ENE 423
Lecture I
Overview: Optical Communication Systems
Communication is the transfer of information over distance between a source
and a user. A basic communication system consists of a transmitter, a receiver, and a
channel.
Source of
Information
Transmitter
Information Channel
Receiver
Users
To make it more precise, an optical communication system can be expressed
by block diagrams shown below
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Modulator – the modulators convert the electrical message into the proper
format and attach this signal onto the light wave.
Photonic source (Carrier source) – There are two types of semiconductor
diodes being used in optical communications.
1. Light Emitting Diode (LED)
2. Laser Diode (LD)
Channel Coupler – Coupling light into fiber creates a large reduction in power
(loss) because of the small diameter of optical fiber.
3
Information Channel – This refers to the path between the transmitter and
receiver. In fiber optic communications, a glass fiber is the channel. Its useful stretch
of length is limited by dispersion and attenuation.
Photodetector – In an electronic system, this is a demodulator, but, in fiber
system, the light wave is converted into electricity by a photo detector. There are 2
conventional semiconductor photo detectors: p-i-n photodiode and avalanche
photodiode (APD).
4
Optical Amplifier – this amplifier rejuvenates the light signal (widely use
EDFA). A schematic of an optical communication system can be illustrated by
y(t)
x(t)
TX
RX
Signal Processing – For analog transmission, the signal processor includes
amplification and filtering of the signal. Some random fluctuations in the received
signal are called “noise”. For digital system, the signal processor includes decision
circuits in addition to amplifiers and filters. This circuit decides if “1” or “0” was
received.
A figure of merit for characterizing communication systems is the product of
information capacity × maximum length of channel without repeaters.
Ex. optical system
(16 channels at 2.4 Gbps/channel) x
(60 - 70 km) = (2 – 3) × 1012 bps.km
Basic System Considerations
Maximum frequency in a modulating message signal is called “baseband”.
Baseband that is adapted for voice message is 4 kHz.
Bandwidth requirements of several analog systems
Message Type
Used bandwidth(B)
Voice (telephone)
4 kHz
Music -- AM
10 kHz
Music -- FM
200 kHz
TV (Video + Audio)
6 MHz
5
Digital transmission – The sampling theorem says that an analog signal can be
accurately transmitted if sampling rate is twice the highest frequency contained in that
signal. Let R be the required transmission rate. R can be expressed by
R  m. f s
where m = number of bits/sample
fs = sampling frequency = 2(f)
Ex. A telephone system has m = 8 bits/sample. Find R.
R = m(2f)
= 8x(2x4x103)
= 64 kbps
Several messages can be combined (multiplexed) onto a single information
channel. Most fundamental multiplexing in telephone network is incorporating 24
voice channels in one line.
1
2
3
Required rate
= 24 x 64 kbps
= 1.536 x 106 bps
T-1
system
.
.
24
Actually, more data bits of 1.536 x 106 are being sent than ones are required
(1.536 Mb/s). This actual rate of 1.544 Mb/s includes the bits to identify input frames
(synchronization and signaling pulses).
 24 x8 bits / sample   1 8 kHz   1.544 Mb / s
Digital transmission rates of US telephone system
Number of Voice
Transmission
Signaling
channels
Designation
Designation
1
-
-
64 kb/s
24
T1
DS-1
1.544 Mb/s
48(2-T1 systems)
T1C
DS-1C
3.152 Mb/s
96(4-T1 systems)
T2
DS-2
6.312 Mb/s
672(7-T2 systems)
T3
DS-3
44.736 Mb/s
1344(2-T3 systems)
T3C
DS-3C
91.053 Mb/s
4032 (6-T3 systems)
T4
DS-4
274.175 Mb/s
Data Rate
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Data transmission hierarchy used in North America.
By putting information in an ATM (asynchronous transfer mode) format, it is
possible to transmit simultaneously narrowband and broadband communication data
such as telephone, videoconferencing, or imaging, on a single subscriber.
A new transmission standard has been developed and it is called SONET
(Synchronous Optical NETwork) in North America and called SDH (Synchronous
Digital Hierarchy) in other parts of the world. The first level of the SONET is called
STS-1 (synchronous transport signal - level 1) where is then scrambled and converted
to an OC-1 (optical carrier – level 1) signal. Whereas, for SDH systems, the
fundamental building block starts from 155.52 Mb/s or STM-1(synchronous transport
module – level 1)
Transmission
Designation
(electrical)
(optical)
STS-1
SDH system
Data Rate(Mb/s)
OC-1
-
51.84
STS-3c
OC-3
STM-1
155.52
STS-12
OC-12
STM-4
622.08
STS-24
OC-24
STM-8
1,244.16
STS-48
OC-48
STM-16
2,488.32
STS-96
OC-96
STM-32
4,976.64
STS-192
OC-192
STM-64
9,953.28
STS-768
OC - 768
STM-128
39,813.12
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Ex. What level of ATM would be adequate for a commercial TV broadcast if 8
bits/sample is used.
Soln
1st approach: An analog signal has a bandwidth of 6 MHz.
R = m(2f) = 8(2×6×10-6) = 96 Mbps
From table: T4 at 274.17 Mbps …too much
2nd approach: Since video information contained in a bandwidth is less than 6 MHz,
the rate can be reduced.
Audio at f = 15kHz
Consider TV signal
Video at f = 4.5MHz
Encoding by using 8 bits/sample for video and keeping 8 bits/sample for audio.
R = (8×2×15×103)+(9×2×4.5×106) = 81.24 Mbps
Therefore, this signal can be transmitted over the T3-C level.
Ex. Assume 10 billion homes on earth each with a telephone. If all phones are to
communicate simultaneously over one transmission line by using frequency-division
multiplexing
(a) What is the minimum bandwidth required? (Assume analog modulation is
used.)
(b) Would a single optical beam of  = 1 μm carry these conversations?
(c) Repeat (a) and (b), if digital modulation is used with time-division
multiplexing and 64 kb/s for each voice message.
Soln
(a) Minimum bandwidth = (10000×106)×4×103
= 4 ×1013 Hz.
(b) f = c/ = 3x108/1×10-6 = 3×1014 Hz.
3×1014 Hz > 4×1013 Hz …O.K.
(c) Digital case:
R = (10000×106)×64000 = 6.4×1014bps
An optical carrier could not handle this rate. It could not be turned on and off
fast enough.
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Spectral-band classification scheme
Band
Descriptor
Range(nm)
O-band
Original
1260 - 1360
E-band
Extended
1360 - 1460
S-band
Short wavelength
1460 - 1530
C-band
Conventional
1530 – 1565
L-band
Long wavelength
1565 - 1625
U-band
Ultra-long wavelength
1625 - 1675
Optical fiber installations: on poles, in ducts, and undersea
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Ex. A cable contains 144 single-mode fibers, each operating at 2.3 Gbps. How many
digitized voice messages can be transmitted simultaneously along this cable?
Soln For 1 message, R  m(2f )  8  2  4 103  64 kb/s (for voice message)
144 single-mode fibers contain 144  2.3 109  3.312 1011 b/s
3.312 1011
The number of digitized voice message =
 5.175 106 messages
3
64 10
= 5.175 million messages
Decibel (dB) for power levels
Loss in fibers can be expressed by attenuation () in units of dB/km
P2 < P1
P1
l
If the power is P1 W at one point and P2 W at another point further along the
link, then P2/P1 is the fraction of the power transmitted between the 2 locations or it is
called the efficiency of transmission between 2 points.
dB  10 log10
P2
P1
Since P2 < P1 without any amplifier placed between the two, dB < 0.

10log10 ( P2 / P1 )
l (km)
For example, P2 = 0.5 P1 and l = 1 km, then  = 10log10(0.5) = -3 dB/km.
Sometimes, it is convenient to express power at any point relative to 1 mW, this is
denoted by the term dBm.
dBm  10log10 P(mW )
Ex. A receiver of sensitivity 1 μW. If 10 mW is transmitted at 100 MHz, what would
be the maximum length of the link when
(a) coaxial cable is used with  = 22.6 dB/km?
(b) Fiber with loss of 5 dB/km is used?
Note: Sensitivity describes the minimum power required at the receiver (user).
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Soln
Allowed loss in dB =
(a) Max. length = 40 dB/ 22.6 dB/km = 1.8 km
(b) Max. length = 40 dB/ (5 dB/km) = 8 km
History of fiber attenuation.
Ex. An LED radiates 2 mW. Compute the dBm value of this radiated power. This
power travels through a group of components having a combined loss of 23 dB.
Compute the output power.
Soln
(a) dBm = 10log10 (2mW/1mW) = 3 dBm
(b) Output power = 3 dBm – 23 dB
= -20 dBm
10log10 P0(mW) = -20 dBm
P0
= 10-2 = 0.01 mW
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As the preceding example, the level of power at the receiver dBmr are related
by
dBmr  dBmt  dBs
where dBmt = transmitted power (dBm)
dBs = system power (“-” for loss and “+” for gain)