CSE4214 Digital Communications
Chapter 3
Baseband
Demodulation/Detection
Baseband Demodulation/Detection
2
CSE4214 Digital Communications
Demodulation and Detection
Why Baseband Demodulation/Detection ?
•
Received pulses are distorted because of the following factors:
1.
Intersymbol Interference causes smearing of the transmitted pulses.
Addition of channel noise degrades the transmitted pulses.
Transmission channel causes further smearing of the transmitted pulses.
2.
3.
•
Demodulation (Detection) is the process of determining the transmitted
bits from the distorted waveform.
Transmitted
waveform
Received waveforms as a function of distance
4
4
Models for Transmitted and Received Signals
•
For binary transmission, the transmitted signal over a symbol interval (0, T) is modeled by
 s1 (t ) 0  t  T
si (t )  
s2 (t ) 0  t  T
•
for bit 1
for bit 0
The received signal is degraded by: (i) noise n(t) and (ii) impulse response of the channel
r (t )  si (t )  hc (t )  n
(t )


transmitted
signal
•
•
channel
impulse
response
AWGN
for i = 1, …, M.
Given r(t), the goal of demodulation is to detect if bit 1 or bit 0 was transmitted.
In our derivations, we will first use a simplified model for received signal
r (t )  si (t )  n
(t )

transmitted
signal
•
AWGN
Later, we will see that degradation due to the impulse response of the channel is eliminated by
equalization.
5
Basic Steps in Demodulation
6
A Vector View of CT Waveforms (1)
1.
Orthonormal Waveforms: Two waveforms y1(t) and y2(t) are orthonormal if they satisfy the
following two conditions
T
ò y (t)y (t)dt = 0
Orthogonality Condition:
1
2
(0 £ t £ T )
0
T
Unit Magnitude Condition:
ò y (t)y (t)dt = K
1
1
1
=1
(0 £ t £ T )
0
T
ò y (t)y (t)dt = K
2
2
2
= 1 (0 £ t £ T )
0
2.
Normalized to have unit energy
Two arbitrary signals s1(t) and s2(t) can be represented by linear combinations of two
orthonormal basis functions f1 (t) and f2 (t) , i.e.
s1 (t) = s11f1 (t) + s12f2 (t)
s2 (t) = s21f1 (t) + s22f2 (t)
where sij =
T
ò s (t)f (t)dt,
i
0
j
i, j Î {1, 2}
7
Geometric Representation of Signals
s1 (t) = s11f1 (t) + s12f2 (t)
s2 (t) = s21f1 (t) + s22f2 (t)
where sij =
T
ò s (t)f (t)dt,
i
j
i, j Î {1, 2}
0
T
ò s (t)f (t)dt is the projection of s
i
0
j
i
on to
A Vector View of CT Waveforms (2)
1. N-dimensional basis functions: consists of a set {yi(t)}, (1 ≤ i ≤ N), of
orthonormal (or orthogonal) waveforms.
T
Orthogonality Condition:
ò y (t)y (t)dt = K d
j
k
j
jjk
(0 £ t £ T, j, k = 1,… N )
0
where the operator:
ì 1
d jk = í
î 0
for j = k
otherwise
2. Given a basis function, any waveform in the represented as a linear
combination of the basis functions.
9
A Vector View of CT Waveforms (3)
3.
Given a basis function, any waveform in the represented as a linear combination of the basis
functions
s1 (t) = a11y1 (t) + a12y2 (t) +
+ a1N y N (t)
s2 (t) = a21y1 (t) + a22y2 (t) +
+ a2 N y N (t)
sM (t) = aM 1y1 (t) + aM 2y2 (t) +
+ aMN y N (t)
or, in a more compact form,
si (t ) 
N
a y
ij
j (t )
i  1,, M
NM
j 1
4.
The coefficient aij are calculated as follows
T
1
aij 
si (t )y j (t )dt i  1,, M
Kj 0

where Kj is the energy present in the basis signal.
j  1,, N
(0  t  T )
10
Activity 1
(a)
(b)
(c)
(d)
(e)
Demonstrate that signals si(t),
i = 1,2,3, are not orthogonal.
Demonstrate that yi(t), i = 1,2,
are orthogonal.
Express si(t), i = 1,2,3, as a
linear combination of the basis
functions yi(t), i = 1,2.
Demonstrate that yi(t), i = 1,2,
are orthogonal.
Express si(t), i = 1,2,3, as a
linear combination of the basis
functions y i(t), i = 1,2.
11
1
1
Activity 1
(a) s1(t), s2(t), and s3(t) are clearly not orthogonal, i.e. the time integrated value
over a symbol duration of the product of any two of the three waveforms is not
zero, e.g.
T
T
T
2
 s (t )s (t )dt   s (t )s (t )dt   s (t ) s (t )dt
1
2
1
0
2
0
T
2
T
0
T
2
1
2
T
2
  (1)( 2)dt   ( 3)( 0)dt  T
(b) Using the same method, we can verify this signal form an orthogonal set.
T
y
1
T
2
T
0
T
2
(t )y 2 (t )dt  y 1 (t )y 2 (t )dt  y 1 (t )y 2 (t )dt
0
T
2
T
0
T
2
  (1)(1)dt   (1)(1)dt  0
Activity 1
(c) We know that
T
1
aij 
si (t )y j (t )dt i  1,, M
Kj 0

j  1,, N
(0  t  T )
T
T


1
2
 1
2

a11   s1 (t )y 1 (t )dt   s1 (t )y 1 (t )dt    (1)(1)dt   (3)( 1)dt   1
T 0
T 0
T
T


2
2




T
T
T
T


1
2
 1
2

a12   s1 (t )y 2 (t )dt   s1 (t )y 2 (t )dt    (1)(1)dt   (3)(1)dt   2
T 0
T 0
T
T


2
2




T
a21  1, a22  1, a31  2, a32  1
s1 (t )  y 1 (t )  2y 2 (t )
s2 (t )  y 1 (t ) y 2 (t )
s3 (t )  2y 1 (t ) y 2 (t )
T
Activity 1
(d) y 1' (t ),y ' (t ) are obviously orthogonal.
(e) Similar to (c), we have
2
s1 (t )  y 1' (t )  3y 2' (t )
s2 (t )  2y 1' (t )
s3 (t )  y 1' (t )  3y 2' (t )
Activity 2

Show that the energy in a signal si(t) is given by
T

Ei  si2 (t )dt 
0
N

j 1
aij2 K j .
Activity 2
2


2
Ei   si (t )dt    aijy j (t ) dt
0
0  j

T
T
T
T
   aijy j (t ) aiky k (t )dt   aij aik y j (t )y k (t )dt
j
0
k
j
N
k
0
  aij aik K j jk   aij2 K j i  1,  , M
j
k
j 1
SNR used in Digital Communications
In digital communications, SNR is defined as the ratio
of the energy (Eb) present in the signal representing a
bit to the power spectral density (N0) of noise.
2. In terms of signal power S and the duration T of bit,
the bit energy is given by Eb = S × T.
3. In terms of noise power N and bandwidth W, the PSD
of noise is given by N0 = N / W.
4. SNR is therefore given by
1.
SNR =
Eb S ´ T S / Rb S W
=
=
=
N0 N / W N / W N R
where Rb is the rate of transmission in bits transmitted
per second (bps).
5. Bit-error probability is the probability of error in a
transmitted bit.
6. ROC curves are plots of Bit-error probability versus
SNR.
17
Detection of binary signal in AWGN
1
8
Maximum Likelihood Detector (1)
1.
The sampled received signal is given by
z (T )  r (t ) t T 
ai (T )

 n0 (T )

2.
where ai(T) represents the signal levels obtained
after sampling.
Level of signal
Gaussian RV For bit 1, ai(T) = a1 and for bit
0, ai(T) = a2.
The pdf of n0 is Gaussian with zero mean
3.
 1  n 2 
p(n0 ) 
exp  0  
 2  0  
 2
 transmitted
 is given by
The conditional pdf of z given that bit 01 or bit 0 was
1
2



1
1 z  a1 
 
p z s1  
exp  
 2  0  
 0 2


2

1
1  z  a2  
 
p z s2  
exp  
 2  0  
 0 2


19
Maximum Likelihood Detector (2)
4.
The conditional pdf of z given that bit 1 or bit 0 was transmitted are referred to as maximum
likelihoods.
2



1
1 z  a1 
  maximum likelihood of s1
p z s1  
exp 
 2  0  
 0 2


2



1
1 z  a2 
  maximum likelihood of s2
p z s2  
exp 
 2  0  
 0 2


5.
The maximum likelihoods have the following distributions
20
Maximum Likelihood Detector (1)
6.
Maximum Likelihood Ratio Test: is given by
pz s1 
H1
 Ps 
pz s   Ps 
2
2
1
H2
6.
where P(s1) and P(s2) are the priori probabilities that s1(t) and s2(t), respectively, are
transmitted.
H1 and H2 are two possible hypotheses. H1 states that signal s1(t) was transmitted and
hypothesis H2 states that signal s2(t) was transmitted.
For P(s1) = P(s2), the maximum likelihood ratio test reduces to
H1
z (T )
a

1
 a2
 0
2
H2
21
Activity 3

Show that the probability of bit error for maximum likelihood ratio test
is given by
æ a1 - a2 ö
PB = Q ç
÷
è 2s 0 ø
æ u2 ö
ò exp çè- 2 ÷ødx
x
æ -x 2 ö
1
exp ç
or Q(x) »
÷ for x > 3
x 2p
è 2 ø
1
where Q(x) =
2p

¥
Values of Q(x) are listed in Table B.1 in Appendix B page 1046.
22
Matched Filtering (1)
23
Matched Filtering (2)
r (t )  si (t )  n(t )
Z (T )  ai (T )  n0 (T )
CTFT
h(t ) 
 H ( f )
Design the Receiving filter h(t)
1.
Design a filter that maximizes the SNR at time (t = T) of the sampled signal
z (T )  r (t ) t T 
ai (T )

Level of signal
2.
 n0 (T )

Gaussian RV
The instantaneous signal power to noise power is given by
ai2
S
   2
 N T  0
where ai(t) is the filtered signal and 02 is the variance of the output noise
3.
The information bearing component is given by

ai (T ) 

H ( f ) S i ( f )e j 2 ft dt

24
Matched Filtering (3)
4.
Given that input noise n(t) is AWGN with Sn(f) = N0/2, the PSD of the output noise is given by
2
Sn0 ( f )  H ( f ) Sn ( f ) 
5.
The output noise power is given by
N
 02  0
2
6.
N0
Sn ( f )
2
The SNR is given by


2
H ( f ) df

2

ai2
S
   2 
 N T  0

H ( f ) S ( f )e j 2 ft df

N0
2


2
H ( f ) df

25
Matched Filtering (4)
Schwartz Inequality:
2


f1 ( x) f 2 ( x)dx 




2
f1 ( x) dx


2
f 2 ( x) dx

with the equality valid if f1(x) = kf2*(x).
7.
Applying the Schwatz inequality to the SNR gives

S
  
 N T


2
H ( f ) df

N0
2


2
S ( f )e j 2 ft df

 H( f )
2
or,
2
S
  
 N T N 0


2
S ( f ) df

df

26
Matched Filtering (5)
8.
The maximum SNR is given by
2
S
max  
 N T N 0

2E
S ( f ) df 
N0


2
which is possible if
H ( f )  S  ( f )e  j 2ft
or h(t )  s(T  t ).
27
Activity 4

Given that the transmitted signal is shown in the
following figure, determine the impulse response of
the matched filter.
28
Matched Filtering (6)
Correlator Implementation of matched filter:
The output of the matched filter is given by
t

z (t )  r ()h(t  )d
0
The output at t = T is given by
T
T


z (T )  r ()h(T  )d  r () si ()d
0
0
which leads to the following correlator implementation for matched filter
si (t )
T
r (t )  si (t )  n(t )

Z (T )  ai (T )  n0 (T )
0
29
Detection of binary signal: Review
r (t )  si (t )  n(t )
Z (T )  ai (T )  n0 (T )
h(t )  H ( f )
H1
a

z (T )

1
CTFT
 a2
 0
2
10111000 
H2
Matched Filter:
h(t )  si (T  t )
Impulse response:
Maximum SNR = (a1)2/20 = 2Es/N0
ML Detector:
Probability of Error:
 a1  a2 

PB  Q
 20 
•
The overall goal of the receiver should be to minimize the probability of bit error PB..
•
In other words, we are interested in maximizing (a1 – a2)2/20.
•
The filter is designed such that it is matched to the difference of [s1(t) – s2(t)].
•
Maximum SNR of the matched filter = (a1 – a2)2/20 = 2(Es1 – Es2)/N0 = 2Ed/N0.
•
Probability of Error:
 Ed
 a  a2 
  Q
PB  Q 1
 2N
0
 2 0 





30
Activity 5
By defining the cross-correlation coefficient as
1

Eb



s1 (t ) s2 (t )dt with

Eb 


s12 (t )dt 


s22 (t )dt

show that the probability of bit error using a filter matched to [s1(t) – s2(t)] is given by
 Eb (1  ) 

PB  Q


N0


Using the above relationship, show that the probability of bit error is given by

1. PB  Q



2. PB  Q



 for antipodal signals : s1 (t )   s2 (t )


Eb 
for orthogonal signals s1 (t )  s2 (t ).

N0 
2 Eb
N0
31
Activity 6
Determine the probability of bit error for a binary communication system,
which receives equally likely signals s1(t) and s2(t) shown in the following
diagram
Assume that the receiving filter is a matched filter and the power spectral
density of AWGN is N0 = 1012 Watt/Hz.
32
Error Probability Performance of Binary Signal
1. Unipolar signaling
In Unipolar signaling, the signal selection to represent bits 1 and 0 is as follows:
s1 (t )  A, (0  t  T ), for bit 1
s2 (t )  0, (0  t  T ), for bit 0.
The receiver is shown by the following diagram.
33
Unipolar Signaling




Unipolar signal forms an orthogonal signal set.
When s1(t) plus AWGN being received, the expected value of z(T), given that
s1(t) was sent, is
ìT 2
ü
a1 (T ) = E { z(T ) s1 (t)} = E í ò (A + An(t))dt ý = A 2T, where E {n(t)} = 0
î0
þ
When s2(t) plus AWGN being received, a2(T)=0.
The optimum decision threshold is:
g0 =

a1 + a2 1 2
= AT
2
2
The bit-error performance is:
2
æ A 2T ö
æ E ö
æ E ö
A
T
d
b
÷ = Q çç
PB = Q çç
,
where
E
=
÷÷ = Q çç
÷
b
÷
÷
2N
2N
N
2
è
è
0 ø
0 ø
0 ø
è
Ed =
T
ò [ s1 (t) - s2 (t)] dt = A2T
0
2
34
Bipolar Signaling
In bipolar signaling, the signal selection to represent bits 1 and 0 is as follows:
s1 (t )  A, (0  t  T ), for bit 1
s2 (t )   A, (0  t  T ), for bit 0.
The receiver is shown by the following diagram.
35
Bipolar Signaling


Bipolar signal is a set of antipodal signal, e.g. s1(t)=-s2(t).
The optimum decision threshold is:
g0 =

a1 + a2
=0
2
The bit-error performance is:
æ 2A 2T
æ E ö
d
PB = Q çç
÷÷ = Q çç
è 2N 0 ø
è N0
Ed =
T
ö
æ
ö
÷ = Q çç 2Eb ÷÷, where Eb = A 2T
÷
è N0 ø
ø
ò [ s1 (t) - s2 (t)] dt = (2A)2 T
2
0
36
Unipolar vs. Bipolar
37