Telecom system and Networks LEcture notes

7/29/2019 Telecom system and Networks LEcture notes

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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

E-mail

Telephone

TV

Imran Khan, 2013 TE 2


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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, ?


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


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


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


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


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


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


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.


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


Transmission Impairments

Communication Channel

Pair of copper wires

Coaxial cable Radio

Light in optical fiber

Light in air

Infrared


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. . .


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. . .


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


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


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)


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


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


Quantizer Performance


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


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


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?


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


Signal attenuation Signal distortion Spurious noise Interference from other

signals Limits accuracy of

measurements on receivedsignal

Transmitted

Signal

Received

Signal Receiver


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


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)


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


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


-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


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


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


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)>]


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


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



36/88

Line Coding


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?



37/88

Line coding examples

NRZ-inverted(differentialencoding)

1 0 1 0 1 1 0 01

UnipolarNRZ

Bipolarencoding

Manchester

encoding

DifferentialManchester

encoding

Polar NRZ


-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



38/88

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


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



39/88

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


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



40/88

Modems and Dig i ta l Modulat ion


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



41/88

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-


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



42/88

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


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


43/88

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)}


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


44/88

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


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


45/88

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)


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))



46/88

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


Propert ies of Media and Digi talTransm iss ion Systems



47/88

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


dmeters


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



48/88

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


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



49/88

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


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



50/88

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


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



51/88

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



52/88

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


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


53/88

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


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



54/88

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)


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



55/88

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


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



56/88

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.


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

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)


57/88

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


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



58/88

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


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



59/88

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


Error Detect ion and Correct io n



60/88

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)


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



61/88

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



62/88

Checkbits & Error Detection

Calculatecheck bits

Channel

Recalculatecheck bits

Compare

Information bits Received information bits

Sentcheckbits

Informationaccepted ifcheck bits

match

Receivedcheck bits

kbits

n kbits


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



63/88

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


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



64/88

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


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



65/88

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


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



66/88

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


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



67/88

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


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

Imran Khan, 2013 TE


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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

Imran Khan, 2013 TE


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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


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


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)


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


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)



74/88

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


75/88

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


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


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



77/88

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


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



78/88

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


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



79/88

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


Mult ip lexing



80/88

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


(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



81/88

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


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



82/88

(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


(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



83/88

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


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



84/88

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


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



85/88

RS-232 Asynchronous Data Transmission


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



86/88

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


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



87/88

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


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


88/88

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)


Documents

Telecom system and Networks LEcture notes