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We now have a look at the block diagram of an optical communication system and its
components. The source of the information (signal) can be a cell phone, a landline phone, a
computer, television or any such device. The source is followed by a signal pro
box. The functionality of this box may vary depending on the system;
conversion, pulse-shaping, encoding etc
transmitter, where it is converted from the
diagram, the red-shaded region represents the electrical domain, and the blue
the optical domain.

The electrical-to-optical conversion
conversion without introducing its own distortion. Typically,
emitting diodes (LED) or laser
conversion :(1) Direct modulation:
varied by the information signal (

Optic Communication Technology, Lecture 8
Lecture 8

Optical communication system

We now have a look at the block diagram of an optical communication system and its
components. The source of the information (signal) can be a cell phone, a landline phone, a
computer, television or any such device. The source is followed by a signal pro
box. The functionality of this box may vary depending on the system; it may be analog

shaping, encoding etc. The information from this box is now fed to the optical
transmitter, where it is converted from the electrical domain to the optical domain

the shaded region represents the electrical domain and the blue-
optical conversion is carried out using a transducer, which can perform this

conversion without introducing its own distortion. Typically, the transducers used are either
laser diodes. There are two methods for the electrical

Direct modulation: In this case, the driving current of the LED or the laser is
varied by the information signal (modulating signal), thereby changing the intensity of the
Page 1
We now have a look at the block diagram of an optical communication system and its
components. The source of the information (signal) can be a cell phone, a landline phone, a

computer, a television or any such device. The source is followed by a signal processing black-
it may be analogue-to-digital

The information from this box is now fed to the optical
domain. In this
-the shaded region is

is carried out using a transducer, which can perform this
the transducers used are either light
. There are two methods for the electrical-to-optical
ent of the LED or the laser is
), thereby changing the intensity of the light.

NPTEL-Fiber Optic Communication Technology, Lecture 8 Page 2
This method can be used only for amplitude/intensity modulation, and not for phase modulation.
(2) External modulation: In this case, the light from the source is passed through a modulator,
which is capable of varying the amplitude or the phase of the light according to the modulating

signal. The source of information, the signal processing (if necessary), and the electrical-to-
optical conversion together constitutes the optical transmitter.

The transmitter may be followed by a coupler, a switch, a router, a multiplexer, depending on the
kind of the network. For example, if the signals from different users are at different carrier
frequencies (f௖ଵand f௖ଶ), a wavelength multiplexer may be required to multiplex the two signals.
Depending on the network requirements, these passive devices may or may not be present in the
network.
Next, the signal passes through the channel, which is the optical fiber. The attenuation in the
commercial standard single-mode fiber, which is the most commonly used fiber in a
communication link, is 0.2 dB/km. This translates to attenuation of 20 dB after transmission
through 100 km, which corresponds to a 100 times reduction in the signal power. Thus, to
facilitate transmission to longer distances, the signal power needs to be boosted in the mid-span,
which is done using optical amplifiers, which amplifies the optical signal directly, without
having to convert to electrical domain. Depending on the link distance, one amplifier or more
than one amplifiers are used at regular intervals in the link. In case of short-haul links, no
amplifiers are used in the system. At the end of the channel, once again, there may be a coupler,
switch, router, filter, demultiplexer depending on the kind of network.
The receiver is basically an optical to the electrical converter, which is based on a photodiode that
senses light and converts it to a proportional electric current. This method, however, is limited to
intensity detection, and hence cannot detect phase-modulated data. There are other receiver
architectures for detecting phase – referred to as coherent receivers- which we will discuss in
detail during the course. Several kinds of amplifiers are used in optical communication links,
such as doped fiber amplifiers, semiconductor amplifiers, Raman amplifiers, out of which, the
most commonly used is the erbium-doped fiber amplifier.
The objective of the course is now to study the various parts of this block diagram, transmitters,
fiber, amplifiers, components, and receivers. We will also see how to set up this link, depending
on the requirement, and the kind of network architectures possible.
Note :
Since we deal with power levels varying over several orders of magnitude, we represent power in
logarithmic units, which is dBm. Power in dBm is defined as 10 × logଵ଴(Power in mW). Thus,
1 mW is represented as 0 dBm. −30dBm is equal to 10ିଷ mW = 1 μW. Similarly, 30 dBm
would correspond to 1 W. Power ratios such as loss and gain are also designated as logarithmic
units in dB as 10 × logଵ଴(
௉೚ೠ೟
௉೔೙
).

NPTEL-Fiber Optic Communication Tech
The intensity I of light is defined as power per unit area, and power
terms of number of photons,
I = number of photons per unit
Energy of a photon is hν = ௛௖

where

We work out this example: An optical trans
OOK signal at an average power, which is launched into a fiber of loss 6 dB. Assuming equal
number of 1’s and 0’s, what is the received number of photons in 1 bit?
Average received power (dBm) =
∴ P௔௩ = 3 dBm − 6 dB = −3 dBm
scale).
Since the signal is in OOK format, the power is
‘0’ bit. Let us assume that the peak power during the ‘1’ bit is

Optic Communication Technology, Lecture 8
of light is defined as power per unit area, and power P is energy per unit time. Thus i

unit area per unit time × photon energy.
where λ is the wavelength of the light.

We work out this example: An optical transmitter of wavelength 0.8 μm generates
OOK signal at an average power, which is launched into a fiber of loss 6 dB. Assuming equal
number of 1’s and 0’s, what is the received number of photons in 1 bit?
received power (dBm) = Average launched power (dBm) – loss (dB)

dBm. (Division in linear scale is subtraction in logarithmic

Since the signal is in OOK format, the power is ‘on’ only during the ‘1’ bit, it is
Let us assume that the peak power during the ‘1’ bit is Pଵ. Since the pulse shape is NRZ,

Page 3
is energy per unit time. Thus in

generates 1 Gbps NRZ-
OOK signal at an average power, which is launched into a fiber of loss 6 dB. Assuming equal

(Division in linear scale is subtraction in logarithmic

only during the ‘1’ bit, it is ‘off’ during the
. Since the pulse shape is NRZ,

NPTEL-Fiber Optic Communication Technology, Lecture 8 Page 4
which is a rectangular pulse, its magnitude is constant over the bit duration T௕. Thus, the energy
in bit ‘1’ is PଵT௕, and the number of photons in 1 bit is ௉భ்್
௛ఔ
=
௉భ்್ఒ
௛௖
.

Bit duration is the reciprocal of bit rate:T௕ =

ଵ଴వ s = 1 ns.

Statistically, it is expected that the number of 1’s and 0’s are the same, and hence,
Pଵ = 2 × P௔௩ = 1 mW.
Substituting, we get the number of photons per bit as ∼ 4 × 10଺

photons. This kind of
calculation of the number of photons per bit is important in the context of receiver sensitivity,