多輸入多輸出雙使用者干擾通道在傳送端 沒有通道狀態的資訊下使用功率劃分的自由度

國 立 交 通 大 學

電信工程研究所

碩 士 論 文

多輸入多輸出雙使用者干擾通道在傳送端沒

有通道狀態的資訊下使用功率劃分的自由度

Degrees of freedom for MIMO interference channel

using power split with no channel state information at

transmitters

研 究 生：羅傑

指導教授：陸曉峯 教授

多輸入多輸出雙使用者干擾通道在傳送端沒有通

道狀態的資訊下使用功率劃分的自由度

Degrees of freedom for MIMO interference channel using

power split with no channel state information at

transmitters

研 究 生：羅傑 Student：Chieh Lo

指導教授：陸曉峯 Supervisor：Prof. Hsiao-feng Lu

Submitted to Institute of Communications Engineering College of Electrical and Computer Engineering

National Chiao Tung Uniersity in Partial Fulfillment of the Requirements

for the Degree of Master in

Communications Engineering June 2013

Hsinchu, Taiwan, Republic of China

i

多輸入多輸出雙使用者干擾通道在傳送端沒有通道狀態

的資訊下使用功率劃分的自由度

學生：羅傑

指導教授

陸曉峯 教授

國立交通大學電信工程研究所碩士班

摘

要

ii

Degrees of freedom for MIMO interference channel

using power split with no channel state information at

transmitters

Student：Chieh Lo

Supervisor：Dr. Hsiao-Feng Lu

Institute of Communications Engineering

National Chiao Tung University

Abstract

In this thesis we investigate the degree-of-freedom (DoF) region of a

two-user MIMO interference channel without channel state information at the

transmitters. Following the work of Huang, Jafar, and Shamai, we distinguish

three types of channels based on the numbers of transmit and receive antennas.

We derive the DoF region for a power-split scheme followed by a SIC decoder.

We also improve an outer bound by Huang et. al. on the DoF region for a

specific type of MIMO interference channels. We extend our discussion to

two-user MIMO interference channels with perfect receiver cooperation and

compare the result to two-user interference without cooperation.

iii

誌 謝

Contents

1.1 MIMO Systems and Degrees of Freedom (DoF) . . . 1

1.2 Two-User MIMO Interference Channel . . . 2

1.3 Motivation . . . 4

1.4 Thesis Outline . . . 4

2 General Two-User MAC Sum Rate 5 2.1 Sum Rate for General Two-User MAC . . . 5

2.2 Proof of Theorem 2.1 . . . 6

3 DoF region of Two-User MIMO Interference Channels 11 3.1 From two-user MAC Sum Rate to Two-User MIMO Interference Channel 11 3.1.1 Types of Two-User MIMO Interference Channel . . . 12

3.2 DoF region when nt2 ≤ nr1 . . . 13

3.2.2 DoF region when nt1 < nr1 < nt1+ nt2 . . . 18

3.2.3 DoF region when nt1+ nt2 ≤ nr1 . . . 21

3.3 DoF region when nr1 < nt2 and nr1 ≤ nt1 . . . 24

3.4 DoF region when nr1 < nt2 and nt1< nr1 . . . 29

4 DoF region of Two-User MIMO Interference Channels with Perfect Receiver Cooperation 33 4.1 DoF region with perfect receiver cooperation when nt2≤ nr1 . . . 34

4.1.1 DoF region with perfect receiver cooperation when nr1 ≤ nt1 . 34 4.1.2 DoF region with perfect receiver cooperation when nt1< nr1 < nt1+ nt2 . . . 38

4.1.3 DoF region with perfect receiver cooperation when nt1+ nt2 ≤ nr1 40 4.2 DoF region with perfect receiver cooperation when nr1 < nt2and nr1 ≤ nt1 . . . 41

4.3 DoF region with perfect receiver cooperation when nr1 < nt2and nt1 < nr1 . . . 47

5 Conclusion 50

List of Figures

1.1 Two-user MIMO interference channel . . . 3

3.1 The DoF region of type 1.(a) . . . 18 3.2 The DoF region of type 1.(a) with nt1 = 3, nt2 = 1, nr1 = 2 and nr2 = 4 19

3.3 The DoF region of type 1.(b) . . . 22 3.4 The DoF region result of type 1.(b) with nt1 = 3, nt2 = 2, nr1 = 4 and

nr2 = 4 . . . 22

3.5 The DoF region of type 1.(c) . . . 25 3.6 The DoF region of type 1.(c) with nt1 = 2, nt2 = 2, nr1 = 4 and nr2 = 5 25

3.7 The DoF region of type 2 . . . 28 3.8 The DoF region of type 2 with nt1= 3, nt2 = 4, nr1 = 2 and nr2 = 4 . 29

3.9 The DoF region of type 3 . . . 31 3.10 The DoF region of type 3 with nt1 = 1, nt2 = 3, nr1 = 2 and nr2 = 4

compared to the new outer bound . . . 32

4.1 The DoF region of the DoF region of type 1.(a) with perfect receiver cooperation compared to the DoF region with no cooperation. . . 36 4.2 The DoF region of type 1.(a) with nt1 = 3, nt2 = 1, nr1 = 2 and nr2 = 4 37

4.3 The DoF region of the DoF region of type 1.(b) with perfect receiver cooperation compared to the DoF region with no cooperation. . . 39 4.4 The DoF region of type 1.(b) . . . 40 4.5 The DoF region of the DoF region of type 1.(c) with perfect receiver

4.6 The DoF region of type 1.(c) . . . 43 4.7 The DoF region of the DoF region of type 2 with perfect receiver

cooperation compared to the DoF region with no cooperation. . . 45 4.8 The DoF region of type 2 . . . 46 4.9 The DoF region of the DoF region of type 3 with perfect receiver

cooperation compared to the DoF region with no cooperation. . . 48 4.10 The DoF region of type 3 . . . 49

Chapter 1

Introduction

1.1

MIMO Systems and Degrees of Freedom (DoF)

In wireless communication, multiple-input-multiple-output (MIMO) systems have the ability to provide remarkable increase of capacity compared to single-input-single-output (SISO) systems. One of the key benefit from MIMO systems is multiplexing signal in space. For a MIMO system, the ability of multiplexing signals is measured by the spatial multiplexing gain [1], also known as degrees of freedom (DoF).

Let us consider a point-to-point MIMO system, the transmitter has nt antennas,

and the receiver has nr antennas. The transmitter can transmit at most nt

indepen-dent streams simultaneously, and the receiver can receiver at most nr independent

streams simultaneously. Thus, it is easy to see that the maximal DoF of a point-to-point MIMO system is min(nt, nr). The standard definition of maximal DoF is

dmax := lim SNR→∞

C(SNR)

log SNR, (1.1)

where C(SNR) is the capacity of the point-to-point MIMO system given by

C(SNR) = E  log det  Inr + SNR nt HH†  , (1.2)

and the SNR is the signal-to-noise power ratio. The elements of H and N are circu-larly symmetric complex Gaussian random variables with zero mean and unit

vari-ance. The equation (1.2) can be written as C(SNR) = E "_nmin X i=1 log  1 + SNR nt λ2_i # , (1.3)

where λ1 ≤ λ2. . . < λnmin are the ordered singular values of H, and nmin = min(nt, nr).

By Jensen’s inequality, we have that

E "_n min X i=1 log  1 + SNR nt λ2_i # ≤ nminlog 1 + SNR nt " 1 nmin nmin X i=1 λ2_i #! . (1.4)

At high SNR, the capacity approximates

C(SNR) ≈ nminlog SNR nt + nmin X i=1 Elog λ2i . (1.5)

So we have the maximal DoF is

dmax = lim SNR→∞

C(SNR)

log SNR = nmin. (1.6) Thus we can know that the maximal DoF of a point-to-point MIMO system is min(nt, nr).

In [2], it has been shown that for the point-to-point MIMO communication, the absence of channel state information at transmitters, i.e., CSIT, does not reduce the DoF. But in a MIMO network with distributed processing units, [3] shows that in the absence of CSIT, the DoF may be lost.

1.2

Two-User MIMO Interference Channel

In this section, we consider the two-user MIMO interference channel. We have two transmitters and two receivers. Transmitter 1 and 2 are equipped with nt1, nt2

anten-nas, respectively. Receiver 1 and 2 are equipped with nr1, nr2 antennas, respectively.

The channel is described as

Y₁[n] = H11[n]X1[n] + H12[n]X2[n] + Z1[n] (1.7)

1.2. Two-User MIMO Interference Channel

Tx1

Tx2 Rx2

Rx1

Figure 1.1: Two-user MIMO interference channel

where at the n-th channel use, Y_j[n] and Z_j[n] are the nrj×1 vectors representing the

channel output and additive white Gaussian noise at receiver j, Hji[n] is the nrj× nti

channel matrix corresponding to receiver j and transmitter i, and X_i[n] is the nti× 1,

i, j ∈ 1, 2. The elements of Hji[n] and Zj[n] are i,i.d. circularly symmetric complex

Gaussian random variables with zero mean and unit variance. The transmit power constraint is

E[X2_i] ≤ SNR, i = 1, 2 (1.9) Fig. 1.1 is an example of two-user MIMO interference channel.

The capacity region C(SNR) for the two-user MIMO interference channel is the set of all rate pairs (R1, R2) for which the probability of error can be driven arbitrarily

close to zero by using suitable long codewords. Thus, the DoF region for two-user MIMO interference channel is defined as

D :=  (d1, d2) ∈ R+2 : ∃(R1(SNR), R2(SNR)) ∈ C(SNR) s.t. di = lim SNR→∞ Ri(SNR) log(SNR), i = 1, 2.  (1.10)

Except in [4], in [5] and [6] also had considered the DoF region of two-user MIMO interference channel with no CSIT. In[7] and [8], the DoF region of two-user MIMO

interference channel with perfect CSIT and CSIR had been analyzed.

1.3

Motivation

The DoF region of a two-user MIMO interference channel is determined by the values of nt1, nt2, nr1 and nr2. In [4], Huang and Jafar distinguish the values of

nt1, nt2, nr1 and nr2 as several types and analyze the DoF region of each type based

on an information theoretic approach by introducing an auxiliary random variable. Yet, there is one type of the two-user MIMO interference channel whose DoF region remains unknown in [4], except for an outer bound.

In this thesis, we follow [4] to distinguish the types of the two-user MIMO in-terference channel by the associated values of nt1, nt2, nr1 and nr2 and determine the

DoF region of each type based on a power-split transmission scheme followed by a successive-interference-cancellation (SIC) decoder. We will also try to reduce the outer bound of the unknown DoF region to tighten the bound.

1.4

Thesis Outline

In this thesis, we will consider the DoF region of two-user MIMO interference channel with no CSIT and perfect CSIR, we will use the power-split scheme and apply SIC decoder to characterize the DoF region. Our technique is closely related to the characterization of the DoF for a general two-user multiple-access channel (MAC), by which we mean the case when the two users can have different number of transmit antennas and different power constraint. The corresponding result will be given in Chapter 2. In Chapter 3, we will apply the results in Chapter 2 to the two-user MIMO interference channel, and investigate the DoF region for two-user MIMO interference channel. Also we will provide an outer bound for a specific two-user MIMO interference channel. In Chapter 4, we will consider two-two-user MIMO interference channel with perfect receiver cooperation. The results provide an outer bound for DoF region derived in Chapter 3. Chapter 5 is the conclusion.

Chapter 2

General Two-User MAC Sum Rate

In this chapter, we consider a two-user multiple access channel (MAC). The two transmitters have nt1, nt2 antennas, respectively. The receiver has nr antennas. We

assume that the transmitters do not have the channel state information, i.e., no CSIT. We develope the theorem of general MAC sum rate, which will be applied to the two-user MIMO interference channel.

2.1

Sum Rate for General Two-User MAC

In this section, we consider a two-user MAC channel, the channel model is de-scribed as

Y[n] = H1[n]X1[n] + H2[n]X2[n] + Z[n] (2.1)

Assuming the power constraints for the two users are given by

E kXi[n]k 2

≤ SNRαi_{, i = 1, 2}

the MAC sum rate at high SNR regime is

C(SNR) = E[log det(Inr + SNR α1_H 1H † 1+ SNR α2_H 2H † 2)] (2.2)

Theorem 2.1 For α1 ≥ α2 ≥ 0, C(SNR) = E[log det(Inr + SNR α1_H 1H † 1+ SNR α2_H 2H † 2)] =          α1nrlog SNR + O(1), if nr ≤ nt1 (α1nt1+ α2(nr− nt1)) log SNR + O(1), if nt1 < nr < nt1+ nt2 (α1nt1+ α2nt2) log SNR + O(1), if nt1+ nt2 ≤ nr

The proof of Theorem 2.1 is given in the next section. From Theorem 2.1, we can simply extend it to the DoF.

Corollary 2.2 For α1 ≥ α2 ≥ 0, the maximum DoF dmax for nonzero diversity gain

is given by dmax =          α1nr, if nr ≤ nt1 α1nt1+ α2(nr− nt1), if nt1 < nr < nt1+ nt2 α1nt1+ α2nt2, if nt1+ nt2 ≤ nr (2.3)

2.2

Proof of Theorem 2.1

We consider a two-user MAC channel, the transmitter 1 and 2 have nt1 and nt2

antennas, respectively. The receiver has nr antennas. The channel model is described

as

Y[n] = H1[n]X1[n] + H2[n]X2[n] + Z[n], (2.4)

where the channel matrices H1 and H2 are complex Gaussian random matrices with

CN (0, 1). The sum rate at high SNR regime is

C = E [log det(Inr + SNR α1_H 1H † 1+ SNR α2_H 2H † 2)] (2.5)

Let we consider two lemmas first.

Lemma 2.3 Let A, B ∈ Mm,n(C), then

log det(Im+ AA†+ BB†) ≤ log det(Im+ AA†) + log det(Im+ BB†) (2.6)

and

2.2. Proof of Theorem 2.1

Proof: The upper bound follows from

log det(Im+ AA†+ BB†) = log det

    In+ A†A A†B B†A In+ B†B     (2.8) ≤ log[det(In+ A†A) det(In+ B†B)] (2.9)

= log det(In+ A†A) + log det(In+ B†B), (2.10)

where the inequality is due to Fischer’s inequality on positive definite matrices. The lower bound simply follows from

det(Im+ AA†+ BB†) ≥ det(Im+ AA†), (2.11)

and

det(Im+ AA†+ BB†) ≥ det(Im+ BB†), (2.12)

since AA† and BB† are nonnegative definite matrices. Lemma 2.4 E[log det(Inr + SNR α1_H 1H†1+ SNR α2_H 2H†2)] ≤ (α1K1+ (α2)+K2) log(SNR) + O(1), (2.13) where K1 = min(nt1, nr) and K2 = min(nt2, nr).

Proof: It follows from previous lemma that

C _{= E[log det(I}nr + SNR α1_H 1H†1+ SNR α2_H 2H†2)] (2.14) ≤ E[log det(Int1+ SNR α1_H 1H†1)] + E[log det(Int2+ SNR α2_H 2H†2)] (2.15) = (α1K1+ (α2)+K2) log(SNR) + O(1). (2.16)

Now we start with the proof of the Theorem 2.1. Proof:

1. Case nr ≤ nt1: By Lemma 1, we observe that for any combination of nt1, nt2, nr>

0 that C _{= E[log det(I}nr + SNR α1_H 1H † 1+ SNR α2_H 2H † 2)] (2.17) ≥ max(α1K1, α2K2) log SNR. (2.18)

If nt1 ≥ nt2, then K1 ≥ K2 and hence max(α1K1, α2K2) = α1K1 = α1nr. On

the othre hand, if nt1 < nt2, we have K2 = K1 = nr. Thus, it follows that

max(α1K1, α2K2) = α1nr. To show the converse, note that log det(.) is strictly

convex, so we have C _{= E[log det(I}nr + SNR α1_H 1H † 1+ SNRα2H2H † 2)] (2.19) ≤ log det(E[Inr + SNR α1_H 1H † 1+ SNRα2H2H † 2]) (2.20) = log det(Inr + nt1SNR α1_I nr + nt2SNR α2_I nr) (2.21) . = α1nrlog SNR. (2.22)

The case of nr≤ nt1 is proved.

2. Case nt1 < nr < nt1+ nt2 Define G =h √SNRα1_H 1 √ SNRα2_H 2 i =h GL GR i (2.23) GL= h H1 H2L i   √ SNRα1Int1 √ SNRα2Inr−nt1  = HL √ Λ, (2.24)

where GLis a submatrix of G of size nr×nr and GRis of size nr×(nt1+nt2−nr).

Note that rank(GL) = nr with probability one. It then follows that

C _{= E[log det(I}nr + SNR α1_H 1H † 1+ SNR α2_H 2H † 2)] (2.25) = E[log det(I + GG†)] (2.26) = E[log det(I + GLG † L+ GRG † R)] (2.27) ≥ E[log det(I + GLG † L)] (2.28) = E[log det(I + HLλH † L)] (2.29)

≥ E[log(1 + det(λ) det(HLH †

L))], (2.30)

where the last inequality follows from det(I + AA†) ≥ 1 + det(AA†). Now let us take expectation of both sides show

C _{= E[log det(I}nr + SNR α1_H 1H † 1+ SNR α2_H 2H † 2)] (2.31)

> log det(λ) + E[log det(HLH †

L)] (2.32)

2.2. Proof of Theorem 2.1

For the converse, we have the following inequalities

C = _{E[log det(I}nr + SNR α1_H 1H † 1+ SNR α2_H 2H † 2)] (2.34) = _{E[log det(I}nr + GLG † L+ GRG † R)] (2.35) (a) = _EGL[EGR[log det(Inr + GLG † L+ GRG † R)]] (2.36) (b) < _EGL[log det(EGR[Inr + GLG † L+ GRG † R])] (2.37) (c) . = _EGL[log det(Inr + GLG † L+ SNR α2_I nr)] (2.38)

= nrlog SNR + EGL[log det(Inr + (1 + SNR

α2₎−1_G

LG †

L)] (2.39)

= α2nrlog SNR + EGL[log det(Inr +

SNRα1 1 + SNRα2 H1H † 1 (2.40) + SNRα2 1 + SNRα2 H2LH † 2L)] (2.41) (d)

= α2nrlog SNR + (α1− α2)nt1log SNR + O(1). (2.42)

(a) follows from conditional expectation.

(b) follows from that log det(.) is strictly concave on positive definite matrices.

(c) follows from EGR[GRG † R] = SNR α2_(2n t2− nr)Inr and (2nt2− nr) . = 1. (d) follows from _1+SNRSNRα2α2 .

= 1 for α2 > 0 and from Lemma 2.

3. Case nr ≥ nt1+ nt2 Define G = h √SNRα1_H 1 √ SNRα2_H 2 i =   GU GD  , (2.43)

where GU is of size ns× ns and GD is of size (nr− ns) × ns. Then note that

C _{= E[log det(I}nr + SNR α1_H 1H † 1+ SNR α2_H 2H † 2)] (2.44) = E[log det(Ins+ G † G)] (2.45) = E[log det(Ins+ G † UGU + G † DGD)] (2.46) ≥ E[log det(Ins+ G † UGU)]. (2.47)

case with nr = ns. So we have C _{= E[log det(I}nr + SNR α1_H 1H † 1+ SNR α2_H 2H † 2)] (2.48) ≥ (α1nt1+ α2nt2) log SNR. (2.49)

The converse follows from lemma 2.

Chapter 3

DoF region of Two-User MIMO

Interference Channels

In this chapter, we will investigate the DoF region of two-user MIMO interference channel with no CSIT.

3.1

From two-user MAC Sum Rate to Two-User

MIMO Interference Channel

In this section, we discuss how to extend Theorem 1 in Chapter 2 to the two-user MIMO interference channel with the following channel model

Y₁[n] = H11[n]X1[n] + H12[n]X2[n] + Z1[n] (3.1)

Y₂[n] = H21[n]X1[n] + H22[n]X2[n] + Z2[n] (3.2)

We can view the above interference channel as two two-user MACs. The receivers of the two MACs are nr1 and nr2, respectively. For simplicity, let us denote the

number of receive antennas at the receiver by nr. Also, we will consider the use of

successive-interference-cancellation (SIC) scheme for decoding. Intuitively, we decode the message with stronger power and treat the message with weaker power as noise first. After we decode the message with stronger power successfully, we can subtract

it out and then decode the message with weaker power. Assume the power constraints for the two users are given by

E kXi[n]k 2

≤ SNRαi_{, i = 1, 2}

For the case of α1 ≥ α2, we shall decode X1 first. Thus the resulting rate R1 for the

first user is given by

R1 ≤ E[log det(Inr + SNR α1_H 1H † 1+ SNR α2_H 2H † 2)] − E[log det(Inr + SNR α1_H 1H † 1]

Hence by Theorem 1 in Chapter 2 the corresponding DoF is

d1 ≤          α1nr− α2min(nt2, nr), if nr≤ nt1 α1nt1+ α2(nr− nt1) − α2min(nt2, nr), if nt1< nr < nt1+ nt2 α1nt1+ α2nt2− α2min(nt2, nr), if nt1+ nt2 ≤ nr (3.3)

and then we subtract X₁ off and decode X₂. We have

d2 ≤ α2min(nt2, nr). (3.4)

it should be noted that (3.3) and (3.4) are the DoF constraint of two-user MAC when we apply SIC at receiver.

3.1.1

Types of Two-User MIMO Interference Channel

We divide the analysis of DoF region into three different types, depending on the values of nt1, nt2, nr1 and nr2. The condition nr1 ≤ nr2 is assumed, and the result can

be easily extended to nr1 > nr2. In the following, we distinguish three types:

1. nt2 ≤ nr1:

In this type, it is known [4] that the absence of CSIT does not reduce the DoF region. To simplify the analysis, we divide this type into three sub-types. From the result of Theorem 1, we can divide type 1 into three sub-types naturally:

(a) nr1 ≤ nt1.

3.2. DoF region when nt2≤ nr1

(c) nt1+ nt2≤ nr1.

2. nr1 < nt2 and nr1 ≤ nt1:

In this type, the DoF region shrinks due to the absence of CSIT. We will use power-split scheme and apply SIC to characterize the DoF region.

3. nt1 < nr1 ≤ nt2:

In this type, the absence of CSIT will also reduce the DoF region. But unfor-tunately, we still do not know the exact DoF region of this type. There is an outer bound given in [4]. We will use a power-split transmission scheme and apply SIC to get a DoF region, and compare it to the outer bound to see the gap between them. Finally, we use a genie-aided two-user interference channel to obtain an outer bound that is tighter than the one in [4].

We discuss the DoF region of each type in the following three sections.

3.2

DoF region when n

≤ n

In this section, we discuss the DoF region of the type 1, where the antenna dis-tribution satisfies nt2 ≤ nr1. The transmit power can be expressed as SNRα1 and

SNRα2_{, where 0 ≤ α}

1 ≤ 1 and 0 ≤ α2 ≤ 1, for transmitter 1 and transmitter 2,

respectively. Intuitively, the receiver decodes the message with stronger power first and treat the other as noise, i.e., if α2 ≤ α1, we decode the X1 first, both at the

receiver 1 and receiver 2, else we reverse the decoding order. After we decode the message with stronger power at both receivers successfully, we can subtract it off from the received signal, hence, the remaining is the message with weaker power and we can easily decode it at its intended receiver. We discuss the three sub-types of type 1 in the following three subsections.

3.2.1

DoF region when n

≤ n

1. α1 ≥ α2.

2. α2 ≥ α1.

We start with the case α1 ≥ α2. At the receiver 1, we just treat the X2 as noise and

decode thew X₁. We have the following DoF constraint of d1:

d1 ≤ α1nr1− α2min(nt2, nr1) (3.5)

= α1nr1− α2nt2. (3.6)

At receiver 2, we also treat the X₂ as noise and decode X₁, then we subtract X₁ off and decode X₂ at receiver 2. Since we do two decoding operations at receiver 2, we have two constraints on DoF, one for d1 and one for d2. We have three sub-cases for

this type depending the the values of nr1 and nt2.

1. nr2 ≤ nt1: d1 ≤ α1nr2− α2min(nt2, nr2) (3.7) = α1nr2− α2nt2. (3.8) 2. nt1 ≤ nr2 ≤ nt1+ nt2: d1 ≤ α1nt1+ α2(nr2− nt1) − α2min(nt2, nr2) (3.9) = α1nt1+ α2(nr2− nt1− nt2). (3.10) 3. nt1+ nt2 ≤ nr2: d1 ≤ α1nt1+ α2nt2− α2min(nt2, nr2) (3.11) = α1nt1. (3.12)

After we decode X₁ at receiver 2 successfully, we can remove it and decode X₂, the constraint of d2 is

d2 ≤ α2min(nt2, nr2) (3.13)

3.2. DoF region when nt2≤ nr1

The above gives the DoF region of type 1.(a) when α1 ≥ α2.

Now we consider the case when α2 ≥ α1. We treat X1 as noise, decode X2 first,

then subtract X₂ off and decode X₁ at receiver 1. At receiver 1, since nt2 ≤ nr1 ≤

nt1+ nt2, we have that

d2 ≤ α2nt2+ α1(nr1− nt2) − α1min(nt1, nr1) (3.15)

= (α2 − α1)nt2 (3.16)

d1 ≤ α1min(nt1, nr1) (3.17)

= α1nr1. (3.18)

At receiver 2, we just treat X₁ as noise and decode X₂ directly, since nt2 ≤ nr2 ≤

nt1+ nt2 and nt1+ nt2≤ nr2 are both possible, we need to consider two conditions of

DoF constraint. 1. nt2 ≤ nr2 ≤ nt1+ nt2: d2 ≤ α2nt2+ α1(nr2+ nt2) − α1min(nt1, nr2). (3.19) 2. nt1+ nt2 ≤ nr2: d2 ≤ α2nt2+ α1nt1− α1min(nt1, nr2) (3.20) = α2nt2. (3.21)

These give the DoF region of type 1.(a) when α2 ≥ α1. Now we have the complete

constraints of the DoF region of type 1.(a), but how to use these constraints to characterize the exact DoF region? Here we provide a simple way to get an insight into the DoF region. First, we consider the case of α1 ≥ α2. We try to find the

corner points on the DoF region, so we consider the extreme values of (α1, α2), that

is, (α1, α2) = (1, 0) and (α1, α2) = (1, 1), respectively. (α1, α2) = (1, 0) means that we

give almost all power to transmitter 1 only when SNR is very high. We start with the case when α1 ≥ α2, and replace the (α1, α2) = (1, 0) to equation (3.5)-(3.14), then,

at receiver 1, the DoF constraint becomes

At receiver 2, we need to decode the X₁ reliably and then decode X₂, the DoF constraints of d1 is following: 1. nr2 ≤ nt1: d1 ≤ nr2. (3.23) 2. nt1 ≤ nr2 ≤ nt1+ nt2: d1 ≤ nt1. (3.24) 3. nt1+ nt2 ≤ nr2 d1 ≤ nt1. (3.25)

And since α2 = 0, d2 ≤ 0. From equation (3.15)-(3.21), we have 4 constraints on

d1, and we know that nr1 ≤ nt1 and nr1 ≤ nt2. We can only choose the lower rate in

order to have reliable communication for both two streams, so here we need

d1 ≤ nr1 (3.26)

The above shows that (d1, d2) = (nr1, 0) is a corner point in the DoF region. Then

we substitute (α1, α2) = (1, 1) into equation (3.5)-(3.14). At receiver 1, we have

d1 ≤ nr1− nt2. (3.27) At receiver 2, we have 1. nr2 ≤ nt1: d1 ≤ nr2− nt2. (3.28) 2. nt1 ≤ nr2 ≤ nt1+ nt2: d1 ≤ nr2− nt2. (3.29) 3. nt1+ nt2 ≤ nr2 d1 ≤ nt1. (3.30)

3.2. DoF region when nt2≤ nr1

From (3.23)-(3.26), we can see that d1 ≤ nr1 − nt2 can yield reliable communication

for both streams. After we decode X₁ successfully, we can subtract it off at receiver 2 then decode X₂. The DoF constraint is:

d2 ≤ nt2. (3.31)

Thus we can achieve (d1, d2) = (nr1 − nt2, nt2). This is also one of the corner point.

Now we consider the case when α2 ≥ α1. If we replace (α1, α2) = (0, 1) in

(3.15)-(3.21), at receiver 1 we have that

d2 ≤ nt2. (3.32)

Then we can decode X₁ by subtracting the component X₂ off at receiver 1, but here the power of X₁ is vanishingly small compared to SNR, so the DoF constraint of d1

is

d1 ≤ 0. (3.33)

At receiver 2, we treat the X₁ as noise and decode the X₂ directly, then we have 1. nt2 ≤ nr2 ≤ nt1+ nt2:

d2 ≤ nt2 (3.34)

2. nt1+ nt2 ≤ nr2:

d2 ≤ nt2 (3.35)

Hence, now we have third corner point, which is (d1, d2) = (0, nt2). Finally, we

consider (α1, α2) = (1, 1). We follow the same procedure before. We replace (α1, α2) =

(1, 1) in (3.15)-(3.21). For receiver 1, we have

d2 ≤ 0. (3.36)

This is a surprising fact. It means that if we want to decode the X₂ at receiver 1 when the powers of both X₁ and X₂ are almost the same, the reliable communication rate will approach 0. With this condition, we can get d1 ≤ nr1. But since the constraint of

d2 ≤ 0, we cannot have a better rate for X2 at receiver 2. The DoF we can achieve in

Figure 3.1: The DoF region of type 1.(a)

points before. We then use a time-sharing scheme to connect the 3 corner points, and see that this is the exact DoF region, which corresponds to the result of [4] as shown In Fig. 3.1.

Furthermore, in Fig. 3.2 we consider the case nt1 = 3, nt2 = 1, nr1 = 2 and

nr2 = 4. For a given d1, we vary the value of α1 and α2 to compute the greatest d2,

where 0 ≤ d1 ≤ min(nt1, nr1), in this example min(nt1, nr1) = 2. The DoF region is

given in Fig. 3.2.

3.2.2

DoF region when n

< n

< n

+ n

In this subsection we discuss the DoF region of type 1.(b). We follow the similar way in the previous subsection. We can use Theorem 1 and power-split scheme to calculate the DoF constraints. Again we start with the case of α1 ≥ α2. So we treat

X₂ as noise and decode X₁. After we extract X₁ at receiver 2, we can apply SIC to decode X₂. At receiver 1, we have

d1 ≤ α1nt1+ α2(nr1− nt1) − α2min(nt2, nr1) (3.37)

3.2. DoF region when nt2≤ nr1

Figure 3.2: The DoF region of type 1.(a) with nt1 = 3, nt2 = 1, nr1 = 2 and nr2 = 4

At receiver 2, we have to consider the distribution of antenna being whether nt1 <

nr2 < nt1+ nt2 and nt1+ nt2 ≤ nr2. We decode X1 and treat X2 as noise. Then we

have 1. nt1 < nr2 < nt1+ nt2: d1 ≤ α1nt1+ α2(nr2− nt1) − α2min(nt2, nr2) (3.39) = α1nt1+ α2(nr2− nt1− nt2) (3.40) 2. nt1+ nt2 ≤ nr2: d1 ≤ (α1nt1+ α2nt2) − α2min(nt2, nr1) (3.41) = α1nt1. (3.42)

After we decode X₁ at receiver 2 successfully, we can apply SIC to decode the X₂. The DoF constraint of d2 is simply

d2 ≤ α2min(nt2, nr2) (3.43)

Now let us consider the case when α2 ≥ α1. In this case we decode the X2 first,

both at receiver 1 and receiver 2. At receiver 1, we have

d2 ≤ α2nt2+ α1(nr1− nt2) − α1min(nt1, nr1) (3.45)

= α2nt2+ α1(nr1− nt1− nt2). (3.46)

Then we use SIC to decode X₁ and have

d1 ≤ α1min(nt1, nr1) = α1nt1. (3.47)

At receiver 2, we deocode X₂ directly, but there are two possible different antenna distributions. 1. nt2 < nr2 < nt1+ nt2: d2 ≤ α2nt2+ α1(nr2− nt2) − α1min(nt1, nr2) (3.48) = α2nt2+ α1(nr2− nt1− nt2). (3.49) 2. nt1+ nt2 ≤ nr2: d2 ≤ (α2nt2+ α1nt1) − α1min(nt1, nr2) = α2nt2. (3.50)

Now, we have the full DoF constraints of the type1.(b), then we can follow the same step in last subsection. We use the extreme values of α1 and α2 to find the corner

point in the DoF region. The results are given below. When α1 ≥ α2, we have

1. (α1, α2) = (1, 0): d1 ≤ nt1 (3.51) d2 ≤ 0. (3.52) 2. (α1, α2) = (1, 1): d1 ≤ nr1− nt2 (3.53) d2 ≤ nt2. (3.54)

3.2. DoF region when nt2≤ nr1 When α2 ≥ α1, we have 1. (α1, α2) = (0, 1): d1 ≤ 0 (3.55) d2 ≤ nt2. (3.56) 2. (α1, α2) = (1, 1): d1 ≤ nt1 (3.57) d2 ≤ nr1− nt1. (3.58)

From (3.51)-(3.58), we have the four corner points of the DoF region for type 1.(b). Then we can apply the time-sharing scheme to connect the four corner points to get the entire DoF region for type 1.(b). This DoF region is the same as in [4]. Fig. 3.3 shows the DoF region of type 1.(b)

We also consider an example of type 1.(b)by setting nt1 = 3, nt2 = 2, nr1 = 4 and

nr2 = 4. Fig. 3.4 shows the DoF region of type 1.(b) by the computer calculation.

3.2.3

DoF region when n

+ n

≤ n

In this subsection, we discuss the DoF region of type 1.(c). We use the similar way as in the previous two subsections. We start with the case α1 ≥ α2, and then

the case α2 ≥ α1. Type 1.(c) is relatively simple since we have both nt1+ nt2 ≤ nr1

and nt1+ nt2 ≤ nr2.

When α1 ≥ α2, we decode X1first and then apply SIC to cancel out the component

of X₁ and decode X₂. At receiver 1, we have

d1 ≤ α1nt1+ α2nt2− α2min(nt2, nr1) (3.59)

= α1nt1. (3.60)

Figure 3.3: The DoF region of type 1.(b)

Figure 3.4: The DoF region result of type 1.(b) with nt1 = 3, nt2 = 2, nr1 = 4 and

3.2. DoF region when nt2≤ nr1

X₁, then we can decode X₂, the DoF constraint at receiver 2 is below.

d1 ≤ α1nt1+ α2nt2− α2min(nt2, nr2) (3.61)

= α1nt1. (3.62)

After we decode X₁ successfully and then apply SIC, we have

d2 ≤ α2min(nt2, nr2) = α2nt2. (3.63)

Now we consider the case when α2 ≥ α1. We decode X2 first and apply SIC at

receiver 1. At receiver 1, we have

d2 ≤ α2nt2+ α1nt1− α2min(nt1, nr1) (3.64)

= α1nt2. (3.65)

Then we apply SIC to decode X₂. The result is simply

d1 ≤ α1min(nt1, nr1) (3.66)

= α1nt1. (3.67)

At receiver 2, we have

d2 ≤ α2nt2+ α1nt1− α2min(nt1, nr2) (3.68)

= α1nt2. (3.69)

Now we have the entire DoF constraint of type 1.(c). We then use the extreme values of α1 and α2 to find the corner points of DoF region. Both the case α1 ≥ α2

and α2 ≥ α1 have the same corner points. The result is below. When α1 ≥ α2, we

have

1. (α1, α2) = (1, 0):

d1 ≤ nt1 (3.70)

2. (α1, α2) = (1, 1): d1 ≤ nt1 (3.72) d2 ≤ nt2. (3.73) When α2 ≥ α1, we have 1. (α1, α2) = (0, 1): d1 ≤ 0 (3.74) d2 ≤ nt2. (3.75) 2. (α1, α2) = (1, 1): d1 ≤ nt1 (3.76) d2 ≤ nt2. (3.77)

From the above result, we can clearly see that (d1, d2) = (nt1, 0), (0, nt2) and (nt1, nt2)

are the three corner points on the DoF region. Thus, the DoF region is a rectangle. This is the same as in [4]. Fig. 3.5 shows the DoF of type 1.(c).

An example of type 1.(c) is a two-user MIMO interference channel with nt1 =

2, nt2 = 2, nr1 = 4 and nr2 = 5. Fig. 3.6 shows the DoF region of type 1.(c) by the

computer calculation.

3.3

DoF region when n

< n

and n

≤ n

In this section, we discuss the type 2, in this type, we still have the exact DoF region when the CSIT is not available. Also, we consider two cases, α1 ≥ α2 and

α2 ≥ α1, respectively. If α1 ≥ α2, we decode X1 first at both and then apply SIC at

receiver 2 to decode X₂. At receiver 1, we have nr1 ≤ nt1, so the DoF constraint of

d1 is

d1 ≤ α1nr1− α2min(nt2, nr1) (3.78)

3.3. DoF region when nr1 < nt2 and nr1 ≤ nt1

Figure 3.5: The DoF region of type 1.(c)

At receiver 2, unfortunately, nr2 ≤ nt1, nt1 < nr2 < nt1+ nt2 and nt1+ nt2 ≤ nr2 are

all possible in type 2, so we need to give DoF constraint for three different antenna distributions. 1. nr2 ≤ nt1: d1 ≤ α1nr2 − α2min(nt2, nr2). (3.80) 2. nt1 < nr2 < nt1+ nt2: d1 ≤ α1nt1+ α2(nr2 − nt1) − α2min(nt2, nr2). (3.81) 3. nt1+ nt2 ≤ nr2: d1 ≤ α1nt1+ α2nt2− α2min(nt2, nr2) (3.82) = α1nt1. (3.83)

After we decode X₁ successfully at receiver 2, we apply SIC and decode X₂. The DoF constraint is

d2 ≤ α2min(nt2, nr2). (3.84)

Note that we have nt2 ≤ nr2 only when nt1 + nt2 ≤ nr2. Then we consider the case

when α2 ≥ α1. We decode X2 first at both receivers. At receiver 1, we have

d2 ≤ α2nr1− α1min(nt1, nr1) (3.85)

= α2nr1− α1nr1. (3.86)

After we decode X₂ successfully at receiver 1, we can apply SIC and decode X₁. The DoF constraint is d1 ≤ α1min(nt1, nr1) (3.87) = α1nr1. (3.88) At receiver 2, we have 1. nr2 ≤ nt2: d2 ≤ α2nr2 − α1min(nt1, nr2). (3.89)

3.3. DoF region when nr1 < nt2 and nr1 ≤ nt1 2. nt2 < nr2 < nt1+ nt2: d2 ≤ α2nt2+ α1(nr2 − nt2) − α1min(nt1, nr2). (3.90) 3. nt1+ nt2 ≤ nr2: d2 ≤ α1nt2+ α1nt1− α1min(nt1, nr2) (3.91) = α2nt2. (3.92)

Note that again we have nt1 ≤ nr2 only when nt1+ nt2 ≤ nr2, otherwise we do not

know whether nt1 ≤ nr2 or nr2 ≤ nt1. So far, we have all DoF constraints of the type

2, then we can replace the extreme values to above DoF constraints to get the corner points in the DoF region. We skip the calculations and show the results directly. When α1 ≥ α2, we have 1. (α1, α2) = (1, 0): d1 ≤ nr1 (3.93) d2 ≤ 0. (3.94) 2. (α1, α2) = (1, 1): d1 ≤ 0 (3.95) d2 ≤ min(nt2, nr2). (3.96) When α2 ≥ α1, we have 1. (α1, α2) = (0, 1): d1 ≤ 0 (3.97) d2 ≤ nr1. (3.98) 2. (α1, α2) = (1, 1): d1 ≤ nr1 (3.99) d2 ≤ 0. (3.100)

Figure 3.7: The DoF region of type 2

We see that (3.96) and (3.98) give different DoF constraints for d2 when d1 ≤ 0. Since

we always try to maximize the rate, we should choose (3.96) as our rate constraint for d2, because min(nt2, nr2) ≥ nr1. Thus we have two corner points in DoF region here,

(d1, d2) = (nr1, 0) and (0, min(nt2, nr2)). We can use time-sharing scheme to connect

these two corner points to get the exact DoF region, which corresponds to the result in [4]

In type 2, we can see that if we want to achieve the maximum value of d1, we

need d2 to be zero, and vice versa. The entirely DoF region is given by time sharing

scheme between the maximum of d1 and d2. The shape of DoF region is a triangle.

Fig. 3.7 shows the DoF region of type 2.

An example of type 2 is a two-user MIMO interference channel with nt1 = 3, nt2=

4, nr1 = 2 and nr2 = 4. Fig. 3.8 shows the DoF region of type 2 by the computer

3.4. DoF region when nr1 < nt2 and nt1 < nr1

Figure 3.8: The DoF region of type 2 with nt1 = 3, nt2= 4, nr1 = 2 and nr2 = 4

3.4

DoF region when n

< n

and n

< n

In this section, we consider the DoF region of type 3, note that type 2 and type 3 are only differ by nr1 ≤ nt1 or nr1 > nt1. From [4], there is only an outer bound for

this type.

In type 3, we no longer decode the stream with higher power first. Instead, we will consider two decoding strategies at both receivers, regardless of the power of the two streams. For user 1, we want to transmit X₁ to receiver 1 reliably, so we have that at receiver 1:

1. Treating X₂ as noise and decoding X₁ directly.

2. First treating X₁ as noise and decoding X₂, then applying SIC to decode X₁. For user 2, the concept is same and at receiver 2 we have

1. Treating X₁ as noise and decoding X₂ directly.

2. First treating X₂ as noise and decoding X₁, then applying SIC to decode X₂. Each receiver should choose the best decoding strategy for a given power allocation and antenna distribution. At receiver 1, if we decode the X₁ directly, we have

1. nt1 < nr1 < nt1+ nt2:

d1 ≤ α1nt1+ α2(nr1 − nt1) − α2min(nt2, nr1). (3.101)

2. nt1+ nt2 ≤ nr1:

d1 ≤ α1nt1+ α2nt2− α2min(nt2, nr1). (3.102)

If we decode the X₂ first and then apply SIC to decode X₁, the DoF constraint of d2 is

d2 ≤ α2nr1− α1min(nt1, nr1) (3.103)

and d1 is simply

d1 ≤ α1min(nt1, nr1). (3.104)

At receiver 2, we also have two decoding strategies, but the situation becomes more complicated since we do not know nr2 ≤ nt2 or nt2 < nr2, thus we need to list

all the possible conditions. We consider the strategy of decoding X₂ directly first: 1. nr2 ≤ nt2: d2 ≤ α2nr2 − α1min(nt1, nr2). (3.105) 2. nt2 < nr2 < nt1+ nt2: d2 ≤ α2nt2+ α2(nr2 − nt2) − α1min(nt1, nr2). (3.106) 3. nt1+ nt2 < nr2: d2 ≤ α2nt2+ α1nt1− α1min(nt1, nr2). (3.107)

The above are the three conditions when we decode X₂ directly at receiver 2. Now we consider the strategy of decoding the X₁ first and then applying SIC to decode X₂. The DoF constraint of d1 is:

1. nt1 < nr2 < nt1+ nt2:

3.4. DoF region when nr1 < nt2 and nt1 < nr1

Figure 3.9: The DoF region of type 3

2. nt1+ nt2 ≤ nr2:

d1 ≤ α1nt1+ α2nt2− α2min(nt2, nr2) (3.109)

and d2 is simply

d2 ≤ α2min(nt2, nr2). (3.110)

The above (3.101)-(3.110) is the whole DoF constraints of the type 3. From the work in [4], we have an outer bound of DoF region, denoted as Dout,

Dout =  (d1, d2) ∈ R+2 : d1 nr1 + d2 min(nt2, nr2)  . (3.111)

But this outer bound seems loose. Consider a genie-aided two-user MIMO interference channel, by which we mean there is a genie at both receivers, informing the receiver the interference from the unintended user completely. The point (d1, d2) = (nr1, 0)

is still not achievable. We can achieve only (d1, d2) = (nt1, 0) since nt1 < nr1. The

DoF region of the two-user MIMO interference channel must be bounded by the one of the genie-aided two-user MIMO interference channel, so we can make the outer bound tighter by the genie-aided 2-user MIMO interference channel. This is a new outer bound for type 3. Fig. 3.9 shows the DoF region of type 3.

Figure 3.10: The DoF region of type 3 with nt1 = 1, nt2 = 3, nr1 = 2 and nr2 = 4

compared to the new outer bound

An example of type 3 is nt1 = 1, nt2 = 3, nr1 = 2 and nr2 = 4. Fig. 3.10 shows

the gap between our DoF region and the new outer bound.

Although we do not know that for a two-user Rayleigh fading channel, the gap between the DoF region and new outer bound is achievable or not, the work in [9] had already shown that for a two-user isotropic and independent (or block-wise independent) fading channel, the exact DoF region of this type is just the same as the DoF region in our work, which implies that the gap between the DoF region and the new outer bound may not be achievable, no matter whether beamforming or interference alignment is used.

Chapter 4

DoF region of Two-User MIMO

Interference Channels with Perfect

Receiver Cooperation

In this chapter, we investigate the DoF region of two-user MIMO interference chan-nel with perfect receiver cooperation and no CSIT. Here, perfect receiver cooperation means that two receivers have a noiseless and interference-free link to communicate with each other. For example, if we decode X₁ successfully at receiver 1, then we can send it to receiver 2, so receiver 2 can know X₁completely and remove the component of X₁ perfectly. The remaining signal is only X₂ plus noise, as the interference from X₁ has been removed. We can also use Theorem 1 in Chapter 2 for the general MAC sum rate to analyze the DoF region in this chapter. We then compare the results in the previous chapter, where we do not have any cooperation. We use the same way to divide the antenna distributions into three types. The three types are listed below.

1. nt2 ≤ nr1

(a) nr1 ≤ nt1

(b) nt1 < nr1 < nt1+ nt2

2. nr1 < nt2 and nr1 ≤ nt1

3. nr1 < nt2 and nt1≤ nr2

The condition nr1 ≤ nr2is assumed, and the result can be easily extended to nr1 > nr2.

We discuss the DoF region of each type in the following three sections.

4.1

DoF region with perfect receiver cooperation

when n

≤ n

In this section, we consider the DoF region of type 1. The transmit power of transmitter 1 and transmitter 2 are SNRα1 _{and SNR}α2_{, respectively. Note that 0 ≤}

α1 ≤ 1 and 0 ≤ α2 ≤ 1. We decode the message with stronger power at its intended

receiver first. Since the two receivers have perfect cooperation, we can transmit the decoded message(the stronger power message) to the other receivers. Thus the interference can be remove easily. We discuss the three sub-types of type 1 in the following three subsections

4.1.1

DoF region with perfect receiver cooperation when n

≤

n

We have two cases in the following

1. α1 ≥ α2.

2. α2 ≥ α1.

We start with the case α1 ≥ α2. At receiver 1, we treat X2 as noise and decode X1

directly. The constraint of d1 is

d1 ≤ α1nr1− α2min(nt2, nr1) (4.1)

4.1. DoF region with perfect receiver cooperation when nt2 ≤ nr1

If d1 satisfies the above constraint, we can decode X1 reliably at receiver 1. Since we

assume that the two receivers have perfect cooperation, the receiver 2 can know X₁. Thus receive 2 can subtract it off, then we can decode X₂ at receiver 2. The DoF constraint of X₂ is

d2 ≤ α2min(nt2, nr2) (4.3)

= α2nt2. (4.4)

Let us consider the case α2 ≥ α1 now. We decode X2 at receiver 2 first. Here,

since nt2 < nr2 < nt1+ nt2 and nt1+ nt2 ≤ nr2 are both possible, we have to consider

both of them. If nt2 < nr2 < nt1+ nt2, at receiver 2, we have the constraint of d2 is

d2 ≤ α2nt2+ α1(nr2 − nt2) − α1min(nt2, nr2). (4.5)

If nt1+ nt2≤ nr2, the constraint of d2 is:

d2 ≤ α2nt2+ α1nt1− α1min(nt1, nr2) (4.6)

= α2nt2. (4.7)

By the receiver cooperation, receiver 1 can easily subtract the component of X₂. The DoF constraint of d1 is

d1 ≤ α1min(nt1, nr1) (4.8)

= α1nr1. (4.9)

We see that if nt1+ nt2 ≤ nr2, we can achieve d1 = nr1 and d2 = nt2 simultaneously.

This DoF pair is the maximum of the two-user MIMO interference channel. If nt2<

nr2 < nt1+ nt2, we can achieve (d1, d2) = (nr1, nr2 − min(nt1, nr2)). This result is

better than no cooperation in section 3.1.1. Fig. 4.1 shows the DoF region of type 1.(a), both no cooperation and perfect receiver cooperation are considered.

An example of type 1.(a) is a two-user MIMO interference channel with nt1 =

3, nt2 = 1, nr1 = 2 and nr2 = 4. Fig. 4.2 shows the DoF region of type 1.(a) with

Figure 4.1: The DoF region of the DoF region of type 1.(a) with perfect receiver cooperation compared to the DoF region with no cooperation.

4.1. DoF region with perfect receiver cooperation when nt2 ≤ nr1

4.1.2

DoF region with perfect receiver cooperation when n

<

n

< n

+ n

In this subsection, we discuss the DoF region of type 1.(b). The method is just the same as previous subsection. Again we start with the case α1 ≥ α2. At receiver

1, we have the constraint of d1 is

d1 ≤ α1nt1+ α2(nr1− nt1) − α2min(nt2, nr1) (4.10)

= α1nt1+ α2(nr1− nt1− nt2). (4.11)

After we decode X₁ at receiver 1 successfully, the receiver 2 can know X₁ by the perfect receiver cooperation. Thus we can subtract X₁ off at receiver 2 and then decode X₂, the constraint of d2 is

d2 ≤ α2min(nt2, nr2) (4.12)

= α2nt2. (4.13)

Now we consider the case α2 ≥ α1. In this case we simply decode X2 at receiver

2 first. But here nt2 < nr2 < nt1+ nt2 and nt1+ nt2 ≤ nr2 are both possible, we need

to consider the two possibilities, respectively. If nt2< nr2 < nt1+ nt2, the constraint

of d2 is d2 ≤ α2nt2+ α1(nr2− nt2) − α1min(nt1, nr2) (4.14) = α2nt2+ α1(nr2− nt1− nt2). (4.15) If nt1+ nt2≤ nr2, the constraint of d2 is d2 ≤ α2nt2+ α1nt1− α1min(nt1, nr2) (4.16) = α2nt2. (4.17)

After we decode X₂ at receiver 2 successfully, receiver 1 can know X₂ by the perfect receiver cooperation. Thus the constraint of d1 is

d1 ≤ α1min(nt1, nr1) (4.18)

4.1. DoF region with perfect receiver cooperation when nt2 ≤ nr1

Figure 4.3: The DoF region of the DoF region of type 1.(b) with perfect receiver cooperation compared to the DoF region with no cooperation.

We can see that the result is almost the same as type 1.(a). The only difference is that when nt2 < nr2 < nt1+ nt2. We know min(nt1, nr2) = nt1in type 1.(b) but do not

know in type 1.(a). If nt2< nr2 < nt1+ nt2, we can achieve (d1, d2) = (nt1, nr2− nt1).

This is better than (d1, d2) = (nt1, nr1−nt1), the result of no cooperation. If nt1+nt2 ≤

nr2, we can achieve (d1, d2) = (nt1, nt2), and both the two stream can achieve their

maximum DoF. Fig. 4.3 shows the DoF region of type 1.(b), with and without receiver cooperation.

Figure 4.4: The DoF region of type 1.(b)

3, nt2 = 2, nr1 = 4 and nr2 = 4. Fig. 4.4 shows the DoF region of type 1.(b) with

perfect receiver cooperation by computer calculation. In this example, we can see that the DoF regions of no cooperation and with perfect receiver cooperation are the same, since nr1 = nr2 = 4.

4.1.3

DoF region with perfect receiver cooperation when n

+

n

≤ n

In this subsection, we discuss the DoF region of type 1.(c). Again we start with the case α1 ≤ α2. At receiver 1, we treat X2 as noise and decode X1 directly. Thus

we have the constraint of d1 is

d1 ≤ α1nt1+ α2nt2− α2min(nt2, nr1) (4.20)

4.2. DoF region with perfect receiver cooperation when nr1 < nt2 and

nr1 ≤ nt1

After we decode X₁ at receiver 1 successfully, the receiver 2 knows X₁ by the perfect receiver cooperation. Then we have the constraint of d2 is

d2 ≤ α2min(nt2, nr2) (4.22)

= α2nt2. (4.23)

Now, let us consider the case α2 ≥ α1. In this case, we decode X2 at receiver 2

first, so we have the constraint of d2 is

d2 ≤ α2nt2+ α1nt1− α1min(nt1, nr2) (4.24)

= α2nt2. (4.25)

After we decode X₂ at receiver 2 successfully, the receiver 1 knows X₂ by the perfect receiver cooperation. So the constraint of d1 is simply

d1 ≤ α1min(nt1, nr1) (4.26)

= α1nt1. (4.27)

From the above constraints of DoF, the corner points of the DoF region are (d1, d2) =

(nt1, 0), (0, nt2) and (nt1, nt2). The DoF region with perfect receiver cooperation is

the same as the DoF region with no cooperation in section 3.1.3. Fig. 4.5 shows the DoF region of type 1.(c), with and without receiver cooperation.

An example of type 1.(c) is two-user MIMO interference channel with nt1 =

2, nt2 = 2, nr1 = 4 and nr2 = 5. Fig. 4.6 shows the DoF region of type 1.(c)

with perfect receiver cooperation by computer calculation.

4.2

DoF region with perfect receiver cooperation

when n

< n

and n

≤ n

In this section, we consider the DoF region of type 2. The decoding strategy is the same as previous section. We start with the case α1 ≥ α2.

Figure 4.5: The DoF region of the DoF region of type 1.(c) with perfect receiver cooperation compared to the DoF region with no cooperation.

4.2. DoF region with perfect receiver cooperation when nr1 < nt2 and

nr1 ≤ nt1

When α1 ≥ α2, we treat X2 as noise and decode X1 at receiver 1. Thus we have

the constraint of d1 is

d1 ≤ α1nr1− α2min(nt2, nr1) (4.28)

= (α1− α2)nr1. (4.29)

After we decode X₁ at receiver 1 successfully, the receiver 2 knows X₁ by the perfect receiver cooperation. The constraint of d2 is simply

d2 ≤ α2min(nt2, nr2) (4.30)

Note that we do not know min(nt2, nr2) = nt2 or nr2 here.

Now let us consider the case α2 ≥ α1. In this case, we treat X1 as noise and

decode X₂ at receiver 2 directly. But in type 2, nr2 ≤ nt2, nt2 < nr2 < nt1+ nt2 are

all possible. So we need to consider these three conditions separately.

1. nr2 ≤ nt2 d2 ≤ α2nr2 − α1min(nt1, nr2). (4.31) 2. nt2 < nr2 < nt1+ nt2 d2 ≤ α2nt2+ α1(nr2 − nt2) − α1min(nt1, nr2). (4.32) 3. nt1+ nt2 ≤ nr2 d2 ≤ α2nt2+ α1nt2− α1min(nt1, nr2) (4.33) = α2nt2. (4.34)

After we decode X₂ successfully at receiver 2, the receiver 1 can subtract the compo-nent of X₂ by the perfect receiver cooperation. Thus, we can decode X₁ at receiver 1 without the interference from X₂ . The constraint of d1 is

d1 ≤ α1min(nt1, nr1) (4.35)

4.2. DoF region with perfect receiver cooperation when nr1 < nt2 and

nr1 ≤ nt1

Figure 4.7: The DoF region of the DoF region of type 2 with perfect receiver cooper-ation compared to the DoF region with no coopercooper-ation.

These are all DoF constraints of type 2 with perfect receiver cooperation. We can see that if we want to achieve the maximum of d2, i.e. min(nt2, nr2), d1 must be zero.

But if we want to achieve the maximum of d1, i.e. nr1, d2 can still be positive. This is

the improvement given by the perfect receiver cooperation. Fig. 4.7 shows the DoF region of type 2, with and without receiver cooperation.

An example of type 2 is a two-user MIMO interference channel with nt1 = 3, nt2=

4, nr1 = 2 and nr2 = 4. Fig. 4.8 shows the DoF region of type 2 with perfect receiver

4.3. DoF region with perfect receiver cooperation when nr1 < nt2 and

nt1 < nr1

4.3

DoF region with perfect receiver cooperation

when n

< n

and n

< n

In this section, we consider the DoF region of type 3. The decoding strategy is the same as previous two sections. Let us start with the case α1 ≥ α2.

When α1 ≥ α2, we treat X2 as noise and decode X1 at receiver 1. The constraint

of d1 is

d1 ≤ α1nt1+ α2(nr1− nt1) − α2min(nt2, nr1) (4.37)

= (α1− α2)nt1. (4.38)

The receiver 2 knows X₁ by the perfect receiver cooperation and can then subtract it off. Thus the constraint of d2 is simply

d2 ≤ α2min(nt2, nr2). (4.39)

Note that we do not know min(nt2, nr2) = nt2 or nr2 here.

Now let us consider the case α2 ≥ α1. In this case, we treat X1as noise and decode

X₂ at receiver 2. But in type 3, nr2 ≤ nt2, nt2 < nr2 < nt1+ nt2 and nt1+ nt2 ≤ nr2

are all possible. So we need to consider these three conditions separately.

1. nr2 ≤ nt2 d2 ≤ α2nr2− α1min(nt1, nr2) (4.40) = α2nr2− α1nt1. (4.41) 2. nt2 < nr2 < nt1+ nt2 d2 ≤ α2nt2+ α1(nr2− nt2) − α1min(nt1, nr2) (4.42) = α2nt2+ α1(nr2− nt1− nt2). (4.43) 3. nt1+ nt2 ≤ nr2 d2 ≤ α2nt2+ α1nt2− α1min(nt1, nr2) (4.44) = α2nt2 (4.45)

Figure 4.9: The DoF region of the DoF region of type 3 with perfect receiver cooper-ation compared to the DoF region with no coopercooper-ation.

The above characterize the DoF region of type 3 with perfect receiver cooperation. In type 3, we cannot have nonzero d1when we achieve the maximum of d2. If we have

perfect receiver cooperation, we can have better d2 when we achieve the maximum of

d1, compared with no cooperation in Section 2.3. Fig. 4.9 shows the DoF region of

type 3, with and without receiver cooperation.

An example of type 3 is a two-user MIMO interference channel with nt1 = 1, nt2=

3, nr1 = 2 and nr2 = 4. Fig. 4.10 shows the DoF region of type 3 with perfect receiver

cooperation by computer calculaton. We can see that if we have perfect receiver cooperation, the DoF region might be larger than the new outer bound of type 3 that we stated in Section 3.3.

4.3. DoF region with perfect receiver cooperation when nr1 < nt2 and

nt1 < nr1

Conclusion

In this thesis, we study the DoF region of the two-user MIMO interference channel without CSIT. By considering the power-split transmission scheme at the transmit-ters, we first view the interference channel as two two-user MAC and develop the DoF for the general two-user MAC. Following [4], we distinguish the two-user MIMO in-terference channel by the values of transmit and receive antennas and consider three types. For type 1 and type 2, we show that the power-split scheme with SIC can achieve the same DoF region as in [4]. For type 3, we reduce the outer bound in [4] by a two-user genie-aided MIMO interference channel. It is seen that for this type the gap between the power-split scheme with SIC and the outer bound is small. Also, from [9] it is seen that the exact DoF region of type 3 in the case of isotropic and independent fading is the same as our power-split scheme with SIC. Finally, we consider the DoF region of two-user MIMO interference channel with perfect re-ceiver cooperation, and compare to the two-user MIMO interference channel without cooperation.

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