Chapter 5: Optimum Receiver for Binary Data Transmission
EE456 – Digital Communications
Professor Ha Nguyen
September 2015
EE456 – Digital Communications
1
Chapter 5: Optimum Receiver for Binary Data Transmission
Block Diagram of Binary Communication Systems
m(t )
ˆ (t )
m
{bk }
{ }
bˆ k
b k = 0 ↔ s1 (t )
b k = 1 ↔ s2 (t )
r( t )
w (t )
Bits in two different time slots are statistically independent.
a priori probabilities: P [bk = 0] = P1 , P [bk = 1] = P2 .
Signals s1 (t) and s2 (t) have a duration of Tb seconds and finite energies:
R
R
E1 = 0Tb s21 (t)dt, E2 = 0Tb s22 (t)dt.
Noise w(t) is stationary Gaussian, zero-mean white noise with two-sided power
spectral density of N0 /2 (watts/Hz):
N0
N0
E{w(t)} = 0, E{w(t)w(t + τ )} =
δ(τ ), w(t) ∼ N 0,
.
2
2
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Chapter 5: Optimum Receiver for Binary Data Transmission
m(t )
ˆ (t )
m
!"
.!*'' !
{bk }
{ }
bˆ k
&%$'
()$!*+ '',
-+%$'
(. /,
b k = 0 ↔ s1 (t )
b k = 1 ↔ s2 (t )
#$!!%
r( t )
w (t )
Received signal over [(k − 1)Tb , kTb ]:
r(t) = si (t − (k − 1)Tb ) + w(t),
(k − 1)Tb ≤ t ≤ kTb .
Objective is to design a receiver (or demodulator) such that the probability of
making an error is minimized.
Shall reduce the problem from the observation of a time waveform to that of
observing a set of numbers (which are random variables).
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Chapter 5: Optimum Receiver for Binary Data Transmission
OOK (Rx)
OOK (Tx)
NRZ (Rx)
NRZ (Tx)
Can you Identify the Signal Sets {s1 (t), s2 (t)}?
1
0
−1
0
0.5
1
1.5
2
2.5
t/Tb
3
3.5
4
4.5
5
0.5
1
1.5
2
2.5
t/Tb
3
3.5
4
4.5
5
0.5
1
1.5
2
2.5
t/Tb
3
3.5
4
4.5
5
0.5
1
1.5
2
2.5
t/Tb
3
3.5
4
4.5
5
5
0
−5
0
1
0
−1
0
5
0
−5
0
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Chapter 5: Optimum Receiver for Binary Data Transmission
Geometric Representation of Signals s1 (t) and s2 (t) (I)
Wish to represent two arbitrary signals s1 (t) and s2 (t) as linear combinations of
two orthonormal basis functions φ1 (t) and φ2 (t).
φ1 (t) and φ2 (t) form a set of orthonormal basis functions if and only if:
φ1 (t) and φ2 (t) are orthonormal if:
Z
Tb
φ1 (t)φ2 (t)dt = 0 (orthogonality),
0
Z
0
Tb
φ21 (t)dt =
Z
0
Tb
φ22 (t)dt = 1 (normalized to have unit energy).
If {φ1 (t), φ2 (t)} can be found, the representations are
s1 (t)
=
s11 φ1 (t) + s12 φ2 (t),
s2 (t)
=
s21 φ1 (t) + s22 φ2 (t).
It can be checked that the coefficients sij can be calculated as follows:
sij =
Z
0
Tb
si (t)φj (t)dt,
i, j ∈ {1, 2},
An important question is: Given the signal set s1 (t) and s2 (t), can one always
find an orthonormal basis functions to represent {s1 (t), s2 (t)} exactly? If the
answer is YES, is the set of orthonormal basis functions UNIQUE?
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Chapter 5: Optimum Receiver for Binary Data Transmission
Geometric Representation of Signals s1 (t) and s2 (t) (II)
Provided that φ1 (t) and φ2 (t) can be found, the signals (which are waveforms) can be
represented as vectors in a vector space (or signal space) spanned (i.e., defined) by the
orthonormal basis set {φ1 (t), φ2 (t)}.
φ 2 (t )
s2 (t )
s1 (t) = s11 φ1 (t) + s12 φ2 (t),
s22
s12
d12 = d 21
E2
s1 ( t )
s2 (t) = s21 φ1 (t) + s22 φ2 (t),
Z Tb
sij =
si (t)φj (t)dt, i, j ∈ {1, 2},
Ei =
E1
s21
0
s11
φ1 (t )
Z
0
Tb
0
s2i (t)dt = s2i1 + s2i2 , i ∈ {1, 2},
d12 = d21 =
s
Z
0
R Tb
0
Tb
[s2 (t) − s1 (t)]2 dt
si (t)φj (t)dt is the projection of signal si (t) onto basis function φj (t).
The length of a signal vector equals to the square root of its energy.
It is always possible to find orthonormal basis functions φ1 (t) and φ2 (t) to
represent s1 (t) and s2 (t) exactly. In fact, there are infinite number of choices!
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Chapter 5: Optimum Receiver for Binary Data Transmission
Gram-Schmidt Procedure
Gram-Schmidt (G-S) procedure is one method to find a set of orthonormal basis
functions for a given arbitrary set of waveforms.
√
s1 (t)
1 Let φ1 (t) ≡ p
. Note that s11 = E1 and s12 = 0.
E1
′
s2 (t)
2 Project s (t) = p
onto φ1 (t) to obtain the correlation coefficient:
2
E2
ρ=
Z
Tb
0
s2 (t)
1
√ φ1 (t)dt = √
E2
E1 E2
′
3
′
Subtract ρφ1 (t) from s2 (t) to obtain φ2 (t) =
Z
Tb
s1 (t)s2 (t)dt.
0
s2 (t)
√
E2
− ρφ1 (t).
′
4
Finally, normalize φ2 (t) to obtain:
′
φ2 (t)
=
=
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′
φ2 (t)
φ (t)
= p2
2
Tb ′
1 − ρ2
φ
(t)
dt
2
0
1
ρs1 (t)
s2 (t)
p
√
− √
.
E2
E1
1 − ρ2
qR
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Chapter 5: Optimum Receiver for Binary Data Transmission
Gram-Schmidt Procedure: Summary
φ2 (t )
s2′ (t )
φ2′ (t )
s2 (t )
s 22
E2
0
=
φ2 (t)
=
s21
=
s22
=
d21
=
d 21
ρφ1 (t )
α
E1
s 21 s1 (t )
−1 ≤ ρ = cos(α ) ≤ 1
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φ1 (t)
φ1 (t )
s1 (t)
√ ,
E1
ρs1 (t)
s2 (t)
1
p
√
− √
,
E2
E1
1 − ρ2
Z Tb
p
s2 (t)φ1 (t)dt = ρ E2 ,
0
p
p
1 − ρ2
E2 ,
s
Z Tb
[s2 (t) − s1 (t)]2 dt
0
=
p
E1 − 2ρ E1 E2 + E2 .
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Chapter 5: Optimum Receiver for Binary Data Transmission
Example 1
s1 (t )
s 2 (t )
V
0
Tb
0
Tb
0
0
123
−V
φ1 (t )
1
Tb
s1 (t )
s 2 (t )
0
Tb
567
4
− E
0
φ1 (t )
E
89:
(a) Signal set. (b) Orthonormal function. (c) Signal space representation.
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Chapter 5: Optimum Receiver for Binary Data Transmission
Example 2
s1 (t )
s 2 (t )
V
V
Tb
;
0
0
;
Tb
−V
<=>
φ 2 (t )
φ1 (t )
1 Tb
1 Tb
Tb
0
φ2 (t )
?
0
s 2 (t )
E
Tb
?
s1 (t )
− 1 Tb
@AB
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0
φ1 (t )
E
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Chapter 5: Optimum Receiver for Binary Data Transmission
Example 3
s1 (t )
s 2 (t )
V
V
0
α
C
Tb
Tb
C
0
−V
φ 2 (t, α )
α=
I
E 3E
G−
G 2 , 2
H
α =0
(−
F
D
D
E
Tb
2
s 2 (t )
ρ
=
increasing α , ρ
=
α=
E ,0
Tb
4
)
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E
=
s1 (t )
0
(
E ,0
)
1
E
Z
Tb
s2 (t)s1 (t)dt
0
1
V 2 Tb
2α
−1
Tb
h
i
V 2 α − V 2 (Tb − α)
φ1 (t )
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Chapter 5: Optimum Receiver for Binary Data Transmission
Example 4
s1 (t )
s 2 (t )
3V
V
0
Tb
J
0
KLM
φ1 (t )
1
Tb
J
φ2 (t )
3
Tb
Tb
0
Tb 2
Tb
Tb
N
0
−
Tb 2
N
3
Tb
OPQ
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Chapter 5: Optimum Receiver for Binary Data Transmission
φ 2 (t )
s2 (t )
E
2
E
60o
0
s1 (t )
φ1 (t )
3E
2
!
√
√
3
2 3
V t V dt =
,
Tb
2
0
0
#
"
√
1
3
s1 (t)
2
s2 (t)
s
(t)
,
φ2 (t) =
−
ρ
s
(t)
−
=
√
√
√
1
2
1
2
E
E
E
(1 − 43 ) 2
√
3√
1√
E, s22 =
E.
s21 =
2
2
ρ=
1
E
Z
d21 =
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Tb
Z
s2 (t)s1 (t)dt =
Tb
0
2
E
Z
Tb /2
r
1
√
2
2
=
2 − 3 E.
[s2 (t) − s1 (t)] dt
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Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5
φ2 (t )
θ = 3π 2
ρ =0
s1 (t)
=
√
E
s
2
cos(2πfc t),
Tb
s2 (t)
=
√
E
s
2
cos(2πfc t + θ).
Tb
where fc =
s1 (t )
0
θ =π
ρ = −1
θ
k , k an integer.
2Tb
φ1 (t )
locus of s2 (t ) as θ
E
varies from 0 to 2π .
s 2 (t )
θ =π 2
ρ =0
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Chapter 5: Optimum Receiver for Binary Data Transmission
Obtaining Different Basis Sets by Rotation
φ2 (t )
φ 2 (t )
s2 (t )
φ 2 (t )
s22
s22
s11
θ
0
s11 = s21
s12
s1 ( t )
φ1 (t )
s11
s21
s22
φ 2 (t )
s21
s12
φ1 (t )
s2 (t )
φ1 ( t )
s22
s1 (t )
θ
s21
0
s12
s11
φ1 (t )
s12
s11 = s21
φ̂1 (t)
φ̂2 (t)
=
cos θ
− sin θ
sin θ
cos θ
φ1 (t)
φ2 (t)
.
(a) Show that, regardless of the angle θ, the set {φ̂1 (t), φ̂2 (t)} is also an orthonormal
basis set.
(b) What are the values of θ that make φ̂1 (t) perpendicular to the line joining s1 (t)
to s2 (t)? For these values of θ, mathematically show that the components of
s1 (t) and s2 (t) along φ̂1 (t), namely ŝ11 and ŝ21 , are identical.
Remark: Rotating counter-clockwise for positive θ and clock-wise for negative θ.
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Chapter 5: Optimum Receiver for Binary Data Transmission
Example: Rotating {φ1 (t), φ2 (t)} by θ = 60◦ to Obtain {φ̂1 (t), φ̂2 (t)}
φ̂1 (t)
φ̂2 (t)
=
cos θ
− sin θ
sin θ
cos θ
φ1 (t)
φ2 (t)
⇒
φ1 (t )
φ̂1 (t) = cos θ × φ1 (t) + sin θ × φ2 (t)
φ̂2 (t) = − sin θ × φ1 (t) + cos θ × φ2 (t)
φ2 (t )
1
1
0 .5
0
1
0
1
−1
φ1 (t )
φ2 (t )
(1 + 3 )
2
(1 − 3 )
2
0
0 .5
1
(1 − 3 )
2
−
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0
0 .5
1
(1 + 3 )
2
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Chapter 5: Optimum Receiver for Binary Data Transmission
Gram-Schmidt Procedure for M Waveforms {si (t)}M
i=1
s1 (t)
φ1 (t)
=
qR
φi (t)
=
qR
′
φi (t)
=
ρij
=
∞
s2 (t)dt
−∞ 1
,
′
φi (t)
2 ,
∞ ′
φi (t) dt
−∞
i = 2, 3, . . . , N,
i−1
si (t) X
√
−
ρij φj (t),
Ei
j=1
Z ∞
si (t)
√ φj (t)dt, j = 1, 2, . . . , i − 1.
Ei
−∞
If the waveforms {si (t)}M
i=1 form a linearly independent set, then N = M . Otherwise
N < M.
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Chapter 5: Optimum Receiver for Binary Data Transmission
Example: Find a Basis Set for the Following M = 4 Waveforms
s1 (t )
s3 ( t )
1
1
0
1
2
0
2
3
2
3
−1
s4 ( t )
s 2 (t )
1
1
0
1
−1
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2
0
−1
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Chapter 5: Optimum Receiver for Binary Data Transmission
Answer Found by Applying the G-S Procedure
φ3 ( t )
φ1 (t )
1
2
0
1
1
0
2
2
3
−1
φ2 ( t )
1
2
−
1
2
0
1
2
√
2
s1 (t)
s2 (t)
= √0
s3 (t)
√2
s4 (t)
2
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√0
2
0
0
0
φ1 (t)
0
φ2 (t)
1
φ3 (t)
1
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Chapter 5: Optimum Receiver for Binary Data Transmission
Representation of Noise with Walsh Functions
x1(t)
5
0
−5
5
x2(t)
0
−5
2
φ1(t)
1
φ2(t)
0
2
0
−2
2
φ3(t)
0
−2
2
φ4(t)
0
−2
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
1
Exact representation of noise using 4 Walsh functions is not possible.
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Chapter 5: Optimum Receiver for Binary Data Transmission
The First 16 Walsh Functions
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
1
Exact representation might be possible by using many more Walsh functions.
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Chapter 5: Optimum Receiver for Binary Data Transmission
The First 16 Sine and Cosine Functions
Can also use sine and cosine functions (Fourier representation).
1.5
−1.5
0
EE456 – Digital Communications
0.25
0.5
t
0.75
1
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Chapter 5: Optimum Receiver for Binary Data Transmission
Representation of Noise
To represent noise w(t), need to use a complete set of orthonormal functions:
Z Tb
∞
X
w(t) =
wi φi (t), where wi =
w(t)φi (t)dt.
0
i=1
The coefficients wi ’s are random variables and understanding their statistical
properties is imperative in developing the optimum receiver.
Of course, the statistical properties of random variables wi ’s depend on the
statistical properties of the noise w(t), which is a random process.
In communications, a major source of noise is thermal noise, which is modelled as
Additive White Gaussian Noise (AWGN):
White: The power spectral density (PSD) is a constant (i.e., flat) over all frequencies.
Gaussian: The probability density function (pdf) of the noise amplitude at any given
time follows a Gaussian distribution.
When w(t) is modelled as AWGN, the projection of w(t) on each basis function,
R
wi = 0Tb w(t)φi (t)dt, is a Gaussian random variable (this can be proved).
For zero-mean and white noise w(t), w1 , w2 , w3 , . . . are zero-mean and
uncorrelated random
variables: o
n
1
2
E{wi } = E
R Tb
E{wi wj } = E
0
w(t)φi (t)dt
nRT
0
b
=
dλw(λ)φi (λ)
R Tb
0
R Tb
0
E{w(t)}φi (t)dt = 0.
(
o
N0 ,
2
dτ w(τ )φj (τ ) =
0,
i=j .
i 6= j
Since w(t) is not only zero-mean and white, but also Gaussian ⇒ {w1 , w2 , . . .}
are Gaussian and statistically independent!!!
EE456 – Digital Communications
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Chapter 5: Optimum Receiver for Binary Data Transmission
Need to Review Probability Theory &
Random Processes – Chapter 3 Slides
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Chapter 5: Optimum Receiver for Binary Data Transmission
Observing a waveform ⇒ Observing a set of numbers
t = Tb
Tb
∫ ( • ) dt
0
r1 = si1 + w1
φ1 ( t )
t = Tb
Tb
∫ ( • ) dt
0
r2 = si 2 + w 2
φ2 ( t )
t = Tb
r (t ) =
Tb
∫ ( • ) dt
si (t ) + w (t )
r3 = 0 + w 3
0
AWGN
φ3 ( t )
The decision can be
based on the
observations
r1 , r2 , r3 , r4 , . . ..
t = Tb
Tb
∫ ( • ) dt
0
φn ( t )
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Choose φ1 (t) and φ2 (t)
so that they can be used
to represent the two
signals s1 (t) and s2 (t)
exactly. The remaining
orthonormal basis
functions are simply
chosen to complete the
set in order to represent
noise exactly.
rn = 0 + w n
Note that rj , for
j = 3, 4, 5, . . ., does not
depend on which signal
(s1 (t) or s2 (t)) was
transmitted.
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Chapter 5: Optimum Receiver for Binary Data Transmission
Optimum Receiver
The criterion is to minimize the bit error probability.
Consider only the first n terms (n can be very very large), ~
r = {r1 , r2 , . . . , rn }
⇒ Need to partition the n-dimensional observation space into two decision
regions, ℜ1 and ℜ2 .
Observation space ℜ
Decide a "1" was transmitted
R
if r falls in this region.
ℜ1
Decide a "0" was transmitted
R
if r falls in this region.
ℜ2
ℜ1
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Chapter 5: Optimum Receiver for Binary Data Transmission
P [error] = P [(“0” decided and “1” transmitted) or
(“1” decided and “0” transmitted)].
=
P [0D , 1T ] + P [1D , 0T ]
=
P [0D |1T ]P [1T ] + P [1D |0T ]P [0T ]
=
=
P [~r ∈ ℜ1 |1T ]P2 + P [~r ∈ ℜ2 |0T ]P1
Z
Z
f (~
r |0T )d~
r
f (~
r |1T )d~
r + P1
P2
ℜ2
ℜ1
=
P2
Z
ℜ−ℜ2
=
P2
Z
ℜ
=
P2 +
f (~
r |1T )d~
r+
Z
ℜ2
=
P1 −
f (~
r |1T )d~
r + P1
Z
ℜ1
Z
ℜ2
Z
ℜ2
f (~
r |0T )d~
r
[P1 f (~
r |0T ) − P2 f (~
r |1T )]d~
r
[P1 f (~
r |0T ) − P2 f (~
r |1T )] d~
r
[P1 f (~
r |0T ) − P2 f (~
r |1T )] d~
r.
The minimum error probability decision rule is
P1 f (~
r |0T ) − P2 f (~
r |1T ) ≥ 0 ⇒
P1 f (~
r |0T ) − P2 f (~
r |1T ) < 0 ⇒
EE456 – Digital Communications
decide “0” (0D )
.
decide “1” (1D )
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Chapter 5: Optimum Receiver for Binary Data Transmission
Pr[error] =
t = Tb
P2 +
Tb
∫ ( • ) dt
∫ [ P f (r | 0
1
ℜ2
) − P2 f ( r |1T )]dr
g(r )
= P1 + ∫ [ P2 f ( r |1T ) − P2 f ( r | 0T )]dr
r1 = si1 + w1
0
T
ℜ1
φ1 ( t )
−g(r )
0D
t = Tb
⇒ Optimal decision rule: g ( r )
Tb
∫ ( • ) dt
0
r2 = si 2 + w 2
φ2 ( t )
t = Tb
r (t ) =
Tb
∫ ( • ) dt
si (t ) + w (t )
r3 = 0 + w 3
0
AWGN,
PSD
g(r )
ℜn
>
0
<
ℜ1 ⇔ 0 D
g (r ) > 0
φ3 ( t )
N0
2
>
0
<
1D
t = Tb
Tb
∫ ( • ) dt
0
φn ( t )
g(r ) < 0
rn = 0 + w n
g (r ) = 0
ℜ 2 ⇔ 1D
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Chapter 5: Optimum Receiver for Binary Data Transmission
Equivalently,
1D
f (~
r |1T )
f (~
r |0T )
The expression
R
0D
P1 .
P2
(1)
f (~
r |1T )
is called the likelihood ratio.
f (~
r |0T )
The decision rule in (1) was derived without specifying any statistical properties
of the noise process w(t).
Simplified decision rule when the noise w(t) is zero-mean, white and Gaussian:
(r1 − s11 )2 + (r2 − s12 )2
1D
R
0D
(r1 − s21 )2 + (r2 − s22 )2 + N0 ln P1 .
P2
For the special case of P1 = P2 (signals are equally likely):
(r1 − s11 )2 + (r2 − s12 )2
1D
R
0D
(r1 − s21 )2 + (r2 − s22 )2 .
⇒ minimum-distance receiver!
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Chapter 5: Optimum Receiver for Binary Data Transmission
Minimum-Distance Receiver
1D
(r1 − s11 )2 + (r2 − s12 )2
|
{z
}
R
0D
d2
1
1D
d21
R
r2 φ 2 (t )
0D
(r1 − s21 )2 + (r2 − s22 )2
{z
}
|
d2
2
d22
s 2 (t )
d2
( s21 , s22 )
r (t )
d1
( r1, r2 )
s1 (t )
( s11 , s12 )
φ1 (t )
0
Choose s2 (t )
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Choose s1 ( t )
r1
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Chapter 5: Optimum Receiver for Binary Data Transmission
Correlation Receiver Implementation
t = Tb
Tb
T (•)dt
r1
Compute
(r1 − si1 )2 + (r2 − si 2 )2
0
r (t ) = si (t ) + w (t )
− N 0 ln( Pi )
φ1 (t )
t = Tb
Tb
S (•)dt
r2
Decision
for i = 1, 2
and choose
the smallest
0
φ 2 (t )
t = Tb
Tb
V (•)dt
r1
Form
0
the
r (t )
φ1 (t )
dot
product
t = Tb
Tb
U (•)dt
r2
a a
N0
E
ln(P1 ) − 1
2
2
WXYYZ[
\X[
]^_`[Z\
Decision
r ⋅ si
0
φ 2 (t )
EE456 – Digital Communications
N0
E
ln( P2 ) − 2
2
2
31
Chapter 5: Optimum Receiver for Binary Data Transmission
Receiver Implementation using Matched Filters
t = Tb
Tb
∫ (•)dt
r1
0
r (t )
Decision
φ1 (t )
t = Tb
Tb
∫ (•)dt
r2
0
φ 2 (t )
t = Tb
r1
h1 (t ) = φ1 (Tb − t )
r (t )
h2 (t ) = φ 2 (Tb − t )
EE456 – Digital Communications
Decision
t = Tb
r2
32
Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.6
s1 (t )
s 2 (t )
1.5
0.5
0
0.5
1
b
cde
φ1 (t )
1
b
1
f
−2
φ 2 (t )
1
1
0
0.5
0
1
f
0.5
0
−1
ghi
EE456 – Digital Communications
33
Chapter 5: Optimum Receiver for Binary Data Transmission
φ2 (t )
s 2 (t )
1
s1 (t )
0.5
−1
EE456 – Digital Communications
− 0.5
0
s1 (t)
=
s2 (t)
=
0.5
1
φ1 (t )
1
φ2 (t),
2
−φ1 (t) + φ2 (t).
φ1 (t) +
34
Chapter 5: Optimum Receiver for Binary Data Transmission
For each value of the signal-to-noise ratio (SNR), Matlab simulation was conducted
for transmitting/receiving 500 equally-likely bits.
(E1 +E2 )/2
N0
=0.99 (dB); P[error]=0.1
3
3
2
2
1
1
φ2 (t)
φ2 (t)
(E1 +E2 )/2
N0
0
−1
−2
−3
=6.17 (dB); P[error]=0.01
0
−1
−2
−1
EE456 – Digital Communications
0
φ1 (t)
1
2
3
−2
−3
−2
−1
0
φ1 (t)
1
2
3
35
Chapter 5: Optimum Receiver for Binary Data Transmission
φ2 (t ) r2
φ2 (t ) r2
s 2 (t )
s 2 (t )
1
1
jkl
mno
s1 (t )
s1 (t )
0.5
0.5
Choose s1 (t )
Choose s2 (t )
−1
− 0.5
0
0.5
1
Choose s1 (t )
Choose s2 (t )
φ1 (t )
r1
−1
− 0.5
0
0.5
1
φ1 (t )
r1
φ2 (t ) r2
s 2 (t )
1
pqr
Choose s2 (t )
s1 (t )
0.5
Choose s1 (t )
−1
− 0.5
0
0.5
1
φ1 (t )
r1
(a) P1 = P2 = 0.5, (b) P1 = 0.25, P2 = 0.75. (c) P1 = 0.75, P2 = 0.25.
EE456 – Digital Communications
36
Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.7
s2 (t) = φ1 (t) + φ2 (t),
s1 (t) = φ1 (t) − φ2 (t).
φ 2 (t )
φ1 (t )
3
1
1
0
−1
EE456 – Digital Communications
1
2
s
0
1
s
37
Chapter 5: Optimum Receiver for Binary Data Transmission
φ2 (t ) r2
1
s 2 (t )
Choose s 2 (t )
N 0 y P1 v
lnw t
4 wx P2 tu
φ1 (t )
0
r1
1
Choose s1 (t )
−1
EE456 – Digital Communications
s1 (t )
38
Chapter 5: Optimum Receiver for Binary Data Transmission
t = Tb
Tb
(•)dt
r (t )
r2
z{|}~~{
0
φ 2 (t )
T=
3
0
Tb
r2
Tb
T=
EE456 – Digital Communications
N 0 P1
ln
4 P2
t = Tb
3
0
r2 < T choose s1 (t )
~
h2 (t )
r (t )
r2 ≥ T choose s2 (t )
r2 ≥ T choose s2 (t )
r2 < T choose s1 (t )
N 0 P1
ln
4 P2
39
Chapter 5: Optimum Receiver for Binary Data Transmission
Implementation with One Correlator/Matched Filter
Always possible by choosing φ̂1 (t) and φ̂2 (t) such that one of the two basis functions
is perpendicular to the line joining the two signals.
φ2 (t )
φ1 (t )
s2 (t )
s22
s =s
11
φ 2 (t )
21
s12
s22
s1 (t )
θ
0
s21
s11
φ1 (t )
s12
The optimum receiver is still the minimum-distance receiver. However the terms
(r̂1 − ŝ11 )2 and (r̂1 − ŝ21 )2 are the same on both sides of the comparison and hence
can be removed. This means that one does not need to compute r̂1 !
1D
(r̂1 − ŝ11 )2 + (r̂2 − ŝ12 )2
{z
}
|
R
0D
d2
1
1D
r̂2
R
0D
ŝ22 + ŝ12
2
| {z }
midpoint of two signals
EE456 – Digital Communications
+
1D
(r̂1 − ŝ21 )2 + (r̂2 − ŝ22 )2 ⇔ (r̂2 − ŝ12 )2 R (r̂2 − ŝ22 )2
{z
}
|
0D
d2
2
N0 /2
ln
ŝ22 − ŝ12
|
{z
P1
P2
equal to 0 if P1 =P2
}
≡ T.
40
Chapter 5: Optimum Receiver for Binary Data Transmission
t = Tb
Tb
¥ (•)dt
r (t )
r̂2
rˆ2 ≥ T ¨ 1D
¡¢£¢¤£
rˆ2 < T ¨ 0 D
0
φˆ2 ( t )
Threshold T
¦¢§
t = Tb
r (t )
r̂2
h (t ) = φˆ2 (Tb − t )
rˆ2 ≥ T ³ 1D
©ª«¬®¯ª®
rˆ2 < T ³ 0D
°±²
φ̂2 (t) =
Threshold T
s2 (t) − s1 (t)
ŝ22 + ŝ12
, T ≡
+
√
1
2
2
(E2 − 2ρ E1 E2 + E1 )
EE456 – Digital Communications
N0 /2
ŝ22 − ŝ12
ln
P1
P2
.
41
Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.8
φ2 (t )
φˆ1 (t )
s2 ( t )
φˆ2 (t )
E
sˆ11 = sˆ21
θ =π /4
E
EE456 – Digital Communications
φ̂1 (t)
=
φ̂2 (t)
=
s1 (t )
φ1 (t )
1
√ [φ1 (t) + φ2 (t)],
2
1
√ [−φ1 (t) + φ2 (t)].
2
42
Chapter 5: Optimum Receiver for Binary Data Transmission
t = Tb
Tb
r (t )
rˆ2 ≥ T ¿ 1D
r̂2
» (•)dt
´µ¶·¸¹¸ºµ¹
rˆ2 < T ¿ 0D
0
φˆ2 ( t )
Threshold T
2
Tb
0
Tb 2
Tb
¾
¼¸½
t = Tb
h(t )
r (t )
2
Tb
0
r̂2
Tb 2
rˆ2 ≥ T Ë 1D
ÀÁÂÃÄÅÄÆÁÅ
rˆ2 < T Ë 0D
Ê
Threshold T
ÇÈÉ
EE456 – Digital Communications
43
Chapter 5: Optimum Receiver for Binary Data Transmission
Receiver Performance
To detect bk , compare r̂2 =
T =
ŝ12 +ŝ22
2
+
N0
2(ŝ22 −ŝ12 )
Z
kTb
r(t)φ̂2 (t)dt to the threshold
b
(k−1)T
P1
.
ln P
2
ÍÎÏÐÑÐÒÓ ÔÒÕÓÖ×ØÙ
f (rˆ2 0T )
f (rˆ2 1T )
ŝ12
choose 0T ⇐
P [error]
ŝ22
r̂2
Ì choose 1T
=
T
P [(0 transmitted and 1 decided) or (1 transmitted and 0 decided)]
=
P [(0T , 1D ) or (1T , 0D )].
EE456 – Digital Communications
44
Chapter 5: Optimum Receiver for Binary Data Transmission
ßàåæ
f (rˆ2 0T ) ßàááàâãä çàèéâêåë
ìàâíë
ìàâíë
f (rˆ2 1T )
çàèéâêåë îãéïåéâä ðññéòàåéâä
ìàâíë óôõæ
ìàâíë
ßàåæ
ðññéòàåéâä îãéïåéâä
çàèéâêåë çàèéâêåë ßàááàâãä
öõàä÷ó
óôõæ
öõàä÷ó
ôæøõñéíéáë îãéïåêéñ ùâéøæää
ôæøõñéíéáë îãéïåêéñ ùâéøæää
ŝ12
ŝ 22
ÚÛÜÝ Þ
r̂2
ÚÛÜÝ û
ú choose 1T
choose 0T ⇐
T
P [error]
=
=
P [0T , 1D ] + P [1T , 0D ] = P [1D |0T ]P [0T ] + P [0D |1T ]P [1T ]
Z T
Z ∞
P1
f (r̂2 |1T )dr̂2
f (r̂2 |0T )dr̂2 +P2
−∞
T
{z
}
|
{z
}
|
Area B
=
EE456 – Digital Communications
P1 Q
T − ŝ12
p
N0 /2
Area A
!
"
+ P2 1 − Q
T − ŝ22
p
N0 /2
!#
.
45
Chapter 5: Optimum Receiver for Binary Data Transmission
Q-function
2
1 − λ2
e
2π
ý
üÿý
ÿ
üýþÿ0
þý
x
1
Q(x) ≡ √
2π
λ
Area = Q ( x )
Z
∞
exp
x
−
λ2
2
!
dλ.
0
10
−2
10
−4
Q(x)
10
−6
10
−8
10
−10
10
EE456 – Digital Communications
0
1
2
3
x
4
5
6
46
Chapter 5: Optimum Receiver for Binary Data Transmission
Performance when P1 = P2
f (r2 0T )
0
f (r2 1T )
s12
r2
s 22
=Q
choose 0T ⇐
T=
P [error] = Q
ŝ22 − ŝ12
p
2 N0 /2
!
(
s22 − s12
2 N 0 /2
)
⇒ choose 1T
s12 + s22
2
=Q
distance between the signals
2 × noise RMS value
.
Probability of error decreases as either the two signals become more dissimilar
(increasing the distances between them) or the noise power becomes less.
To maximize the distance between the two signals one chooses them so that they
are placed 180◦ from each other ⇒ s2 (t) = −s1 (t), i.e., antipodal signaling.
The error probability does not depend on the signal shapes but only on the
distance between them.
EE456 – Digital Communications
47
Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.9
1
φ2 (t )
1
0
t
2
s1 (t ) ⇔ 0T
1
1
−2
−1
0
1
2
−1
1T ⇔ s 2 (t )
1
φ1 (t )
0
t
−1
−2
(a) Determine and sketch the two signals s1 (t) and s2 (t).
EE456 – Digital Communications
48
Chapter 5: Optimum Receiver for Binary Data Transmission
(b) The two signals s1 (t) and s2 (t) are used for the transmission of equally likely bits 0 and 1,
respectively, over an additive white Gaussian noise (AWGN) channel. Clearly draw the
decision boundary and the decision regions of the optimum receiver. Write the expression for
the optimum decision rule.
(c) Find and sketch the two orthonormal basis functions φ̂1 (t) and φ̂2 (t) such that the optimum
receiver can be implemented using only the projection r̂2 of the received signal r(t) onto the
basis function φ̂2 (t). Draw the block diagram of such a receiver that uses a matched filter.
(d) Consider now the following argument put forth by your classmate. She reasons that since the
component of the signals along φ̂1 (t) is not useful at the receiver in determining which bit
was transmitted, one should not even transmit this component of the signal. Thus she
modifies the transmitted signal as follows:
(M)
s1 (t) = s1 (t) − component of s1 (t) along φ̂1 (t)
(M)
s2 (t) = s2 (t) − component of s2 (t) along φ̂1 (t)
(M)
(M)
Clearly identify the locations of s1 (t) and s2 (t) in the signal space diagram. What is the
average energy of this signal set? Compare it to the average energy of the original set.
Comment.
EE456 – Digital Communications
49
Chapter 5: Optimum Receiver for Binary Data Transmission
s 2 (t )
s1 (t )
3
1
t
1
0
0
−1
−1
t
−3
EE456 – Digital Communications
50
Chapter 5: Optimum Receiver for Binary Data Transmission
1
φ2 (t )
0
!"#
0D
1D
2
1
t
φˆ2 ( t )
s1M (t )
s1 (t ) ⇔ 0T
1
1
−2
−1
0
1
θ =−
s2M (t )
1T ⇔ s 2 (t )
EE456 – Digital Communications
π
2
−1
4
−2
φˆ1 (t )
1
φ1 (t )
0
t
−1
51
Chapter 5: Optimum Receiver for Binary Data Transmission
φ̂1 (t)
φ̂2 (t)
=
cos(−π/4)
− sin(−π/4)
sin(−π/4)
cos(−π/4)
φ1 (t)
φ2 (t)
=
"
√1
2
√1
2
− √1
2
1
√
2
#
φ1 (t)
φ2 (t)
.
1
φ̂2 (t) = √ [φ1 (t) + φ2 (t)].
2
1
φ̂1 (t) = √ [φ1 (t) − φ2 (t)],
2
φˆ1 (t )
φˆ2 (t )
2
1/ 2
1
t
0
0
1/ 2
t
− 2
h (t ) = φˆ2 (1 − t )
t =1
r (t )
2
0
EE456 – Digital Communications
1/ 2
1
rˆ2 ≥ 0 $ 0D
rˆ2 < 0 $ 1D
t
52
Chapter 5: Optimum Receiver for Binary Data Transmission
Antipodal Signalling
w (t ) PSD
N0
2
t = kTb
(1)
Tb
2Tb
0
t
( −1) ( −1)
hT (t ) = p( t )
x (t )
s2 (t ) = − s1 ( t ) = p ( t )
E1 = E2 = E
+
hR (t ) = κ p (Tb − t )
y (t )
0
y ( k ) = ± (κ E ) + w ( k )
(0, N20 E )
The pulse shaping filter hT = p(t) defines the power spectrum density of the
transmitted signal, which can be shown to be proportional to |P (f )|2 .
The error performance, P [error] only depends on the energy E of p(t)
√and noise
PSD level N0 . Specifically, the distance between s1 (t) and s2 (t) is 2 E (you
should show this for yourself, algebraically or geometrically). Therefore
s
!
2E
P [error] = Q
.
N0
For antipodal signalling, the optimum decisions are performed by comparing the
samples of the matched filter’s output (sampled at exactly integer multiples of
the bit duration) with a threshold 0. Of course such an optimum decision rule
does not change if the impulse response of the matched filter is scaled by a
positive constant.
EE456 – Digital Communications
53
Chapter 5: Optimum Receiver for Binary Data Transmission
Scaling the matched filter’s impulse response hR (t) does not change the receiver
performance because it scales both signal and noise components by the same
factor, leaving the signal-to-noise ratio (SNR) of the decision variable unchanged!
√
In the above block diagram, hR (t) = κp(Tb − t). We have been using κ = 1/ E
in√
order to represent the signals on the signal space diagram (which would be at
± E) and to conclude that the variance of the noise component is exactly N0 /2.
For an arbitrary scaling factor κ, the signal component becomes ±κE, while the
variance of the noise component is N20 κ2 E. Thus, the SNR is
SNR =
(±κE)2
Signal power
2E
= N
,
=
0 κ2 E
Noise power
N0
2
(indepedent of κ!)
In terms of the SNR, the error performance of antipodal signalling is
P [error] = Q
s
2E
N0
!
=Q
√
SNR
In fact, it can be proved that the receive filter that maximizes the SNR of the
decision variable must be the matched filter. It is important to emphasize that
the matching property here concerns the shapes of the impulse responses of the
transmit and receive filters.
EE456 – Digital Communications
54
Chapter 5: Optimum Receiver for Binary Data Transmission
Outputs of the Matched/Mismatched Filters (No-Noise Scenario)
Clean received signals for rect and half-sine (HS) shaping filters
Output of a matched filter: rect/rect matching
Outputs of HS/HS matched filter (red) and HS/rect mismatched filter (pink)
When the matched filter is used, sampling at exact multiples of the bit duration maximizes the power of the signal component in the
decision variable, hence maximizing the SNR. A timing error (imperfect sampling) would reduce the power of the signal component, hence
reducing the SNR, hence degrading the performance, i.e., increasing P [error].
When the receive filter is not matched to the transmit filter, the power of the signal component and the SNR are not maximized, even
under perfect sampling!
EE456 – Digital Communications
55
Chapter 5: Optimum Receiver for Binary Data Transmission
Antipodal Baseband Signalling with Rectangular Pulse Shaping
bk
h(t ) = p(Tb − t )
EE456 – Digital Communications
56
Chapter 5: Optimum Receiver for Binary Data Transmission
Antipodal Baseband Signalling with Half-Sine Pulse Shaping
bk
h(t ) = p (Tb − t )
EE456 – Digital Communications
57
Chapter 5: Optimum Receiver for Binary Data Transmission
PSD Derivation of Arbitrary Binary Modulation
Applicable to any binary modulation with arbitrary a priori probabilities, but
restricted to statistically independent bits.
sT (t )
s1 ( t − 2Tb )
s1 (t )
−2Tb
−Tb
0
2Tb
Tb
3Tb
4Tb
t
s2 (t − 3Tb )
sT (t) =
∞
X
k=−∞
gk (t),
gk (t) =
s1 (t − kTb ),
s2 (t − kTb ),
with probability P1
.
with probability P2
The derivation on the next slide shows that:
∞
P 1 S1
X
P1 P2
SsT (f ) =
|S1 (f ) − S2 (f )|2 +
Tb
n=−∞
EE456 – Digital Communications
n
Tb
+ P 2 S2
Tb
n
Tb
2
δ
f−
n
Tb
.
58
Chapter 5: Optimum Receiver for Binary Data Transmission
sT (t) = E{sT (t)} + sT (t) − E{sT (t)} = v(t) + q(t)
| {z } |
{z
}
DC
AC
∞
X
v(t) = E{sT (t)} =
k=−∞
Sv (f ) =
∞
X
n=−∞
Sv (f ) =
|Dn |2 δ
∞
X
n=−∞
[P1 s1 (t − kTb ) + P2 s2 (t − kTb )]
f−
P1 S1
n
Tb
n
Tb
1
Tb
, Dn =
+ P2 S2
Tb
n
Tb
P1 S1
2
δ
n
Tb
+ P2 S2
n
Tb
,
.
n
.
f−
Tb
To calculate Sq (f ), apply the basic definition of PSD:
Sq (f ) = lim
T →∞
SsT (f ) =
P1 P2
E{|GT (f )|2 }
= ··· =
|S1 (f ) − S2 (f )|2 .
T
Tb
∞
P1 S1
X
P1 P2
|S1 (f ) − S2 (f )|2 +
Tb
n=−∞
n
Tb
+ P2 S2
Tb
n
Tb
2
δ
f−
n
Tb
For the special, but important case of antipodal signalling, s2 (t) = −s1 (t) = p(t), and equally
likely bits, P1 = P2 = 0.5, the PSD of the transmitted signal is solely determined by the
Fourier transform of p(t):
|P (f )|2
SsT (f ) =
Tb
EE456 – Digital Communications
59
Chapter 5: Optimum Receiver for Binary Data Transmission
Baseband Message Signals with Different Pulse Shaping Filters
Information bits or amplitude levels
1
0.5
0
−0.5
−1
2
4
6
8
10
12
Output of the transmit pulse shaping filter − Rectangular
14
PSD - Rectangular
Magnitude (dB)
2
1
0
−1
−2
2
4
6
8
10
Half-sine pulse shaping filter
12
Magnitude (dB)
1
0
−1
2
4
6
8
10
SRRC pulse shaping filter (β = 0.5)
12
Magnitude (dB)
1
0
−1
2
EE456 – Digital Communications
4
6
8
t/T b
10
12
−40
−60
14
−2
0
2
f × Tb
PSD - Half-sine
0
−20
−40
−60
14
2
−2
−20
14
2
−2
0
−2
0
2
f × Tb
PSD - SRRC (30 symbols long)
0
−20
−40
−60
−2
0
2
Normalized frequency, f × T b
60
Chapter 5: Optimum Receiver for Binary Data Transmission
Building A Binary (Antipodal) Comm. System in Labs #4 and #5
low-rate sequence, index k
high-rate sequences, index n
...
...
...
t
s[n ] = i[n ] * h[n ] = å a[k ] p(nT - kTsym )
data
bits
k
S/P
bits-tolevels
mapping
N
N=
Tsym
T
pulse
shaping
filter
hT [n ] = p(nT )
i[n ] = i (nT ) = å a[k ]d (nT - kTsym )
k
DAC
correction
filter
DAC
cos( wˆ c n )
NCO
fˆc
T
wc = wˆ c Fs
æ
ö
s(t ) = ç å a ( k ) p(t - kTs ) ÷ cos(wct )
è k
ø
Figure 1: Block diagram of the transmitter.
EE456 – Digital Communications
61
Chapter 5: Optimum Receiver for Binary Data Transmission
s (t )
∑
x[ n − d ]
x[ n]
hR [n ] = p( − nT )
xc [ n ]
k =n
Tsym
T
cos( ω c n − θ )
θ
fc
T
sin( ω c n − θ )
ωc d = θ
x s [n ]
Figure 2: Block diagram of the receiver.
EE456 – Digital Communications
62