zero-forcing precoding and generalized inverses
TRANSCRIPT
Zero-Forcing Precoding and Generalized Inverses
Daniel Tai
12/09/2013
Daniel Tai Zero-Forcing Precoding and Generalized Inverses 12/09/2013 1 / 23
Reference
1 Ami Wiesel et al., “Zero-Forcing Precoding and Generalized Inverses”, IEEE Trans. SignalProcessing, Vol. 56, No. 9, 2008
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Introduction
Goal of this paper
Show the relationship between ZF precoding and generalized inverses:Nullspace of H gives freedom to adapt to non-total power constraints with betterperformance
The paper considers
2 objectives:
Max sum rate: max∑
k log(1 + SINRk)Max fairness: max min SINRk
2 power constraints:
Total TX power: tr(TTH) =∑
k tHk t ≤ P
Per-Antenna: [TTH ]n,n ≤ P/N
Methodology: Consider SINRk as a variable and solve it together/separately with T
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System Model and Problem Formulation
Outline
1 System Model and Problem Formulation
2 Solving for Total Power Constraint
3 Per-Antenna Power Constraint
4 Simulation Results
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System Model and Problem Formulation
System Model
A MISO system with K users and N TX antennas. K ≤ N
yk = hHk x + wk , ∀ky = Hx + w
where wk is zero-mean unit-variance Gaussian noisey = [y1, . . . , yK ]T , H = [h1, . . . ,hK ]H , w = [w1, . . . ,wK ]T .
Assume H is always full row-rank
Precoding:x = Ts, E{ssH} = I
SNR:
pk =|[HT]k,k |2∑
j 6=k |[HT]k,j |2 + 1, ∀k (4)
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System Model and Problem Formulation
Zero-forcing
When ZF applies, [HT]k,j = 0, ∀k 6= j
pk =|[HT]k,k |2∑
j 6=k |[HT]k,j |2 + 1= |[HT]k,k |2
yk =√pksk + wk
Equivalent constraint:
HT = diag{√p},
√p = [
√p1, . . . ,
√pk ]T
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System Model and Problem Formulation
Power Constraints
Total TX power:E{‖x‖2} = tr{TTH} = ‖T‖2F ≤ P
Per-Antenna power:
E{|xn|2} = [TTH ]n,n ≤P
N
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System Model and Problem Formulation
Objectives
Two objectives are considered in this paper
1 Fairness:f (p) = min
kpk
2 Throughput:
f (p) =∑k
log(1 + SNRk) =∑k
log(1 + pk)
where SNRk = pk due to zero-forcing.
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System Model and Problem Formulation
Generalized Inverses
To achieve ZF, HT = IK (K ≤ N and H is full row rank)
Generalized inverse H− :
H− = H† + P⊥U (9)
where H† is pseudo-inverse H† = HH(HHH)−1, P⊥ = I−H† is a projection matrix to thenull space of H. U is something to be calculated. Uopt is related to power constraint.
ZF precoder:
T = H− diag{√p} =
[H† + P⊥U
]diag{
√p} (10)
Uopt = 0,T = H† diag{√p} for total power constraint. Not necessarily zero for PAconstraint.
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Solving for Total Power Constraint
Outline
1 System Model and Problem Formulation
2 Solving for Total Power Constraint
3 Per-Antenna Power Constraint
4 Simulation Results
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Solving for Total Power Constraint
Total Power Constraint (1)
maxp≥0,T
f (p)
s.t. HT = diag{√p}
tr{TTH} ≤ P
Topt = H† diag{√p} (which is U = 0, proven in paper)
So the power becomes:
tr{TTH} =∑k
pk
[H†HH†
]k,k
=∑k
pk
[(HHH
)−1]k,k
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Solving for Total Power Constraint
Total Power Constraint (2)
Solving p:
maxp≥0
f (p)
s.t.∑k
pk
[(HHH
)−1]k,k
≤ P
It can be solved using water-filling algorithm
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Per-Antenna Power Constraint
Outline
1 System Model and Problem Formulation
2 Solving for Total Power Constraint
3 Per-Antenna Power Constraint
4 Simulation Results
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Per-Antenna Power Constraint
Problem Formulation
maxp≥0,T
f (p)
s.t. HT = diag{√p}
[TTH ]n,n ≤P
N, ∀n
Under PA power constraint, P⊥U, the null-space component can contribute to additionalperformance.
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Per-Antenna Power Constraint
Performance Bounds
L ≤ f (popt) ≤ U
where
L =
{maxp≥0,T f (p)
s.t.∑
k pk
∣∣∣H†n,k ∣∣∣2 ≤ PN , ∀n
U =
{maxp≥0,T f (p)s.t.
∑k pk
[H†HH†
]k,k≤ P
L and U are the performances when U = 0 complying to PA power constraint (smallestfeasible region) and total power constraint (largest feasible region) respectively.
For some performance metrices f (p), the popt might be hard to solve. If the bound bound istight enough (by examining U − L or U/L), the solution using bound can be used instead.
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Per-Antenna Power Constraint
Using Fairness as Objective (1)
maxp≥0,T
mink
pk
s.t. HT = diag{√p} [TTH ]n,n ≤ P/N, ∀n
The fairness criterion implies that
p = p1
=⇒ T =√p[H† + P⊥U
]So the problem becomes
maxp,T
p s.t. p
∥∥∥∥[H† + P⊥U]n,:
∥∥∥∥2 ≤ P
N,∀n
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Per-Antenna Power Constraint
Using Fairness as Objective (2)
The U can be solved by
maxU,t
t
s.t. p
∥∥∥∥[H† + P⊥U]n,:
∥∥∥∥2 ≤ t,∀n
(29)
,which is a SOCP problem. Then,
p =P
N maxn
∥∥∥[H† + P⊥U]n,:
∥∥∥2 (28)
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Per-Antenna Power Constraint
Using Sum Rate as Objective (1)
maxp≥0,T
∑k
log(1 + pk)
s.t. HT = diag{√p} [TTH ]n,n ≤ P/N, ∀n
Letting T = [t1, . . . , tK ], the problem can be written as:
maxtk
log∣∣∣I + diag
{|hHk tk |2
}∣∣∣s.t. |hHj tk | = 0∀k 6= j∑
k
[tktHk ]n,n ≤ P/N,∀n
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Per-Antenna Power Constraint
Using Sum Rate as Objective (2)
Letting Qk = tktHk � 0, the problem can be rewritten as
maxQk
log∣∣∣I + diag
{hHk Qkhk
}∣∣∣s.t. hHk Qkhj = 0,∀k 6= j∑
k
[Qk ]n,n ≤ P/N, ∀n
Qk � 0,∀krank(Qk) = 1, ∀k (33)
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Per-Antenna Power Constraint
Using Sum Rate as Objective (3)
The rank constraint can be dropped. It is proven that after Qk is solved(without the rankconstraint), tk can always be obtained using:
maxtk
<{hHk tk
}s.t. hHj tk = 0, ∀k 6= j
|[t]n|2 ≤ [Qk,opt ]n,n,∀n (35)
(33) without the rank constraint is a standard maximization (MAXDET) program subjectto linear matrix inequalities. The software provided in the paper usually gives rank-onesolution. If not, further solve (35) to get the solution.(L. Vandenberghe, S. Boyd, and S. P. Wu, “Determinant maximization with linear matrix inequality
constraints,” SIAM J. Matrix Anal. Appl., vol. 19, no. 2, pp. 499533, 1998.)
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Simulation Results
Outline
1 System Model and Problem Formulation
2 Solving for Total Power Constraint
3 Per-Antenna Power Constraint
4 Simulation Results
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Simulation Results
Simulation Environment
Elements of H are iid zero-mean, unit-variance complex Gaussian
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Simulation Results
Simulation Results
Maximize fairness (K = 3, P = 1) Maximize sum rate (N = 4)
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