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Imran Khan, 2013 TE
The Digital Telephone Channel
Telecommunication Systems and Networks
1
Digital Networks
Digital transmission enables networks to support manyservices
Telephone
TV
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Questions of Interest
How long will it take to transmit a message? How many bits are in the message (text, image)?
How fast does the network/system transfer information?
Can a network/system handle a voice (video) call? How many bits/second does voice/video require? At what
quality?
How long will it take to transmit a message withouterrors?
How are errors introduced?
How are errors detected and corrected?
What transmission speed is possible over radio,copper cables, fiber, infrared, ?
Imran Khan, 2013 TE 3
Digital Transmission Fundamentals
Digi tal Representat ion of Inform ation
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Bits, numbers, information
Bit: number with value 0 or 1 nbits: digital representation for 0, 1, , 2n
Byte or Octet, n = 8
Computer word, n = 16, 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 content Imran Khan, 2013 TE 5
Block vs. Stream Information
Block
Information that occursin a single block
Text message
Data file
JPEG image
MPEG file Size = Bits / block
or bytes/block
1 kbyte = 210 bytes
1 Mbyte = 220 bytes
1 Gbyte = 230 bytes
Stream
Information that isproduced & transmittedcontinuously
Real-time voice
Streaming video
Bit rate = bits / second 1 kbps = 103 bps
1 Mbps = 106 bps
1 Gbps =109 bps
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Transmission Delay
Use data compression to reduce LUse higher speed modem to increase R
Place server closer to reduce d
L number of bits in message Rbps speed of digital transmission system
L/R time to transmit the information
tprop time for signal to propagate across medium
d distance in meters
c speed of light (3x108 m/s in vacuum)
Delay = tprop + L/R = d/c + L/Rseconds
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Compression
Information usually not represented efficiently
Data compression algorithms
Represent the information using fewer bits
Noiseless: original information recovered exactly E.g. zip, compress, GIF, fax
Noisy: recover information approximately
JPEG Tradeoff: # bits vs. quality
Compression Ratio
#bits (original file) / #bits (compressed file)
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H
W
= + +H
W
H
W
H
W
Colorimage
Redcomponentimage
Greencomponentimage
Bluecomponentimage
Total bits = 3 H W pixels B bits/pixel = 3HWB bits
Example: 810 inch picture at 400 400 pixels per inch2
400 400 8 10 = 12.8 million pixels8 bits/pixel/color12.8 megapixels 3 bytes/pixel = 38.4 megabytes
Color Image
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Type Method Format Original Compressed(Ratio)
Text Zip,compress
ASCII Kbytes-Mbytes
(2-6)
Fax CCITT
Group 3
A4 page
200x100pixels/in2
256
kbytes
5-54 kbytes
(5-50)
ColorImage
JPEG 8x10 in2 photo
4002 pixels/in2
38.4Mbytes
1-8 Mbytes(5-30)
Examples of Block Information
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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)
Stream Information
A real-time voice signal must be digitized &transmitted as it is produced
Analog signal level varies continuously in time
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Digitization of Analog Signal
Sample analog signal in time and amplitude Find closest approximation
D/2
3D/2
5D/2
7D/2
-D/2
-3D/2
-5D/2
-7D/2
Original signalSample value
Approximation
Rs = Bit rate = # bits/sample x # samples/second
3bits/sample
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Bit Rate of Digitized Signal
Bandwidth WsHertz: how fast the signal changes Higher bandwidth more frequent samples
Minimum sampling rate = 2 x Ws
Representation accuracy: range of approximationerror
Higher accuracy
smaller spacing between approximation values
more bits per sample
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Example: Voice & Audio
Telephone voice
Ws= 4 kHz 8000samples/sec
8 bits/sample
Rs=8 x 8000 = 64 kbps
Cellular phones use morepowerful compressionalgorithms: 8-12 kbps
CD Audio
Ws= 22 kHertz 44000samples/sec
16 bits/sample
Rs=16 x 44000= 704 kbpsper audio channel
MP3 uses more powerfulcompression algorithms:50 kbps per audiochannel
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Video Signal
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 for
conventional broadcast TV HDTV frames
30 fps
Rate = M bits/pixel x (WxH) pixels/frame x Fframes/second Imran Khan, 2013 TE 15
Video Frames
Broadcast TVat 30 frames/sec =
10.4 x 106
pixels/sec
720
480
HDTV at 30 frames/sec =
67 x 106pixels/sec
1080
1920
QCIF videoconferencing
at 30 frames/sec =
760,000pixels/sec
144
176
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Digital Video Signals
Type Method Format Original Compressed
VideoConfer-ence
H.261 176x144 or352x288 pix
@10-30fr/sec
2-36Mbps
64-1544kbps
FullMotion
MPEG2
720x480 pix@30 fr/sec
249Mbps
2-6 Mbps
HDTV MPEG2 1920x1080@30 fr/sec 1.6Gbps 19-38 Mbps
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Transmission of Stream Information
Constant bit-rate Signals such as digitized telephone voice produce a
steady stream: e.g. 64 kbps
Network must support steady transfer of signal, e.g.64 kbps circuit
Variable bit-rate Signals such as digitized video produce a stream that
varies in bit rate, e.g. according to motion and detailin a scene
Network must support variable transfer rate of signal,e.g. packet switching or rate-smoothing with constantbit-rate circuit
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Stream Service Quality Issues
Network Transmission Impairments
Delay: Is information delivered in timely fashion? Jitter: Is information delivered in sufficiently smooth
fashion?
Loss: Is information delivered without loss? If lossoccurs, is delivered signal quality acceptable?
Applications & application layer protocols developed todeal with these impairments
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Distortion
On metallic transmission links, such as coaxial cable and wire-pair cable, linecharacteristics distort and attenuate the digital signal as it traverses the medium.
There are three cable characteristics that create this distortion:
loss, amplitude distortion (amplitude-frequency response), and delay distortion.
When considering digital voice transmission, imperfections in the conversionprocess (A/D and D/A) add to the signal quality.
The first potential source of signal distortion during A/D conversion is imperfectfiltering (band limiting) of the signal to be digitized.
The process of filtering may also introduce delay distortion, which fortunately as
discussed earlier, is not an important issue in voice transmission. Sampling does not influence voice signal quality, provided the condition set by
Nyquists sampling theorem is satisfied, and there is no jitter in the sampling clock.
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Noise
As discussed in the analog voice channel, here also thermal noise, impulse noise,crosstalk etc. affect system design.
Because of the nature of a digital system, these impairments need only beconsidered on a per-repeater-section basis because noise does not accumulatedue to the regenerative process carried out at repeaters and nodes.
But bit errors do accumulate, and this impairment family is one of several thatcreate these errors.
One way of limiting error accumulation is to specify a stringent BER for eachrepeater section.
Repeater sections are often specified with a median BER of 1 in 10 9.
It is interesting to note that PCM provides reasonable voice performance for a BERas poor as 1 in 102.
However, the worst tolerable BER is 1 in 103 at system end points. This value isrequired to ensure the correct operation of supervisory signaling. It should benoted that such degraded BER values are completely unsuitable for datatransmission.
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Communication Networks and Services
Why Dig i tal Communicat ions?
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A Transmission System
Transmitter
Converts information into signalsuitable for transmission Injects energy into communications medium or channel
Telephone converts voice into electric current Modem converts bits into tones
Receiver
Receives energy from medium
Converts received signal into form suitable for delivery to user Telephone converts current into voice Modem converts tones into bits
ReceiverCommunication channel
Transmitter
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Transmission Impairments
Communication Channel
Pair of copper wires
Coaxial cable Radio
Light in optical fiber
Light in air
Infrared
Transmission Impairments
Signal attenuation
Signal distortion Spurious noise
Interference from othersignals
Transmitted
SignalReceived
Signal Receiver
Communication channel
Transmitter
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Analog Long-Distance Communications
Each repeater attempts to restore analog signal toits original form
Restoration is imperfect Distortion is not completely eliminated Noise & interference is only partially removed
Signal quality decreases with # of repeaters
Communications is distance-limited Still used in analog cable TV systems Analogy: Copy a song using a cassette recorder
Source DestinationRepeater
Transmission segmentRepeater. . .
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Analog vs. Digital Transmission
Analog transmission: all details must be reproduced accurately
Sent
Sent
Received
Received
Distort ion
Attenuation
Digital transmission: only discrete levels need to be reproduced
Distort ion
AttenuationSimple Receiver:Was original pulsepositive ornegative?
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Digital Long-Distance Communications
Regenerator recovers original data sequence andretransmits on next segment
Can design so error probability is very small Then each regeneration is like the first time! Analogy: copy an MP3 file
Communications is possible over very long distances Digital systems vs. analog systems Less power, longer distances, lower system cost Monitoring, multiplexing, coding, encryption, protocols
Source DestinationRegenerator
Transmission segmentRegenerator. . .
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Bit Rates of Digital Transmission Systems
System Bit Rate Observations
Telephonetwisted pair
33.6-56 kbps 4 kHz telephone channel
Ethernettwisted pair
10 Mbps, 100 Mbps 100 meters of unshieldedtwisted copper wire pair
Cable modem 500 kbps-4 Mbps Shared CATV return channel
ADSL twistedpair
64-640 kbps in, 1.536-6.144 Mbps out
Coexists with analogtelephone signal
2.4 GHz radio 2-11 Mbps IEEE 802.11 wireless LAN
28 GHz radio 1.5-45 Mbps 5 km multipoint radio
Optical fiber 2.5-10 Gbps 1 wavelength
Optical fiber >1600 Gbps Many wavelengths Imran Khan, 2013 TE 28
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Examples of Channels
Channel Bandwidth Bit Rates
Telephone voicechannel
3 kHz 33 kbps
Copper pair 1 MHz 1-6 Mbps
Coaxial cable 500 MHz(6 MHz channels)
30 Mbps/channel
5 GHz radio(IEEE 802.11)
300 MHz(11 channels)
54 Mbps /channel
Optical fiber Many TeraHertz 40 Gbps /wavelength
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Digi tal Representat ion of An alog Signals
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Digitization of Analog Signals
1. Sampling: obtain samples ofx(t) at uniformly spacedtime intervals
2. Quantization: map each sample into an approximationvalue of finite precision
Pulse Code Modulation: telephone speech
CD audio
3. Compression: to lower bit rate further, apply additionalcompression method
Differential coding: cellular telephone speech
Subband coding: MP3 audio
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Sampling Rate and Bandwidth
A signal that varies faster needs to be sampledmore frequently
Bandwidth measures how fast a signal varies
What is the bandwidth of a signal?
How is bandwidth related to sampling rate?
1 ms
1 1 1 1 0 0 0 0
. . . . . .
t
x2(t)1 0 1 0 1 0 1 0
. . . . . .
t
1 ms
x1(t)
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Periodic Signals
A periodic signal with period Tcan be representedas sum of sinusoids using Fourier Series:
DClong-termaverage
fundamentalfrequency f0=1/Tfirst harmonic
kth harmonic
x(t) = a0 + a1cos(2pf0t+ f1) + a2cos(2p2f0t+ f2) +
+ akcos(2pkf0t+ fk) +
|ak| determines amount of power in kth harmonic
Amplitude specturm |a0|, |a1|, |a2|, Imran Khan, 2013 TE 33
Example Fourier Series
T1 = 1 ms
1 1 1 1 0 0 0 0
. . . . . .
t
x2(t)1 0 1 0 1 0 1 0
. . . . . .
t
T2 =0.25 ms
x1(t)
Only odd harmonics have power
x1(t) = 0 + cos(2p4000t)
+ cos(2p3(4000)t)
+ cos(2p5(4000)t) +
4p
45p
43p
x2(t) = 0 + cos(2p1000t)
+ cos(2p3(1000)t)
+ cos(2p5(1000)t) +
4p
45p
43p
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Spectra & Bandwidth
Spectrum of a signal:magnitude of amplitudes asa function of frequency
x1(t) varies faster in time &has more high frequencycontent thanx2(t)
Bandwidth Ws is defined asrange of frequencies wherea signal has non-negligiblepower, e.g. range of bandthat contains 99% of totalsignal power
0
0.2
0.4
0.6
0.8
1
1.2
0 3 6 9 12 15 18 21 24 27 30 33 36 39 42
frequency (kHz)
0
0.2
0.4
0.6
0.8
1
1.2
0 3 6 9 12 15 18 21 24 27 30 33 36 39 42
frequency (kHz)
Spectrum ofx1
(t)
Spectrum ofx2(t)
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Bandwidth of General Signals
Not all signals are periodic E.g. voice signals varies
according to sound
Vowels are periodic, s isnoiselike Spectrum of long-term signal
Averages over many sounds,many speakers
Involves Fourier transform Telephone speech: 4 kHz CD Audio: 22 kHz
s (noisy ) | p (air stopped) | ee (periodic) | t (stopped) | sh (noisy)
X(f)
f0 Ws
speech
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Samplert
x(t)
t
x(nT)
Interpolationfilter t
x(t)
t
x(nT)
(a)
(b)
Nyquist: Perfect reconstruction if sampling rate 1/T> 2Ws
Sampling Theorem
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Digital Transmission of Analog Information
Interpolationfilter
Displayorplayout
2Wsamples / sec
2W m bits/secx(t)Bandwidth W
Sampling(A/D)
QuantizationAnalogsource
2Wsamples / sec m bits / sample
Pulsegenerator
y(t)
Original
Approximation
Transmission
or storage
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inputx(nT)
output y(nT)
0.5D1.5D
2.5D3.5D
-0.5D-1.5D
-2.5D-3.5D
D 2D 3D D-D2D3DD
Quantization error:noise =x(nT)y(nT)
Quantizer maps inputinto closest of 2mrepresentation values
D/2
3D/25D/27D/2
-D/2-3D/2-5D/2-7D/2
Original signalSample value
Approximation
3bits/s
ample
Quantization of Analog Samples
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Quantizer Performance
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0
a
thlevel
Aj+a/2
Aj
Aj-a/2
Aj+
Aj- receiver output at quantized voltage, Aj+- instantaneous voltage of the
signal, - equally likely error voltage
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Quantizer Performance
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Mean squared error
12
1)(
22/
2/22
adaE
a
a
- the average quantizing noise power
qPa
12
22
, 2 3
aRMS
SQR for sinusoidal modulation: amplitude of sine wave mA , average signal power 2
2
mA
peak-to-peak excursion is Am2 L
Ama
2
,
No of levels isL 2
2
3LAP mq 2
3
3
2)(
2
2
2
2
L
L
A
A
SQRm
m
ou t
LdBSQR out log208.1)( n68.1 dB forn
L 2
W= 4KHz, so Nyquist sampling theorem
2W = 8000 samples/second
Suppose error requirement 1% error
SNR = 10log(1/.01)2 = 40 dB
Assume V/x then
40 dB = 6m 7 m = 8 bits/sample
PCM (Pulse Code Modulation) TelephoneSpeech:
Bit rate= 8000 x 8 bits/sec= 64 kbps
Example: Telephone Speech
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Character izat ion of Comm unicat ionChannels
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Communications Channels
Aphysical medium is an inherent part of acommunications system Copper wires, radio medium, or optical fiber
Communications system includes electronic or opticaldevices that are part of the path followed by a signal Equalizers, amplifiers, signal conditioners
By communication channelwe refer to the combined
end-to-end physical medium and attached devices Sometimes we use the term filterto refer to a channel
especially in the context of a specific mathematicalmodel for the channel
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How good is a channel?
Performance: What is the maximum reliable transmissionspeed? Speed: Bit rate, Rbps Reliability: Bit error rate, BER=10-k
Cost: What is the cost of alternatives at a given level ofperformance?
Wired vs. wireless? Electronic vs. optical? Standard A vs. standard B?
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Communications Channel
Signal Bandwidth
In order to transfer datafaster, a signal has to varymore quickly.
Channel Bandwidth A channel or medium has
an inherent limit on how fastthe signals it passes canvary
Limits how tightly inputpulses can be packed
Transmission Impairments
Signal attenuation Signal distortion Spurious noise Interference from other
signals Limits accuracy of
measurements on receivedsignal
Transmitted
Signal
Received
Signal Receiver
Communication channel
Transmitter
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Channel
t t
x(t)= Aincos 2pft y(t)=Aoutcos (2pft + (f))
Aout
AinA(f) =
Frequency Domain Channel
Characterization
Apply sinusoidal input at frequency f Output is sinusoid at same frequency, but attenuated & phase-shifted Measure amplitude of output sinusoid (of same frequency f) Calculate amplitude response
A(f) = ratio of output amplitude to input amplitude IfA(f) 1, then input signal passes readily IfA(f) 0, then input signal is blocked
Bandwidth Wcis range of frequencies passed by channel Imran Khan, 2013 TE 47
Ideal Low-Pass Filter
Ideal filter: all sinusoids with frequency f
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Example: Low-Pass Filter
Simplest non-ideal circuit that provides low-pass filtering Inputs at different frequencies are attenuated by different amounts Inputs at different frequencies are delayed by different amounts
f
1A(f)= 1
(1+4p2f2)1/2
Amplitude Response
f0
(f)= tan-1 2pf
-45o
-90o
1/2p
Phase Response
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Example: Bandpass Channel
Some channels pass signals within a band thatexcludes low frequencies Telephone modems, radio systems,
Channel bandwidth is the width of the frequency bandthat passes non-negligible signal power
f
Amplitude Response
A(f)
Wc
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Channel Distortion
Channel has two effects: If amplitude response is not flat, then different frequency
components ofx(t) will be transferred by different amounts If phase response is not flat, then different frequency
components ofx(t) will be delayed by different amounts
In either case, the shape ofx(t) is altered
Letx(t) corresponds to a digital signal bearing datainformation
How well does y(t) followx(t)?
y(t) = A(fk) akcos(2pfkt+ k+ (fk))
Channel y(t)x(t) =akcos(2pfkt+ k)
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Example: Amplitude Distortion
Letx(t) input to ideal lowpass filter that has zero delay andWc= 1.5 kHz, 2.5 kHz, or 4.5 kHz
1 0 0 0 0 0 0 1. . . . . .
t1 ms
x(t)
Wc= 1.5 kHz passes only the first two terms
Wc= 2.5 kHz passes the first three terms
Wc= 4.5 kHz passes the first five terms
p
x(t) = -0.5 + sin( )cos(2p1000t)
+ sin( )cos(2p2000t) + sin( )cos(2p3000t) +
4p
p4
4p
4p
2p4
3p4
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-1.5
-1
-0.5
0
0.5
1
1.5
0
0.1
25
0.2
5
0.3
75
0.5
0.6
25
0.7
5
0.8
75 1
-1.5
-1
-0.5
0
0.5
1
1.5
0
0.1
25
0.2
5
0.3
75
0.5
0.6
25
0.7
5
0.8
75 1
-1.5
-1
-0.5
0
0.5
1
1.5
0
0.1
25
0.2
5
0.3
75
0.5
0.6
25
0.7
5
0.8
75 1
(b) 2 Harmonics
(c) 4 Harmonics
(a) 1 Harmonic
Amplitude Distortion
As the channelbandwidthincreases, theoutput of thechannelresembles theinput moreclosely
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Channel
t0t
h(t)
td
Time-domain Characterization
Time-domain characterization of a channel requires
finding the impulse response h(t) Apply a very narrow pulse to a channel and observe
the channel output
h(t) typically a delayed pulse with ringing Interested in system designs with h(t) that can be
packed closely without interfering with each other Imran Khan, 2013 TE 54
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Nyquist Pulse with Zero Intersymbol
Interference
For channel with ideal lowpass amplitude response ofbandwidth Wc, the impulse response is a Nyquist pulseh(t)=s(tt), where T= 1/2 Wc, and
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
t
s(t) = sin(2pWc t)/2pWct
T T T T T T T T T T T T T T
s(t) has zero crossings at t = kT, k= +1, +2, Pulses can be packed every T seconds with zero interference
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-2
-1
0
1
2
-2 -1 0 1 2 3 4
tT T T T TT
-1
0
1
-2 -1 0 1 2 3 4
tT T T T TT
Example of composite waveform
Three Nyquist pulsesshown separately
+ s(t)
+ s(t-T)
- s(t-2T)
Composite waveform
r(t) = s(t)+s(t-T)-s(t-2T)
Samples at kT
r(0)=s(0)+s(-T)-s(-2T)=+1
r(T)=s(T)+s(0)-s(-T)=+1
r(2T)=s(2T)+s(T)-s(0)=-1
Zero ISI at sampling
times kT
r(t)
+s(t) +s(t-T)
-s(t-2T)
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0f
A(f)
Nyquist pulse shapes
If channel is ideal low p ass with Wc, then pulses maxim umrate puls es can be transm it ted witho ut ISI isT =1/2Wc sec.
s(t) is one example of class of Nyquist pulses with zero ISI Problem: sidelobes in s(t) decay as 1/twhich add up quickly when
there are slight errors in timing
Raised cosine pulse below has zero ISI Requires slightly more bandwidth than Wc
Sidelobes decay as 1/t3, so more robust to timing errors
1
sin(pt/T)pt/T cos(pt/T)1 (2t/T)2
(1)Wc Wc (1 + )Wc Imran Khan, 2013 TE 57
Fundamenta l L imi ts in Dig i talTransmiss ion
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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
-A0 T 2T 3T 4T 5T 6T
1 1 1 10 0
Bit rate = 1 bit / T seconds
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Pulse Transmission Rate
Objective: Maximize pulse rate through a channel,that is, make Tas small as possible
Channel
t t
If input is a narrow pulse, then typical output is aspread-out pulse with ringing
Question: How frequently can these pulses betransmitted without interfering with each other?
Answer: 2 x Wcpulses/second
where Wc is the bandwidth of the channel
T
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Bandwidth of a Channel
If input is sinusoid of frequency f,then
output is a sinusoid of same frequency f
Output is attenuated by an amountA(f)that depends on f
A(f)1, then input signal passes readily
A(f)0, then input signal is blocked
Bandwidth Wc is range offrequencies passed by channel
ChannelX(t) = a cos(2pft) Y(t) =A(f) a cos(2pft)
Wc0f
A(f)1
Ideal low-pass
channel
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TransmitterFilter
CommunicationMedium
ReceiverFilter Receiver
r(t)
Received signal
+A
-A 0 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 (takesinto account pulse shape at input, transmitter & receiver filters,and communications medium)
r(t) waveform that appears in response to sequence of pulses Ifs(t) is a Nyquist pulse, then r(t) has zero intersymbol
interference (ISI) when sampled at multiples ofT
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Multilevel Signaling
Nyquist pulses achieve the maximum signalling rate with zero ISI,2Wcpulses per second or
2Wcpulses / WcHz = 2 pulses / Hz
With two signal levels, each pulse carries one bit of informationBit rate = 2Wcbits/second
With M= 2m signal levels, each pulse carries m bitsBit rate = 2Wcpulses/sec. * m bits/pulse = 2Wcm bps
Bit rate can be increased by increasing number of levels r(t) includes additive noise, that limits number of levels that can be
used reliably. Imran Khan, 2013 TE 63
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
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Four signal levels Eight signal levels
Typical noise
Noise Limits Accuracy
Receiver makes decision based on transmitted pulse level + noise Error rate depends on relative value of noise amplitude and spacingbetween signal levels
Large (positive or negative) noise values can cause wrong decision Noise level below impacts 8-level signaling more than 4-level signaling
+A
+A/3
-A/3
-A
+A
+5A/7
+3A/7
+A/7
-A/7
-3A/7
-5A/7
-A
Imran Khan, 2013 TE 65
222
2
1
p
xe-
x0
Noise distribution
Noise is characterized by probability density of amplitude samples Likelihood that certain amplitude occurs Thermal electronic noise is inevitable (due to vibrations of electrons) Noise distribution is Gaussian (bell-shaped) as below
t
x
Pr[X(t)>x0] = ?
Pr[X(t)>x0] =Area undergraph
x0
x0
2 = Avg Noise Power
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1.00E-12
1.00E-11
1.00E-10
1.00E-09
1.00E-081.00E-07
1.00E-06
1.00E-05
1.00E-04
1.00E-03
1.00E-02
1.00E-01
1.00E+00
0 2 4 6 8/2
Probability of Error
Error occurs if noise value exceeds certain magnitude Prob. of large values drops quickly with Gaussian noise Target probability of error achieved by designing system so
separation between signal levels is appropriate relative toaverage noise power
Pr[X(t)>]
Imran Khan, 2013 TE 67
signal noise signal + noise
signal noise signal + noise
HighSNR
Low
SNR
SNR =Average Signal Power
Average Noise Power
SNR (dB) = 10 log10 SNR
virtually error-free
error-prone
Channel Noise affects Reliability
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If transmitted power is limited, then as Mincreases spacingbetween levels decreases
Presence of noise at receiver causes more frequent errorsto occur as Mis increased
Shannon Channel Capacity:
The maximum reliable transmission rate over an ideal channelwith bandwidth WHz, with Gaussian distributed noise, andwith SNR S/Nis
C= Wlog2
( 1 + S/N) bits per second
Reliable means error rate can be made arbitrarily small byproper coding
Shannon Channel Capacity
Imran Khan, 2013 TE 69
Example
Consider a 3 kHz channel with 8-level signaling. Comparebit rate to channel capacity at 20 dB SNR
3KHz telephone channel with 8 level signalingBit rate = 2*3000 pulses/sec * 3 bits/pulse = 18 kbps
20 dB SNR means 10 log10S/N= 20Implies S/N= 100
Shannon Channel Capacity is thenC= 3000 log ( 1 + 100) = 19, 963 bits/second
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Line Coding
Imran Khan, 2013 TE 71
What is Line Coding?
Mapping of binary information sequence into the digitalsignal that enters the channel Ex. 1 maps to +A square pulse; 0 to A pulse
Line code selected to meet system requirements: Transmitted power: Power consumption = $ Bittiming: Transitions in signal help timing recovery Bandwidthefficiency: Excessive transitions wastes bw Lowfrequencycontent: Some channels block low frequencies
long periods of +A or ofA causes signal to droop Waveform should not have low-frequency content
Errordetection: Ability to detect errors helps Complexity/cost: Is code implementable in chip at high speed?
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Line coding examples
NRZ-inverted(differentialencoding)
1 0 1 0 1 1 0 01
UnipolarNRZ
Bipolarencoding
Manchester
encoding
DifferentialManchester
encoding
Polar NRZ
Imran Khan, 2013 TE 73
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
00.2
0.4
0.6
0.8 1
1.2
1.4
1.6
1.8 2
fT
pow
erd
ensity
NRZ
Bipolar
Manchester
Spectrum of Line codes
Assume 1s & 0s independent & equiprobable
NRZ has highcontent at lowfrequencies
Bipolar tightlypacked around T/2
Manchester wastefulof bandwidth
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Unipolar & Polar
Non-Return-to-Zero (NRZ)
Unipolar NRZ
1 maps to +A pulse
0 maps to no pulse
High Average Power
0.5*A2 +0.5*02=A2/2
Long strings of A or 0 Poor timing
Low-frequency content
Simple
Polar NRZ
1 maps to +A/2 pulse
0 maps to A/2 pulse
Better Average Power
0.5*(A/2)2 +0.5*(-A/2)2=A2/4
Long strings of +A/2 orA/2 Poor timing
Low-frequency content
Simple
1 0 1 0 1 1 0 01
Unipolar NRZ
Polar NRZ
Imran Khan, 2013 TE 75
Bipolar Code
Three signal levels: {-A, 0, +A} 1 maps to +A or A in alternation 0 maps to no pulse
Every +pulse matched bypulse so little content at lowfrequencies
String of 1s produces a square wave Spectrum centered at T/2
Long string of 0s causes receiver to lose synch Zero-substitution codes
1 0 1 0 1 1 0 01
BipolarEncoding
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Manchester code
1 maps into A/2 first T/2, -A/2 lastT/2
0 maps into -A/2 first T/2, A/2 lastT/2
Every interval has transition inmiddle
Timing recovery easy
Uses double the minimum
bandwidth Simple to implement
Used in 10-MbpsEthernet & otherLAN standards
1 0 1 0 1 1 0 01Manchester
Encoding
Imran Khan, 2013 TE 77
Differential Coding
Errors in some systems cause transposition in polarity, +A becomeA and vice versa
All subsequent bits in Polar NRZ coding would be in error Differential line coding provides robustness to this type of error 1 mapped into transition in signal level 0 mapped into no transition in signal level Same spectrum as NRZ Errors occur in pairs Also used with Manchester coding
NRZ-inverted(differentialencoding)
1 0 1 0 1 1 0 01
DifferentialManchester
encoding
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Modems and Dig i ta l Modulat ion
Imran Khan, 2013 TE 79
Bandpass Channels
Bandpass channels pass a range of frequenciesaround some center frequency fc Radio channels, telephone & DSL modems
Digital modulators embed information into waveformwith frequencies passed by bandpass channel
Sinusoid of frequency fc is centered in middle ofbandpass channel
Modulators embed information into a sinusoid
fcWc/2 fc0 fc+ Wc/2
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Information1 1 1 10 0
+1
-10 T 2T 3T 4T 5T 6T
AmplitudeShiftKeying
+1
-1
Frequency
ShiftKeying 0 T 2T 3T 4T 5T 6T
t
t
Amplitude Modulation and Frequency
Modulation
Map bits into amplitude of sinusoid: 1 send sinusoid; 0 no sinusoidDemodulator looks for signal vs. no signal
Map bits into frequency: 1 send frequency fc+ ; 0 send frequency fc- Demodulator looks for power around fc+ orfc-
Imran Khan, 2013 TE 81
Phase Modulation
Map bits into phase of sinusoid: 1 send A cos(2pft), i.e. phase is 0
0 send A cos(2pft+p), i.e. phase is p
Equivalent to multiplying cos(2pft) by +A or -A 1 send A cos(2pft), i.e. multiply by 1 0 send A cos(2pft+p) = - A cos(2pft), i.e. multiply by -1
We will focus on phase modulation
+1
-1
PhaseShiftKeying 0 T 2T 3T 4T 5T 6T t
Information 1 1 1 10 0
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Modulatecos(2pfct)by multiplying byAkforTseconds:
Ak x
cos(2pfct)
Yi(t) =Akcos(2pfct)
Transmitted signalduring kth interval
Demodulate (recoverAk) by multiplying by 2cos(2pfct)forTseconds and lowpass filtering (smoothing):
x
2cos(2pfct)2Akcos
2(2pfct) =Ak{1 + cos(2p2fct)}
Lowpass
Filter(Smoother)
Xi(t)Yi(t) =Akcos(2pfct)
Received signalduring kth interval
Modulator & Demodulator
Imran Khan, 2013 TE 83
1 1 1 10 0
+A
-A0 T 2T 3T 4T 5T 6T
Information
BasebandSignal
ModulatedSignalx(t)
+A
-A0 T 2T 3T 4T 5T 6T
Example of Modulation
A cos(2pft) -A cos(2pft) Imran Khan, 2013 TE 84
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1 1 1 10 0RecoveredInformation
Basebandsignal discernableafter smoothing
After multiplicationat receiverx(t) cos(2pfct)
+A
-A0 T 2T 3T 4T 5T 6T
+A
-A
0 T 2T 3T 4T 5T 6T
Example of Demodulation
A {1 + cos(4pft)} -A {1 + cos(4pft)}
Imran Khan, 2013 TE 85
Signaling rate and Transmission Bandwidth
Fact from modulation theory:
Baseband signalx(t)with bandwidth B Hz
If
then B
fc+B
f
f
fc-B fc
Modulated signalx(t)cos(2pfct) hasbandwidth 2B Hz
If bandpass channel has bandwidth WcHz,
Then baseband channel has Wc/2 Hz available, so
modulation system supports Wc/2 x 2 = Wc pulses/second
That is, Wcpulses/second perWcHz = 1 pulse/Hz
Recall baseband transmission system supports 2 pulses/Hz Imran Khan, 2013 TE 86
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Ak x
cos(2pfct)
Yi(t) =Akcos(2pfct)
Bk x
sin(2pfct)
Yq(t) = Bksin(2pfct)
+ Y(t)
Yi(t) and Yq(t) both occupy the bandpass channel
QAM sends 2 pulses/Hz
Quadrature Amplitude Modulation (QAM)
QAM uses two-dimensional signaling
Akmodulates in-phase cos(2pfct) Bkmodulates quadrature phase cos(2pfct+ p/4) = sin(2pfct) Transmit sum of inphase & quadrature phase components
TransmittedSignal
Imran Khan, 2013 TE 87
QAM Demodulation
Y(t) x
2cos(2pfct)2cos2(2pfct)+2Bk cos(2pfct)sin(2pfct)
=Ak{1 + cos(4pfct)}+Bk{0 + sin(4pfct)}
Lowpassfilter(smoother)
Ak
2Bksin2(2pfct)+2Ak cos(2pfct)sin(2pfct)
= Bk{1 - cos(4pfct)}+Ak{0 + sin(4pfct)}
x
2sin(2pfct)
BkLowpassfilter(smoother)
smoothed to zero
smoothed to zero Imran Khan, 2013 TE 88
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Signal Constellations
Each pair (Ak, Bk) defines a point in the plane Signal constellation set of signaling points
4 possible points perTsec.2 bits / pulse
Ak
Bk
16 possible points perTsec.4 bits / pulse
Ak
Bk
(A, A)
(A,-A)(-A,-A)
(-A,A)
Imran Khan, 2013 TE 89
Ak
Bk
4 possible points perTsec.
Ak
Bk
16 possible points perTsec.
Other Signal Constellations
Point selected by amplitude & phase
Akcos(2pfct) + Bksin(2pfct) = Ak2+ Bk2cos(2pfct+ tan-1(Bk/Ak))
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Telephone Modem Standards
Telephone Channel for modulation purposes hasWc= 2400 Hz 2400 pulses per second
Modem Standard V.32bis
Trellis modulation maps m bits into one of 2m+1 constellation points 14,400 bps Trellis 128 2400x6 9600 bps Trellis 32 2400x4 4800 bps QAM 4 2400x2
Modem Standard V.34 adjusts pulse rate to channel
2400-33600 bps Trellis 960 2400-3429 pulses/sec
Imran Khan, 2013 TE 91
Propert ies of Media and Digi talTransm iss ion Systems
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Fundamental Issues in Transmission Media
Information bearing capacity Amplitude response & bandwidth
dependence on distance Susceptibility to noise & interference
Error rates & SNRs
Propagation speed of signal
c = 3 x 108 meters/second in vacuum n = c/ speed of light in medium where >1 is the dielectric constant of
the medium
n = 2.3 x 108 m/sec in copper wire; n = 2.0 x 108 m/sec in optical fiber
t= 0t = d/c
Communication channel
dmeters
Imran Khan, 2013 TE 93
Communications systems &
Electromagnetic Spectrum
Frequency of communications signals
Analogtelephone
DSL Cellphone
WiFiOpticalfiber
102 104 106 108 1010 1012 101410161018 102010221024
Frequency (Hz)
Wavelength (meters)
106 104 102 10 10-2 10-4 10-6 10-8 10-10 10-12 10-14
Poweran
d
telephon
e
Broadcas
t
radio
Microwave
radio
Infraredlig
ht
Visibleligh
t
Ultravioletlig
ht
X-rays
Gammara
ys
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Wireless & Wired Media
Wireless Media Signal energy propagates in
space, limited directionality
Interference possible, sospectrum regulated
Limited bandwidth
Simple infrastructure:antennas & transmitters
No physical connectionbetween network & user
Users can move
Wired Media Signal energy contained &
guided within medium
Spectrum can be re-used inseparate media (wires orcables), more scalable
Extremely high bandwidth
Complex infrastructure
Imran Khan, 2013 TE 95
Attenuation
Attenuation varies with media Dependence on distance of central importance
Wired media has exponential dependence Received power at d meters proportional to 10-kd
Attenuation in dB = k d, where kis dB/meter
Wireless media has logarithmic dependence
Received power at d meters proportional to d-n Attenuation in dB = n log d, where n is path loss exponent; n=2
in free space
Signal level maintained for much longer distances
Space communications possible
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Twisted Pair
Twisted pair Two insulated copper wiresarranged in a regular spiralpattern to minimizeinterference
Various thicknesses, e.g.0.016 inch (24 gauge)
Low cost Telephone subscriber loop
from customer to CO Old trunk plant connecting
telephone COs Intra-building telephone
from wiring closet todesktop
Attenuation(dB/mi)
f (kHz)
19 gauge
22 gauge
24 gauge
26 gauge
6
12
18
24
30
110 100 1000
Lower attenuation rateanalog telephone
Higher attenuation ratefor DSL
Imran Khan, 2013 TE 97
Twisted Pair Bit Rates
Twisted pairs can providehigh bit rates at shortdistances
Asymmetric DigitalSubscriber Loop (ADSL) High-speed Internet Access Lower 3 kHz for voice Upper band for data 64 kbps outbound
640 kbps inbound Much higher rates possible atshorter distances Strategy for telephone
companies is to bring fiberclose to home & then twistedpair
Higher-speed access +video
Table 3.5 Data rates of 24-gauge twisted pair
Standard Data Rate Distance
T-1 1.544 Mbps 18,000 feet, 5.5 km
DS2 6.312 Mbps 12,000 feet, 3.7 km
1/4 STS-1 12.960Mbps
4500 feet, 1.4 km
1/2 STS-1 25.920Mbps
3000 feet, 0.9 km
STS-1 51.840Mbps
1000 feet, 300 m
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Ethernet LANs
Category 3 unshielded twisted pair(UTP): 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 FastEthernet 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
Imran Khan, 2013 TE 99
Coaxial Cable
Twisted pair
Cylindrical braided outerconductor surroundsinsulated inner wireconductor
High interference immunity Higher bandwidth than
twisted pair
Hundreds of MHz Cable TV distribution Long distance telephone
transmission
Original Ethernet LANmedium
35
30
10
25
20
5
15
Attenuation(dB/km)
0.1 1.0 10 100
f (MHz)
2.6/9.5 mm
1.2/4.4 mm
0.7/2.9 mm
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Upstream Downstream
5MHz
42MHz
54MHz
500MHz
550MHz
750MHz
Downstream
Cable Modem & TV Spectrum
Cable TV network originally unidirectional Cable plant needs upgrade to bidirectional 1 analog TV channel is 6 MHz, can support very high data rates Cable Modem: sharedupstream & downstream
5-42 MHz upstream into network; 2 MHz channels; 500 kbps to 4Mbps
>550 MHz downstream from network; 6 MHz channels; 36 Mbps Imran Khan, 2013 TE 101
Cable Network Topology
Headend
Upstream fiber
Downstream fiber
Fibernode
Coaxialdistribution
plant
Fibernode
= Bidirectionalsplit-bandamplifier
Fiber Fiber
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Optical Fiber
Light sources (lasers, LEDs) generate pulses of light that aretransmitted on optical fiber Very long distances (>1000 km) Very high speeds (>40 Gbps/wavelength) Nearly error-free (BER of 10-15)
Profound influence on network architecture Dominates long distance transmission Distance less of a cost factor in communications Plentiful bandwidth for new services
Optical fiber
Opticalsource
ModulatorElectricalsignal
Receiver Electricalsignal
Imran Khan, 2013 TE 103
Core
Cladding JacketLight
c
Geometry of optical fiber
Total Internal Reflection in optical fiber
Transmission in Optical Fiber
Very fine glass cylindrical core surrounded by concentric layer of glass(cladding)
Core has higher index of refraction than cladding Light rays incident at less than critical angle c is completely reflected
back into the core Imran Khan, 2013 TE 104
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Multimode: Thicker core, shorter reach
Rays on different paths interfere causing dispersion & limiting bit rate Single mode: Very thin core supports only one mode (path)
More expensive lasers, but achieves very high speeds
Multimode fiber: multiple rays follow different paths
Single-mode fiber: only direct path propagates in fiber
Direct path
Reflected path
Multimode & Single-mode Fiber
Imran Khan, 2013 TE 105
Optical Fiber Properties
Advantages
Very low attenuation Noise imm uni ty Extremely high bandw idth Security: Very difficult to
tap without breaking
No corrosion
More compact & lighter thancopper wire
Disadvantages
New types of optical signalimpairments & dispersion Polarization dependence Wavelength dependence
Limited bend radius If physical arc of cable too
high, light lost or wont reflect
Will break Difficult to splice Mechanical vibration becomes
signal noise
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100
50
10
5
1
0.5
0.1
0.05
0.01 0.8 1.0 1.2 1.4 1.6 1.8 Wavelength (m)
Loss(dB/km)
Infrared absorption
Rayleigh scattering
Very Low Attenuation
850 nmLow-cost LEDsLANs
1300 nmMetropolitan AreaNetworksShort Haul
1550 nmLong Distance NetworksLong Haul
Water Vapor Absorption(removed in new fiberdesigns)
Imran Khan, 2013 TE 107
100
50
10
5
1
0.5
0.1
0.8 1.0 1.2 1.4 1.6 1.8
L
oss(dB/km)
Huge Available Bandwidth
Optical range from 1to1 contains bandwidth
Example: 1= 1450 nm1 =1650 nm:
B = 19 THz
B = f1 f2 = v1 +
v1
v12
= / 11 + / 1
v1
2(108)m/s 200nm(1450 nm)2
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Wavelength-Division Multiplexing
Different wavelengths carry separate signals Multiplex into shared optical fiber
Each wavelength like a separate circuit
A single fiber can carry 160 wavelengths, 10 Gbpsper wavelength: 1.6 Tbps!
1
2
mopticalmux
1
2
mopticaldemux
1 2. m
opticalfiber
Imran Khan, 2013 TE 109
Coarse & Dense WDM
Coarse WDM
Few wavelengths 4-8 with verywide spacing
Low-cost, simple
Dense WDM
Many tightly-packedwavelengths
ITU Grid: 0.8 nm separationfor 10Gbps signals
0.4 nm for 2.5 Gbps
1550
1560
1540
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Fiber Span Analysis
Span analysis is the calculation and verification of a fiber-optic system'soperating characteristics. fiber routing, electronics, wavelengths, fiber type, and circuit length. Attenuation and
nonlinear considerations are the key parameters for loss-budget analysis.
Both the passive and active components of the circuit have to beincluded in the loss-budget calculation. Passive loss is made up of fiber loss, connector loss, splice loss, and losses involved
with couplers or splitters in the link.
Active components are system gain, wavelength, transmitter power, receiversensitivity, and dynamic range.
Imran Khan, 2013 TE 111
Regenerators & Optical Amplifiers
The maximum span of an optical signal is determined by the availablepower & the attenuation:
Ex. If 30 dB power available, then at 1550 nm, optical signal attenuates at 0.25 dB/km, so max span = 30 dB/0.25 km/dB = 120 km
Optical amplifiers amplify optical signal (no equalization, no regeneration) It is made by doping a length of fiber with the rare-earth
mineral erbium, and pumping it with light from a laser with a shorterwavelength than the communications signal (typically 980 nm).
Amplifiers have largely replaced repeaters in new installations. Impairments in optical amplification limit maximum number of opticalamplifiers in a path
Optical signal must be regenerated when this limit is reached Requires optical-to-electrical (O-to-E) signal conversion, equalization,
detection and retransmission (E-to-O)
Expensive Severe problem with WDM systems
Imran Khan, 2013 TE 112
http://en.wikipedia.org/wiki/Doping_(semiconductors)http://en.wikipedia.org/wiki/Erbiumhttp://en.wikipedia.org/wiki/Laser_pumpinghttp://en.wikipedia.org/wiki/Laserhttp://en.wikipedia.org/wiki/Nanometerhttp://en.wikipedia.org/wiki/Nanometerhttp://en.wikipedia.org/wiki/Laserhttp://en.wikipedia.org/wiki/Laser_pumpinghttp://en.wikipedia.org/wiki/Erbiumhttp://en.wikipedia.org/wiki/Doping_(semiconductors)7/29/2019 Telecom system and Networks LEcture notes
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RegeneratorR R R R R R R R
DWDMmultiplexer
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
Imran Khan, 2013 TE 113
R
R
R
R
Opticalamplifier
RR
R
R
OA OA OA OA
Optical Amplifiers
Optical amplifiers can amplify the composite DWDM signalwithout demuxing or O-to-E conversion
Erbium Doped Fiber Amplifiers (EDFAs) boost DWDM signalswithin 1530 to 1620 range
Spans between regeneration points >1000 km Number of regenerators can be reduced dramatically
Dramatic reduction in cost of long-distance communications
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Radio Transmission
Radio signals: antenna transmits sinusoidal signal(carrier) that radiates in air/space
Information embedded in carrier signal using modulation,e.g. QAM
Communications without tethering (connecting onedevice to another)
Cellular phones, satellite transmissions, Wireless LANs
Multipath propagation causes fading
Interference from other users
Spectrum regulated by national & international regulatoryorganizations
Imran Khan, 2013 TE 115
104 106 107 108 109 1010 1011 1012
Frequency (Hz)
Wavelength (meters)
103 102 101 1 10-1 10-2 10-3
105
Satellite and terrestrialmicrowave
AM radio
FM radio and TV
LF MF HF VHF UHF SHF EHF
104
Cellularand PCS
Wireless cable
Radio Spectrum
Omni-directional applications Point-to-Point applications
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Examples
Cellular Phone Allocated spectrum First generation:
800, 900 MHz Initially analog voice
Second generation: 1800-1900 MHz Digital voice, messaging
WirelessLAN Unlicenced ISM spectrum
Industrial, Scientific, Medical 902-928 MHz, 2.400-2.4835
GHz, 5.725-5.850 GHz
IEEE 802.11 LAN standard 11-54 Mbps
Point-to-Multipoint Systems Directional antennas atmicrowave frequencies
High-speed digitalcommunications between sites
High-speed Internet AccessRadio backbone links for ruralareas
SatelliteCommunications Geostationary satellite @ 36000
km above equator Relays microwave signals from
uplink frequency to downlinkfrequency
Long distance telephone Satellite TV broadcast
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Error Detect ion and Correct io n
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Error Control
Digital transmission systems introduce errors Applications require certain reliability level Data applications require error-free transfer Voice & video applications tolerate some errors
Error control used when transmission system does notmeet application requirement
Error control ensures a data stream is transmitted to acertain level of accuracy despite errors
Two basic approaches:
Errordetect ion& retransmission (ARQ) Forward errorcorrect ion(FEC)
Imran Khan, 2013 TE 119
Key Idea
All transmitted data blocks (codewords) satisfy apattern
If received block doesnt satisfy pattern, it is in error Redundancy: Only a subset of all possible blocks
can be codewords
Blindspot: when channel transforms a codewordinto another codeword
ChannelEncoderUserinformation
Patternchecking
All inputs to channelsatisfy pattern or condition
Channeloutput
Deliver userinformation orset error alarm
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Single Parity Check
Append an overall parity check to kinformation bits
Info Bits: b1, b2, b3, , bk
Check Bit: bk+1= b1+ b2+ b3+ + bk modulo 2
Codeword: (b1, b2, b3, , bk,, bk+1)
All codewords have even # of 1s
Receiver checks to see if # of 1s is even
All error patterns that change an odd # of bits aredetectable
All even-numbered patterns are undetectable
Parity bit used in ASCII code Imran Khan, 2013 TE 121
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 1s =5, odd Error detected
If errors in bits 3 and 5: (0, 1, 1, 1, 0, 0, 0, 1) # of 1s =4, even Error not detected
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Checkbits & Error Detection
Calculatecheck bits
Channel
Recalculatecheck bits
Compare
Information bits Received information bits
Sentcheckbits
Informationaccepted ifcheck bits
match
Receivedcheck bits
kbits
n kbits
Imran Khan, 2013 TE 123
How good is the single parity check code?
Redundancy: Single parity check code adds 1 redundantbit perkinformation bits: overhead = 1/(k+ 1)
Coverage: all error patterns with odd # of errors can bedetected An error patten is a binary (k+ 1)-tuple with 1s where errors
occur and 0s elsewhere Of 2k+1 binary (k+ 1)-tuples, are odd, so 50% of error
patterns can be detected
Is it possible to detect more errors if we add more checkbits?
Yes, with the right codes
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What if bit errors are random?
Many transmission channels introduce bit errors at random,independently of each other, and with probabilityp Some error patterns are more probable than others:
In any worthwhile channelp < 0.5, and sop/(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 error pattern
occurs?
P[10000000] =p(1p)7 and
P[11000000] =p2(1p)6
Imran Khan, 2013 TE 125
Single parity check code with random bit errors
Undetectable error pattern if even # of bit errors:
Example: Evaluate above forn = 32,p = 10-3
For this example, roughly 1 in 2000 error patterns isundetectable
P[error detection failure] = P[undetectable error pattern]= P[error patterns with even number of 1s]
= p2(1p)n-2 + p4(1p)n-4+ n
2
n
4
P[undetectable error] = (10-3)2 (1 10-3)30 + (10-3)4 (1 10-3)28
496 (10-6) + 35960 (10-12) 4.96 (10-4)
322
324
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x = codewords
o = noncodewords
x
x x
x
x
x
x
o
oo
oo
oo
o
oo
o
o
ox
x x
x
xx
x
o o
oo
oooo
o
o
oPoordistanceproperties
What is a good code?
Many channels havepreference for error patternsthat have fewer # of errors
These error patterns maptransmitted codeword tonearby n-tuple
If codewords close to eachother then detection failureswill occur
Good codes should maximizeseparation betweencodewords Gooddistance
properties
Imran Khan, 2013 TE 127
Two-Dimensional Parity Check
1 0 0 1 0 0
0 1 0 0 0 1
1 0 0 1 0 0
1 1 0 1 1 0
1 0 0 1 1 1
Bottom row consists ofcheck bit for each column
Last column consistsof check bits for eachrow
More parity bits to improve coverage Arrange information as columns Add single parity bit to each column Add a final parity column Used in early error control systems
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1 0 0 1 0 0
0 0 0 1 0 1
1 0 0 1 0 0
1 0 0 0 1 0
1 0 0 1 1 1
1 0 0 1 0 0
0 0 0 0 0 1
1 0 0 1 0 0
1 0 0 1 1 0
1 0 0 1 1 1
1 0 0 1 0 0
0 0 0 1 0 1
1 0 0 1 0 0
1 0 0 1 1 0
1 0 0 1 1 1
1 0 0 1 0 00 0 0 0 0 1
1 0 0 1 0 0
1 1 0 1 1 0
1 0 0 1 1 1
Arrows indicate failed check bits
Two errorsOne error
Three
errors Four errors(undetectable)
Error-detecting capability
1, 2, or 3 errors canalways be detected; Notall patterns >4 errorscan be detected
Imran Khan, 2013 TE 129
Other Error Detection Codes
Many applications require very low error rate
Need codes that detect the vast majority of errors
Single parity check codes do not detect enough errors
Two-dimensional codes require too many check bits
The following error detecting codes used in practice: Internet Check Sums
CRC Polynomial Codes
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Internet Checksum
Several Internet protocols (e.g. IP, TCP, UDP) use checkbits to detect errors in the IP header(or in the header anddata for TCP/UDP)
A checksum is calculated for header contents and includedin a special field.
Checksum recalculated at every router, so algorithmselected for ease of implementation in software
Let header consist ofL, 16-bit words,
b0, b1, b2, ..., bL-1
The algorithm appends a 16-bit checksum bL
Imran Khan, 2013 TE 131
The checksum bL is calculated as follows:
Treating each 16-bit word as an integer, find
x = b0 + b1 + b2+ ...+ bL-1 modulo 216-1
The checksum is then given by:
bL = - x modulo 216-1
Thus, the headers must satisfy the following pattern:
0 = b0 + b1 + b2+ ...+ bL-1 + bL modulo 216-1
The checksum calculation is carried out in softwareusing ones complement arithmetic
Checksum Calculation
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Internet Checksum Example
Use Modulo Arithmetic Assume 4-bit words
Use mod 24-1arithmetic
b0=1100 = 12
b1=1010 = 10
b0+b1=12+10=7 mod15
b2 = -7 = 8 mod15
Therefore
b2=1000
Use Binary Arithmetic Note 16 =1 mod15
So: 10000 = 0001 mod15
leading bit wraps around
b0 + b1 = 1100+1010=0110+1=0111 (1s complement)=7
Take 1s complementb2 = -0111 =1000
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134
Checksum Example
Simple example:
e.g., on the transmitter side the DATA= 10111110 Split into two four bit 1011 1110 Calculate the checksum by taking the 1s complement sum as:
1 1011
1110
--------------
1001
1-------------
1010
Take 1s complement of the results 10100101 (Checksum) On the receivers side: Add the checksum and the data by using
the 1s complement arithmetic and take the complement of theresult:
If result =0000, OK ; otherwise, there is an error.
Carry
bit
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Internet Checksum Algorithm
On the senders side: View message as a sequence of 16-bit integers
Sum using 16-bit ones-complement arithmetic
Take ones-complement of the result; its thechecksum.
On the receiver side:
Sum the data and the checksum using ones-complement arithmetic
Take the 1s complement of the result
If result is:000000000000000, the entire packet isOK
Otherwise, there is an error.
135 Imran Khan, 2013 TE
136
Error Detection or Correction ?
Detection implies discardingmessage and waiting forretransmission
Uses bandwidthIntroduces latency
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137
Error Detection or Correction ?
Error correction requires moreredundant bits to send al l thet ime: Forward Error-correctingCode (FEC)
Error correction is useful when: Errors are quite probable (wireless
links) Retransmission cost is too high(latency in satellite link, multicast)
Imran Khan, 2013 TE
Polynomial Codes
Polynomials instead of vectors for codewords
Polynomial arithmetic instead of check sums
Implemented using shift-register circuits
Also called cyclic redundancy check (CRC) codes
Most data communications standards use polynomial codesfor error detection
Polynomial codes also basis for powerful error-correctionmethods
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Addition:
Multiplication:
Binary Polynomial Arithmetic
Binary vectors map to polynomials(ik-1 ,ik-2 ,, i2 , i1 , i0) ik-1xk-1 + ik-2xk-2+ + i2x2 + i1x+ i0
(x7 +x6 + 1) + (x6 + x5) =x7 +x6 + x6 +x5 + 1
=x7 +(1+1)x6+x5 + 1
=x7 +x5 + 1 since 1+1=0 mod2
(x+ 1) (x2 + x + 1) =x(x2 +x + 1) + 1(x2 +x+ 1)
=x3 +x2+ x+ (x2+x+ 1)
=x3 + 1
Imran Khan, 2013 TE 139
Binary Polynomial Division
Division with Decimal Numbers
32
35 ) 1222
3
105
17 2
4
140divisor
quotient
remainder
dividend1222 = 34 x 35 + 32
dividend = quotient x divisor +remainder
Polynomial Division
x3 +x+ 1 )x6 +x5x6 + x4 +x3
x5 +x4 +x3
x5 + x3 +x2
x4 + x2
x4 + x2 +x
x
= q(x) quotient
= r(x) remainder
divisordividend
+x+x2x3
Note: Degree of r(x) is less than
degree of divisor
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Polynomial Coding
Code has binary generating polynomialof degree nk
k information bits define polynomial of degree k 1
Find remainder polynomialof at most degree nk 1
g(x) )xn-ki(x)
q(x)
r(x)
xn-ki(x) = q(x)g(x) + r(x)
Define the codeword polynomialof degree n 1b(x) = xn-ki(x) + r(x)
n bits kbits n-kbits
g(x) =xn-k+ gn-k-1xn-k-1+ + g2x2 + g1x+ 1
i(x) = ik-1xk-1 + ik-2x
k-2+ + i2x2 + i1x+ i0
Imran Khan, 2013 TE 141
Transmitted codeword:b(x) = x6+ x5+ x
b = (1,1,0,0,0,1,0)
1011 ) 11000001110
1011
1110
1011
1010
1011
010
x3 + x+ 1 ) x6+ x5
x3 + x2+ x
x6+ x4 + x3
x5+ x4 + x3
x5+ x3 + x2
x4 + x2
x4 + x2+ x
x
Polynomial example: k= 4, nk= 3
Generator polynomial: g(x)= x3 + x + 1
Information: (1,1,0,0) i(x) = x3 + x2
Encoding: x3i(x) = x6+ x5
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The Patternin Polynomial Coding
All codewords satisfy the following pattern:
All codewords are a multiple ofg(x)!
Receiver should divide received n-tuple by g(x) andcheck if remainder is zero
If remainder is nonzero, then received n-tuple is not a
codeword
b(x) = xn-ki(x) + r(x) = q(x)g(x) + r(x) + r(x) = q(x)g(x)
Imran Khan, 2013 TE 143
Shift-Register Implementation
1. Accept information bits ik-1,ik-2,,i2,i1,i0
2. Append nkzeros to information bits
3. Feed sequence to shift-register circuit that performspolynomial division
4. After n shifts, the shift register contains the remainder
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Clock Input Reg 0 Reg 1 Reg 2
0 - 0 0 0
1 1 = i3 1 0 0
2 1 = i2 1 1 0
3 0 = i1 0 1 1
4 0 = i0 1 1 1
5 0 1 0 1
6 0 1 0 07 0 0 1 0
Check bits: r0= 0 r1= 1 r2= 0
r(x) = x
Division Circuit
Reg 0 ++
Encoder forg(x) =x3 +x+ 1
Reg 1 Reg 2
0,0,0,i0,i1,i2,i3g0 = 1 g1 = 1 g3 = 1
Imran Khan, 2013 TE 145
Undetectable error patterns
e(x) has 1s in error locations & 0s elsewhere Receiver divides the received polynomial R(x) by g(x) Blindspot: Ife(x) is a multiple ofg(x), that is, e(x) is a
nonzero codeword, thenR(x) = b(x) + e(x) = q(x)g(x) + q(x)g(x)
The set of undetectable error polynomials is the set ofnonzero code polynomials
Choose the generator polynomial so that selectederror patterns can be detected.
b(x)
e(x)
R(x)=b(x)+e(x)+
(Receiver)(Transmitter)
Error polynomial(Channel)
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Designing good polynomial codes
Select generator polynomial so that likely errorpatterns are not multiples ofg(x)
Detecting Single Errors e(x) = xi for error in location i + 1
Ifg(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) is aprimitive polynomial, it cannot dividexm+1 for allm
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Standard polynomials C(x)
Name Polynomial Application
CRC-8 x8 +x2 +x+ 1 ATM header
CRC-10 x10+x9 +x5 +x4 +x2+ 1 ATM AAL
ITU-16 x16+x12+x5 + 1 HDLC
CRC-32x
32+x26 +x23+x22+x16+x12+x11+x10+
x8 +x7+x5+x4+x2+x+ 1LANs
Imran Khan, 2013 TE 149
Hamming Codes
Class oferror-correctingcodes
Capable of correcting all single-errorpatterns
For each m > 2, there is a Hamming code of lengthn = 2m 1 with nk= m parity check bits
m n = 2m1 k= nm m/n
3 7 4 3/7
4 15 11 4/15
5 31 26 5/31
6 63 57 6/63
Redundancy
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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
Imran Khan, 2013 TE 151
Hamming (7,4) code
Information Codeword Weight
b1 b2 b3 b4 b1 b2 b3 b4 b5 b6 b7 w(b)
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 1 1 1 1 4
0 0 1 0 0 0 1 0 1 0 1 3
0 0 1 1 0 0 1 1 0 1 0 3
0 1 0 0 0 1 0 0 0 1 1 3
0 1 0 1 0 1 0 1 1 0 0 3
0 1 1 0 0 1 1 0 1 1 0 4
0 1 1 1 0 1 1 1 0 0 1 4
1 0 0 0 1 0 0 0 1 1 0 3
1 0 0 1 1 0 0 1 0 0 1 3
1 0 1 0 1 0 1 0 0 1 1 4
1 0 1 1 1 0 1 1 1 0 0 4
1 1 0 0 1 1 0 0 1 0 1 4
1 1 0 1 1 1 0 1 0 1 0 4
1 1 1 0 1 1 1 0 0 0 0 3
1 1 1 1 1 1 1 1 1 1 1 7
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Parity Check Equations
Rearrange parity check equations:
All codewords must satisfythese 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 = Hbt = 0
0 = 0 1 1 1 0 0 1 b5
b6
b7
Imran Khan, 2013 TE 153
0010000
s = H e= =101
Single error detected
0100100
s = H e= = + =01
1
Double error detected10
0
1 0 1 1 1 0 01 1 0 1 0 1 00 1 1 1 0 0 1
1110000
s = H e= = + + = 0110
Triple error notdetected
011
101
1 0 1 1 1 0 01 1 0 1 0 1 0
0 1 1 1 0 0 1
1 0 1 1 1 0 01 1 0 1 0 1 00 1 1 1 0 0 1
11
1
Error Detection with Hamming Code
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Minimum distance of Hamming Code
Previous slide shows that undetectable error patternmust have 3 or more bits
At least 3 bits must be changed to convert one codewordinto another codeword
b1 b2o o
o
o
o oo
o
Set ofn-tupleswithindistance 1of b1
Set ofn-tuples withindistance 1 ofb2
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
Imran Khan, 2013 TE 155
General Hamming Codes
Form > 2, the Hamming code is obtained through the checkmatrix H:
Each nonzero m-tuple appears once as a column of H The resulting code corrects all single errors
For each value ofm, there is a polynomial code with g(x) ofdegree m that is equivalent to a Hamming code and correctsall single errors
Form = 3, g(x) = x3
+x+1
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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 decoderassumes 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
Imran Khan, 2013 TE 157
Mult ip lexing
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Circuit Switching Networks
End-to-end dedicated circuits between clients Client can be a person or equipment (router or switch) Circuit can take different forms
Dedicated path for the transfer of electrical current Dedicated time slots for transfer of voice samples Dedicated frames for transfer of Nx51.84 Mbps signals Dedicated wavelengths for transfer of optical signals
Circuit switching networks require: Multiplexing & switching of circuits Signaling & control for establishing circuits
These are the subjects covered in this chapter
Imran Khan, 2013 TE 159
(a) A switch provides the network to a cluster of users, e.g.,a telephone switch connects a local community
(b) A multiplexer connects two access networks, e.g., a highspeed line connects two switches
Accessnetwork
Network
How a network grows
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Metropolitan network Aviewed as Network A of
Access Subnetworks
National network viewedas Network of RegionalSubnetworks (including A)
A
National &International
Network of RegionalSubnetworks
(a)
(b)
A
Network ofAccessSubnetworks
dc
ba
A
Metropolitan
1*
a
c
b
d
2
34
A Network Keeps Growing
Very high-speed lines
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Multiplexing involves the sharing of a transmission channel(resource) by several connections or information flows
Channel = 1 wire, 1 optical fiber, or 1 frequency band
Significant economies of scale can be achieved by combiningmany signals into one
Fewer wires/pole; a fiber replaces thousands of cables
Implicit or explicit information is required to demultiplex theinformation flows.
Multiplexing
B B
C C
A A
B
C
A
B
C
A(a) (b)
MUX MUX
SharedChannel
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(b) Combinedsignal fits intochannelbandwidth
Frequency-Division Multiplexing
A channel divided into frequency slots
Guard bandsrequired
AM or FM radiostations
TV stations in air orcable
Analog telephonesystems
Cf
Bf
Af
Wu
Wu
0
0
0 Wu
A CBf
W0
(a) Individualsignals occupyWu Hz
Imran Khan, 2013 TE 163
(a) Each signaltransmits 1 unit
every 3T
seconds
(b)Combinedsignaltransmits 1
Time-Division Multiplexing
tA1 A2
3T0T 6T
tB1 B2
3T0T 6T
tC1 C2
3T0T 6T
B1 C1 A2 C2B2A1 t0T 1T 2T 3T 4T 5T 6T
High-speed digital channel divided into time slots
Framingrequired
Telephone
digitaltransmission
Digitaltransmission inbackbonenetwork
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T-Carrier System
Digital telephone system uses TDM. PCM voice channel is basic unit for TDM 1 channel = 8 bits/sample x 8000 samples/sec. = 64 kbps
T-1 carrier carries Digital Signal 1 (DS-1) thatcombines 24 voice channels into a digital stream:
Bit Rate = 8000 frames/sec. x (1 + 8 x 24) bits/frame= 1.544 Mbps
2
24
1 1
2
24
24 b1 2 . . .b2322
Frame
24.
.
.
.
.
.
MUX MUX
Framing bit
Imran Khan, 2013 TE 165
North American Digital Multiplexing
Hierarchy
DS0, 64 Kbps channel DS1, 1.544 Mbps channel DS2, 6.312 Mbps channel DS3, 44.736 Mbps channel DS4, 274.176 Mbps channel
1
24
1
4
1
7
1
6
.
.
.
.
.
.
.
.
Mux
Mux
Mux
Mux
DS1 signal, 1.544Mbps
DS2 signal, 6.312Mbps
DS3 signal, 44.736Mpbs
DS4 signal
274.176Mbps
24 DS04 DS1
7 DS2
6 DS3
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CCITT Digital Hierarchy
1
30
1
4
1
1
4
.
.
.
.
.
.
.
.
Mux
Mux
Mux
Mux
2.048 Mbps
8.448 Mbps
34.368 Mpbs
139.264 Mbps
64 Kbps
CCITT digital hierarchy based on 30 PCM channels
E1, 2.048 Mbps channel
E2, 8.448 Mbps channel E3, 34.368 Mbps channel
E4, 139.264 Mbps channel
Imran Khan, 2013 TE 167
Wavelength-Division Multiplexing
Optical fiber link carries several wavelengths From few (4-8) to many (64-160) wavelengths per fiber
Imagine prism combining different colors into single beam
Each wavelength carries a high-speed stream Each wavelength can carry different format signal
e.g., 1 Gbps, 2.5 Gbps, or 10 Gbps
1
2
m
OpticalMUX 1
2
m
OpticaldeMUX
1 2. m
Opticalfiber
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RS-232 Asynchronous Data Transmission
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Recommended Standard (RS) 232
Serial line interface between computer and modem or similardevice
Data Terminal Equipment (DTE): computer
Data Communications Equipment (DCE): modem
Mechanical and Electrical specification
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DTE DCE
Protective Ground (PGND)
Transmit Data (TXD)
Receive Data (RXD)
Request to Send (RTS)
Clear to Send (CTS)
Data Set Ready (DSR)
Ground (G)Carrier Detect (CD)
Data Terminal Ready (DTR)
Ring Indicator (RI)
1
2
3
4
5
6
7
8
20
22
1
2
3
4
5
6
7
8
20
22
(b)
13
(a)
1
2514
Pins in RS-232 connector
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Synchronization
Synchronization ofclocks in transmittersand receivers. clock drift causes a loss
of synchronization
Example: assume 1and 0 are representedby V volts and 0 voltsrespectively
Correct reception Incorrect reception due
to incorrect clock (slowerclock)
Clock
Data
S
T
1 0 1 1 0 1 0 0 1 0 0
Clock
Data
S
T
1 0 1 1 1 0 0 1 0 0 0
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Synchronization (contd)
Incorrect reception (faster clock) How to avoid a loss of synchronization?
Asynchronous transmission
Synchronous transmission
Clock
Data
S
T
1 0 1 1 1 0 0 1 0 0 0
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Asynchronous Transmission
Avoids synchronization loss by specifying a short maximumlength for the bit sequences and resetting the clock in thebeginning of each bit sequence.
Accuracy of the clock?
Startbit
Stopbit
1 2 3 4 5 6 7 8
Data bits
Lineidle
3T/2 T T T T T T T
Receiver samples the bits Imran Khan, 2013 TE 174
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Synchronous Transmission
Voltage
1 0 0 0 1 1 0 1 0
time
Sequence contains data + clock information (line coding) i.e. Manchester encoding, self-synchronizing codes, is used.
R transition forRbits per second transmission R transition contains a sine wave with RHz. RHz sine wave is used to synch receiver clock to the
transmitters clock using PLL (phase-lock loop)
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