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Institute for Experimental Mathematics Ellernstrasse 29 45326 Essen - Germany DATA COMMUNICATION 2-dimensional transmission A.J. Han Vinck May 1, 2003

DATA COMMUNICATION 2-dimensional transmission

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DATA COMMUNICATION 2-dimensional transmission. A.J. Han Vinck May 1, 2003. we describe orthogonal signaling 2-dimensional transmission model. Content. „orthogonal“ binary signaling. 2 signals S 1 (t) S 2 (t) in time T Example: Property :orthogonal energy E. T. T. - PowerPoint PPT Presentation

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Page 1: DATA COMMUNICATION 2-dimensional transmission

Institute for Experimental Mathematics

Ellernstrasse 2945326 Essen - Germany

DATA COMMUNICATION2-dimensional transmission

A.J. Han VinckMay 1, 2003

Page 2: DATA COMMUNICATION 2-dimensional transmission

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University Duisburg-Essen digital communications group

Contentwe describe

orthogonal signaling 2-dimensional transmission model

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University Duisburg-Essen digital communications group

„orthogonal“ binary signaling2 signals S1 (t) S2 (t) in time T

Example:

Property: orthogonal

energy E

T22

T11

T21

Edt)t(S)t(S

Edt)t(S)t(S

0dt)t(S)t(S

T T

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Quadrature Amplitude Modulation: QAM

)tf2sin(

T/kf

)tf2cos(

cT2

2

c

cT2

1

bE

bE

S(t)

1

1

0 0

0

1

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QAM receiver

T)1i(

iTdt

)tf2sin(

T/kf

)tf2cos(

cT2

2

c

cT2

1

T)1i(

iTdt

+/-1/0

+/-1/0

r(t)bE)/(

r(t) = S(t) + n(t) Note: sin(x)sin(x) = ½ (1 – cos (2x) )

sin(x)cos(x) = ½ sin (2x)

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about the noise

211

cc2

T2

T)1i(

iT

cc2

T2

T)1i(

iT

T)1i(

iT

ccT2

T)1i(

iT

T)1i(

iT

cT2

T)1i(

iTcT

2T)1i(

iT21

i

cT2

T)1i(

iT2cT

2T)1i(

iT1

)nn(Ethatseetoeasyisit

0

dt)tf2sin()tf2cos(

dtd)f2sin()tf2cos()t(

dtd)f2sin()tf2cos()](n)t(n[E

]d)f2sin()(ndt)tf2cos()t(n[E)nn(E

2,1i;0)n(E

dt)tf2sin()t(nn;dt)tf2cos()t(nn

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about the noise

Conclusion: n1 and n2 are Gaussian Random Variables

zero mean

uncorrelated (and thus statistically independent (f(x,y) =f(x)f(y) )

with variance 2.

.e2

1),(p)(p

0)nn(E

.)n(E.2,1i;0)n(E

;dt)tf2sin()t(nn

;dt)tf2cos()t(nn

22/)22

21(

2212n,1n

21

221i

cT2

T)1i(

iT2

cT2

T)1i(

iT1

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Geometric presentation (1)

b1 E

b2 E

b2 E

b1 E

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Geometric presentation (2)

ML receiver: find maximum p(r|s) min p(n) decision regions

1110

00 01

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performance

bE

bE

b

b

Ebit.inf/energyE2symbol/energy

224/2d

21 e)2/d(Q

From Chapter 1: P(error) =

bE2/d

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extension

4-QAM 2 bits 16-QAM

4 bits/s

Channel 1 Channel 2

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Geometric presentation (2)1

2

transmitted

received

noise vector n

The noise vector n has length |n| = ( 12

+ 22) ½

n has a spherically symmetric distribution!

equal density

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Geometric presentation (1)

r‘

rd/2

Prob (error) = Prob(length noise vector > d/2)

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Error probability for coded transmissionThe error probabiltiy is similar to the 1-dimensional situation:

We have to determine

the minimum d2Euclidean between any two codewords

Example:

sEC

C‘sE2

sE2

d2Euclidean = sE6

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Error probability

The two-code word error probability is then given by:

'candcbetween

cetandisEuclideansquaredtheis)'c,c(d

where

)2/)'c,c(d(QPe

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modulation schemes

On-off FSK8-PSK

3 bits/s

16-QAM

4 bits/s

4-QAM 2 bits

1 bit/symbol 1 bit/symbol

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transmitted symbol energyenergy: per information bit must be the same

FSK

bE

bE

bE

b

b

Ebit.inf/energyE2symbol/energy

bit.inf/energyEsymbol/energy b

b

b

Ebit.inf/energyE3symbol/energy

bE2

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performance

224/2d

21 e)2/d(Q

From Chapter 1: P(error) =

d/2

FSK b21 E2/d

bE

bE2/d bE

bE

bE3

b21 E2/d

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Coding with same symbol speed

2/Eb

In k symbol transmissions, we transmit k information bits. We use a rate ½ code

bE

In k symbol transmissions, we transmit k bits

ML receiver:

2/Eb

k

1i

2i

k

1i

2ii n)cr(:imizemin

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Famous Ungerböck coding

bE2In k symbol transmissions transmit

We can transmit

2k information bits

and k redundant digits

In k symbol transmissions transmit 2k digits

Hence, we can use a code with rate 2/3 with the same energy per info bit!

bE

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modulator

info

encoder

ci

ci {000,001,010,...111}

bE2

Signal mapper

23

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exampletransmit

or

Parity even

Parity odd

Decoder:

1) first detect whether the parity is odd or even

2) do ML decoding given the parity from 1)

Homework: estimate the coding gain

00

11

0110

00 00

11 11

11

00 01

10

00

10 11

01

10

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Example: Frequency Shift Keying-FSK

Transmit: s(1):=

s(0):= .0i;T/iff;Tt0tdtf2cos

tf2cos

100TbE2

1TbE2

1iforE

0ifor0

Tt0tdtf2costf2coss

b

iTbE2

1TbE2T

0i1

Note:

FSK

bE

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Modulator/demodulator

)tf2cos(

)tf2cos(

0TbE2

0

1TbE2

1

S(t)

m

m

0

1

r(t)

m

Select largest

demodulator

modulator

T

0dt*

T

0dt*

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Ex: Binary Phase Shift Keying-BPSK

Transmit: s(1):=

s(0):= .Tt0tdtf2cos

tf2cos

cTbE2

cTbE2

m

> or < 0?

m‘T

0dt*

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On-off

BFSK

BPSK

Modulation formats

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10-1

10-2

10-3

10-4

10-5

10-6

10-7

On-off

BPSKQPSK

Eb/N0 dB

Error rate

PERFORMANCE

5 10 15