Transcript Systematic Design of Space-Time Trellis Codes for Wireless

ECE 4371, Fall, 2014
Introduction to Telecommunication
Engineering/Telecommunication Laboratory
Zhu Han
Department of Electrical and Computer Engineering
Class 15
Oct. 22nd, 2014
Outline

BER and Decision

Digital Carrier System
– Carrier band vs. baseband
– Baud rate, bit rate, bandwidth efficiency
– Spectrum
– Coherent, noncoherent receiver
– BER
– Comparison

Homework 4
– 7.2.6, 7.3.4, 7.4.2, 7.5.1, 7.7.4, 7.8.1, Due 11/18/13

4117 Lab #4 and #5 11/10,

4117 Lab #6 due at last class,
Bit Error Probability
Noise na(t)
d(i)
We assume:
gTx(t)
i T
gRx(t)
r0 (i T )  n(iT )
• binary transmission with d (i )  {d 0 ,d1}
• transmission system fulfills 1st Nyquist criterion
• noise
n(i T), independent of data source
p N (n )
Probability density function (pdf) of
n(i T)
Mean and variance
n
Conditional pdfs
The transmission system induces two conditional pdfs depending on d (i )
• if d (i )  d 0
• if d (i )  d1
p0 (x )  pN ( x d 0 )
p1 ( x )  pN ( x  d1 )
p0 (x )
p1 ( x )
d0
x
d1
x
Example of samples of matched filter output
for some bandpass modulation schemes
Figure 5.8 Illustrating the
partitioning of the observation
space into decision regions for
the case when N  2 and M  4;
it is assumed that the M
transmitted symbols are equally
likely.
Probability of wrong decisions
S
Placing a threshold
p1 ( x )
p0 (x )
Probability of
wrong decision
x
x
S
 d0
S
Q0   p0 ( x) dx
Q1 
S
S
d1
 p1 ( x)dx
When we define P0 and P1 as equal a-priori probabilities of d 0 and d1
1
(
P

P

0
1
2)
we will get the bit error probability

Pb  P0Q0  P1Q1 
1
2
S
s p ( x)dx   p ( x)dx 
0
1

S
S
1
1
2
 p0 ( x ) dx

S
1
2



1
2
p1 ( x)  12 p0 ( x )  dx
Conditions for illustrative solution
With 
P1  P0 
1
and
2
 pN (  x )  pN ( x )

S
d 0  d1
2
S
S

1
Pb  1   p1 ( x ) dx   p0 (x ) dx 
2  


S
d d
S 0 1
2
p1 ( x) dx   pN ( x d1 )dx
S
p1 ( x) dx
S 

d 0 d1
2
 pN ( x )d x
equivalently
S
with 
substituting x d1  x 
 p0  x  dx 
d1  d 0
d 0 d1
d 0 d1
2
x


1 2
1
S
for
d1  d 0
   p N ( x  )d x     p N ( x  )d x 
2
2
2 0
2 0
1
d 0  d1
d 0 d1
  p N  x '  dx '
S 
 d1
d
d

0
2
1
2
2
0
2

1
Pb  1  2  p N ( x )dx 
2
0

Special Case: Gaussian distributed noise
Motivation: • many independent interferers
• central limit theorem
• Gaussian distribution
pN ( n ) 

1
2  N
e
n
2
2 N2
d1 d 0
2 

x2 
1
2

2 N2
e
d x
 Pb  1 

2
2  N 0 0


 

no closed solution
Definition of Error Function and Error Function Complement
erf( x ) 
2
x
e

 0

x
2
d x
erfc( x )  1  erf( x )
Error function and its complement
function y = Q(x)
y = 0.5*erfc(x/sqrt(2));
2.5
erf(x)
erfc(x)
2
1.5
erf(x), erfc(x)

1
0.5
0
-0.5
-1
-1.5
-3
-2
-1
0
x
1
2
3
Bit error rate with error function complement

1
2
Pb  1 
2


1
2 N
d1  d0
2

0


2 N2
e
d x 


x2
 d  d0 
1
Pb  erfc 1

2
 2 2 N 
Expressions with ES and N 0
antipodal: d1   d ; d 0  d
 d d
1
Pb  erfc  1 0
2
 2 2 N
 d2
1
 erfc 
 2 N2
2

 1
 d
  erfc 
 2
 2 N
unipolar d1  d ; d 0  0



 1
 SNR 
  erfc 

 2
2



d2
ES
SNR  2 
 N matched N0 / 2
 ES 
1

Pb  erfc
2
 N0 
 d
1
Pb  erfc 
2
 2 2 N
 d2
 1
  erfc 
2
 2
 8 N




 d2 / 2  1
 SNR 
1
 erfc 

erfc



 4 N2  2
2
4




SNR 
Q function
d2 / 2
 N2
ES
matched N / 2
0

 ES
1
Pb  erfc 
 2 N0
2





Bit error rate for unipolar and antipodal transmission
BER vs. SNR
theoretical
10
-1
simulation
unipolar
10
-2
BER

antipodal
10
10
-3
-4
-2
0
2
4
ES
in dB
N0
6
8
10
Digital Carrier System Baseband analysis

Signal in baseband: xTp (t )  T   d (l )  jd (l )  gTx (t  lT )
l 

mean symbol energy:
ES  T 2 D


gT2x (t )dt

signal in carrier band:
xBp (t )  2 Re xTp (t )e j 2 f0t
2





 2T cos(2 f 0 )  d (l ) gTx (t  lT )  sin(2 f 0 )  d (l ) gTx (t  lT ) 
l 
l 


mean symbol energy:
 D 2 D 2  
  gTx2 (t )dt ES
EX  T 2  2  


2  
 2


D2
Conclusion: analysis of carrier band = base band. Fc=0 in project
Baud Rate, Bit Rate, Bandwidth Efficiency

Remember channel capacity C=Wlog2 (1+ SNR)> fb
Power Spectrum, ASK

Baseband

Sy(W)=Sx(W) P(W)

ASK: Sy(t)=b Acoswct, Square wave convolute with sinusoid.
FSK Spectrum

FSK: two sinc added together

 A cos2f1t 
s t   

 A cos2f 2t 
binary1
binary 0
BPSK Spectrum

BPSK: Sx(W): NRZ. P(t): raised cosine function. Sy(W)= P(W)

Rb
baud rate
QPSK Spectrum

Same Rb
Narrow BW
Pulse Shaped M-PSK

Different 
Bandwidth vs. Power Efficiency

Bandwidth efficiency high, required SNR is high and low power efficiency
QAM efficiencies


For l =1  PSD for BPSK
For l =2  PSD for QPSK, OQPSK …

PSD for complex envelope of the bandpass multilevel signal is
same as the PSD of baseband multilevel signals

Same baud rate, higher bit rate.

Same bit rate, less bandwidth. But higher power
Minimum Shift Keying spectra

Continuous phase and constant envelop. So narrow spectrum
GMSK spectral shaping
Coherent Reception

An estimate of the channel phase and attenuation is recovered. It
is then possible to reproduce the transmitted signal, and
demodulate. It is necessary to have an accurate version of the
carrier, otherwise errors are introduced. Carrier recovery
methods include:
Coherent BER

PSK
– BPSK
– MPSK
QPSK
Coherent BER performance

 1
Eb 

Pb  2(1  )Q

 L 1 N 
ASK
1
L
 1
Eb 

Pb  2(1  L1 )Q

L

1
N


 1.217Eb
Pb  Q
N






FSK

MSK: less bandwidth but the same BER

MQAM
Non-coherent detection

Non-coherent detection
– does not require carrier phase recovery (uses differentially encoded mod.
or energy detectors) and hence, has less complexity at the price of higher
error rate.

No need in a reference in phase with the received carrier

Differentially coherent detection
– Differential PSK (DPSK)


The information bits and previous symbol, determine the phase of the
current symbol.
Energy detection
– Non-coherent detection for orthogonal signals (e.g. M-FSK)


Carrier-phase offset causes partial correlation between I and Q
braches for each candidate signal.
The received energy corresponding to each candidate signal is used
for detection.
Differential Reception
Differential Coherent

DBPSK

3dB loss
Non-coherent detection of BFSK
2 / T cos(1t )

T

T

T
z11
 2
0
z11  z12
2
2 / T sin(1t )
r (t )
z12
0
 2
+
2
z (T )
2 / T cos(2t )
z 21
 2
-
0
z21  z22
2
2 / T sin(2t )

T
0
z 22
 2
2
Decision stage:
if z (T )  0, mˆ  1
if z (T )  0, mˆ  0
ˆ
m
Non-coherent detection BER

Non-coherent detection of BFSK
1
1
PB  Pr(z1  z2 | s 2 )  Pr(z2  z1 | s1 )
2
2
 Pr(z1  z2 | s 2 )  EPr(z1  z2 | s 2 , z2 )


0
0
  Pr(z1  z2 | s 2 , z2 ) p( z2 | s 2 )dz2  
 Eb 
1

PB  exp  
2
 2N0 

  p( z | s )dz  p( z | s )dz
1
2
1
2
2
2
z2

Rayleigh pdf
Similarly, non-coherent detection of DBPSK
 E 
1
PB  exp  b 
2
 N0 
Rician pdf
BER Example
Example of samples of matched filter output
for some bandpass modulation schemes
Comparison of Digital Modulation
Comparison of Digital Modulation
Spectral Efficiencies in practical radios

GSM- Digital Cellular
– Data Rate = 270kb/s, bandwidth = 200kHz
– Bandwidth Efficiency = 270/200 =1.35bits/sec/Hz
– Modulation: Gaussian Minimum Shift Keying (FSK with
orthogonal frequencies).
– “Gaussian” refers to filter response.

IS-54 North American Digital Cellular
– Data Rate = 48kb/s, bandwidth = 30kHz
– Bandwidth Efficiency = 48/30 =1.6bits/sec/Hz
– Modulation: pi/4 DPSK
Modulation Summary

Phase Shift Keying is often used, as it provides a highly
bandwidth efficient modulation scheme.

QPSK, modulation is very robust, but requires some form of
linear amplification. OQPSK and p/4-QPSK can be
implemented, and reduce the envelope variations of the signal.

High level M-ary schemes (such as 64-QAM) are very
bandwidth efficient, but more susceptible to noise and require
linear amplification.

Constant envelope schemes (such as GMSK) can be employed
since an efficient, non-linear amplifier can be used.

Coherent reception provides better performance than
differential, but requires a more complex receiver.