This document summarizes key aspects of designing an optimum receiver for binary data transmission presented in Chapter 5. It begins by representing signals using orthonormal basis functions to reduce the problem from waveforms to random variables. It then describes representing noise using a complete orthonormal set and how the noise coefficients are statistically independent Gaussian variables. Finally, it outlines how the optimum receiver works by projecting the received signal onto the basis functions, resulting in random variables that can be used for binary decision making to minimize the bit error probability.
Optimum Receiver corrupted by AWGN ChannelAWANISHKUMAR84
Optimum Receiver corrupted by AWGN Channel
This topic is related to Advance Digital Communication Engineering. In this ppt, you will get all details explanations of the receiver how to get affected by white Noise.
These slides deal with the basic problem of channel equalization and exposes the issue related to it and shows how it can be balanced by the usage of effective and robust algorithms.
Sampling is a Simple method to convert analog signal into discrete Signal by using any one of its three methods
if the sampling frequency is twice or greater than twice then sampled signal can be convert back into analog signal easily......
Optimum Receiver corrupted by AWGN ChannelAWANISHKUMAR84
Optimum Receiver corrupted by AWGN Channel
This topic is related to Advance Digital Communication Engineering. In this ppt, you will get all details explanations of the receiver how to get affected by white Noise.
These slides deal with the basic problem of channel equalization and exposes the issue related to it and shows how it can be balanced by the usage of effective and robust algorithms.
Sampling is a Simple method to convert analog signal into discrete Signal by using any one of its three methods
if the sampling frequency is twice or greater than twice then sampled signal can be convert back into analog signal easily......
Cyclostationary analysis of polytime coded signals for lpi radarseSAT Journals
Abstract In Radars, an electromagnetic waveform will be sent, and an echo of the same signal will be received by the receiver. From this received signal, by extracting various parameters such as round trip delay, doppler frequency it is possible to find distance, speed, altitude, etc. However, nowadays as the technology increases, intruders are intercepting transmitted signal as it reaches them, and they will be extracting the characteristics and trying to modify them. So there is a need to develop a system whose signal cannot be identified by no cooperative intercept receivers. That is why LPI radars came into existence. In this paper a brief discussion on LPI radar and its modulation (Polytime code (PT1)), detection (Cyclostationary (DFSM & FAM) techniques such as DFSM, FAM are presented and compared with respect to computational complexity.
Keywords—LPI Radar, Polytime codes, Cyclostationary DFSM, and FAM
A signal is a pattern of variation that carry information.
Signals are represented mathematically as a function of one or more independent variable
basic concept of signals
types of signals
system concepts
the presentation consists of a brief description about ADAPTIVE LINEAR EQUALIZER , its classification and the associated attributes of ZERO FORCING EQUALIZER and MMSE EQUALIZER
Communication Systems_B.P. Lathi and Zhi Ding (Lecture No 31-39)Adnan Zafar
Lecture No 31: https://youtu.be/UaBW0d2o9eY
Lecture No 32: https://youtu.be/X5rwgJjajy0
Lecture No 33: https://youtu.be/NbH_iysZss8
Lecture No 34: https://youtu.be/85e7cJXidM0
Lecture No 35: https://youtu.be/8BCnKJkEXqA
Lecture No 36: https://youtu.be/aFOVu8TOF9U
Lecture No 37: https://youtu.be/_PUG203mg6s
Lecture No 38: https://youtu.be/52pHpqvP2BA
Lecture No 39: https://youtu.be/ljeWa2XWjfA
The presentation covers sampling theorem, ideal sampling, flat top sampling, natural sampling, reconstruction of signals from samples, aliasing effect, zero order hold, upsampling, downsampling, and discrete time processing of continuous time signals.
Blind equalization is a digital signal processing technique in which the transmitted signal is inferred (equalized) from the received signal, while making use only of the transmitted signal statistics. Hence, the use of the word blind in the name.
Cyclostationary analysis of polytime coded signals for lpi radarseSAT Journals
Abstract In Radars, an electromagnetic waveform will be sent, and an echo of the same signal will be received by the receiver. From this received signal, by extracting various parameters such as round trip delay, doppler frequency it is possible to find distance, speed, altitude, etc. However, nowadays as the technology increases, intruders are intercepting transmitted signal as it reaches them, and they will be extracting the characteristics and trying to modify them. So there is a need to develop a system whose signal cannot be identified by no cooperative intercept receivers. That is why LPI radars came into existence. In this paper a brief discussion on LPI radar and its modulation (Polytime code (PT1)), detection (Cyclostationary (DFSM & FAM) techniques such as DFSM, FAM are presented and compared with respect to computational complexity.
Keywords—LPI Radar, Polytime codes, Cyclostationary DFSM, and FAM
A signal is a pattern of variation that carry information.
Signals are represented mathematically as a function of one or more independent variable
basic concept of signals
types of signals
system concepts
the presentation consists of a brief description about ADAPTIVE LINEAR EQUALIZER , its classification and the associated attributes of ZERO FORCING EQUALIZER and MMSE EQUALIZER
Communication Systems_B.P. Lathi and Zhi Ding (Lecture No 31-39)Adnan Zafar
Lecture No 31: https://youtu.be/UaBW0d2o9eY
Lecture No 32: https://youtu.be/X5rwgJjajy0
Lecture No 33: https://youtu.be/NbH_iysZss8
Lecture No 34: https://youtu.be/85e7cJXidM0
Lecture No 35: https://youtu.be/8BCnKJkEXqA
Lecture No 36: https://youtu.be/aFOVu8TOF9U
Lecture No 37: https://youtu.be/_PUG203mg6s
Lecture No 38: https://youtu.be/52pHpqvP2BA
Lecture No 39: https://youtu.be/ljeWa2XWjfA
The presentation covers sampling theorem, ideal sampling, flat top sampling, natural sampling, reconstruction of signals from samples, aliasing effect, zero order hold, upsampling, downsampling, and discrete time processing of continuous time signals.
Blind equalization is a digital signal processing technique in which the transmitted signal is inferred (equalized) from the received signal, while making use only of the transmitted signal statistics. Hence, the use of the word blind in the name.
Contemporary communication systems 1st edition mesiya solutions manualto2001
Contemporary Communication Systems 1st Edition Mesiya Solutions Manual
Download:https://goo.gl/DmVRQ4
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contemporary communication systems mesiya download
contemporary communication systems pdf
contemporary communication systems mesiya solutions
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7076 chapter5 slides
1. Chapter 5: Optimum Receiver for Binary Data Transmission
A First Course in Digital Communications
Ha H. Nguyen and E. Shwedyk
February 2009
A First Course in Digital Communications 1/58
3. Chapter 5: Optimum Receiver for Binary Data Transmission
$%'() 0%1234 05633)7
8%1769%'
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( )tm { }kb 10 ( )k s t= ↔b
21 ( )k s t= ↔b
ˆ ( )tm
{ }ˆ
kb ( )tr
( )tw
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).
A First Course in Digital Communications 3/58
4. 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) are orthonormal if:
Tb
0
φ1(t)φ2(t)dt = 0 (orthogonality),
Tb
0
φ2
1(t)dt =
Tb
0
φ2
2(t)dt = 1 (normalized to have unit energy).
The representations are
s1(t) = s11φ1(t) + s12φ2(t),
s2(t) = s21φ1(t) + s22φ2(t).
where sij =
Tb
0
si(t)φj(t)dt, i, j ∈ {1, 2},
A First Course in Digital Communications 4/58
5. Chapter 5: Optimum Receiver for Binary Data Transmission
Geometric Representation of Signals s1(t) and s2(t) (II)
)(1 tφ
)(2 tφ
11s
)(2 ts
)(1 ts
22s
21s
12s
0
s1(t) = s11φ1(t) + s12φ2(t),
s2(t) = s21φ1(t) + s22φ2(t),
sij =
Tb
0
si(t)φj(t)dt, i, j ∈ {1, 2},
Tb
0 si(t)φj(t)dt is the projection of si(t) onto φj(t).
How to choose orthonormal functions φ1(t) and φ2(t) to
represent s1(t) and s2(t) exactly?
A First Course in Digital Communications 5/58
6. Chapter 5: Optimum Receiver for Binary Data Transmission
Gram-Schmidt Procedure
1 Let φ1(t) ≡
s1(t)
E1
. Note that s11 =
√
E1 and s12 = 0.
2 Project s
′
2(t) =
s2(t)
E2
onto φ1(t) to obtain the correlation
coefficient:
ρ =
Tb
0
s2(t)
√
E2
φ1(t)dt =
1
√
E1E2
Tb
0
s1(t)s2(t)dt.
3 Subtract ρφ1(t) from s
′
2(t) to obtain φ
′
2(t) = s2(t)
√
E2
− ρφ1(t).
4 Finally, normalize φ
′
2(t) to obtain:
φ2(t) =
φ
′
2(t)
Tb
0 φ
′
2(t)
2
dt
=
φ
′
2(t)
1 − ρ2
=
1
1 − ρ2
s2(t)
√
E2
−
ρs1(t)
√
E1
.
A First Course in Digital Communications 6/58
8. Chapter 5: Optimum Receiver for Binary Data Transmission
Gram-Schmidt Procedure for M Waveforms {si(t)}M
i=1
φ1(t) =
s1(t)
∞
−∞ s2
1(t)dt
,
φi(t) =
φ
′
i(t)
∞
−∞ φ
′
i(t)
2
dt
, i = 2, 3, . . . , N,
φ
′
i(t) =
si(t)
√
Ei
−
i−1
j=1
ρijφj(t),
ρij =
∞
−∞
si(t)
√
Ei
φj(t)dt, j = 1, 2, . . . , i − 1.
If the waveforms {si(t)}M
i=1 form a linearly independent set, then
N = M. Otherwise N M.
A First Course in Digital Communications 8/58
9. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 1
H
)(2 ts
0
bT
V−
H
)(1 ts
V
0 bT
IPQ
R
)(1 tφ
bT1
0 bT
STU
)(1 tφ
0
)(1 ts)(2 ts
E− E
VWX
(a) Signal set. (b) Orthonormal function. (c) Signal space representation.
A First Course in Digital Communications 9/58
10. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 2
V−
Y
)(1 ts
V
0
bT Y
)(2 ts
V
0 bT
`ab
c
)(1 tφ
bT1
0
bT
bT1−
c
)(2 tφ
0 bT
bT1
def
A First Course in Digital Communications 10/58
11. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 3
g
)(2 ts
0
bT
V−
g
)(1 ts
V
0 bT
V
α
)(1 tφ
0
)(1 ts
)(2 ts
),(2 αφ t
0=α
2
bT
=α
ρα,increasing
( )0,E( )0,E−
E
hh
i
p
qq
r
s
−
2
3
,
2
EE
4
bT
α =
ρ =
1
E
Tb
0
s2(t)s1(t)dt
=
1
V 2Tb
V 2
α − V 2
(Tb − α)
=
2α
Tb
− 1
A First Course in Digital Communications 11/58
12. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 4
t
)(2 ts
0 bT
V3
t
)(1 ts
V
0 bT 2bT
uvw
x
)(1 tφ
bT1
0 bT
x
)(2 tφ
0
bT
2bT
bT
3
−
bT
3
y€
A First Course in Digital Communications 12/58
13. Chapter 5: Optimum Receiver for Binary Data Transmission
)(1 tφ
0
)(1 ts
)(2 ts
( )2,23 EE
( )0,E
)(2 tφ
ρ =
1
E
Tb
0
s2(t)s1(t)dt =
2
E
Tb/2
0
2
√
3
Tb
V t V dt =
√
3
2
,
φ2(t) =
1
(1 − 3
4 )
1
2
s2(t)
√
E
− ρ
s1(t)
√
E
=
2
√
E
s2(t) −
√
3
2
s1(t) ,
s21 =
√
3
2
√
E, s22 =
1
2
√
E.
d21 =
Tb
0
[s2(t) − s1(t)]2
dt
1
2
= 2 −
√
3 E.
A First Course in Digital Communications 13/58
14. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5
)(1 tφ
0
)(1 ts
)(2 ts
E
)(2 tφ
1−=
=
ρ
πθ
0
23
=
=
ρ
πθ
0
2
=
=
ρ
πθ
2locus of ( ) as
varies from 0 to 2 .
s t θ
π
θ
s1(t) =
√
E
2
Tb
cos(2πfct),
s2(t) =
√
E
2
Tb
cos(2πfct + θ).
where fc = k
2Tb
, k an integer.
A First Course in Digital Communications 14/58
15. Chapter 5: Optimum Receiver for Binary Data Transmission
Representation of Noise with Walsh Functions
−5
0
5
x
1
(t)
−5
0
5
x2
(t)
0
1
2
φ1
(t)
−2
0
2
φ
2
(t)
−2
0
2
φ3
(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−2
0
2
φ4
(t)
t
Exact representation of noise with 4 Walsh functions is not possible.
A First Course in Digital Communications 15/58
16. Chapter 5: Optimum Receiver for Binary Data Transmission
The First 16 Walsh Functions
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
t
Exact representations might be possible with many more Walsh functions.
A First Course in Digital Communications 16/58
17. Chapter 5: Optimum Receiver for Binary Data Transmission
The First 16 Sine and Cosine Functions
Can also use sine and cosine functions (Fourier representation).
0 0.25 0.5 0.75 1
−1.5
1.5
t
A First Course in Digital Communications 17/58
18. Chapter 5: Optimum Receiver for Binary Data Transmission
Representation of the Noise I
To represent the random noise signal, w(t), in the time
interval [(k − 1)Tb, kTb], need to use a complete orthonormal
set of known deterministic functions:
w(t) =
∞
i=1
wiφi(t), where wi =
Tb
0
w(t)φi(t)dt.
The coefficients wi’s are random variables and understanding
their statistical properties is imperative in developing the
optimum receiver.
A First Course in Digital Communications 18/58
19. Chapter 5: Optimum Receiver for Binary Data Transmission
Representation of the Noise II
When w(t) is zero-mean and white, then:
1 E{wi} = E
Tb
0 w(t)φi(t)dt =
Tb
0 E{w(t)}φi(t)dt = 0.
2 E{wiwj} = E
Tb
0 dλw(λ)φi(λ)
Tb
0 dτw(τ)φj (τ) =
N0
2 , i = j
0, i = j
.
{w1, w2, . . .} are zero-mean and uncorrelated random variables.
If w(t) is not only zero-mean and white, but also Gaussian ⇒
{w1, w2, . . .} are Gaussian and statistically independent!!!
The above properties do not depend on how the set
{φi(t), i = 1, 2, . . .} is chosen.
Shall choose as the first two functions the functions φ1(t) and φ2(t)
used to represent the two signals s1(t) and s2(t) exactly. The
remaining functions, i.e., φ3(t), φ4(t), . . . , are simply chosen to
complete the set.
A First Course in Digital Communications 19/58
20. Chapter 5: Optimum Receiver for Binary Data Transmission
Optimum Receiver I
Without any loss of generality, concentrate on the first bit interval.
The received signal is
r(t) = si(t) + w(t), 0 ≤ t ≤ Tb
=
s1(t) + w(t), if a “0” is transmitted
s2(t) + w(t), if a “1” is transmitted
.
= [si1φ1(t) + si2φ2(t)]
si(t)
+ [w1φ1(t) + w2φ2(t) + w3φ3(t) + w4φ4(t) + · · · ]
w(t)
= (si1 + w1)φ1(t) + (si2 + w2)φ2(t) + w3φ3(t) + w4φ4(t) + · · ·
= r1φ1(t) + r2φ2(t) + r3φ3(t) + r4φ4(t) + · · ·
A First Course in Digital Communications 20/58
21. Chapter 5: Optimum Receiver for Binary Data Transmission
Optimum Receiver II
where rj =
Tb
0
r(t)φj(t)dt, and
r1 = si1 + w1
r2 = si2 + w2
r3 = w3
r4 = w4
...
Note that rj, for j = 3, 4, 5, . . ., does not depend on which
signal (s1(t) or s2(t)) was transmitted.
The decision can now be based on the observations
r1, r2, r3, r4, . . ..
The criterion is to minimize the bit error probability.
A First Course in Digital Communications 21/58
22. Chapter 5: Optimum Receiver for Binary Data Transmission
Optimum Receiver III
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 decision regions.
1ℜ
1ℜ
2ℜ
ℜspacenObservatioDecide a 1 was transmitted
if falls in this region.r
‚
Decide a 0 was transmitted
if falls in this region.r
‚
A First Course in Digital Communications 22/58
23. Chapter 5: Optimum Receiver for Binary Data Transmission
Optimum Receiver IV
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 ]
= P2
ℜ1
f(r|1T )dr + P1
ℜ2
f(r|0T )dr
= P2
ℜ−ℜ2
f(r|1T )dr + P1
ℜ2
f(r|0T )dr
= P2
ℜ
f(r|1T )dr +
ℜ2
[P1f(r|0T ) − P2f(r|1T )]dr
= P2 +
ℜ2
[P1f(r|0T ) − P2f(r|1T )] dr.
A First Course in Digital Communications 23/58
24. Chapter 5: Optimum Receiver for Binary Data Transmission
Optimum Receiver V
The minimum error probability decision rule is
P1f(r|0T ) − P2f(r|1T ) ≥ 0 ⇒ decide “0” (0D)
P1f(r|0T ) − P2f(r|1T ) 0 ⇒ decide “1” (1D)
.
Equivalently,
f(r|1T )
f(r|0T )
1D
0D
P1
P2
. (1)
The expression
f(r|1T )
f(r|0T )
is called the likelihood ratio.
The decision rule in (1) was derived without specifying any
statistical properties of the noise process w(t).
A First Course in Digital Communications 24/58
25. Chapter 5: Optimum Receiver for Binary Data Transmission
Optimum Receiver VI
Simplified decision rule when the noise w(t) is zero-mean,
white and Gaussian:
(r1 − s11)2 + (r2 − s12)2
1D
0D
(r1 − s21)2 + (r2 − s22)2 + N0ln P1
P2
.
For the special case of P1 = P2 (signals are equally likely):
(r1 − s11)2 + (r2 − s12)2
1D
0D
(r1 − s21)2 + (r2 − s22)2.
⇒ minimum-distance receiver!
A First Course in Digital Communications 25/58
26. Chapter 5: Optimum Receiver for Binary Data Transmission
Minimum-Distance Receiver
(r1 − s11)2 + (r2 − s12)2
1D
0D
(r1 − s21)2 + (r2 − s22)2.
)(1 tφ
0
)(1 ts
)(2 ts
)(2 tφ
),( 1211 ss
),( 2221 ss
)(Choose 2 ts )(Choose 1 ts 1r
2r
( )r t
( )1 2,r r
1d
2d
A First Course in Digital Communications 26/58
27. Chapter 5: Optimum Receiver for Binary Data Transmission
Correlation Receiver Implementation
( )
ƒ
•
bT
t
0
d
)(2 tφ
( )
„
•
bT
t
0
d
)(1 tφ
bTt =
bTt =
Decision
1r
2r
( ) ( ) ( )it s t t= +r w
2 2
1 1 2 2
0
Compute
( ) ( )
ln( )
for 1, 2
and choose
the smallest
i i
i
r s r s
N P
i
− + −
−
=
( )
…
•
bT
t
0
d
)(2 tφ
( )
†
•
bT
t
0
d
)(1 tφ
bTt =
bTt =
Decision
‡ˆ‰‰‘
’ˆ‘
“”•–‘’
2
)ln(
2
1
1
0 E
P
N
−
2
)ln(
2
2
2
0 E
P
N
−
( )tr
1r
2r
Form
the
dot
product
ir s⋅
— —
A First Course in Digital Communications 27/58
28. Chapter 5: Optimum Receiver for Binary Data Transmission
Receiver Implementation using Matched Filters
˜™defegh
iejdkel
)()( 22 tTth b −= φ
)()( 11 tTth b −= φ
bTt =
bTt =
Decision( )tr
1r
2r
A First Course in Digital Communications 28/58
29. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.6 I
m
)(2 ts
0
m
)(1 ts
0 5.0
5.0
1
1
5.0
5.1
2−
nop
q
)(1 tφ
1
0 1
q
)(2 tφ
5.0
0
1−
1
1
rst
A First Course in Digital Communications 29/58
30. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.6 II
)(1 tφ
0
)(1 ts
)(2 ts
)(2 tφ
5.0−
5.0
5.0 11−
1
s1(t) = φ1(t) +
1
2
φ2(t),
s2(t) = −φ1(t) + φ2(t).
A First Course in Digital Communications 30/58
32. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.7 I
s2(t) = φ1(t) + φ2(t),
s1(t) = φ1(t) − φ2(t).
~
)(1 tφ
0
1
~
)(2 tφ
0 1
1−
1
3
2
1
A First Course in Digital Communications 32/58
33. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.7 II
)(1 tφ
0
)(1 ts
)(2 ts1
1−
1
)(Choose 2 ts
)(Choose 1 ts
)(2 tφ
€
‚‚
ƒ
„
2
10
ln
4 P
PN
2r
1r
A First Course in Digital Communications 33/58
34. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.7 III
…†‡ˆ‰Š‰‹†Š( )
Œ
•
bT
t
0
d
)(2 tφ
bTt =
Ž
‘
’
=
2
10
ln
4 P
PN
T
“
0 bT
3
”‰•
)(tr 2r
2 2
2 1
choose ( )
choose ( )
r T s t
r T s t
≥ –
–
—˜™š›œ›˜œ
bTt =
žž
Ÿ
¡¡
¢
£
=
2
10
ln
4 P
PN
T
)(2 th
¤
0 bT
3
¥¦§
2 2
2 1
choose ( )
choose ( )
r T s t
r T s t
≥ ¨
¨
2r)(tr
A First Course in Digital Communications 34/58
37. Chapter 5: Optimum Receiver for Binary Data Transmission
¯°±²³´³µ°´( )
¶
•
bT
t
0
d
)(ˆ
2 tφ
bTt =
TThreshold·³¸
)(tr 2
ˆr 2
2
ˆ 1
ˆ 0
D
D
r T
r T
≥ ¹
¹
º»¼½¾¿¾À»¿
bTt =
TThreshold
)(ˆ)( 2 tTth b −= φ
ÁÂÃ
2
ˆr)(tr
2
2
ˆ 1
ˆ 0
D
D
r T
r T
≥ Ä
Ä
ˆφ2(t) =
s2(t) − s1(t)
(E2 − 2ρ
√
E1E2 + E1)
1
2
, T ≡
ˆs22 + ˆs12
2
+
N0/2
ˆs22 − ˆs12
ln
P1
P2
.
A First Course in Digital Communications 37/58
38. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.8 I
)(1 tφ
)(1 ts
)(2 ts
)(2 tφ
)(ˆ
1 tφ
)(ˆ
2 tφ
2111
ˆˆ ss =
4/πθ =
E
E
ˆφ1(t) =
1
√
2
[φ1(t) + φ2(t)],
ˆφ2(t) =
1
√
2
[−φ1(t) + φ2(t)].
A First Course in Digital Communications 38/58
39. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.8 II
ÅÆÇÈÉÊÉËÆÊ( )
Ì
•
bT
t
0
d
)(ˆ
2 tφ
bTt =
TThreshold
ÍÉÎÏ
0 bT
bT
2
2bT
)(tr 2
ˆr 2
2
ˆ 1
ˆ 0
D
D
r T
r T
≥ Ð
Ð
ÑÒÓÔÕÖÕ×ÒÖ
bTt =
TThreshold
ØÙÚ
Û
0
bT
2
2bT
)(th
)(tr 2
ˆr 2
2
ˆ 1
ˆ 0
D
D
r T
r T
≥ Ü
Ü
A First Course in Digital Communications 39/58
40. Chapter 5: Optimum Receiver for Binary Data Transmission
Receiver Performance
To detect bk, compare ˆr2 =
kTb
(k−1)Tb
r(t)ˆφ2(t)dt to the threshold
T = ˆs12+ˆs22
2 + N0
2(ˆs22−ˆs12) ln P1
P2
.
12ˆs 22ˆs
( )Trf 0ˆ2 ( )Trf 1ˆ2
TT 1choose0choose Ý⇐
2
ˆr
Þßàáâáãä åãæäçèéê
T
P[error] = P[(0 transmitted and 1 decided) or (1 transmitted and 0 decided)]
= P[(0T , 1D) or (1T , 0D)].
A First Course in Digital Communications 40/58
42. Chapter 5: Optimum Receiver for Binary Data Transmission
Q-function
! #$% '())
0
12)30
42 2(#) 42%
526(%!
72(!
526(%!
72(!
#$%() 892%()
0
λ
x
)(Area xQ=
2
2
e
2
1 λ
π
−
Q(x) ≡
1
√
2π
∞
x
exp −
λ2
2
dλ.
0 1 2 3 4 5 6
10
−10
10
−8
10
−6
10
−4
10
−2
10
0
x
Q(x)
A First Course in Digital Communications 42/58
43. Chapter 5: Optimum Receiver for Binary Data Transmission
Performance when P1 = P2
P[error] = Q
ˆs22 − ˆs12
2 N0/2
= 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.
A First Course in Digital Communications 43/58
44. Chapter 5: Optimum Receiver for Binary Data Transmission
Relationship Between Q(x) and erfc(x).
The complementary error function is defined as:
erfc(x) =
2
√
π
∞
x
exp(−λ2
)dλ
= 1 − erf(x).
erfc-function and the Q-function are related by:
Q(x) =
1
2
erfc
x
√
2
erfc(x) = 2Q(
√
2x).
Let Q−1(x) and erfc−1
(x) be the inverses of Q(x) and
erfc(x), respectively. Then
Q−1
(x) =
√
2erfc−1
(2x).
A First Course in Digital Communications 44/58
45. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.9 I
)(1 tφ
Tts 0)(1 ⇔
)(1 2 tsT ⇔
)(2 tφ
1−
2
2−
1
1−
1
2
2−
0 1
1
t
0
1
1
t
1−
0
A First Course in Digital Communications 45/58
46. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.9 II
(a) Determine and sketch the two signals s1(t) and s2(t).
(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 ˆφ1(t) and
ˆφ2(t) such that the optimum receiver can be implemented using
only the projection ˆr2 of the received signal r(t) onto the basis
function ˆφ2(t). Draw the block diagram of such a receiver that uses
a matched filter.
A First Course in Digital Communications 46/58
47. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.9 III
(d) Consider now the following argument put forth by your classmate.
She reasons that since the component of the signals along ˆφ1(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:
s
(M)
1 (t) = s1(t) − component of s1(t) along ˆφ1(t)
s
(M)
2 (t) = s2(t) − component of s2(t) along ˆφ1(t)
Clearly identify the locations of s
(M)
1 (t) and s
(M)
2 (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.
A First Course in Digital Communications 47/58
48. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.9 IV
0
t
3−
)(2 ts
1−
1
)(1 ts
0
1
3
t
1−
A First Course in Digital Communications 48/58
49. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.9 V
)(1 tφ
Tts 0)(1 ⇔
)(1 2 tsT ⇔
)(2 tφ
1−
2
2−
1
1−
1
2
2−
0
2
ˆ ( )tφ
4
π
θ = −
@ABCDCEF GEHFIPQR
D0
D1
1
ˆ ( )tφ
2 ( )M
s t
1 ( )M
s t
0
1
1
t
1−
0 1
1
t
A First Course in Digital Communications 49/58
50. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.9 VI
ˆφ1(t)
ˆφ2(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)
.
ˆφ1(t) =
1
√
2
[φ1(t) − φ2(t)],
ˆφ2(t) =
1
√
2
[φ1(t) + φ2(t)].
A First Course in Digital Communications 50/58
51. Chapter 5: Optimum Receiver for Binary Data Transmission
Example 5.9 VII
0
t
0
1 t
2−
1/2
2
1/2
1
ˆ ( )tφ 2
ˆ ( )tφ
2
ˆ( ) (1 )h t tφ= −
1t =
2
2
ˆ 0 0
ˆ 0 1
D
D
r
r
≥ S
S
)(tr
0
t
2
1/ 2 1
A First Course in Digital Communications 51/58
52. Chapter 5: Optimum Receiver for Binary Data Transmission
PSD of Digital Amplitude Modulation I
TUVVWXY `a
bcUd Xefgcbh
{ }0,1k ∈b
kcInformation bits
( )p t
( )ts
Amplitude
modulation
ck is drawn from a finite set of real numbers with a
probability that is known.
Examples: ck ∈ {−1, +1} (antipodal signaling), {0, 1} (on-off
keying), {−1, 0, +1} (pseudoternary line coding) or
{±1, ±3, · · · , ±(M − 1)} (M-ary amplitude-shift keying).
p(t) is a pulse waveform of duration Tb.
A First Course in Digital Communications 52/58
53. Chapter 5: Optimum Receiver for Binary Data Transmission
PSD of Digital Amplitude Modulation II
The transmitted signal is
s(t) =
∞
k=−∞
ckp(t − kTb).
To find PSD, truncate the random process to a time interval
of −T = −NTb to T = NTb:
sT (t) =
N
k=−N
ckp(t − kTb).
Take the Fourier transform of the truncated process:
ST (f) =
∞
k=−∞
ckF{p(t − kTb)} = P(f)
∞
k=−∞
cke−j2πfkTb
.
A First Course in Digital Communications 53/58
54. Chapter 5: Optimum Receiver for Binary Data Transmission
PSD of Digital Amplitude Modulation III
Apply the basic definition of PSD:
S(f) = lim
T→∞
E |ST (f)|2
2T
= lim
N→∞
|P(f)|2
(2N + 1)Tb
E
N
k=−N
cke−j2πfkTb
2
.
=
|P(f)|2
Tb
∞
m=−∞
Rc(m)e−j2πmfTb
.
where Rc(m) = E {ckck−m} is the (discrete) autocorrelation
of {ck}, with Rc(m) = Rc(−m).
A First Course in Digital Communications 54/58
55. Chapter 5: Optimum Receiver for Binary Data Transmission
PSD of Digital Amplitude Modulation IV
The output PSD is the input PSD multiplied by |P(f)|2
, a
transfer function.
bT0
i
t
bkT
LTI System
( ) ( )
( )
h t p t
P f
=
↔
2
in
Transmitted signal ( )
( ) ( ) ( )
t
S f S f P f=
s
( )kc
2
in
1
( ) ( )e bj mfT
mb
S f R m
T
π
∞
−
=−∞
= p c
S(f) =
|P(f)|2
Tb
∞
m=−∞
Rc(m)e−j2πmfTb
.
A First Course in Digital Communications 55/58
56. Chapter 5: Optimum Receiver for Binary Data Transmission
PSD Derivation of Arbitrary Binary Modulation I
Applicable to any binary modulation with arbitrary a priori
probabilities, but restricted to statistically independent bits.
⋯ ⋯
t
0
( )Ts t
bT 2 bT
3 bT 4 bT
1( )s t 2 ( 3 )bs t T−
bT−2 bT−
sT (t) =
∞
k=−∞
gk(t), gk(t) =
s1(t − kTb), with probability P1
s2(t − kTb), with probability P2
.
A First Course in Digital Communications 56/58
57. Chapter 5: Optimum Receiver for Binary Data Transmission
PSD Derivation of Arbitrary Binary Modulation II
Decompose sT (t) into a sum of a DC and an AC component:
sT (t) = E{sT (t)}
DC
+ sT (t) − E{sT (t)}
AC
= v(t) + q(t)
v(t) = E{sT (t)} =
∞
k=−∞
[P1s1(t − kTb) + P2s2(t − kTb)]
Sv(f) =
∞
n=−∞
|Dn|2
δ f −
n
Tb
, Dn =
1
Tb
P1S1
n
Tb
+ P2S2
n
Tb
,
where S1(f) and S2(f) are the FTs of s1(t) and s2(t).
Sv(f) =
∞
n=−∞
P1S1
n
Tb
+ P2S2
n
Tb
Tb
2
δ f −
n
Tb
.
A First Course in Digital Communications 57/58
58. Chapter 5: Optimum Receiver for Binary Data Transmission
PSD Derivation of Arbitrary Binary Modulation III
To calculate Sq(f), apply the basic definition of PSD:
Sq(f) = lim
T→∞
E{|GT (f)|2}
T
= · · · =
P1P2
Tb
|S1(f) − S2(f)|2
.
Finally,
SsT
(f) =
P1P2
Tb
|S1(f) − S2(f)|2
+
∞
n=−∞
P1S1
n
Tb
+ P2S2
n
Tb
Tb
2
δ f −
n
Tb
.
A First Course in Digital Communications 58/58