Telecomunicazioni Docente: Andrea Baiocchi DIET - Stanza 107, 1° piano palazzina “P. Piga” Via Eudossiana 18 E-mail: [email protected]University of Roma “Sapienza” Corso di Laurea in Ingegneria Gestionale A.A. 2014/2015 Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi About bit and numbers… There are 10 types of people. Those who understand binary numbers and those that do not. [anonymous joke] But let your communication be Yea, yea; Nay, nay. For whatsoever is more than these, cometh of evil. [Matthew, 5:37]
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Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
About bit and numbers…
There are 10 types of people.Those who understand binary numbers and thosethat do not.
[anonymous joke]
But let your communication be Yea, yea; Nay, nay.For whatsoever is more than these, cometh of evil.
[Matthew, 5:37]
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Programma
1. SERVIZI E RETI DI TELECOMUNICAZIONE
2. ARCHITETTURE DI COMUNICAZIONE EMODI DI TRASFERIMENTO
3. FONDAMENTI DI COMUNICAZIONI
4. ACCESSO MULTIPLO
5. LO STRATO DI COLLEGAMENTO
6. LO STRATO DI RETE IN INTERNET
7. LO STRATO DI TRASPORTO IN INTERNET
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Fundamentals of communicationsA roadmap! Digital Representation of Information! Digital Representation of Analog Signals! Why Digital Communications?! Characterization of Communication Channels! Fundamental Limits in Digital Transmission! Line Coding! Modems and Digital Modulation! Properties of Media and Digital Transmission Systems! Error Detection and Correction
Chapter 3
Fundamentals ofcommunications
Digital Representation ofInformation
Adapted from slides of the book:A. Leon Garcia, I. Widjaja, “Communication
networks”, McGraw Hill, 2004
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Bits, numbers, information
! Bit: BInary digiT! Either symbol belonging to a set of two elements or
number with value 0 or 1! n bits: digital representation for 0, 1, … , 2n–1! Byte or Octet, n = 8! Computer word, typically n = 32, or 64
! n bits allows enumeration of 2n possibilities! n-bit field in a header! n-bit representation of a voice sample! Message consisting of n bits
! The number of bits required to represent a messageis a measure of its information content! More bits -> More information
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Block vs. Stream Information
Block
! Information that occurs in a single, delimited data unit(bit string)! Text message, Data file, JPEG image, MPEG file
! Size = bits / block
Stream
! Information that is produced and possibly conveyedover a communication system continuously! Real-time voice (e.g. telephony)
! Streaming video
! Bit rate = bits / second
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Information and …
! Information is a logical concept! Made precise by Shannon theory: more later on
! Information can appear under two basic forms! A SEQUENCE of values taken from a discrete and
finite set (DIGITAL INFORMATION)
! A REAL-VALUED FUNCTION of one or more REAL-VALUED variables (ANALOG INFORMATION)
! Examples! A book, a file, a digitally encoded speech/photo/movie
! Voice in the air, images as captured by camera sensors
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
… signals
! In communications engineering “signal” refersto e physically measurable entity that can beused to carry information! E.g. e.m. field, voltage and current in lumped
circuits, air pressure…
! An analog signal is a function x(t) defined overthe real axis and taking values in an interval ofthe real line
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Examples of analog information sources
! Images and sounds come natively in the formon analog information carried by analog signals! Images: e.m. field in the visible bandwidth, i.e., the
range of frequencies that produces a reaction inhuman sight sensors.
! Sounds: variation of air pressure that produces areaction in human hearing sensors.
! Why analog?! for our purposes all these phenomena are well
described by classical physics models (Newtonmechanics, Maxwell e.m. field theory)
! classical physics rests on classical analysis todescribe its models (variables taking values on acontinuum, namely the real axis).
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Th e s p ee ch s i g n al l e v el v a r ie s w i th t i m(e)
Voice
! Voice is carried by a signal (air pressure orcurrent induced in a microphone by airpressure)
! Analog voice signal level varies continuously intime
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
H
W
= + +H
W
H
W
H
W
Color
image
Red
component
image
Green
component
image
Blue
component
image
Color image
! A color image can be described by the linearcombination of three basic e.m. signals withgiven frequencies in the visible range! Example: intensity of Red, Green, Blue
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
H
W
= + +H
W
H
W
H
W
Color
image
Red
component
image
Green
component
image
Blue
component
image
Total bits = 3 ! H ! W pixels ! B bits/pixel = 3HWB bits
Example: 8!10 inch picture at 400 ! 400 pixels per inch2
400 ! 400 ! 8 ! 10 = 12.8 million pixels
8 bits/pixel/color
12.8 megapixels ! 3 bytes/pixel = 38.4 megabytes
Color image
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Moving picture
! Sequence of picture frames! Each picture digitized &
compressed
! Frame repetition rate! 10-30-60 frames/second
depending on quality
! Frame resolution! Small frames for
videoconferencing
! Standard frames forconventional broadcast TV
! HDTV frames
30 fps
Rate = M bits/pixel x (WxH) pixels/frame x F frames/second
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Video Frames
Broadcast TV at 30 frames/sec =
10.4 x 106 pixels/sec
720
480
HDTV at 30 frames/sec =
67 x 106 pixels/sec1080
1920
QCIF videoconferencing at 30 frames/sec =
760,000 pixels/sec
144
176
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Stream Service Quality Issues
Network Transmission Impairments
! Delay: Is information delivered in timely fashion?! E.g. mean e2e delay
! Jitter: Is information delivered in smooth fashion?! E.g. delay standard deviation
! Loss: Is information delivered without loss? If lossoccurs, is delivered signal quality acceptable?! Errored data
! Undelivered data
! Mis-ordered data
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Transmission Delay on a link
Use data compression to reduce LUse higher speed modem to increase RPlace far end system closer to reduce d
L number of bits in messageR bps speed of digital communication systemL/R time to transmit the messagetprop time for signal to propagate across mediumd distance in metersc speed of light (3x108 m/s in vacuum)
Delay = tprop + L/R = d/c + L/R seconds
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Compression
! Information usually not representedefficiently
! Data compression algorithms! Represent the information using fewer bits than
provided natively! Noiseless: original information recovered exactly
! E.g. zip, compress, GIF
! Noisy: recover information approximately.Tradeoff: # bits vs. quality! E.g. JPEG, MPEG
s (noisy ) | p (air stopped) | ee (periodic) | t (stopped) | sh (noisy)
“speech”
f
W
X(f)
0
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Sampling theorem
Sampling theorem (Nyquist):
Perfect reconstruction if sampling rate 1/T ! 2W
Interpolation
filtert
x(t)
t
x(nT)
(b)
Samplert
x(t)
t
x(nT)(a)
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Digitization of Analog Signal
! Sample analog signal in time and amplitude
! Find closest approximation
#/2
3#/2
5#/2
7#/2
$#/2
$3#/2
$5#/2
$7#/2
Original signal
Sample value
Approximation
Rs = Bit rate = # bits/sample x # samples/second
3 b
its /
sa
mp
le
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Quantization error:
“noise” = y(nT) – x(nT)
Quantizer maps input
into closest of 2m
representation values
#/2
3#/2
5#/2
7#/2
-#/2
-3#/2
-5#/2
-7#/2
Original signal
Sample value
Approximation
3 b
its /
sa
mp
le
Quantization of Analog Samples
input x(nT)
output y(nT)
0.5#
1.5#
2.5#
3.5#
-0.5#
-1.5#
-2.5#
-3.5#
# 2# 3# 4#
$#$2#$3#$4#
Uniform
quantizer
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
M = 2m quantization levels
Dynamic range (–V, V)Quantization interval # = 2V/M (uniform quantization)
If the number of levels M is large, the error e(x)=y(x)–x isapproximately uniformly distributed between (–#/2, #/2) in
each quantization interval
Quantizer error
!
2
...
error = y(nT)–x(nT) = e(nT)
input...
!"
2
3## #$2# 2#
x(nT) V-V
y(nT)
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Quantizer performance
! Power of quantization error signal = average ofsquared error
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Signal-to-Noise Ratio (SNR) =
Let %x2 be the signal power, then
The SNR is usually stated in decibels:
SNR dB = 10 log10(#x2/#e
2) = 6m + 10 log10(3#x2/V2)
Example: SNR dB = 6m – 7.27 dB for V/#x = 4.
Signal-to-quantization noise ratio
Average signal power
Average noise power
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Digital Transmission of AnalogInformation
Interpolationfilter
Displayor
playout
2W samples / sec
2W m bits/secx(t)
Bandwidth W
Sampling(A/D)
QuantizationAnalogsource
2W samples / sec m bits / sample
Pulse
generator
y(t)
Original
Approximation
Transmission
or storage
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
W = 4 kHz, so Nyquist sampling theorem
& 2W = 8000 samples/second
Suppose error requirement = 1% error
SNR = 10 log10(1/.01)2 = 40 dB
Assume V/#x = 4, then
40 dB = 6m – 7.27 & m = 8 bits/sample
PCM (“Pulse Code Modulation”):
Bit rate= 8000 x 8 bits/sec= 64 kbps
Example: Telephone Speech
Why Digital Communications?
Fundamentals ofcommunications
Adapted from slides of the book:A. Leon Garcia, I. Widjaja, “Communication
networks”, McGraw Hill, 2004
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
A Transmission System
Transmitter! Converts information into signal suitable for transmission
! Signal = measurable physical quantity that can be modifiedaccording to the value of the data to be transmitted, conveyed overa transmissin medium and detected by a receiving device.
! Injects energy into communications medium or channel
Receiver! Receives energy from medium! Converts received signal into form suitable for delivery to user
Receiver
Communication channel
Transmitter
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Transmission Impairments
Communication Channel! Pair of copper wires
! Coaxial cable
! Optical fiber
! Radio! Including infrared
Transmission Impairments! Attenuation
! Distortion
! Noise
! Interference
! Timing errors
Transmitted
SignalReceived
Signal Receiver
Communication channel
Transmitter
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Analog vs. Digital Transmission
Analog transmission: all details must be reproduced accurately
Sent
Sent
Received
Received
Distortion
Attenuation
Digital transmission: only discrete levels need to be reproduced
Distortion
AttenuationSimple Receiver:
Was original
pulse positive or
negative?
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Analog Long-Distance Communications
! Each repeater attempts to restore signal to its original form! Attenuation is removed (amplifier)
! Distortion is not completely eliminated
! In-band noise & interference can be removed only in part (out of band)
! Signal quality decreases with # of repeaters
Source DestinationRepeater
Transmission segment
Repeater. . .
Attenuated and
distorted signal + noise
Equalizer
Recovered signal +
residual noiseRepeater
Amp
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Digital Long-Distance Communications
! Regenerator recovers original data (bit) sequence fromdegraded signal and retransmits on next segment byusing a “clean” signal! But timing recovery is required!
! All impairments are “condensed” into bit errors
Source DestinationRegenerator
Transmission segment
Regenerator. . .
Amplifier
equalizer
Timing
recovery
Decision circuit
and signal
regenerator
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Digital Binary Signal
For a given communications medium:! How do we increase transmission speed?! How do we achieve reliable communications?! Are there limits to speed and reliability?
+A
-A
0 T 2T 3T 4T 5T 6T
1 1 1 10 0
Bit rate = 1 bit / T seconds
Signal is meaningless without associated clock
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Clock signal
Message bits
Baseband signalwith NRZ coding
+d
-d
1
0
0 1 0 1 1 0 1 0 0 0 1 1 0
Example of clock
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Many wavelengths>1600 GbpsOptical fiber
1 wavelength2.5-10 GbpsOptical fiber
5 km multipoint radio1.5-45 Mbps28 GHz radio
IEEE 802.11b/a/g wireless LANFrom 1 to 54 Mbps2.4 GHz radio
Coexists with analog telephonesignal
Up 8 Mbps down (ADSL)20 Mbps down (ADSL2+)50 Mbps down (VDSL)
Adapted from slides of the book:A. Leon Garcia, I. Widjaja, “Communication
networks”, McGraw Hill, 2004
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Communications Channels
! A physical medium is an inherent part of acommunications system! Copper wires, radio medium, or optical fiber
! Communications system includes electronic oroptical devices that are part of the pathfollowed by a signal! Transmitter, equalizers, amplifiers, filters, couplers,
detector, clocks and carrier generators
! By communication channel we refer to thecombined end-to-end physical medium andattached devices
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Communication system block scheme
SourceSource
encoder
Channel
encoder
Line
encoder
Modulator &
tx front end
Information source
Channel
Line
decoder
A/D
converter
Rx
front end
Timing
recovery
Channel
decoder
Information
rendering
Final
user
Information destination
Demodulation,
equalization,
symbol decision
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
How good is a channel?
! Performance: What is the maximum reliabletransmission speed?! Speed: Bit rate, R bps
! Reliability: Bit error rate, BER=10–k
! Focus of this section
! Cost: What is the cost of alternatives at a given levelof performance?! Wired vs. wireless?
! Electronic vs. optical?
! Standard A vs. standard B?
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Communications Channel
Bandwidth
! In order to transfer datafaster, a signal has to varymore quickly.
! A channel or medium has aninherent limit on how fastthe signals it passes can vary
! Apply sinusoidal input at frequency f! Measure amplitude of output sinusoid (of same frequency f) and
calculate amplitude response A(f) = ratio of output amplitude toinput amplitude! If A(f) " 1, then input signal passes readily! If A(f) " 0, then input signal is blocked
! Bandwidth Wc is range of frequencies passed by channel
H(f) = A(f)·ei'(f)Transfer function:
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Ideal Low-Pass Filter
! Ideal filter: all sinusoids with frequency f<Wc arepassed without attenuation and delayed by ( seconds;sinusoids at other frequencies are blocked
! 10 Gbit/s signal aftertransmission througha dispersive channel(with non negligibleISI)
Eye diagram
Fundamental Limits in DigitalTransmission
Fundamentals ofcommunications
Adapted from slides of the book:A. Leon Garcia, I. Widjaja, “Communication
networks”, McGraw Hill, 2004
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Transmitter
Filter
Communication
Medium
Receiver
Filter Receiver
r(t)
Received signal
+A
-A0 T 2T 3T 4T 5T
1 1 1 10 0
t
Signaling with Nyquist Pulses
! p(t) pulse at receiver in response to a single input pulse (takes intoaccount pulse shape at input, transmitter & receiver filters, andcommunications medium)
! r(t) waveform that appears in response to sequence of pulses
! If s(t) is a Nyquist pulse, then r(t) has zero intersymbolinterference (ISI) when sampled at multiples ofT
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Multilevel Signaling! Nyquist pulses achieve the maximum signalling rate with
zero ISI! 2Wc pulses per second or 2Wc pulses / Wc Hz = 2 pulses / Hz
! With two signal levels, each pulse carries one bit of thesource bit stream! Bit rate = 2Wc (bit/s)
! With M = 2m signal levels, each pulse carries m bit! Bit rate = 2Wc (pulse/s) · m (bit/pulse) = 2Wcm (bit/s)
In the absence of noise, the bit rate can be increasedwithout limit by increasing m
BUTAdditive noise limits # of levels that can be used reliably.
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Example of Multilevel Signaling
! Four levels {-1, -1/3, 1/3, +1} for {00,01,10,11}
! Waveform for 11,10,01 sends +1, +1/3, -1/3
! Zero ISI at sampling instants
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-1 0 1 2 3
Composite waveform
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Noise & Reliable Communications
! All physical systems have noise! Electrons always vibrate at non-zero temperature:
motion of electrons induces noise
! Presence of noise limits accuracy ofmeasurement of received signal amplitude
! Errors occur if signal separation is comparableto noise level
! Noise places a limit on how many amplitudelevels can be used in pulse transmission
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Four signal levels Eight signal levels
Noise Limits Accuracy! Receiver makes decision based on (sampled) received
signal level = source pulse level + noise! Error rate depends on relative value of noise amplitude and
spacing between signal levels
! Large (positive or negative) noise values can cause wrong decision
Typical noise
+A
+A/3
-A/3
-A
+A
+5A/7
+3A/7
+A/7
-A/7
-3A/7
-5A/7
-A
Decision thresholds
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Noise! Noise signal is usually a zero-mean process z(t)
characterized by! probability distribution of amplitude samples, i.e.
Pr(z(t) > u)
! Time auto-correlation, i.e. Rzz(t)=E[z(h)z(h+t)]
! Thermal electronic noise is inevitable (due tovibrations of electrons); thermal Noise can bemodeled as a “white” Gaussian process! Probability distribution is Gaussian zero mean
! Time auto-correlation is a Dirac pulse at 0
! Often interference from a large number ofscattered and similar sources can be modeledas white Gaussian “noise”
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
22
2
2
1 !
!"
xe#
x0
Gaussian noise
t
x
Pr(X(t)>x0) = area under graph on the right of x0
x0
x0
%2 = Avg Noise Power
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Probability of Error! Error occurs if noise value exceeds the
information signal magnitude over the decisionthreshold
! With two-level signalling, +A and –A, probabilityof error is Q(A/%)
1.00E-121.00E-11
1.00E-101.00E-091.00E-08
1.00E-071.00E-061.00E-05
1.00E-041.00E-031.00E-02
1.00E-011.00E+00
0 2 4 6 8
x
Q(x)
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Role of SNR
! With M=2m levels per symbol, the tx symbolvalues are ak=–A+(2k–1)A/M, with k=1,…,M.
! With equiprobable symbols:E[ak
2]=(M2–1)A2/(3M2)=PTX
! Received sample is (dispersive channel, zeroISI): yn=h0xn+zn.
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
signal noise signal + noise
signal noise signal + noise
High
SNR
Low
SNR
SNR = Average Signal Power
Average Noise Power
SNR (dB) = 10 log10 SNR
virtually error-free
error-prone
Channel Noise affects Reliability
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
! If transmitted power is limited, then as M increasesspacing between levels decreases
! Presence of noise at receiver causes more frequenterrors to occur as M is increased
Shannon Channel Capacity:! The maximum reliable transmission rate over an AWGN
bandlimited channel with bandwidth Wc Hz is
Cb = Wc log2(1+SNR) bit/s
Shannon Channel Capacity
X
Input symbol
(Gaussian)
Y=X+Z
Output symbol
Z
White gaussian noise
2Wc symbols/s
Bandlimited channel (Wc)
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Capacity of AWGN channel
! It can be shown that in case of real inputsignal the optimal source is gaussian and theAWGN capacity is
C = 0.5 log2(1+PR/PN) [bit/symbol]
where PN is the additive noise power, PR is theuseful signal received power
! A dispersive, additive noise channel can bereduced to AWGN if zero ISI is provided; tothat end it must be symbol rate < 2Wc. Then
Cb = Wclog2(1+(Eb/N0)Rb/Wc) [bit/s]
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
! Reliable communications is possible if the tx rate Rb<Cb.
! If Rb > Cb, then reliable communications is not possible.
“Reliable” means the bit error rate (BER) can be madearbitrarily small through sufficiently complex coding.
! Bandwidth Wc & SNR determine Cb
! Cb can be used as a measure of how close a system design is tothe best achievable performance.
! SNR=P/(N0Wc), with
! P= average power of input signal
! N0=noise power spectral density=k-F, k=1.38·10–23 J/K, -=noisetemperature, typically 300 K, F=noise figure, typically 6 dB
Shannon Channel Capacity
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Example
! Find the Shannon channel capacity for a telephonechannel with Wc = 3400 Hz and SNR = 10000
C = 3400 log2 (1 + 10000)
= 3400 log10 (10001)/log102 = 45200 bps
Note that SNR = 10000 corresponds to
SNR (dB) = 10 log10(10000) = 40 dB
Modems and Digital Modulation
Fundamentals ofcommunications
Adapted from slides of the book:A. Leon Garcia, I. Widjaja, “Communication
Codifiche di linea: (a) Bit, (b) NRZ, (c)NRZI, (d) Manchester, (e) Bipolare o AMI.
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Bandpass Channels
! Bandpass channels pass a range of frequencies aroundsome center frequency fc
! Radio channels, telephone & DSL modems
! Digital modulators embed information into waveformwith frequencies passed by bandpass channel! A sinusoidal signal with frequency fc centered in middle of
ordinary telephone wires! Category 5 UTP: tighter twisting to improve
signal quality! Shielded twisted pair (STP): to minimize
interference; costly! 10BASE-T Ethernet
! 10 Mbps, Baseband, Twisted pair! Two Cat3 pairs! Manchester coding, 100 meters
! 100BASE-T4 Fast Ethernet! 100 Mbps, Baseband, Twisted pair! Four Cat3 pairs! Three pairs for one direction at-a-time! 100/3 Mbps per pair;! 3B6T line code, 100 meters
! Cat5 & STP provide other options
! ! ! ! ! !
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
! Non linear effect: Brillouin, Raman, Kerr (Self-PhaseModulation (SPM), Cross-Phase Modulation (CPM),Four Wave Mixing (FWM))
! Launched power of 20'mW (13 dBm) in the cross section of amonomodal fiber (50 (m2) corresponds to 40 kW/cm2.
! Crosstalk in (D)WDM systems
! Noise of optical amplifiers (ASE)
! Laser phase noise
Transmission impairments of o.f.
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
! Optical fiber response to an input pulse x(t) is the sumof two terms:! a linear dispersive term yL(t) = #x(()h(t–()d(! an instantaneous cubic non linear term yNL(t) = Ax3(t)
! Let x(t) be the sum of three sinusoidal terms withfrequencies f1, f2, f3! Model for DWDM channels (extremely narrowband!!!)
! Besides linear terms, at the output we can findsinusoidal signals at frequencies ±f1±f2±f3! Part of these spurious harmonics falls within signal band
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Power spectrum of an 8x10 Gbps DWDM signal with UnequalChannel Spacing after 50 km of G.653 fiber (2 mW/ch).
Effect of FWM
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Standard fiber (G.652)
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Dispersion shifted fiber (G.653)
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Non-zero dispersion fiber (G.655)
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Regenerators & Optical Amplifiers
! The maximum span of an optical signal is determined bythe available power & the attenuation:! If 30 dB attenuation are allowed, then at 1550 nm, optical
signal attenuates at 0.25 dB/km, so max span = 30 dB/0.25km/dB = 120 km
! Optical amplifiers increase optical signal power (noequalization, no regeneration): Pout=G•Pin+PASE.! Noise is added by each amplifier
! SNRout < SNRin, so there is a limit to the number of OAs in anoptical path
! Optical signal must be regenerated when this limit isreached! Requires optical-to-electrical (O-to-E) signal conversion,
equalization, detection and retransmission (E-to-O)
! Expensive
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Regenerator
R R R R R R R R
DWDM
multiplexer
… …R
R
R
R
…R
R
R
R
…R
R
R
R
…R
R
R
R
…
DWDM & Regeneration
! Single signal per fiber requires 1 regenerator per span
! DWDM system carries many signals in one fiber
! At each span, a separate regenerator required per signal
! Very expensive
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
R
R
R
R
Optical
amplifier
… … …R
R
R
R
OA OA OA OA… …
Optical Amplifiers
! Optical amplifiers can amplify the composite DWDMsignal without demuxing or O-to-E conversion
! Erbium Doped Fiber Amplifiers (EDFAs) boost DWDMsignals within 1530 to 1620 range
! Spans between regeneration points >1000 km
! Number of regenerators can be reduced dramatically withdramatic reduction in cost of long-distance communications
Adapted from slides of the book:A. Leon Garcia, I. Widjaja, “Communication
networks”, McGraw Hill, 2004
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Error Control
! Digital transmission systems introduce errors
! Applications require certain reliability level
! Data applications require error-free transfer
! Voice & video applications tolerate some errors, the less themore source coding removes redundancy from original signal
! Error control used when transmission system does notmeet application requirement
! Two basic approaches:
Error detection & retransmission (ARQ)
Forward error correction (FEC)
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Key Idea
! All transmitted data blocks (“codewords”) satisfy apattern
! If received block doesn’t satisfy pattern, it is in error
! If it satisfies pattern, it is assumed to be correct
! Redundancy: Only a subset of all possible blocks can becodewords
! Example: in the set of all possible strings made of alphabetcharacters of length less or equal to 20 chars (to make the setfinite), current English dictionary words are “codewords”
ChannelEncoderUser
information
Pattern
checking
All inputs to channel
satisfy pattern or condition
Channel
output
Deliver user
information or
set error alarm
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Codewords communication
All n-bit strings
2n
2k
n-bit strings thatare codewords
All n-bit strings
2n
2k
TX RX
Transfer of a givencodeword, received
as a corruptedstring
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Single Parity Check! Given a data block of k bits, add one more bit so as to
make the number of 1’s even! Same as making 0 the xor of the coded (k+1)-bit block
Info Bits: b1, b2, b3, …, bk
Check Bit: bk+1 = b13b23b33 …3bk
Codeword: (b1, b2, b3, …, bk,, bk+1)
! Receiver checks to see if # of 1’s is even! All error patterns that change an odd # of bits are detectable;
all even-numbered patterns are undetectable
! Parity bit used in ASCII code
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Example of Single Parity Code
! Information (7 bits): (0, 1, 0, 1, 1, 0, 0)
! Parity Bit: b8 = 0 + 1 +0 + 1 +1 + 0 = 1
! Codeword (8 bits): (0, 1, 0, 1, 1, 0, 0, 1)
! If single error in bit 3 : (0, 1, 1, 1, 1, 0, 0, 1)! # of 1’s =5, odd
! Error detected
! If errors in bits 3 and 5: (0, 1, 1, 1, 0, 0, 0, 1)! # of 1’s =4, even
! Error not detected
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
How good is the single parity checkcode?
! Redundancy: Single parity check code adds 1 redundantbit per k information bits: overhead = 1/(k+1)
! Coverage: all error patterns with odd # of errors canbe detected
! An error patten is a binary (k + 1)-tuple with 1s where errorsoccur and 0’s elsewhere
! Of 2k+1 binary (k+1)-tuples, 1/2 have odd weight, so 50% of errorpatterns can be detected
! Is it possible to detect more errors if we add morecheck bits?
! Yes, with the right codes
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
What if bit errors are random?
! Many transmission channels introduce bit errors atrandom, independently of each other, with probability p
! Some error patterns are more probable than others:
! In any worthwhile channel p < 0.5, and so p/(1–p) < 1! It follows that patterns with 1 error are more likely than
patterns with 2 errors and so forth
! What is the probability that an undetectable errorpattern occurs?
P[10000000] = p(1 – p)7 = (1 – p)8 and
P[11000000] = p2(1 – p)6 = (1 – p)8
p
1 – p p 2
1 – p
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Single parity check code withrandom bit errors
! Undetectable error pattern if even # of bit errors:
! Example: Evaluate above for n = 32, p = 10–3
! For this example, roughly 1 in 2000 error patterns isundetectable
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Undetectable error patterns
! e(x) has 1s in error locations & 0s elsewhere! Receiver divides the received polynomial R(x) by g(x)! Blindspot: If e(x) is a multiple of g(x), that is, e(x) is
a nonzero codeword, then R(x) = b(x) + e(x) = q(x)g(x) + q’(x)g(x)! The set of undetectable error polynomials is the set
of nonzero code polynomials! Choose the generator polynomial so that most
common error patterns can be detected.
b(x)
e(x)
R(x)=b(x)+e(x)+
(Receiver)(Transmitter)
Error polynomial(Channel)
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Designing good polynomial codes
! Select generator polynomial so that likely errorpatterns are not multiples of g(x)
! Detecting Single Errors
! e(x) = xi for error in location i+1
! If g(x) has more than 1 term, it cannot divide xi
! Detecting Double Errors
! e(x) = xi + xj = xi(xj-i+1) where j > i
! If g(x) has more than 1 term, it cannot divide xi
! If g(x) is a primitive polynomial, it cannot divide xm+1 for allm<2n-k-1 (Need to keep codeword length less than 2n-k-1)
! Primitive polynomials can be found by consulting coding theory books
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Designing good polynomial codes
! Detecting Odd Numbers of Errors
! Define g(x) to be a multiple of x+1
! This implies x+1 must be a factor of all codewords b(x)
! For e(x) to be an undetectable error pattern, itmust be e(x) = q’(x)g(x) for some q’(x)
! Evaluate this identity at x=1 and get e(1) = 0 (itmust be g(1)=0; why?); then e(x) cannot have ad oddnumber of 1’s (why?)
! All odd numbers of errors are detectable!
! Pick g(x)=(x+1)p(x), where p(x) is primitive
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Standard Generator Polynomials
! CRC-8:
! CRC-16:
! CCITT-16:
! CCITT-32:
CRC = cyclic redundancy check
HDLC, XMODEM, V.41
IEEE 802, DoD, V.42
Bisync
ATM= x8 + x2 + x + 1
= x16 + x15 + x2 + 1
= (x + 1)(x15 + x + 1)
= x16 + x12 + x5 + 1
= x32 + x26 + x23 + x22 + x16 + x12 + x11
+ x10 + x8 + x7 + x5 + x4 + x2 + x + 1
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Hamming Codes
! Class of error-correcting codes
! Capable of correcting all single-error patterns
! For each m > 2, there is a Hamming code of lengthn = 2m – 1 with n – k = m parity check bits
5726114
k = n–m
6/636365/313154/151543/773m/nn = 2m–1m
Redundancy
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
m = 3 Hamming Code
! Information bits are b1, b2, b3, b4
! Equations for parity checks b5, b6, b7
! There are 24 = 16 codewords
! (0,0,0,0,0,0,0) is a codeword
b5 = b1 + b3 + b4
b6 = b1 + b2 + b4
b7 = + b2 + b3 + b4
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Hamming (7,4) code
71 1 1 1 1 1 11 1 1 1
31 1 1 0 0 0 01 1 1 0
41 1 0 1 0 1 01 1 0 1
41 1 0 0 1 0 11 1 0 0
41 0 1 1 1 0 01 0 1 1
41 0 1 0 0 1 11 0 1 0
31 0 0 1 0 0 11 0 0 1
31 0 0 0 1 1 01 0 0 0
40 1 1 1 0 0 10 1 1 1
40 1 1 0 1 1 00 1 1 0
30 1 0 1 1 0 00 1 0 1
30 1 0 0 0 1 10 1 0 0
30 0 1 1 0 1 00 0 1 1
30 0 1 0 1 0 10 0 1 0
40 0 0 1 1 1 10 0 0 1
00 0 0 0 0 0 00 0 0 0
w(b)b1 b2 b3 b4 b5 b6 b7b1 b2 b3 b4
WeightCodewordInformation
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Parity Check Equations
! Rearrange parity check equations:
! All codewords mustsatisfy these equations
! Note: each nonzero 3-tuple appears once as acolumn in check matrix H
! In matrix form:
0 = b5 + b5 = b1 + b3 + b4 + b5
0 = b6 + b6 = b1 + b2 + b4 + b6
0 = b7 + b7 = + b2 + b3 + b4 + b7
b1
b2
0 = 1 0 1 1 1 0 0 b3
0 = 1 1 0 1 0 1 0 b4 = H bt = 0
0 = 0 1 1 1 0 0 1 b5
b6
b7
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
0
0
1
0
0
0
0
s = H e = =1
0
1
Single error detected
0
1
0
0
1
0
0
s = H e = = + =0
1
1
Double error detected1
0
0
1 0 1 1 1 0 0
1 1 0 1 0 1 0
0 1 1 1 0 0 1
1
1
1
0
0
0
0
s = H e = = + + = 0 1
1
0
Triple error not
detected
0
1
1
1
0
1
1 0 1 1 1 0 0
1 1 0 1 0 1 0
0 1 1 1 0 0 1
1 0 1 1 1 0 0
1 1 0 1 0 1 0
0 1 1 1 0 0 1
1
1
1
Error Detection with Hamming Code
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Minimum distance of Hamming Code
! With Hamming (7,4) code undetectable error patternmust have 3 or more bits, i.e. at least 3 bits must bechanged to convert one codeword into another codeword
b1 b2o o
o
o
oo
o
o
Set of n-tuples
within distance
1 of b1
Set of n-tuples
within distance
1 of b2
! Spheres of distance 1 around each codeword do notoverlap
! If a single error occurs, the resulting n-tuple will be in aunique sphere around the original codeword
Distance 3
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
General Hamming Codes
! For m > 2, the Hamming code is obtainedthrough the check matrix H:! Each nonzero m-tuple appears once as a column of H
! The resulting code corrects all single errors
! For each value of m, there is a polynomial codewith g(x) of degree m that is equivalent to aHamming code and corrects all single errors! For m = 3, g(x) = x3+x+1
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Error-correction using Hamming Codes
! The receiver first calculates the syndrome:s = HR = H (b + e) = Hb + He = He
! If s = 0, then the receiver accepts R as the transmittedcodeword
! If s is nonzero, then an error is detected! Hamming decoder assumes a single error has occurred! Each single-bit error pattern has a unique syndrome! The receiver matches the syndrome to a single-bit error pattern
and corrects the appropriate bit
b
e
R+ (Receiver)(Transmitter)
Error pattern
Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi
Error correcting codes
! Very powerful correcting codes have beendevised over time! Used especially for wireless communications,