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WIRELESS COMMUNICATIONSlecture notes (part 1)

c Prof. Giorgio Taricco c

Politecnico di Torino

2013/2014

1 c Prof. Giorgio Taricco c WIRELESS COMMUNICATIONS

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Copyright Notice

All material in this course is the property of the Author. Copyrightand other intellectual property laws protect these materials.Reproduction or retransmission of the materials, in whole or inpart, in any manner, without the prior written consent of the

copyright holder, is a violation of copyright law. A single printedcopy of the materials available through this course may be made,solely for personal, noncommercial use. Individuals must preserveany copyright or other notices contained in or associated withthem. Users may not distribute such copies to others, whether or

not in electronic form, whether or not for a charge or otherconsideration, without prior written consent of the copyright holderof the materials.


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Basic concepts

Outline

1 Basic concepts

2 Digital modulations over the AWGN channel


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Basic concepts

Reference books in Wireless Communications

S. Benedetto and E. Biglieri, Principles of Digital Transmission:With Wireless Applications. Kluwer.

A. Goldsmith, Wireless Communications. Cambridge UniversityPress.

U. Madhow, Fundamentals of Digital Communication. Cambridge

University Press.A. Molisch, Wireless Communications. Wiley.

J. Proakis and M. Salehi, Digital Communications (4th Edition).McGraw-Hill.

T. Rappaport, Wireless Communications: Principles and Practice(2nd Edition). Prentice-Hall.

D. Tse and P. Viswanath, Fundamentals of WirelessCommunication. Cambridge University Press.


B i

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Basic concepts

Reference books in Wireless Communications

S. Benedetto and E. Biglieri, Principles of Digital Transmission:With Wireless Applications. Kluwer.

A. Goldsmith, Wireless Communications. Cambridge UniversityPress.

U. Madhow, Fundamentals of Digital Communication. Cambridge

University Press.A. Molisch, Wireless Communications. Wiley.

J. Proakis and M. Salehi, Digital Communications (4th Edition).McGraw-Hill.

T. Rappaport, Wireless Communications: Principles and Practice(2nd Edition). Prentice-Hall.

D. Tse and P. Viswanath, Fundamentals of WirelessCommunication. Cambridge University Press.


B i t M d l f di it l i ti t

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Basic concepts Model of a digital communication system

Model of a digital communication system

CHANNEL MODELThe main subject of this course is the study of digitalcommunications over a transmission channel.

To this purpose, it is useful to characterize the model of adigital communication system in order to get acquainted withits different constituent parts.

TOP LEVEL CLASSIFICATION

The model can be divided into three sections, as illustrated in

the following picture:1 Theusersection2 Theinterfacesection3 Thechannelsection


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CHANNEL MODELThe main subject of this course is the study of digitalcommunications over a transmission channel.

To this purpose, it is useful to characterize the model of adigital communication system in order to get acquainted withits different constituent parts.

TOP LEVEL CLASSIFICATION

The model can be divided into three sections, as illustrated in

the following picture:1 Theusersection2 Theinterfacesection3 Thechannelsection


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TX ENCODER MODULATOR

CHA

NNEL

DEMODULATORDECODERRX

user section interface section channel section



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p g y


TX ENCODER MODULATOR

CHA

NNEL

DEMODULATORDECODERRX

user section interface section channel section

D D W

WDD

D=digital

W=waveform



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p g y


ENCODER

Implementssource encodingto limit the amount oftransmitted data (for example, voice can be encoded at 4kbit/s or sent at 64 kbit/s with conventional telephony).

Implementschannel encodingto limit the effect of channeldisturbances

MODULATOR

Converts the digital signal into a waveform to be transmittedover the channel

CHANNEL

Reproduces the transmitted waveform at the receiver

Its operation is affected by frequency distortion, fading,additive noise, and other disturbances


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ENCODER

Implementssource encodingto limit the amount oftransmitted data (for example, voice can be encoded at 4kbit/s or sent at 64 kbit/s with conventional telephony).

Implementschannel encodingto limit the effect of channeldisturbances

MODULATOR

Converts the digital signal into a waveform to be transmittedover the channel

CHANNEL

Reproduces the transmitted waveform at the receiver

Its operation is affected by frequency distortion, fading,additive noise, and other disturbances



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DEMODULATOR

Converts the received waveform into a sequence of samples tobe processed by the decoder

DECODER

Implements channel decoding to limit the effect of the errorsintroduced by the channel

Implements source decoding



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DEMODULATOR

Converts the received waveform into a sequence of samples tobe processed by the decoder

DECODER

Implements channel decoding to limit the effect of the errorsintroduced by the channel

Implements source decoding


Basic concepts Band-pass signalling

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Band-pass signals

A band-pass signal has spectral components in a limited range

of frequencies f (f2, f1) (f1, f2) provided that0< f1< f2.

A certain frequency in the range (f1, f2) (usually the middlefrequency) is called carrier frequency and denoted by fc.

The signal bandwidth is Bx =f2 f1.

f

Bx

f1 f1

f2 f2

fc fc

It is often convenient to representband-pass signalsasequivalent complex signals with low-pass frequency spectrum(i.e., including the zero frequency).



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The analytic signal

A real band-pass signal x(t) can be mapped to a complexanalytic signal x(t) by passing through a linear filter withtransfer function 2u(f) = 2 1f>0:

x(t) 2u(f) x(t)

The indicator function 1A= 1 ifA is true, 0 otherwise.

Summarizing:

The analytic signal is a complex representation of a real signal.

It is used to simplify the analysis of modulated signals.It generalizes the concept of phasor used in electronics.

The basic properties of the analytic signal derive from theFourier transform.



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The analytic signal (cont.)

Ifx(t) is a real signal, then its Fourier transform is aHermitian functionsince:

X(f) =

x(t)e +j 2ftdt=

x(t)ej 2(f)tdt

= X(f)

Therefore, the spectrum is completely determined by its

positive frequency (or negative frequency) part.



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Hilbert transform

Since:

X(f) = 2u(f)X(f) =X(f) +sgn(f)X(f) X(f) + jX(f),applyingF1 yields:

x(t) =x(t) +jx(t).The signalx(t) is called theHilbert transformofx(t):x(t) x(t) 1

t=

1

x()

t d



( )

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Hilbert transform (cont.)

Here, the Cauchy principal part of the integral has been taken,namely,

lim0,T

tT

+ Tt+

x()t d



S

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Spectral properties

Assuming x(t) a zero-mean stationary real random process,

we have

Gx(f) = |2u(f)|2Gx(f) = 4u(f)Gx(f)

Therefore,

Gx(f) =1

4[Gx(f) + Gx(f)]

Moreover,

E[|x(t)|2] =

04Gx(f)df= 2

Gx(f)df= 2E[x(t)2]



B d i lli

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Band-pass signalling

Assume that x(t) is a zero-mean stationary band-pass randomprocess with bandwidth Bx and carrier frequency fc so thatits power density spectrum is nonzero over the frequencies

f

(

fc

Bx/2,

fc+ Bx/2)

(fc

Bx/2, fc+ Bx/2)

where fc> Bx/2> 0.

We define the basebandcomplex envelopeofx(t) as

x(t) = x(t) ej 2fct

The complex envelope is sometimes calledbasebandequivalent signal.



Th l l (t)

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The complex envelope x(t)

Then, we derive the autocorrelation function and the powerdensity spectrum ofx(t):

Rx() = E

x(t + ) ej 2fc(t+)x(t) ej 2fct

= Rx() ej 2fc= Gx(f) =Gx(f+ fc).

Then, the power density spectrum ofx(t) is nonzero over thefrequencies f (Bx/2, Bx/2), i.e., it is a baseband signalwith bandwidth Bx/2.



I h d d t t

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In-phase and quadrature components

The real and imaginary parts ofx(t) =xc(t) + j xs(t) arecalledin-phaseandquadraturecomponents of the signal.

They can be expressed in terms of the signal itself and of itsHilbert transform:

xc(t) = Re[x(t)ej 2fct] =x(t) cos(2fct) +x(t) sin(2fct)xs(t) =Im[x(t)ej 2fct] =x(t) cos(2fct) x(t) sin(2fct)

The previous relationships can be inverted and yield:

x(t) = Re[x(t)e j 2fct] =xc(t) cos(2fct) xs(t) sin(2fct)x(t) =Im[x(t)e j 2fct] =xs(t) cos(2fct) + xc(t) sin(2fct)



F i d l ti

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Frequency up-conversion = modulation

Themodulation(or frequency up-conversion) of a real signal

x(t) consists in the following operation:

x(t) x(t) cos(2fct).

The modulation of a couple of real signals xc(t) and xs(t)consists in the following operation:

[xc(t), xs(t)] xc(t) cos(2fct) xs(t) sin(2fct).

In the analytic signal domain, modulation can be representedas follows:

x(t) x(t) = x(t)e +j 2fct.



F d si d d l ti

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Frequency down-conversion = demodulation

Therefore, demodulation(or frequency down-conversion) inthe analytic signal domain is represented as follows:

x(t) x(t) = x(t)ej 2fct.

Correspondingly, in the real signal domain, demodulation canbe represented by:

xc(t) =x(t) cos(2fct) +x(t) sin(2fct)xs(t) =x(t) cos(2fct) x(t) sin(2fct)



Frequency down conversion demodulation (cont )

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Frequency down-conversion = demodulation (cont.)

In a real system, demodulation can be implemented byobserving that

MULTIPLICATION BY IN-PHASE CARRIER

x(t) cos(2fct + ) 2 cos(2fct + )= x(t) + x(t) cos(4fct + 2)

In other words, multiplication of the signal x(t) cos(2fct + )by the phase-coherent sinusoid 2 cos(2fct + ) returns thesuperposition of

the modulating signal x(t);another modulated signal with carrier frequency 2fc.

Low-pass filtering with bandwidth Bx eliminates themodulated signal with carrier frequency 2fc.



Demodulator

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Demodulator

The following picture illustrates the block diagram of ademodulator with input:

x(t) =xc(t) cos(2fct) xs(t) sin(2fct).

x(t)

2 sin(2fct) = 2 cos(2fct + 2 )

2 cos(2fct)

LOW-PASSFILTER

LOW-PASSFILTER

xs(t)

xc(t)


Basic concepts Problem set 1

Problem set 1

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Problem set 1

1 Calculate the analytic signal corresponding to

x(t) = cos(2fct + ).

x(t) = sinc(t) cos(20t).

x(t) = sinc(t)2

cos(20t).2 Calculate the baseband equivalent signal corresponding to

x(t) = cos(41t) + 2 sin(39t), fc= 20.

x(t) = sinc(t) cos(20t), fc= 10.

x(t) = sinc(t) cos(20t), fc= 9.


Basic concepts Probability

Topics on Probability

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Topics on Probability

Probability is based on the concept ofprobability space.

A probability space consists of three parts:

1 A set of all possible outcomes.2 A set ofevents,F, which aresets of outcomes.3 A probability function P : F [0, 1], assigning a probability to

every event.

The probability function is a normalized measure:


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Topics on Probability (cont )

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Topics on Probability (cont.)

3 In all cases, the followingnormalizationholds:

P() = 1,

that is, the probability of the set of all possible outcomes is 1.Given two events A, B, we can build the union A B and theintersection A B:

A B = set of outcomes in A orin B.A

B = set of outcomes in A andin B.



Topics on Probability (cont )

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The probability of the intersection is called joint probability

and allows to define the conditional probability as

P(A | B) = P(A B)P(B)

.

Commonly, we write P(A, B) P(A B).Conditional probabilities satisfy the Bayes rule:

P(A | B) = P(A, B)P(B)

=P(B, A)

P(A)

P(A)

P(B)

= P(B| A) P(A)

P(B) .




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A useful result is thetotal probability law:

If the events Bi, i= 1, 2, . . . form apartitionof (i.e.,iBi= and Bi Bj = fori =j), then:

P(A) =

iP(A, Bi) =

iP(A | Bi)P(Bi).

By the total probability law, one can obtain the conditionalprobabilitiesP(Bi| A) from the P(A | Bi):

P(Bi| A) = P(A | Bi)P(Bi)jP(A | Bj)P(Bj)

.




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The above result finds application in the design of digitalcommunication receivers where the event A represents the

received signal and the events Bi represent all the possibletransmitted data in a given framework.




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p b b y ( )

The probability of the union has also some special properties.

For illustration, we interpret events as two-dimensional regionsand their probabilities as the areas of the regions.

A

B

P(A) P(A B)P(B) P(A B)P(A B) P(A) + P(B)

=P(A) + P(B) P(A B)

The inequalities derive from the fact that the area of the unionis always greater than or equal to the areas of each event.




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p y ( )

Moreover, the sum of the areas is equal to the area of theunion plus that of the intersection, which is counted twice.This yields the last inequality.

The previous results can be generalized to the case ofmevents:

Lower and upper union bounds

Given a set of events{A1, . . . , Am}, the following inequalities hold:

max1im

P(Ai)

Pi Ai

m

i=1 P(Ai). (1)



Random variables

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A discrete real random variable X is characterized by itsprobability distribution

pX(xn) =P(X=xn)

forn= 1, 2, . . . , N (where Nmay become infinity).

Theexpectation operator E[] is defined by

E[(X)] = n (xn)pX(xn)for an arbitrary function ().



Random variables (cont.)

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( )

Since the expected value of every constant is the constantitself, we obtain by definition:

E[1] =n

pX(xn) = 1.

This property holds for all probability distributions.

ThemeanofX is X= E[X] =

n xnpX(xn).

Thesecond momentofX is (2)X = E[X

2] =

n x

2npX(xn).

ThevarianceofX is 2X= E[(X X)2] =(2)X 2X.

The square root of the variance is calledstandard deviation.




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( )

Continuous random variables are characterized by a pdffX(x)defining the expectation operator as

E[(X)] =

I

(x)fX(x)dx,

whereI is the support of the random variable, i.e., the set ofvalues where fX(x)>0.

Again, the expected value of every constant is the constantitself, so that:

E[1] = IfX(x)dx= 1.This property holds for all pdfs.




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The mean, second moment, and variance are given by

X =

I

xfX(x)dx

(2)

X = Ix2fX(x)dx2X =

(2)X 2X,

respectively.

Random variables can also be complex. Their propertiesderive from the properties of real random variables.




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A complex random variable Z=X+ j Y has mean

Z= E[X] + j E[Y]

and variance

2Z E[|Z Z|

2

]= E[(X X)2 + (Y Y)2]= 2X+

2Y

= E[X2 + Y2] (2X+ 2Y)= E[|Z|2] |Z|2.


Basic concepts Gaussian random variables

Gaussian random variables

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We will be particularly concerned with Gaussian randomvariables whose distribution is given by

fX(x) = 1

22e(x)

2/(22)

and denoted byN(, 2).The parameters and are the mean and the standarddeviation of a Gaussian random variable with distributionN(, 2).



Gaussian random variables (cont.)

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We will often be interested in calculating the probability

P(N(, 2)> x), i.e., the probability that a Gaussian randomvariable with mean and standard deviation exceeds thereal value x.

These probability can be calculated by using the function

Q(x) (referred to as Q-function), which is the countercumulative probability distribution function of the normalizedGaussian random variableN(0, 1).The Q-function is defined as:

Q(x) P(N(0, 1)> x) = x

12

eu2/2du. (2)




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By using the Q-function we can see that

P(N(, 2)> x) = P(N(0, 2)> x )= PN(0, 1)>

x

= Qx

.

The Q-function cannot be calculated in terms of elementaryfunctions (such as exp, ln, and trigonometric functions).




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However, by applying integration by parts

(udv=uv vdu), we obtain the following simpleinequalities:

ex2/22x

1 1x2

Q(x) ex2/2

2x

.

The upper bound is a good approximation for x 3 andyields the asymptotic behavior:

Q(x) ex2

/22x

.




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In some cases, the following crude approximation is used:

Q(x) ex2/2.

The following diagram compares the Q-function and the two

approximations, which are plotted as thered( ex2/2

2x) and

green(ex2/2) dashed curves, respectively.

We can see that the approximation of the red curve is betterthan 10% forx 3.




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4 2 0 2 4

0.2

0.4

0.6

0.8

1.0


Basic concepts Complex Gaussian random variables

Complex Gaussian random variables

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We will also considercomplexGaussian random variables withaspecial property, namely, that of having zero mean andindependent and identically distributed (iid) real and

imaginary parts.

In other words, ifZ=X+ j Y, we assume thatX N(0, 2/2) and Y N(0, 2/2), where X and Y arestatistically independent.



Complex Gaussian random variables (cont.)

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These assumptions lead to the following equivalent pdf

expressions:

fZ(z) = fXY(x, y)

= 1

2ex

2/2 12

ey2/2

= 1

2e(x

2+y2)/2

= 1

2e|z|

2/2 .

The concept can be extended to vectors of complex Gaussianrandom variables.



Complex Gaussian random variables (cont.)

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Ifz = (Z1, . . . , Z n)T is a vector of complex Gaussian random

variables with zero mean and covariance matrixz = E[zz

H], then the pdf ofz can be expressed as follows:

fz(z) = det(z)1ez

H1z z.

In the special case of iid components ofz, corresponding toz =

2In, the pdf simplifies to

fz(z) = (2)nez

2/2 ,

which is the product of the individual (marginal) pdfs of theZis: (

2)1e|zi|2/2 .


Basic concepts Signal spaces

Signal spaces

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Signal spacesare linear (or vector) spaces built upon theconcept ofHilbert space, i.e., finite or infinite-dimensionalcomplete inner product spaces.

The elements of a signal space are real or complex signals x(t)

defined over a support intervalI, for exampleI= (0, T).The inner product of two elements (signals) xandy is definedas

(x, y) = Ix(t)y(t)dt. (3)



Signal spaces (cont.)

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Correspondingly, theinduced normofx is given by

x (x, x)1/2. (4)

Accordingly, the set

L2(I) = {x: x < }

is defined as asignal space.




I d i f h C h S h i li

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Inner products satisfy theCauchy-Schwarzinequality

|(x, y)

| x

y

.

If|(x, y)| = x y, then the two signals are proportional,i.e., y(t) =x(t) for some C.A signal x(t) H L2(I) ifx is finite.The squared norm of a signal x(t) is thesignal energy:

E(x) (x, x) =

I

|x(t)|2dt= x2.


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Th ffi i t i thi i b l l t d b

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The coefficients xn in this expansion can be calculated by

xn = (x, n) = Ix(t)n(t)dt.In many cases, a signal spaceH is defined as the set of allpossible linear combinations of a set of signals:

H = L(s1, . . . , sM)= {x(t) =1s1(t) + + MsM(t),

(1, . . . , M) CM}.

The setL

(s1, . . . , sM)is calledlinear spanofs1(t), . . . , sM(t).

In general, the signal set{s1, . . . , sM} is not a base, but abase can be found by using the Gram-Schmidt algorithm.




Th G S h idt l ith fi d b ( )N f

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The Gram-Schmidt algorithm finds a base (n)Nn=1 of

H= L(s1, . . . , sM) by the following set of iterative equations:For k= 1, . . . , n:

dk = sk k1i=1

(sk, i)i (projection step)

k =

dk

dk (normalization step)




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At every projection step such that dk = 0 the correspondingk is not assigned and not accounted for in the remaining

steps.

The number of signals in the base is the number ofdimensions ofH.




The Gram Schmidt algorithm works since at every step the

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The Gram-Schmidt algorithm works since, at every step, thesignal dk(t) is orthogonal to all previously generated signals

i(t), i= 1, . . . , k 1. In fact,

(dk, i) =

sk

k1=1

(sk, ), i

= (sk, i) k

1

=1

(sk, )(, i)

= (sk, i) (sk, i) = 0.




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The projection of a signal x(t) H over the subspaceY= L(1, . . . , N) H is a signal xY(t) with the followingproperties:

It can be expressed through the base ofYas follows:

xY(t) =N

n=1(x, n)n(t).It is theclosestsignal inY to x(t):

xY= arg minyY

x y.The minimum distance is:

minyY

x y2 = x2 xY2.


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The minimization over|

yn|

is straightforward and gives|yn| = |(x, n)|. In this case, n= |(x, n)|2.Summarizing, the yn minimizingx y2 overy Y is:

yn= |yn|ejyn = |(x, n)|ej(x,n) = (x, n).

The minimum is:

minyY

x y2 = x2 N

n=1|(x, n)|2.



A matrix version of the GS algorithm

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An alternative approach to the Gram-Schmidt algorithm is

based on standard linear algebra methods.It can be noticed that the GS algorithm leads to expressionsof the orthogonal base signals of the following type:

i(t) = ji Cijsj(t).These equations can be written in matrix form as follows:

=Cs,

where = (1(t), . . . , N(t))T and s= (s1(t), . . . , sM(t))

T.

C is a lower triangular matrix.



A matrix version of the GS algorithm (cont.)

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From the previous equation we get

(,T) = (Cs, sTCT) =C(s, sT)CT.

The lhs is the identity matrix IN since (i, j) =ij.

The rhs can be written as CsCT where the matrix s isthe Gram matrixof the signals in s(t).

The elements ofs are(s)ij = (si, sj).

If a signal si(t) is linearly dependent from the signals

s1(t), . . . , si1(t), the corresponding row and column ofsmust be eliminated because linear combinations of theprevious row and columns.



A matrix version of the GS algorithm (cont.)

A reduced Cholesky factorization can be applied to s leading

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y pp s gto the following matrix expression:

EsET =LLT

where the N M matrix Eremoves the redundant rows andcolumns and L is a square nonsingular lower triangularmatrix.

The previous result leads to C=L1E.As an example, if we have four signals but only s3 is a linearcombination ofs1, s2,

E= 1 0 0 00 1 0 00 0 0 1

The matrix product E(s1, s2, s3, s4)

T = (s1, s2, s4)T

eliminates s3.



Problem set 2

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1 Calculate the mean and variance of the discrete random

variableXwith probability distributionpX(1) = 0.5, pX(2) = 0.25, pX(4) = 0.25.

2 Calculate the mean and variance of the continuous randomvariableXwith probability distribution fX(x) = e

x1x>0.

3 Calculate the mean and variance of the continuous randomvariableXwith probability distribution fX(x) = 0.5 1|x| a) for a Gaussian random

variableX N(, 2).5 Show that the variance identity holds: 2

X= E[X2]

E[X]2.

6 Assume support intervalI= (0, 1) from now on.LetHbe the linear span ofcos(2t) and sin(2t). Determineif the signal cos(2t + /4) belongs toH.



Problem set 2 (cont.)

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7 Given the signals s1(t) =u(t) u(t 0.6),s2(t) =u(t 0.4) u(t 1), and s3(t) =u(t) u(t 1),apply the Gram-Schmidt algorithm to find a base ofL(s1, s2, s3) [u(t) = 0 fort 0 is the unitstep function].

8 Given the signalss1(t) = cos(2t)ands2(t) = sin(3t), apply

the Gram-Schmidt algorithm to find a base ofL(s1, s2).9 Check Schwarzs inequality for the signals

s1(t) =u(t) u(t 0.6), s2(t) =u(t 0.4) u(t 1), ands3(t) =u(t) u(t 1).

10 LetY= L(s1= sin(t), s2= cos(3t)). Find the projectionofx(t) =u(t) u(t 1) overYand calculatex xY2(notice that s1 and s2 are orthogonal).




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11 Apply the matrix GS algorithm to find the matrix Cdetermining the orthogonal base to the set of signals

s1= 10

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Digital modulations over the AWGN channel Additive White Gaussian Noise (AWGN) Channel

AWGN channel

Thi h l d l i ifi d b th ti

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This channel model is specified by the equation

y(t) =Ax(t) + z(t) (6)

where:

Channel parameters

A is the real channel gain.

x(t) and y(t) are the channel input and output signals.

z(t) is the zero-mean additive white Gaussian noise process. It hasautocorrelation function and power density spectrum:

Rz() = E[z(t + )z(t)] = N02 (),

Gz(f) = N0

2 .


Digital modulations over the AWGN channel Linear digital modulation

Linear modulations

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We consider the following modulated signal:

x(t;a) =N

n=1

ann(t) (7)

where

N is the number ofdimensionsof the modulation scheme.The vector a= (an)

Nn=1 represents amodulation symbol

vectorand is taken from a finite setA = {1, . . . ,M}.A is calledmodulation alphabetorsignal constellation.n(t) is the nthshaping pulseof the modulated signal.We assume that each n(t) = 0 only for t (0, T).We also assume that (m, n) =mn= 1 ifm= n and 0otherwise (Kroneckers delta).



Linear modulations (cont.)

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The signal x(t;a) allows us to send one symbol vector everyTtime units so that T is calledsymbol time,symbol interval,orsignalling interval.

The signalx(t;a)belongs to the Hilbert spaceHgenerated byall possible linear combinations of the base signals (n)

Nn=1.

The corresponding received signal

y(t) =AN

n=1

ann(t) + z(t) (8)

does not necessarily belong toH= L(1, . . . , N).



Linear modulations (cont.)

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The projection ofy(t) onHis given by:

yH(t) =N

n=1

ynn(t) =N

n=1

(Aan+ zn)n(t). (9)

In this expression we find the nth received signal and noise

components:yn= (y, n) =

T0

y(t)n(t)dt and

zn= (z, n) =T0

z(t)n(t)dt

Here, zn, is a Gaussian random variable.

A receiver calculating the vector y= (y1, . . . , yN) fromy(t) iscalledcorrelation receiver.


Digital modulations over the AWGN channel Digital receiver design

Receiver design

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The goal of a digital receiver is to recover the transmittedsymbol vector a from the received signal y(t).

The correlation receiver projects y(t) onHto obtain(9)andoutputs the coefficients yn=Aan+ zn forn= 1, . . . , N .

In the absence of noise, the correlation receiver outputs ascaled version (by the channel gain A) of the transmittedsymbol vector.

When noise is present, the receiverguesseswhich symbolvector from

Awas transmitted with the goal of minimizing

theerror probability.This process is calleddetectionordecision.



Receiver design (cont.)

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The correlation receiver can be interpreted as amatched filterby observing that:

yn =

T0

y(t)n(t)dt

= T0

y(t)hn(T t)dt= [y(t) hn(t)]t=T,

where we defined the impulse response of the matched filter

as hn(t) =n(T t).



Sequential receiver design

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The signalling model described can be repeated over many

symbol times.

We can write the sequential modulated signal as:

x(t;a0, . . . ,aL) =L

i=0N

n=1 ai,nn(t iT).The corresponding received signal over the AWGN channel is:

y(t) =A

Li=0

Nn=1

ai,nn(t iT) + z(t).



Sequential receiver design (cont.)

The matched filter structure lends itself to a sequential

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The matched filter structure lends itself to a sequentialimplementation accounting for the transmission of successive

modulated symbols in time.

The bank of matched filters receiver is illustrated as follows:

y(t) 1(T

t)

2(T t)

...

N(T t)

yi,1

yi,2

yi,N

t= (i + 1)T

73 c Prof. Giorgio Taricco c WIRELESS COMMUNICATIONS Digital modulations over the AWGN channel Digital receiver design


The output of the nth matched filter at time t= (i + 1)T isgiven by:

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given by:

yi,n = [y(t) n(T t)]t=(i+1)T=

y(t1)n(T (i + 1)T+ t1)dt1

= T

0 y(t2+ iT)n(t2)dt2

= AL

j=0

Nm=1

T0

aj,mm(t2+ iTjT)n(t2)dt2

+ T0

z(t2)n(t2)dt2

= A ai,n+ zi,n

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The previous expression holds since the shaping pulses areorthogonal and time-limited to the signalling interval (0, T).Therefore,

T

0 m(t2+ iTjT)n(t2)dt2=m,ni,j.

This is just an extension of the digital receiver operationdescribed over the first signalling interval (0, T).

75 c Prof Giorgio Taricco c WIRELESS COMMUNICATIONS

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Digital modulations over the AWGN channel Baseband digital modulation

Baseband digital modulation (cont.)

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Another example (same data symbols).

x(t)

t

+1 +1 1 1 1 +1 1 +1 +1 1

Baseband digital modulations are represented in aone-dimensional space.

77 c Prof Giorgio Taricco c WIRELESS COMMUNICATIONS Digital modulations over the AWGN channel Band-pass digital modulation

Band-pass digital modulation

Next, we deal with band-pass digital modulations.

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We start from the baseband equivalent signal:

x(t; a) =a g(t),

where g(t) is a real baseband signal of bandwidth lower thanfc and we assume that a

A=

{a1, . . . , aM

} is acomplex

modulation symbol.

Then, we obtain the corresponding band-pass signal as:

x(t; a) = Re

x(t; a)ej 2fct

= Re(a) [g(t) cos(2fct)]+ Im(a) [g(t) sin(2fct)],

78 c Prof Giorgio Taricco c WIRELESS COMMUNICATIONS Digital modulations over the AWGN channel Band-pass digital modulation

Band-pass digital modulation (cont.)

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The signal x(t; a) can be interpreted as a two-dimensionallinear modulation.

In fact, it can be represented as a linear combination of:

1(t) =g(t) cos(2fct)

2(t) = g(t) sin(2fct) (10)

with coefficientsRe(a) andIm(a).Now, assume that the bandwidth ofg(t) is Bg < fc, i.e., itsFourier transform G(f) =

F[g(t)] is equal to 0 for every

f fc.Then, G2(f) F[g(t)2] =G(f) G(f) has bandwidth2Bg

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G2(2fc) = g(t)2ej 4fctdt= 0.Using the above result,

1,2

2 =

g(t)2

1 cos(4fct)2

dt=1

2g

2

and

(1, 2) = 12

g(t)2 sin(4fct)dt= 0.

Thus, ifg2 = 2, the signals (1, 2) are orthogonal andform the base of a two-dimensional signal spaceHprovidedthat the bandwidth ofg(t) is smaller than the carrierfrequency fc.

80 c Prof Giorgio Taricco c WIRELESS COMMUNICATIONS Digital modulations over the AWGN channel Signal detection

Detection of transmitted symbols

The receiver outputs an estimate of the transmitted symbol a

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based on the received signal y(t) over t (0, T).The first stage of the receiver converts y(t) into the vectory=Aa + z

where a= (a1, . . . , aN) and z = (z1, . . . , zN).

We define a generic decision rule (or detection rule):

a(y) = (a1(y), . . . , aN(y)). (11)

a(y) mapsH into the modulation alphabetA.The decision rule can be optimized according to somegoodness criterion.

Typically, the goal isminimizing the error probability.


Optimum digital receiver

We can write the (average) error probability as follows:

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( g ) p y

P(e) =M

m=1

P(m)P(e | m) (12)

where

P(m) is thea priori probability of transmitting m.P(e | m) is the probability of error conditioned on thetransmission ofm.

We notice that

P(e | m) =Pa(y=Am+ z) =m,i.e., the probability that the decision rule returns a symboldifferent from the transmitted one.


Optimum digital receiver (cont.)

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It is plain to see that minimizing the (average) errorprobability is equivalent to maximizing the (average)probability of correct decision P(c) since P(c) = 1 P(e).Let us define

The pdf ofy given the transmitted symbol : f(y

|).

The decision regions

Rm {y:a(y) =m), m= 1, . . . , M .Notice that there is a one-to-one correspondence between the

set of decision regions and the decision rule.



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Since the decision rulea(y) is a well defined (i.e.,single-valued) function for all y H= RN (the signal space),the decision regions do not intersect and their union fillsHitself:

Mm=1

Rm= H

(

denotes the union of disjoint sets).



Now, we can write the probability of correct decision as af f h b b l ( ) f ( ) d h

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function of the a prioriprobabilities P(m), f(y

|), and the

decision rule. Sincea(y) =m fory Rm, we get:P(c) =

Mm=1

P(m)P(

a(y) =m| m)

=

Mm=1

yRm

P(m)f(y|m)dy

=Mm=1

yRm

P(

a(y))f(y|

a(y))dy

= H=RN

P(a(y))f(y|a(y))dy (13)



Maximizing P(c) requires to maximize the integrand in(13),which can be accomplished by selecting the symbol A

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that maximizes

P()f(y| )for all possible received vectors y.

The resulting optimum decision rule is:

aopt(y) = arg maxA P()f(y| ). (14)



Since applying the Bayes rule, we have

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P(m| y) = P(m)f(y| m)f(y)

,

the optimum decision rule is equivalent to maximizing the aposteriori probability P(m

|y).

Thus, the optimum decision rule is calledmaximuma-posteriori (MAP)decision.



When transmitted symbols are equiprobable, i.e.,P ( ) 1/M h MAP l d i

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P(m) = 1/M, the MAP rule reduces to amaximum

likelihood (ML)rule:

a(y) = arg maxm

f(y| m)

This name comes from the name of the functions f(y| m)(likelihood functionsin radar theory).



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The decision regions can be represented as follows:

Rm= {y: P(m)f(y| m)> P(n)f(y| n)n =m} MAP{y: f(y| m)> f(y| n)n =m} ML

89 P f Gi i T i WIRELESS COMMUNICATIONS Digital modulations over the AWGN channel Signal detection

Special case: The AWGN channel

Proposition. The additive noise components of an AWGN

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channel areiid Gaussian random variables with zero mean andvarianceN0/2.

Proof.We have, by definition,

zn= T0

z(t)n(t)dt

forn= 1, . . . , N .

Then,

E[zn] = T0

E[z(t)]n(t)dt= 0

since the additive noise random process has zero mean.

90 P f Gi i T i WIRELESS COMMUNICATIONS Digital modulations over the AWGN channel Signal detection

Special case: The AWGN channel (cont.)

Moreover,

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E[znzn ] = E T0

z(t)n(t)dt T0

z(t)n(t)dt=

T0

T0

E[z(t)z(t)]n(t)n(t)dtdt

= T0

T0

N02

(t t)n(t)n(t)dtdt

=

T0

N02

n(t)n(t)dt

= N02 n,n .

91 c

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WIRELESS COMMUNICATIONS Digital modulations over the AWGN channel Signal detection


In other words, different components of the noise vector z areuncorrelated (and hence independent since Gaussian), and

h h i N /2

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each one has variance N0/2.

As a result, the conditional pdf of the received vector y is

f(y| ) = fz(y A)= (N0)

N/2eyA2/N0 . (15)

It is worth noting that the joint pdf(15)depends only on thedistance of the received signal from the transmitted onescaled by the channel gain A.

Using(15), the logarithms of the likelihood functions arereadily obtained as follows:

ln f(y| m) = N2

ln(N0) 1N0

y Am2.

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WIRELESS COMMUNICATIONS Digital modulations over the AWGN channel Signal detection

Special case: The AWGN channel (cont.)Since these functions depend on a distance, they are calleddecision metrics.

Th MAP d ML d i i l b d i f

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The MAP and ML decision rules can be expressed in terms ofdecision metrics for the AWGN channel as follows:

a(y) =

arg minm

y Am2 N0ln P(m)

MAP

arg minm y Am2 ML

As a result, the ML decision rule for the AWGN channel isoften referred to asminimum distance decision.


Digital modulations over the AWGN channel Signal detection


The decision regions on the AWGN channel can bet d f ll

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represented as follows:

Rm =

{y: y Am2 N0ln P(m) P(m)f(y| m) | m

.

forn= 1, . . . , M and n

=m.

Notice that all the pairwise error events contain theconditioning clause | m. This clause is equivalent to theassumption that m was transmitted.


Digital modulations over the AWGN channel Error probability

Error probability (cont.)

Thus, the error probability, conditioned on the transmission ofm, is given by

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m, is given by

P(e | m)= P

n=m

P(n)f(y| n)> P(m)f(y| m)

m

whererepresents the union of events.




The above expression of the error probability is too complex

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to calculate analytically, whereas thepairwise errorprobabilities (PEPs)

Pairwise Error Probability

P(m n) P

P(n)f(y| n)> P(m)f(y| m)

m

can be calculated very simply!

Thus, lower and upper bounds are used to approximateP(e|m).



Error probability (cont.)Applying the bounds(1)to the conditional probabilitiesP(e|m), we obtain

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Error probability lower and upper bounds

Mm=1

P(m)maxn=m

P(m n) P(e) Mm=1

P(m)n=m

P(m n)

(16)




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Assuming theMAP decision ruleover theAWGN channel, andletting A= 1, the PEPs are given by

Pairwise error probability

P(m n) =Qm n2 + N0ln[P(m)/P(n)]2N0 m n .

(17)

Equation(17)is based on the Q-function(2)and will bederived in detail in a problem.



Error probability of binary modulations

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The inequalities(16)yield the exact error probability in thecase of for binary modulations (M= 2):

P(e) = P(1)P(1 2) + P(2)P(2 1)

= P(1)Q1 22 + N0ln[P(1)/P(2)]2N0 1 2

+ P(2)Q

1 22 N0ln[P(1)/P(2)]2N0 1 2

.



Error probability of binary modulations (cont.)

With equiprobable signals i e P (m) = 1/M inequalities

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With equiprobable signals, i.e., P(m) 1/M, inequalities(16)yield:

1

M

M

m=1maxn=m

P(m n) P(e)

1M

Mm=1

n=m

P(m n)

Here, P(m

n) =Q(

m

n

/

2N0).



High SNR approximation

In most situations, one is mostly interested to the high SNR(and then low N0) case.

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Since the Q function decreases very quickly, we can keep inthe bounds only the terms with minimum distance:

dmin minm=n

m n (18)

and disregard the others which are very small.

To be conservative, we use the upper bound to P(e) andobtain this approximation:

P(e) NminQ dmin2N0 (19)where Nmin=

1M

m

n 1mn=dmin.


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Bit error probability

In some applications, it is better considering the bit errorprobability Pb(e) than the symbol error probability P(e).

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Typically, modulation symbols are assigned to bit vectors (bitmapping), so that a symbol error corresponds to having areceived bit vector different from the transmitted one.

The bit error probability is the average number of errors in the

received bit vector divided by the vector size:

Pb(e) = E[Nb]

log2 M,

where Nb

denotes the number of bit errors.Of course, Pb(e) depends on the bit mapping.

Assuming high SNR, most errors occur between minimumdistance symbols.


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Digital modulations over the AWGN channel Standard digital modulations

PAM = Pulse Amplitude Modulation

The alphabet ofM-PAM is = (2m M 1) Mm=1.

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A { }For example, the constellation of8-PAM is as follows:

7

5

3

+ +3 +5 +7

The error probability ofM-PAM is:

P(e) = 2M 1

M

Q6log2 M

M2

1Eb

N0. (20)


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QAM = Quadrature Amplitude Modulation (cont.)

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The error probability ofM-QAM is:

P(e)

4

M 1

MQ

3log2 M

M 1

Eb

N0. (21)



PSK = Phase Shift Keying

The alphabet ofM-PSK isA = {Esej (2m1)/M}Mm=1.

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For example, the constellation of8-PSK is as follows:



PSK = Phase Shift Keying (cont.)

The error probability ofM-PSK is:

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P(e) 2 Q

2sin2

M log2 M

EbN0

. (22)

In the special case ofM= 4 we have:

P(e) 2 Q

2EbN0

.



Orthogonal modulations

The alphabet of an orthogonal modulation consists ofMvectors in RM with a single nonzero coordinate equal to

Es.

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For example, a quaternary orthogonal modulation isrepresented by the following four signals:

1= (

Es, 0, 0, 0), 2= (0,

Es, 0, 0),

3= (0, 0,

Es, 0), 4= (0, 0, 0,

Es).

The error probability of an M-ary orthogonal modulation is:

P(e) (M 1)Q

log2 MEbN0

. (23)



Orthogonal modulations (cont.)

Two examples of orthogonal modulations are given as follows.

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1 Pulse position modulation (M-PPM): Given the signal pulse(t), the modulated signals are:

xm(t) =

M (M t (m 1)T), (24)

i.e.,(t) is contracted in time to (0,T/M) and shifted by(m 1)T /M.

2 Frequency shift keying (M-FSK):

xm(t) =

2 cos[2(fc+ mf)t] (25)

where fc T and f Tare integer numbers.



Asymptotic comparison of digital modulations

Consider two digital modulation schemes with approximateunion bounds to the error probability

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P(e) iQ

iEbN0

fori= 1, 2.

The asymptotic behavior of the error probability (when Eb/N0is very large) is dominated by the Q-function term and can beapproximated by

P(e) e(i/2)Eb/N0

fori= 1, 2 (we used Q(x) exp(x2/2)).If1> 2, the first modulation is better than the secondsince its error probability is smaller at the same Eb/N0.



Asymptotic comparison of digital modulations (cont.)

The same asymptotic error probability is obtained when

Eb Eb

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1EbN01 =2EbN02,disregarding thei.

Hence, we define theasymptotic gainof the first modulation

with respect to the second one as the dB-difference between(Eb/N0)2 and (Eb/N0)1, which are the Eb/N0 ratios requiredto have the same asymptotic error probability:

G= 10 log10EbN02 10log10EbN01 = 10log10 12


Digital modulations over the AWGN channel Problem set 3

Problem set 3

1 Derive the PEP(17).

2 Derive the union bound approximation (19)along with the

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expression ofNmin by keeping only those terms from theupper bound

1

M

M

m=1 n=mP(m n)

corresponding to minimum distance errors, i.e., such thatm n =dmin.

3 Derive the error probability in(20).

4

Derive the error probability in(21).5 Derive the error probability in(22).

6 Check the orthogonality of the signals in(24)and(25).



Problem set 3 (cont.)7 Derive the error probability in(23).

8 Calculate the error probability of the 32-QAM modulationcharacterized by the following signal set:

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5 3 1 +1 +3 +55

3

1

+1

+3

+5




9 Find the error probability of the binary modulation whosesignals are

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s1(t) = 10

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s1(t) = A[u(t) u(t T)]s2(t) = A[u(t) u(t T /4) + u(t T /2) u(t 3T /4)]s3(t) = A[u(t) u(t T /4) u(t T /2) + u(t 3T /4)]

s4(t) = A[u(t) 2u(t T /2) + u(t T)]Calculate i) the average energy per bit Eb, ii) the minimumdistance d2min, and iii) the average symbol error probability(high-SNR approximation) in the form Q(

Eb/N0).




11 Calculate the error probability of a 4-PSK signal set assumingthat the receiver has a constant phase offset that rotates the

decision regions b an angle

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decision regions by an angle .12 Calculate the error probability of an octonary signal set whose

signals are located over two concentric circles with rays 1 and0.5 +

1.5. The signals are equally spaced over each circle

and have a phase offset of/4 radians between thecorresponding signals over different circles.

13 Calculate the error probability of the digital modulation basedon the following four signals:

sm(t) = sin 52Tt (m 1) T5 1|tmT/5|T/5form= 1, 2, 3, 4.


Digital modulations over the AWGN channel Power density spectrum of digital modulated signals

Power density spectrum of digital modulations

The power density spectrum ofx(t) = n an(t nT),where an is a wide-sense stationary sequence with

autocorrelation function Ra(p) E[an+pan] can be expressed

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autocorrelation function Ra(p) = E[an+pan], can be expressedas the product of two terms:

Power density spectrum: Gx(f) =Sa(f) G(f)

S

a(f) p Ra(p)ej 2pfT (data spectrum)

G(f) 1

T|(f)|2 (pulse spectrum)

In many circumstances, the bandwidth of a digital signal isapproximated by an expression depending only on thesignalling interval T.



Shannon bandwidth

A common approximation to the bandwidth of a digital signal

is the Shannon bandwidth:

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is theShannon bandwidth:

Wsh Nd 12T

where Nd is the signal space dimension and T is the symbol

interval.It can be shown that this approximation is very good whenthe number of dimensions is large.

However, even with Nd= 1, the bandwidth overhead is

limited for suitably chosen pulses, as illustrated in thefollowing example.


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Bandwidth of antipodal signals (cont.)

0.8

1

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0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

WT

(W)

square

sine



Bandwidth of antipodal signals (cont.)One way to limit the bandwidth occupation of a digitalmodulation signal consists of extending the duration of themodulation pulse beyond the signalling interval (0, T).

When the signalling pulse is limited to the signalling interval

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When the signalling pulse is limited to the signalling interval,the modulation signal is calledfull response. When itsduration exceeds T, it is calledpartial response.

Stretching in time the signalling pulse by a factor

corresponds to an equivalent stretching in the frequencydomain by the inverse of:

(t) (t/) (f) (f) (W) (W).

The price to be payed is related to the generation ofintersymbol interference.


Digital modulations over the AWGN channel Comparison of digital modulations

Key parameters

The performance of different modulation schemes is describedby three system parameters:

1 Error probability (symbol or bit).

2 Spectral efficiency, i.e., the ratio between the bit rate Rb and

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Spectral efficiency, i.e., the ratio between the bit rate R andthe occupied bandwidth W.

3 The signal-to-noise ratio Eb/N0.

For Nd-dimensional signal sets, the occupied bandwidth isapproximately equal to the Shannon bandwidth

Wsh=Nd 12T

= Nd Rb2log2 M

Hence, the spectral efficiency is given by

b RbWsh

= 2log2 M

Nd.



Spectral efficiency

The spectral efficiency b grows slowly (logarithmically) with

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The spectral efficiency grows slowly (logarithmically) withthe constellation size Mand decreases rapidly (linearly) withthe number of dimensions Nd.

For a fixed M, PAM modulations have higher spectralefficiency than orthogonal modulations. Therefore,

PAM modulations are used in channels with limited bandwidth(bandwidth limited channels) and high power.

Orthogonal modulations are used in channels with limitedpower (power limited channels) and large bandwidth.



Shannons bound

Shannons theorem yields the maximum bit rate that can be

sustained with arbitrarily low error probability by an

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sustained with arbitrarily low error probability by anNd-dimensional digital modulation with symbol interval Tover an AWGN channel:

Rb =

Nd

2T log21 + SN. (26)Here, S is the received power, N is the noise power, and S/Nis calledsignal-to-noise ratio.

We assume that the signal bandwidth is Wsh=Nd/(2T).



Shannons bound (cont.)Since the noise power is N=N0Wsh and the signal power isS=Eb/Tb =RbEb,(26)can be written as:

Rb Wshlog21 + RbEbWshN0.

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g + WshN0Since the spectral efficiency is b=Rb/Wsh, we obtain:

b log21 + b EbN0 EbN0 2b 1b .Finally, for b

0, we have

EbN0

ln 2

EbN0

dB

10log10(ln2) 1.6 dB.



Shannons bound (cont.)

50

60

1024-PAMShannon's bound

Pb(e)=1e-4

Pb(e)=1e-6

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10-2

10-1

100

101

0

10

20

30

40

Rb/W [bit/s/Hz]

Eb

/N0

[dB]

4-PPM

1024-PPM

2-PAM

-1.6 dB

b( )

Pb(e)=1e-8



Problem set 4

1 Derive the power density spectrum formulaGx(f) =Sa(f) G(f) for the signal

(t)

(t T ) (27)

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x(t) =n

an(t nT), (27)

where

an is a wide-sense stationary sequence with autocorrelationfunction Ra(p) = E[an+pan];

Sa(f)

p Ra(p)ej 2pfT is the data spectrum;

G(f) 1T|(f)|2 is the pulse spectrum.

Hint: Consider the (randomly delayed and stationary) signal

x(t ), with uniformly distributed in (0, T), and calculatethe Fourier transform of its autocorrelation function to obtainthe power density spectrum.




2 Calculate the power density spectrum of the signal (27)

assuming that the symbols an are uncorrelated with mean ad i 2

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g y and variance 2a.

3 Calculate the power density spectrum of the signal (27)assuming that the transmitted symbols an have zero meanand correlation Ra(m) =

|m| (where

(0, 1)), (t) hasunit energy, and the average signal power is P.

4 Calculate the power density spectrum of(27)assuming thatthe transmitted symbols are iid and taken from a 4-PSKsignal set with probabilities (0.7, 0.1, 0.1, 0.1).


notes 2013a

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