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Wire vs. Wireless Communication
2
y[m] =X
l
hlx[m� l] + w[m] y[m] =X
l
hl[m]x[m� l] + w[m]
Wired Channel Wireless Channel
•Deterministic channel gains•Main issue: combat noise•Key technique: coding to
exploit degrees of freedom and increase data rate
(coding gain)
•Random channel gains•Main issue: combat fading•Key technique: coding to
exploit diversity and increase reliability
(diversity gain)
• Remark: In wireless channel, there is still additive noise, and hence the techniques developed in wire communication are still useful.
Plot• Study detection in flat fading channel to learn- Communication over flat fading channel has poor performance
due to significant probability that the channel is in deep fade- How the performance scale with SNR
• Investigate various techniques to provide diversity across- Time- Frequency- Space
• Key: how to exploit additional diversity efficiently
3
Outline• Detection in Rayleigh fading channel vs. static AWGN
channel
• Code design and degrees of freedom
• Time diversity
• Antenna (space) diversity
• Frequency diversity
4
Detection in Rayleigh Fading Channel
Baseline: AWGN Channel
6
y = x+ w, w ⇠ CN�0,�2
�
BPSK: x = ±a
Transmitted constellation is real, it suffices to consider the real part:
ML rule:
Probability of error: Pr {E} = Pr
⇢<{w} >
a� (�a)
2
�= Q
ap�2/2
!= Q
⇣p2SNR
⌘
bx =
(a, if |<{y}� a| < |<{y}� (�a)|�a, otherwise
SNR :=
average received signal energy per (complex) symbol time
noise energy per (complex) symbol time
a2
�2
<{y} = x+ <{w}, <{w} ⇠ N�0,�2
/2�
Gaussian Scalar Detection
7
504 Appendix A Detection and estimation in additive Gaussian noise
Figure A.4 The ML rule is tochoose the symbol that isclosest to the received symbol.
y
If y < (uA +
uB)
/
2
choose uA
If y > (uA + uB) / 2choose uB
uA2
uB(uA+uB)
!{y | x = uA} !{y | x = uB}
Thus, the error probability only depends on the distance between the twotransmit symbols uA!uB.
A.2.2 Detection in a vector space
Now consider detecting the transmit vector u equally likely to be uA or uB
(both elements of !n). The received vector is
y= u+w! (A.33)
and w " " "0! "N0/2#I#. Analogous to (A.30), the ML decision rule is tochoose uA if
1"$N0#n/2
exp!#$y#uA$2
N0
"% 1
"$N0#n/2exp
!#$y#uB$2
N0
"! (A.34)
which simplifies to, analogous to (A.31),
$y#uA$< $y#uB$! (A.35)
the same nearest neighbor rule. By the isotropic property of the Gaussiannoise, we expect the error probability to be the same for both the transmitsymbols uA!uB. Suppose uA is transmitted, so y = uA +w. Then an erroroccurs when the event in (A.35) does not occur, i.e., $w$> $w+uA#uB$.So, the error probability is equal to
!%$w$2 > $w+uA#uB$2&= !#"uA#uB#
tw <#$uA#uB$22
$' (A.36)
• Sufficient statistic for detection: projection on to
• Since w is circular symmetric,
Gaussian Vector Detection
8
506 Appendix A Detection and estimation in additive Gaussian noise
We observe that the transmit symbol (a scalar x) is only in a specific direction:
v != "uA!uB#/"uA!uB"$ (A.40)
The components of the received vector y in the directions orthogonal to vcontain purely noise, and, due to the isotropic property of w, the noise inthese directions is also independent of the noise in the signal direction. Thismeans that the components of the received vector in these directions areirrrelevant for detection. Therefore projecting the received vector along thesignal direction v provides all the necessary information for detection:
y != vt!y! 1
2"uA+uB#
"$ (A.41)
We have thus reduced the vector detection problem to the scalar one.Figure A.6 summarizes the situation.More formally, we are viewing the received vector in a different orthonor-
mal basis: the first direction is that given by v, and the other directions areorthogonal to each other and to the first one. In other words, we form anorthogonal matrix O whose first row is v, and the other rows are orthogonalto each other and to the first one and have unit norm. Then
O!y! 1
2"uA+uB#
"=
#
$$$%
x"uA!uB"0$$$0
&
'''(+Ow$ (A.42)
Since Ow #! "0% "N0/2#I# (cf. (A.8)), this means that all but the first com-ponent of the vector O"y! 1
2 "uA + uB## are independent of the transmitsymbol x and the noise in the first component. Thus it suffices to make adecision on the transmit symbol x, using only the first component, which isprecisely (A.41).
Figure A.6 Projecting thereceived vector y onto thesignal direction v reduces thevector detection problem tothe scalar one.
y
y
uA
uB
UA
UB
y2
y1
v :=uA � uB
||uA � uB ||
ey := v⇤✓y � uA + uB
2
◆= ex+ ew
ex := v
⇤✓x� uA + uB
2
◆
=
(||uA�uB ||
2 , x = uA
� ||uA�uB ||2 , x = uB
y = x+w, w ⇠ CN�0,�2
I
�
=) ew ⇠ CN�0,�2
�
Binary Detection in Gaussian Noise
9
Binary signaling:
It suffices to consider the projection onto
Probability of error:
y = x+w, w ⇠ CN�0,�2
I
�
x = uA,uB
(uA � uB)
Pr
⇢<{w} >
||uA � uB ||2
�= Q
||uA � uB ||2p
�2/2
!
ey = x||uA � uB ||+ ew, x = ±1
2, ew ⇠ CN
�0,�2
�
Rayleigh Fading Channel
• Note: |h| is an exponential random variable with mean 1- Fair comparison with the AWGN case (same avg. signal power)
• Coherent detection:- The receiver knows h perfectly (channel estimation through pilots)- For a given realization of h, the error probability is
• Probability of error:
10
y = hx+ w, h ⇠ CN (0, 1) , w ⇠ CN�0,�2
�
Pr {E | h} = Q
a|h|p�2/2
!= Q
⇣p2|h|2SNR
⌘
Check!Hint: exchange the order in the double integral
Pr {E} = EhQ⇣p
2|h|2SNR⌘i
=1
2
1�
rSNR
1 + SNR
!
BPSK: x = ±a SNR =E⇥|h|2
⇤a2
�2=
a2
�2
Non-‐coherent Detection
• If Rx does not know the realization of h:- Scalar BPSK (! ! ) completely fails - Because the phase of h is uniform over [0, 2π]
• Orthogonal modulation:- Use two time slots m = 0,1
- Modulation:
11
y = hx+ w, h ⇠ CN (0, 1) , w ⇠ CN�0,�2
�
x = ±a
xA =
a0
�or xB =
0
a
�
51 3.1 Detection in a Rayleigh fading channel
This is a simple hypothesis testing problem, and it is straightforward toderive the maximum likelihood (ML) rule:
!"y#!<
XA
XB
0$ (3.5)
where !"y# is the log-likelihood ratio
!"y# %= ln!f"y"xA#f"y"xB#
"& (3.6)
It can be seen that, if xA is transmitted, y'0( # !" "0$a2 +N0# and y'1( #!" "0$N0# and y'0($y'1( are independent. Similarly, if xB is transmitted,y'0( # !" "0$N0# and y'1( # !" "0$a2 +N0#. Further, y'0( and y'1( areindependent. Hence the log-likelihood ratio can be computed to be
!"y#=#"y'0("2$"y'1("2
$a2
"a2+N0#N0& (3.7)
The optimal rule is simply to decide xA is transmitted if "y'0("2 > "y'1("2 anddecide xB otherwise. Note that the rule does not make use of the phases ofthe received signal, since the random unknown phases of the channel gainsh'0($h'1( render them useless for detection. Geometrically, we can interpretthe detector as projecting the received vector y onto each of the two possibletransmit vectors xA and xB and comparing the energies of the projections(Figure 3.1). Thus, this detector is also called an energy or a square-lawdetector. It is somewhat surprising that the optimal detector does not dependon how h'0( and h'1( are correlated.We can analyze the error probability of this detector. By symmetry, we
can assume that xA is transmitted. Under this hypothesis, y'0( and y'1( are
Figure 3.1 The non-coherentdetector projects the receivedvector y onto each of the twoorthogonal transmitted vectorsxA and xB and compares thelengths of the projections.
m = 1
m = 0
y
xB
|y[1]|
|y[0]|
xA
=) y :=
y[0]y[1]
�= h
x[0]x[1]
�+
w[0]w[1]
�
:= hx+w
Non-‐coherent Detection
• ML rule: - Given
- LLR:
• Energy detector:
12
Orthogonal modulation: xA =
a0
�or xB =
0
a
�y = hx+w, h ⇠ CN (0, 1) , w ⇠ CN
�0,�2
I2
�
x = xA =) y ⇠ CN✓0,
a2 + �2 0
0 �2
�◆
x = xB =) y ⇠ CN✓0,
�2 00 a2 + �2
�◆
⇤(y) := lnf(y | xA)
f(y | xB)=
a2
(a2 + �2)�2
�|y[0]|2 � |y[1]|2
�
���2 + a2
�|y[0]|2 + �2|y(1)|2
���2|y(0)|2 +
��2 + a2
�|y[0]|2
(a2 + �2)�2
bx = xA () |y[0]| > |y[1]|bx = xB () |y[0]| < |y[1]|
SNR =a2
2�2
Non-‐coherent Detection
• Probability of error:- Given
- Hence
13
Orthogonal modulation: xA =
a0
�or xB =
0
a
�y = hx+w, h ⇠ CN (0, 1) , w ⇠ CN
�0,�2
I2
�
x = xA =) y ⇠ CN✓0,
a2 + �2 0
0 �2
�◆
=) |y[0]|2 ⇠ Exp
�(a2 + �2
)
�1�, |y[1]|2 ⇠ Exp
�(�2
)
�1�
|y[0]|2 and |y[1]|2 are independent
Check!
Pr {E} = Pr
�Exp
�(�2
)
�1�> Exp
�(a2 + �2
)
�1�
=
(a2 + �2)
�1
(�2)
�1+ (a2 + �2
)
�1=
1
2 + a2/�2
=
1
2(1 + SNR)
SNR =a2
2�2
Comparison: AWGN vs. Rayleigh• AWGN: Error probability decays faster than e-SNR
• Rayleigh fading: Error probability decays as SNR-1 - Coherent detection:
- Non-coherent detection:
14
Pr {E} = Q⇣p
2SNR⌘⇡ 1p
2SNRp2⇡
e�SNR at high SNR
Q (x) := Pr {N (0, 1) > a} ⇡ 1
x
p2⇡
e
�x
2/2 when x � 1
rx
1 + x
=
✓1� 1
1 + x
◆1/2
⇡ 1� 1
2(1 + x)⇡ 1� 1
2xwhen x � 1
Pr {E} =1
2
1�
rSNR
1 + SNR
!⇡ (4SNR)�1 at high SNR
Pr {E} =1
2(1 + SNR)⇡ (2SNR)�1 at high SNR
Comparison: AWGN vs. Rayleigh
15
54 Point-to-point communication
Detection of x from y can be done in a way similar to that in the AWGNcase; the decision is now based on the sign of the real sufficient statistic
r != !"#h/"h"$#y%= "h"x+ z& (3.17)
where z$ N#0&N0/2$. If the transmitted symbol is x=±a, then, for a givenvalue of h, the error probability of detecting x is
Q
!a"h""N0/2
#
=Q$"
2"h"2SNR%
(3.18)
where SNR = a2/N0 is the average received signal-to-noise ratio per symboltime. (Recall that we normalized the channel gain such that !'"h"2( = 1.)We average over the random gain h to find the overall error probability. ForRayleigh fading when h$ "# #0&1$, direct integration yields
pe = !&Q$"
2"h"2SNR%'
= 12
(
)1%*
SNR1+ SNR
+
, ) (3.19)
(See Exercise 3.1.) Figure 3.2 compares the error probabilities of coherentBPSK and non-coherent orthogonal signaling over the Rayleigh fading chan-nel, as well as BPSK over the AWGN channel. We see that while the errorprobability for BPSK over the AWGN channel decays very fast with theSNR, the error probabilities for the Rayleigh fading channel are much worse,
Figure 3.2 Performance ofcoherent BPSK vs.non-coherent orthogonalsignaling over Rayleigh fadingchannel vs. BPSK over AWGNschannel.
0 10 20 30 40
Non-coherentorthogonalCoherent BPSK
BPSK over AWGN
pe
SNR (dB)
10–8
–10–20
1
10–2
10–4
10–6
10–10
10–12
10–14
10–16
Pr {E}
15 dB
3 dB
Coherent Detection under QPSK• BPSK only makes use of the real dimension (I channel)• Rate can be increased if an additional bit is sent on the
imaginary dimension (Q channel)• QPSK:
• (Bit) Probability of error- Simply a product of two BPSK- Analysis is the same- Simply replace SNR by SNR/2
16
57 3.1 Detection in a Rayleigh fading channel
is twice that of BPSK since both the I and Q channels are used. Equiv-alently, for a given SNR, the bit error probability of BPSK is Q!
!2SNR"
(cf. (3.13)) and that of QPSK is Q!!SNR". The error probability of QPSK
under Rayleigh fading can be similarly obtained by replacing SNR by SNR/2in the corresponding expression (3.19) for BPSK to yield
pe =12
!
"1"#
SNR2+ SNR
$
%# 12SNR
# (3.28)
at high SNR. For expositional simplicity, we will consider BPSK modulationin many of the discussions in this chapter, but the results can be directlymapped to QPSK modulation.
One important point worth noting is that it is much more energy-efficientto use both the I and Q channels rather than just one of them. For example,if we had to send the two bits carried by the QPSK symbol on the I channelalone, then we would have to transmit a 4-PAM symbol. The constellation is$"3b%"b%b%3b& and the average error probability on the AWGN channel is
32Q
b2
N0
'
# (3.29)
To achieve approximately the same error probability as QPSK, the argumentinside the Q-function should be the same as that in (3.25) and hence b shouldbe the same as a, i.e., the same minimum separation between points in the twoconstellations (Figure 3.3). But QPSK requires a transmit energy of 2a2 persymbol, while 4-PAM requires a transmit energy of 5b2 per symbol. Hence,for the same error probability, approximately 2.5 times more transmit energyis needed: a 4 dB worse performance. Exercise 3.4 shows that this loss is evenmore significant for larger constellations. The loss is due to the fact that it ismore energy efficient to pack, for a desired minimum distance separation, a
Figure 3.3 QPSK versus4-PAM: for the same minimumseparation betweenconstellation points, the 4-PAMconstellation requires highertransmit power.
Re
b
–b
b–b
QPSKIm
Re–3b –b b 3b
4-PAMIm
Pr {E}AWGN = Q⇣p
SNR⌘
Pr {E}Rayleigh =1
2
1�
rSNR
2 + SNR
!
Double of BPSK!SNR =2b2
�2
x 2 {b(1 + j), b(1� j), b(�1 + j), b(�1� j)}
⇡ (2SNR)�1 at high SNR
Typical Error Event: Deep Fade• In Rayleigh fading channel, regardless of constellation
size and detection method (coherent/non-coherent),
• For BPSK, - If - If - Hence,
17
Pr {E} ⇠ 1
SNR
Pr {E | h} = Q⇣p
2|h|2SNR⌘
probability of deep fade
|h|2 � SNR�1=) the conditional probability is very small
|h|2 < SNR�1=) the conditional probability is very large
/ Pr�|h|2 < SNR�1 = 1� eSNR
�1
⇡ SNR�1
Pr {E} = Pr�|h|2 > SNR�1 Pr
�E | |h|2 > SNR�1
+ Pr�|h|2 < SNR�1 Pr
�E | |h|2 < SNR�1
