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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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Institute for Experimental Mathematics
Ellernstrasse 2945326 Essen - Germany
DATA COMMUNICATION2-dimensional transmission
A.J. Han VinckMay 1, 2003
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University Duisburg-Essen digital communications group
Contentwe describe
orthogonal signaling 2-dimensional transmission model
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„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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University Duisburg-Essen digital communications group
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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University Duisburg-Essen digital communications group
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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University Duisburg-Essen digital communications group
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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University Duisburg-Essen digital communications group
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