Digital communication - vector approach Dr. Uri Mahlab 1 Digital Communication Vector Space concept

Digital communication - vector approach1Dr. Uri Mahlab

DigitalCommunicationVector Space

concept


Signal space Signal Space Inner Product Norm Orthogonality Equal Energy Signals Distance Orthonormal Basis Vector Representation Signal Space Summary


Signal Space

S(t) S=(s1,s2,…)

•Inner Product (Correlation)•Norm (Energy)•Orthogonality•Distance (Euclidean Distance)•Orthogonal Basis


Energy dttsEsT

)(0

2

ONLY CONSIDER SIGNALS, s(t)

Tt

tifts

0

0)( Tt


T

dttytxtytx0

)()()(),(

Similar to Vector Dot Product

x

yyx

cosyxyx

Inner Product - (x(t), y(t))


A

-A2A

A/2

T

Tt

t

Example

TAT

AATA

Atytx 2

4

3

2)2)((

2)

2)(()(),(


Norm - ||x(t)||

T

ExEnergydttxtxtxtx0

22)()(),()(

Extx )(

Similar to norm of vector

T

A

-A

x

xxx 2

ExT

AdttT

AtxT

2)

2cos()(

0

2


Orthogonality

0)(),( tytx T

dttytx0

0)()(

Similar to orthogonal vectors

T

A

-Ax

0yx

T

Y(t)B

y


•ORTHONORMAL FUNCTIONS

{

1)()(

0)(),(

tytx

and

tytx

TT

T

dttydttx

dttytx

0

2

0

2

0

1)()(

0)()(

T

T

X(t)

Y(t)

T/2

T/2

1

1

x

y

1)()(

0)(),(

tytx

tytx


Correlation Coefficient

EyEx

dttytx

tytx

tytx

T

0

)()(

)()(

)(),(

•In vector presentation

1 -1=1 when x(t)=ky(t) (k>0)

yx

yx cos

x

y


Example

T

TAdttytxtytx0

2

4

5)()()(),(

Now,

14.0

)87)(10(

45)(),(

2

TATA

TA

EyEx

tytx

shows the “real” correlation

t tA

-AT/2 7T/8

T

10A

X(t) Y(t)


Distance, d

ExEy2EyExd

dt)t(y)t(x)t(y)t(xd

2

T

0

222

• For equal energy signals

)1(E2d2 • =-1 (antipodal) E2d

• 3dB “better” then orthogonal signals

• =0 (orthogonal) E2d


Equal Energy Signals)1(2 Ed

E2d

E

y

x

•PSK (phase Shift Keying)

tfAty

Tt

tfAtx

0

0

2cos)(

)0(

2cos)(

•To maximize d

)()(

1

tytx (antipodal signals)

E2d


•EQUAL ENERGY SIGNALS ORTHOGONAL SIGNALS (=0)

Ed 2E

y

x

Ed 2

PSK (Orthogonal Phase Shift Keying)

tfAty

Tt

tfAtx

0

1

2cos)(

)0(

2cos)(

(Orthogonal if ...),2

3,1,

2

1)( 01 Tff


Signal Space summary• Inner Product

T

dttytxtytx0

)()()(),(

•Norm ||x(t)||

EnergydttxtxtxtxT

0

22)()(),()(

•Orthogonality

)(1)()(

0)(),(

Orthogonaltytx

if

tytx


• Corrolation Coefficient,

ExEy

dttytx

tytx

tytx

T

0

)()(

)()(

)(),(

•Distance, d

ExEy2EyExd

dt)t(y)t(x)t(y)t(xd

2

T

0

222


Modulation


Modulation Modulation

BPSK

QPSK

MPSK

QAM

Orthogonal FSK

Orthogonal MFSK

Noise

Probability of Error


)()(

)()(

so,

2cos2

)(

define We

Tt0

2cos2

)(

2cos2

)(

11

10

01

01

00

tEtx

tEtx

tfT

t

tfT

Etx

tfT

Etx

EE-)(t

Ed 2

sec

1 bit

TRbit

Binary Phase Shift Keying – (BPSK)


Binary antipodal signals vector presentationConsider the two signals:

Tt0 tf2cosT

E2)t(s)t(s c21

The equivalent low pass waveforms are:

Tt0 T

E2)t(u)t(u 21


The vector representation is – Signal constellation.

E E


The cross-correlation coefficient is:

1ss

ss)Re(

21

2112

The Euclidean distance is:

E2)Re(1E2d 2/11212

Two signals with cross-correlation coefficient of -1 are called antipodal


Multiphase signals

Consider the M-ary PSK signals:

tf2sin)1m(M

2sin

T

E2tf2cos)1m(

M

2cos

T

E2

Tt0 , M1,2,...,m )1m(M

2tf2cos

T

E2)t(s

cc

cm

The equivalent low pass waveforms are:

Tt0 , M1,2,...,m eT

E2)t(u M/)1m(2j

m


The vector representation is:

M1,2,...,m )1m(M

2sinE),1m(

M

2cosEsm

Or in complex-valued form as:

M/)1m(2jm eE2u

E

1s

4s

3s

2s

E

1s

4s

5s

3s

2s

8s

7s6s

4M 8M


Their complex-valued correlation coefficients are :

k)/M-(mj2

T

0

m*kkm

e

M1,2,...,m , M 1,2,...,k dt)t(u)t(uE2

1

and the real-valued cross-correlation coefficients are:

)km(M

2cos)Re( km

The Euclidean distance between pairs of signals is:

2/1

2/1kmkm

)km(M

2cos1E2

)Re(1E2d


The minimum distance dmin corresponds to the case which| m-k |=1

M

2cos1E2dmin


)(1 tx)(4 tx

)(3 tx )(2 tx

sE(00)

(01)(11)

(10)

sEd 2min

*

Quaternary PSK - QPSK


tf2sinT

E2)t( 02

X(t)

E

1a

2a

tf2cosT

E2)t( 01

sinT/E2a 2

T

EA

2

)tf2cos(T

E2)t(x 0

cosT/E2a1


)t(a)t(a

tf2sinsinT

E2tf2coscos

T

E2)t(x

Tt0 A0

)tf2cos(A)t(x

2211

00

0

)(1 tx)(4 tx

)(3 tx )(2 tx

sE(00)

(01)(11)

(10)

sEd 2min mind


2

E

4log

E

Mlog

EE

2/RR

sec

bits

T

1R

E2d

)4

3tf2cos(

T

E2)t(x

s

2

s

2

sb

bitsymbol

bbit

smin

04


mind

)(1 t

)(2 t

MEd

sin2

MPSK


3

E

8log

EE

8M

g.esec

bits

T

1)M(logR

s

2

sb

2bit


Consider the M-ary PAM signals

]e)t(uRe[A

tf2cosT

2A)t(s

tf2jm

cmm

c

m=1,2,….,M

Where this signal amplitude takes the discrete values (levels)

mA

M1m2Am m=1,2,….,M

The signal pulse u(t) , as defined is rectangular

U(t)= T

2Tt0

But other pulse shapes may be used to obtain a narrower signal spectrum .

Multi-amplitude Signal


Clearly , this signals are one dimensional (N=1) and , hence, are represented by the scalar components

m1m As M=1,2,….,M

The distance between any pair of signal is

|AA|)ss(d km

2

1k1mmk

2

0

M=2

M=4

)t(f1

1s

3s1s

2s

2s 4s

222

)t(f1

0

Signal-space diagram for M-ary PAM signals .


The minimum distance between a pair signals

2dmin


tf2sinT

2Atf2cos

T

2A)t(s cmscmcm

]e)t(u)jAARe[( tf2jmsmc

c

Where and are the information bearing signal amplitudes of the quadrature carriers and u(t)= .

mcA msA

T

2 Tt0

A quadrature amplitude-modulated (QAM) signal or a quadrature-amplitude-shift-keying (QASK) is represented as

Multi-Amplitude MultiPhase signalsQAM Signals


QAM signals are two dimensional signals and, hence, they are represented by the vectors

)AA(s ms,mcm The distance between a pair of signal vectors is

2kmmk |ss|d

])()[( 22ksmskcmc AAAA k,m=1,2,…,M

When the signal amplitudes take the discrete values

M,....,2,1m,M1m2 In this case the minimum distance is 2dmin


QAM (Quadrature Amplitude Modulation)

)(2 t

)(1 td


Mlog/TT

E5

2d

d2

5E

2

d3

2

d

2

1

2

d32

4

1

2

d2

4

1E

QAM16

2bitsysmbol

AVG

2AVG

2222

AVG

QAM=QASK=AM-PM

)(2 t

)(1 td


M=256M=128M=64M=32M=16M=4

+


For an M - ary QAM Square Constellation

22

S

Sn2

n

2S

d12

1ME

signal lDimentiona - One aFor

E12

6d

lbits/symbon 2M

d6

1ME

AVG

AVG

In general for large M - adding one bit requires 6dB more energy to maintain same d .


Binary orthogonal signals

Tt0 tf2sinT

E2)t(s

Tt0 tf2cosT

E2)t(s

c2

c1

Consider the two signals

Where either fc=1/T or fc>>1/T, so that

T

0

2112 dt)t(s)t(sE

1)Re(

Since Re(p12)=0, the two signals are orthogonal.


The equivalent lowpass waveforms:

Tt0 T

E2j)t(u

Tt0 T

E2)t(u

2

1

The vector presentation:

E,0s 0,Es 22

Which correspond to the signal space diagram

E

E

12d

1s

2s

Note that

E2d12


We observe that the vector representation for the equivalent lowpass signals is

]u[u

]u[u

212

111

Where

E2j0u

0jE2u

21

11


]ftm2tf2cos[T

2)t(s cm

]e)t(uRe[ tf2jm

c m=1,2,….,M Tt0

This waveform are characterized as having equal energy and cross-correlation coefficients

dte2T2

ftm2jkm

f)km(Tjef)km(T

f)km(Tsin

Let us consider the set of M FSK signals

M-ary Orthogonal Signal


r

0

T2

1

T

1

T2

3T

2

The real part of iskm

f)km(Tcosf)km(T

f)km(Tsin)Re( kmr

f)km(T2

f)km(T2sin

f


First, we observe that =0 when and . Since |m-k|=1 corresponds to adjacent frequency slots , represent the minimum frequency separation between adjacent signals for orthogonality of the M signals.

)Re( kmT2

1f

km

T2

1f


2

1s

3s

2s

2

2

Orthogonal signals for M=N=3signal space diagram

For the case in which ,the FSK signalsare equivalent to the N-dimensional vectors

1s

2s

=( ,0,0,…,0)

=(0, ,0,…,0)

Ns =(0,0,…,0, )

Where N=M. The distance between pairs of signals is

2dkmall m,k

Which is also the minimum distance.

T2/1f


Orthogonal FSK(Orthogonal Frequency Shift Keying)

,...2

3,1,

2

1)f(f

Tt0 2cos

2)(

2cos2

)(

01

11

00

T

tfT

Etx

tfT

Etx

T

tfT

tfT

tfT

tfT

010

10

02cos2

2cos2

02cos2

,2cos2


Ed 2

)(1 t

)(2 t

“0”

“1”

tfT

t

tfT

t

12

01

2cos2

)(

2cos2

)(

sec

1 bits

TRbit


ORTHOGONAL MFSK

2cos2

)(

2cos2

)(

2cos2

)(

33

22

11

tft

Etx

tft

Etx

tft

Etx


All signals are orthogonal to each other

)(1 t

)(2 t

)(3 t

Ed 2

E

E

E

sec

1)(log2

bits

TMRbit


How togeneratesignals


0 T 2T 3T 4T 5T 6T

0 T 2T 3T 4T 5T 6T

+

tfEb 02cos2

tfEb 02sin2

tf2sinT

2Atf2cos

T

2A)t(s cmscmcm


0 T 2T 3T 4T 5T 6T

0 T 2T 3T 4T 5T 6T

+

tfEb 02cos2

tfEb 02sin2

tf2sin)t(Qtf2cos)t(I)t(s ccm

)t(sm


0 T 2T 3T 4T 5T 6T

0 T 2T 3T 4T 5T 6T

+

tfEb 02cos2

tfEb 02sin2

tf2sin)t(Qtf2cos)t(I)t(s ccm

)t(sm

)t(I

)t(Q


+

tfEb 02sin2

)t(sm

)t(I

)t(Q

tfEb 02cos2

IQ Modulator


+

tfEb 02sin2

)t(sm

)t(I

)t(Q

tfEb 02cos2

IQ ModulatorPulse shaping filter


NOISE


What about Noise•White Gaussian Noise

T T

)(1 tn )(2 tn

1i

ii1 )t(a)t(n

1i

ii2 )t(a)t(n

•The coefficients are random variables !


WHITE GAUSSIAN NOISE (WGN)

Hz

Watts

2

0N)f(pn

We write

)t(n)t(n i1i

i

•All are gaussian variables•All are independent

in

in

)n(f

)...n(f)n(f,...)n,n(f)n(f

i1i

2121


•All have same probability distribution

e

2

N2

1)f(n

2

NnE

0nE

0

2i

N

n

0

i02

i

i


)t(n

1n

3n

2n

•White Gaussian Noise has energy in every dimension

0

2i

N

n

01i

i1i

e

2N

2

1)n(f)n(f


Probability of Error for Binary

SignalingThe two signal waveforms are given as

These waveforms are assumed to have equal energy E and their equivalent lowpass um(t), m=1,2 are characterized by the complex-valued correlation coefficient ρ12 .

]e)t(uRe[)t(s tf2jmm

cTt0 1,2m


The optimum demodulator forms the decision variables

Or,equivalently

And decides in favor of the signal corresponding to the larger decision variable .

T

0

*mm dt)t(u)t(rReU

*m

jm ureRe)u(

1,2m

1,2m


Lets see that the two expressions yields the same probability of error .

Suppose the signal s1(t) is transmitted in the interval 0tT . The equivalent low-pass received signal is

Substituting it into Um expression obtain

Where Nm, m=1,2, represent the noise components in the decision variables,given by

r111 NE2)NE2Re(U

r2r22 NE2)NE2Re(U

)t(z)t(ue)t(r 1j Tt0

T

0

*m

jm dt)t(u)t(ZeN


And .

The probability of error is just the probability that the decision variable U2 exceeds the decision variable u1 . But

Lets define variable V as

N1r and N2r are gaussian, so N1r-N2r is also gaussian-distributed and, hence, V is gaussian-distributed with mean value

)NRe(N mmr

)0UU(P)0UU(P)UU(P 211212

r2r1r21 NN)1(E2UUV

)1(E2)v(Em rv


And variance

Where N0 is the power spectral density of z(t) .

The probability of error is now

r0

2r2r2r1

2r1

2r2r1

2v

1EN4

NENNE2NE

NNE

r0

2

02/)mv(

v

0

1N2

Eerfc

2

1

dve2

1

dv)v(p)0V(P

2v

2v


Where erfc(x) is the complementary error function, defined as

It can be easily shown that

x

t dte2

)x(erfc2

r

0

2

2 1N2

Eerfc

2

1p


Distance, d

ExEy2EyExd

dt)t(y)t(x)t(y)t(xd

2

T

0

222

• For equal energy signals

)1(E2d2 • =-1 (antipodal) E2d

• 3dB “better” then orthogonal signals

• =0 (orthogonal) E2d


It is interesting to note that the probability of error P2 is expressed as

Where d12 is the distance of the two signals . Hence,we observe that an increase in the distance between the two signals reduces the probability of error .

0

212

2

2 N2

derfc

2

1p


0

212

2

2 N2

derfc

2

1p

r

0

2

2 1N2

Eerfc

2

1p


0

212

2

2 N2

derfc

2

1p

M=256M=128M=64M=32M=16M=4

+

2

1s

3s

2s

2

2

mindM

Ed

sin2 E E

Documents

Digital communication - vector approach Dr. Uri Mahlab 1 Digital Communication Vector Space concept