Baseband Digital TransmissionBaseband Digital Transmission

Multidimensional Signals

Multidimensional Vs. MultiamplitudeMultidimensional Vs. Multiamplitude

Multiamplitude signal is One-dimensional

With One basis signal: g(t)

Multidimensional signal Can be defined using multidimensional

orthogonal signals

-3d -d d 3d

00 01 10 11

Multidimensional Orthogonal SignalsMultidimensional Orthogonal Signals

Many ways to construct In this text, Construction of a set of M=2k

waveforms si(t), i=1,2,…,M-1 Mutual Orthogonality Equal Energy Using Mathematics

where Kronecker delta is

0( ) ( ) , , 0,1,..., 1

T

i k iks t s t dt i k M 1,

0,ik

i k

i k

Example of M=2Example of M=222

Multiamplitude Vs. Multidimensional

All signal has identical energy

See figure 5.27, page 202, for signal constellation

s3

s2

s1

s0

T

3dd

-d-3d

s0 s1 s2 s3

T

A

22

0( ) , , 0,1,..., 1

T

i

A Ts t dt i k M

M

Receiver for AWGN ChannelReceiver for AWGN Channel

Received signal from AWGN channel

Receiver decides which of M signal waveform was transmitted by observing r(t) Optimum receiver minimize Probability of Error

( ) ( ) ( ), 0,1,..., 1, 0ir t s t n t m M t T

White Gaussian processWith power spectrum N0/2

Optimum Receiver Optimum Receiver for AWGN Channelfor AWGN Channel

Needs M Correlators (or Matched Filters)

0( ) ( ) , 0,1,..., 1

T

i ir r t s t dt i M

0( )td

r(t)

s0(t) Samples at t=T

0( )td

sM-1(t) Detector OutputDecision

r0

rM-1

Signal CorrelatorsSignal Correlators

Let s0(t) is transmitted

Noise component (Gaussian Process) Zero mean and Variance

0 0 0 00 0

20 0 00 0

( ) ( ) { ( ) ( )} ( )

( ) ( ) ( )

T T

T T

r r t s t dt s t n t s t dt

s t dt s t n t dt n

0 00 0

00 0

( ) ( ) { ( ) ( )} ( )

( ) ( ) ( ) ( ) , 0

T T

i i

T T

i i i

r r t s t dt s t n t s t dt

s t s t dt s t n t dt n i

Orthogonal

2 2 20 0

0( ) ( )

2 2

T

i i

N NE n s t dt

pdf of correlator outputpdf of correlator output

Let s0(t) is transmitted

Mean of correlator output E[r0] = , E[ri] = 0

0( )td

s0(t)+n(t)

s0(t) Samples at t=T

0( )td

sM-1(t)

r0

rM-1

2 20

0 0

( ) / 2

( | ( ))

1

2r

P r s t

e

2 2

0

/ 2

( | ( ))

1

2i

i

r

P r s t

e

0

Optimum DetectorOptimum Detector

Let s0(t) is transmitted The probability of correct decision

Is probability that r0 > ri

The average probability of symbol error

For M=2 case (Binary Orthogonal signal)

0 1 0 2 0 1( , ,..., )c MP P r r r r r r

20

0 1 0 2 0 1

( 2 / ) / 21

1 1 ( , ,..., )

1{1 [1 ( )] }

2

M c M

y NM

P P P r r r r r r

Q y e dy

20

( ),bbP Q for binary

N

More on Probability errorMore on Probability error

Average probability of symbol error Same even if si(t) is changed Numerically evaluation of integral

Converting probability of symbol error to probability of a binary digit error

Figure 5.29 page 206 As M64, we need small SNR to get a given probability of error

12

2 1

k

b MkP P

Symbol error probabilityBit error probability

Biorthogonal SignalsBiorthogonal Signals

Biorthogonal For M=2k multidimensional signal

M/2 signals are orthogonal M/2 signals are negative of above signals M signals having M/2 dimensions

Example of M = 22

A

T/2

s0(t)

A

T/2 T

s1(t)

-A

T/2

s3(t)= -s0(t)

-A

T/2 T

s4(t)= -s1(t)

Symbol interval T=kTb

Biorthogonal SignalsBiorthogonal Signals

Signal Constellation

Examples M=2 : Antipodal signal M=4

0 / 2

1 / 2 1

/ 2 1 1

( ) ( ,0,0,...,0) ( ,0,0,...,0)

( ) (0, ,0,...,0) (0, ,0,...,0)

( ) (0,0,0,..., ) (0,0,0,..., )

M

M

M M

s t s

s t s

s t s

( ,0)( ,0) ( ,0)( ,0)

(0, )

(0, )

Receiver for AWGN ChannelReceiver for AWGN Channel

Received signal from AWGN channel

Receiver decides which of M signal waveform was transmitted by observing r(t) Optimum receiver minimize Probability of Error

( ) ( ) ( ), 0,1,..., 1, 0ir t s t n t m M t T

White Gaussian processWith power spectrum N0/2

Optimum Receiver Optimum Receiver for AWGN Channelfor AWGN Channel

Needs M/2 correlator (or matched filter)

0( ) ( ) , 0,1,..., 1

2

T

i i

Mr r t s t dt i

0( )td

r(t)

s0(t) Samples at t=T

0( )td

sM/2-1(t) Detector OutputDecision

r0

rM/2-1

Signal CorrelatorsSignal Correlators

Let s0(t) is transmitted (i = 0,1, … ,M/2 – 1)

Noise component (Gaussian Process) Zero mean and Variance

0 0 0 00 0

20 0 00 0

( ) ( ) { ( ) ( )} ( )

( ) ( ) ( )

T T

T T

r r t s t dt s t n t s t dt

s t dt s t n t dt n

0 00 0

00 0

( ) ( ) { ( ) ( )} ( )

( ) ( ) ( ) ( ) , 0

T T

i i

T T

i i i

r r t s t dt s t n t s t dt

s t s t dt s t n t dt n i

Orthogonal

2 2 20 0

0( ) ( )

2 2

T

i i

N NE n s t dt

pdf of correlator outputpdf of correlator output

Let s0(t) is transmitted (for M/2 –1 correlator)

Mean of correlator output E[r0] = , E[ri] = 0

0( )td

s0(t)+n(t)

s0(t) Samples at t=T

0( )td

sM/2-1(t)

r0

rM/2-1

2 20

0 0

( ) / 2

( | ( ))

1

2r

P r s t

e

2 2

0

/ 2

( | ( ))

1

2i

i

r

P r s t

e

0

The detectorThe detector

Select correlator output whose magnitude |ri| is largest

But, we don’t distinguish si(t) and –si(t) By observing |ri | We need sign information

si(t) if ri > 0 - si(t) if ri < 0

max{ }, 0,1,..., 12j i

i

Mr r i

Probability of ErrorProbability of Error

Probability of correct decision

Probability of symbol error

See figure 5.33 page 213 Note M=2 and M=4 case

Symbol error probability and Bit-error probability

2 220 00

0 0

/ / 2 ( ) / 2/ 2 100 / / 2

1 1[ ]2 2

r N rx Mc r NP e dx e dr

1M cP P

HomeworkHomework

Illustrative problem 5.10, 5.11

Problems 5.11, 5.13