Embed Size (px)
Citation preview
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 ℜn). 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 != 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-efficient
to 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
(√2b2
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
Degrees of Freedom
• A complex scalar channel has 2 degrees of freedom- BPSK only uses 1 but QPSK uses 2 ⟹ QPSK rate is doubled
• QPSK is 2.5× more energy efficient than 4-PAM- QPSK avg. Tx energy = 2b2
- 4-PAM avg. Tx energy = 5b2
17
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-efficient
to 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
(√2b2
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
symbol error probability =
3
2
Q
r2b2
�2
!4-PAM
symbol error probability ⇡ 2Q
r2b2
�2
!QPSK
Typical Error Event: Deep Fade• In Rayleigh fading channel, regardless of constellation
size and detection method (coherent/non-coherent),
• For BPSK, - If - If - Hence,
18
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
Diversity
• Reception only relies on a single “path” h
• If h is in deep fade ⟹ trouble (low reliability)
• Increase the number of “paths” ⟺ Increase diversity- If one path is in deep fade, other paths can compensate!
• Diversity over time, space, and frequency
19
y = hx+ w, h ⇠ CN (0, 1) , w ⇠ CN�0,�2
�
Deep Fade Event:�|h|2 < SNR�1
Outlook• Time diversity- Coding + Interleaving: obtain diversity over time- Repetition coding- Rotation coding: utilize degrees of freedom better
• Space (Antenna) diversity- Receive diversity: multiple Rx antennas- Transmit diversity: multiple Tx antennas- Space-time codes
• Frequency diversity- ISI mitigation- Time-domain equalization- Direct-sequence spread spectrum- OFDM
20
Time Diversity
Repetition Coding + Interleaving• A simple idea: Repetition Coding- Repeat the symbol over L time slots (note: L is NOT the # of taps)
- As long as the channels {hl | l = 1, 2, ... L} are not ALL in deep fade, there is a good probability that we can decode the symbol
• Interleaving: - Channels within coherence time are highly correlated- Diversity is obtained if we interleave the codeword across multiple
coherence time periods
22
Info. Symbol b ! ENC ! Codeword x :=
⇥b b · · · b
⇤
yl = hlxl + wl, l = 1, 2, . . . , L
Coding + Interleaving Increases Diverirsit
23
61 3.2 Time diversity
Figure 3.5 The codewords aretransmitted over consecutivesymbols (top) and interleaved(bottom). A deep fade willwipe out the entire codewordin the former case but onlyone coded symbol from eachcodeword in the latter. In thelatter case, each codeword canstill be recovered from theother three unfaded symbols.
Interleaving
x2
Codewordx3
Codewordx0
Codewordx1
Codeword
| hl |
L = 4
l
No interleaving
Consider now coherent detection of x1, i.e., the channel gains are knownto the receiver. This is the canonical vector Gaussian detection problem inSummary A.2 of Appendix A. The scalar
h∗
"h"y= "h"x1+h∗
"h"w (3.33)
is a sufficient statistic. Thus, we have an equivalent scalar detection problemwith noise !h∗/"h""w∼ !" !0#N0". The receiver structure is a matched filterand is also called a maximal ratio combiner: it weighs the received signal ineach branch in proportion to the signal strength and also aligns the phasesof the signals in the summation to maximize the output SNR. This receiverstructure is also called coherent combining.Consider BPSK modulation, with x1 = ±a. The error probability, condi-
tional on h, can be derived exactly as in (3.18):
Q!"
2"h"2SNR#
(3.34)
where as before SNR= a2/N0 is the average received signal-to-noise ratio per(complex) symbol time, and "h"2SNR is the received SNR for a given channelvector h. We average over "h"2 to find the overall error probability. UnderRayleigh fading with each gain hℓ i.i.d. !" !0#1",
"h"2 =L$
ℓ=1
$hℓ$2 (3.35)
h1 h2 h3 h4
h1 h2 h4h3
All are bad
Only one is bad
Analysis of Repetition Coding• Equivalent vector channel- Original channel:- Sufficient interleaving - Repetition coding- Vector channel:
• Analysis of error probability: under BPSK x = ±a,- After projection we get a scalar equivalent channel
- Probability of error:
24
=) {hl | l = 1, 2, . . . , L} : i.i.d. CN (0, 1)
=) xl = x, l = 1, 2, . . . , L
y = hx+w
y :=⇥y1 y2 · · · yL
⇤Th :=
⇥h1 h2 · · · hL
⇤Tw :=
⇥w1 w2 · · · wL
⇤T
yl = hlxl + wl, l = 1, 2, . . . , L, wl ⇠ CN�0,�2
�
ey = ||h||x+ ew, x = ±a, ew ⇠ CN�0,�2
�
EhQ⇣p
2||h||2SNR⌘i
⇡✓2L� 1
L
◆1
(4SNR)L
Probability of Deep Fade• Deep fade event:
- Chi-squared distribution with 2L degrees of freedom
• Probability of deep fade:- Approximation:
25
�||h||2 < SNR�1
||h||2 =
LX
l=1
|hl
|2 : sum of i.i.d. Exp(1) RV’s
=) density of ||h||2 : f(x) =
1
(L� 1)!
x
L�1e
�x
, x � 0
f(x) =1
(L� 1)!x
L�1e
�x ⇡ 1
(L� 1)!x
L�1, 0 x ⌧ 1
=) Pr�||h||2 < SNR�1 ⇡
Z SNR�1
0
1
(L� 1)!x
L�1dx =
1
L!
1
SNRL
Deep Fades Become Rarer
26
63 3.2 Time diversity
Furthermore,
L−1!
ℓ=0
"L−1+ℓ
ℓ
#="2L−1
L
#" (3.40)
Hence,
pe ≈"2L−1
L
#1
#4SNR$L(3.41)
at high SNR. In particular, the error probability decreases as the Lth power ofSNR, corresponding to a slope of −L in the error probability curve (in dB/dBscale).To understand this better, we examine the probability of the deep fade
event, as in our analysis in Section 3.1.2. The typical error event at high SNRis when the overall channel gain is small. This happens with probability
!%#h#2 < 1/SNR&" (3.42)
Figure 3.7 plots the distribution of #h#2 for different values of L; clearly thetail of the distribution near zero becomes lighter for larger L. For small x, theprobability density function of #h#2 is approximately
f#x$≈ 1#L−1$!x
L−1 (3.43)
and so
!%#h#2 < 1/SNR&≈$ 1
SNR
0
1#L−1$!x
L−1dx = 1L!
1SNRL
" (3.44)
Figure 3.7 The probabilitydensity function of !h!2 fordifferent values of L. Thelarger the L, the faster theprobability density functiondrops off around 0.
0
0.7
0.8
0.9
1.0
0 5 7.5 10
0.5
0.4
0.3
0.2
0.1
0.6
22L
2.5
χ
L = 1
L = 2
L = 3L = 4
L = 5
Pr�||h||2 < SNR�1 ⇡ 1
L!
1
SNRL
Diversity Gain: 1 → L• Comparison of probabilities of deep fade- Without coding and interleaving:- With coding and interleaving:- Diversity: increase from 1 to L
27
⇠ SNR�1
⇠ SNR�L
62 Point-to-point communication
is a sum of the squares of 2L independent real Gaussian random variables,each term !hℓ!2 being the sum of the squares of the real and imaginary partsof hℓ. It is Chi-square distributed with 2L degrees of freedom, and the densityis given by
f"x#= 1"L−1#!x
L−1e−x$ x ≥ 0% (3.36)
The average error probability can be explicitly computed to be (cf. Exer-cise 3.6)
pe =! $
0Q"√
2xSNR#f"x#dx
=$1−&
2
%L L−1&
ℓ=0
$L−1+ℓ
ℓ
%$1+&
2
%ℓ
$ (3.37)
where
& '='
SNR1+ SNR
% (3.38)
The error probability as a function of the SNR for different numbers of diver-sity branches L is plotted in Figure 3.6. Increasing L dramatically decreasesthe error probability.At high SNR, we can see the role of L analytically: consider the leading
term in the Taylor series expansion in 1/SNR to arrive at the approximations
1+&
2≈ 1$ and
1−&
2≈ 1
4SNR% (3.39)
Figure 3.6 Error probability asa function of SNR for differentnumbers of diversitybranches L.
–10
L = 1
L = 2
L = 3
L = 4
L = 5
–5 0 5 10 15 25 3530 4020
1
10–5
10–10
10–15
10–20
10–25
pe
SNR (dB)
Error Prob.
Beyond Repetition Coding• Repetition coding:- Achieves full diversity gain L- Only one symbol per L symbol times- Does not fully exploit the degrees of freedom
• How to do better?
28
Rotation Code (L=2)• 2 BPSK symbols (x1, x2 = ±a) over two time slots (L = 2)- No diversity, as each BPSL symbol experiences only one “path”
• Rotation:
- 4 codewords:
29
x1
x2
(a, a)
(a,�a)(�a,�a)
(�a, a)
x1
x2xA
xB
xC
xD
xA = R
aa
�, xB = R
�aa
�, xC = R
�a�a
�, xD = R
a�a
�
x = R
x1
x2
�, R :=
cos ✓ � sin ✓
sin ✓ cos ✓
� unitary matrix
Rotation vs Repetition Coding
• Again, like QPSK vs 4-PAM, rotation code uses the available DoF better.
• Coding Gain: saving power by 3.5 dB (!! )
30
65 3.2 Time diversity
Figure 3.8 (a) Codewords ofrotation code. (b) Codewordsof repetition code.
xC
xD
xB = (b, b)
xA = (3b, 3b)
xC = (–b, –b)
x2
x1
(–a, a)
(–a, –a) (a, –a)
(a, a)
xA
x2
xB
x1
xD = (–3b, –3b)
It is difficult to obtain an explicit expression for the exact error probability.So, we will proceed by looking at the union bound. Due to the symmetryof the code, without loss of generality we can assume xA is transmitted. Theunion bound says that
pe ≤ !!xA → xB"+!!xA → xC"+!!xA → xD"# (3.49)
where !!xA → xB" is the pairwise error probability of confusing xA withxB when xA is transmitted and when these are the only two hypotheses.Conditioned on the channel gains h1 and h2, this is just the binary detectionproblem in Summary A.2 of Appendix A, with
uA =!h1xA1h2xA2
"and uB =
!h1xB1h2xB2
"$ (3.50)
Hence,
!!xA→xB#h1#h2"=Q
#$uA−uB$2$N0/2
%
=Q
#&SNR%#h1#2#d1#2+ #h2#2#d2#2&
2
%
#
(3.51)
where SNR= a2/N0 and
d '= 1a%xA−xB&=
!2 cos(2 sin (
"(3.52)
is the normalized difference between the codewords, normalized such that thetransmit energy is 1 per symbol time. We use the upper bound Q%x&≤ e−x2/2,for x > 0, in (3.51) to get
!!xA → xB#h1#h2"≤ exp'−SNR%#h1#2#d1#2+ #h2#2#d2#2&
4
($ (3.53)
Rotation Code Repetition Code
p5
Vector Channel
31
y = Hx+w = u+w
y =
y1
y2
�, H =
h1 00 h2
�, x =
x1
x2
�, w =
w1
w2
�
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
Probability of error (given channel): Pr {xA ! xB | H,xA is sent}
= Pr
⇢<{w} >
||uA � uB ||2
�
= Q
||uA � uB ||2p
�2/2
!
Pairwise Error Probability• Hard to compute the exact error probability • Union bound: WLOG assume xA is sent.
• Pairwise error probability:
32
Pr {E} Pr {xA ! xB}+ Pr {xA ! xC}+ Pr {xA ! xD}
x1
x2xA
xB
xC
xD
= Q
r|h1|2(ad1)2 + |h2|2(ad2)2
2�2
!
ad1
ad2
= Q
rSNR (|h1|2|d1|2 + |h2|2|d2|2)
2
!
exp
�SNR
�|h1|2|d1|2 + |h2|2|d2|2
�
4
!Q(x) e
�x
2/2
=) Pr {xA ! xB}
1
1 + SNR|d1|24
! 1
1 + SNR|d2|24
!⇡ 16
|d1|2|d2|21
SNR2
Pr {xA ! xB | h1, h2} = Q
||uA � uB ||2p
�2/2
!uA =
h1xA,1
h2xA,2
�
uB =
h1xB,1
h2xB,2
�
conditional probability
Product Distance• Diversity Order = 2- Intuition:
• Squared product distance:
33
Pr {xA ! xB | h1, h2} = Q
rSNR (|h1|2|d1|2 + |h2|2|d2|2)
2
!
=) Pr {Deep Fade}
⇡ Pr
⇢|h1|2 <
1
|d1|2SNR, |h2|2 <
1
|d2|2SNR
�
⇡ 1
|d1|2|d2|21
SNR2
�AB := |d1d2|2
=) Pr {xA ! xB} / 16
�AB
1
SNR2dAB :=
xA � xB
a=
2 cos ✓2 sin ✓
�=
d1d2
�
x1
x2xA
xB
xC
xD
ad1
ad2
Rotation Code Achieves Full Diversity• Total probability of error: upper bounded by
• Diversity Order = 2• Coding Gain: maximize the minimum product distance
- max-min is achieved when
34
Pr {E} Pr {xA ! xB}+ Pr {xA ! xC}+ Pr {xA ! xD}
/ 16
✓1
�AB+
1
�AC+
1
�AD
◆SNR�2
48
min {�}SNR�2
x1
x2xA
xB
xC
xD
�AB = �AD = 4 sin
22✓, �AC = 16 cos
22✓
4 sin
22✓ = 16 cos
22✓ =) ✓ =
1
2
tan
�12
Summary: Time-‐Diversity Code• Code:• Union bound on error probability:
• Pairwise error probability:
• Diversity order:
• Squared product distance
35
x 2 {x1, x2, · · · ,xM}, xi 2 CL
Pr {E} 1
M
X
i 6=j
Pr {xi ! xj}
Pr {xi ! xj} LY
l=1
1
1 + SNR|xi,l � xj,l|2/4
mini 6=j
{Lij} , Lij =LX
l=1
I {xi,l 6= xj,l}
�ij =LY
l=1
|xi,l � xj,l|2
If full diversity L is obtained
/ 4L
M
X
i 6=j
✓1
�ij
◆SNR�L
Antenna Diversity
Multiple Antennas
37
ReceiveDiversity
TransmitDiversity Both
SIMO MISO MIMO
Typical antenna separation for space diversity ~ �c
Receive Diversity
• Same as repetition coding in time diversity• Except that there is a further power gain- Receive SNR in repetition coding =
- Receive SNR in SIMO =
• Probability of Error:
38
hx
y
a2
�2
E"Q
s
2
✓1
L||h||2
◆La2
�2
!#tends to 1 as L tends to ∞: Diversity GainL
a2
�2
L-fold Power Gain
Diversity Order = LL-fold Power Gain
y = hx+w 2 CL, x = ±a
ey = ||h||x+ ew, ew ⇠ CN�0,�2
�
#
MRC,h⇤
||h||#
Transmit Diversity
• SIMO: Rx beamforming• MISO: if Tx knows the channel, it can send
• Same as SIMO: diversity order = L; L-fold power gain
• What if Tx does not know the channel?
39
hx
y
y = h
⇤x+ w 2 C, x,h 2 CL
Tx Beamformingx = x
h
||h||
#Tx Beamform, x = xh/||h||
#h⇤ =
⇥h1 h2
⇤y = x||h||+ w
Space-‐Time Codes• Transmit the same symbol at all antennas
simultaneously won’t work: (diversity order = 1)
• Time-diversity code can be used to get full Tx diversity: - Idea: use just one antenna at one time (let x = [x1 x2]T be the time-
diversity codeword)
• Space-time codes
40
x = x1 =) y = x
LX
l=1
hl + w,
LX
l=1
hl ⇠ CN (0, L)
x1
x20
0⇥y1 y2
⇤= h⇤
x1 00 x2
�+⇥w1 w2
⇤
X, space-time codeword
Space-‐Time Codes: Simple Examples
• Convert a time-diversity code x to a space-time code X:- Spatial coding: turning one antenna on per time
• Achieves full diversity; waste available DoF
• Better design is out there!
41
⇥y1 y2
⇤= h⇤
x1 00 x2
�+⇥w1 w2
⇤
Space
Time
x =
⇥x1 x2
⇤: time-diversity codeword
l
X =
x1 0
0 x2
�: space-time codeword
yT = h⇤X+wT
Alamouti Scheme
42
Time 1 Time 2u1
u2
�u⇤2
u⇤1
X =
u1 �u⇤
2
u2 u⇤1
�u1, u2 2 C space-time codeword
Equivalent Channel:y1y⇤2
�=
h1 h2
h⇤2 �h⇤
1
� u1
u2
�+
w1
w2
�= u1
h1
h⇤2
�+ u2
h2
�h⇤1
�+
w1
w2
�
h1
h2
h1
h2
Projection onto the two column vectors respectively, we can get two clean channels for u1 and u2!
eh1eh2
eh1 ? eh2
Performance of Alamouti Scheme
• Projection onto two orthogonal directions
- Double the rate of repetition coding- Diversity order = 2! ! Full diversity- 3dB loss in Tx power compared to Tx beamforming
43
y1y⇤2
�= u1
h1
h⇤2
�+ u2
h2
�h⇤1
�+
w1
w2
�
eh1 ? eh2ey = u1eh1 + u2
eh2 + ew
eh⇤1
||eh1||ey = u1||eh1||+ ew1 = u1||h||+ ew1
eh⇤2
||eh2||ey = u2||eh2||+ ew2 = u2||h||+ ew2
Two parallel channels, each for one symbol!
Space-‐Time Code Design• In general we can extract L Tx diversity by using an L×L
space-time code in an L×1 MISO channel
- Similar to time-diversity code!
• Channel: • Pairwise error probability:
44
X 2 {XA, XB , · · · }, X 2 CL⇥L
yT = h⇤X+wT
= Q
sSNR
2
X
l=1L
|ehl|2�2l
!(XA �XB) (XA �XB)
⇤ = U⇤U⇤
eh := U⇤h ⇤ = diag��21, · · · ,�2
L
�
Pr {XA ! XB | h} = Q
||h⇤ (XA �XB) ||
2p
�2/2
!= Q
rSNR
2h⇤ (XA �XB) (XA �XB)
⇤ h
!
Pr {XA ! XB} LY
l=1
1
1 + SNR|�l|2/4/ 4L
QLl=1 |�l|2
SNR�L
det�(XA �XB) (XA �XB)
⇤
Design Criteria• Time-diversity code:- Maximize the min squared product distance
• Space-time code- Maximize the min determinant
• Full diversity ⟺ (Xi - Xj) is full rank for all i, j
45
mini,j
�ij , �ij :=LY
l=1
|xi,l � xj,l|2
mini,j
det�(Xi �Xj) (Xi �Xj)
⇤
Frequency Diversity
Diversity in Frequency-‐Selective Channel
• Resolution of multipaths provides diversity.• Full diversity is achieved by sending one symbol every L
symbol times.• But this is inefficient (like repetition coding).• Sending symbols more frequently may result in
intersymbol interference.• Challenge is how to mitigate the ISI while extracting the
inherent diversity in the frequency-selective channel.
47
y[m] =X
l
hlx[m� l] + w[m]
Approaches• Time-domain equalization (eg. GSM)
• Direct-sequence spread spectrum (eg. CDMA)
• Orthogonal frequency-division multiplexing OFDM (eg. 802.11a, Flash-OFDM, LTE)
48
ISI Equalization• Suppose a sequence of uncoded symbols are
transmitted.• Maximum likelihood sequence detection is performed
using the Viterbi algorithm.• full diversity can be achieved.• Complexity grows exponentially with number of taps L.
49
Reduction to Transmit Diversity
50
86 Point-to-point communication
Figure 3.14 The MISOscenario equivalent to thefrequency- selective channel.
h0
h0
h1
h0
h1
h2
h0
h1
h2
x [1]
y[1]
y[2]
y[3]
y[4]
x [3]
x [3]
x [4]
x [2]
x [2]
x [2]
Increasing time
x [1]
x [3]
Error probability analysisConsider the maximum likelihood detection of the sequence x based on thereceived vector y (MLSD). With MLSD, the pairwise error probability ofconfusing xA with xB, when xA is transmitted is, as in (3.85),
!!xA → xB"≤L!
ℓ=1
11+ SNR$2
ℓ/4% (3.106)
where $2ℓ are the eigenvalues of the matrix &XA−XB'&XA−XB'
∗ and SNR isthe total received SNR per received symbol (summing over all paths). This
MLSD Achieves Full Diversity
51
Space-time code matrix for input sequence
Difference matrix for two sequences first differing at
is full rank.
Direct Sequence Spread Spectrum
• Information symbol rate is much lower than chip rate (large processing gain).
• Signal-to-noise ratio per chip is low.• ISI is not significant compared to interference from other
users and match filtering (Rake) is near-optimal.
52
92 Point-to-point communication
Channel decoder
ModulatorChannel encoder
Pseudorandom pattern
generator
Pseudorandom pattern
generator
Informationsequence
OutputdataDemodulatorChannel
that hℓ"m# does not vary with m during the transmission of the sequence,Figure 3.18 Basic elements of adirect sequence spread-spectrum system.
i.e., the channel is considered time-invariant. This holds if n≪ TcW , whereTc is the coherence time of the channel. We also assume that there is negli-gible interference between consecutive symbols, so that we can consider thebinary detection problem in isolation for each symbol. This assumption isvalid if n≫ L, which is quite common in a spread-spectrum system with highprocessing gain. Otherwise, ISI between consecutive symbols becomes signif-icant, and an equalizer would be needed to mitigate the ISI. Note however weassume that simultaneously n≫ TdW and n≪ TcW , which is possible only ifTd ≪ Tc. In a typical cellular system, Td is of the order of microseconds andTc of the order of tens of milliseconds, so this assumption is quite reasonable.(Recall from Chapter 2, Table 2.2 that a channel satisfying this condition iscalled an underspread channel.)With the above assumptions, the output is just a convolution of the input
with the LTI channel plus noise
y"m#= $h∗x%"m#+w"m#& m= 1& ' ' ' &n+L (3.118)
where hℓ is the ℓth tap of the time-invariant channel filter response, withhℓ = 0 for ℓ < 0 and ℓ > L− 1. Assuming the channel h is known to thereceiver, two sufficient statistics, rA and rB, can be obtained by projectingthe received vector y (= "y"1#& ' ' ' &y"n+L##t onto the n+L dimensionalvectors vA and vB, where vA (= "$h∗xA%"1#& ' ' ' & $h∗xA%"n+L##t and vB (="$h∗xB%"1#& ' ' ' & $h∗xB%"n+L##t, i.e.,
rA (= v∗Ay& rB (= v∗By) (3.119)
The computation of rA and rB can be implemented by first matched filteringthe received signal to xA and to xB. The outputs of the matched filters arepassed through a filter matched to the channel response h and then sampledat time n+L (Figure 3.19). This is called the Rake receiver. What the Rakeactually does is taking inner products of the received signal with shiftedversions at the candidate transmitted sequences. Each output is then weightedby the channel tap gains at the appropriate delays and summed. The signalpath associated with a particular delay is sometimes called a finger of theRake receiver.
Frequency Diversity via Rake Receiver• Considered a simplified situation (uncoded).• Each information bit is spread into two pseudorandom
sequences xA and xB (xB = -xA).
• Each tap of the match filter is a finger of the Rake.
53
93 3.4 Frequency diversity
Figure 3.19 The Rake receiver.Here, h is the filter matched toh, i.e., hℓ = h∗−ℓ . Each tap of hrepresents a finger of the Rake.
XA
XB
h
w[m]
XA
h
XB
h
Decision
Estimate h
+
As discussed earlier, we are continuing with the assumption that the channelgains hℓ are known at the receiver. In practice, these gains have to be estimatedand tracked from either a pilot signal or in a decision-directed mode usingthe previously detected symbols. (The channel estimation problem will bediscussed in Section 3.5.2.) Also, due to hardware limitations, the actualnumber of fingers used in a Rake receiver may be less than the total numberof taps L in the range of the delay spread. In this case, there is also a trackingmechanism in which the Rake receiver continuously searches for the strongpaths (taps) to assign the limited number of fingers to.
Performance analysisLet us now analyze the performance of the Rake receiver. To simplify ournotation, we specialize to antipodal modulation (i.e., xA = −xB = u); theanalysis for other modulation schemes is similar. One key aspect of spread-spectrum systems is that the transmitted signal "±u# has a pseudonoise char-acteristic. The defining characteristic of a pseudonoise sequence is that itsshifted versions are nearly orthogonal to each other. More precisely, if wewrite u= $u$1%& ' ' ' &u$n%%, and
u"ℓ# (= $0& ' ' ' &0&u$1%& ' ' ' &u$n%&0& ' ' ' &0%t (3.120)
as the n+L dimensional version of u shifted by ℓ chips (hence there areℓ zeros preceding u and L− ℓ zeros following u above), the pseudonoiseproperty means that for every ℓ= 0& ' ' ' &L−1,
""u"ℓ##∗"u"ℓ′##" ≪n!
i=1
"u$i%"2& ℓ = ℓ′) (3.121)
To simplify the analysis, we assume full orthogonality: "u"ℓ##∗"u"ℓ′## = 0 ifℓ = ℓ′.We will now show that the performance of the Rake is the same as that
in the diversity model with L branches for repetition coding described inSection 3.2. We can see this by looking at a set of sufficient statistics for the
ISI vs Frequency Diversity• In narrowband systems, ISI is mitigated using a complex
receiver.
• In asynchronous CDMA uplink, ISI is there but small compared to interference from other users.
• But ISI is not intrinsic to achieve frequency diversity.
• The transmitter needs to do some work too!
54
OFDM: Basic Concept• Most wireless channels are underspread - Delay spread ≪ Coherence time.
• Can be approximated by a linear time invariant channel over a long time scale.
• Complex sinusoids are the only eigenfunctions of linear time-invariant channels.
• Should signal in the frequency domain and then transform to the time domain.
55
OFDM
56
99 3.4 Frequency diversity
d[N–1]
y0
x [N + L – 1] = d[N – 1]
Cyclic prefix
y [N + L – 1]dN–1
IDFT DFT
Remove prefix
yN–1
y[L]
y[N + L – 1]
y[1]
y[L – 1]y[L]
x [L – 1] = d[N – 1]x [L] = d[0]
x [1] = d[N – L + 1]
d0 d[0]Channel
This representation suggests a natural rotation at the input and at the outputFigure 3.22 The OFDMtransmission and receptionschemes.
to convert the channel to a set of non-interfering channels with no ISI.In particular, the actual data symbols (denoted by the length Nc vector d)in the frequency domain are rotated through the IDFT (inverse DFT) matrixU−1 to arrive at the vector d. At the receiver, the output vector of lengthNc (obtained by ignoring the first L symbols) is rotated through the DFTmatrix U to obtain the vector y. The final output vector y and the actual datavector d are related through
yn = hndn+ wn! n= 0! " " " !Nc−1# (3.145)
We have denoted w $= Uw as the DFT of the random vector w and we seethat since w is isotropic, w has the same distribution as w, i.e., a vector ofi.i.d. !" %0!N0& random variables (cf. (A.26) in Appendix A).These operations are illustrated in Figure 3.22, which affords the following
interpretation. The data symbols modulate Nc tones or sub-carriers, whichoccupy the bandwidth W and are uniformly separated by W/Nc. The datasymbols on the sub-carriers are then converted (through the IDFT) to timedomain. The procedure of introducing the cyclic prefix before transmissionallows for the removal of ISI. The receiver ignores the part of the output signalcontaining the cyclic prefix (along with the ISI terms) and converts the lengthNc symbols back to the frequency domain through a DFT. The data symbolson the sub-carriers are maintained to be orthogonal as they propagate throughthe channel and hence go through narrowband parallel sub-channels. Thisinterpretation justifies the name of OFDM for this communication scheme.Finally, we remark that DFT and IDFT can be very efficiently implemented(using Fast Fourier Transform) whenever Nc is a power of 2.
OFDM block lengthThe OFDM scheme converts communication over a multipath channel intocommunication over simpler parallel narrowband sub-channels. However, thissimplicity is achieved at a cost of underutilizing two resources, resulting ina loss of performance. First, the cyclic prefix occupies an amount of timewhich cannot be used to communicate data. This loss amounts to a fraction
OFDM
57
OFDM transforms the communication problem into the frequency domain:
a bunch of non-interfering sub-channels, one for each sub-carrier.
Can apply time-diversity techniques.
Cyclic Predix• The Nc data symbols constitute one OFDM symbol:
• Cyclic prefix prevents inter-OFDM-symbol interference.
• It also converts linear convolution into circular convolution.
58
ed0, ed1, . . . , edNc�1
Cyclic Predix Overhead• OFDM overhead • ! ! = length of cyclic prefix / OFDM symbol time• Cyclic prefix dictated by delay spread.• OFDM symbol time limited by channel coherence time.• Equivalently, the inter-carrier spacing should be much
larger than the Doppler spread.• Since most channels are underspread, the overhead can
be made small.
59
Example 1: Flash OFDM• Bandwidth = 1.25 Mz
• OFDM symbol = 128 samples = 100 μ s ! !
• Cyclic prefix = 16 samples = 11 μ s delay spread
• 11 % overhead.
60
Example 2: Long-‐term Evolution (LTE) • Bandwidth = 1.25 - 20MHz
• OFDM symbol = 128 – 2048 samples (100 μ s) ! !
• Inter-carrier spacing = 15 kHz
• Cyclic prefix = 9 – 144 samples = 5 μ s delay spread
• 5 % overhead.
61
Channel Uncertainty• In fast varying channels, tap gain measurement errors
may have an impact on diversity combining performance.
• The impact is particularly significant in channel with many taps each containing a small fraction of the total received energy. (eg. Ultra-wideband channels)
• The impact depends on the modulation scheme.
62
Summary• Fading makes wireless channels unreliable.
• Diversity increases reliability and makes the channel more consistent.
• Smart codes yields a coding gain in addition to the diversity gain.
• This viewpoint of the adversity of fading will be challenged and enriched in later parts of the course.
63
