etsf15 physical layer communication - eit.lth.se · pam (pulse amplitude modulation) § nrz and...
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Physicallayer
• Analogvsdigital(Previouslecture)
• Transmissionmedia
• Modulation
• Represent digitaldatainacontinuousworld
• Disturbances,Noiseanddistortion
• Information
2
TransmissionmediaGuidedmedia
• Fibre opticcable
• Twistedpaircoppercables
• Coaxcable
Unguidedmedia
• Radio
• Microwave
• Infrared
Fibreoptic
• Transmissionisdonebylightin
aglasscore(verythin)
• Totalreflectionfromcoreto
cladding
• Bundleseveralfibres inone
cable
• Notdisturbedbyradiosignals
❚4
Opticalnetworkarchitekture
Pointtopoint
• Twonodesareconnected byonededicated fibre
Pointtomulti-point
• Onepoint isconnected toseveralendnodes
• PON(PassiveOpticalNetwork)
❚5
Twistedpaircoppercables
Twocopperlinestwistedaroundeachother
• Twistingdecreasesdisturbances(andemission)
• Usedfor
• Telephony grid(CAT3)
• Ethernet (CAT5,CAT6andsometimes CAT7)
On-offkeying• SendonebitduringTb secondsandusetwo
signallevels,“on”and“off”,for1and0.
Ex.
t
s(t)
T 2T 3T
x=10010010101111100
A
0
a(t) = A ⋅ x 0 ≤ t ≤ Tb
10
Non-returntozero(NRZ)
• SendonebitduringTb secondsandusetwo
signallevels,+A and-A,for0and1.
Ex.
t
s(t)
T 2T 3T
x=10010010101111100
A
-A
a(t) = A ⋅(−1)x 0 ≤ t ≤ Tb
11
DescriptionofgeneralsignalWiththepulseform! " = $, 0 ≤ " < )*,thesignalscanbedescribedas
, " =- ./!(" − 2)*)/
12
Twosignalalternatives
On-offkeying
./ = 4/ ⇒ ,6 " = 0 and,7 " = !(")NRZ
./ = −1 9: ⇒ ,6 " = ! " and,7 " = −!(")
Manchestercoding
• Togetazeropassingineachsignaltime,splitthe
pulseshapeg(t) intwopartsanduse+/- as
amplitude.
Ex.
T
g(t)
t
A
T/2
-A
13
DifferentialManchestercoding§ Useazerotransition atthestarttoindicatethedata.
§ Foratransmitted 0thesamepulseasprevious slotis
used,while foratransmitted 1theinvertedpulse is
used,i.e../ = ./;7 −1 9:
14
PAM (Pulse Amplitude Modulation)
§ NRZandManchesterareformsofbinaryPAM
§ Thedataisstoredintheamplitudeand
transmittedwithapulseshapeg(t)
§ Graphicalrepresentation
15
a(t) = an ⋅g(t) an = (−1)x
M-PAM
§UseM amplitudelevelstorepresentk=log2(M)bits
t
s(t)
T 2T 3T
x=10 01 00 10 10 11 11 10 00
A
-A
3A
-3A
01:
10:
00:
11:
16
§Ex.Twobitspersignal(4-PAM)
Bandwidth of signal
§ Thebandwidth,W,isthe(positiveside)frequency
bandoccupiedbythesignal
§ Sofaronlybase-bandsignals(centeredaroundf=0)
18
Pass-band signal
§ Frequencymodulatethesignaltoacarrier
frequency<6
19
§ Thefollowingmultiplicationcentersthesignal
aroundthecarrierfrequency<6, " = . " = cos 2B<6"where .(") isabase-bandsignal
Modulationinfrequency⇥ =
# F
s(t)
t
�3A
�1A
1A
3A
cos(2⇡f0t)
t
sf0(t)
t
�3A
�1A
1A
3A
⇤ =S(f)
f
12 (�(f + f0) + �(f � f0))
f�f0 f0
12 (S(f + f0) + S(f � f0))
f�f0 f0
20
BPSK(BinaryPhaseShiftKeying)
§ UseNRZatfrequencyf0,butviewinformationin
phase
s(t) = (−1)xn g(t − nT )cos(2π f0t) = g(t − nT )cos(2π f0t + xnπ )n∑
n∑
t
s(t)
T 2T 3T
x=10010010101111100
A
-A22
M-QAM(QuadratureAmplitudeModulation)Usethatcos(2B<6") andsin(2B<6") areorthogonal(forhigh<0)tocombinetwoorthogonalM-PAM
constellations
g(t)cos
g(t)sin
g(t)cos
g(t)sin
23
OFDMOrthogonalFrequencyDivisionMultiplexing
§ NQAMsignalscombinedinanorthogonalmanner
§ Usedine.g.ADSL,VDSL,WiFi,DVB-C&T&H,LTE,etc
f
W
24
Idea of OFDM implementation
25 (x1,..., xN )∈!
N (y1,..., yN ) = IFFT (
!x)
QAM
QAM
QAM
(a1,...,aN )∈Z16N
QAMmapping
IFFT
Frequencydomain Timedomain
Some important parameters
§ Ts timepersymbol
§ Rs symbolpersecond
§ Es energypersymbol
26
k=bitpersymbol
§ Tb=Ts /k timeperbit
§ Rb=kRs bitpersecond[bps]
§ Eb=Es /k energyperbit
§ SNR,Signaltonoiseratio
♦averagesignalpowerrelativetonoisepower
§ W Bandwidth,frequencybandoccupiedbysignal
§ Bandwidthutilisation:bitspersecondperHz[bps/Hz]
ρ = RbW
Impairments on the communication channel (link)§ Attenuation
§ Multipathpropagation(fading)
§ Noise
27
y(t) = x(t)∗h(t)+ n(t)
Noisedisturbances• Thermalnoise(Johnson-Nyquist)
• Generated bycurrentinaconductor
• -174dBm/Hz(=3.98*10-18 mW/Hz)
• Impulsenoise(Oftenusergenerated,e.g.electricalswitches)
• Intermodulation noise(Fromothersystems)
• Cross-talk (Users inthesamesystem)
• Backgroundnoise(Misc disturbances)
https://en.wikipedia.org/wiki/Johnson-Nyquist_noise
28
Some Information TheoryEntropy
§ Discretecase:X discreterandomvariable
Entropyisuncertaintyofoutcome (fordiscrete case)
§ Continuouscase:X continuousrandomvariable
29
H (X) = E[− log2 p(X)]= − p(x)log2 p(x)x∑
H (X) = E[− log2 f (X)]= − f (x)log2 f (x)R∫ dx
Example Entropy
LetX beabinaryrandom
variablewith
P(X=0)=p
P(X=1)=1-p.
Thebinaryentropy
functionis
30
h(p) = − p log2 p − (1− p)log2(1− p)
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3.2. Entropy 45
p
h(p)
1
1/2 1
Figure 3.1: The Binary entropy function.
The definition of the entropy is also valid for vectorised random variables, suchas (X, Y) with the joint probability function p(x, y).
DEFINITION 3.5 The joint entropy for a pair of random variables with the jointdistribution p(x, y) is
H(X, Y) = EXY⇥
� log p(X, Y)⇤
= �Âx,y
p(x, y) log p(x, y) (3.23)
⇤
Similarly, in the general case with an n-dimensional vector X = (X1, . . . , Xn),the joint entropy function is
H(X1, . . . , Xn) = EX
⇥
� log p(X)⇤
= �Âx
p(x) log p(x) (3.24)
EXAMPLE 3.6 Let X and Y be the outcomes from two independent fair dice.Then the joint probability is P(X, Y = x, y) = 1/36 and the joint entropy
H(X, Y) = �Âx,y
136
log1
36= log 36 = 2 log 6 ⇡ 5.1699 (3.25)
Clearly, the uncertainty of the outcome of two dice is twice the uncertainty ofone die.
Let Z the sum of the dice, Z = X + Y. The probabilities are shown in thefollowing table
Z 2 3 4 5 6 7 8 9 10 11 12P(Z) 1
362
363
36436
536
636
536
436
336
236
136
Compression
Theentropysetsalimitonthecompressionratio
§ ConsiderasourceforX withN symbolsandthe
distributionP(N).Inaverageasymbolmustbe
representedbyatleastH(P) bits.
§ Wellknowncompressionalgorithmsarezip,gz,
png,Huffman
§ Lossy compressione.g.jpeg andmpeg
31
Some more Information TheoryMutual information
§ LetX andY betworandomvariables
§ TheinformationaboutX byobservingY isgiven
by
32
I(X;Y ) = E log2P(X,Y )P(X)P(Y )
⎡⎣⎢
⎤⎦⎥
§ Thisgives
I(X;Y ) = H (X)+ H (Y )− H (X,Y )
Example Mutual Information
P(X,Y) Y=0 Y=1X=0 0 3/4X=1 1/8 1/8
33
TherandomvariablesX andY hasthejointdistribution
That givesandand
P(X = 0) = 3 / 4 P(X = 1) = 1/ 4P(Y = 0) = 1/ 8 P(Y = 1) = 7 / 8
Entropies: H (X) = h( 14 ) = 0.8114H (Y ) = h( 18 ) = 0.5436H (X,Y ) = − 3
4 log 34 − 1
8 log 18 − 1
8 log 18 = 1.0613
Information: I(X;Y ) = H (X)+ H (Y )− H (X,Y ) = 0.2936
Some Information TheoryChannel capacity
§ Thechannelisamodelofthetransmissionlink.
§ TransmitX andreceiveY.Howmuchinformation
canthereceivergetfromthetransmitter?
§ Thechannelcapacity isdefinedas
34
C = maxp(x )
I(X;Y )
AWGNAdditive White Gaussian Noise channel
§ LetX bebandlimited inbandwidthW
§ ,where
§ Thecapacityis
[bps]
§ whereP isthepowerofX, i.e.E[X2]=P.
§ Itisnotpossibletogethigherdatarateonthis
channel!
35
N ∼ N 0, N0 / 2( )Y = X + N
C =W log2 1+P
N0W⎛⎝⎜
⎞⎠⎟
AWGN Example (VDSL)
§ Considerachannelwith
E = 17GHIJK = −60MNO/HIQ6 = −145MNO/HI
§ PowermW
§ NoisemW/Hz
§ CapacityMbps
36
P = 10−60/10 ⋅17 ⋅106
N0 = 10−145/10
C =W log 1+ PN0W( ) =W log 1+ 10−60/10
10−145/10( ) = 480
Shannon’s fundamental limit
§ PlotcapacityvsW
§ Istherealimit?
37
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9.3. Fundamental Shannon limit 253
W
C
Figure 9.7: Capacity as a function of the bandwidth W for a fixed total signalpower P.
where Es is the average energy per transmitted symbol, Eb the average energyper information bit and Tb = Ts/k the average transmission time for each infor-mation bit. The variable Eb is very important since it can be compared betweendifferent systems, without having the same number of bits per symbol or eventhe same coding rate. Hence, a system independent signal to noise ratio can bederived as SNR = Eb/N0. From (9.53) the energy per information bit can bewritten as
Eb =PTs
k(9.54)
Since the achieved bit rate can be expressed as Rb = kTs
, the ration between C•and the bit rate Rb can be written
C•Rb
=P/N0ln 2
Tsk
=Eb/N0
ln 2> 1 (9.55)
where it is used that reliable communication requires C• > Rb. Rewriting theabove, concludes that for reliable communication
EbN0
> ln 2 = 0.69 = �1.59 dB (9.56)
The value �1.6 dB is a well known bound in communication theory, and isoften referred to as the Shannon limit or the fundamental limit. It constitutea hard limit for when it is possible to achieve reliable communication. If theSNR is less than this limit it is not possible to reach error probability that tendsto zero, independent of what system is used. To see the demands this limitputs on the system, the capacity formula will be rewritten to show a limit onthe bandwidth efficiency Rb/W in the used system. Often the bandwidth is alimited resource and then to reach a high data rate the system should have ahigh bandwidth efficiency.
§ LetW->∞
§ WithEb=PTb andTs=kTb
C∞ = limW→∞
W log 1+ P/N0W( )
= limW→∞
log 1+ P/N0W( )W = logeP/N0 = P / N0
ln2
C∞
Rb= Eb / N0
ln2>1
Eb
N0
> ln2 = −1.59dB
§ Whichgivesthefundamental
limit
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