Wireless Communications - Lecture slides · Rice distribution Rayleigh distribution Nakagami...

RUHR-UNIVERSITY BOCHUM

Wireless CommunicationsLecture slides

Karlheinz Ochs

Chair of Digital Communication Systems

Communications Systems

Chair ofDigital

Faculty of

Electrical Engineering and

Information Technology

www.dks.rub.de WS 2019/20

Wireless Communications

Contents

1 Motivation

2 Wireless Communication Channel

3 Single Input Single Output Systems

4 Multiple Input Multiple Output Systems

5 Optimal Transmission Strategies

6 Multiple Access Channel

7 X Channel

Lehrstuhl fürDigitale Kommunikationssysteme

K. Ochs Wireless Communications WS 2019/20

Wireless Communications Motivation

Wireless CommunicationsMotivation

Karlheinz Ochs




Motivation

Contents

1 Preliminaries

2 Transmission Scenario

3 Challenges



Motivation Preliminaries

Contents

1 Preliminaries


3 Challenges



Preliminaries 1 / 126

Preliminaries

Trends in communication systems

mobile communication

high data rates

low latency

Constraints on mobile communication systems

expensive and limited bandwidth

limited transmitter signal power

time-variant transfer behavior

Problem-solving approaches

orthogonal frequency-division multiplexing (OFDM)

multiple input multiple output systems

multiple antenna systems

cooperative communication



Motivation Transmission Scenario

Contents

1 Preliminaries


3 Challenges



Transmission Scenario 2 / 126

Transmission scenario

Multipath Propagation Channel

cellular phone

echos

noise

base station

Time-variancemultipath propagation due to mobile objectssample and hold devices, modulators, HF amplifiers, . . .

Transmission conditions are changing with time!Lehrstuhl fürDigitale Kommunikationssysteme


Motivation Challenges

Contents

1 Preliminaries


3 Challenges



Challenges 3 / 126

Challenges

Communication theorie

How to design and synthesize digital communication systems?

Information theorie

What is the maximum data rate of a reliable transmission?

Digital signal processing

What is the optimal processing strategy?

Programable hardware

How can a digital communication system be verified?

Determine and reach the limits of communications!



Wireless Communications Wireless Communication Channel

Wireless CommunicationsWireless Communication Channel

Karlheinz Ochs




Wireless Communication Channel

Contents


2 Passband Transmission

3 Baseband Transmission

4 Time-Discrete Transmission



Wireless Communication Channel Transmission Scenario

Contents








Wireless Communications

Transmission Scenario

cellular phone

echos

noise

base station

Time-variancemultipath propagation due to mobile objectssample and hold devices, modulators, HF amplifiers, . . .

Transmission conditions are changing with time!Lehrstuhl fürDigitale Kommunikationssysteme



Time-Invariant Multipath Propagation

Input-Output Relation

y(t) =n∑ν=0

cνxν(t)

transmitted signal arrives at the receiver on different paths

xν(t) = x(t − Tν)

different durationsTν = T0 + νT

attenuation and change of phase

cν = |cν | e j arccν

x(t) T0 T T

c0 c1 cn−1 cn

y(t)





Impulse Response

definitionx(t) = δ(t − Tx) → y(t) = h(t − Tx)

impulse response of the multipath channel

h(t) =n∑ν=0

cνδ(t − Tν)

input-output relation

y(t) =∫ ∞−∞

h(t − t′)x(t′)dt′

x(t) T0 T T

c0 c1 cn−1 cn

y(t)





Transfer function

definition

x(t) = e jΩx t → y(t) = H(jΩx)e jΩx t

transfer function of the multipath channel

H(jω) =n∑ν=0

cν e−jωTν

input-output relation

Y(jω) = H(jω)X(jω)

x(t) T0 T T

c0 c1 cn−1 cn

y(t)




Time-Variant Multipath Propagation

Input-output Relation

y(t) =n∑ν=0

cν(t)x(t − Tν) , with cν(t) ∈ C

Transfer behavior

x(t) = e jΩx t → y(t) = H(t, jΩx) e jΩx t , with H(t, jω′) =n∑ν=0

cν(t)e−jω′Tν

x(t) T0 T T

c0(t) c1(t) cn−1(t) cn(t)

y(t)

Time-variant transfer behavior!Lehrstuhl fürDigitale Kommunikationssysteme



Time-Variant Multipath Propagation

Stochastic Modeling

y(t) = c(t)x(t) + w(t) , with c(t),w(t) ∈ C

c(t) associated with probability density function

Rice distributionRayleigh distributionNakagami distribution

additive white noise w(t)

Transmission Scheme

x(t)

c(t) w(t)

y(t)digitalsource

digitalmodulator

digitaldemodulator

digitalsink

baseband channel



Wireless Communication Channel Passband Transmission

Contents







Passband Transmission 10 / 126

Passband Transmission

Channel

real

center radian frequency ωc

bandwidth Bc

Bc

−ωc ωc ω

availablefrequency range

availablefrequency range

x0(t) y0(t)source transmitter

transmissionchannel

receiver sink





Transmission Signal

real

center radian frequency ω0

bandwidth Bx

X0(jω)

Bx

−ω0 ω0 ω


transmissionchannel

receiver sink





Transmission Signal

real

bandwidth Bx ≤ Bc

carrier radian frequency ω0

X0(jω) Bc

Bx

−ω0 ω0 ω


transmissionchannel

receiver sink





Equivalent Baseband

channel is complex-valued

transmitter signal x(t) is complex-valued

receiver signal y(t) is complex-valued

X(jω)

Bc

Bxω

x0(t) y0(t)x(t) y(t)digitalsource

digitalmodulator

analogmodulator

transmissionchannel

analogdemodulator

digitaldemodulator

digitalesink

baseband channel



Wireless Communication Channel Baseband Transmission

Contents







Baseband Transmission 14 / 126

Baseband Transmission

Channel Resources

bandwidth Bc

dynamic Dc

duration Tc

x(t) y(t)

dynamic duration

bandwidth

Dc Tc

Bc

digitalsource

digitalmodulator

basebandchannel

digitaldemodulator

digitalsink





Transmitter Signal

bandwidth Bx

dynamic Dx

duration Tx

x(t) y(t)

dynamic dynamicduration

duration

bandwidth bandwidth

Dx DxDcTx Tx Tc

Bx Bx

Bc

digitalsource

digitalmodulator

basebandchannel

digitaldemodulator

digitalsink





Limitations

bandwidth Bx ≤ Bc

dynamic Dx ≤ Dc

duration Tx ≤ Tc

x(t) y(t)

dynamic dynamic dynamicduration

duration

duration

bandwidth bandwidth bandwidth

Dx Dx DxDc DcTx Tx TxTc Tc

Bx Bx Bx

Bc Bc

digitalsource

digitalmodulator

basebandchannel

digitaldemodulator

digitalsink





Symbol Mapping

matching of the signal dynamic Dx ≤ Dc

finite alphabet A

information in symbols u(tk) ∈ A

u(t) x(t) y(t) v(t)


duration

duration



Bx Bx Bx

Bc Bc

digitalsource

impulseshaping

basebandchannel

symbolrecovery

digitalsink

symbolmapping

inversesymbolmapping





Impulse Shaping

matching of the signal bandwidth Bx ≤ Bc

real pulse with finite energy q(t) ∈ R , Eq <∞crucial for symbol recovery

u(t) x(t) y(t) v(t)


duration

duration



Bx Bx Bx

Bc Bc

digitalsource

impulseshaping

basebandchannel

symbolrecovery

digitalsink

symbolmapping

inversesymbolmapping




Transmission Scheme

Transmission Scheme1 Nyquist pulse

z(tk) = u(tk)

2 separation into transmit and receive filter

S(jω) = R(jω)Q(jω)

3 channel with additive white noise

y(t) = x(t) + w(t)

4 optimal signal to noise ratio at decider

r(t) = q(−t)/Eq

u(t)

δT (t− t0)

s(t)z(t)

δT (t− t0)

PAM v(t)

transmitter receiver




Transmission Scheme


z(tk) = u(tk)




y(t) = x(t) + w(t)


r(t) = q(−t)/Eq

u(t)

δT (t− t0)

q(t)x(t)

r(t)z(t)

δT (t− t0)

PAM v(t)

transmitter receiver




Transmission Scheme


z(tk) = u(tk)




y(t) = x(t) + w(t)


r(t) = q(−t)/Eq

u(t)

δT (t− t0)

q(t)x(t)

w(t)

y(t)r(t)

z(t)

δT (t− t0)

PAM v(t)

transmitter additive noise receiver




Transmission Scheme


z(tk) = u(tk)




y(t) = x(t) + w(t)


r(t) = q(−t)/Eq

u(t)

δT (t− t0)

q(t)x(t)

w(t)

y(t) q(−t)

Eq

z(t)

δT (t− t0)

PAM v(t)





Transmission Scheme

Transmission Scheme

5 low-pass band-limited pulse

Bx ≤ Bc

6 moderate timing jitter

|τ | T

u(t)

δT (t− t0)

q(t)x(t)

w(t)

y(t) q(−t)

Eq

z(t)

δT (t− t0)

PAM v(t)





Transmission Scheme

Transmission Scheme

5 low-pass band-limited pulse

Bx ≤ Bc

6 moderate timing jitter

|τ | T

u(t)

δT (t− t0)

q(t)x(t)

w(t)

y(t) q(−t)

Eq

z(t)

δT (t− t0 − τ)

PAM v(t)





Transmission Scheme

Transmission Scheme

7 stochastic model for flat fading and noise

8 minimizing decision error probability

transmitted u(tk) ∈ A

received z(tk − τ) ∈ C

decided v(tk) = Qz(tk − τ) ∈ A

u(t)

δT (t− t0)

q(t)x(t)

c(t) w(t)

y(t) q(−t)

Eq

z(t)

δT (t− t0 − τ)

PAM v(t)

transmitter basebandchannel

receiver




Transmission Scheme

Transmission Scheme

7 stochastic model for flat fading and noise8 minimizing decision error probability

transmitted u(tk) ∈ A

received z(tk − τ) ∈ C

decided v(tk) = Qz(tk − τ) ∈ A

u(t)

δT (t− t0)

q(t)x(t)

c(t) w(t)

y(t) q(−t)

Eq

z(t)

δT (t− t0 − τ)

PAM v(t)


receiver



Wireless Communication Channel Time-Discrete Transmission

Contents







Time-Discrete Transmission 22 / 126

Time-Discrete Channel

Preliminary Considerations

1 perfect timing synchronization

τ = 0

2 no receiver filter, transmitter filter is Nyquist filter

x(tk) = u(tk)

3 time-discrete channel

y(tk) = c(tk)x(tk) + w(tk)

4 decision

v(tk) = Qy(tk)

u(t)

δT (t− t0)

q(t)x(t)

c(t) w(t)

y(t) q(−t)

Eq

z(t)

δT (t− t0)

PAM v(t)


receiver







τ = 0


x(tk) = u(tk)



4 decision

v(tk) = Qy(tk)

u(t)

δT (t− t0)

s(t)x(t)

c(t) w(t)

y(t)

δT (t− t0)

PAM v(t)


receiver







τ = 0


x(tk) = u(tk)



4 decision

v(tk) = Qy(tk)

replacements

u(t)

δT (t− t0)

s(t)x(t)

c(t) w(t)

y(t)

δT (t− t0)

PAM v(t)

time-discrete channel







τ = 0


x(tk) = u(tk)



4 decision

v(tk) = Qy(tk)

u(tk) = x(tk)

c(tk) w(tk)

y(tk)v(tk)

time-discrete channel







Communication scenariodynamic du

ration

bandwidth

DxDc Tx Tc

Bx

Bc

u(tk)x(tk)

c(tk)w(tk)

y(tk)v(tk)Tx Rx

Find an optimal strategy to exploit the communication resources!Lehrstuhl fürDigitale Kommunikationssysteme






Communication limits

U

equivo

cation

mutual information

irrelev

ance

V

u(tk)x(tk)

c(tk)w(tk)

y(tk)v(tk)Tx Rx

Use information theory to determine the communication limits!Lehrstuhl fürDigitale Kommunikationssysteme


Wireless Communications Single Input Single Output Systems

Wireless CommunicationsSingle Input Single Output Systems

Karlheinz Ochs




Single Input Single Output Systems

Contents

1 Signal Space

2 AWGN Channel

3 Flat Fading Channel



Single Input Single Output Systems Signal Space

Contents

1 Signal Space

2 AWGN Channel




Signal Space 25 / 126

Signal Space

Digital Modulator

u(t)digital

modulatorx(t)

1 Retrieve transmitted symbols

x(tk) = u(tk) , with tk = t0 + kT

2 Nyquist criterion

s(kT) =

1 for k = 00 for k ∈ Z \ 0

3 Minimal bandwidth

s(t) = si (Ωt/2) , with ΩT = 2π




Signal Space

Digital Modulator

u(t)

δT (t− t0)

s(t) x(t)



2 Nyquist criterion

s(kT) =

1 for k = 00 for k ∈ Z \ 0

3 Minimal bandwidth





Signal Space

Digital Modulator

u(t)

δT (t− t0)

si(Ωt2

)x(t)



2 Nyquist criterion

s(kT) =

1 for k = 00 for k ∈ Z \ 0

3 Minimal bandwidth





Signal Space

Digital Modulator

u(t)

δT (t− t0)

si(Ωt2

)x(t)



2 Nyquist criterion

s(kT) =

1 for k = 00 for k ∈ Z \ 0

3 Minimal bandwidth


Signaling with Nyquist rate!Lehrstuhl fürDigitale Kommunikationssysteme



Signal Space

Digital modulator

u(t)

δT (t− t0)

si(Ωt2

)x(t)

Digitally modulated signal

x(t) =∞∑

k=−∞

u(tk) si

(Ω

2[t − tk]

)




Signal Space

Digital modulator

u(t)

δT (t− t0)

si(Ωt2

)x(t)

Digitally modulated signal

x(t) =∞∑

k=−∞

xkϕk(t)

Definitions

1 samples

xk = u(tk)

2 base functions

ϕk(t) = si

(Ω

2[t − tk]

)−−• Φk(jω) = T rect

(2ωΩ

)e−jωtk




Signal Space

Scalar Product

〈ϕk(t), ϕ`(t)〉 =∫ ∞

−∞ϕk(t)ϕ∗` (t)dt

Orthogonal base functions

〈ϕk(t), ϕ`(t)〉 = T

1 for k = `0 for k 6= `

Proof

1 〈ϕk(t), ϕ`(t)〉 =∫ ∞−∞ ϕk(t)ϕ∗` (t)dt

2 〈ϕk(t), ϕ`(t)〉 = 12π

∫ ∞−∞ Φk(jω)Φ∗` (jω)dω

3 〈ϕk(t), ϕ`(t)〉 = 12π

∫ ∞−∞ T2 rect

( 2ωΩ

)e−jω[tk−t`]dω

4 〈ϕk(t), ϕ`(t)〉 = T 1Ω

∫ Ω/2−Ω/2 e−jω[k−`]T dω




Signal Space

Energy

Ex =

∫ ∞

−∞|x(t)|2dt = T

∞∑k=−∞

|xk|2

Proof

1 Ex = 〈x(t), x(t)〉 = ‖x(t)‖2

2 Ex = 〈∞∑

k=−∞xkϕk(t),

∞∑`=−∞

x`ϕ`(t)〉

3 Ex =∞∑

k=−∞

∞∑`=−∞

xkx∗` 〈ϕk(t), ϕ`(t)〉

4 Ex = T∞∑

k=−∞|xk|2




Signal Space

Signal Vector (finite number of symbols)

x =[

x1, x2, . . . , xK]T

power

Px =Ex

T= ‖x‖2 , with ‖x‖2 = xHx =

K∑k=1

|xk|2

law of large numbers

1K

K∑k=1

|xk|2 ≈ E|X |2

relation to stochastic power

Px ≈ KPx , with Px = E|X |2Lehrstuhl fürDigitale Kommunikationssysteme


Single Input Single Output Systems AWGN Channel

Contents

1 Signal Space

2 AWGN Channel




AWGN Channel 30 / 126

Channel with Additive White Gaussian Noise

AWGN Channel

U

equivo

cation

mutual information

irrelev

ance

V

u(k)x(k)

1z(k)

y(k)v(k)Tx Rx

Remarkstransmitter sends message U to the receiver

K channel uses: Tx = KT ≤ Tc , k ∈ 1, . . . ,K

transmitter signal x(k) ∈ C with limited power Px ≤ P

additive noise z(k) ∈ C

independent and identically distributed

normal distribution, with zero mean and variance σ2z = Pz

receiver signal y(k) ∈ C





AWGN Channel

u(k)x(k)

1z(k)

y(k)v(k)Tx Rx

Remarks

transmitter sends message U to the receiver

K channel uses: Tx = KT ≤ Tc , k ∈ 1, . . . ,K


additive noise z(k) ∈ Cindependent and identically distributed







AWGN Channel

Signal flow diagram

x(k)

z(k)

y(k)

Mathematical model

y(k) = x(k) + z(k) , 1 ≤ k ≤ K

Px =1K

K∑k=1

|x(k)|2 ≤ P

z(k) ∼ N (0,Pz)

Communication

encoding of message U

x(k) ∈ C is a symbol of a finite alphabet A

y(k) ∼ N (x(k),Pz)

Highest data rate for a reliable transmission?Lehrstuhl fürDigitale Kommunikationssysteme




AWGN Channel

Signal flow diagram

x

z

y

Mathematical model

y = x+ z

‖x‖ =√

KPx

‖z‖ =√

KPz

Remarks

transmitted signal vector x =[

x(1), . . . , x(K)]T

noise vector z =[

z(1), . . . , z(K)]T

received signal vector y =[

y(1), . . . , y(K)]T

Received vector y lies in a hypersphere with center x and radius√

KPz !Lehrstuhl fürDigitale Kommunikationssysteme




AWGN Channel

Illustration

xz

y0

noise hypersphere

Mathematical model

y = x+ z

E‖y‖2 = E‖x‖2+ E‖z‖2x and z are independent

Remarks

E‖y‖ ≤√E‖y‖2 , Jensen’s inequality

E‖y‖2 = E‖x‖2+ E‖z‖2 , independence

E‖x‖2 ≈ KPx , E‖z‖2 ≈ KPz , law of large numbers





AWGN Channel

Illustration

√ KPx

√KPz

E‖y‖≤√

K[Px +Pz]

noise hypersphere

Mathematical model

y = x+ z

E‖y‖2 = E‖x‖2+ E‖z‖2x and z are independent

Upper bound

E‖y‖ ≤√

K[Px + Pz]

All received signal vectors lie in a hypersphere of radius√

K[Px + Pz]!





AWGN Channel

Illustration

1

2

3

M

√KPz

√KPz

√KPz

√KPz

√K[Px + Pz ]

Decoding

transmitted vector x is center of thehypersphere

M nonoverlapping hypersheres

received vector y within hypershperebelongs to the center x

How many noise hyperspheres fit into the received vector hypersphere?





AWGN Channel

Hypersphere

K (real) dimensions

radius r

volume VK(r) ∼ rK

Nonoverlapping hyperspheres

K complex dimensions

2K real dimensions

Upper bound for nonoverlapping hyperspheres

M ≤ [1 + ΓSNR]K , with ΓSNR =

Px

Pz=E|x(k)|2E|z(k)|2

Proof

M ≤ V2K(√

2K[Px + Pz])

V2K(√

2KPz)=

VK(2K[Px + Pz])

VK(2KPz)= [1 + Px/Pz]

K





AWGN Channel

Upper bound for the data rate

R =ld(M)

K≤ ld (1 + ΓSNR)

M different messages can be reliably distinguished

each message can be encoded with ld(M) bits

data rate R is the ratio of bits per channel use

Maximum data rate for a reliable communication?





AWGN Channel


R =ld(M)

K≤ ld (1 + ΓSNR)





AWGN channel capacity

C = ld(1 + ΓSNR) in bits/channel use





AWGN Channel


R =ld(M)

K≤ ld (1 + ΓSNR)





AWGN channel capacity

C = ld(1 + ΓSNR) in bits/channel use

This is only a heuristic approach!Information theory provides a fundamental derivation!



Single Input Single Output Systems Flat Fading Channel

Contents

1 Signal Space

2 AWGN Channel




Flat Fading Channel 37 / 126

Flat Fading Channel

Flat Fading Channel

u(k)x(k)

h(k)z(k)

y(k)v(k)Tx Rx

Remarks


channel gain h(k) ∈ Csmall-scale fading caused by echoes of the transmitted signaltransmitted signal period is larger than multi-path delay spread

additive noise z(k) ∈ Cindependent and identically distributed (i.i.d.)



Channel capacity?Lehrstuhl fürDigitale Kommunikationssysteme



Flat Fading Channel

Flat Fading Channel

x(k)

h(k) z(k)z(k)

y(k)y(k) h(k)x(k)

Idea

equivalent to AWGN channel

surrogate input signal is product of input signal times channel gain

time-variant signal to noise ratio

γSNR(k) =E|h(k)x(k)|2E|z(k)|2 = ΓSNR|h(k)|2

Channel capacity

C(k) = ld(1 + ΓSNR|h(k)|2) , with ΓSNR =E|x(k)|2E|z(k)|2 =

Px

Pz




Outage Probability

Transmission system is in outage

ld

(1 +

Px,fading

Pz|h(k)|2

)< R0 ,

with target rate R0

Outage Probability

Pout(R0) = Pld

(1 +

Px,fading

Pz|h(k)|2

)< R0

Power Penalty

ηp =Px,fading

Px,AWGN

Extra effort to achieve a certain outage probability for a target rate!Lehrstuhl fürDigitale Kommunikationssysteme



Flat Fading Channel

Rayleigh Fading

Channel gain

distribution of real and imaginary part

Reh, Imh ∼ N (0, σ2)

transformation to polar coordinates h = re jϕ

magnitude r > 0 has Rayleigh distribution

fr(r) = u(r)rσ2 e−

r2

2σ2

phase ϕ ∈ (−π, π] has uniform distribution

fϕ(ϕ) =1

2πrect

(ϕπ

)Real- and imaginary part have zero-mean, which indicates no line of sight!




Flat fading Channel

Rayleigh Distribution (σ = 1)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

r

f r(r)




Flat Fading Channel

Rician Fading

Channel gain

distribution of real and imaginary part

Reh ∼ N (µRe, σ2) , Imh ∼ N (µIm, σ

2)

transformation to polar coordinates h = re jϕ

magnitude r > 0 has Rice distribution

fr(r) = u(r)rσ2 I0

( rµσ2

)e−

r2+µ2

2σ2 , with µ =√µ2

Re + µ2Im

and modified Bessel function of the first kind with order zero I0

Rician factor K =µ

2σ2

K = 0 yields Rayleigh fading

Line of sight if Rician factor is greater than zero!Lehrstuhl fürDigitale Kommunikationssysteme



Flat Fading Channel

Rician Distribution (σ = 1)

0 1 2 3 4 5 6 7 80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

r

f r(r)

µ = 0µ = 0.5µ = 1µ = 2µ = 4




Flat Fading Channel

Nakagami Fading

Channel gainsum of multiple i.i.d. Rayleigh fading signals has Nakagami distributedmagnitudemagnitude r > 0 has Nakagami distribution

fr(r) = u(r)2

Γ (m)

[mΩ

]mr2m−1 e−

mr2Ω ,

with

gamma function Γ (m) =

∫ ∞0

e−rrm−1dr

received signal average power Ω = Er2

shape factor m =Ω2

E[r −Ω]2≥

12

m = 1 yields Rayleigh fading

Useful to model urban radio multipath channels!Lehrstuhl fürDigitale Kommunikationssysteme



Flat Fading Channel

Nakagami Distribution

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 30

0.2

0.4

0.6

0.8

1

1.2

1.4

r

f r(r

)

Ω = 1,m = 0.5Ω = 1,m = 1Ω = 2,m = 1Ω = 3,m = 1Ω = 1,m = 2Ω = 2,m = 2Ω = 1,m = 3



Wireless Communications Multiple Input Multiple Output Systems

Wireless CommunicationsMultiple Input Multiple Output Systems

Karlheinz Ochs




Multiple Input Multiple Output Systems

Contents


2 MIMO Detectors

3 Random Channels

4 Eigenmode Decomposition

5 Capacity and Degrees of Freedom



Multiple Input Multiple Output Systems Transmission Scenario

Contents


2 MIMO Detectors

3 Random Channels






MIMO Passband Transmitter

MIMO Passband Transmitter

u(t)

S

P

symbolmapping

symbolmapping

δT (t− t0)

δT (t− t0)

q(t)

q(t)

ejω0t

ejω0t

Re

Re

u1(t)

um(t)

x1(t)

xm(t)

x10(t)

xm0(t)




MIMO Passband Receiver

MIMO Passband Receiver

PAM

PAM

e−jω0t

e−jω0t

LP

LP

r(t)

r(t)

δT (t− t0)

δT (t− t0)

detection

v(t)

w10(t)

wn0(t)

y10(t)

yn0(t)

y1(t)

yn(t)

v1(t)

vn(t)

Instead of single decisions a combined detection of transmitted symbols!Lehrstuhl fürDigitale Kommunikationssysteme



MIMO passband transmission

MIMO Passband Transmitterm transmit antennas

symbol vector u(tk) ∈ Am , with u(t) =[

u1(t), . . . , um(t)]T

equivalent low-pass signal vector x(t) =∞∑

k=−∞

u(tk)q(t − tk)

baseband signal vector x0(t) = Rex(t)e jω0t

,

with x0(t) =[

x10(t), . . . , xm0(t)]T

MIMO Passband Receivern receive antennas

additive noise w0(t) =[

w10(t), . . . ,wn0(t)]T

baseband signal y0(t) =[

y10(t), . . . , yn0(t)]T




MIMO Baseband Transmission

MIMO Baseband Transmission

AM

AM

u(t)

S

P

P

P

symbolmapping

symbolmapping

δT (t− t0)

δT (t− t0)

δT (t− t0)

δT (t− t0)

q(t)

q(t)

u1(t)

um(t)

x1(t)

xm(t)

c11(t)

cn1(t)

c1m(t)

cnm(t)

r(t)

r(t)

detection

v(t)

w1(t)

wn(t)

y1(t)

yn(t)

v1(t)

vn(t)




MIMO Baseband Transmission (simplified)

MIMO Baseband Transmission (simplified)

AM

AM

u(t)

S

P

P

P

symbolmapping

symbolmapping

δT (t− t0)

δT (t− t0)

δT (t− t0)

δT (t− t0)

s(t)

s(t)

u1(t)

um(t)

x1(t)

xm(t)

h11(t)

hn1(t)

h1m(t)

hnm(t)

detection

v(t)

z1(t)

zn(t)

y1(t)

yn(t)

v1(t)

vn(t)




MIMO Digital Baseband Transmission


u(k)

x1(k)

xm(k)

h11(k)

hn1(k)

h1m(k)

hnm(k)

z1(k)

zn(k)

y1(k)

yn(k)

v(k)Tx Rx

Flat fading channel

y1(k)...

yn(k)

=

h11(k) · · · h1m(k)...

. . ....

hn1(k) · · · hnm(k)

x1(k)

...xm(k)

+

z1(k)...

zn(k)






u(k) Txx(k)

H(k)z(k)

y(k)Rx v(k)

Transmitter

maps message U to signal x(k)

transmitter signal has limited power

Flat fading channel

y(k) =H(k)x(k) + z(k)

channel matrix has almost sure full rankReceiver

knows the channel state from estimationretrieves message V from signal y(k)

Channel capacity?Lehrstuhl fürDigitale Kommunikationssysteme


Multiple Input Multiple Output Systems MIMO Detectors

Contents


2 MIMO DetectorsZero-Forcing DetectorMinimum Mean-Squared Error DetectorMaximum Likelihood Detector

3 Random Channels





MIMO Detectors 53 / 126

MIMO Detectors

MIMO Detector

x(k)

H(k) z(k)

y(k) MIMO-detector

x(k)

Remarks

m transmit and n receive antennas

flat fading channel

y(k) =H(k)x(k) + z(k)

channel matrix has (almost sure) full rank

H(k) ∈ Cn×m , with rankH(k) = minm, n

MIMO detector estimates transmitted signalLehrstuhl fürDigitale Kommunikationssysteme


MIMO Detectors Zero-Forcing Detector 54 / 126

Zero-Forcing Detector

Scenario 1same number of antennas at transmitter and receiverrank(H(k)) = m = n

equation system

x(k)H(k) z(k)y(k) +=

Zero-Forcing Detection

x =H−1y ⇒ x = x+H−1z

Remarkssimple case

multiplication with H−1 can significantly amplify the noiseLehrstuhl fürDigitale Kommunikationssysteme




Scenario 2

more transmit antennas than receive antennas

rank(H(k)) = n < m

m− n times underdetermined equation system

x(k)H(k) z(k)z(k)y(k) +=


x =H sy + [1−H sH]v , v arbitrary ⇒ Hx =Hx+HH sz





Remarks

semi-inverse H s , with HH sH =H and H sHH s =H s

Moore-Penrose right pseudoinverse can be used

H s =H+ =HH[HHH]−1

1−H sH is projection matrix to null space of H

Hx =HH sy

multiplication with HH s can significantly amplify the noise

Improper approach because of infinite many solutions!

Remedy

time variance of the channel is helpful

sent x again to increase number of linearly independent equationsLehrstuhl fürDigitale Kommunikationssysteme




Scenario 3

more receive antennas than transmit antennas

rank(H(k)) = m < n

n− m times overdetermined equation system

x(k)H(k) z(k)z(k)y(k) +=


x =H+y ⇒ x = x+H+z





Remarks

Moore-Penrose left pseudoinverse H+ = [HHH]−1HH

multiplication with H+ can significantly amplify the noise

Solution is an optimal approximation!

Optimization problem

x = argminxJ

, with J = ‖y −Hx‖2





Solution

necessary and sufficient conditions

∂J∂x

= 0T and∂J∂xH = 0

Wirtinger derivatives, x and xH independent

J = J∗ ⇒ ∂J∂x

=

[∂J∂xH

]H

,∂J∂x

= 0T ⇔ ∂J∂xH = 0

J = [yH − xHHH][y −Hx]

∂J∂xH = −HH[y −Hx]

HHy =HHHx

x =[HHH

]−1HHy



MIMO Detectors Minimum Mean-Squared Error Detector 60 / 126

Minimum Mean-Squared Error Detector

Detection

x =Dy , with y =Hx+ z

A minimum mean-squared error detector considers noise!

Optimization problem

D = argminDJ

, with J = E‖x− x‖2

Approach

J = J∗

necessary and sufficient conditions

∂J∂D

= 0T and∂J∂DH = 0

Wirtinger derivatives, D and DH independentLehrstuhl fürDigitale Kommunikationssysteme




Realness

J = J∗

necessary and sufficient condition

∂J∂DH = 0

Some basics

‖x‖2 = xHx = tr(xxH)

Kxy = ExyH

E‖x‖2

= tr (Kxx)

∂

∂Mtr (AMB) = ATBT





Error reformulation

x− x = x−Dy

[x−Dy][xH − yHDH] = xxH −DyxH +DyyHDH − xyHDH

J = tr(Kxx −DKyx +DKyyDH −KxyD

H)

J = tr(Kxx)− tr(DKyx) + tr(DKyyDH)− tr(KxyD

H)

Solution

∂J∂DH = [DKyy]

T −KTxy

DKyy =Kxy , regularity of Kyy is assumed!

Minimum Mean-Squared Error Detection

x = Dy , with D =KxyK−1yy





Uncorrelation

x,z uncorrelated

Kxy = Ex[xHHH + zH] =KxxH

H

Kyy = E[Hx+ z]

[xHHH + zH] =Kzz +HKxxH

H

Detection matrix

D =KxxHH [Kzz +HKxxH

H]−1

Minimum Mean-Squared Error Detection

x =KxxHH [Kzz +HKxxH

H]−1[Hx+ z]

Solution plausible?





Simplifying assumptions

1 Symbols and noise are spatially uncorrelated

Kxx = Px1 , Kzz = Pz1 , with Pz = σ2z

Detection matrix

D =HH[HHH +

1ΓSNR

1]−1

, with ΓSNR =Px

Pz

2 No noise

z = 0 , Pz = 0 ,1

ΓSNR= 0 , Kyy = PxHH

H

Detection matrix

D =H+ , with H+ =HH [HHH]−1

This MMSE detector is weak for high SNR and m < n!Lehrstuhl fürDigitale Kommunikationssysteme


MIMO Detectors Maximum Likelihood Detector 65 / 126

Maximum Likelihood Detector

Stochastic Channel Model

y =Hx+ z , with known fy|x(y|x)

Detection

x = argmaxx∈Am fy|x(y|x)

Interpretation

fy|x(y|x) ≥ fy|x(y|x) for all x ∈ Am

A maximum likelihood detector chooses the most likely sent x!





Reformulation

z = y −Hx

fy|x(y|x) = fz(y −Hx)

Maximum Likelihood Detection

x = argmaxx∈Am fz(y −Hx)

Simplifying assumptions

1 circularly symmetric complex Gaussian random variables

fz =1

det(πKzz)exp

(−zHK−1

zzz)

2 spatially uncorrelated noise

Kzz = Pz1 , with Pz = σ2z





Intermediate results

fz =1

[πσ2z ]n

exp

(− 1σ2

z‖z‖2

)

x = argmaxx∈Am

1

[πσ2z ]n

exp

(− 1σ2

z‖y −Hx‖2

)Exponential function is strictly monotonically increasing!

Maximum Likelihood Detection

x = argminx∈Am

‖y −Hx‖

Remarksgeometrical task to find x ∈ Am, such that Hx has minimal distance to ythere are |A|m possible vectors x ∈ Am

Effort increases exponentially with the number of transmit antennas!Lehrstuhl fürDigitale Kommunikationssysteme


Multiple Input Multiple Output Systems Random Channels

Contents


2 MIMO Detectors

3 Random ChannelsSpatially Uncorrelated ChannelSpatially Correlated Channel





Random Channels 68 / 126

Random MIMO Channel

Random MIMO Channel

x(k)

H(k) z(k)

y(k)

flat fading (frequency-non-selective)

channel matrix H(k) has (almost sure) full rank

elements hµν(k) of H(k) are random variables

Random channel matrix!

Example

no line of sight between transmit antenna µ and receive antenna ν

modeled e. g. with Rayleigh-distributed |hνµ(k)|Lehrstuhl fürDigitale Kommunikationssysteme


Random Channels Spatially Uncorrelated Channel 69 / 126

Spatially Uncorrelated Channel

Independent and Identically Distributed Random MIMO Channel

Elements hµν(k) of channel matrix

independent and identically distributed random variableszero-mean

Ehµν(k) = 0

spatially uncorrelated

Ehνµ(k)hλκ(k) =σ2 for ν = λ and µ = κ0 otherwise

Channel Correlation Matrix

KHH = σ21 , with KHH = E

vec(H)vec(H)H =KHHH

Spatially uncorrelated MIMO channels are of minor practical importance!Lehrstuhl fürDigitale Kommunikationssysteme


Random Channels Spatially Correlated Channel 70 / 126

Spatially Correlated Channel

Dense antenna array

improved directional characteristiccorrelation between antenna signals


x(k)

√Kx(k) HIID(k)

√Ky(k)

Hz(k)

y(k)

at transmitter Kx =KHx ≥ 0 , with Kx =

√Kx

H√Kx

at receiver Ky =KHy ≥ 0 , with Ky =

√Ky

H√Ky

Random channel matrix

H =√Ky

HHIID√Kx



Random Channels Spatially Correlated Channel 71 / 126


Intermediate calculations

vec(H) = vec(√

KyHHIID√Kx

)=[√Kx

T ⊗√Ky

H]

vec(HIID)

vec(H)H = vec(HIID)H[√Kx∗ ⊗

√Ky

]Derivation of the channel correlation matrix

KHH =[√Kx

T ⊗√Ky

H]E

vec(HIID)vec(HIID)H [√Kx

∗ ⊗√Ky

]= σ2

[√Kx

T ⊗√Ky

H] [√

Kx∗ ⊗

√Ky

]= σ2

[√Kx

T√Kx∗]⊗[√Ky

H√Ky

]= σ2

[√Kx

H√Kx

]∗⊗[√Ky

H√Ky

]Channel Correlation Matrix

KHH = σ2K∗x ⊗Ky



Multiple Input Multiple Output Systems Eigenmode Decomposition

Contents


2 MIMO Detectors

3 Random Channels

4 Eigenmode DecompositionSingular Value DecompositionEigenmodes of a MIMO Channel




Eigenmode Decomposition Singular Value Decomposition 72 / 126

Singular Value Decomposition

Singular Value Decomposition

H = UΣV H =[U1 U2

] [ Σr 00 0

] [V H

1

V H2

],

with

H ∈ Cn×m , r = rank(H) ≤ minm, n

U ∈ Cn×n , UHU = UUH = 1n , U1 ∈ Cn×r , U2 ∈ Cn×n−r

V ∈ Cm×m , V HV = V V H = 1m , V1 ∈ Cm×r , V2 ∈ Cm×m−r

Σ ∈ Cn×m , Σr = diag(σ1, . . . , σr) > 0

For every matrix H ∈ Cn×m with rank r there exists asingular value decomposition with positive singular values σ1, . . . , σr!



Eigenmode Decomposition Eigenmodes of a MIMO Channel 73 / 126

Eigenmodes of a MIMO Channel

Scenario

PSfrag

x(k)Pre-

Encoder

x′(k)

H(k) z′(k)

y′(k)Post-

Encodery(k)

Transmitter has channel state information (CSIT)!

Channel

y′ =Hx′ + z′

Encoders

use encoders to decouple transmission paths

exploit singular value decomposition

H = UΣV H ⇔ UHHV = Σ





Scenario

PSfrag

x(k)

V (k)

x′(k)

H(k) z′(k)

y′(k)

UH(k)

y(k)

Encoders

pre-encoding

x′ = V x

post-encoding

y = UHy′

Encoded channel

y = UH [HV x+ z′]





Scenario

x(k)

V (k)

x′(k)

H(k) UH(k) UH(k)z′(k)

y(k)

Equivalent channel

y = UHHV x+UHz′





Scenario

x(k)

Σ(k) z(k)

y(k)

Equivalent channel

y = Σx+ z ,

with

encoded channel

Σ = UHHV

encoded noise

z = UHz′ , with ‖z‖2 =∥∥z′∥∥2

Noise power conserved after unitary transformation!Lehrstuhl fürDigitale Kommunikationssysteme




Scenario

x(k)

Σ(k) z(k)

y(k)

Equivalent channel

y = Σx+ z

[yr

yn−r

]=

[Σr 00 0

] [xr

xm−r

]+

[zr

zn−r

]

yν = σνxν + zν for ν = 1, . . . , ryν = zν for ν = r + 1, . . . , n

Encoding yields r (relevant) parallel SISO channels!





Unitary transformation

Noise with circularly-symmetric and zero mean complex normal distribution

z = UHz′

z′ ∼ N (0,Kz′z′) i. e. f ′z(z) =exp

(−zHK−1

z′z′z)

|πKz′z′ |

z ∼ N (0,Kzz) , Kzz = UHKz′z′U

In addition, spatially uncorrelated with identical power

Kz′z′ = σ2z 1

z ∼ N(

0, σ2z 1)

, Kzz =Kz′z′

Stochastic properties conserved after unitary transformation!Lehrstuhl fürDigitale Kommunikationssysteme


Multiple Input Multiple Output Systems Capacity and Degrees of Freedom

Contents


2 MIMO Detectors

3 Random Channels


5 Capacity and Degrees of FreedomCapacityDegrees of Freedom



Capacity and Degrees of Freedom Capacity 77 / 126

Capacity of SISO Channels

Capacity of SISO Channels

AWGN channel

x(k)

z(k)

y(k)

C = ld(1 + ΓSNR) , with ΓSNR =Px

Pz=E|x(k)|2E|z(k)|2

Flat fading channel

x(k)

h(k) z(k)

y(k)

C(k) = ld(1 + ΓSNR|h(k)|2)

Capacity of a MIMO channel?Lehrstuhl fürDigitale Kommunikationssysteme



Capacity of MIMO Channels

Capacity of a Decomposed MIMO Channel

Parallel SISO channels

x (k)

σ (k) z(k)

y(k)

y%(k) = σ%(k) x%(k) + z%(k) for % = 1, . . . , r

Simplifying assumption

ΓSNR =E|x%(k)|2E|z%(k)|2

Capacity

C(k) =r∑

%=1

C%(k) , with C%(k) = ld(

1 + ΓSNRσ2% (k)

)Reformulation independent from singular value decomposition?





Capacity of a Decomposed MIMO Channel

C(k) =r∑

%=1

ld(

1 + ΓSNRσ2% (k)

)

Refomulation

1r∑

%=1ld(ξ%) = ld

(r∏

%=1ξ%

)

C(k) = ld

(r∏

%=1

[1 + ΓSNRσ

2% (k)

])

2r∏

%=1ξ% = |diag(ξ%)|

C(k) = ld(∣∣∣diag

(1 + ΓSNRσ

2% (k)

)∣∣∣)Lehrstuhl fürDigitale Kommunikationssysteme




Refomulation

3 diag(1 + ξ%) = 1 + diag(ξ%) , diag(αξ%) = α diag(ξ%) ,

C(k) = ld(∣∣∣1 + ΓSNR diag

(σ2% (k)

)∣∣∣)4 diag(ξ2

%) = diag(ξ%)2

C(k) = ld(∣∣∣1 + ΓSNRΣ

2r (k)

∣∣∣)

5

∣∣∣∣[ A 00 1

]∣∣∣∣ = |A|C(k) = ld

(∣∣1m + ΓSNRΣH(k)Σ(k)

∣∣)6 HHH = V ΣHΣV H i. e. ΣHΣ = V HHHHV

C(k) = ld(∣∣V H(k)

[1m + ΓSNRH

H(k)H(k)]V (k)

∣∣)Lehrstuhl fürDigitale Kommunikationssysteme




Refomulation

7 |AB| = |A| |B| ,∣∣A−1

∣∣ = |A|−1

C(k) = ld(∣∣1m + ΓSNRH

H(k)H(k)∣∣)

8 |1n +AB| = |1m +BA| for A ∈ Cn×m , B ∈ Cm×n

C(k) = ld(∣∣1n + ΓSNRH(k)HH(k)

∣∣)Hint

m ≤ n

C(k) = ld(∣∣1m + ΓSNRH

H(k)H(k)∣∣)

n ≤ m


∣∣)No need for singular value decomposition!





Capacity of a MIMO Channel

C(k) = maxKxx

ld(∣∣Kzz +HKxxH

H∣∣)− ld (|Kzz|)

s. t. trace(Kxx) ≤ P

This is a (convex) optimization problem!

Special case

Kxx = Px1 , Kzz = σ2z 1 , ΓSNR =

Px

σ2z


∣∣)Capacity of r parallel SISO channels with constant signal to noise ratio!



Capacity and Degrees of Freedom Degrees of Freedom 83 / 126

Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)

average of symbols per channel use

synonymous DoF

closely related to multiplexing gain

SISO channel

η = 1

MIMO channel

η = r

The degrees of freedoms are equal to the rank of the channel matrix!Lehrstuhl fürDigitale Kommunikationssysteme



Degrees of Freedom

Proof

1 SISO is a special case of MIMO with r = 1

2 η = limΓSNR→∞

r∑%=1

ld(1+σ2%ΓSNR)

ld(ΓSNR)=

r∑%=1

limΓSNR→∞

ld(1+σ2%ΓSNR)

ld(ΓSNR)

3 limΓSNR→∞

ld(1+σ2%ΓSNR)

ld(ΓSNR)= limΓSNR→∞

ld(σ2%ΓSNR)

ld(ΓSNR)

4 limΓSNR→∞

ld(1+σ2%ΓSNR)

ld(ΓSNR)= limΓSNR→∞

[ld(σ2

%)

ld(ΓSNR)+ 1]

5 limΓSNR→∞

ld(1+σ2%ΓSNR)

ld(ΓSNR)= 1

6 η =r∑

%=11 = r




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)

Interpretation

η = limΓSNR→∞

CCSISO

, with CSISO = ld(1 + |σ|2ΓSNR)

Multiplexing Gain

C ≈ ηCSISO for ΓSNR →∞

Asymptotic measurement for the high signal to noise ratio regime!



Wireless Communications Optimal Transmission Strategies

Wireless CommunicationsOptimal Transmission Strategies

Karlheinz Ochs




Optimal Transmission Strategies

Contents

1 Maximum Ratio Combining

2 Maximum Ratio Transmission

3 Water-Filling



Optimal Transmission Strategies Maximum Ratio Combining

Contents



3 Water-Filling



Maximum Ratio Combining 86 / 126

Maximum Ratio Combining

Scenario

w(k) Txx(k)

h1(k)

hn(k)

z1(k)

zn(k)

y1(k)

yn(k)

Rx w(k)

Transmitter has no channel state information (no CSIT)

Channel

y(k) = h(k)x(k) + z(k) , with Kzz = σ2z 1





Strategy

x(k) = hH(k)y(k)

Channel with strategy

x(k)

‖h(k)‖2 hH(k)z(k)

x(k)

x(k) = ‖h(k)‖2x(k) + hH(k)z(k)

Maximum achievable data rate?





Signal to noise ratio

γSNR = E|‖h‖2x|2E|hHz|2 = ‖h‖4E|x|2

EhHzzHh = ‖h‖4PxhHKzzh

= ‖h‖4Pxσ2

zhHh

= ‖h‖2ΓSNR

Achievable data rate

R(k) ≤ Rmax(k) = ld(

1 + ‖h(k)‖2ΓSNR

)Capacity of the (MIMO) channel

C(k) = ld(1 + hH(k)h(k)ΓSNR

)Performance of strategy

Rmax(k) = C(k)

Maximum ratio combining is optimal!Lehrstuhl fürDigitale Kommunikationssysteme


Optimal Transmission Strategies Maximum Ratio Transmission

Contents



3 Water-Filling



Maximum Ratio Transmission 89 / 126

Maximum Ratio Transmission

Scenario

w(k) Tx

x1(k)

xm(k)

h1(k)

hm(k)

z(k)

y(k)Rx w(k)

feed-back channel

Transmitter has perfect channel state information (CSIT)

Channel

y(k) = hH(k)x(k) + z(k) , with Pz = σ2z





Strategy

x(k) =h(k)‖h(k)‖w(k)

Channel with strategy

w(k)

‖h(k)‖ z(k)

y(k)

y(k) = ‖h(k)‖w(k) + z(k)

Maximum achievable data rate?





Signal to noise ratio

γSNR = E|‖h‖w|2E|z|2 = ‖h‖2E|w|2

σ2z

=‖h‖2E

∥∥∥ h‖h‖ w

∥∥∥2

σ2z

=‖h‖2E‖x‖2

σ2z

= ‖h‖2ΓSNR

Achievable data rate

R(k) ≤ Rmax(k) = ld(

1 + ‖h(k)‖2ΓSNR

)Capacity of the (MIMO) channel

C(k) = ld(1 + hH(k)h(k)ΓSNR

)Performance of strategy

Rmax(k) = C(k)

Maximum ratio transmission is optimal!Lehrstuhl fürDigitale Kommunikationssysteme


Optimal Transmission Strategies Water-Filling

Contents



3 Water-Filling



Water-Filling 92 / 126

Water-Filling

Scenario

w(k)

x1(k)

xm(k)

h11(k)

hn1(k)

h1m(k)

hnm(k)

z1(k)

zn(k)

y1(k)

yn(k)

w(k)Tx Rx

feed-back channel


Problem

Power allocation strategy to achieve maximum data rate?




Water-Filling

Prerequisite

CSIT allows for singular value decomposition

exploit SVD to reduce problem to parallel SISO channels

r = rank(H) , with H = UΣV H ,

noise power at each receiver antenna

Pz% = σ2z% for % = 1, . . . , r = rank(H)

variable transmit signal power

Px% = α%Px , with α% ≥ 0

limited total transmit power

E‖x(k)‖2 ≤ mPx , respectivelyr∑

%=1

Px% ≤ mPx




Water-Filling

Equivalent Scenario

w(k)

α1

αr

x1(k)

xr(k)

σ1(k)

σr(k)

z1(k)

zr(k)

y1(k)

yr(k)

w(k)Tx Rx

feed-back channel


Reduced problem

Power allocation strategy to achieve maximum data rate?Lehrstuhl fürDigitale Kommunikationssysteme



Water-Filling

Equivalent parallel SISO channels

y%(k) = σ%(k)√α%(k)x%(k) + z%(k) for % = 1, . . . , r = rank(H)

Capacity

C = maxα1,...,αr

r∑

%=1

ld(1 + α%γSNR%

), with γSNR% = σ2

%Px

Pz%

r∑%=1

α% ≤ m

α% ≥ 0

How to compute this maximum?




Water-Filling

Optimization problem formulation

minα f (α) s. t. g(α) ≤ 0 , α 0

Objective function

f (α) = −r∑

%=1

ld(1 + α%γSNR%

)differentiable, convex

Inequality constraint function

g(α) =r∑

%=1

α% − m

differentiable, convex

Convex optimization problem!Lehrstuhl fürDigitale Kommunikationssysteme



Water-Filling

Karush-Kuhn-Tucker conditions In particular , with % = 1, . . . , r

1 µ ≥ 0 1 µ = 0 or µ > 0

2 f ′(α) + µg′(α) = 0T 2 − γSNR%

ln(2)[1 + α%γSNR%

] + µ = 0

3 µg(α) = 0 3 µ

[r∑

%=1

α% − m

]= 0

4 g(α) ≤ 0 4

r∑%=1

α% − m ≤ 0

5 α ≥ 0 5 α% ≥ 0

For this convex optimization problem theKarush-Kuhn-Tucker conditions are necessary and sufficient!




Water-Filling

Consequences

2 ⇒ µ =γSNR%

ln(2)[1 + α%γSNR%

] > 0

1 is feasible

3 ⇒r∑

%=1

α% = m

4 is feasible

2 ⇒ α% =1

µ ln(2)− 1γSNR%

Solution

5 ⇒ α% =

(1

µ ln(2)− 1γSNR%

)+

for % = 1, . . . , r




Water-Filling

Water-Filling Algorithm γSNR1 ≥ γSNR2 ≥ · · · ≥ γSNRr > 0

1 2 3 4 • • • r − 1 r

1γSNR1

1γSNR2

1γSNR3

1µ ln(2)

1γSNR4

1γSNRr−1

1γSNRr

α1 α

2

α3

α4=0

αr−1=0

αr=0

•••

Who has will be given more!Lehrstuhl fürDigitale Kommunikationssysteme


Wireless Communications Multiple Access Channel

Wireless CommunicationsMultiple Access Channel

Karlheinz Ochs




Multiple Access Channel

Contents

1 Scenario

2 Time Division Multiple Access

3 Time Sharing

4 Successive Interference Cancelation



Multiple Access Channel Scenario

Contents

1 Scenario


3 Time Sharing




Scenario 100 / 126

Scenario

Scenario

w1(k)

w2(k)

Tx1

Tx2

x1(k)

x2(k)

h∗1

h∗2

z(k)

y(k)Rx w1(k), w2(k)

Transmitters have no channel state information (no CSIT)

Channel

y(k) = hHx(k) + z(k) , with Kxx = diag(Px1 ,Px2) , Pz = σ2z



Scenario 101 / 126

Scenario

Signal flow diagram

x1(k)

x2(k)

h∗1

h∗2

z(k)

y(k)

Channel

y(k) = h∗1 x1(k) + h∗2 x2(k) + z(k)

Objective

R1 + R2 → max

Optimal transmission strategy?



Multiple Access Channel Time Division Multiple Access

Contents

1 Scenario


3 Time Sharing




Time Division Multiple Access 102 / 126

Time Division Multiple Access

Time Division Multiple Access Strategy

x1(k)

x2(k)

k ∈ K1

k ∈ K2

h∗1

h∗2

z(k)

y(k)

Transmitter Tx1

k ∈ K1 = 1, . . . ,κ

1K

κ∑k=1

P1 ≤ Px1

worst case κ = K, (K2 = ∅)P1 ≤ Px1

Transmitter Tx2

k ∈ K2 = κ + 1, . . . ,K

1K

K∑k=κ+1

P2 ≤ Px2

worst case κ = 0, (K1 = ∅)P2 ≤ Px2





Channel usage proportions

Tx1: α =κK Tx2: 1− α

Maximum achievable data rates

Tx1: R1 ≤ αC1 ,

with C1 = ld

(1 + |h1|2 Px1

σ2z

) Tx2: R2 ≤ [1− α]C2 ,

with C2 = ld

(1 + |h2|2 Px2

σ2z

)

Rate region

R2 ≤ C2 − C2

C1R1





TDMA Rate Region

0 C10

C2

TDMA

α = 0

α = 1

R1

R2

Optimal strategy?Lehrstuhl fürDigitale Kommunikationssysteme


Multiple Access Channel Time Sharing

Contents

1 Scenario


3 Time Sharing




Time Sharing 105 / 126

Time Sharing

Time Sharing Strategy

x1(k)

x2(k)

k ∈ K1

k ∈ K2

1√α

1√1−α

h∗1

h∗2

z(k)

y(k)

Transmitter Tx1

k ∈ K1 = 1, . . . ,κ

1K

κ∑k=1

P1 ≤ Px1

average power constraintαP1 ≤ Px1 , α 6= 0

Transmitter Tx2

k ∈ K2 = κ + 1, . . . ,K

1K

K∑k=κ+1

P2 ≤ Px2

average power constraint[1− α]P2 ≤ Px2 , α 6= 1




Time Sharing

Time Sharing Rate Region

0 C10

C2

TDMA

Time-Sharing

α = 0

α = 1

R1

R2

Optimal strategy?Lehrstuhl fürDigitale Kommunikationssysteme


Multiple Access Channel Successive Interference Cancelation

Contents

1 Scenario


3 Time Sharing




Successive Interference Cancelation 108 / 126

Successive Interference Cancelation

Upper Bounds

1 Tx1 transmits only

R1 ≤ C1 , with C1 = ld

(1 + |h1|2 Px1

σ2z

)


R2 ≤ C2 , with C2 = ld

(1 + |h2|2 Px2

σ2z

)

3 Tx1, Tx2 are cooperating (MISO)

R2 ≤ −R1 + C , with C = ld

(1 + |h1|2 Px1

σ2z+ |h2|2 Px2

σ2z

)

Are the upper bounds achievable?





Successive Interference Cancelation Rate Region

0 R12 C10

R21

C2Tx2 transmits only

Tx1 , Tx

2 are cooperating

Tx1

transmits

only

R1

R2

Upper bounds achievable?






0 R12 C10

R21


Tx1 , Tx

2 are cooperating

Tx1

transmits

only

TDMA

Time-Sharing

(0, C2)

(C1, 0)

(R12, C2)

(C1, R21)

R1

R2

Upper bounds achievable?Lehrstuhl fürDigitale Kommunikationssysteme




Known upper bound

3 R1 ≤ ld

(1 + |h1|2 Px1

σ2z+ |h2|2 Px2

σ2z

)− R2

Achievability

(R1,R2) = (R12,C2)





Known upper bound

3 R1 ≤ ld

(1 + |h1|2 Px1

σ2z+ |h2|2 Px2

σ2z

)− R2

Achievability

(R1,R2) = (R12,C2)

R12 ≤ ld

(1 + |h2|2 Px2

σ2z+ |h1|2 Px1

σ2z

)− ld

(1 + |h2|2 Px2

σ2z

)





Known upper bound

3 R1 ≤ ld

(1 + |h1|2 Px1

σ2z+ |h2|2 Px2

σ2z

)− R2

Achievability

(R1,R2) = (R12,C2)

R12 ≤ ld

(1 +

|h1|2Px1

σ2z + |h2|2Px2

)





Known upper bound

3 R1 ≤ ld

(1 + |h1|2 Px1

σ2z+ |h2|2 Px2

σ2z

)− R2

Achievability

(R1,R2) = (R12,C2)

R12 ≤ ld

(1 +

|h1|2Px1

σ2z + |h2|2Px2

)

Successive interference cancelation strategy

1 receiver decodes x1 treating x2 as noise : R1 = R12

2 receiver cancels x1 by decoding x2 from y− h∗1 x1 : R2 = C2

Point (R12,C2) is achievable!Lehrstuhl fürDigitale Kommunikationssysteme





0 R12 C10

R21


Tx1 , Tx

2 are cooperating

Tx1

transmits

only

TDMA

Time-Sharing

(0, C2)

(C1, 0)

(R12, C2)

(C1, R21)

R1

R2





Known upper bound

3 R2 ≤ ld

(1 + |h1|2 Px1

σ2z+ |h2|2 Px2

σ2z

)− R1

Achievability

(R1,R2) = (C1,R21)

R21 ≤ ld

(1 +

|h2|2Px2

σ2z + |h1|2Px1

)

Successive interference cancelation strategy

1 receiver decodes x2 treating x1 as noise : R2 = R21

2 receiver cancels x2 by decoding x1 from y− h∗2 x2 : R1 = C1

Point (C1,R21) is achievable!Lehrstuhl fürDigitale Kommunikationssysteme





0 R12 C10

R21


Tx1 , Tx

2 are cooperating

Tx1

transmits

only

TDMA

Time-Sharing

(0, C2)

(C1, 0)

(R12, C2)

(C1, R21)

R1

R2





Achievable Points


(R1,R2) = (C1, 0)


(R1,R2) = (0,C2)

3 receiver successively cancels interference

(R1,R2) = (R12,C2)

4 receiver successively cancels interference

(R1,R2) = (C1,R21)

Use TDMA of the particular strategiesto achieve points on the connecting line!






0 R12 C10

R21

C2

TDMA

Time-Sharing

(0, C2)

(C1, 0)

(R12, C2)

(C1, R21)

R1

R2

Successive interference cancelation is an optimal strategy!Lehrstuhl fürDigitale Kommunikationssysteme


Wireless Communications X Channel

Wireless CommunicationsX Channel

Karlheinz Ochs




X Channel

Contents


2 Interference Alignment

3 Interference Cancelation



X Channel Transmission Scenario

Contents







X Channel


u(t) h(t, t′) v(t)

Requirement specifications

k even [v1(kT)v2(kT)

]=

[u1(kT)u2(kT)

]

k odd [v1(kT)v2(kT)

]=

[u2(kT)u1(kT)

]




X Channel


u(t) h(t, t′) v(t)



]=

[u1(kT)u2(kT)

]or H(kT, jω′) =

[1 00 1

]

k odd [v1(kT)v2(kT)

]=

[u2(kT)u1(kT)

]or H(kT, jω′) =

[0 11 0

]




X Channel


u(t) h(t, t′) v(t) u(t) Encoderx(t)

Channely(t)

Decoder v(t)



]=

[u1(kT)u2(kT)

]or H(kT, jω′) =

[1 00 1

]

k odd [v1(kT)v2(kT)

]=

[u2(kT)u1(kT)

]or H(kT, jω′) =

[0 11 0

]

Optimal usage of channel resources?




X Channel


u1(t)

u2(t)

x1(t)

x2(t)

h11(t)

h21(t)

h12(t)

h22(t)

z1(t)

z2(t)

y1(t)

y2(t)

v1(t)

v2(t)Encoder

Encoder

Decoder

Decoder

Frequency-invariant channel with additive noise

y1(t) = h11(t)x1(t) + h12(t)x2(t) + z1(t)

y2(t) = h21(t)x1(t) + h22(t)x2(t) + z2(t)

Optimal usage of channel resources?Lehrstuhl fürDigitale Kommunikationssysteme



X Channel

Simplified Transmission Scenario

u11, u21

u22, u12

x1(k)

x2(k)

h11(k)

h21(k)

h12(k)

h22(k)

y1(k)

y2(k)

u11, u12

u22, u21Encoder

Encoder

Decoder

Decoder

Frequency-invariant channel

y1(k) = h11(k)x1(k) + h12(k)x2(k)

y2(k) = h21(k)x1(k) + h22(k)x2(k)

Transmission of 4 symbols with 3 channel usages?Lehrstuhl fürDigitale Kommunikationssysteme



X Channel

Transmission Scenario for 3 Channel Usages

u11, u21

u22, u12

x1

x2

H11

H21

H12

H22

y1

y2

u11, u12

u22, u21Encoder

Encoder

Decoder

Decoder

Frequency-invariant channel µ, ν ∈ 1, 2

yν =

yν(1)yν(2)yν(3)

, Hνµ =

hνµ(1) 0 00 hνµ(2) 00 0 hνµ(3)

, xµ =

xµ(1)xµ(2)xµ(3)

hνµ(1) 6= hνµ(2) 6= hνµ(3) 6= hνµ(1)Lehrstuhl fürDigitale Kommunikationssysteme


X Channel Interference Alignment

Contents






Interference Alignment 120 / 126

X Channel


u11, u21

u22, u12

x1

x2

H11

H21

H12

H22

y1

y2

u11, u12

u22, u21Encoder

Encoder

Decoder

Decoder

Transmitted signals

x1 = q11u11 + q21u21

x2 = q12u12 + q22u22

Vectors qνµ are design parameters!Lehrstuhl fürDigitale Kommunikationssysteme



X Channel


u11, u21

u22, u12

x1

x2

H11

H21

H12

H22

y1

y2

u11, u12

u22, u21Interference-Alignment

Interference-Alignment

Decoder

Decoder

Received signals

y1 =H11q11u11 +H12q12u12 +H11q21u21 +H12q22u22

y2 =H21q21u21 +H22q22u22 +H21q11u11 +H22q12u12

Reasonable choice of vectors qνµ?Lehrstuhl fürDigitale Kommunikationssysteme



X Channel

Received signals

y1 =H11q11u11 +H12q12u12 +H11q21u21 +H12q22u22

y2 =H21q21u21 +H22q22u22 +H21q11u11 +H22q12u12

Interference Alignment

q21 =H−111 H12q22 , q12 =H−1

22 H21q11

Aligned received signals

y1 =H11q11u11 +H12H−122 H21q11u12 +H12q22[u21 + u22]

y2 =H21H−111 H12q22u21 +H22q22u22 +H21q11[u11 + u12]

Interferences lie in 1-dimensional subspaces!Lehrstuhl fürDigitale Kommunikationssysteme


X Channel Interference Cancelation

Contents






Interference Cancelation 123 / 126

X Channel


u11, u21

u22, u12

x1

x2

H11

H21

H12

H22

y1

y2

u11, u12



Decoder

Decoder


y1 =H11q11u11 +H12H−122 H21q11u12 +H12q22[u21 + u22]

y2 =H21H−111 H12q22u21 +H22q22u22 +H21q11[u11 + u12]

Choice of q11 and q22, such that decoding is possible?Lehrstuhl fürDigitale Kommunikationssysteme



X Channel


y1 = R1

u11

u12

u21 + u22

and y2 = R2

u21

u22

u11 + u12

with matrices depending on vectors q11 and q22

R1 =[H11q11 H12H

−122 H21q11 H12q22

]R2 =

[H21H

−111 H12q22 H22q22 H21q11

]Decoding of transmitted symbols

Choice of q11 and q22, such that R1 and R2 are regular?




X Channel


u11, u21

u22, u12

x1

x2

H11

H21

H12

H22

y1

y2

u11, u12



Decoder

Decoder

Interference Cancelation u11

u12

u21 + u22

= R−11 y1 and

u22

u21

u11 + u12

= R−12 y2




X Channel


u(t) Encoderx(t)

Channely(t)

Decoder v(t)

Remarks

encoder and decoder are time-variant transmission systems

channel state information at transmitter required

transmission of 4 symbols with 3 channel uses

strategy achieves DoF

DoF is a first order approximation of channel capacity

Information theory

Interference alignment and cancelation is a DoF-optimal strategy!



Wireless Communications - Lecture slides · Rice distribution Rayleigh distribution Nakagami...

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