet the band gap
requirement. Leads to
concept of cut-off
wavelength.

Carriers absorbed here must

Light intensity as a
diffuse to the intrinsic layer
function of distance below
before they recombine if they
the surface
are to contribute to the
photocurrent. Slow diffusion
can lead to slow “tails” in the
temporal response.
ECE 455 Lecture 10 11

Typical current-voltage characteristics

Id - Dark current
Unwanted component
generated under no light
Leads to noise

Operate under reverse bias

Increased incident power leads to increased
photocurrent
ECE 455 Lecture 10 12

Optical radiation is absorbed in the

semiconductor material according
to:

Px   PO 1   e  x

Incident Front-facet reflectivity

optical Reduced with anti-
power reflection coating

Absorption coefficient vs.

wavelength for selected
semiconductors
ECE 455 Lecture 10 13

• The upper wavelength cut-off λc is determined by the band-gap energy Eg (eV):

hc 1.24
C (μm)  
Eg Eg (eV)

• Longer wavelengths do not have high enough photon energies to excite

electrons from the valence to the conduction band.

• If the depletion width is w, the total power absorbed is:


PA  PO 1    1  e   w 
• In some devices this can be increased through back reflection from the bottom
metal contact:

PA  PO 1    1   B e  w
1  e
  w

ECE 455 Lecture 10 14

Basic figures of merit

• Internal quantum efficiency
numberof e  h pairs collected
i   1  e   w
no.of photons enteringdevice
• External quantum efficiency

number of e  h pairs collected

e 
no.of photons incidenton device
 1    1  e  w
 IP q
PO hf

• Responsivity
photocurrent IP
R  Units of A/W
incidentopticalpower PO

q
R 
hc
ECE 455 Lecture 10 15
Responsivity vs. wavelength

• Increases with wavelength until cut-off

• Maximum value of quantum efficiency is unity, which places an upper limit
on responsivity
ECE 455 Lecture 10 16

PIN bandwidth and quantum efficiency versus absorption layer thickness

for different area diameters (10, 20, 40 & 100 μm).
ECE 455 Lecture 10 17

Trade-off between
bandwidth and
quantum efficiency

Bandwidth is limited by both transit time and RC product, which are both dependent
on thickness, as is quantum efficiency.
1
1 
 1 

  2d  2  2AR  R  2  2

  
2
1
f 3dB   2  2   S L

 fT f RC    3.5veh   d  
 

ECE 455 Lecture 10 18

Vertically illuminated photodiode

photon
-ve bias
• Photons enter through top layer of device
p • Absorption throughout device
• Only depletion region absorption useful
absorption region

• Long depletion region:

i
– High absorption (efficient)
– Transit time limited
n

+ve bias
ECE 455 Lecture 10 19

Edge coupled photodiode

• Light enters from edge

-ve bias • Coupled into absorption region optical
waveguide
p

absorpti on regi on
• Can be long but narrow
photon i – Short transit times/good absorption
– High device capacitance, 3dB
n bandwidth given by:
+ve bias f CR
f 3dB 
2
 f CR 
1   
 ft 
ECE 455 Lecture 10 20

PHOTODIODE & PHOTODETECTOR

NOISE
ECE 455 Lecture 10 21

Photodetector Noise

• Photodiodes must detect very weak optical signals. Must

maintain an adequate signal to noise ratio:

• SNR = signal power from photocurrent

photodiode noise power + amp noise power

Photodiode noise is caused by statistical nature of photon-

electron conversion process, while the amplifier noise is due
to thermal noise
ECE 455 Lecture 10 22

• To achieve high SNR:

(1) The photodiode must have a high quantum efficiency to

generate a large signal power
(2) Photodiode and amplifier noises must be minimised.

• Photodiode efficiencies are normally high, hence it is the noise

currents that determine the minimum optical power level that
can be detected.

• Minimum detectable optical power = optical power needed to

produce a photocurrent of same magnitude as rms of the total
noise current. This is equivalent to having an SNR of unity.
ECE 455 Lecture 10 23

Sources of noise in a photoreceiver:

bias voltage

photodiode
dark current noise
input (photon stream)
multiplication noise (only for APDs)

quantum noise output

AMP

RL

amplifier noise
thermal noise

APD = avalanche photodiode

ECE 455 Lecture 10 24

Impact on digital reception

photodiode
A

AMP B vout B
Comparator
C
RL
VREF C

A: ideal signal (no noise)

B: output of photoreceiver (noise due to amplifier and photodiode)
C: output of comparator; noise can generate bit errors
ECE 455 Lecture 10 25

SNR performance of PIN photoreceiver

• Let detected signal (photocurrent) be IP :

IP = Im + ip

mean value (“DC”) signal component

• signal power is proportional to: ip2 (normalise to RL)

N.B. this is a mean square current, units A2

ECE 455 Lecture 10 26

• quantum noise: due to random arrival of photons hence

detected current = mean value + random fluctuations

Can be modelled as a current source with mean square given by:

iQ2 = 2qBIm

B = bandwidth
q = electron charge

N.B. Also known as shot noise.

ECE 455 Lecture 10 27

• dark current noise: extra shot noise component due to dark

current (i.e. current that is present in the absence of optical
illumination)

iD2 = 2qBID

ID = mean value of dark current

• thermal noise: due to bias resistor RL

iT2 = 4kTB / RL

k = Boltzmann’s constant = 1.38110-23 J/K

T = absolute temperature in K
ECE 455 Lecture 10 28

• amplifier noise: introduced by amplifier circuitry.

We can combine iT and iAMP as follows:

iT2 = 4kTBFn / RL

Fn = amplifier noise figure

• Hence: SNR = average signal power

average noise power

= ip2
2qB(Im + ID) + 4kTBFn / RL
ECE 455 Lecture 10 29

• SNR is maximised by:

– low receiver noise figure
– low bandwidth
– large load/bias resistance, although this
tends to increase receiver time constant,
reducing the bandwidth
ECE 455 Lecture 10 30

Typical SNR
plots for APDs
and PINs

APD = avalanche
photodiode
ECE 455 Lecture 10 31

AVALANCHE PHOTODETECTION
http://impact-ionisation.group.shef.ac.uk/tools/
ECE 455 Lecture 10 32

APD structure and electric field variation

(provides current gain)

© Keiser, McGraw Hill

photons absorbed here to give
primary photocurrent

Optical input
ECE 455 Lecture 10 33

Device is operated under reverse bias; relatively high voltages (20 V or more) needed to achieve the high
electric field in the avalanche region.

Most photons are absorbed in the depletion region, where they generate electron-hole pairs in much
the same way as in a pin photodiode. The resulting photocurrent is known as the primary photocurrent.

In the high field region, photo-generated carriers are accelerated and gain enough energy to ionise
covalent electrons in the valance band if they collide, thus releasing more e-h pairs. This process of
carrier multiplication is termed impact ionisation. Newly created carriers are also accelerated by the high
electric field, gaining enough energy to cause further impact ionisation. This phenomenon leads to the
avalanche effect. In most devices, impact ionisation is confined to electrons alone.

The multiplication factor M for all carriers generated in the photodiode is:

• IM is the average value of the total

I
M M multiplied current and IP is the primary (i.e.
IP un-multiplied) current
q
• Responsivity of an APD is: R M  R0 M
hf

where R0 is the unity gain responsivity.

ECE 455 Lecture 10 34

Application of sufficient reverse bias leads to avalanche

multiplication, i.e. internal gain.

However, the dark current also increases.

In addition, the avalanche effect is random in nature, and this

introduces a new source of noise:

Variable gain m: m  M  mn

Random “arrival” of carriers tMean gain Random variable,

m5 zero mean
m3
m1 m2 m4
These are multiplied through
avalanche process, which is
noisy (random nature);
i.e. there is multiplication noise

Photocurrent

Quantum and shot noise

are increased by excess noise
of APD (multiplication noise)
t
ECE 455 Lecture 10 35

• It has been found experimentally that: m 2  M 2 x

where 0 < x < 1, and this value depends on the material,
e.g. for silicon, 0.1 < x < 0.5, for germanium, 0.85 < x < 1.0

Excess Noise Factor

Definition:
The ratio of the actual noise generated to the noise generated if all
carrier pairs were multiplied by M.

m 2 M 2 x
Fe  2   M x

M M2

Note: M m m2  m  2
ECE 455 Lecture 10 36

APD Signal-to-Noise Ratio (SNR)

Both a PIN and APD will have contributions from shot noise, dark current noise and
thermal noise, while an APD will also exhibit excess noise. However, for high values of
multiplication M, an APD will achieve the shot noise limit (or quantum limit).

M 2 i p2
SNRAPD 
2qBI m  I D  M 2 x 
4kTBFn
Improvement in SNR (dB)

RL
(SNR)APD - (SNR)pin

i p2

2qBI m  I D  M x 
4kTBFn
M 2 RL

i p2
SNRPIN 
Avalanche multiplication factor M
2qBI m  I D  
4kTBFn
RL
ECE 455 Lecture 11 1

Optical Sources and Modulation of Light

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 455
• Lecture 10
• Fall Semester 2016
ECE 455 Lecture 11 2

ELECTRO-OPTIC CONVERSION
ECE 455 Lecture 11 3

• In electrical-to-optical (E/O) conversion, our aim is to convert an electronic

waveform (current or voltage) to corresponding variations of optical power.

In addition to having sufficient optical output power,

adequate conversion efficiency and good impedance matching, we
require that the output is an accurate copy of the input
– i.e. nonlinearity and noise can be bad news.

• If we choose to use coherent detection, then we might want to modulate the

optical frequency (or phase) instead of the power.

• In thise case we must use external modulators.

ECE 455 Lecture 11 4

Simplistic model of E/O & O/E conversion

Electrical-to- Optical fibre Optical

Optical -to-Electrical
Modulation ωOPT ωOPT Demodulation

ωRF ωRF

ωOPT
ωRF ωRF

ωOPT ωOPT

E/O “upconversion” O/E “downconversion”

ECE 455 Lecture 11 5

Optical Optical RF
Photodiode
input coupler out
Source and detector RF
Optical out
options input CW
+ Square-law
Detection
Laser Photo-
& LPF
(LO) diode

Direct detection Coherent detection

Direct intensity modulation
Intensity
RF input / Direct detection (IM/DD)
modulated
optical
Directly • Simple technique, cheap
signal
Modulated • Problems can include:
(“AM”)
Laser Diode • Chirp
• Nonlinearity
Direct modulation

Modulated External Intensity Coherent detection of:

optical modulation / Direct • Amplitude
External Signal detection (IM/DD) • Phase
modulator • or Frequency
Intensity, • No chirp problems
Phase, • Larger bandwidth Offers better sensitivity,
CW or Frequency compared to direct but increased receiver
Laser RF input
modulation complexity compared to
External modulation • Relatively expensive direct detection
ECE 455 Lecture 11 6

If we apply a sinusoidal microwave signal of frequency m to an E/O component, the

resulting optical field will contain a central optical frequency 0 and multiple sidebands. The
exact form of the spectrum depends on the E/O device and bias conditions. For a directly
modulated laser diode, we have:

Drive current: I L  I B 1  m cos(mt  m ) IB = bias current m = modulation index

I
Emitted electric field at a fixed point in space:
Forward
biased E (t )  E0 1  m cosmt   m  exp  jot  o 
laser
diode
Chirp is neglected (0 is fixed)

For small-signal modulation m <<1 so we can use Bessel functions to expand the electric field
expression. We can show that this contains multiple frequency components of the formo  nm

However, the optical intensity will be given by the square of the electric field magnitude:

E (t )  E0 1  m cosmt   m  
2 2

and so the optical output power is of the form: P0 1  m cosmt   m 

where P0 is the average power, m is the microwave modulation frequency and m the phase.
ECE 455 Lecture 11 7

Impact of microwave modulation on the optical spectrum of a laser

diode as the modulation frequency is varied
ECE 455 Lecture 11 8

If we consider the light-current characteristic of a laser diode, the above result makes
sense. The L-I characteristic is a plot of optical output power versus drive current:
PL (mW) L-I characteristic
IL
saturation
Drive Optical power
current PL sL (W/A)

IL (mA)

Threshold current

The L-I characteristic is similar to the I-V characteristic of a diode. Above threshold and
below saturation, the L-I characteristic can be approximated very well by a straight line
segment with a slope given by:
PL
sL  Slope efficiency in W/A
I L
Ideally we the slope efficiency to be as high as possible but it is fundamentally limited
by the quantum efficiency of the laser.
ECE 455 Lecture 11 9

PL (mW)
Hence if we ensure that
the drive current does not
PL  P0 1  m cos mt   P0  p(t )
go below threshold or into PL (t )  sL I L (t )
saturation, the optical
power will follow the drive P0
current. The “DC”
components are related
via: P0  sL I B IL (mA)

This is also the average I L  I B 1  m cos mt   I B  i(t )

optical power
IB

Although it is not obvious from the L-I curve, the slope efficiency is actually frequency-
dependent. At a given frequency, the sinusoidal components of the current and
optical power can be described using phasors, and they are related via:
pL ( jm ) This is referred to as the intensity
sL ( jm ) 
iL ( jm ) modulation response
where iL(jm) is the modulation current phasor and pL(jm) is the corresponding
output optical power phasor.
ECE 455 Lecture 11 10

The intensity modulation response of a directly modulated laser diode is a low-pass second-
order response which places a limit on the bandwidth they can support:

p L ( j m ) Resonance peak: The laser is modulated at

i L ( j m ) frequencies below this point

m

We can model our E/O component as a linear two-port with a transfer function:

i( jm ) E/O
sL ( jm ) p( jm )

i (t ) p(t )
Modulation current
t Optical power has same
frequency but with an
amplitude and phase change t
ECE 455 Lecture 11 11

Although we have used a directly modulated laser diode to look at E/O

conversion, a similar approach can be used with a modulator.

In this case, the device is driven with voltage instead of current, and the light-
voltage characteristic has a sinusoidal shape as opposed to a diode-like curve.

Bias point and

modulation depth
Optical chosen to give
External modulation power incrementally linear
External slope P0  p(t )
modulator Modulated This will
depend on the
light CW laser
CW output output power
RF input as well as drive
Laser V V
+ Bias conditions

VB  v(t )
ECE 455 Lecture 11 12

OPTICAL SOURCES FOR

COMMUNICATIONS
ECE 455 Lecture 11 13

• Optoelectronic sources convert electrical energy to light energy (i.e. they

“convert” electrons to photons).

• There are two main sources for optical communications:

– monochromatic incoherent sources (LEDs: light emitting diodes)
– monochromatic coherent sources (laser diodes)

– Both LEDs and laser diodes are semiconductor optoelectronic

devices that can be modulated at high-speeds (laser diodes much
more so than LEDs).

– For high-speed long distance links, laser diodes are used. These can
be modulated directly or externally.
– Direct modulation is achieved by varying the drive current, external
through varying the optical power with an external device (a
modulator).
ECE 455 Lecture 11 14

• Optical fibre sources should have the following properties:

(a) compatibility for launching light into fibre

(b) linearity
(c) emit light at wavelengths where fibre is low loss and has low
dispersion
(d) wide modulation bandwidth (i.e. small rise time)
(e) deliver sufficient optical power to overcome losses
(f) narrow spectral linewidth to minimise chromatic dispersion
(g) maintain stable optical output against environmental changes and
ageing
(h) be reliable, low cost and compatible with drive electronics
ECE 455 Lecture 11 15

Surface-emitting LED

Laser Diode
(Fabry-Perot)
resonator cavity
ECE 455 Lecture 11 16

(a) Compatibility for launching light into fibre

• LED gives very poor

coupling into single-
mode fibre, but is OK
for multimode

• Laser diode power is

more efficiently coupled
into single-mode fibre
(directional beam)
ECE 455 Lecture 11 17

(b) Linearity

Laser diode
LED
ECE 455 Lecture 11 18

(c) Operating
wavelengths:
ECE 455 Lecture 11 19

(d) Bandwidth:

• LEDs: 3 dB bandwidth up to a few hundreds of MHz can be achieved

• Laser diodes: up to tens of GHz (approx. 30 GHz is max.)

• Laser diodes can also be externally modulated, up to at least 100 GHz

(depending on the external modulator material)

• However, in addition to the frequency response as described in slide 14,

for digital applications the time-domain characteristics are important, and
in particular the rise time:
Drive current

Optical power
ECE 455 Lecture 11 20

Gaussian profile
(f) Spectral width:

Typical for a long

wavelength LED:

Typical for a Fabry-Perot laser

diode (gives multimode output). Spectral width, at
Mode spacing is determined by FWHM (full width-half
cavity length (mirror-mirror maximum)
spacing)
ECE 455 Lecture 11 21

In a distributed feedback (DFB) laser diode, feedback is provided by a grating, which

selects a single mode whilst suppressing all the others

Relative optical
DFB lasers give a single-mode
power spectrum. (Only one “line”)

Spectral linewidth
(In the case of a DFB, this
is the same as the spectral
width)


ECE 455 Lecture 11 22

(g) Temperature dependence: in laser diodes, the threshold current has a distinct
temperature dependence; this means that temperature control circuits are
required, which adds to the cost of laser transmitters.
ECE 455 Lecture 11 23

LEDs:

• Good points: cheap, easy to drive (no thermal or optical power stabilisation
needed)

• Bad points: low bandwidth, large spectral width, high source-to-fibre coupling
loss for single mode fibres

• Conclusions: best used with multimode fibres in LAN-type applications for low
bit rates
ECE 455 Lecture 11 24

Laser diodes:

• Good points: large bandwidths, narrow spectral linewidth, can couple several
mW into single mode fibre

• Bad points: relatively expensive, most need power and temperature stabilisation
circuits, source-to-fibre coupling can be difficult.

• However, VCSELs (vertical cavity surface emitting lasers are cheap, and
are used in many multimode fibre links, in some cases up to several Gb/s)

• Conclusions: best used with single-mode fibres in high-speed (often 10 Gb/s plus)
long distance applications, and VCSELs have now become popular for multimode
fibres (e.g. active optical cables for data centres).
ECE 455 Lecture 11 25
Basic laser structures: Summary

FABRY-PEROT DIODE LASER DISTRIBUTED-FEEDBACK VERTICAL-CAVITY

DIODE LASER (DFB) SURFACE-EMITTING
LASER (VCSEL)

•Simple structure •Low laser noise •Testable at wafer

•High power available •High linearity level
•Major application: •Major application: •Circular, low-divergence
CD players CATV distribution beam
•Cost: $10 - 500 •Cost: $500 – 5,000 •Major application:
LAN links
•Cost: $10 -500

OPTICAL SPECTRUM OPTICAL SPECTRUM OPTICAL SPECTRUM

ECE 455 Lecture 11 26
ECE 455 Lecture 11 27
ECE 455 Lecture 11 28

EXTERNAL MODULATION
ECE 455 Lecture 11 29

External modulation

In addition to direct modulation of a External

laser we can also modulate the optical modulator
Modulated
power with the following arrangement, light
known as external modulation: output
CW
Laser Bias
+ modulation

Laser emits constant optical power. This then passes through an optical modulator
(external modulator) – this is a voltage driven device. As we adjust the voltage, the
amount of optical power absorbed will vary. In this way, we achieve modulation of the
optical power coming out of the modulator:
Optical
power
P0  p(t )
This will
depend on
the CW laser
output power
as well as
drive V V
conditions
VB  v(t )
ECE 455 Lecture 11 30

O
Microwave signals
reside as sidebands
on an optical carrier
RF input Modulated RF output
Photoreceiver
optical source
Single-mode [O/E demodulation]
[E/O modulation]
optical fibre
m
m
Source options: Receiver options:

Direct modulation (intensity modulation) Direct detection

RF input Intensity RF
modulated Modulated output
Directly optical optical
Modulated signal input Photo
Laser Diode (“AM”) -diode

OR OR
External modulation Modulated Coherent detection
optical
Modulated Optical RF
External Signal
optical coupler output
modulator
Intensity, input
Phase, + Square-law
or Frequency CW Detection
CW Laser Photo-
RF input Modulation & LPF
Laser (LO) diode
ECE 455 Lecture 11 31

Light from a laser can be described by its electric field. To keep things simple we consider
a purely monochromatic laser (i.e. a “perfect” laser), for which the emitted field at some
fixed distance from the laser is given by:

E (t )  Eo (t )e j (o (t )t o (t ))
Optical phase

Amplitude (complex quantity)

Optical frequency (i.e. 100’s of THz)

In analogy with electronic communications, we are able to modulate amplitude,

frequency or phase.

Amplitude modulation in optical communications is known as intensity modulation, and

this is the most common approach. It can be achieved either through direct or external
modulation.

Frequency and phase modulation can only be achieved with an external modulator, and
can only be detected with a coherent photoreceiver.
ECE 455 Lecture 11 32

The optical intensity is directly proportional to the square of the electric field magnitude.
The optical power emitted by the laser is, in turn, directly proportional to the intensity.
So we can write:

optical power  E (t )  Eo (t )
2 2

So the optical power varies only with variations in the amplitude of the electric field, and
this is achieved either through direct modulation or an external modulator.

We will now consider the operation of an external modulator based on the principle of
an interferometer:

Electrical input (modulation)

Unmodulated
light from laser

Modulator
ECE 455 Lecture 11 33

External modulators that are based on the interferometer principle are known as Mach-
Zehnder modulators (MZM). To understand the basic principle, we need to remember
something about superposition (and constructive and destructive interference).

Consider some examples:

Constructive Destructive "Quadrature phase" ±90°

interference: interference: interference:

1.0 1.0 1.0

+ + +
0.2 -0.2 -0.2i
= = =
1.2 0.8 1-0.2i

time time time

ECE 455 Lecture 11 34

Now consider the optical waveguide structure of a MZM:

The two waveguide arms have equal

length, so the delay and hence phase
shift is equal for both paths.
Input light

Output light
Y-junction. The incoming light is
split equally into two paths at
this point. So the light on each Second Y-junction. Here light from the two arms
of these paths for an ideal is combined in phase. However, the optical power
device will be 3 dB less in of the output will be lower than that of the input
optical power compared to the due to losses in the waveguides and at the Y-
input light. junctions. We refer to this as the insertion loss of
the MZM
ECE 455 Lecture 11 35

The waveguides are formed from titanium on a layer of lithium niobate, which forms the
substrate. Lithium niobate is a material that has a strong electro-optic effect – if we apply a
voltage to it, then its refractive index changes. We can show that this is equivalent to
introducing a phase shift.

Ti-diffused optical waveguide

CW light Electrodes

Lithium
niobate
substrate Modulated light

In the MZM shown above, a voltage applied to the electrodes will introduce a phase shift into
the upper arm.

For zero volts there is no phase shift and we have constructive interference, but if we increase
the voltage to some value (called V) then there is a  radians relative phase shift leading to
total extinction. Values in between will lead to varying levels of absorption.
ECE 455 Lecture 11 36

The light-voltage (L-V) characteristic for a MZM:

Po Pi
Po  T ff Pi
1 T ff  1
Reduction due to
optical insertion loss
of modulator
T ff

0 Vm V
0 1 2 3 4

Vm  V
ECE 455 Lecture 11 37

The transfer characteristic is given by:

Po T ff   Vm 
 1  cos 
Pi 2   V 

We can use this to find bias points at which the slope of the L-V characteristic is
maximized:

dPo T ff Pi    Vm 
  sin    0
dVm 2  V  V 

This gives:
Vm 1 3 5
 , , ,.....
V 2 2 2
ECE 455 Lecture 11 38

So if we use an appropriate bias point (say 3V/2), and then apply modulation, we have
the following:
Bias point and modulation depth chosen to give
Po Pi incrementally linear slope. We can show that at
this point,
dPo T ff 
 Pi
1 dvm 2V
Reduction due to So by increasing the optical power from the CW laser,
optical insertion loss we can increase the efficiency of the modulator.
of modulator
T ff
Po  p(t )
Pi

0 Vm V
0 1 2 3 4
VB  vm (t )
V
ECE 455 Lecture 12 1

Introduction to Optical Link Design

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 445
• Lecture 12
• Fall Semester 2016
ECE 455 Lecture 12 2

OPTICAL FIBER SYSTEMS – BASIC

CONSIDERATIONS
ECE 455 Lecture 12 3

Optical fibre communication systems vary according to application, and can be categorised
in many ways, e.g.

• Topology – in local area networks, for example, rings and stars are possible, while
transatlantic links will be point-to-point.

• Length – from a few hundred metres in data centres through to thousands of km

• Modulation – analogue or digital

• If digital, this can be binary or multilevel pulse amplitude modulation
• Modulation can be direct (applies to LEDs and lasers), or it can be external
modulation of a laser
• Direct modulation simply leads to modulation of optical power, and is called
intensity modulation. Detection is then called direct detection.
• External modulation can be used to modulate the optical phase, which then
allows the use of coherent detection.

• Wavelength – can be a single wavelength system, or a wavelength division multiplexed

(WDM) system (CWDM – coarse WDM, DWDM – dense WDM).
ECE 455 Lecture 12 4

We will keep things simple for the moment, by considering links that are:

• Point-to-point
• Use direct intensity modulation and direct detection (IM/DD)
• Single wavelength
ECE 455 Lecture 12 5

Information Optical Optical Information

source transmitter receiver recipient

Optical
fibre

POINT-TO-POINT DIGITAL LINK

DESIGN
ECE 455 Lecture 12 6

• In digital communications, three key specifications are:

1. The length of the link (in km)

2. The bit rate (in Mb/s or Gb/s)
3.The bit error rate (BER).

• In addition, such considerations as component cost (per

subscriber), environmental conditions and reliability also
have to be taken care of. In satisfying the link
specifications, a designer has a number of decisions to
take, because ....
ECE 455 Lecture 12 7

Bandwidth (bit rate) and repeater spacing are

determined by:

Transmitter Transmission Receiver

(e.g. laser) Medium (photodiode)
(Fibre)

 
Power Attenuation Sensitivity

Modulation   Modulation
bandwidth Dispersion bandwidth
ECE 455 Lecture 12 8

The bit rate-transmission length grid

1-10 m 10-100 m 100-1000 m 1-3 km 3-10 km 10-50 km 50-100 km >100 km
<10 Kb/s
10-100 Kb/s VII
100-1000 Kb/s V
1-10 Mb/s
I V
10-50 Mb/s
50-500 Mb/s
II
500-1000 Mb/s VI
>1 Gb/s
III IV

I Region: BL  100 Mb/s SLED with SI MMF

II Region: 100 Mb/s  BL  5 Gb/s LED or LD with SI or GI MMF
III Region: BL  100 Mb/s ELED or LD with SI MMF
IV Region: 5 Mb/s  BL  4 Gb/s ELED or LD with GI MMF
V Region: 10 Mb/s  BL  1 Gb/s LD with GI MMF
VI Region: 100 Mb/s  BL  100 Gb/s LD with SMF
VII Region: 5 Mb/s  BL  100 Mb/s LD with SI or GI MMF
SI: step index, GI: graded index, MMF: multimode fiber, SMF: single mode fiber
ECE 455 Lecture 12 9

A. Choice of operating wavelength

• Short haul links (e.g. LANs) :- use short wavelengths (e.g.

0.85 mm). Moderate fibre losses can be tolerated and the
technology is cheap. By using multimode fibre, connectors
are more rugged than for single mode.

• Long haul links (e.g. transatlantic) :- use long wavelengths

where attenuation and dispersion are low. (e.g. 1.3 μm -
gives dispersion minimum, or 1.55 μm - has attenuation
minimum and is compatible with optical amplifiers;
dispersion-shifted fibre also available).
ECE 455 Lecture 12 10

Loss of modern fibres

ECE 455 Lecture 12 11

20

10
Dispersion
(ps/(nm.km))0

-10

-20

Dispersion for a standard silica single-mode fibre

ECE 455 Lecture 12 12

B. Choice of source

• Power :- laser couples more power into single mode fibre

than LED, but high-bit rate versions can be expensive and
require temperature and optical power control. This makes
them unsuitable for short links, unless VCSELs are considered
(vertical cavity surface emitting laser).

• Spectral width :- at short wavelengths (high material

dispersion) LEDs with large spectral widths might cause
problems with inter-symbol interference. At 1.3 μm, we have
very low dispersion fibre, which combined with low spectral
width lasers allows high bit rates (e.g. 10 Gb/s and above),
while dispersion management is possible at 1.55 μm
ECE 455 Lecture 12 13

Comparison of spectral widths

Single-Mode
Laser Diode
Relative Power Density

Fabry-Perot
Laser Diode
< 1 pm

3 to 6 nm
LED

50 to 100 nm

Wavelength
ECE 455 Lecture 12 14

Source Bandwidth:

• LEDs: 3 dB bandwidth of a few hundreds of MHz

available from commercial devices.

• Laser diodes: up to a few tens of GHz

• External modulation (e.g. Mach-Zehnder

modulator plus laser) to more than 100 GHz has
been demonstrated.

Typical LD frequency
response (second-order)
ECE 455 Lecture 12 15

C. Choice of fibre
• Multimode :-
– modal dispersion limited
– can be used with LEDs and laser diodes (esp. VCSELs)
– graded index multimode fibre can achieve reasonable
reduction in modal dispersion.

• Single-mode :
– no modal dispersion problems
– only used with laser diodes (high tolerance coupling)
– can support > 1 Tb/s (using WDM)
– small core diameter (8μm) leads to high tolerance (high
price) connectors.
ECE 455 Lecture 12 16

Example:
ECE 455 Lecture 12 17

D. Choice of photodetctor

• PIN :-
– simpler construction than APD
– relatively low sensitivity
– available for short and long wavelengths
– higher bandwidths achievable compared to APDs (up to
100 GHz)

• APD :-
– better receiver sensitivity
– temperature sensitive
– high bias voltages
ECE 455 Lecture 12 18

Typical receiver sensitivities vs bit

rate:
ECE 455 Lecture 12 19

LINK POWER BUDGET

ECE 455 Lecture 12 20

Link Power Budget

• Put simply, the link power budget is an "accounting"

procedure in which one calculates how much power can
be lost between the transmitter and the receiver for a
given receiver sensitivity (which depends on the bit rate)
and transmitter power output. The resulting budget is
allocated to connector losses, splice losses, fibre losses
and a safety margin (system margin).

• dB and dBm units are used in the link power budget.

ECE 455 Lecture 12 21

dB and dBm

• Advantage of this approach is that it replaces multiplication

& division with addition/subtraction in calculation of link
gain/link loss.

Pin (dBm) Pout (dBm) = Pin (dBm) + G (dB)

G (dB)

L (dB)
Pin (dBm) Pout (dBm) = Pin (dBm) - L (dB)
ECE 455 Lecture 12 22

• power link budget

LASER FIBRE PHOTODIODE

PS (dBm) L (dB) PR (dBm)

Lmax = PS - PR
ECE 455 Lecture 12 23

OPTICAL PHOTODIODE
LASER AMPLIFIER
FIBRE

Fibre loss
 dB/km

Loss due to fibre

connector
Fibre splice (permanent connection)
Fibre pigtail; v. short, Introduces splice loss
negligible loss

Laser-to-fibre coupling loss;

can be minimised using lenses
ECE 455 Lecture 12 24

OPTICAL
LASER PHOTODIODE
AMPLIFIER
FIBRE
Power level (dBm)

Total link
loss

Ma

Distance along link (km)

ECE 455 Lecture 12 25

• In practical applications, we often use components that have

connectors attached. Fibre with one connector is known as a
fibre pigtail. A length of fibre with connectors on both ends is
called a patchcord.

• In many link budgets, the splice loss is often combined

together with the fibre loss.

• We also include a safety factor known as the system margin

(Ma) to account for component degradation. A typical value
for Ma is 6 dB.
ECE 455 Lecture 12 26

Example
• Calculate maximum link length for a system with:
– a connectorised laser transmitter (PS = 3 dBm)
– a connectorised receiver with sensitivity PR = -40 dBm
– a fibre patchcord (F = 0.5 dB/km, including splice losses)
– connector losses of C = 1 dB and system margin of 6 dB

C C
Fibre Receiver
Laser (PS) (PR)
F L

Total link loss (dB) = PS - PR = F L + 2 C + Ma

ECE 455 Lecture 12 27

F L = PS - PR - 2 C - Ma = 35 dB

Hence Lmax = 35 / 0.5 = 70 km

C F L C Receiver
Laser (PS) (PR)
PS = 3
2
C
Power
level (dBm) F Lmax
N.B. -33
Not to
C
-34
scale!
Ma
PR = -40 Distance
(km)
0 70
ECE 455 Lecture 12 28

LINK RISE-TIME BUDGET

ECE 455 Lecture 12 29

Digital link design: Rise time budget

• In the previous section, we saw how the maximum link

distance is affected by the fibre attenuation and also the
source power and the photoreceiver sensitivity for a given
bit rate; this gave us the link power budget.

Information Optical Optical Information

source transmitter receiver recipient
Optical
fibre
ECE 455 Lecture 12 30

However, recall that bit rate and repeater spacing are also
determined by rise-time considerations:

Transmitter Transmission Receiver

(e.g. laser) Medium (photodiode)
(Fibre)

 
Power Attenuation Sensitivity

Modulation   Modulation
bandwidth Dispersion bandwidth
ECE 455 Lecture 12 31

• rise-time budget

LASER FIBRE PHOTODIODE

tmat tmod
tTX tRX
t sys  t
2
TX t 2
mat t 2
mod t 2
RX
ECE 455 Lecture 12
32

Concept of rise-time
• Any real-life system with an input/output will have a
finite bandwidth.
• For example, consider typical modulation response of
a laser diode:
ECE 455 Lecture 12 33

• Previous diagram relates to sinusoidal (j) response.

• The corresponding step-response shows that it takes a
finite time to reach the steady-state, and that in some
cases there may even be relaxation oscillations:

iin(t) pLD (t)

iin(t)
pLD (t)
0 0 t 0 t
0

Typical step-response of a laser diode, showing turn-on delay and

relaxation oscillations (due to low damping factor)
ECE 455 Lecture 12 34

• Similarly, a photodiode will take a finite time to respond to

step-changes in the incident optical power, as shown
below for the case of a pulse input:

pPD(t) iout(t) iout (t)

pPD(t)

0 0 t 0 t
0

Note: Output current pulse shape depends on the device capacitance and
also the width of the depletion region. The above response is quite poor
due to large junction capacitance.
ECE 455 Lecture 12 35

• Finally, the optical fibre itself will exhibit its own rise time due to the
effects of dispersion.

• In the case of single-mode fibres, this is entirely due to intramodal

dispersion, with the main contribution to this being being material
dispersion.
• In multimode fibres, the dominant effect is intermodal dispersion.
(Although material dispersion also exists, it is negligible in
comparison).
• Although attenuation is important, it does not have an impact on rise-
time. It affects the link power budget instead.
ECE 455 Lecture 12 36

• So, not surprisingly, the optical fibre link as a whole will

have a rise-time (and fall-time) in response to a rectangular
pulse input:

iin(t) iout(t)
pLD (t) FIBRE
pPD(t)

iin(t) iout (t)

0 0 t 0 t
0
ECE 455 Lecture 12 37

Definition of rise-time and fall-time

N.B. y-axis is voltage, current or optical power as appropriate

• Rise-time: time taken to rise from 10% to 90% of the

steady-state value of the pulse.
• Fall-time: time taken to fall from 90% to 10% of the steady-
state value of the pulse.
ECE 455 Lecture 12
38

• Put simply, the rise-time budget is an "accounting"

procedure in which one calculates how much pulse
spreading can be tolerated between the transmitter
and the receiver for a given transmitter rise-time,
photoreceiver rise-time and dispersion due to the fibre
(both modal and chromatic, as appropriate).
ECE 455 Lecture 12 39

Rise-time Budget
• The total rise-time of the fibre-optic link is known as the
system rise time tsys.
• It depends on the rise-times of the individual systems
components, and assuming these are independent of
one another, they affect tsys as follows:

t sys  t
2
TX t 2
mat t 2
mod t 2
RX
ECE 455 Lecture 12 40

• tTX = optical transmitter rise-time

• tRX = optical receiver rise-time
• tmat = material dispersion rise-time
• tmod = modal dispersion rise-time (for multimode
fibre only)

• The usual requirement on tsys is:

tsys < 0.7 

where  is the pulse duration.

ECE 455 Lecture 12 41

• The pulse duration depends on the data format.

• Two main data formats are used: NRZ and RZ

ECE 455 Lecture 12 42

Consider an NRZ stream composed of alternating “0”s and “1”s,

and an RZ stream composed entirely of “1”s:
NRZ

NRZ T= 101010.....

RZ 1/BT

RZ 111111.....

0.7
For NRZ signalling, NRZ = 1/BT , hence: t sys 
BT

For RZ signalling, RZ = 1/2BT , hence: 0.35

t sys 
BT
ECE 455 Lecture 12 43

tmat : material dispersion rise-time

• due to material dispersion

• significant in single-mode fibres

tmat = Dmat  L

• Dmat = material dispersion parameter (ps/nm.km)

•  = spectral width of optical source (in nm or m)

 tmat can usually be neglected if the spectral width is

narrow (e.g. in DFB lasers) or if operation is at 1.3 mm.
ECE 455 Lecture 12 44

tmod : modal dispersion rise-time

• due to (inter)modal dispersion

• dominant in multimode fibres

• In theory tmod is proportional to fibre length. In a real

system, pulse distortion increases less rapidly after a
certain initial length because of mode coupling.
ECE 455 Lecture 12 45

• It can be shown that tmod in ns is given by:

0.44Lq
t mod 
BO

where q is between 0.5 and 1.0 (depends on

amount of mode coupling), L is the fibre length
(km) and BO is the 3dB electrical bandwidth (in
GHz) of 1 km of fibre.
ECE 455 Lecture 12 46

tRX : photoreceiver rise-time

• Assuming a simple low-pass RC characteristic for the

frequency response of a photoreceiver, then we can relate
tRX (in ns) to the 3 dB receiver bandwidth (BRX in units of
GHz) as follows:

0.35
t RX 
BRX

isignal inoise A simple photodiode model;

Cd Rd
high frequency versions also
include parasitics due to the
packaging
ECE 455 Lecture 12 47

tTX : optical transmitter rise-time

• This is a function of both the intrinsic frequency

response (of either the LED or the laser diode) along
with any drive electronics.
• LED and laser diode data sheets usually specify the
device rise time.
ECE 455 Lecture 12

48

Finally....

Although the above equations/analysis may appear to be

straightforward, be VERY careful in using units.

Bandwidths of laser diodes, for example, tend to be in the

GHz range, so rise-times tend to be quoted in ns.

However, it is possible to encounter a mix of units when

performing rise-time calculations.

If in any doubt, convert all quantities to SI units, and

perform calculations in SI units, before converting to ns at the
end.
ECE 455 Lecture 13 1

Bit Error Rate for Photoreceivers: Part 1

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 455
• Lecture 13
• Fall Semester 2016
ECE 455 Lecture 13 2

BIT ERROR RATE IN DIGITAL

PHOTORECEIVERS
ECE 455 Lecture 13 3

• Although SNR is a very useful figure of merit, in a binary

digital system we are concerned with accurately
detecting only two types of signal level: binary zeros and
ones.

• The photoreceiver therefore makes a decision as to

whether the recovered waveform is above (“1”) or below
(“0”) the threshold level. (This decision is usually made
halfway through the bit duration TB).

• When noise is present, it can lead to a wrong decision,

i.e. a bit error.
ECE 455 Lecture 13 4
The effect of noise can be deleterious.....

Analogue
IM waveform

Digital
IM waveform

© Wiley
ECE 455 Lecture 13 5

Probability of Error
• The bit error rate (BER) is obtained by dividing the
number of errors (Ne) occurring over a time interval t
by the number of pulses (ones and zeros) transmitted
during this interval (Nt):

Ne Ne
BER  
N t BT t

• BT is the bit rate (bits/sec) and is equivalent to 1/TB

where TB is the bit duration. We assume that a “1” is
“high” and “0” is low for the duration of TB, but other
line coding schemes are possible.
ECE 455 Lecture 13 6
Common data (line coding) formats

© Keiser,
McGraw-Hill

N.B. We will assume NRZ coding here.

ECE 455 Lecture 13 7

• Given that most of the noise is added at the

photoreceiver itself, it is desirable to maximise the
signal level (at the input to the receiver) so that the
probability of bit errors can be reduced.

• However, this conflicts with the need to maximise

transmission distances.

• The ability to calculate BER for a given SNR is useful in

calculating the receiver sensitivity.

• This requires use of probability theory.

ECE 455 Lecture 13 8

REVIEW OF BASIC PROBABILITY

THEORY
ECE 455 Lecture 13 9

• Suppose we have an experiment that gives rise to an

event A. If the experiment is repeated N times and A
occurs nA times, then the probability of A occurring is
given by:
nA
Pr ( A)  lim
N  N

• It follows that:
0  Pr ( A)  1
• Example:
Experiment: Tossing a coin
Event A: Appearance of tails
Pr(A) = 0.5
ECE 455 Lecture 13 10

• In an experiment giving rise to events A1, A2, .., An,

– The event (A1 OR A2 ..... OR An) can be written as

A1 + A2 ..... + An

– The event (A1 AND A2 ..... AND An) can be written as

A1 . A2 ..... . An
Whole event space E
Pr(E) = 1
• Consider a Venn diagram:
Area is
 Pr(A1) A1
A2
ECE 455 Lecture 13 11

A1 A2 A1 OR A2

A1 + A2 = A1  A2

A1 A2 A1 AND A2

A1 . A2 = A1  A2
ECE 455 Lecture 13 12

Example:
Experiment: Tossing a coin A2
Event A1: Appearance of heads: Pr(A1) = 0.5
Event A2: Appearance of tails: Pr(A2) = 0.5 A1

Pr( A1+ A2) = 1

A2
A1

Pr( A1 . A2) = 0
A2
A1 In this case, A1 and A2 are mutually
exclusive events; their areas do not
intersect one another.
ECE 455 Lecture 13 13

Mutually exclusive events

e.g. event A1 = today is Sunday, event A2 = today is Monday; cannot
have the joint event A1. A2 (= today is Sunday AND today is Monday)

A3
A1
An
A2

By inspection of the Venn diagram, for mutually exclusive events

we have:
Pr( A1+ A2 + ...... + An) = Pr(A1) + Pr(A2) + ...... + Pr(An)

Pr( A1 . A2 . ...... An) = 0

ECE 455 Lecture 13 14

Joint probability

A1 A2 A1 A2

= Pr(A1) = Pr(A2)

A1 A2 A1 A2

= Pr(A1 . A2) = Pr(A1 + A2)

Pr ( A1  A2 )  Pr  A1   Pr  A2   Pr  A1 . A2 
ECE 455 Lecture 13 15
Conditional probability

• A conditional probability is one such as:

Pr(A2/A1) = probability that A2 occurs given that A1 has occurred

• If A1 has occurred, the event space is effectively reduced to the area

shown in the Venn diagram below:

A1 A2

= Pr(A1) = new event space

ECE 455 Lecture 13 16

= Pr(A1) = new event space

A1 A2

= Pr(A1 . A2)
A1 A2

Pr(A2/A1) = probability that A2 occurs given that A1 has occurred

= purple area/gold area

Pr  A1 . A2  Baye’s
Pr ( A2 / A1 ) 
Pr  A1  formula
ECE 455 Lecture 13 17

Random Variables
• A random variable is a function X whose values
depend on events
Example:

Toss a coin;
if a head appears, let X(head) = 1,
if a tail appears, let X(tail) = 0.

In other words, a random variable assigns numerical values (real

numbers) to events (such as the appearance of heads, tails) which
themselves may or may not be numerical in nature.

The above example is a discrete random variable; X takes on

a discrete number (in this case two) of values.
ECE 455 Lecture 13 18

Continuous Random Variables

• A continuous random variable is one where X takes on
a continuous range of (real number) values.

Example:
Experiment: switch on a sinusoidal oscillator and measure its
phase  at an instant in time. The phase can take on any value
in the range -    

If we choose the random variable X() = cos , then every value in

the range -1  X  1 is a possible outcome of this experiment.

Alternatively, we could simply choose X () = , in which case

every value in the range -   X   is a possible outcome of this
experiment.
ECE 455 Lecture 13 19

Cumulative Distribution Function (CDF)

• A continuous random variable, by its very nature,
will have an uncountable number of possible
values, so that the probability of observing a
specific value of X is vanishingly small.
• However, the probability that X is between some
range does exist; if we take the above example,
with a random variable X = , then the probability
Pr(X  0) = 0.5.
• In general, we are interested in the probability
Pr(X  x). This is known as the cumulative
distribution function (CDF) and is denoted by F(x).
ECE 455 Lecture 13 20
Properties of the CDF F(x)

• F(-) = Pr(X  -) = 0

• F() = Pr(X  ) = 1

• 0  Pr(X  x)  1 for all x, i.e.

0  F(x)  1 for all x

• F(x) does not decrease as x increases

• If x1 < x2, then:

Pr(x1 < X  x2) = F(x2) - F(x1)
ECE 455 Lecture 13 21

• This last property follows from the mutual exclusivity

shown below: x1 < X  x2
EVENT C
- 
x1 x2
X  x1 EVENT B

X  x2 EVENT A

• EVENT A = EVENT B + EVENT C

• EVENT B and EVENT C are mutually exclusive
• Hence Pr(A) = Pr(B) + Pr(C)  Pr(C) = Pr(A) - Pr(B)

Pr(x1 < X  x2) = Pr(X  x2) - Pr(X  x1) = F(x2) - F(x1)

ECE 455 Lecture 13 22

Probability Density Function (PDF)

• From basic calculus, we have:
x2

x dx dx  F ( x)x1  F ( x2 )  F ( x1 )
dF ( x) x2

• The derivative dF(x)/dx is given the symbol p(x) and is

known as the probability density function (PDF).

x2

 p( x) dx  F ( x )  F ( x )  P ( x
x1
2 1 r 1  X  x2 )
ECE 455 Lecture 13 23

Properties of the Probability Density Function

• From the properties for F(x), we have:
– F(-) = 0, F() = 1



 p( x) dx  F ()  F ()  1
• Since F(x) never decreases with increasing x, it follows
that its slope dF(x)/dx  0, hence p(x)  0.
ECE 455 Lecture 13 24

Example
• Consider the sinusoidal oscillator mentioned earlier.
The output will be of the form:

v(t )  V0 cos (t   )  V0 cos  (t )

/2
• Hence the phase will vary
with time as:
(t0)
2 t 
 (t )   -
0 rads
T
(t0): value of phase 
measured at time t = t0 - /2
ECE 455 Lecture 13  25

time

-
• Because the phase varies linearly with time, there is an
equal likelihood of measuring the phase to be any
value in the range -    . So, for example, if X() is
the random variable whose value is given by X() = ,
then:
Pr(a < X   a + ) = Pr(b < X   b + )

• a and b are any two arbitrary values in -    

ECE 455 Lecture 13 26

• So p() will have a constant value over -    , and

will be zero elsewhere. Since we must have


 p( ) d  1


it follows that p() will be uniformly distributed as shown,

i.e. this is an example of a uniform probability distribution:

p()

1/2

- 0 
ECE 455 Lecture 13 27

p()
Uniform PDF

1/2

- 0 
F()
Corresponding CDF
1

0 
- 0 
ECE 455 Lecture 14 1

Bit Error Rate for Photoreceivers: Part 2

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 455
• Lecture 13
• Fall Semester 2016
ECE 455 Lecture 14 2

APPLICATION TO DIGITAL OPTICAL

COMMUNICATIONS
ECE 455 Lecture 14 3

Only two types of bit can be sent in a

S0 binary system: “1”s and “0”s. These
events are mutually exclusive, so we
have Pr(S0) + Pr(S1) = 1.
S0 is the event “0” was sent
S1
S1 is the event “1” was sent

Only two types of decision can be made:

D0 the detected signal is above or below the
threshold level, i.e. either a “1” or a “0”
is detected. These events are mutually
exclusive, so we have Pr(D0) + Pr(D1) = 1.
D1 D0 is the event “0” was detected
D1 is the event “1” was detected
ECE 455 Lecture 14 4

Conditional probabilities

S0

D0 .S0
S1
D1 .S0
D0 .S1
D0 D1 .S1
A total of four mutually exclusive
D1 outcomes are possible in a binary
communications system
ECE 455 Lecture 14 5

Conditional probabilities
The shaded regions represent events that
D0 .S0 give a bit error:

D1 .S0 • D1. S0 = a “1” is detected and a “0”

was sent
D0 .S1
• D0 .S1 = a “0” is detected and a “1”
D1 .S1 was sent

• These two events are mutually exclusive, hence:

Pr (bit error )  Pr ( D1. S0 )  Pr ( D0 . S1 )

ECE 455 Lecture 14 6

• From Baye’s formula, we have:

Pr D0 . S1 
Pr ( D0 / S1 ) 
Pr S1 

• Rearranging gives: Pr D0 . S1   Pr (S1 ) Pr ( D0 / S1 )

• Similarly, we have: Pr D1 . S0   Pr (S0 ) Pr ( D1 / S0 )

• Thus the bit error probability can be written as:

Pr (bit error )  Pr (S0 ) Pr ( D1 / S0 )  Pr (S1 ) Pr ( D0 / S1 )

ECE 455 Lecture 14 7

• The previous formula can be used to calculate the bit

error probability provided:
– we know what the probabilities of sending “0”s and “1”s are
(often we have Pr(S0) = Pr(S1) = 0.5)
– and we can obtain the conditional probabilities Pr(D1/S0) and
Pr(D0/S1).

• We can obtain Pr(D1/S0) and Pr(D0/S1) if we know what

the PDFs associated with reception of the bits “0” and
“1” in the presence of noise are.

• These processes can be very accurately approximated by

gaussian random variables; the gaussian PDF is plotted on
the next slide.
ECE 455 Lecture 14 8

Gaussian PDF
 ( x  m)2
1
p ( x)  e 2 2
   x  
2 2

p(x)

m
ECE 455 Lecture 14 9

• The gaussian PDF occurs very widely in many

applications (and for that reason is also called the
Normal distribution).

• One reason for this is the central limit theorem. This theorem
tells us that if we take the sum of a large number of
independent variables X1, X2, .... Xn, and if each of these makes
a small contribution to the sum X = X1 + X2 + .... + Xn, then the
PDF of X will approach a gaussian shape as n  .

• The proof is beyond the scope of this course, but the idea can
be illustrated best by an example, e.g. roll n dice and add their
values. If this event is repeated enough times, you get a
gaussian distribution.
• http://www.mathsisfun.com/data/quincunx.html
ECE 455 Lecture 14 10

Properties of the gaussian PDF



 p(x)  1


Pr ( X  m)  Pr ( X  m)  0.5 by symmetry

mean: X  m

 is the standard deviation: when p(x) is used to describe

the probability of detecting a noise current (or voltage)
then  represents the rms value of the noise current
(or voltage).
ECE 455 Lecture 14 11

Obtaining probabilities from the gaussian PDF

• When calculating the bit error probability later on, we
will have to evaluate probabilities such as:

Pr ( X  x1 )   p( x) dx
x1

• This expression cannot be calculated analytically, we

must use numerical techniques. We define:

1
e
 y2 2
Q( k )  dy
2 k

• This can be obtained numerically and then plotted:

ECE 455 Lecture 14 12

Q(k)
ECE 455 Lecture 14 13

• To calculate:

e   dx
1
Pr ( X  x1 )    ( x  m ) 2 2 2

x1 2 2

• Let: xm
y


1  y2 / 2
Pr ( X  x1 )  
x1  m 2
e dy


 x1  m 
Pr ( X  x1 )  Q  
  
ECE 455 Lecture 14 14


Pr ( X  x1 )   p( x) dx
p(x)
x1

 x1  m 
Pr ( X  x1 )  Q  
  

m x1
ECE 455 Lecture 14 15

Towards BER .....

• In the context of our digital photoreceiver, we can say
that output voltage v(t) generated immediately after
the amplifier stage in response to the transmission of
“0” and “1” will have mean values of Vm0 and Vm1 for
these two pulses. The threshold level (Vth) will be set
between these two values.
• However, noise (due e.g. to thermal and amplifier
contributions) will be superimposed on these mean
values, and the distributions will follow that of a
gaussian PDF. Hence the received voltages for “0”
and “1” have PDFs given by p0(v) and p1(v)
respectively:
ECE 455 Lecture 14 16

detected voltage, v

p1(v)

Vm1

Pr(D1/S0) Vth

Pr(D0/S1) Vm0

p0(v)

Assume 0 = 1 = 
ECE 455 Lecture 14 17

• We saw earlier that the bit error probability is:

Pe  Pr (S0 ) Pr ( D1 / S0 )  Pr (S1 ) Pr ( D0 / S1 )

• If we assume that “ones” and “zeros” are equally

likely to be sent, then Pr(S0) = Pr(S1) = 0.5 and:

Pe  12 Pr ( D1 / S0 )  Pr ( D0 / S1 )

• We will consider a NRZ waveform with Vm0 = 0, and

pick a threshold midway between this and Vm1, i.e.
Vth= Vm1 /2. We refer to this as a unipolar waveform.
ECE 455 Lecture 14 18

v 2
Pr ( D1 / S0 ) 1
p0 (v)  e 2 2

0 Vth Vm1 2 2

p0(v) p1(v)


Pr ( D1 / S0 )  Pr (v  Vth )   p (v) dv
Vth
0
ECE 455 Lecture 14 19

• Using the relationship:

 x1  m 
Pr ( X  x1 )  Q  
  

we have:

Pr ( D1 / S 0 )  Pr (v  Vth )
 Vth 
 Q 
 
ECE 455 Lecture 14 20

 v Vm1  2
Pr ( D0 / S1 ) 1
p1 (v)  e 2 2

0 Vth Vm1
2 2

p0(v) p1(v)

Vth

Pr ( D0 / S1 )  Pr (v  Vth ) 

 p1 ( v ) dv
ECE 455 Lecture 14 21

• By symmetry, we have:
0 Vth Vm1 3Vth

Green area = black area

p1(v)


Pr ( D0 / S1 )   p (v) dv
3Vth
1
ECE 455 Lecture 14 22

• Using the relationship:

 x1  m 
Pr ( X  x1 )  Q  
  
we have:

Pr ( D0 / S1 )  Pr (v  3Vth )
 3Vth  Vm1 
 Q 
  
 Vth 
 Q 
 
ECE 455 Lecture 14 23

• Hence:

Pe  1
2 Pr ( D1 / S0 )  Pr ( D0 / S1 )
 Vth 
 Q 
 
 Vm1 
 Q 
 2 
• Now, remember that  is the rms noise voltage, so:

mean square noise power  2

ECE 455 Lecture 14 24

• Also, if ones and zeros are equally likely,

mean square signal power  1

2
V
2
m0 V 2
m1  1
2 V2
m1

2
• Hence the SNR is: V m1
2 2

• Comparing with the bit error probability,

 Vm1   SNR 
Pe  Q   Q  

 2   2 
ECE 455 Lecture 14 25

From plot of Q function,

for Pe = 10-9, need to
find Q(k) = 10-9, which
gives k = 6.0.
ECE 455 Lecture 14 26

• Hence we have for Pe = 10-9:

 SNR 

Pe  Q    10 9

2 
 
• From the plot of Q(k) versus k, we have k = 6.0,
i.e.:
SNR
 6.0  SNR  72.0
2

• In dB, we have SNR = 10 log10(72.0) = 18.6 dB

ECE 455 Lecture 14 27
BER versus SNR for unipolar NRZ
0
1 10

10-510-5
Bit error probability

10-1010-10

10-1510-15

10-2010-20

10-25 10-25
-10 -5 0 5 10 15 20 25
-10 -5 0 5 10 15 20 25

SNR (dB)
ECE 455 Lecture 14 28

• Note that we have used Q(k) in these

calculations; most textbooks make use of the
complementary error function erfc(x) defined
as: 
2

u2
erfc ( x)  e du
 x
• It is easy to show this is related to Q(k) as
follows:
1  k 
Q(k )  erfc  
2  2
ECE 455 Lecture 14 29

• Also, note that this analysis was carried out for a

unipolar format, i.e. one in which the “0”s
produced a zero mean voltage, while the “1”s
have a positive mean voltage.

• Had a bipolar scheme been chosen (i.e. Vm0 = - A

and Vm1 = A) with a decision threshold set to 0
volts, then:
Pe  Q  SNR 
• This represents a 3 dB improvement in the SNR
required for a given BER!
ECE 455 Lecture 14 30

• Finally....

– This treatment has taken into account the various

sources of noise that we may encounter and treated
their impact as making the distributions for “0” and
“1” gaussian. This is accurate provided thermal and
amplifier noise dominate shot noise.

– In the next lecture, we will see what happens in an

ideal receiver, where these noise sources are
“turned off” and the only source of noise is the
statistical nature of photon detection itself.
ECE 455 Lecture 15 1

The Quantum Limit

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 455
• Lecture 15
• Fall Semester 2016
ECE 455 Lecture 15 2

PHOTON COUNTING
ECE 455 Lecture 15 3

A photon noise simulation, using a sample image as a source and a per-pixel Poisson process to model an otherwise perfect camera
(quantum efficiency = 1, no read-noise, no thermal noise, etc). Going from left to right, the mean number of photons per pixel over the
whole image is (top row) 0.001, 0.01, 0.1 (middle row) 1.0, 10.0, 100.0 (bottom row) 1,000.0, 10,000.0 and 100,000.0. Note the rapid
increase in quality past 10 photons/pixel. (The source image was collected with a camera with a per-pixel well capacity of about
40,000 photons.) Photon noise is the dominant source of noise in the images that are collected by most digital cameras on the market
today. Better cameras can go to lower levels of light -- specialized, expensive, cameras can detect individual photons -- but ultimately
photon shot noise determines the quality of the image.

Eric Bajart
ECE 455 Lecture 15 4

Quantum Limit on Optical Receiver Sensitivity

Part A: Digital Communications
• Imagine that you can observe the arrival of photons at
a detector. The detector counts the number of
electron-hole pairs that are generated in an interval t.
• The following assumptions can be made:
1. The probability of one photon being detected in t is
proportional to t when t is very small.
2. The probability that more than one photon is detected in t
is negligible when t is very small.
3. The number of photons detected in any one interval is
independent of the number of photons detected in any other
separate interval.
ECE 455 Lecture 15
5

• Under these conditions, we can show that the

probability of detecting L photons in a time period T
obeys the Poisson distribution:
L N
N e
PN ( L) 
L!
• N is the expected number (i.e. mean number) of
detected photons (i.e. generated electron-hole pairs) in
the time period T.
ECE 455 Lecture 15 6

• Ideally, optical energy would only be sensed (and e-h

pairs generated) if a “1” is sent. Therefore, the ideal
receiver would then be an electron-hole pair counter,
and would make a decision based on a threshold
current.

• The most efficient situation is when the threshold is

set between no pairs generated (a “0” was sent) and
“at least one pair generated” (a “1” was sent).

• Therefore no errors occur if a “0” was sent (because

no carriers can be generated).
ECE 455 Lecture 15 7

• However, errors may occur for a “1” if the incident

optical power fails to generate any carrier pairs at all
(when N could be expected).

• The probability of this occurring is:

N 0 e N
PN (0)   e N
0!
ECE 455 Lecture 15 8

• Since “0”s are received with no errors (P0(0) = 0), and

“1”s and “0”s are equally likely, the overall error
probability is:
N
Pe  e1
2

• Now:
N hf
Popt 
T
where N = mean no. of photons received in T seconds
when an optical power Popt is incident, and Popt is the
optical power received for a stream of “1”s.
ECE 455 Lecture 15 9

NRZ (non-return zero format):

 for NRZ format, the mean optical power is:

Popt N hf
Pm  
2 2 T
ECE 455 Lecture 15

10

• If we require:
1
2 e  N  Per

where Per is the required error probability, then:

N   ln(2 Per )

ln( 2 Per ) hf
 Pm  
2 T
ECE 455 Lecture 15 11

• Now, the bit rate is BT = 1/T, and f = c/.

ln( 2 Per ) BT hc
 Pm  
2

• This sets the lower limit on the optical power needed to

achieve binary transmission at BT bits per second with
an error rate of Per , i.e. this is the receiver sensitivity PR.
ECE 455 Lecture 15 12

ln( 2 Per ) BT hc
PR  
2

• Note that receiver sensitivity worsens (i.e. becomes larger)

with increasing bit rate. In dBm, we have:

 PR (mW ) 
PR (dBm)  10 log10  1 mW 
 
 PR ( W ) 
 10 log10  1 W   30
 
ECE 455 Lecture 15 13

 ln( 2 Per ) hc 
 PR (dBm)  10 log    30
 2 
 10 log ( BT )

Slope of 10 dBm per decade increase in BT

ECE 455 Lecture 15 14

For example:

PR(dBm)

-100

-110

-120

1 100 1000
BT (Gb/s)
10
ECE 455 Lecture 15 15

Typical receiver sensitivities vs bit rate:

ECE 455 Lecture 15 16

Quantum Limit on Optical Receiver Sensitivity

Part B: Analogue Communications
• Recall SNR expression for a pin photoreceiver:
2
i
SNR 
p

2qBI m  I D 
4kTBFn

RL

• If we only consider the quantum noise, then there will be

no dark current, thermal or amplifier noise. (Note: if we
take an APD photoreceiver, we would set M =1 to minimise
SNR for the shot-noise limited case in any case).
ECE 455 Lecture 15 17

• Assuming that quantum noise is the only type of noise

present, we have:
2
i
SNR 
p

2qBI m
• From the definition for quantum efficiency, we can
substitute ip and Im with the optical signal power p(t) and
the mean optical power Pm:

q
2

Im   q  2
Pm i    p
2
p
hf  hf 
ECE 455 Lecture 15 18

• Hence:
2
 q  2
  p
1  
hf  p 2
SNR  
2qB q P 2 B hf Pm
m
hf
• SNR will be maximised if we have a signal component
whose rms value is Pm. This then gives:


SNR  Pm
2 B hf
ECE 455 Lecture 15 19

• Rearranging gives the required mean optical power to

maintain a given SNR for analogue signals:

2 B hc
Pm  SNR


• Compare this with the mean optical power needed to

maintain a given BER for digital communications:

ln( 2 Per ) BT hc
Pm  
2
ECE 455 Lecture 15 20

• We can therefore compare the fundamental (i.e. quantum)

limit to receiver sensitivity for both digital and analogue
communications:
PR digital  ln( 2 Per ) BT

PR analogue 4 SNR B

• Normally, we choose the ratio BT/B (bit rate to

analogue bandwidth) to be 16. From previous lecture,
we have an SNR of 15.6 dB for a bipolar NRZ scheme and
a Per of 10-9. Substituting, we get:

PR digital (dBm) - PR analogue (dBm) = 3.43 dB

ECE 455 Lecture 16 1

Coherent Optical Communications

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 455
• Lecture 16
• Fall Semester 2016
ECE 455 Lecture 16 2

COHERENT COMMUNICATIONS
ECE 455 Lecture 16

Coherent Optical Communications

• In the optical domain, coherent refers to systems in which
mixing of optical signals occurs.
• Consider mixing in an electronic receiver:

Mixer
m(t) = A cos (mt  m ) . B cos ( LO t )
A cos (m t + m) AB
 cos [(m   LO )t  m ]
2
AB
Local oscillator  cos [(m   LO )t  m ]
2
B cos (LO t)
ECE 455 Lecture 16

• Mixing creates sum (m + LO ) and difference (m - LO )

frequency terms.
• The sum term is then removed by low-pass filtering:

LPF
Mixer AB
m(t) = cos [(m   LO )t  m ]
2
A cos (m t + m)

• Downconverted signal:
at lower frequency than the
Local oscillator original message m(t)
• Still “contains” information on
amplitude A, frequency m
B cos (LO t) and phase m of message.
ECE 455 Lecture 16

• The optical equivalent of a receiver

IP
using mixing is:

Incoming signal
~ ~
ER  ER cos  R t  R 
~ ELO  ER Load
resistor
RL
Optical
combiner Photodiode

ELO  ELO cos  LO t 

~

Phase-locked
local oscillator
laser Coherent optical detector
ECE 455 Lecture 16

• We will consider digital modulation, in which a

baseband digital signal can modulate the:

– amplitude ER [ASK - amplitude shift keying]

– frequency R [FSK]
– phase R [PSK]

ER  ER cos  R t  R 
– of a sinusoidal optical signal:
~
ECE 455 Lecture 16 7

Modulated optical carrier waveforms:

ASK
(OOK)

FSK

PSK
ECE 455 Lecture 16 8

Constellation maps for PSK

ECE 455 Lecture 16

• Now, consider a direct detection (DD) receiver:

IP
Incoming signal (from an intensity
modulated laser):
Load
ER  ER cos  R t  R 
~
resistor
RL
Photodiode

optical phase

optical frequency (of order 100 THz) = c / R

electric field amplitude; optical power is  ER2

In other words, this quantity is modulated by the
laser drive current if the laser is intensity modulated.

Electric field of incident optical signal

ECE 455 Lecture 16

However, the frequency response of a photodiode is limited. Consider

an ASK waveform:

Envelope Optical carrier

The photodiode cannot detect the fast variations of the

optical carrier, it can only respond to the modulation
envelope, i.e. it acts as an envelope detector.
ECE 455 Lecture 16
IP
Incoming signal (from an intensity
modulated laser):
Load
ER  ER cos  R t  R 
~
resistor
RL
Photodiode

The incident optical power is proportional to the square

of the E-field, i.e.:

Pincident  ER  ER cos  R t  R 
~2 2 2

 Pincident  12 ER2 1  cos 2 R t  2R 

ECE 455 Lecture 16

• We saw in earlier lectures that the photocurrent generated

in a photodiode is proportional to the incident optical power, i.e. we
might expect:

I P , DD  12 ER2 1  cos 2 R t  2R 

• However, even state-of-the-art photodiodes have frequency

responses that extend no more than 100 GHz, i.e. the photodiode
cannot detect the term 2R. Hence for intensity modulation/direct
detection schemes (IM/DD),

I P , DD  E1
2
2
R
ECE 455 Lecture 16

Coherent detection theory

• Photodiodes are only sensitive to intensity
fluctuations, i.e. they can only detect modulation of
ER.

• So the key to implementing optical FSK and PSK

detection is to convert the optical frequency and
phase fluctuations into optical intensity fluctuations.

• This is achieved using coherent optical detectors:

ECE 455 Lecture 16

Coherent optical detector

IP
Incoming signal
~ ~
ER  ER cos  R t  R 
~ ELO  ER Load
resistor
Optical
RL
combiner Photodiode

ELO  ELO cos  LO t 

~

Phase-locked
local oscillator
laser
ECE 455 Lecture 16

• Basic theory: the incoming light beam, i.e. the electric field:

ER (t )  ER cos  R t  R 
~

• is added to a beam produced by a stable local oscillator laser:

ELO (t )  ELO cos  LO t 

~

• The “mixing” process, between the information bearing and

local oscillator fields is done before photodetection.
ECE 455 Lecture 16

• As before, the photodiode current is directly proportional to

the square of the incident electric field, which in this case is
given by:

I P ,coh
~
 ~
 ER (t )  ELO (t )
2

 ER cos( R t  R )  ELO cos( LO t )
2

• Expanding:

I P ,coh  E cos ( R t  R )
2
R
2

E 2
LO cos ( LO t )
2

 2 ER ELO cos ( R t  R ) cos ( LO t )

ECE 455 Lecture 16

• Using trigonometric identities for cos x cos y and cos2x, we get:

I P ,coh  E
1
2
2
R1  cos(2 Rt  2R ) 
 12 ELO
2
1  cos(2 LOt ) 
 ER ELO cos [( R   LO t )  R ]
 ER ELO cos [( R   LO t )  R ]
ECE 455 Lecture 16

• Notice that we have components at:

 DC
The photodiode cannot respond to the
 2R
terms in 2R , 2LO and R + LO because
 2LO they are outside the detection bandwidth.
 R + LO
 R - LO = IF = intermediate frequency

• If R  LO then the frequency IF will be in the microwave range

or less, i.e. it can be detected by the photodiode. Hence the actual
photocurrent is:

I P ,coh  E  E
1
2
2
R
1
2
2
LO  ER ELO cos [( IF t )  R ]
ECE 455 Lecture 16
• Since optical power varies with the square of electric
field,


I P ,coh   PR  PLO  2 PR PLO cos [( IF t )  R ] 
•  is a responsivity term.
• The first two terms, i.e. PR and PLO , are DC terms.
• Hence the signal component is:

iP ,coh  2 PR PLO cos [( IF t )  R ]

• Usually, PLO >> PR , and:

I P ,coh   PLO  2 PR PLO cos [( IF t )  R ]
ECE 455 Lecture 16

Advantages of Coherent Detection

 Detection of amplitude, frequency and phase
modulated signals:


I P ,coh   PR  PLO  2 PR PLO cos [( IF t )  R ]

Can detect ASK, FSK and PSK

• In contrast, IM/DD systems can only detect ASK.

ECE 455 Lecture 16

 Improved sensitivity (compared to IM/DD systems):


I P ,coh   PLO  2 PR PLO cos [( IF t )  R ]
• Since the signal component is:

iP ,coh  2 PR PLO cos [( IF t )  R ]

then by increasing PLO we can increase the value of iP ,

hence the LO laser acts as the equivalent of an optical
amplifier, giving greater receiver sensitivity.
ECE 455 Lecture 16
 Better channel selectivity:

 
I P ,coh   PR  PLO  2 PR PLO cos [( IF t )  R ]

IF = R - LO

• In other words, modulation of R is “downconverted” to

lower frequencies (microwave range), allowing channels
to be filtered using microwave filters instead of optical
filters.
•. Because microwave filters have sharper selectivity than
optical filters, more channels can be accommodated
in a given wavelength range.
ECE 455 Lecture 16
Comparison of typical optical and microwave filter
transfer functions
ECE 455 Lecture 16

10 GHz channel
spacing
Attenuation (dB/km)

1000
1500
channels
channels

Wavelength (nm)
ECE 455 Lecture 16

Homodyne detection

• When the incoming and local oscillator frequencies are different

(i.e. R  LO ), we refer to the system as being heterodyne.

• If the frequencies are locked to one another (R = LO ), the

system is homodyne. The locking is achieved using an optical
phase-locked loop. In this case, the IF frequency is zero (IF = 0),
and:

I P ,coh  E  E
1
2
2
R
1
2
2
LO  ER ELO cos (R )

Note that homodyne systems can detect ASK & PSK, but not FSK
ECE 455 Lecture 17 1

Coherent Optical Communications

Stavros Iezekiel
Department of Electrical and
Computer Engineering
University of Cyprus

• HMY 455
• Lecture 17
• Fall Semester 2016
ECE 455 Lecture 17 2

SNR FOR COHERENT

COMMUNICATIONS
ECE 455 Lecture 17

• Recall that: SNR = average signal power

average noise power

= ip2
2qB(Im + ID) + 4kTBFn / RL

Shot noise
Thermal and
Dark current amplifier noise
ECE 455 Lecture 17

• How can the SNR be improved by, for example, choosing

homodyne PSK?

• Homodyne  R = LO  IF = 0, so the detected

photocurrent is:


I P   PR  PLO  2 PR PLO cos (R )
• NB, in PSK, we modulate the phase (R = 0 for binary “0”
and R = 1 for binary “1”), i.e. the envelope is constant
(PR = const.).
ECE 455 Lecture 17
• Hence:

I P  PR  PLO   2 R PR PLO cos (R )

“DC” current  Im signal current  ip

• In binary PSK, we want the two phases to be as far apart

as possible, i.e. to be  radians apart. Hence we might
choose 0 =  and 1 = 0.

• cos 0 = -1 and cos 1 = 1

ECE 455 Lecture 17

i  4 PR PLO cos (R )

2
p
2 2

 4 PR PLO
2
for either “0” or “1”

• The mean square signal photocurrent is:

i  4 PR PLO
2
p
2

• If we normalise our currents to an impedance of 1 ,

then the above is equivalent to the signal power.
• Notice that increasing the LO laser’s power will boost
the received signal power. What does it do to the
noise?
ECE 455 Lecture 17

• For homodyne PSK,

2
i
SNRPSK 
p

2qBI m  I D 
4kTBFn

RL
4 PR PLO
2

4kTBFn
2qB [( PR  PLO )  I D ] 
RL
ECE 455 Lecture 17

• We can now examine the effect of PLO ,

4 PR PLO
2
SNRPSK 
4kTBFn
2qB [( PR  PLO )  I D ] 
RL

• Thermal and
• Shot noise component is amplifier noise
amplified by PLO , but so is independent of PLO
the signal component!
ECE 455 Lecture 17

• So in the limit, if PLO   ,

4 PR PLO
2
SNRPSK 
2qBPLO
2PR

qB

• Hence by choosing a suitably large value of PLO we can

make the SNR independent of thermal and amplifier
noise, i.e. we can make it shot-noise limited.
ECE 455 Lecture 17

• Remember that responsivity  and quantum

efficiency  are related via:

IP q
 
PO hf

2PR 2 PR
 SNRPSK  
qB B hf
ECE 455 Lecture 17
• We saw that the ideal (quantum) limited SNR for
intensity modulation/direct detection was:


SNRIM / DD  Pm
2 B hf
• This is the best we can get with direct detection. Now,

SNRPSK 4 PR

SNRIM / DD Pm

• In other words, going to homodyne PSK gives a fourfold

(i.e. 6 dB) improvement in SNR! (Assumes PR = Pm).
ECE 455 Lecture 17

Quantum Limit on Optical Receiver Sensitivity

For Coherent PSK (Homodyne)

• Similar to ideas for quantum limit of direct detection

receiver:
– A steady light beam incident on a photodiode
generates electron-hole pairs in independent
random events
– The probability of detecting L photons in a time
period T obeys the Poisson distribution:

L N
N e
PN ( L) 
L!
ECE 455 Lecture 17

• N is the expected number of detected photons (i.e.

generated electron-hole pairs) in the time period T.

• In an ideal system (no thermal noise, no dark current),

the variation in the numbers of generated e-h pairs per
time interval (i.e. shot noise) would be the only source of
noise.

• This implies that e-h pairs could only be generated in a

photodiode provided that light was incident upon it.

• Remember that the production of photons is statistical in

nature.
ECE 455 Lecture 17

• Start by reconsidering the direct detection case using on-off

keying (OOK).
– “1” represented by power level PR
– “0” represented by no power (i.e. no photons sent)

IP

Load
ER  E cos  R t  R 
~ resistor
RL
Photodiode
E = 0 for “0” E = ER for “1”
ECE 455 Lecture 17

• Ideally, optical energy would only be sensed (and e-h

pairs generated) if a “1” is sent. Therefore, the ideal
receiver would then be an electron-hole pair counter,
and would make a decision based on a threshold
current.

• The most efficient situation is when the threshold is

set between no pairs generated (a “0” was sent) and
“at least one pair generated” (a “1” was sent).

• Therefore no errors occur if a “0” was sent (because

no carriers can be generated).
ECE 455 Lecture 17

• Errors may occur for a “1” if the incident optical power

fails to generate any carrier pairs at all (when N could be
expected).
– The probability of this occurring is:
N 0 e N N
PN (0)  e
0!
• Since “0”s are received with no errors (P0(0) = 0), and
“1”s and “0”s are equally likely, the overall error
probability is:
N
Pe  e
1
2 For direct-detection
ECE 455 Lecture 17
• Using the previous expression, we find that 10 photons per bit are
needed to achieve a BER of 10-9 if we have a direct detection
scheme using OOK.
– Can we do any better with coherent communications?
IP

Incoming signal
~ ~
E R  E R cos  R t   R 
~ ELO  ER
Load
resistor
Optical RL
combiner Photodiode

E LO  E LO cos  LO t 
~

Phase-locked
local oscillator
laser
ECE 455 Lecture 17
• Consider homodyne PSK:


I P   PR  PLO  2 PR PLO cos (R ) 
• If we look at this in terms of incident optical power, we
have:
P(t )  PR  PLO  2 PR PLO cos (R )
 PR  PLO  2a PR PLO

• a = -1 for “0” { cos 0 = cos  = -1 }

• a = 1 for “1” { cos 1 = cos 0 = 1 }
ECE 455 Lecture 17
• If we choose PLO = PR , then:

P(t )  PR  PLO  2a PR PLO

 2 PR (1  a )

• For “0”, a = -1, hence: P(t )  2 PR (1  a)  0

• For “1”, a = 1, hence: P (t )  2 PR (1  a )  4 PR

• In other words, for this special case, the system behaves

like the OOK one, but the binary “1” pulse has four times
the power !
ECE 455 Lecture 17
• In the OOK case, we had:
N hf
Popt 
T
where N = mean no. of photons received in T sec when
an optical power Popt is incident, and Popt is the optical
power received for a stream of “1”s.

• For this case, we have four times the power, hence four
times as many detected photons per “1”:

4 N hf
4 Popt 
T
ECE 455 Lecture 17
• So for homodyne PSK with PLO = PR, we have:

4 N
Pe  e1
2
Homodyne PSK

Super quantum limit

• Compare this with direct detection of binary OOK:

N
Pe  e
1
2
Quantum limit
for direct-detection OOK
ECE 455 Lecture 17
• Put another way, to achieve the same BER, then direct-
detection of OOK needs four times as many photons per
binary “1” as homodyne PSK does in the super-quantum
limit:

N 9
Pe  e 1
2  10  N  20.03 DD/OOK

4 N 9 Homodyne
Pe  e 1
2  10  N  5.01 PSK

• If we assume equal numbers of “0”s and “1”s, then in

the super-quantum limit, we only need 2.5 photons per
bit.