Receivers

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Outline
• Transmitters (Chapters 3 and 4, Source Coding and
Modulation) (week 1 and 2)
• Receivers (Chapter 5) (week 3 and 4)
•
•
•
•
Received Signal Synchronization (Chapter 6) (week 5)
Channel Capacity (Chapter 7) (week 6)
Error Correction Codes (Chapter 8) (week 7 and 8)
Equalization (Bandwidth Constrained Channels) (Chapter
10) (week 9)
• Adaptive Equalization (Chapter 11) (week 10 and 11)
• Spread Spectrum (Chapter 13) (week 12)
• Fading and multi path (Chapter 14) (week 12)
Digital Communication System:
Transmitter
Receiver
Receivers (Chapter 5) (week 3 and 4)
• Optimal Receivers
• Probability of Error
Optimal Receivers
• Demodulators
• Optimum Detection
Demodulators
• Correlation Demodulator
• Matched filter
Correlation Demodulator
• Decomposes the
signal into
orthonormal
basis vector
correlation terms
• These are
strongly
correlated to the
signal vector
coefficients sm
Correlation Demodulator
• Received Signal model
– Additive White Gaussian Noise (AWGN)
r (t )  sm (t )  n(t )
– Distortion
• Pattern dependant noise
– Attenuation
• Inter symbol Interference
– Crosstalk
– Feedback
Additive White Gaussian Noise
(AWGN)
r (t )  sm (t )  n(t )
1
 rr ( f )   ss ( f )  N 0
2
1
 nn ( f )  N 0
2
i.e., the noise is flat in Frequency domain
Correlation Demodulator
• Consider each
demodulator
output
T
rk   r (t ) f k (t )dt
0
T
  sm (t ) f k (t )dt
0
T
  n(t ) f k (t )dt
0
 smk  nk
Correlation Demodulator
• Noise components
E (nk nm )  
T
0

T
0
E[n(t )n( )] f k (t ) f m ( )dtd
T
1
 N 0  f k (t ) f m ( )dt
0
2
1
 N 0 m  k {nk} are uncorrelated
 2
m  k Gaussian random
0
variables
Correlation Demodulator
• Correlator outputs
E (rk )  E ( smk  nk )  smk Have mean = signal
 N (rk  smk ) 2 
1
p(r | s m ) 
exp  

N /2
(N 0 )
N0
 k 1

m  1,2, , M
For each of the M codes
N  Number of basis functions (=2 for QAM)
Matched filter Demodulator
• Use filters whose
impulse response is
the orthonormal
basis of signal
• Can show this is
exactly equivalent to
the correlation
demodulator
Matched filter Demodulator
• We find that this
Demodulator
Maximizes the SNR
• Essentially show that
any other function
than f1() decreases
SNR as is not as well
correlated to
components of r(t)
The optimal Detector
• Maximum Likelihood (ML):

 N (rk  smk ) 2  
1
max  p (r | s m )  max 
exp  

N /2
N0
 k 1

 (N 0 )
N
1
(rk  smk ) 2 
 max  N ln N 0  

N0
k 1
2

 min
N
2
(
r

s
)
 k mk
k 1

 min r  2r  s m  s m
2
2

The optimal Detector
• Maximum Likelihood (ML):

min r  2r  s m  s m
2
2
  max 2r  s
m
 sm


m
 max r  s m 

2


m  1,2,  M
2

Optimal Detector
• Can show that
N
N T
T
n 1
n 1 0
0
r  s m   rn smn    r (t ) f n (t )dt  sm (t ) f n (t )dt
T
so
  r (t ) sm (t )dt
0
T





m
m
max r  s m 
 max   r (t ) sm (t )dt 


2 
2 

0
Optimal Detector
• Thus get new type of correlation demodulator
using symbols not the basis functions:
Alternate Optimal rectangular QAM
Detector
• M level QAM = 2 x M level PAM signals
• PAM = Pulse Amplitude Modulation
sm (t )  Am g (t ) cos 2f c t
 sm f (t )
1

g
2
sm  Am
f (t ) 
(e)
d min
d
2

g
g (t ) cos 2f c t
sm 
2
g
1

g ( 2m  1  M ) d
2
m  1,2,, M
The optimal PAM Detector
 g A2 

m




max r  s m  m   max r  s m  2

2 
2 





 g d 2 ( 2m  1  M ) 2 



 max r  s m  2
2





r  sm  d

g
2

d 2 g
2
For PAM
The optimal PAM Detector
r  sm 
d 2 g
2
(e)
d min

2
sm
r
r  si r  si 1
(e)
d min
2
Optimal rectangular QAM Demodulator
• d = spacing of rectangular grid
si 
s
f1 (t ) 
2

g (t ) cos 2f ct


0
f 2 (t ) 
2



s
g
T
M
1
()dt
1
 g (2i 1  M )d
2
Select si
for which

d 2 g
sm1  si
2

g (t ) sin 2f ct
g
T

0
()dt

s
1

Select si
for which

d 2 g
2
sm 2  si
Probability of Error for rectangular
M-ary QAM
• Related to error probability of M PAM
PM
d 2 g
M 1 

P  r  sm 
2
M

Accounts for ends
sm
r



Probability of Error for rec. QAM
• Assume Gaussian noise

d 2 g
P  r  sm 
2

0

2

N 0



d g /2
2
e
 d 2 
g

 erfc 
 2 N 0 
 d 2
g

 2Q
 N 0



d 2 g
2
 x2 / N0
r  sm
dx
Probability of Error for rectangular
M-ary QAM
• Error probability of M PAM
PM 
2

d g 
M 1

2 Q
M
 2 N 0 
SNR for M-ary QAM
• Related to M PAM
• For M PAM find average energy in equally
probable signals

av
1

M

d


M
m 1
2
g
m
M
 ( 2m  1 
2 M m 1
1
2
 ( M  1)d  g
6
M)
2
SNR for M-ary QAM
• Related to M PAM
Find average Power
Pav


av
T
d g
1
 ( M  1)
6
T
2
SNR for M-ary QAM
• Related to M PAM
Find SNR
Then SNR per bit

SNR 
av
(ratio of powers)
N0
Tb  av
SNRb 

T N0


 M
av
N 0 log 2
d 2 g
1
 ( M  1)
6
N 0 log 2
 M
SNR for M-ary QAM
• Related to M PAM
d

2
g
6 N 0 SNRb log 2

( M  1)
PM 
 M
 
 6 log M SNR
M 1
2
b
2 Q
( M  1)
M




SNR for M-ary QAM
• Related to M PAM
• Now need to get
M-ary QAM from
PAM
M½=16
M½=8
M½=4
M½=2
SNR for M-ary QAM
• Related to M PAM
PM  1  (1  P M ) 2
(1- probability of no QAM error)
 

 3 log M SNR
M 1
2
b
 1  1 
2 Q
( M  1)
M




SNR b PAM


 
2
SNR b QAM
2
(Assume ½ power in each PAM)
SNR for M-ary QAM
• Related to M PAM


 
1.E-01
2
Probabilty of symbol Error PM
 

 3 log M SNR
M 1
2
b
PM  1  1 
2 Q
( M  1)
M



Probability of Symbol Error for QAM
1.E-02
M=
1.E-03
256
1.E-04
64
16
4
1.E-05
1.E-06
24
22
20
18
16
14
12
10
8
6
4
2
0
-2
-4
-6
SNR per bit (dB)
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