1

How much feedback is required in MIMO Broadcast Channels? Alireza Bayesteh, and Amir K. Khandani Dept. of Electrical Engineering University of Waterloo Waterloo, ON, N2L 3G1 alireza, [email protected]

Abstract In this paper, a downlink communication system, in which a Base Station (BS) equipped with M antennas communicates with N users each equipped with K receive antennas is considered. We study the minimum required amount of feedback at the BS, in order to achieve the maximum sum-rate capacity. First, we define the amount of feedback as the average number of users who send information to the BS. In the asymptotic case of N → ∞, we show that with finite amount of feedback, it is not possible to achieve the maximum sum-rate. Indeed, in order to reduce the gap between the achieve sum-rate and the optimum value to zero, a minimum feedback of ln ln ln N is asymptotically necessary. Then, we consider a practical scenario, in which the amount of feedback is defined as the average number of bits which is sent to the BS. We show that to achieve the maximum sum-rate, infinite amount of feedback February 9, 2006

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is required. Moreover, the minimum amount of feedback, in order to reduce the gap to the optimum sum-rate to zero, scales as Θ(ln ln ln N ), which is achievable by the random beam-forming scheme proposed in [12]. We also study the minimum required amount of feedback when vector quantization is used. It is shown that using Random Vector Quantization (RVQ) and applying zero-forcing beamforming, the minimum number of feedback bits scales as Θ(ln ln N ). Indeed, we consider the variable SNR scenario, and derive the minimum amount of feedback in the low SNR and high SNR regimes.

I. I NTRODUCTION Multiple-Input Multiple-Output (MIMO) systems have proved their ability to achieve high bit rates in a scattering wireless network. In a point-to-point scenario, it has been shown that the capacity scales linearly with the minimum number of transmit and receive antennas, regardless of the availability of Channel State Information (CSI) at the transmitter [1] [2]. This linear increase is so-called multiplexing gain. In a MIMO Broadcast Channel (MIMO-BC), a BS equipped with multiple antennas communicates with several multiple-antenna users. Recently, there has been a lot of interest in characterizing the capacity region of this channel [3], [4], [5], [6]. In these works, it has been shown that the sum-rate capacity of MIMO-BC grows linearly with the minimum number of transmit and receive antennas, provided that both transmitter and receiver sides have perfect CSI. Indeed, in a network with a large number of users, the BS can increase the throughput by selecting the best set of users to communicate with. This results in the so-called multiuser diversity gain [7], [8]. Unlike the point-to-point scenario, in MIMO-BC, it is crucial for the transmitter to have CSI. It has been shown that MIMO-BC without CSI at the BS is degraded [9]. Moreover, for the case of single antenna users, multiplexing gain reduces to one, and multiuser diversity gain disappears February 9, 2006

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[10] [11]. Due to the weak performance of having no CSI at the BS, some authors have considered MIMO-BC with partial CSI [12] [13] [10] [14] [15] [16]. In [12], the authors have proposed a user selection strategy in a single-antenna broadcast channel, which exploits the maximum sum-rate capacity with only one bit feedback per user. This idea has been generalized for MIMO-BC in [13], using the idea of antenna selection. Reference [14] proposes a downlink transmission scheme based on random beam-forming relying on partial CSI at the transmitter. In this scheme, the BS randomly constructs M orthogonal beams and transmits data to the users with the maximum Signal to Interference plus Noise Ratio (SINR) for each beam. Therefore, only the value of maximum SINR, and the index of the beam for which the maximum SINR is achieved, are fed back to the BS for each user. This significantly reduces the amount of feedback. Reference [14] shows that when the number of users tends to infinity, the optimum sum-rate throughput can be achieved. Reference [10] considers a downlink channel where the receivers have perfect CSI, but the transmitter only has quantized information regarding the channel instantiation. This reference shows that the full multiplexing gain can be achieved with partial CSI if the quality of the CSI is increased linearly with SNR. A similar result is obtained in [15], by showing that achieving the asymptotic sum-rate capacity of the MIMO-BC requires the transmitter to know the fading levels with infinite precision. More precisely, this reference shows that with finite feedback rate, the maximum achievable multiplexing gain is 1. In fact, both [10] and [15] study the performance degradation of MIMO-BC due to the imperfect CSI, at high SNR regime. The size of the network (the number of users) is assumed to be fixed in these references. In [16], we have considered a downlink scheme based on zero-forcing beam-forming and have

February 9, 2006

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proved that when the number of users, N , tends to infinity, the maximum sum-rate capacity is achievable with the amount of feedback scaling as [ln N ]M . Two essential questions arises here; i) Is it possible to achieve the maximum sum-rate capacity with a finite feedback in a large network (N → ∞), at a fixed SNR? ii) If not, what is the minimum feedback rate (in terms of N ), in order to achieve the sum-rate capacity of the system? In this paper, we aim to answer the above questions. First, we define the amount of feedback as the average number of users who send information to the BS. In the fixed SNR regime, our results show that it is not possible to achieve the maximum sum-rate with a finite amount of feedback. Moreover, to reduce the gap between the achieved sum-rate and the optimum value to zero, the amount of feedback must be greater than ln ln ln N . In the second part, we define the amount of feedback as the number of information bits sent to the BS. Our analysis shows that the minimum amount of feedback, in order to reduce the gap to the optimum sum-rate to zero, scales as Θ(ln ln ln N ), which can be achieved using random beam-forming scheme proposed in [14]. We also show that using Random Vector Quantization (RVQ) and applying zero-forcing beam-forming, the minimum number of feedback bits scales as Θ(ln ln N ). Indeed, we consider the variable SNR scenario, and derive the minimum amount of feedback in the low SNR and high SNR regimes. The rest of the paper is organized as follows. In section II, we introduce the system model. Sections III is devoted to asymptotic analysis of the amount of feedback. Throughout this paper, the norm of the vectors and the Frobenius norm of the matrices are denoted by k.k, the Hermitian operation is denoted by (.)∗ , and the determinant and the trace operations are denoted by det(.) and Tr(.), respectively. E{.} represents the expectation, notation “ln” is used for the natural logarithm, and the rates are expressed in nats. RH(.) represents the

February 9, 2006

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right hand side of the equations. Indeed, for any functions f (N ) and g(N ), f (N ) = O(g(N )) ¯ ¯ ¯ ¯ ¯ (N ) ¯ ¯ f (N ) ¯ is equivalent to limN →∞ ¯ fg(N < ∞, f (N ) = o(g(N )) is equivalent to lim ¯ ¯ N →∞ g(N ) ¯ = ) 0, f (N ) = Ω(g(N )) is equivalent to limN →∞ limN →∞

f (N ) g(N )

f (N ) g(N )

> 0, f (N ) = ω(g(N )) is equivalent to

= ∞, and f (N ) = Θ(g(N )) is equivalent to limN →∞

f (N ) g(N )

= c, where 0 < c < ∞.

II. S YSTEM M ODEL In this work, a MIMO-BC in which a BS equipped with M antennas communicates with N users, each equipped with K antennas, is considered. The channel between each user and the BS is modeled as a zero-mean circularly symmetric Gaussian matrix (Rayleigh fading). The received vector by user k can be written as y k = H k x + nk ,

(1)

where x ∈ CM ×1 is the transmitted signal, H k ∈ CK×M is the channel matrix from the transmitter to the kth user (assumed to be known at the receiver side), and nk ∈ CK×1 ∼ CN (0, I K ) is the noise vector at this receiver. We assume that the transmitter has an average power constraint P , i.e. E {Tr(xx∗ )} ≤ P . We consider a block fading model in which each H k is constant for the duration of a frame. The frame itself is assumed to be long enough to allow communication at rates close to the capacity. III. A SYMPTOTIC A NALYSIS A. The average number of users send feedback to the BS In this section, we define the amount of feedback as the average number of users who send feedback to the BS. It is assumed that the SNR (P ) is fixed. In the following theorems, we provide the necessary and sufficient condition, in order to achieve limN →∞

RS ROpt

= 1, and limN →∞ ROpt −

RS = 0, for any user selection strategy S, respectively: February 9, 2006

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Theorem 1 Consider a MIMO-BC with N users (N → ∞), which utilizes a fixed user selection strategy S. Let NS be the number of users who send information to the BS in this strategy. Then, the necessary and sufficient condition to achieve limN →∞

RS ROpt

= 1, is having

E{NS } ∼ ω(1).

(2)

Proof- Necessary Condition- Let us denote GS as the set of users who send information to the BS, when using strategy S. Define pS (k) as the probability that user k belongs to GS . Since we consider a homogenous network, this probability is independent of k, and we denote it by pS . Therefore, NS = |GS | is a Binomial random variable with parameter pS , and we have E{NS } = N pS . Let us define

  R1 = E

and

Ã

P

max

Qn Tr(Qn )=P

  R2 = E

ln det I M +

n=1

Ã

P

max

Qn Tr(Qn )=P

N X

ln det I M

!¯  ¯  ¯ H ∗n Qn H n ¯ AS , ¯ 

 !¯ ¯  ¯ + H ∗n Qn H n ¯ AC , S ¯  n=1 N X

where AS is the event that |GS | = 0, and AC S is the complement of AS . We have RS ≤ Pr{AS }RNCSI + Pr{AC AS S }R2 £ ¤ = (1 − pS )N RNCSI + 1 − (1 − pS )N R2 , AS

(3)

where RS denotes the achievable sum-rate by the strategy S, RNCSI stands for the sum-rate of AS MIMO-BC when no CSI is available at the BS, conditioned on AS . The above equation comes from the fact that with probability (1 − pS )N no users are selected and hence, the resulting sum-rate is upper-bounded by RNCSI AS . Using (3), and having ROpt = Pr{AS }R1 + Pr{AC S }R2 , February 9, 2006

(4) DRAFT

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where ROpt is the maximum achievable sum-rate in MIMO-BC, we can write ROpt − RS ≥ (1 − pS )N (R1 − RNCSI AS ).

(5)

It can be easily shown that ½

¶¯ ¾ ¯ ln 1 + P max kH j,k k ¯¯ AS , j,k µ

2

R1 ≥ E

(6)

where H j,k denotes the jth row of H k . The right hand side of (6) can be lower-bounded as, ½ RH(6) ≥ E

¾ ¶¯ ¯ ¯ ln 1 + P max kH j,k k ¯ AS , Mt Pr {Mt |AS } , j,k µ

2

(7)

where Mt is the event that maxj,k kH j,k k2 > t, for a chosen t. Hence, RH(6) ≥ ln(1 + P t)

Pr{AS , Mt } Pr{AS }

C 1 − Pr{AC S } − Pr{Mt } Pr{AS } µ ¶ Pr{MC t } = ln(1 + P t) 1 − , Pr{AS }

≥ ln(1 + P t)

(8)

C where MC t is the complement of Mt . Pr{Mt } can be computed as

Pr{MC t }

½ ¾ 2 = Pr max kH k,j k ≤ t k,j

= (1 − e−t )N K .

(9)

Now, assume that E{NS } = N pS  ω(1), i.e., N pS ∼ O(1). Choosing t =

ln N , 2

(10)

from (9), we obtain

−K Pr{MC t } ∼ e

√ N

.

(11)

Indeed, noting Pr{AS } = (1 − pS )N and N pS ∼ O(1), we have Pr{AS } ∼ Θ(1). February 9, 2006

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Substituting (11) and (12) in (8) yields µ

P RH(6) ≥ ln 1 + ln N 2

¶³

1 − Θ(e−K

√ N

´ )

∼ ln ln N + O(1).

(13)

Indeed, similar to [9], it can be shown that ¸¯ ¾ · ¯ P ∗ ¯ EH k |AS ln det I + H k H k ¯ AS M ½ µ ¶¯ ¾ ¯ P 2 ¯ M EH k |AS ln 1 + kH k k ¯ AS M µ ¶ ¯ ª © P 2¯ M ln 1 + EH k |AS kH k k AS M µ ¶ P EH k |AS {kH k k2 } M ln 1 + M Pr{AS } µ ¶ PK M ln 1 + Pr{AS } ½

RNCSI AS

= ≤ ≤ ≤ = (12)

∼

Θ(1).

(14)

Combining (6), (13), and (14), and substituting in (5), under the assumption of (10), we get ROpt − RS

µ ¶N O(1) ≥ 1− [ln ln N + O(1)] N ∼ e−O(1) ln ln N.

⇒

RS e−O(1) ln ln N ≤ 1− . ROpt ROpt

(15)

As a result, noting that ROpt ∼ M ln ln N , we obtain RS 6= 1. N →∞ ROpt

E{NS }  ω(1) ⇒ lim

(16)

Sufficient Condition- Let us define the strategy S as selecting M users randomly among the following set: GS = {k|λmax (H k ) > t}, February 9, 2006

(17) DRAFT

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where λmax (H k ) is the maximum singular value of H k H ∗k , and t is a threshold value. After selecting the users, the BS performs zero-forcing beam-forming, where the coordinates are chosen as the eigenvectors, corresponding to the maximum singular values of the selected users. In [17], it has been shown that for a K × M matrix A, whose elements are i.i.d Gaussian, we have pS , Pr{λmax (A) > t} =

tM +K−2 e−t (1 + O(t−1 )) . Γ(M )Γ(K)

(18)

Hence, E{NS } = N pS = N

tM +K−2 e−t (1 + O(t−1 )) . Γ(M )Γ(K)

(19)

Having E{NS } ∼ ω(1), yields, t ∼ ln N + (M + K − 2) ln ln N − ω(1).

(20)

Utilizing zero-forcing beam-forming at the BS, and defining ( Ã !¯ ) ¯ P ¯ R∗ , M EH ln 1 + © ∗ −1 ª ¯ |GS | ≥ M , ¯ Tr [H H] we can write RS ≥ R∗ Pr{|GS | ≥ M }, p ¤T £ where H = g Ts1 ,max | g Ts2 ,max | · · · | g Tsm ,max in which g si ,max = λmax (H si )V ∗si ,max , 1, · · · , m

(21) i =

(m ≤ M ), and V si ,max is the eigenvector corresponding to maximum singular value

of the ith selected user (si ).

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ηS , Pr{|GS | ≥ M } can be computed as follows: ηS = 1 − Pr{|GS | < M } M −1 µ ¶ X N m = 1− pS (1 − pS )N −m m m=0 ≥ 1−

M −1 X m=0

(N pS )m −(N −m)pS e . m!

(22)

Since N pS ∼ ω(1), we have ηS ∼ 1 − o(1). Indeed, we can lower-bound R∗ as R∗ ≥ M ln P − M EH {X(H)| |GS | ≥ M } ,

(23)

¡ © ª¢ where X(H) , ln Tr [H∗ H]−1 . In [18], it has been shown that EH { X(H)| |GS | ≥ M } ≤ ln

M + (M − 1) ln(2M 2 ). t

(24)

Using the above equation and (23), and selecting t > ln N , yields, R∗ ≥ M ln(

P ln N ) − M (M − 1) ln(2M 2 ). M

(25)

Substituting R∗ and ηS in (21), and having the fact that ROpt ∼ M ln ln N [14], yields lim

N →∞

RS = 1. ROpt

(26) ¥

Theorem 2 For any user selection strategy S, the necessary condition to achieve limN →∞ ROpt − RS = 0 is having E{NS } ∼ ln ln ln N + ω(1).

(27)

E{NS }  ln ln ln N + ω(1).

(28)

Proof - Assume that

February 9, 2006

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In other words, E{NS } ∼ ln ln ln N + O(1), or E{NS } < ln ln ln N . Similar to (5), we can write ROpt − RS ≥ (1 − pS )N [R1 − RNCSI Opt ].

(29)

Following the same approach as in Theorem 1, under the assumption of (28), we can show that R1 & ln ln N + O(1), and RNCSI Opt ∼ O(ln ln ln N ). Hence, ROpt − RS ≥ (1 − pS )N [ln ln N + O(ln ln ln N )] ∼ e−E{NS }[1+O(pS )] [ln ln N + O(ln ln ln N )] ∼ e−(E{NS }−ln ln ln N ) [1 + o(1)] .

(30)

In the case of E{NS } ∼ ln ln ln N + O(1), we have RH(30) ∼ e−O(1) [1 + o(1)], and in the case of

E{NS } < ln ln ln N , we have RH(30) ∼ Υ [1 + o(1)], where Υ > 1. As a result, E{NS }  ln ln ln N + ω(1) ⇒ lim ROpt − RS 6= 0. N →∞

(31) ¥

Theorem 3 The sufficient condition to achieve limN →∞ Ropt − RS = 0 is having E{NS } = M ln ln ln N + ω(1).

(32)

Proof - Consider the random beam-forming strategy, introduced in [14]. In this strategy, the BS randomly constructs M orthogonal beams and transmits data to the users with the maximum SINR for each beam. Assuming each user’s antenna as a separate user, we define the following set: (m)

GRBF = {k|∃i,

February 9, 2006

(m)

SINRk,i > t},

m = 1, · · · , M,

(33)

DRAFT

12

(m)

where SINRk,i is the received SINR over the ith antenna of the kth user, for the mth transmitted beam. GRBF =

SM m=1

(m)

GRBF is the set of users who send feedback to the BS. The achievable sum-

rate by this scheme, denoted by RRBF , is lower-bounded as, ) (M \ RRBF ≥ M ln(1 + t)Pr Am Ã

m=1

≥ M ln(1 + t) 1 −

M X

! Pr{AC m} ,

(34)

m=1 (m)

where Am is the event that |GRBF | ≥ 1, and AC m is the complement of Am . For a randomly chosen user k, we define (m)

pk

(m)

, Pr{k ∈ GRBF } (K ) [ (m) = Pr Bk,i i=1

≤

K X

(m)

ηk,i ,

(35)

i=1 (m)

(m)

(m)

(m)

where Bk,i is the event that SINRk,i > t, and ηk,i , Pr{Bk,i }, which is independent of k, i, (m)

m, and we denote it by η. Indeed, pk

is independent of k, m, and is denoted by p. Hence,

p ≤ Kη. To evaluate the right hand side of (34), first we compute Pr{AC m } as follows: KN Pr{AC m } = (1 − η)

³ ≤

1−

p ´KN . K

(36)

Therefore, ·

p ´KN RH(34) ≥ M (1 + t) 1 − M 1 − K ³

≥ M (1 + t)[1 − M e−N p ].

February 9, 2006

¸

(37)

DRAFT

13

Under the condition of (32), and knowing the fact that p =

e−M t/P (1+t)M −1

[14], we can write

µ

¶ P ≥ M ln 1 + ln N + O(ln ln N ) × M ¡ ¢ 1 − M e−N p .

RRBF

(38)

¡ Using the above equation and having the facts that ROpt ∼ M ln 1 +

P M

¢ ln N + O(ln ln N )

[14], E{NRBF } ≤ M N p, and E{NRBF } ∼ M ln ln ln N + ω(1), we can write µ ROpt − RRBF ≤ O

ln ln N ln N

¶ + M 2 e−(

E{NRBF } −ln ln ln N ) M

[1 + o(1)]

∼ o(1).

(39)

Consequently, limN →∞ ROpt − RRBF = 0. ¥

B. Amount of bits fed back to the BS In this section, we study the minimum amount of feedback required at the BS (in terms of bits), in order to achieve the optimum sum-rate capacity. It is assumed that the SNR (P ) is fixed, and the number of bits per each user is an integer value. Theorem 4 The necessary and sufficient condition to achieve limN →∞

RS ROpt

= 1 for any users

selection strategy S, is having E{FS } ∼ ω(1),

(40)

where FS is the total number of bits fed back to the BS. Proof- Necessary condition- From Theorem 1, it is realized that the average number of users in order to achieve limN →∞

February 9, 2006

RS ROpt

= 1 must be infinite. Since the number of bits sent to the BS by

DRAFT

14

each user in GS is at least Θ(1), to achieve limN →∞

RS ROpt

= 1, the total number of bits sent to

the BS must be infinite. Sufficient Condition- Consider the scheme described in the proof of the sufficient condition, in Theorem 1. Assume that each selected user quantizes the eigenvector corresponding to their maximum singular value, using Random Vector Quantization (RVQ), and sends this quantized value to the BS. In Appendix A, it has been shown that RS − R Q S ≤ M ln(1 +

P ∗ − NMF−1B t .2 ), M

(41)

where RQ S is the achievable sum-rate of the proposed strategy, when quantization is used, NF B is the number of quantized bits for each selected user, and t∗ = ln N + (M + K − 2) ln ln N . The total number of bits sent to the BS is computed from FS = M NS NF B .

(42)

Assume that E{FS (N )} = ϕ(N ) ∼ ω(1). Let E{NS (N )} = NF B (N ) = Ψ(N ), where Ψ(N ) , q ϕ(N ) . It follows from Theorem 1 that M RS = 1. N →∞ ROpt lim

(43)

Indeed, from (41), it is realized that having NF B ∼ ω(1) yields RS = 1. N →∞ RQ S

(44)

RQ S = 1, N →∞ ROpt

(45)

lim

Combining (43) and (44), we have lim

for any function ϕ(N ) ∼ ω(1). ¥

February 9, 2006

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Theorem 5 The necessary and sufficient condition to achieve limN →∞ ROpt −RS = 0, is having E{FS } ∼ Θ(ln ln ln N ) + ω(1).

(46)

Proof- Necessary condition- The necessary condition, easily follows from Theorem 2, and the fact that each user in the set GS must send at least Θ(1) bits to the BS. Sufficient condition- Consider the random beam-forming scheme. In the proof of Theorem 3, it can be observed that given any function f (N ) , E{NS } ∼ M ln ln ln N + ω(1), one can find a threshold t, which is the solution to the following equation: e−M t/P f (N ) = . M −1 (1 + t) MN

(47)

(m)

Since the users in GRBF only need to send the index m to the BS, for each user dlog2 (M )e bits are required. Consequently, it is possible to achieve limN →∞ ROpt − RS = 0 with the average number of feedback bits scaling as M dlog2 (M )e ln ln ln N + ω(1). ¥ In many applications, the users feed back their channel information via quantizing their channel matrices, using a quantization code book which is known by both the transmitter and the receiver, and sending the quantization index to the BS. Efficient algorithms for quantization have been proposed in [19] [20] [21] [22]. Theorem 6 Consider a MIMO-BC with large number of users (N → ∞), in which the BS utilizes zero-forcing beam-forming, using the quantized eigenvectors (based on RVQ), corresponding to the maximum singular values of the selected users. Then, the necessary condition to achieve limN →∞ ROpt − RBF = 0, is having E{FZF } ∼ ln(2)M (M − 1) ln ln N + ω(1),

February 9, 2006

(48)

DRAFT

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where FZF and RZF are the total number of bits fed back to BS, and the achievable sum-rate of this scheme, respectively. Proof - Let s1 , · · · , sM be the selected users, and V s1 ,max , · · · , V sM ,max be their eigenvectors corresponding to their maximum singular values, which are denoted by Φ1 , · · · , ΦM , for simplicity of notation. Each of selected users quantizes its corresponding eigenvector to the closest vector in the set W = {w1 , · · · , w2NFB }, where NFB is the number of feedback bits sent by 2 each user. The BS uses the beam-forming vectors {Ψm }M m=1 (kΨm k = 1,

m = 1, · · · , M ),

b j = 0, for i 6= j, where Φ b j is the quantized vector of Φj (Φ b j ∈ W). The sum-rate such that Ψ∗i Φ capacity can be written as, RZF = max E{ P Pm Pm =P

where SINRm =

Pm ηm |Φ∗m Ψm |2 P , 1+ j6=m Pj ηm |Φ∗m Ψj |2

M X

ln(1 + SINRm )},

(49)

m=1

ηm = λmax (H sm ), and the expectation is taken over W,

M {Φm }M m=1 , and {ηm }m=1 . The sum-rate can be upper-bounded as

RZF ≤

max

M X

P Pm Pm =P m=1 M X

ln(1 + E{SINRm }) Ã

(

Pm ηm P ≤ max ln 1 + E P 1 + j6=m Pj ηm |Φ∗m Ψj |2 P m m=1 Pm =P Ã ( )! P η max M P ≤ M ln 1 + E P 1+ M ηmax j6=m |Φ∗m Ψj |2 Ã ( )! P η max M ≤ M ln 1 + E , P 1+ M ηmax |Φ∗m Ψj |2

)!

(50)

where ηmax , maxk λmax (H k ). The first inequality comes from the concavity of ln function. The second inequality comes from the fact that |Φ∗m Ψm |2 ≤ 1. The third inequality results from substituting all ηm ’s by ηmax . Since {|Φ∗m Ψj |2 )}M j=1 are i.i.d. random variables, by symmetry, j6=m

February 9, 2006

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Pm =

P M

maximizes the right hand side of above equation. Finally, the fourth inequality comes

from keeping one term of the summation

P j6=m

|Φ∗m Ψj |2 , and removing the rest.

Φm can be written as k b Φm = αm Φm + Φ⊥ m,

(51)

k b m. b ∗ Φm , and Φ⊥ is the projection of Φm over the space perpendicular to Φ where αm , Φ m m

b ∗ Ψj = 0, Θm,j , |Φ∗ Ψj |2 can be written as, Having the fact that Φ m m 2 Θm,j = |Ψ∗j Φ⊥ m| ∗ 2 2 = kΦ⊥ m k |Ξm Ψj | ,

(52)

Φ⊥ m . Ξm and Ψj are two independent unit vectors in the (M − 1)-dimentional kΦ⊥ mk b m . Indeed, It can be easily shown that Ξm has uniform distribution. space perpendicular to Φ

where Ξm ,

Hence, the pdf of zm,j , |Ξ∗m Ψj |2 can be computed from Appendix A of [18] as fzm,j (z) = (M − 2)(1 − z)M −3 .

(53)

2 Using the above equation, and defining µm = kΦ⊥ m k , we can evaluate the expectation in the

last line of (50) as ¯ ( ) ( ( )) ¯ P P η η max max ¯ M M E = Eηmax ,µm Ezm,j ¯ ηmax , µm P P 1+ M ηmax |Φ∗m Ψj |2 1+ M ηmax µm zm,j ¯ " ( #) M −2 1 P ηmax + ≤ Eηmax ,µm P M M − 1 (M − 1)(1 + M ηmax µm ) ½ ¾ ξ 1 M −2 ξ+ Eµ m , (54) ≤ M −1 M −1 1 + ξµm where ξ ,

P E{ηmax }. M

1 P ηmax µm zm,j 1+ M

function

η , 1+ηµ

February 9, 2006

≤ 1−

The first inequality in the above equation, comes from the fact that

P η µ M max m z . P ηmax µm m,j 1+ M

The second inequality comes from the concavity of the

with respect to η. DRAFT

18

µm can be computed as follows: k 2 µm = 1 − kαm k

b m |2 = 1 − |Φ∗m Φ = 1 − max |Φ∗m w|2 . w

(55)

w∈W

Since w has a uniform distribution, similar to (53), the pdf of r , |Φ∗m w|2 is given as fr (r) = (M − 1)(1 − r)M −2 .

(56)

Hence, the cdf of µm can be computed as Fµm (µ) = 1 − [1 − µM −1 ]L ,

(57)

where L = |W| = 2NFB . From the above equation, we can upper-bound Eµm ½ Eµm

1 1 + ξµm

¾ ≤ Pr{µm > t}

n

1 1+ξµm

1 + Pr{µm ≤ t}, 1 + ξt

o as, (58)

for any chosen t. 1

Selecting t = L− M −1 , and substituting in (58), we can write ½ Eµm

1 1 + ξµm

¾ ≤

µ

1 1

1 1− L

¶L

"

1 + 1− 1− L

1 + ξL− M −1 µ ¶L 1 1 ξL− M −1 ≤ 1− 1− . 1 L 1 + ξL− M −1

Using (50), (54), (59), and having the facts that ROpt ∼ M ln ξ∼

P M

µ

¶L #

(59) ¡P M

¢ ln N + O(ln ln N ) , and

ln N + O(ln ln N ), we have #! Ã " ¶ 1 (1 − L1 )L ξL− M −1 P ≥ M ln ln N + O(ln ln N ) − M ln ξ 1 − M M − 1 1 + ξL− M1−1 ! Ã µ ¶ 1 (1 − L1 )L L− M −1 ln N ln ln N +O ∼ −M ln 1 − . (60) M − 1 1 + L− M1−1 ln N ln N µ

ROpt − RZF

February 9, 2006

DRAFT

19

Now, assume that NFB  ln(2)(M −1) ln ln N +ω(1). Then, it is easy to see that

1 L (1− L ) M −1

1 M −1 ln N − 1 1+L M −1 ln N

L

−

∼

Θ(1). As a result, lim ROpt − RZF 6= 0.

N →∞

(61)

In order to achieve the maximum sum-rate, at least M users must sent information to BS. Assuming each of them sends NFB bits, and using the above result, it follows that the necessary condition to have limN →∞ ROpt −RZF = 0, is having E{FZF } ∼ ln(2)M (M −1) ln ln N +ω(1). ¥ Theorem 7 Consider the set up of Theorem 5. Then, the sufficient condition to achieve limN →∞ ROpt − RS = 0 is having E{FZF } ∼ ln(2)M (M − 1) ln ln N + ω(ln ln ln N ).

(62)

Proof - Consider the following algorithm: 1. Set the thresholds t and β. 2. Define S0 = {(k, j)|

λj (k) > t},

where λj (k) is the jth eigenvalue of the kth user 1 . All the users in the set S0 , send one bit to the BS. The BS selects one user in S0 at random, and inform this user (s1 ) to feed back its corresponding eigenvector in S0 , which is indexed by d1 . 3. User s1 quantizes its eigenvector based on RVQ, and feeds back the quantization index to b s1 ,d1 , to all the users in the set the BS. The BS sends the quantized vector, denoted by Φ S0 − {(s1 , d1 )}. 1

It is assumed that the eigenvalues are ordered, i.e., λ1 > λ2 > · · · > λK .

February 9, 2006

DRAFT

20

4. Define γk,j (0) = 0 for all (k, j) ∈ S0 . For m = 1 to M − 1 the following steps are repeated: o n ∗ 2 b – Define Sm = (k, j)|(k, j) ∈ Sm−1 , |Φsm ,dm Φk,j | < β and γk,j (m) = ∗

2 b |Φ sm ,dm Φk,j | +γk,j (m−1), for all (k, j) ∈ Sm , where Φk,j denotes the jth eigenvector

b k,j represents the quantized vector of Φk,j . All of the kth user’s channel matrix, and Φ users in Sm send one bit to the BS. – The BS selects one user in Sm at random, and inform this user (sm+1 ) to feed back its corresponding eigenvector in Sm , which is indexed by dm . – User sm+1 quantizes its eigenvector based on RVQ, and feeds back the quantization b sm+1 ,dm+1 , to all the index to the BS. The BS sends the quantized vector, denoted by Φ users in the set Sm − {(sm , dm )}. After selecting the coordinates, the BS constructs the beam-forming vectors {Ψm }M m=1 , such that b s ,d = 0, for i 6= m. Similar to (49), the sum-rate can be written as Ψ∗m Φ i i (M à !) ¯ ¯2 X Pm λdm (sm ) ¯Φ∗sm ,dm Ψm ¯ RZF = max E ln 1 + ¯ ∗ ¯2 P ¯ ¯ P Pm 1 + P λ (s ) Φ Ψ j d m j m m=1 sm ,dm j6=m Pm =P ½ µ ¶¾ M X ¯ ∗ ¯2 P ≥ E ln 1 + λdm (sm ) ¯Φsm ,dm Ψm ¯ − M m=1 !) ( à M X X P ¯ ∗ ¯2 E ln 1 + λdm (sm ) ¯Φsm ,dm Ψj ¯ M m=1 j6=m ¶ X ½ µ ¶¾ M ¯2 ¯ ∗ M P ¯ ¯ t + E ln Φsm ,dm Ψm + − ≥ M ln M P ηmax m=1 ! à M X ¯2 o ¯ ∗ P X n ln 1 + E λdm (sm ) ¯Φsm ,dm Ψj ¯ M m=1 j6=m µ

¶ X µ ¶ M M X P (M − 1) P t + χm − ln 1 + [E{ηmax } × E{Θm,j }] , (63) ≥ M ln M M m=1 m=1 n ³¯ ´o ¯ ¯2 ¯ 2 where χm , E ln ¯Φ∗sm ,dm Ψm ¯ + P ηMmax , ηmax , maxk λmax (H k ), and Θm,j , ¯Φ∗sm ,dm Ψj ¯ . µ

It can be easily shown that E{ηmax } ≤ ln N + (M + K − 2) ln ln N . Moreover, in Appendix A, February 9, 2006

DRAFT

21

it has been shown that µ E{Θm,j } ≤ L

− M1−1

e−1 1+ M −1

¶ .

(64)

¡ ¢ Hence, having L ∼ ω [ln N ]M −1 results in E{ηmax } × E{Θm,j } ∼ o(1). As a result, the third term in the right hand side of (63) approaches zero, as N → ∞. To evaluate the second term in (63), we define P m , as the sub-space defined by the vectors b s ,d }i6=m . We can write {Φ i i k

Φsi ,di = Φsi ,di + Φ⊥ si ,di ,

(65)

k

⊥ where Φsi ,di is the projection of Φsi ,di over P m , and Φ⊥ si ,di is the projection of Φsi ,di over P m ,

and P ⊥ m denotes the sub-space perpendicular to P m . Since Ψm is perpendicular to all vectors

February 9, 2006

DRAFT

22

b s ,d }i6=m , it belongs to P ⊥ , and we have in the set {Φ m i i ¯³ ¯2 ´∗ ¯2 ¯ k ¯ ⊥ ¯ Ψ + Φ = Φ Ψ ¯ m m¯ si ,di si ,di m ,dm

¯ ∗ ¯ Φs

¯ ¯2 ¯ = ¯Ψ∗m Φ⊥ si ,di 2 = kΦ⊥ si ,di k k

= 1 − kΦsi ,di k2 ≥ 1−

¯2 X ¯¯ b s ,d ¯¯ ¯Φ∗sm ,dm Φ i i

(a)

m−1 X

i6=m

≥ 1−

i=1

M ¯ ¯2 X ¯ ∗ b s ,d ¯¯ β− ¯Φsm ,dm Φ i i i=m+1

M ¯³ ´∗ ³ ´¯2 X ⊥ ¯ k b ¯ k ⊥ b = 1 − (m − 1)β − γi Φsi ,di + Φsi ,di ¯ ¯ αm Φsm ,dm + Φsm ,dm

(b)

i=m+1 (c)

≥ 1 − (m − 1)β −

M ¯ ³¯ ∗ ´2 X ¯b ¯ ⊥ b⊥ k k + kΦ ¯Φsm ,dm Φsi ,di ¯ + 2kΦ sm ,dm si ,di i=m+1

(d)

≥ 1 − (m − 1)β −

M ³p X

√ √ ´2 β + 2 µm + µi

i=m+1 (e)

≥ 1 − (m − 1)β − 3

M X

(β + 4µm + µi )

i=m+1

= 1 − (3M − 2m − 1)β − 12(M − m)µm − 3

M X

µi .

(66)

i=m+1

¯ ¯2 ¯ b s ,d ¯¯ < β, for In the above equation, the inequality (a) comes from the fact that ¯Φ∗sm ,dm Φ i i k b b⊥ i < m, by the algorithm. The equality (b) comes from writing Φsm ,dm as αm Φ sm ,dm + Φsm ,dm , ∗ ⊥ b s ,d as γ k Φs ,d + Φ⊥ , with the assumption of Φ b∗ b⊥ and Φ si ,di sm ,dm Φsm ,dm = 0, and Φsi ,di Φsi ,di = 0. i i i i i ¯ ¯2 ⊥ ¯ k¯ k k ∗ 2 b b b∗ = Φ Φ , k Φ k = 1 − Hence, it easily follows that αm = Φ Φ , γ ¯αm ¯ , s ,d s ,d m m si ,di sm ,dm sm ,dm i i i ¯ ¯ ¯ ¯ ¯ ¯2 ¯ k¯ ¯ k¯ ¯ k¯ 2 < 1, γ . Inequality (c) comes from the facts that and kΦ⊥ γ k = 1 − ¯αm ¯ < 1, ¯ ¯ ¯ ¯ si ,di i i ° ° ¯ ¯ ¯ ∗ ¯ ° ° ¯ °b⊥ ¯ ∗ b⊥ ¯b ¯ ° ⊥ ° ° ¯Φsm ,dm Φ⊥ si ,di ¯ < Φsi ,di , and ¯Φsi ,di Φsm ,dm ¯ < °Φsm ,dm °. Inequality (d) comes from the fact

February 9, 2006

DRAFT

23

¯ ∗ ¯2 ° ⊥ °2 ¯b ¯ °b ° that ¯Φ Φ < β, for i > m, by the algorithm, and defining µ , Φ ¯ ° m sm ,dm si ,di sm ,dm ° and µi , ° ⊥ °2 °Φs ,d ° . Finally, inequality (e) comes from the fact that ∀a, b, c, (a + b + c)2 ≤ 3(a2 + b2 + c2 ). i i using the above equation, and defining D as the event that {µm < g(N ), µm+1 < g(N ), · · · , µM < q ¡ ¢ N g(N )}, where g(N ) , M −1 ln ln , and L ∼ ω [ln N ]M −1 , we can lower-bound χm as L χm

¶¯ ¾ ½ µ ¯ C ¯ ∗ ¯2 M ¯ ¯ ¯ D Pr{DC } + = E ln Φsm ,dm Ψm + P ηmax ¯ ½ µ ¶¯ ¾ ¯ ¯ ∗ ¯2 M ¯ D Pr{D} E ln ¯Φsm ,dm Ψm ¯ + P ηmax ¯ ½ µ ¶¾ P ηmax C ≥ −Pr{D }E ln + ln (1 − (3M − 2m − 1)β − 15(M − m)g(N )) Pr{D} M µ ¶ P C ≥ −Pr{D } ln E{ηmax } + ln (1 − (3M − 2m − 1)β − 15(M − m)g(N )) Pr{D}, M (67)

where DC denotes the complement of D. From (57) and independence of µi ’s, Pr{D} can be written as h Pr{D} =

¡ ¢L iM −m+1 1 − 1 − g(N )M −1

h

−Lg(N )M −1

≥

iM −m+1

1−e · ¸M −m+1 1 = 1− ln N M −m+1 ∼ 1− . ln N

(68)

¢ ¡ Using (67) and (68), and having β ∼ o(1) and L ∼ ω [ln N ]M −1 , and noting E{ηmax } ∼ ln N + O(ln ln N ), we have µ χm ∼ O

ln ln N ln N

Ã

¶ − O(β) − o

√

M −1

ln ln N ln N

∼ o(1).

!

(69)

In other words, the second term in (63) approaches zero, as N → ∞. February 9, 2006

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24

From [18], Lemma 4, it is realized that selecting t = ln N + (M + K − 2) ln ln N − ρ(N ), and β = e−

ρ(N )−ln ln ln ln N M

, where ρ(N ) satisfies: ρ(N ) ∼ o(ln N ), ρ(N ) ∼ ln ln ln ln N + ω(1),

(70)

¡ ¢ −1 M −1 with probability one, the sets {Si }M , and i=0 are non-empty. Indeed, having L ∼ ω [ln N ] using (63), (64) and (69), and noting ROpt ∼ M ln µ RZF

∼ M ln ⇒

¡P M

¢ ln N + O(ln ln N ) , we have

¶ P ln N + O(ln ln N ) + o(1) M

lim ROpt − RZF = 0.

N →∞

(71)

The total number of bits sent to the BS can be expressed as (M −1 ) X E{FZF } = E |Sm | + M NF B m=0 (a)

∼ E {|S0 |}

M −1 X

Θ(β m ) + M NF B

m=0

= E {|S0 |} [1 + O(β)] + M NF B (b)

∼

eρ(N ) [1 + O(β)] + M NF B Γ(M )Γ(K)

∼ ω(ln ln ln N ) + M NF B ,

(72)

where inequality (a) results from [18], Lemma 4, and inequality (b) results from [18], Lemma ¢ ¡ 2. Having the facts that NF B = log2 (L), and L ∼ ω [ln N ]M −1 , it is realized that using the proposed algorithm and having E{FZF } ∼ ω(ln ln ln N ) + ln(2)M (M − 1) ln ln N , results in limN →∞ ROpt − RZF = 0, which completes the proof of Theorem 7. ¥

February 9, 2006

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25

C. Variable SNR Scenario In the previous section, the SNR (P ) has been assumed to be fixed. In this section, we study the scaling law of the minimum amount of feedback, in order to achieve the optimum sum-rate, when the SNR itself is a function of N . To this end, we consider two special regimes of low SNR and high SNR. since achieving the optimum sum-rate requires the square magnitudes of the selected coordinates to behave as ln N , the effective SNR of the selected links scales as P ln N . Hence, low SNR and high SNR regimes are defined by the regions of P ln N ∼ o(1), and P ln N ∼ ω(1), respectively. 1) Low SNR Regime: In this regime, it can be easily shown that [23] Ropt ∼ P E{ηmax },

(73)

where ηmax , maxk λmax (H k ). In other words, the optimum strategy requires the BS to perform beam-forming on the eigenvector corresponding to the maximum largest eigenvalue. Since in the low SNR regime Ropt ∼ P ln N ∼ o(1), achievability of the optimum sum-rate, for a given strategy S, is defined by limN →∞

RS Ropt

= 1.

Theorem 8 The necessary and sufficient condition, in order to achieve the optimum sum-rate throughput in the low SNR regime, is having E{NS } ∼ ω(1), and E{FS } ∼ ω(1).

February 9, 2006

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26

Proof - Following the approach of Theorem 1, and using the equations (5), (8), (9), and (14), we have ¡ ¢ ROpt − RS ≥ (1 − pS )N R1 − RNCSI , AS µ ¶ (1 − e−t )N K R1 ≥ ln(1 + P t) 1 − (1 − pS )N ¶ µ (1 − e−t )N K ∼ Pt 1 − (1 − pS )N

(74)

(75)

and µ

RNCSI AS

PK ≤ M ln 1 + (1 − pS )N

¶ .

Under the assumption of E{NS } = N pS ∼ O(1), and choosing t =

(76) ln N , 2

we have R1 ∼

P ln N , 2

and RNCSI ∼ Θ(P ). Noting that ROpt ∼ P ln N , we can write AS R1 − RNCSI RS AS N ≤ 1 − (1 − pS ) ROpt ROpt ∼ Θ(1).

(77)

As a result, RS < 1. N →∞ ROpt

E{NS }  ω(1) ⇒ lim

(78)

The necessity of E{FS } ∼ ω(1) directly follows from the above equation. Sufficient condition - In this part, we prove that for any given function f (N ) ∼ ω(1), one can achieve the optimum sum-rate such that E{NS } ≤ f (N ), and E{FS } ≤ f (N ). Assume that the users in the following set: GS , {k|λmax (H k ) > t}

(79)

where t = ln N + (M + K − 2) ln ln N − 12 ln f (N ), quantize the eigenvector corresponding to √

their maximum singular value, using a quantization code book W, which consists of L = 2 February 9, 2006

f (N ) 2

DRAFT

27

randomly selected unit vectors in the M -dimensional space. The BS selects one of the users in GS at random and serves this user, performing beam-forming on the direction of its singular value, using the corresponding quantized eigenvector. The achievable sum-rate of this scheme can be lower-bounded as n o£ ¤ b 2 1 − (1 − pS )N RS & P t E |Φ∗ Φ| n o£ ¤ b 2 1 − e−N pS , ≥ P t E |Φ∗ Φ|

(80)

where pS , Pr{k ∈ GS }, for a randomly chosen k. Using (18), we can write ¤ tM +K−2 e−t £ 1 + O(t−1 ) Γ(M )Γ(K) p · µ ¶¸ f (N ) 1 ∼ 1+O N ln N √ 1 ∼ e− f (N )[1+O( ln N )] .

pS ∼

⇒ e−N pS

(81)

Indeed, using (55), (57), and Appendix B, we have n

∗b 2

E |Φ Φ|

µ

o ≥ 1−L = 1−2

− M1−1

√ f (N ) − 2(M −1)

¶ e−1 1+ . M −1 ¶ µ e−1 1+ M −1

(82)

Combining (73), (81), and (82), and the fact that E{ηmax } < ln N + (M + K − 2) ln ln N , yields, · √ ´¸ h i √ f (N ) ³ 1 − 2(M −1) e−1 Pt 1 − 2 1 + M −1 1 − e− f (N )[1+O( ln N )] RS = lim lim N →∞ N →∞ ROpt P E{ηmax } = 1.

(83)

Moreover, we have E{NS } = N pS · µ ¶¸ p 1 ∼ f (N ) 1 + O ln N < f (N ), February 9, 2006

(84) DRAFT

28

and E{FS } = E{NS } log2 (L) · µ ¶¸ p p f (N ) 1 ∼ f (N ) 1 + O ln N 2 < f (N ),

(85)

which completes the proof of Theorem 8. ¥ 2) High SNR Regime: The sum-rate capacity in this regime can be written as, µ ¶ P ROpt ∼ M ln ln N + O(P ln ln N ) . M

(86)

With a similar approach as in the proof of Theorems 1 and 2, we can show that the necessary and sufficient condition to achieve limN →∞

RS ROpt

= 1 is having E{NS } = ω(1), and

E{FS } = ω(1). Moreover, the necessary condition to get limN →∞ ROpt − RS = 0, is having (1 − pS )N ln(P ln N ) ∼ o(1), which incurs E{NS } ∼ ln ln(P ln N ) + ω(1),

(87)

√ for the values of P satisfying ln ln(P ln N ) ∼ O( N ). ¡ Remark 1- Since ln ln(P ln N ) can be written as ln ln ln N + ln 1 +

ln P ln ln N

¢

, it turns out that

for the values of P such that ln P ∼ O(ln ln N ), the condition E{NS } ∼ ln ln(P ln N ) + ω(1) is equivalent to E{NS } ∼ ln ln ln N +ω(1). Also, For the values of P satisfying ln P ∼ ω(ln ln N ), the condition (87) reduces to E{NS } ∼ ln ln P + ω(1). In the previous section we observed that the Random Beam-forming scheme introduced in [14] is optimal in the sense of achieving the optimum sum-rate with the minimum required amount of feedback, by a factor. The question here, is whether or not this optimality holds for all range of SNRs. The following theorem answers this question: February 9, 2006

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29

Theorem 9 The necessary and sufficient condition to achieve limN →∞ ROpt − RRBF = 0 is having ln P  Ω(ln N ).

(88)

Proof - Necessary condition - The sum-rate throughput of Random Beam-forming scheme can be upper-bounded as ( RRBF = E

M X

³

ln 1 + SINR(m) max

´

)

m=1

³ ´ ≤ M ln 1 + E{SINR(m) } , max

(89)

where SINR(m) max denotes the maximum received SINR over the mth transmitted beam. Defining Xmax , SINR(m) max , for all values of t, we can write Z

∞

E{Xmax } =

xfXmax (x)dx Z

0

Z

t

∞

xfXmax (x)dx Z ∞ ≤ tFXmax (t) + t [1 − FXmax (t)] + [1 − FXmax (x)] dx t Z ∞ = t+ [1 − FXmax (x)] dx. xfXmax (x)dx +

=

t

0

(90)

t

Having the fact that FX (x) = 1 −

Mx

e− P (1+x)M −1

[14], we can write



Z

∞

E{Xmax } ≤ t + t

1 −

à 1−

(1

e− MPx + x)M −1

!N K   dx.

ln P Now, assume that ln P ∼ Ω(ln N ), i.e., limN →∞ ln = c, where c > 0. We define N     P [ln N − 1 ln P ], c < 1; M 2 t,   P  2M ln N , c ≥ 1.

February 9, 2006

(91)

(92)

DRAFT

30

Substituting t in (91) yields, E{Xmax } . ≤ ≤ (a)

∼

≤ (b)

∼

≤

Ã

(

" à !#)! Mx Mx N Ke− P e− P t+ 1 − exp − 1+O dx (1 + x)M −1 (1 + x)M −1 t " à !# Z ∞ Mx Mx e− P N Ke− P 1+O dx t+ (1 + x)M −1 (1 + x)M −1 t " à !# Z ∞ Mx Mt N Ke− P e− P t+ dx 1 + O (1 + x)M −1 (1 + t)M −1 t ¶¸ · µ Z ∞ Mx N Ke− P 1 t+ dx 1 + O √ (1 + x)M −1 N t µ ¶2−M · µ ¶¸ t P 1 −M P N Ke t+ 1+O √ M N · µ ¶¸ Mt 1 t + N Ke− P 1 + O √ N  h ³ ´i    P [ln N − 1 ln P ] 1 + O √1 , c < 1; M 2 P (93) h ³ √ ´i   P N  2M ln N 1 + O P , c ≥ 1, Z

∞

where (a) comes from the fact that t ≥ the fact that M ≥ 2, and as a result, (89) and (91), we can write ROpt − RRBF

Noting that in the case of c ≥ 1,

P 2M

ln N , and hence

¡ P ¢2−M M

Mt

e− P (1+t)M −1

√1 , N

and (b) comes from

≤ 1. Noting that ROpt ∼ M ln

 ¡    − ln 1 −

¡ P ln N ¢ M

, and using

³ ´ + O √1P , c < 1; ≥ h ³ √ ´i    ln(2) − ln 1 + O PN , c ≥ 1. √ N P

ln P 2 ln N

¢

(94)

∼ o(1), it follows from the above equation that

ln P ∼ Ω(ln N ) ⇒ lim ROpt − RRBF 6= 0. N →∞

February 9, 2006

≤

(95)

DRAFT

31

Sufficient condition - Assume that ln P  Ω(ln N ). RRBF can be lower-bounded as n o (M ) RRBF ≥ M ln(1 + t)Pr SINR(1) > t, · · · , SINR > t max max " # M n o X ) ≥ M ln(1 + t) 1 − Pr SINR(M max ≤ t £

m=1

= M ln(1 + t) 1 − M (1 − η)N K

¤

£ ¤ ≥ M ln(1 + t) 1 − M e−N Kη , (M )

where η , Pr{SINRi,k ≤ t} = ln N , N

is easy to show that η ≥

Mt

e− P . (1+t)M −1

Setting t =

P M

£

(96)

¤ P ln N − (M − 1) ln M − M ln ln N , it

and hence

· ¸¶ µ ¶ P M P ln N − (M − 1) ln − M ln ln N 1− K . ≥ M ln 1 + M M N µ

RRBF

(97)

Since ln P  Ω(ln N ), it follows from the above equation that limN →∞ ROpt − RRBF = 0. ¥ Theorem 9 implies that the Random Beam-forming scheme is not capable of achieving the optimum sum-rate when ln P ∼ Ω(N ), regardless of the amount of feedback sent to the BS. In other words, the Random Beam-forming scheme is not efficient in the high SNR regime. In fact, it is not hard to convince ourselves that the multiplexing gain of this scheme is zero. In the region of ln P  Ω(ln N ), following the approach of Theorem 3, it can be shown that with the number of feedback bits scaling as M dlog2 M e ln ln(P ln N ) + ω(1), the maximum sum-rate capacity can be achieved.

IV. C ONCLUSION In this paper, a downlink communication system, in which a Base Station (BS) equipped with M antennas communicates with N users each equipped with K receive antennas has been considered. We have studied the minimum required amount of feedback at the BS, in order February 9, 2006

DRAFT

32

to achieve the maximum sum-rate capacity. First, in the fixed SNR scenario, we have defined the amount of feedback as the average number of users who send information to the BS. In the asymptotic case of N → ∞, we have shown that with finite amount of feedback, it is not possible to achieve the maximum sum-rate. Moreover, in order to reduce the gap between the achieve sum-rate and the optimum value to zero, a minimum feedback of ln ln ln N is asymptotically necessary. Then, we have considered a practical scenario, in which the amount of feedback is defined as the average number of bits which is sent to the BS. It is shown that to achieve the maximum sum-rate, infinite amount of feedback bits is required. Moreover, the minimum amount of feedback, in order to reduce the gap to the optimum sum-rate to zero, scales as Θ(ln ln ln N ), which is achievable by the random beam-forming scheme proposed in [12]. We have also studied the minimum required amount of feedback when vector quantization is used. It is shown that using Random Vector Quantization (RVQ) and applying zero-forcing beam-forming, the minimum number of feedback bits scales as Θ(ln ln N ). Finally, we have considered the variable SNR scenario, and based on the parameter P ln N , have defined the low SNR and high SNR regimes. It has been shown that in the low SNR regime, the minimum amount of feedback scales as ω(1), while in the high SNR regime scales as Θ(ln ln(P ln N )), which is achievable by the Random Beam-Forming scheme, only in the region of ln P ∼ o(ln N ).

A PPENDIX A Let us define {Φm }M m=1 , as the eigenvectors, corresponding to the maximum eigenvalues of b m }M , as the quantized vectors obtained by applying RVQ on {Φm }M , the selected users, {Φ m=1 m=1 and {Ψm }M m=1 , as the beam-forming vectors constructed by the BS. Since zero-forcing beamb j = 0, j 6= m. Similar to forming is used, the beam-forming vectors are chosen such that Ψ∗m Φ

February 9, 2006

DRAFT

33

[10], we can write !) P X 2 E ln 1 + λm |Φ∗m Ψj | M m=1 j6=m µ ¶ © ∗ P 2ª M ln 1 + (M − 1)E |Φm Ψj | E{ηmax } M µ ¶ © ∗ P 2ª ∗ M ln 1 + (M − 1)t E |Φm Ψj | M µ ¶ P ∗ − NMF−1B M ln 1 + t 2 , M (

RS −

RQ S

≤ (a)

≤

(b)

≤

(c)

≤

Ã

M X

(98)

where λm , λmax (H sm ), ηmax , maxi λmax (H si ), and t∗ , ln N + (M + K − 2) ln ln N . Inequality (a) comes from the concavity of ln function, and replacing λm by ηmax . Inequality (b) comes from the fact that E{ηmax < ln N + (M + K − 2) ln ln N }. Finally, inequality (c) © ª rsults from the fact that E |Φ∗m Ψj |2 ≤

−

2

NF B M −1

M −1

, which is proved in [10].

V. A PPENDIX B It follows from (52) and (57) that E{Θm,j } ≤ E{µm } Z 1 = µfµm (µ)dµ Z

0

Z

0

1

=

(1 − Fµm (µ))dµ 1

= Z

(1 − µM −1 )L dµ

0 1

≤

M −1

e−Lµ

0 1

L− M −1 = M −1 − M1−1

L ≤ M −1 1

= L− M −1

February 9, 2006

Z

L

dµ 2−M

u M −1 e−u du

0

·Z

Z

1

u

du + µ ¶ e−1 1+ . M −1 0

∞

2−M M −1

¸ −u

e

du

1

(99)

DRAFT

34

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“Opportunistic beamforming with limited feedback,”

Presented in Asilomar

Conference on Signals, Systems and Computers, Nov. 2005. [14] M. Sharif and B. Hassibi, “On the capacity of mimo broadcast channel with partial side infonnation,” IEEE Trans. on Inform. Theory, vol. 51, pp. 506–522, Feb. 2005.

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35

[15] A. Lapidoth, S. Shamai, and M. Wigger, “On the capacity of a mimo fading broadcast channel with imperfect transmitter side-information,” in Proceedings of Allerton Conf. on Commun., Control, and Comput., Sept. 2005. [16] Alireza bayesteh and Amir K. Khandani, “On the user selection in mimo broadcast channels,” in Proc. of International Symposium on Information Theory, Adelaide, Australia, Sept. 2005, pp. 2325 – 2329. [17] Mohammad A. Maddah-Ali, Mehdi Ansari, and Amir K. Khandani, “An efficient algorithm for user selection and signalling over mimo multiuser systems,” Technical Report UW-ECE 2005-09, 2004. [18] Alireza Bayesteh and Amir K. Khandani, “On the user selection in mimo broadcast channels,” To be submitted to IEEE Trans. Inform. Theory. [19] D. J. Love, R. W. Heath Jr., W. Santipach, and M. L. Honig , “What is the value of limited feedback for mimo channels?,” IEEE Commun. Mag., vol. 42, no. 10, pp. 54–59, Oct. 2004. [20] Mehdi Ansari, Amir K. Khandani and Farshad Lahouti, “Channel feedback quantization for high data rate multiple antenna systems,” To be appeared in IEEE Trans. Wireless Commun., 2006. [21] J.C. Roh and B. D. Rao, “Multiple antenna channels with partial channel state information at the transmitter,” IEEE Trans. Wireless Commun., vol. 3, no. 2, pp. 677–688, March 2004. [22] V. Lau, Y. Liu, and T.-A. Chen, “On the design of MIMO block-fading channels with feedback-link capacity constraint,” IEEE Trans. Commun., vol. 52, no. 1, pp. 62–70, Jan. 2004. [23] N. Jindal and A. Goldsmith, “Dirty paper coding vs. tdma for mimo broadcast channels,” in IEEE Int. Conf. Commun., June 2004, vol. 2, pp. 682–686.

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DRAFT