Zero-Forcing Precoding and Generalized Inverses

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


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


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



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)



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



Power Constraints

Total TX power:E{‖x‖2} = tr{TTH} = ‖T‖2F ≤ P

Per-Antenna power:

E{|xn|2} = [TTH ]n,n ≤P

N



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.



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.


Solving for Total Power Constraint

Outline







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



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


Per-Antenna Power Constraint

Outline







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.



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.



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



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)



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




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)




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.)


Simulation Results

Outline






Simulation Results

Simulation Environment

Elements of H are iid zero-mean, unit-variance complex Gaussian


Simulation Results

Simulation Results

Maximize fairness (K = 3, P = 1) Maximize sum rate (N = 4)


Zero-Forcing Precoding and Generalized Inverses

Technology

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