Communication Theory/2 I. Frigyes 2009-10/II.. Frigyes: Hírkelm2 frigyes/hirkelm frigyes/hirkelm...

Communication Theory/2

I. Frigyes

2009-10/II.

Frigyes: Hírkelm 2

http://docs.mht.bme.hu/~frigyes/hirkelmhirkelm01bEnglish

http://docs.mht.bme.hu/~frigyes/hirkelm




2. Transmission of digital signals over analog channels: effect of

noise

Frigyes: Hírkelm 4

Introductory comments

• Theory of digital transmission is (at least partly) application of decision theory

• Definition of digital signals/transmission:• Finite number of signal shapes (M)• Each has the same finite duration (T)• The receiver knows (a priori) the

signal shapes (they are stored)• So the task of the receiver is hypothesis

testing.

Frigyes: Hírkelm 5


• Quality parameter: error probability• (I.e. the costs are:• )• Erroneous decision may be caused by:• additíve noise• linear distortion• nonlinear distortion• additive interference (CCI, ACI)• false knowledge of a

parameter• e.g. synchronizing error

MkiCC kiii ,...,2,1, ;1 ;0

Frigyes: Hírkelm 6

Introductory comments – degrading effects causing false decision

DECISIONMAKER

BANDPASSFILTER

BANDPASSFILTER

FADINGCHANNEL +

n(t)

NONLINEARAMPLIFIER

s(t)

INTER-FERENCE

INTER-FERENCE

INTER-FERENCE

ωc

ωcz0(t)

z1(t)

z2(t)

ω1

ω2

CCI

ACI

ACI

Frigyes: Hírkelm 7


• Often it is not one signal of which the error probability is of interest but of a group of signals – e.g. of a frame.

• (A second quality parameter: erroneous recognition of T : the jitter.:

• )

T

Tˆ

pp

TTtJ ˆ

Frigyes: Hírkelm 8

Introductory comments – improvement of performance parameters

Both performance parameters can be improved

Error probability:

ÁTVITELI CSATORNAFORRÁS KÓDOLÓ

DE-KÓDOLÓ NYELŐ

PE

PE,dec

Transmission ChannalSOURCE ENCODER

DE-CODER SINK

JitterTransmission

Channal

data

clock

ELASTICSTORE

JITTERFREE

CLOCK

data

clock

DIGITALSOURCE

Transmission Channal

SINK

Frigyes: Hírkelm 9

Introductory comments – degrading effects causing false decision

• Comments:• 1. These effects canot be described by the

two performance parameters. The channel, at this level is an analog channel producing the effects seen in slide #6

• 2. Behavior of radio and optical channels are rather different. First we deal with the first and then show differences in the second

Frigyes: Hírkelm 10

0.Transmission of single signals in additive Gaussian noise

• Among the many sources of error now we regard only this one

• Model to be investigated:

SOURCESIGNAL

GENERATOR +DECISION

MAKERSINK

TIMING (T)

n(t)

mi

{mi}, Pi

si(t) r(t)= si(t)+n(t)

ˆm



• Specifications:• Pi a-priori probabilities are known• Support of the real time functions • is (0,T)• their energy is finite (E: square

integral of the time functions)

• mutual unique relationship (i.e.: their is no error in the transmitter)

tsi

tsm ii

T

ii dttsE0

2ˆ



• Noise: Gaussian

• 0-mean

• stationary

• additive it’s drown so

• white

• Comment.: in white noise: σn=

2

2

1

2

1

n

n

n

n ep

const;0 nn

2

ˆ2

00 NTFkS B

n



• Com.: white is an approximation More exact: Planck-formule:

• If hf/kBT0<<1:

• If hf/kBT0>>1:

• f =300 GHz,T0=30K:FkBT0 -0,1dB

• f=200 THz,T0=270K:FkBT0-127 dB

10

TBkhfn

e

hfFS

00 11

TFkTkhf

hfFS B

Bn

00 Tkhfn

BehfS



• Decision: based on r(t)=si(t)+n(t).• Application of the general method:

independent samples would result in too high noise; correlated samples yield less information;

• they don’t specify bandwidth.• Contionouos investigation and appropriate

processing of signals yield most information. This is the subject of our next investigations.



• Questions to be looked for:• 0. Vectorial representation of digital

signals• 1. The optimal receiver• 2. Error probability• 3. Coherent – non-coherent• 4. Optimal signal set• 5. Bandwidth occupation



• Given the – somehow chosen – signal set

• We chose an orthonormal base:

• (ortonormal:• So that

• Of course

Mitsi ...1;

MDDjtj ;...1);(

D

lllii Mitats

1, ...1;

T

jiij dtttsa0

T

jllj lj

ljdttt

0;0

;1.



• Thus: time functions are uniquely representet by D numbers (ai,1, ai,2 … ai,D)

• But: any structer represented by D numbers can be regarded as a vector of D dimensions

• I.e.• So we defined a vector space: signal

space• D is the dimensionality of the signal space

iDiiT

i aaats ...,, 21 is


0. Comment

• As said: DM • Earlier we saw: in the general case

dimensionality of the decision space is D=M-1.In the case of concretly defined signal waveforms (like now) decision can be made in the signal space; then D<M-1 is possible.

• (We had also the observation space, with D=N. In the case of continouos obser-vation D=∞, is not too important


0. How to chose the base?

1.1

21

212

1

1

11

dts

dtsdt

E

tst

0.

ˆ.2

122

1121

21122

22

121212122

dtsdtdtsE

dtEdth

tht

dtttsttstatsth

h

h

0

0

;ˆ.3

232

213212332

2321132

11331

23

3323213133

dtsdtdtsdtdtsdt

dtsdtdtsdtdtsdt

dtth

thttatatsth


0. How to chose the base?

• This can be done as long as we have signals (Gram-Schmidt ortogonalization)

• We see: there areM base functions at most. • But if some signal waveforms are linear

combinations of others these don’t introduce new dimensions

• E.g. dimensionality of M-ary PAM signal set is 1, of QAM signal set it is 2.


0. Scalar product

• Schalar product of two signal-space-vectors is the integral of their product:

• By the way from that: |si|2 = Ei

D

jkjij

D

j

D

l

T

ljklij

T T D

j

D

llkljijki

aadtttaa

dttatadttsts

11 1 0

0 0 1 1

kiss


0. Single signals: vectorial form of noise

• After the signal: noise should also be given in vectorial form.

• Of course: compo-nents of the noise vector can be written

• And by that: the noise vector

• But it is not true for the noise process that

T

jj dtttnn0

DT nnn ..., 21n

D

jjj tntn

1


0. Single signals: vectorial form of noise

• (A Gaussian process can not be linear combination of a finite number of functions.)

• So• We know that is orthogonal to the signal

space; as the signal is in the signal space, an efficient receiver can filter out this part of the noise – it is thus irrelevant from the point of view of reception. I.e. n contains that part of the noise what is relevant. (We’ll briefly come back to that.)

D

jjj tntntn

1

`

tn


0. Single signals: vectorial form of the link

• Thus we can investigate the vectorial model of this connection

SOURCESIGNALVECTOR

GEN.+ DECISION SINK

n

mi

{mi}, Pi

si r= si+n

ˆm


0. Single signals: vectorial representation

• Pdf of the noise in the signal space: σ2 was doubtful: of white noise is infinite.

• Without details: in the interval [0,T] Gs noise can be expanded according to any complete orthogonal series.

• Individual terms are independent and have equal σ2.

• Base of the signal space is part of such a complete base (we are interested only in that part).


0. Single signals: vectorial representation

• Thus pdf-s can be written:

0

2

20

1 ND

eN

p

n

n n

0

2

20

|

1| N

i

Diis eN

p

sr

r sr


0.The signal space

D

jjjii Mitats

1, ...1;

T

jiij dtttsa0

iDiiT

i aaats ...,, 21 is


0.The signal space - examples

• 1. (Antipodal) baseband NRZ signals:

s1(t) A=(E/T)1/2T

s2(t)=-s1(T)

φ(t)1/T1/2T

M=2D=1

s2 s1



• 2. BPSK signals

tAts

tAts

c

c

cos2

cos2

2

1

M=2D=2

Φ s1

s2

If Φ=π: antipodalD=1

tT

t

tT

t

c

c

sin2

cos2

2

1



• 3. QPSK signals tststAts

tststAts

c

c

242

131

;sin2

;cos2

M=4D=2

tT

t

tT

t

c

c

sin2

cos2

2

1

s1

s2

s3

s4



• 4. Ortogonal QFSK signals

3,2,1,0;2

;cos2

iT

titAts ci

M=4D=4

ttT

t ci cos2

s1

s2

s3

s4



• 5 Biortogonal signals

tststAts

tststAts

c

c

242

131

;cos2

;cos2

M=4D=2

tT

t

tT

t

c

c

cos2

cos2

2

1

s1

s2

s3

s4Note: just like QPSK



• 6. MQAM jelek

MiM

iqa

tAqtAats

ii

cjcii

....2,1;1

12,

sincos

MD=2

tT

t

tT

t

c

c

sin2

cos2

1

1

Example: M=16


1. Single signals: the optimal decision rule

• Decision rule now: a jeltér optimal partitioning of the signal space (resulting in minimal error probability)

• Pl: D=2s1

s2

s3 s4

s5

n

r

Before we had to partition the decision space

nr



• We’ve seen: risk is minimal if the a-posteriori probability is maximl. We the decide on what is the most likely, i.e.

• To proceed apply Bayes theorem:

Miii ,...2,1;max|Pr rs



• Logarithm: of the a-posteriory pdf:

• Finally

MiPN

NN

DP

i

i

,...2,1max;2ln

thus;ln2

ln

0

0

2

0

2ii

2

i

srsr

sr

MiEPNH iii ,...2,1max;2

1ln

2

1:ˆ

0 irs



• For an instant come back to the noise vector• We’ve seen:• • Details of the decision noise, taking the whole noise into

considerationl

• I.e. ‘n(t) in the optimal receiver is really irrelevant

tntntntnD

jjj `""`

1

n

0

0

0 10 1

`

`

T

i

T D

jjji

T

i

D

jjji

dttnts

dttntsdttntstnts irs


1. Optimal decider – vectorial form

× +

s1

½(N0lnP1-E1)

× +

s2

½(N0lnP2-E2)

× +

sM

½(N0lnPM-EM)

rCOMPARATOR

max

If E-s are equal itcan be omittedfrom the bias. Ifin addition Pi=1/M,the whole bias can be omitted.


1. Optimal decider (correlation)

T

dt0

×

s1(t)

+

½(N0lnP1-E1)

r(t)

COMPARATOR

max×

s2(t)

+

½(N0lnP2-E2)

T

dt0

×

sM(t)

+

½(N0lnPM-EM)

T

dt0

Sense of scalar productis known

Question: are all elements of the model

needed?


SOURCESIGNAL

GENERTOR + DECISION SINK

Timing (T)

n(t)

mi

{mi}, Pi

M


ˆm

s(t) known



• Comment: if Pi≡1/M (equal a-priori prob.)

• I.e. we have to decide on which is closest

MiN

H

MiN

H

i

i

,...2,1min;:ˆ

,...2,1max;:ˆ

0

2

0

2

i

i

sr

sr


1. Optimal decider (matched filter)

• Correlation is a linear operation (multiplication by a signal independent of r(t)and integration).

• But: a linear operation can also be done with a linear filter thus an equivalent filter can also be found – its impulse response is h(t).

Tti

Tt

i

i

T

i

tTstrdtTsr

dsrdttstr

0

h(t)=si(T-t)↓

It is causal!


1. Optimal decider (matched filter)

r(t)

COMPARATOR

max

+

½(N0lnP1-E1)

s1(T-t)

+

½(N0lnP2-E2)

s2(T-t)

+

½(N0lnPM-EM)

sM(T-t)

t=T


1.Some of the previous examples together with decision boundaries

• 1. (Antipodal) NRZ baseband signals

s2(t)=-s1(T)

s1(t) AT

φ(t)1/T1/2T

M=2D=1

s1 s2



• MQAM signals

MjiM

iqa

tAqtAats

ji

cjcii

....2,1,;1

12,

sincos

MD=2

tT

t

tT

t

c

c

sin2

cos2

1

1



• 3. QPSK tststAts

tststAts

c

c

242

131

;sin2

;cos2

M=4D=2

tT

t

tT

t

c

c

sin2

cos2

2

1

s1

s2

s3

s4


1. Optimal reception in the optical band

• We’ve seen that there is no termal noise in the optical band.

• On the other hand there is shot noise. (We’ve seen – without refering to optics – the effect of Poisson noise.)

• To some detail later.


2. Error probability in the optimal detector

• Based on the precedings: conditional probability of correct decision (condition: si is transmitted):

• Total probability of correct decision:

• And the error probability

iV

iC dvpP iisrsrs ||

iV

M

iiC dvpPP iisr

sr ||1

CE PP 1



• If the a-priori probabilities are equal

• Or, if the constellation is in addition symmetric

M

i iVC dvp

MP

1| |

1iisrsr

1

1| |V

C dvpP srisr



• Important comment: we see: error probability depends only on the vectorial constellation and it does not depend on the signal wave forms.

• Thus e.g. antipodal NRZ and antipodal PSK, QPSK and biortogonal FSK yield identical error probability

• Thus choice of wave forms can be based on different criteria.


2. Example: antipodal signal set

• PE can be computed directly (PC is not needed):

• And: it is symmetricEd

0

0

20

0

erfc2

11

N

Edre

NP N

Er

E


2. Example: QPSK (and biortogonal) signal set

2Ed

0

2

0 2erfc

4

1

2erfc

N

E

N

EPE

2

EPP antipod

EQPSK

E

You see: the signal space is rotatedby 450; this doesn’t change the situation

How did we get it?


2. Example: QAM signal set

There are 3 different signalvectors, also if M>16).After much calculation(nearly exact formula):

12

;erfc/1120

M

Ed

N

dMP

peak

E

d


2. Example for preparing a bound : general binary signal set

2212

211222

1

021

211

111

;

;;

aE

attstts

E

dttsts

aE

tstts

T

General binary signal set:

d1+d2

s1

s2

φ1

φ2

ψ2

ψ1

s2-s1



• The original (φ1,φ2) coordinate system can be transformed into (ψ1,ψ2)-be. We see that, decision result depends only on co-ordinate ψ2 of the transformed system. (r.s) – what is a (scalar) Gauss random variable. So it is a 1D situation

• This is a slightly modified version of (H0: noise, H1: signal+noise) case: H0:r=s1+n; H1:r=s2+n

• Thus the optimal decision:


2. Példa, korlát előkészítésére: általános bináris jelkészlet

In the new co-ordinate system

2:ˆ

2:ˆ

1222

1221

ssrss

ssrss

Comments: we assumed Pi=0,5; if notη must also be taken into account.Vector notation was not used as now s is 1D

d1+d2

s1

s2

φ1

φ2

ψ2

s2-s1

ψ1

Ed



• But so: the role of di2 is the same as of E in the

antipodal signal set. Thus error probability:

• Application to M-ary signal set:

0

erfc2

1

N

dPE

dmin


2. Error probability in the optimal detector: a bound

• From that we get a bound for the M-ary signal set:

• As Gaussian pdf changes very fast false decision between closest is of decisive proportion. I.e. error rate is nearly equal to an antipodal case with energy :

• (Miért <?)

0

minerfc2

1

N

dPE

2dE


3. Special case: carrier wave transmission

tAtv ccos2• We know: a sinusoid

has 3 parameters:• It is fully known if we

know all:• A: AGC

• ωc: cristall control

• Φ: ?

Φ can be transmitted on aseparate route: coherent. Or we accept that it is a(uniformly distributed) random variable: non-coherent


3. Special case: carrier transmission, coherent case

SOURCESIGNAL

GEN. + DECIDER SINK

Timing (T)

n(t)

mi

{mi}, Pi


mˆ

Phase (Φ)


3. Special case: carrier wave transmission – coherent system• In this case everything is valid. Optimal detector

can be simplified

T

dt0

×

cos ωct

×

1

cos ωct0 KOMP

BPSK


3. Special case: carrier wave transmission – coherent system

QPSK

T

dt0

×

cos ωct

×

1

cos ωct0 COMP

T

dt0

×

sin ωct

×

1

sin ωct0 COMP

+ LOGICS



64QAM

T

dt0

×

cos ωct

×

1/7, 3/7, 5/7, 1

cos ωctQUANTIZER

T

dt0

×

sin ωct

×

1/7, 3/7, 5/7, 1

sin ωctQUANTIZER

+ LOGICS



BFSK – orthogonal

T

dt0

×

cos ωc1t0 COMP

T

dt0

×

cos ωc2t

1

VCO +

+

-


3/a A few words on noncoherent detection

• First: in this case phase cannot contain information. Further

• Phase is unknown. The criterion that the receiver knows the signals must be eased: it knows it, except one parameter.

• According to precedings, PE is minimal if the a-posteriori probability is maximal. Again it will bw assumed that Pi =1/M.

• Real signal-waveform #i is



sc

iciici

iciir

icii

ss

tttattta

tttats

tttats

sinsincoscos

cos

then waveformreceived the;cos

• Pdf of r (to be maximized) first in conditional form:

sincos21

exp1

sincos1

exp1

,|,

0

0

0

22

00

sci

2sr

2cr

isicisc

rrs

srsrsrr

Ne

N

NNp

NiE

D

D



• We assume that Φ is of uniform distribution.• The total pdf is then

• Omitting terms not depending on si we have

• This integral is known:

dpp

2

0

,|,2

1|, iscisc srrsrr

0

22

0

2

0 0

0

22I

sincos21

exp2

1|,

N

dN

ep NiE

sici

sciisc

rsrs

rrssrr



• Simplification possibilities: if signals have equal energy multiplier e-Ei/N0 can be omitted.

• Further in this case: as I0 is monotonously increasing növekszik, we have to maximize its argument; thus for that case the optimal noncoherent decision rule is:

Mi

Mi

s

s

...,2,1max; or

...,2,1max;

22

22

iic

iic

srsr

srsr


3/a A few words on noncoherent detection: optimal detector

• This is formed by an envelope detector. Thus an optimal noncoherent system

r(t)

COMPARATOR

max

s1(T-t)

s2(T-t)

sM(T-t)

t=T

ENVELOPEDETECTOR

ENVELOPEDETECTOR

ENVELOPEDETECTOR


3/a A few words on noncoherent detection: error probability

• Only only for binary orthogonal signal set :

• Note: it does not differ much from orthogonal:

02

2

1 NEE eP

000

2exp2

1~

2erfc

2

1NE

NEN

EPE


3/a A few words on noncoherent detection; example: FSK

BFSK – orthogonal

0 COMP

1

VCO +

+

-

ENVELOPEDETECTOR

ENVELOPEDETECTOR

BANDPASSω1

BANDPASSω2



• Then (find out): • Why for coherent?• (In practice it is nearly exclusively used.)

000

2exp2

1~

2erfc

2

1NE

NEN

EPE


4. Optimizing the signal set

• Up to now signal set was assumed to be known. The receiver was optimized, analyzed based on that knowledge, etc.

• Of course: the transmitter is also designed by ourselves so the signal set must also be optimized

• As we know that performance depends only on the vector constellation, that must be optimized



• We’ve seen that error performance improves if the signals are more fare away. Thus signal vectors must be chosen according to that: signals should be far away from each other.

• One way is to increase energy – but this is no true optimizing .

• Thus optimization criterion: • signal vectors should be as far as possible• but none of them should be of energy

higher than a specified Emax.



• A more precise drafting in the signal space:

• Put M points on the surface or in the interior of a sphere of dimensionality DM and radius r so, that their distance should be as large as possible; (r2 is the prescribed maximal energ). (Sphere packaging problem.)



• Solution of this problem of variation calcule is the regular simplex signal set:

• Characteristics are:

• In practice only the case M=2, s2=-s1 is of significance

1

1

max

max2

M

E

E

MD

ki

i

ss

s

M = 2

M = 3

M = 4



• A practical problem: the case D≤2

• I.e.: PSK is the best if M6; if M>6, joint modulation of amplitude and phase is better (of course: a surface rather than a line is available)

M=2 M=4 M=6 M=7


4. Optimizing the signal setWith larger M : along a regular triangular grid.(Honeycomb modulation)

Not much applied, for practical reasons; nearly as good: a rectangular grid(i.e. QAM)


5. Frequency band occupation, etc

• 3 questions arise• occupied band?• physical meaning of dimensionality?• why to talk about M>2)?• Frequency band: if a function is of finite support,

that of its Fourier transform is infinit (meaning?)• Theoretically optimized band is thus infinite• but practically: band where most of the

energy is (e.g. 90%) is of course finite. (Of the order of 1/T)



• Dimensionality: dimension-theorem: practically occupied bans (whatever is definition) is proportional to dimensionality.

• To verify examples: PPM and orthogonal FSK)

τ

T M=4, D=4

s1(t)

s2(t)

s3(t)

s4(t)

W=1/τ=4/T=D/T

M=4, D=4

1/T

W=4/T=D/T

fHz



• Decrease of W: increase ofT

• Only possibility: not bit-by-bit, rather several bits together (united into 1 symbol)

• Then: M=2n; and: Ts=nT

• If D remains constant:Wn=W1/n

• Cost of that: we have to distinguish M rather than 2 signals (i.e. points)

• For what higher power is needed



2

11

11

12 if

12

1;

ME

Edd

MEdEd

nn

nn

En

dn

d1= E1

•Detailed: QAM vs. antipodError prob (nearly) equal ifd1=dn

or powers

n

122

TE

nTE

P

P22n

1

n

1

n



• Conclusion: occupied band can arbitrarily be increased if

• n bits are unied into one symbol;• And this applied to a 2D modulation• But its cost: exponental increase of signal

energy and nearly exponental increase of signal power

• Locsl optimum: 4QAM (QPSK): band decreases to its half while power remains unchanged [2(22/2-1)/2=1]



• n=2; M=4 (QPSK) Pn/P1= 0 dB

• n=4 M=16 (16QAM) 7 dB

• n=6 M=64 (64QAM) 12 dB

• n=8 M=256 (256QAM) 17 dB

• n=10 M=1024 (1024QAM) 23dB


Some special characteristics of the optical frequency band

• We’ve seen that tere is no thermal noise in the optical band.

• On the other hand ther is shot noise. (Without refering to optics we saw the effect of Poisson-distributed noise.)

• Some more detailed in what follows.


Some special characteristics of the optical frequency band

Somewhat modified model:

SOURCEOPTICAL

GEN+MOD +PHOTO

DETECTORDECIDER

Időzítés (T)

b(t)

NYESINKLŐ+

N(t)

b(t): optical background noiseN(t): noise of the amplifier(being electrical amp.)

M=2

Modulation: nearly alwaysOn-Off intensitymodulation


Optical frequency band: some elements of the model

• b(t): has at least three elements:

• noise really of the background (free space optical systems,

FSO)

• dark current of the photodetectos

• non-ideal extinction of the modulator

• Photodetector: electrons are generated by the entering photons



• Two different types:

• PIN dióda: 1 photon1 electron

but not sure

• APD: 1 photon1 electron but not sure however secondery electrons to each electron

(APD: Avalanche Photo Diode)



PMhf

eI

Phf

eI

hf

TP

e

TI

hf

E

e

Qnn

c

q

c

q

c

q

c

qphotqel

det

detdet

:APD

:PIN

Detected current is proportional to optical power(!)


Some specialities of the optical frequency band

0

: sintensitie

0 ;cos2

0

112

21

P

PPA

tstAts

s

c

• These are averageintensities. The lightsource emits photons in Poisson-distribution (?). Some of these arrives to the detector

• The fotodetector is an electron- counter

• m: average number ofelectrons during T seconds

Signal waveforms:

!

Prk

emk

mk


Optimal optical detection (we’ve seen in an example)

• First assume that there is no thermal noise• The two hypotheses:

• H0: there is only b(t) van – power: Pb (i.e. s0 is transmited)

• H1: (s1) is transmiited – power: Ps1+Pb

c

bq

c

bsq hf

TPm

hf

TPPm

0

11 ;

A két hipotézis: !

Pr: ;!

Pr:0

00

11

1 k

emkH

k

emkH

mkmk



• The likelihood-ratio:

• Decision rule:

• For being precise:

01

0

1 mm

k

em

mk

001

01

101

01

:ln

ln if

;:ln

ln if

Hmm

mmk

Hmm

mmk

5,0)1Pr()0Pr(

;1:ln

ln if 01

01

01

QQ

QHQHmm

mmk



b

s

s

c

qk

P

P

P

hf

Tk

1

1

1ln

1:H;0:H :then

0:0

10

kk

kP kb

•Like before, we assume thatln η=0. Threshold is then

1. special case: Pb=0(No background noise)

1|0Pr2

10|1Pr

2

1EP

=0

=

!! exp 1

hf

TP qs

Rather unpleasant: it depends both on Ps1/Pb and on Ps1



0|Pr4

11|Pr

4

1

0|Pr2

11|Pr

2

1

0|1Pr2

11|0Pr

2

1

kk

kk

E

kkkk

kkkk

P

2. Pb0

1 2

00

110

01

0

11

!4

1

!2!2 kkk

mkkmkkkmkk

k

km

E k

emem

k

me

k

meP


Some (further) specialities of the optical band

• The case of shot+thermal noise

• Noise then: Poisson + Gaussian – i.e. discrete +continuous . (Discrete RV will be called x)

PD

RL

+

N(t)

DECISION

Udiscrete

x(t)

r(t)=N(t)+x(t)



0

110

1,0;Pr;!

Pr

;;1,0;;

k

ik

ikix

imkiLi

k

sbbc

qii

L

ixxxxxpk

em

T

eRxx

PPPPPihf

TPm

T

keRtx

0

1,0;Pr;ˆix

ikiix ixxxmxpxp

PDF of x(discrete!)

H0 and H1 are distinguished in mi only; thus:

T

RNBRkFTNNp L

LBN0

022 2

4;,0; N is Gaussian:



2

0

2

0

2

0

,;Pr;

i.e.,,0;Pr; so And

,0;Pr;ˆ

i

x

iir

i

x

iir

i

x

iirr

Nxr

xrxmrp

xrxmrp

rxrxmrprp

rprprpNxr

i

i

i

Decision rule: threshold uk, where pr(r;m1)=pr(r;m0)and the error prob.:

ku rku

rE drmrpdrmrpP 10 ;2

1;

2

1



Relevant pdf: sum of Gaussians (of course: different for Pb és Pb+ Ps1).

Consequently PE: sum of erfc-s



• This is rather complex.

• But: if mi is rather large, Poisson distribution is approximately Gaussian

• And then E/N0 yields PE-t (erfc) – i.e. only SNR has to be known

EDDIG!!


Conditional Poisson distribution

• We know: if time between two events is independent,

• but the expected value of the events/sec is constant,

• the number of events is of Poisson- distribution.

• But: if there is noise too: then the expected value is of random variation

• Distribution then is conditional Poisson (CP)


Conditional Poisson distribution

• CP distribution:• Not too much

can be said in general Although characteristic function can be given.

• With Poisson:

dmmpmkk

k

emmk

m

mk

0|PrPr

!|Pr

k

0

Pr RV discretefor or

E

juk

juxx

juxx

ekuU

dxexpeuU

10

0

Finally

!ˆ

!

juemk

k

myk

mk

ju

k

jukmk

k

euU

ek

yeuUmey

ek

emuU

NB: Note that Ux(u)=[F(px(x)]


Feltételes Poisson eloszlás

• If CP: conditional characteristic function

• Thus• On the other hand

(definition)

• Comparing these

1

|juem

k emuU

dmmpeuU m

juem

k

0

1

0

duempuU jmum

1juejumk uUuU


Is bacground noise really Poisson?

• Some complicated formulas:• Példa az előzőkre: amiből kiindultunk, vagyis jel

(modulálatlan szinusz) + háttérzaj (tekintsük most Gaussi-fehér-nek)

• Szinusz+zaj: komplex burkolója: komplex Gs-folyamat, aminek a várható értéke >0. Ennek az absz. négyzet-integrálja ami a (val.vált) m (várható érték).

• Ennek az (ismert) karakt. függv-e.:

00 1

exp1

1

juN

juE

juNuU

BT

E

Itt B az optikai sávszélességT a megfigyelési idő – ez alatti elektron-számra vagyunk kíváncsiak


Jogos volt-e Poisson-eloszlásúnak tekinteni a háttérzajt?

• Alkalmazzuk az helyettesítést;

közvetlenül ez a detektált opt. energiának volna a kar. függv.-e; ha az energiákat (E és N0) még megszorozzuk ηq/hfc-vel, az elektronszám-ét kapjuk:

ju

c

q

ju

c

q

BT

ju

c

qk

eNhf

eEhf

eNhf

uU

11

1

exp

11

1

00

1jueju

Emlékszünk: uUxpxpuU xxxx1 FF



• Így a karakterisztikus függvényt vissza kell transzformálni

• Ez a szép Laguerre- eloszláshoz vezet:

00

1

00

0

11

exp

1

Pr

Nhf

Nhf

Ehf

Nhf

Ehf

Nhf

Nhf

k

c

q

c

q

sc

q

BT

k

c

q

sc

q

BTk

c

q

k

c

q

L



• Definíciók és alkalmazás a spec. esetre:

• Laguerre-polinóm: M-ed rendű, k-ad fokú)

• Továbbá: • Akkor:

!0 j

x

jk

kMx

jk

j

M

kL

1

1

0

c

q

elektromos

optikai

hf

N

B

BBT

!

;

1

0

0

0

k

BT

k

kBT

hf

N

hf

Nhf

Nkk

c

q

k

c

q

c

q

chf

NqBT

BT

c

q

e

hf

N

0

01

1



• Ezekkel a Laguerre polinóm

• És a Laguerre eloszás közelítőleg

• Ez bizony Poisson:

• Ez éppen az, amit szerettünk volna:

!Pr

0

0 k

eBTNE

hfk

BTNsEchf

qk

sc

q

k

s

c

q

c

q

sc

q

BT

kBT

N

E

kN

hfN

hf

Ehf

L

000

1

!

1

1

00 ;; BNPTPEBTNEhf

m bsssc

q

bsc

q PPhf

Tm



• Még egy kérdés: igaz-e, hogy ?

• Adatok: PE=10-9 (ideálisan 20 foton/(s1 bit); gyak.: 15 dB-lel

nagyobb) 2,5 Gbit/sec (T=400 psec)

fc=200 THzPs/Pb: 10 dBB(opt) =1 THz: 1 psec alatt 0,15 zaj-foton keletkezik

10 cq hfN

Communication Theory/2 I. Frigyes 2009-10/II.. Frigyes: Hírkelm2 frigyes/hirkelm frigyes/hirkelm...

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Transcript of Communication Theory/2 I. Frigyes 2009-10/II.. Frigyes: Hírkelm2 frigyes/hirkelm frigyes/hirkelm...