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Multi-‐User MIMO System• So far we discussed how multiple antennas increase the
capacity and reliability in point-to-point channels
• Question: how do multiple antennas help in multi-user uplink and downlink channels?
• Spatial-Division Multiple Access (SDMA): - Multiple antennas provide spatial resolvability for distinguishing
different users’ signals- More spatial degrees of freedom for multiple users to share
2
Plot• First study uplink/downlink scenarios with single-antenna
mobiles and a multi-antenna base station
• Achieve uplink capacity with MMSE and successive interference cancellation
• Achieve downlink capacity with uplink-downlink duality and dirty paper precoding
• Finally extend the results to MIMO uplink and downlink
3
Outline• Uplink with multiple Rx antennas- MMSE-SIC
• Downlink with multiple Tx antennas- Uplink-downlink duality- Dirty paper precoding
• MIMO uplink and downlink
4
5
Uplink with Multiple Rx Antennas
Spatial Division Multiple Access
6
= h1x1 + h2x2 + w
x1 x2
h1 h2
User 1 User 2
yRx: decodes both users’ data
• Equivalent to the point-to-point MIMO using V-BLAST with identity precoding matrix
• Rx beamforming (linear filtering without SIC ) distinguishes two users spatially (and hence the name spatial division multiple access (SDMA))
- MMSE: the optimal filter that maximizes the Rx SINR- As long as the users are geographically far apart ⟹ H := [h1 h2]
is well-conditioned ⟹ 2 spatial DoF for the 2 users to share
Capacity Bounds
7
• Individual rates: each user is faced with a SIMO channel
• Sum rate: viewed as a MIMO channel with V-BLAST and identity precoding matrix: (! ! ! ! ! ! ! ! ! ! )
=) Rk log
�1 +
Pk�2 ||hk||2
�, k = 1, 2
H =⇥h1 h2
⇤, ⇤ = diag (P1, P2)
=) R1 +R2 log det
⇣Inr +
H⇤H⇤
�2
⌘
= h1x1 + h2x2 + w
x1 x2
h1 h2
User 1 User 2
yRx: decodes both users’ data
= log det (Inr + P1h1h⇤1 + P2h2h
⇤2)
Capacity Region of the UL Channel
8
R1
R2
CUplink
CUplink =
[8><
>:(R1, R2) � 0 :
8><
>:
R1 log
�1 +
P1�2 ||h1||2
�
R2 log
�1 +
P2�2 ||h2||2
�
R1 +R2 log det
�Inr +
1�2H⇤H⇤�
9>=
>;
H =⇥h1 h2
⇤, ⇤ = diag (P1, P2)
How to achieve the corner points?From the study of V-BLAST we know the answer:
MMSE-SIC!
Decoding order: User 2 → User 1
Decoding order: User 1 → User 2
K-‐user Uplink Capacity Region• The idea can be easily extended to the K-user case
• Again, can be achieved using MMSE-SIC architectures
9
HS :=⇥hl1 hl2 · · · hl|S|
⇤, l1, . . . , l|S| 2 S
⇤S := diag�Pl1 , Pl2 , . . . , Pl|S|
�, l1, . . . , l|S| 2 S
CUplink =
[
8>>>>><
>>>>>:
(R1, . . . , RK) � 0 :
8S ✓ [1 : K],Pk2S
Rk log det
�Inr +
1�2HS⇤SH
⇤S�
= log det
✓Inr +
1�2
Pk2S
Pkhkh⇤k
◆
9>>>>>=
>>>>>;
Comparison with Orthogonal Access• Orthogonal multiple access can achieve
• Unlike the single-antenna case, it’s cannot achieve the sum capacity
• In total only 1 spatial DoF
10
8<
:R1 = ↵ log
⇣1 +
P1||h1||2↵�2
⌘
R2 = (1� ↵) log⇣1 +
P2||h2||2(1�↵)�2
⌘ ↵ 2 [0, 1]
432 MIMO IV: multiuser communication
Figure 10.4 The two-useruplink with multiple receiveantennas at the base-station:performance of orthogonalmultiple access is strictlyinferior to the capacity.
A
B
R2
R1log 1+|| h1||2P1
N0
log 1 + || h2||2P2N0
other hand, the available degrees of freedom are limited by the number ofreceive antennas, nr , and so there is no further degree-of-freedom gain beyondhaving nr users performing SDMA simultaneously. This suggests a nearlyoptimal multiple access strategy where the users are divided into groups of nr
users with SDMA within each group and orthogonal multiple access betweenthe groups. Exercise 10.5 studies the performance of this scheme in greaterdetail.On the other hand, at low SNR, the channel is power-limited rather than
degrees-of-freedom-limited and SDMA provides little performance gain overorthogonal multiple access. This can be observed by an analysis as in the char-acterization of the capacity of MIMO channels at low SNR, cf. Section 8.2.2,and is elaborated in Exercise 10.6.In general, multiple receive antennas can be used to provide beamforming
gain for the users. While this power gain is not of much benefit to thenarrowband systems, both the wideband CDMA and wideband OFDM uplinkoperate at low SNR and the power gain is more beneficial.
Summary 10.1 SDMA and orthogonal multiple access
The MMSE–SIC receiver is optimal for achieving SDMA capacity.
SDMA with nr receive antennas and K users provides min!nr"K# spatialdegrees of freedom.
R2
R1
Because the rate expressions are the same as those in the single-antenna case!
Total Available Spatial DoF• With K single-antenna mobiles and nr antennas at the
base station, the total # of spatial DoF is min{K, nr} .
• When K ≤ nr , the multi-antenna base station is able to distinguish all K users with SDMA
• When K > nr , the multi-antenna base station cannot distinguish all K users
• Instead, divide the users into nr groups: in each group, users share the single DoF by orthogonalization
11
12
Downlink with Multiple Tx Antennas
Downlink with Multiple Tx Antennas
13
• Superposition of two data streams: x = u1x1+u2x2
- uk: Tx beamforming signature for user k
• Downlink SDMA:- Design goal: given a set of SINR’s, find the power allocation & the
beamforming signatures s.t. the total Tx power is minimized
• Achieve 2 spatial DoF with u1⟂h2 & u2⟂h1 .- Similar to zero forcing (decorrelator) in point-to-point and uplink
y2 = h2*x + w2y1 = h1*x + w1
h1 h2
User 1 User 2
xTx: encodes both users’ data
Downlink SDMA: Power Control Problem• Finding the optimal Tx signatures & power allocation:- SINR of each user depends on all the Tx signatures (and the
power allocation); in contrast to the uplink case- Hence maximizing all SINR is not a meaningful design goal
• Our design goal is to solve a power control problem:- Given a set of SINR’s, find the power allocation & a set of Tx
signatures such that the total amount of Tx power is minimized- It turns out that the power control problem is dual to a power
control problem in a dual uplink channel
• Through the uplink-downlink duality, the downlink problem can be solved
14
Uplink-‐Downlink Duality (1)• Primal downlink:- Superposition of data streams:- Received signals and SINR:
- Vector channel:
• Vector SINR: let- Let the matrix A have entry - Then we have
• For given {uk}, we can compute the power vector p:
15
xdl =PK
k=1 ukxk
SINRdl,k = Pk|h⇤kuk|2
�2+P
j 6=k Pj |h⇤kuj |2 , k = 1, . . . ,K
ydl = H⇤xdl +wdl
ydl,k = (h⇤kuk)xk +
Pj 6=k (h
⇤kuj)xj + wdl,k, k = 1, . . . ,K
Ak,j = |h⇤kuj |2
(IK � diag (a)A)p = �2a
ak := 1|h⇤
kuk|2SINRdl,k
1+SINRdl,k, k = 1, . . . ,K
p = �2 (IK � diag (a)A)�1 a = �2 (Da �A)�1 1
Da := diag (1/a1, . . . , 1/aK)
16
451 10.3 Downlink with multiple transmit antennas
User Kydl, K
x dl
uK
H*
User 1ydl,1
wdl
u1~x1
~xK
User K
User 1
x̂K
x̂1
yul
wul
uK
u1
H
xul,1
xul, K
subscript “dl” to emphasize that this is the downlink. The dual uplink channelhas K users (each with a single transmit antenna) and nt receive antennas:
yul!m"=Hxul!m"+wul!m"# (10.40)
where xul!m" is the vector of transmitted signals from the K users, yul!m" is thevector of received signals at the nt receive antennas, and wul!m"∼ !N $0#N0%.To demodulate the kth user in this uplink channel, we use the receive filter uk,which is the transmit filter for user k in the downlink. The two dual systemsare shown in Figure 10.16.In this uplink, the SINR for user k is given by
Figure 10.16 The originaldownlink with linear transmitstrategy and its uplink dual withlinear reception strategy.
SINRulk &= Qk " u∗khk "2
N0+!
j ̸=k Qj " u∗khj "2
# (10.41)
where Qk is the transmit power of user k. Denoting b &= $b1# ' ' ' #bK%t where
bk &=SINRulk
$1+ SINRulk % " u∗khk "2
#
we can rewrite (10.41) in matrix notation as
$IK −diag(b1# ' ' ' #bK)At%q= N0b* (10.42)
Here, q is the vector of transmit powers of the users and A is the same as in(10.38).
451 10.3 Downlink with multiple transmit antennas
User Kydl, K
x dl
uK
H*
User 1ydl,1
wdl
u1~x1
~xK
User K
User 1
x̂K
x̂1
yul
wul
uK
u1
H
xul,1
xul, K
subscript “dl” to emphasize that this is the downlink. The dual uplink channelhas K users (each with a single transmit antenna) and nt receive antennas:
yul!m"=Hxul!m"+wul!m"# (10.40)
where xul!m" is the vector of transmitted signals from the K users, yul!m" is thevector of received signals at the nt receive antennas, and wul!m"∼ !N $0#N0%.To demodulate the kth user in this uplink channel, we use the receive filter uk,which is the transmit filter for user k in the downlink. The two dual systemsare shown in Figure 10.16.
In this uplink, the SINR for user k is given by
Figure 10.16 The originaldownlink with linear transmitstrategy and its uplink dual withlinear reception strategy.
SINRulk &= Qk " u∗khk "2
N0+!
j ̸=k Qj " u∗khj "2
# (10.41)
where Qk is the transmit power of user k. Denoting b &= $b1# ' ' ' #bK%t where
bk &=SINRulk
$1+ SINRulk % " u∗khk "2
#
we can rewrite (10.41) in matrix notation as
$IK −diag(b1# ' ' ' #bK)At%q= N0b* (10.42)
Here, q is the vector of transmit powers of the users and A is the same as in(10.38).
Uplink-‐Downlink Duality (2)• Dual uplink:- Vector channel:- Filtered output SINR:
• Vector SINR: let- Let the matrix B have entry - Then we have! ! ! ! ! ! ! ! ! since B = AT
• For given {uk}, we can compute the power vector q:
17
yul = Hxul +wul
bk := 1|h⇤
kuk|2SINRul,k
1+SINRul,k, k = 1, . . . ,K
�IK � diag (b)AT
�q = �2b
Db := diag (1/b1, . . . , 1/bK)
q = �2�IK � diag (b)AT
��1b = �2
�Db �AT
��11
SINRul,k = Qk|u⇤khk|2
�2+P
j 6=k Qj |u⇤khj |2 , k = 1, . . . ,K
Bk,j = |u⇤khj |2
Uplink-‐Downlink Duality (3)• For the same {uk}, to achieve the same set of SINR
(a=b), the total Tx power of the UL and DL are the same:
• Hence, to solve the downlink power allocation and Tx signature design problem, we can solve the dual problem in the dual uplink channel
• Tx signatures will be the MMSE filters in the virtual uplink
18
KPk=1
Pk = �21T (Da �A)�1 1 = �21T�Da �AT
��11 =
KPk=1
Qk
Beyond Linear Strategies• Linear receive beamforming strategies for the uplink map
to linear transmit beamforming strategies in the downlink
• But in the uplink we can improve performance by doing successive interference cancellation at the receiver
• Is there a dual to this strategy in the downlink?
19
Transmit Precoding• In downlink Tx beamforming, signals for different users
are superimposed and interfere with each other• With a single Tx antenna, users can be ordered in terms
of signal strength- A user can decode and cancel all the signals intended for the
weaker user before decoding its own
• With multiple Tx antennas, no such ordering exists and no user may be able to decode information beamformed to other users
• However, the base station knows the information to be transmitted to every user and can precode to cancel at the transmitter
20
Symbol-‐by-‐Symbol Precoding• A generic problem: y = x + s + w- x : desired signal- s : interference known to Tx but unknown to Rx- w : noise
• Applications:- Downlink channel: s is the signal for other users- ISI channel: s is the intersymbol interference
21
Naive Pre-‐Cancellation Strategy• Want to send point u in a 4-PAM constellation
• Transmit x = u – s to pre-cancel the effect of s
• But this is very power inefficient if s is large
22
455 10.3 Downlink with multiple transmit antennas
u s
x
precoding scheme which performs better. The idea is to replicate the PAMFigure 10.17 The transmittedsignal is the difference betweenthe PAM symbol and theinterference. The larger theinterference, the more thepower that is consumed.
constellation along the entire length of the real line to get an infinite extendedconstellation (Figures 10.18 and 10.19). Each of the 2M information symbolsnow corresponds to the equivalence class of points at the same relative positionin the replicated constellations. Given the information symbol u, the precodingscheme chooses that representation p in its equivalence class which is closest tothe interference s. We then transmit the difference x = p− s. Unlike the naivescheme, thisdifferencecanbemuchsmalleranddoesnotgrowunboundedwiths.A visual representation of the precoding scheme is provided in Figure 10.20.
One way to interpret the precoding operation is to think of the equivalenceclass of any one PAM symbol u as a (uniformly spaced) quantizer qu!·" ofthe real line. In this context, we can think of the transmitted signal x to be thequantization error: the difference between the interference s and the quantizedvalue p= qu!s", with u being the information symbol to be transmitted.The received signal is
y = !qu!s"− s"+ s+w = qu!s"+w#
The receiver finds the point in the infinite replicated constellation that isclosest to s and then decodes to the equivalence class containing that point.
Let us look at the probability of error and the power consumption of thisscheme, and how they compare to the corresponding performance when thereis no interference. The probability of error is approximately6
2Q! a
2$
"% (10.47)
When there is no interference and a 2M-PAM is used, the error probability ofthe interior points is the same as (10.47) but for the two exterior points, theerror probability is Q !a/2$", smaller by a factor of 1/2. The probability oferror is larger for the exterior points in the precoding case because there is an
6 The reason why this is not exact is because there is a chance that the noise will be so largethat the closest point to y just happens to be in the same equivalence class of the informationsymbol, thus leading to a correct decision. However, the probability of this event isnegligible.
• Replicate the PAM constellation to tile the whole real line
• Represent information u by an equivalent class of constellation points instead of a single point
Tomlinson-‐Harashima Precoding (1)
23
456 MIMO IV: multiuser communication
Figure 10.18 A four-pointPAM constellation.
–3a2
– a2
a2
3a2
– 5a2
– 7a2
– 9a2
– 11a2
3a2
– a2
– 3a2
11a2
9a2
7a2
5a2
a2
additional possibility of confusion across replicas. However, the difference isFigure 10.19 The four-pointPAM constellation is replicatedalong the entire real line. Pointsmarked by the same signcorrespond to the sameinformation symbol (one of thefour points in the originalconstellation).
negligible when error probabilities are small.7
What about the power consumption of the precoding scheme? The distancebetween adjacent points in each equivalence class is 2Ma; thus, unlike in thenaive interference pre-cancellation scheme, the quantization error does notgrow unbounded with s:
!x! ≤Ma!
If we assume that s is totally random so that this quantization error is uniformbetween zero and this value, then the average transmit power is
!"x2#= a2M2
3! (10.48)
In comparison, the average transmit power of the original 2M-PAM constel-lation is a2M2/3−a2/12. Hence, the precoding scheme requires a factor of
Figure 10.20 Depiction of theprecoding operation for M = 2and PAM information symbolu =−3a/2. The crosses formthe equivalence class for thissymbol. The difference betweens and the closest cross p istransmitted.
4M2
4M2−1
more transmit power. Thus, there is still a gap from AWGN detection per-formance. However, this power penalty is negligible when the constellationsize M is large.Our description is motivated from a similar precoding scheme for the
point-to-point frequency-selective (ISI) channel, devised independently by
transmitted signal x
s
– 11a2
– 9a2
– 7a2
– 5a2
– 3a2
– a2
a2
3a2
5a2
7a2
9a2
11a2
p
7 This factor of 2 can easily be compensated for by making the symbol separation slightlylarger.
456 MIMO IV: multiuser communication
Figure 10.18 A four-pointPAM constellation.
–3a2
– a2
a2
3a2
– 5a2
– 7a2
– 9a2
– 11a2
3a2
– a2
– 3a2
11a2
9a2
7a2
5a2
a2
additional possibility of confusion across replicas. However, the difference isFigure 10.19 The four-pointPAM constellation is replicatedalong the entire real line. Pointsmarked by the same signcorrespond to the sameinformation symbol (one of thefour points in the originalconstellation).
negligible when error probabilities are small.7
What about the power consumption of the precoding scheme? The distancebetween adjacent points in each equivalence class is 2Ma; thus, unlike in thenaive interference pre-cancellation scheme, the quantization error does notgrow unbounded with s:
!x! ≤Ma!
If we assume that s is totally random so that this quantization error is uniformbetween zero and this value, then the average transmit power is
!"x2#= a2M2
3! (10.48)
In comparison, the average transmit power of the original 2M-PAM constel-lation is a2M2/3−a2/12. Hence, the precoding scheme requires a factor of
Figure 10.20 Depiction of theprecoding operation for M = 2and PAM information symbolu =−3a/2. The crosses formthe equivalence class for thissymbol. The difference betweens and the closest cross p istransmitted.
4M2
4M2−1
more transmit power. Thus, there is still a gap from AWGN detection per-formance. However, this power penalty is negligible when the constellationsize M is large.
Our description is motivated from a similar precoding scheme for thepoint-to-point frequency-selective (ISI) channel, devised independently by
transmitted signal x
s
– 11a2
– 9a2
– 7a2
– 5a2
– 3a2
– a2
a2
3a2
5a2
7a2
9a2
11a2
p
7 This factor of 2 can easily be compensated for by making the symbol separation slightlylarger.
Tomlinson-‐Harashima Precoding (2)• Given u and s, find the point in its equivalent class
closest to s and transmit the difference
24
456 MIMO IV: multiuser communication
Figure 10.18 A four-pointPAM constellation.
–3a2
– a2
a2
3a2
– 5a2
– 7a2
– 9a2
– 11a2
3a2
– a2
– 3a2
11a2
9a2
7a2
5a2
a2
additional possibility of confusion across replicas. However, the difference isFigure 10.19 The four-pointPAM constellation is replicatedalong the entire real line. Pointsmarked by the same signcorrespond to the sameinformation symbol (one of thefour points in the originalconstellation).
negligible when error probabilities are small.7
What about the power consumption of the precoding scheme? The distancebetween adjacent points in each equivalence class is 2Ma; thus, unlike in thenaive interference pre-cancellation scheme, the quantization error does notgrow unbounded with s:
!x! ≤Ma!
If we assume that s is totally random so that this quantization error is uniformbetween zero and this value, then the average transmit power is
!"x2#= a2M2
3! (10.48)
In comparison, the average transmit power of the original 2M-PAM constel-lation is a2M2/3−a2/12. Hence, the precoding scheme requires a factor of
Figure 10.20 Depiction of theprecoding operation for M = 2and PAM information symbolu =−3a/2. The crosses formthe equivalence class for thissymbol. The difference betweens and the closest cross p istransmitted.
4M2
4M2−1
more transmit power. Thus, there is still a gap from AWGN detection per-formance. However, this power penalty is negligible when the constellationsize M is large.
Our description is motivated from a similar precoding scheme for thepoint-to-point frequency-selective (ISI) channel, devised independently by
transmitted signal x
s
– 11a2
– 9a2
– 7a2
– 5a2
– 3a2
– a2
a2
3a2
5a2
7a2
9a2
11a2
p
7 This factor of 2 can easily be compensated for by making the symbol separation slightlylarger.
Writing on Dirty Paper• Can extend this idea to block precoding- Problem is to design codes which are simultaneously good
source codes (vector quantizers) as well as good channel codes
• Somewhat surprising, information theory guarantees that one can get to the capacity of the AWGN channel with the interference completely removed
• Applying this to the downlink, can perform SIC at the transmitter
• The pre-cancellation order in the downlink is the reverse order of the SIC in the dual uplink
25
26
MIMO Uplink and Downlink
27
MIMO Uplink• Channel model: ! y = H1x1 + H2x2 + w
• Now the mobiles (Tx) have multiple antennas, and hence can form their own Tx covariance matrices
• For the two-user case, capacity bounds become- Individual rate bounds:
- Sum rate bound:
• Note: in general there are no single Kx1 and Kx2 that can simultaneously maximize the three rate constraints
Rk log det
�Inr +
1�2HkKxkH
⇤� , k = 1, 2
R1 +R2 log det
✓Inr +
1�2
2Pk=1
HkKxkH⇤◆
Capacity Region (Two Users)
28
445 10.2 MIMO uplink
log det(Inr +H2Kx2H2
*)N0
log det(Inr +H1Kx1H1
*)N0
R2
R1
log det (Inr +H1Kx1H1 + H2Kx2H2
* )N0R1 + R2 =
*B
A
are nr receive antennas at the base-station. Suppose the system designer wantsFigure 10.13 The rate region ofthe two-user MIMO uplink withtransmitter strategies (powerallocations to the data streamsand the choice of rotationbefore sending over thetransmit antenna array) givenby the covariance matrices Kx1
and Kx2.
to add one more transmit antenna at each mobile. How does this translate toincreasing the number of spatial degrees of freedom?
If we look at each user in isolation and think of the uplink channel as a setof isolated SIMO point-to-point links from each user to the base-station, thenadding one extra antenna at the mobile increases by one the available spatialdegrees of freedom in each such link. However, this is misleading. Due tothe sum rate constraint, the total number of spatial degrees of freedom islimited by the minimum of K and nr . Hence, if K is larger than nr , then thenumber of spatial degrees of freedom is already limited by the number ofreceive antennas at the base-station, and increasing the number of transmitantennas at the mobiles will not increase the total number of spatial degreesof freedom further. This example points out the importance of looking at
Figure 10.14 The achievablerate region for the two-userMIMO MAC with two specificchoices of transmit filtercovariances: Kxk for user k,for k = 1! 2.
R1
R2
A2
B1
A1B2
Two pentagon regions are achieved by different choices of Kx1 and Kx2
Hence the capacity region is NOT a pentagon region anymore
MMSE-SIC can achieve all corner points in each pentagon region
CUplink = conv
[
k=1,2, K
xk⌫0
Tr(
K
xk)Pk
8>><
>>:
(R1, R2) � 0 :
Rk log det
�Inr +
1�2HkKxkH
⇤k
�, k = 1, 2
R1 +R2 log det
✓Inr +
1�2
2Pk=1
HkKxkH⇤k
◆
9>>=
>>;
MIMO Uplink with Fast Fading (1)• Full CSI: need to solve a joint optimization problem
regarding power allocation and precoding matrix design - Cannot use SVD because the two channel matrices may not have
the same factoring left matrix U- The problem can be solved by iterative water-filling efficiently- Reference: W. Yu et al, “Iterative Water-Filling for Gaussian
Vector Multiple-Access Channels,” IEEE Transactions on Information Theory, vol.50, no.1, pp.145 – 152, January 2004
29
MIMO Uplink with Fast Fading (2)• Receiver CSI: - Capacity region is the convex hull of the collection of rate pairs
(R1,R2) that satisfy the following inequalities for some covariance matrix Kx1 and Kx2 with Tr(Kx1) < P1 and Tr(Kx2) < P2 :
• For i.i.d. Rayleigh {Hk}, it is straightforward to see that uniform power allocation and identity precoding matrices maximize all bounds simultaneously
• ⟹ the capacity region is a pentagon with
30
Rk Elog det
✓Inr +
1
�2HkKxkH
⇤k
◆�, k = 1, 2
R1 +R2 E"log det
Inr +
1
�2
2X
k=1
HkKxkH⇤k
!#
Kxk = Pk
nt,kInt,k
Nature of Performance Gains• For the uplink MIMO, regardless of CSIT, the total # of
spatial DoF is
• CSIT is NOT crucial in obtaining multiplexing gain in the uplink, as long as receiver CSI is available
• Power gain is increased with CSIT• Multi-user diversity gain is limited
31
min
⇢KP
k=1nt,k , nr
�
MIMO Downlink• Compared to the case with single-antenna users, one
needs to further design the receive filters at the users• Uplink-downlink duality can be naturally extended to the
case with multiple Rx antennas:- The Rx linear filters are the Tx linear precoding filters in the dual
uplink channel
• Hence in the case without fading, the sum capacity of the MIMO downlink channel is equal to that of the dual uplink channel (with total power constraint)
32
MIMO Downlink with Fast Fading• With full CSI one can assort to the uplink-downlink
duality to solve the joint optimization problem• However, with receiver CSI:- Not possible for the base station to carry out Tx beamforming- In the symmetric downlink channel with CSIR, time-sharing is
optimal in achieving the capacity region- DoF drops significantly from min{nt ,K} to 1
• Some partial CSI at Tx could recover the spatial DoF:- Channel quality of its own link rather than the entire channel- No phase information
33
Opportunistic Beamforming• Opportunistic beamforming with multiple beams:- Form nt orthogonal beams- Whenever a user falls inside the beam, it feedback this piece of
information back to the base station- As the number of users get large, one is able to find a user in
each beam with high probability
• Hence the full DoF could be recovered• Still, need instantaneous channel information
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469 10.3 Downlink with multiple transmit antennas
the receiver CSI model is more natural. The major difference here is thatnow the transmit signatures of the users cannot be allocated as a functionof the channel variations. Furthermore, the base-station is not aware of theinterference caused by the other users’ signals for any specific user k (sincethe channel to the kth user is unknown) and Costa precoding is ruled out.
Exercise 10.21 discusses how to use the multiple antennas at the base-station without access to the channel fluctuations. One of the important con-clusions is that time sharing among the users achieves the capacity region inthe symmetric downlink channel with receiver CSI alone. This implies thatthe total spatial degrees of freedom in the downlink are restricted to one,the same as the degrees of freedom of the channel from the base-station toany individual user. On the other hand, with full CSI at the base-station wehave seen (Section 10.3.1) that the spatial degrees of freedom are equal tomin!nt"K#. Thus lack of CSI at the base-station causes a drastic reduction inthe degrees of freedom of the channel.
Partial CSI at the base-station: opportunistic beamforming with multiple beamsIn many practical systems, there is some form of partial CSI fed back to thebase-station from the users. For example, in the IS-856 standard discussed inChapter 6 each user feeds back the overall SINR of the link to the base-stationit is communicating with. Thus, while the base-station does not have exactknowledge of the channel (phase and amplitude) from the transmit antennaarray to the users, it does have partial information: the overall quality of thechannel (such as !hk$m%!2 for user k at time m).In Section 6.7.3 we studied opportunistic beamforming that induces time
fluctuations in the channel to increase the multiuser diversity. The multipletransmit antennas were used to induce time fluctuations and the partial CSI
user 2
user 1
Figure 10.27 Opportunisticbeamforming with twoorthogonal beams. The user“closest” to a beam isscheduled on that beam,resulting in two parallel datastreams to two users.
was used to schedule the users at appropriate time slots. However, the gainfrom multiuser diversity is a power gain (boost in the SINR of the userbeing scheduled) and with just a single user scheduled at any time slot,only one of the spatial degrees of freedom is being used. This basic schemecan be modified, however, allowing multiple users to be scheduled and thusincreasing the utilized spatial degrees of freedom.
The conceptual idea is to have multiple beams, each orthogonal to oneanother, at the same time (Figure 10.27). Separate pilot symbols are intro-duced on each of the beams and the users feedback the SINR of each beam.Transmissions are scheduled to as many users as there are beams at each timeslot. If there are enough users in the system, the user who is beamformed withrespect to a specific beam (and orthogonal to the other beams) is scheduled onthe specific beam. Let us consider K ≥ nt (if K<nt then we use only K of thetransmit antennas), and at each time m, let Q$m% = $q1$m%" & & & "qnt
$m%% bean nt ×nt unitary matrix, with the columns q1$m%" & & & "qnt
$m% orthonormal.The vector qi$m% represents the ith beam at time m.
Transmitter CSI Affects DoF Drastically• Behaviors of power gain and multiuser diversity gain are
similar to those in uplink• For the downlink MIMO, CSIT is critical for obtaining
spatial multiplexing gain (assuming i.i.d. Rayleigh below)
- Full CSI: Total DoF =- CSIR: Total DoF =- For the case of single-antenna users, DoF with CSIR is merely 1
• Hence it might be beneficial to spend some resource for estimating the channel and predict the current channel from the past observations
• Under i.i.d. Rayleigh, prediction is not possible. • Question: can past CSI at Tx still help?
35
minn
PKk=1 nr,k , nt
o
min
�maxk2[1:K] nr,k , nt
Two-‐user MISO Downlink
36
• Without transmitter CSI (CSIT), DoF = 1
• With instantaneous CSIT H[m] at time m, DoF = 2
• With delayed CSIT H[1:m –1] at time m, DoF = ?
• Prediction is useless ⟹ delayed CSIT is useless?
h1[m]
h2[m]Delay
Delay h1[m], h2[m] : i.i.d. Rayleigh
DoF = 4/3 with Delayed CSIT
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h1[m]
h2[m]Delay
Delaym = 1 m = 2
h1[m]
h2[m]
m = 3
u2
v2
u1
v1
m = 1 m = 2L1(u1,v1)
L2(u1,v1)
L2(u1,v1)+L3(u2,v2)0
L3(u2,v2)
L4(u2,v2)
h1,1(L2+L3)
h2,1(L2+L3)
m = 3
DoF = 4/3 with Delayed CSIT
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h1[m]
h2[m]Delay
Delay L1(u1,v1), L3(u2,v2)L2(u1,v1)+L3(u2,v2)
L2(u1,v1), L4(u2,v2)L2(u1,v1)+L3(u2,v2)
• L1(u1,v1) ∦ L2(u1,v1) & L3(u2,v2) ∦ L4(u2,v2) with prob. 1
• Both decode their desired 2 symbols over 3 time slots• DoF = (2+2)/3 = 4/3
Exploiting Side Information
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h1[m]
h2[m]Delay
DelayL1(u1,v1), L3(u2,v2)
L2(u1,v1), L4(u2,v2)
m = 3
L2(u1,v1)+L3(u2,v2)0
• L2(u1,v1) is useful for user 1 but shipped to user 2
• L3(u2,v2) is useful for user 2 but shipped to user 1
• With delayed CSIT, Tx forms u12 := L2(u1,v1)+L3(u2,v2) - User 1 can extract L2 because it has L3 as useful side info.- User 2 can extract L3 because it has L2 as useful side info.
Hierarchy of Messages• One can view u12 := L2(u1,v1)+L3(u2,v2) as a common
message for both users• Order-1 message: aimed at only 1 user- User 1: u1 , v1 ; User 2: u2 , v2
• Order-2 message: aimed at 2 users- User {1,2}: u12
• Define DoFk := the DoF for sending all order-k messages
• Then we see that
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DoF1 =4
2 + 1DoF2
=4
2 + 11
=4
3
Two Transmission Phases
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• Transmission is divided into two phases• Phase 1: time m = 1,2:- Transmit two order-1 messages using two time slots
• End of Phase 1: - From the delayed CSI, Tx is able to form one order-2 message
• Phase 2: time m = 3:- Transmit this order-2 message using one time slot
• Only phase 1 sends fresh data; the rest is to refine the reception by exploiting delayed CSIT and Rx side info.
• The idea can be extended to scenarios with more users and more Tx antennas, where higher-order messages have to be formed to achieve optimality
Optimality of 4/3• It is remarkable that delayed CSIT is useful and we can
achieve DoF = 4/3 > 1• Can we do better?• The answer is no
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Enhance user 1 by feeding user 2’s signal to user 1
2 Rx antennasStronger! Now we have a natural
ordering of the users and we find again time-sharing is DoF optimal
Hence d1 + 2d2 ≤ 2
Similarly d2 + 2d1 ≤ 2
⟹ 3DoF ≤ 4
