Wireless Communications - Lecture slides · Wireless Communications Contents 1 Motivation 2 Linear...

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 2016/17

Wireless Communications

Contents

1 Motivation

2 Linear Systems

3 Wireless Communication Channel

4 Single Input Single Output Systems

5 Multiple Input Multiple Output Systems

6 Optimal Transmission Strategies

7 Multiple Access Channel

Lehrstuhl fürDigitale Kommunikationssysteme

K. Ochs Wireless Communications WS 2016/17

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 / 140

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 / 140

Transmission scenario

Multipath Propagation Channel

cellular phone

echos

noise

base station

Time-variance

multipath 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 / 140

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 Linear Systems

Wireless CommunicationsLinear Systems

Karlheinz Ochs




Linear Systems

Contents

1 Signals

2 Fourier Series and Transformation

3 Linear Systems



Linear Systems Signals

Contents

1 Signals


3 Linear Systems



Signals 4 / 140

Unit Step Function

Definition

u(ξ) =

1 for ξ > 00 for ξ < 0

Graph

u(ξ)

1

ξ

Remarks

argument can have a physical unit

at discontinuity we may define u(0) = 1/2



Signals 5 / 140

Sign Function

Definition

sgn(ξ) =

1 for ξ > 0−1 for ξ < 0

Graph

sgn(ξ)

1

−1

ξ

Remarks

is a superposition of two unit step functions

argument can have a physical unit

at discontinuity we may define sign(0) = 0



Signals 6 / 140

Rectangular Function

Definition

rect(ξ) =

1 for |ξ| < 10 for |ξ| > 1

Graph

rect(ξ)

−1 1

1

ξ

Remarks

is a superposition of two unit step functions

argument has no physical unit

at discontinuities we may define rect(±1) = 1/2



Signals 7 / 140

Triangular Function

Definition

4(ξ) =

1− |ξ| for |ξ| ≤ 1

0 for |ξ| > 1

Graph

−1 1

1

(ξ)

ξ

Remarks

is the convolution of two rectangular functions




Signals 8 / 140

si-Function

Definition

si(ξ) =

1 for ξ = 0sin(ξ)/ξ for ξ 6= 0

Graph

−4π −3π −2π −π π 2π 3π 4π

1

si(ξ)

ξ

Remarks


decays reciprocal to its argument

zeros si(νπ) = 0 for ν ∈ Z\0



Signals 9 / 140

Dirac Delta Function

Definition

∫ t

−∞δ(ξ)dξ = u(t)

Especially ∫ ∞−∞

δ(ξ)dξ = 1

Properties

even

δ(t) = δ(−t)

scaling

δ(αt) =1|α|δ(−t) for α ∈ R \ 0

sifting property

f (t)δ(t − t0) = f (t0)δ(t − t0)

Remarks

argument can have physical unit

function has reciprocal physical unit of its argument

briefly called δ-function



Linear Systems Fourier Series and Transformation

Contents

1 Signals

2 Fourier Series and TransformationFourier SeriesFourier Transformation

3 Linear Systems



Fourier Series and Transformation Fourier Series 10 / 140

Fourier Series

Periodic signal

s(t) = s(t − T)

fundamental periodsmallest possible positive T

Examples

sine

s(t) = sin(Ωt) , with ΩT = 2π

cosine

s(t) = cos(Ωt) , with ΩT = 2π

periodically repeated function

s(t) =∞∑

n=−∞

f (t − nT)

Fourier series

s(t) =∞∑

n=−∞

Sn e jnΩt , with Sn =1T

∫ T

0s(t)e−jnΩtdt and ΩT = 2π




Often Used Fourier Series

Cosine

s(t) = cos(Ωt) =12

[e jΩt + e−jΩt

]corresponds to real part

Sine

s(t) = sin(Ωt) =12j

[e jΩt − e−jΩt

]corresponds to imaginary part

Periodically repeated function

s(t) =∞∑

n=−∞

f (t − nT) =1T

∞∑n=−∞

F(jnΩ)e jnΩt ,

where f (t) −−•F(jω) and ΩT = 2πLehrstuhl fürDigitale Kommunikationssysteme



Fourier Series of the Sampling Function

Definition

δT(t) =∞∑

k=−∞

δ(t − kT)

Graph

δT (t)

−3 −2 −1 0 1 2 3 t/T

Remarks

also called Dirac comb or impulse trainT-periodic functionδ(t) −−• 1

Fourier series

δT(t) =1T

∞∑n=−∞

e jnΩt ,

where ΩT = 2π.Lehrstuhl fürDigitale Kommunikationssysteme


Fourier Series and Transformation Fourier Transformation 13 / 140

Fourier Transformation

Correspondence

x(t) −−•X(jω)

Time domain

x(t) = F−1X(jω) = 12π

∫ ∞−∞

X(jω)e jωtdω

Frequency domain

X(jω) = Fx(t) =∫ ∞−∞

x(t)e−jωtdt

Remark

X(jω) is called Fourier transform or spectrum of x(t)




Properties of the Fourier Transformation

Complex conjugation

x∗(t) −−•X∗(−jω)

Time shift

x(t − t0) −−• e−jωt0 X(jω) , with t0 ∈ R

Modulation

x(t)e jω0t −−•X(jω − jω0) , with ω0 ∈ R

Time scaling

x(αt) −−• 1|α|X

(jωα

), with α ∈ R\0





Differentiation

time domain

dx(t)dt−−• jωX(jω)

frequency domain

tx(t) −−• jdX(jω)

dω

Integration

time domain∫ t

−∞x(τ)dτ −−• X(jω)

jω+ πX(0)δ(ω) for X(0) <∞

frequency domain

jx(t)

t+ πx(0)δ(t) −−•

∫ ω

−∞X(jv)dv for x(0) <∞





Convolution

time domain

x(t) ∗ y(t) =∫ ∞−∞

x(τ)y(t − τ)dτ −−•X(jω)Y(jω)

frequency domain

x(t)y(t) −−• 12π

X(jω) ∗ Y(jω) =1

2π

∫ ∞−∞

X(jv)X(jω − jv)dv

Parseval’s theorem

∫ ∞−∞

x(t)y∗(t)dt =1

2π

∫ ∞−∞

X(jω)Y∗(jω)dω

energy of x(t) if y(t) = x(t)




Often Used Correspondences

Rectangular function

rect( t

T

)−−• 2T si(ωT) ,

Ω

πsi(Ωt) −−• rect

( ωΩ

),

with T, Ω > 0

Triangular function

4( t

T

)−−• T si2

(ωT2

),

Ω

πsi2(Ωt) −−•4

( ω

2Ω

),

with T, Ω > 0

Gaussian function

e−αt2/2 −−•√

2πα

e−ω2/[2α] , with Reα > 0




Often Used Correspondences

Unit step function

u(t) −−•πδ(ω) +[

1jω

], δ(t) +

[jπt

]−−• 2u(ω)

Signum function

sgn(t) −−•[

2jω

],

[jπt

]−−• sgn(ω)

Delta function

δ(t) −−• 1 , 1 −−• 2πδ(ω)




Correspondence of the Sampling Function

Correspondence

δT(t) −−•∆T(jω)

Fourier transform

∆T(jω) =∞∑

k=−∞

e−jωkT = Ω

∞∑n=−∞

δ(ω − nΩ)

Time shift∞∑

k=−∞

δ(t − tk) =1T

∞∑n=−∞

e jnΩ[t−t0]

−−•

∞∑k=−∞

e−jωtk = Ω∞∑

n=−∞

δ(ω − nΩ)e−jnΩt0 ,

with tk = t0 + kT.Lehrstuhl fürDigitale Kommunikationssysteme


Linear Systems Linear Systems

Contents

1 Signals


3 Linear SystemsTime-Invariant SystemsTime-Variant SystemsExamples



Linear Systems Time-Invariant Systems 20 / 140

Linear Time-Invariant System

System

x(t) S y(t)

Reaction

x(t)→ y(t) = h(t) ∗ x(t) =∫ ∞−∞

h(t − t′)x(t′)dt′

Impulse response

δ(t) S h(t)

Reaction

δ(t)→ y(t) = h(t) ∗ δ(t) = h(t)





Transfer function

ejΩxt S H(jΩx)ejΩxt

Reaction

e jΩx t → y(t) = H(jΩx)e jΩx t

Proof

1 y(t) = e jΩx t ∗ h(t)

2 y(t) =∫∞−∞ e jΩx[t−t′]h(t′)dt′

3 y(t) =∫∞−∞ h(t′)e−jΩx t′dt′ e jΩx t

4 y(t) = H(jΩx)e jΩx t

with

h(t) −−•H(jω)Lehrstuhl fürDigitale Kommunikationssysteme




Periodic excitation

∞∑n=−∞

XnejnΩxt S∞∑

n=−∞H(jnΩx)Xne

jnΩxt

Reaction

∞∑n=−∞

Xn e jnΩx t → y(t) =∞∑

n=−∞

H(jnΩx)Xn e jnΩx t

Proof

1 e jnΩx t → H(jnΩx)e jnΩx t

2 Xn e jnΩx t → H(jnΩx)Xn e jnΩx t

3∞∑

n=−∞Xn e jnΩx t → y(t) =

∞∑n=−∞

H(jnΩx)Xn e jnΩx t





Aperiodic excitation

X(jω) S Y (jω)

Reaction

X(jω)→ Y(jω) = H(jω)X(jω)

Proof

1 x(t) =1

2π∫∞−∞ X(jω)e jωtdt = lim

Ω→0xΩ(t)

2 xΩ(t) =∞∑

n=−∞

Ω

2πX(jnΩ)e jnΩt → yΩ(t) =

∞∑n=−∞

Ω

2πH(jnΩ)X(jnΩ)e jnΩt

3 limΩ→0

xΩ(t)→ limΩ→0

yΩ(t)

4 x(t)→ y(t) = F−1H(jω)X(jω)Lehrstuhl fürDigitale Kommunikationssysteme


Linear Systems Time-Variant Systems 24 / 140

Linear Time-Variant System

Transfer function

ejΩxt St H(t, jΩx)ejΩxt

Reaction

e jΩx t → y(t) = H(t, jΩx)e jΩx t

Definition

H(t, jΩx) = y(t)e−jΩx t for all Ωx ∈ R

Time-variant transfer function

H(t, jω′)





Periodic excitation

∞∑n=−∞

XnejnΩxt St

∞∑n=−∞

H(t, jnΩx)XnejnΩxt

Reaction

∞∑n=−∞

Xn e jnΩx t → y(t) =∞∑

n=−∞

H(t, jnΩx)Xn e jnΩx t

Proof

1 e jnΩx t → H(t, jnΩx)e jnΩx t

2 Xn e jnΩx t → H(t, jnΩx)Xn e jnΩx t

3∞∑

n=−∞Xn e jnΩx t → y(t) =

∞∑n=−∞

H(t, jnΩx)Xn e jnΩx t





Aperiodic excitation

X(jω) St y(t)

Reaction

X(jω)→ y(t) =1

2π

∫ ∞

−∞H(t, jω′)X(jω′)e jω′tdω′

Proof

1 x(t) =1

2π∫∞−∞ X(jω)e jωtdt = lim

Ω→0xΩ(t)

2 xΩ(t) =∞∑

n=−∞

Ω

2πX(jnΩ)e jnΩt → yΩ(t) =

∞∑n=−∞

Ω

2πH(t, jnΩ)X(jnΩ)e jnΩt

3 limΩ→0

xΩ(t)→ limΩ→0

yΩ(t)

4 x(t)→ y(t) = F−1H(t, jω)X(jω)Lehrstuhl fürDigitale Kommunikationssysteme




Impulse response

δ(t− Tx) St h(t, Tx)

Reaction

δ(t − Tx)→ h(t, Tx) with h(t, t′) =1

2π

∫ ∞

−∞H(t, jω′)e jω′[t−t′]dt

Proof

1 x(t) = δ(t − Tx) −−•X(jω) = e−jωTx

2 y(t) =1

2π∫ ∞−∞ H(t, jω′)e−jω′Tx e jω′tdt

3 h(t, Tx) =1

2π∫ ∞−∞ H(t, jω′)e jω′[t−Tx]dt





System

x(t) St y(t)

Reactions

x(t)→ y(t) =∫ ∞

−∞h(t, t′)x(t′)dt′

X(jω)→ y(t) =1

2π

∫ ∞

−∞H(t, jω′)X(jω′)e jω′tdt

Important

Y(jω) 6= H(t, jω)X(jω)



Linear Systems Examples 29 / 140

Examples of Linear Time-Variant Systems

Linear time-invariant system

x(t) S y(t)

Special excitation

δ(t − Tx)→ h(t − Tx)

e jΩx t → H(jΩx)e jΩx t

System functions

pulse response h(t, t′)= h

(t − t′

)transfer function H

(t, jω′

)= H

(jω′)

Transfer function is constant with respect to time!



Linear Systems Examples 30 / 140

Examples of Linear Time-Variant Systems

Linear frequency-invariant system

x(t)

c(t)

y(t)

Special excitation

δ(t − Tx)→ c(Tx)δ(t − Tx)

e jΩx t → c(t)e jΩx t

System functions

pulse response h(t, t′) = c(t′)δ(t − t′)

transfer function H(t, jω′) = c(t)

Transfer function is constant with respect to frequency!



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-variance

multipath 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 pathsxν(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

definition

x(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) =

[n∑ν=0

cν(t)e−jΩxTν

]e jΩx t

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 functionRice 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 37 / 140

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





Reception requires matched 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 41 / 140

Baseband Transmission

Resources of the channel

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





Transmitter signal matched to channel resources

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 Scheme

1 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

Transmission Scheme

1 Nyquist pulse

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

Transmission Scheme

1 Nyquist pulse

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

Transmission Scheme

1 Nyquist pulse

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 probabilitytransmitted 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 49 / 140

Time-Discrete Channel

Preliminary Considerations

1 perfect timing synchronizationτ = 0

2 no receiver filter, transmitter filter is Nyquist filterx(tk) = u(tk)

3 time-discrete channely(tk) = c(tk)x(tk) + w(tk)

4 decisionv(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










u(t)

δT (t− t0)

s(t)x(t)

c(t) w(t)

y(t)

δT (t− t0)

PAM v(t)


receiver










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










u(tk) = x(tk)

c(tk) w(tk)

y(tk)v(tk)

time-discrete channel






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

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





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

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 52 / 140

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(Ωt2

), 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

s(t) = si(Ωt2

), with ΩT = 2π




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

s(t) = si(Ωt2

), with ΩT = 2π




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

s(t) = si(Ωt2

), with ΩT = 2π

Signaling at the Nyquist rate!Lehrstuhl fürDigitale KommunikationssystemeK. Ochs Wireless Communications WS 2016/17


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 |2



Single Input Single Output Systems AWGN Channel

Contents

1 Signal Space

2 AWGN Channel




AWGN Channel 57 / 140

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 receiverK 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 distributednormal distribution, with zero mean and variance σ2

z = 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 distributednormal distribution, with zero mean and variance σ2

z = Pz






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)

What is the highest reliable data rate over this channel?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 !





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

Illustration

1

2

3

M

√KPz

√KPz

√KPz

√KPz

√K[Px + Pz ]

Hypersphere

K (real) dimensions

radius r

volume VK(r) ∼ rK

Nonoverlapping hyperspheres

K complex dimensions





AWGN channel

Illustration

1

2

3

M

√KPz

√KPz

√KPz

√KPz

√K[Px + Pz ]

Hypersphere

K (real) dimensions

radius r

volume VK(r) ∼ rK


2K real dimensions

M ≤ V2K(√

2K[Px + Pz])

V2K(√

2KPz)





AWGN channel

Illustration

1

2

3

M

√KPz

√KPz

√KPz

√KPz

√K[Px + Pz ]

Hypersphere

K (real) dimensions

radius r

volume VK(r) ∼ rK


2K real dimensions

M ≤ VK(2K[Px + Pz])

VK(2KPz)





AWGN channel

Illustration

1

2

3

M

√KPz

√KPz

√KPz

√KPz

√K[Px + Pz ]

Hypersphere

K (real) dimensions

radius r

volume VK(r) ∼ rK


2K real dimensions

M ≤ [1 + Px/Pz]K





AWGN channel

Illustration

1

2

3

M

√KPz

√KPz

√KPz

√KPz

√K[Px + Pz ]

Hypersphere

K (real) dimensions

radius r

volume VK(r) ∼ rK


2K real dimensions

M ≤ [1 + Px/Pz]K

Upper bound for nonoverlapping hyperspheres

M ≤ [1 + ΓSNR]K , with ΓSNR =

Px

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





AWGN channel

Upper bound for the data rate

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





AWGN channel


R =ld(M)

K≤ ld (1 + ΓSNR)




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 64 / 140

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.)normal distribution, with zero mean and variance σ2

z = Pz





Flat fading Channel

Flat fading Channel

x(k)

h(k) z(k)

y(k)

Channel capacity?




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




Flat fading Channel

Flat fading Channel

x(k)

h(k) z(k)

y(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




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 gain

sum 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


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


PAM

PAM

ejω0t

ejω0t

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


m 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


n 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)






Eu(k) Txx(k)

H(k)z(k)

y(k)v(k)

Transmitter

maps message U to signal x(k)

transmitter signal has limited powerFlat 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 79 / 140

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 80 / 140

Zero-Forcing Detector

Scenario 1

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

equation system

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

Detection

x =H−1y ⇒ x = x+H−1z

Remarks

simple 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) +=

Detection

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 usedH 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) +=

Detection

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 = arg

minxJ

, 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 86 / 140

Minimum Mean-Squared Error Detector

Detection

x =Dy , with y =Hx+ z

A minimum mean-squared error detector considers noise!

Optimization problem

D = arg

minDJ

, 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)

E‖x‖2

= tr

(ExxH)

∂

∂Mtr (AMB) = ATBT

Kxy = ExyH





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

D =KxyK−1yy

Regularity of Kyy is assumed!





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

Solution

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 91 / 140

Maximum Likelihood Detector

Stochastic channel model

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

Detection

x = arg

maxx∈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 −Hxfy|x(y|x) = fz(y −Hx)

Intermediate result

x = arg

maxx∈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 = arg

maxx∈Am

1

[πσ2z ]n

exp(− 1σ2

z‖y −Hx‖2

)Exponential function is strictly monotonically increasing!

Solution

x = arg

minx∈Am

‖y −Hx‖

Remarks

geometrical 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 94 / 140

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 95 / 140

Spatially Uncorrelated Channel

Independent and Identically Distributed Random MIMO Channel

Elements hµν(k) of channel matrix

independent and identically distributed random variables

zero-mean

Ehµν(k) = 0

spatially uncorrelated

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

Correlation matrix of the channel 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 96 / 140

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 97 / 140


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 98 / 140

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 99 / 140

Eigenmodes of a MIMO Channel

Scenario

PSfrag

x(k)Pre-

Encoder

x′(k)

H(k) z′(k)

y′(k)Post-

Encodery(k)

Receiver and transmitter have 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!





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!



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 103 / 140

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% = σ% x% + z% 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% )

Reformulation independent from singular value decomposition?Lehrstuhl fürDigitale Kommunikationssysteme




Capacity of a Decomposed MIMO Channel

C(k) =r∑

%=1

ld(1 + ΓSNRσ2% )

Refomulation

1r∑

%=1ld(ξ%) = ld

(r∏

%=1ξ%

)

C(k) = ld

(r∏

%=1

[1 + ΓSNRσ

2%

])

2 ld

(r∏

%=1ξ%

)= |diag(ξ%)|

C(k) = ld(∣∣∣diag(1 + ΓSNRσ

2% )∣∣∣)





Refomulation

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

C(k) = ld(∣∣∣1 + ΓSNR diag(σ2

% )∣∣∣)

4 diag(ξ2%) = diag(ξ%)2

C(k) = ld(∣∣∣1 + ΓSNRΣ

2r )∣∣∣)

5

∣∣∣∣[ A 00 1

]∣∣∣∣ = |A|C(k) = ld

(∣∣1m + ΓSNRΣHΣ∣∣)

6 HHH = V ΣHΣV H i. e. ΣHΣ = V HHHHV

C(k) = ld(∣∣V H [1m + ΓSNRH

HH]V∣∣)





Refomulation

7 |AB| = |A| |B| ,∣∣A−1

∣∣ = |A|−1

C(k) = ld(∣∣1m + ΓSNRH

HH∣∣)

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

C(k) = ld(∣∣1n + ΓSNRHH

H∣∣)Hint

n ≤ m

C(k) = ld(∣∣1n + ΓSNRH(k)HH(k)

∣∣)m ≤ n

C(k) = ld(∣∣1m + ΓSNRH

H(k)H(k)∣∣)

No need for singular value decomposition!Lehrstuhl fürDigitale Kommunikationssysteme




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

C(k) = ld(∣∣1n + ΓSNRH(k)HH(k)

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



Capacity and Degrees of Freedom Degrees of Freedom 109 / 140

Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)

average of symbols per channel use

synonymous DoF

closely related to multiplexing gain




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)


synonymous DoF


SISO channel

η = limΓSNR→∞

ld(1+|σ|2ΓSNR)ld(ΓSNR)




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)


synonymous DoF


SISO channel

η = limΓSNR→∞

ld(|σ|2ΓSNR)ld(ΓSNR)




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)


synonymous DoF


SISO channel

η = limΓSNR→∞

[ld(|σ|2)ld(ΓSNR)

+ 1]




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)


synonymous DoF


SISO channel

η = 1




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)


synonymous DoF


SISO channel

η = 1

MIMO channel

η = limΓSNR→∞

r∑%=1

ld(1+σ2%ΓSNR)

ld(ΓSNR)




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)


synonymous DoF


SISO channel

η = 1

MIMO channel

η =r∑

%=1lim

ΓSNR→∞

ld(1+σ2%ΓSNR)

ld(ΓSNR)




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)


synonymous DoF


SISO channel

η = 1

MIMO channel

η =r∑

%=11




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)


synonymous DoF


SISO channel

η = 1

MIMO channel

η = r




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)


synonymous DoF


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

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)

Interpretation




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)

Interpretation

η = limΓSNR→∞

Cld(ΓSNR)




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)

Interpretation

η = limΓSNR→∞

Cld(1+ΓSNR)




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)

Interpretation

η = limΓSNR→∞

Cld(1+|σ|2ΓSNR)




Degrees of Freedom

Degrees of Freedom

η = limΓSNR→∞

Cld(ΓSNR)

Interpretation

η = limΓSNR→∞

CCSISO

Multiplexing gain

C ≈ ηCSISO for ΓSNR →∞

Asymptotic measurement for the high signal to noise ratio regime!Lehrstuhl fürDigitale Kommunikationssysteme


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 111 / 140

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(k) = E|‖h(k)‖2x(k)|2E|hH(k)z(k)|2






γSNR(k) = ‖h(k)‖4E|x(k)|2EhH(k)z(k)zH(k)h(k)






γSNR(k) = ‖h(k)‖4PxhH(k)Kzzh(k)






γSNR(k) = ‖h(k)‖4Pxσ2

zhH(k)h(k)






γSNR(k) = ‖h(k)‖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 114 / 140

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?






γSNR(k) = E|‖h(k)‖w(k)|2E|z(k)|2






γSNR(k) = ‖h(k)‖2E|w(k)|2σ2

z






γSNR(k) =‖h(k)‖2E

∥∥∥ h(k)‖h(k)‖ w(k)

∥∥∥2

σ2z






γSNR(k) =‖h(k)‖2E‖x(k)‖2

σ2z










)












)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 117 / 140

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?Lehrstuhl fürDigitale Kommunikationssysteme



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 ,

same noise power at each receiver antenna

Pz% = 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 , respectively Px

r∑%=1

α% ≤ 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%σ2%ΓSNR

)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%σ2%ΓSNR

)differentiable, convex

Inequality constraint

g(α) =r∑

%=1

α% − m

differentiable, convex

Convex optimization problem!Lehrstuhl fürDigitale Kommunikationssysteme



Water Filling

Karush-Kuhn-Tucker conditions In particular

1 µ ≥ 0 1 µ = 0 or µ > 0

2 f ′(α) + µg′(α) = 0T2 − σ2

%ΓSNR

ln(2)[1 + α%σ2

%ΓSNR] + µ = 0

for % = 1, . . . , r

3 µg(α) = 0 3 µ

[r∑

%=1

α% − m

]= 0

4 g(α) ≤ 0 4

r∑%=1

α% − m ≤ 0

For this convex optimization problem the Karush-Kuhn-Tucker conditionsare not only necessary but also sufficient!




Water Filling

Consequences

2 ⇒ µ =σ2%ΓSNR

ln(2)[1 + α%σ2

%ΓSNR] > 0

1 is feasible

3 ⇒r∑

%=1

α% = m

4 is feasible

2 ⇒ α% =1

µ ln(2)− 1σ2%ΓSNR

α 0 ⇒ α% ≥ 0

Solution

α% =

(1

µ ln(2)− 1σ2%ΓSNR

)+

for % = 1, . . . , r




Water Filling

Water Filling Algorithm σ1 ≥ σ2 ≥ · · · ≥ σr > 0

1 2 3 4 • • • r − 1 r

1σ21ΓSNR

1σ22ΓSNR

1σ23ΓSNR

1µ ln(2)

1σ24ΓSNR

1σ2r−1

ΓSNR

1σ2rΓSNR

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 125 / 140

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

Channel

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



Scenario 126 / 140

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 127 / 140

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 P1

σ2z

) Tx2: R2 ≤ [1− α]C2 ,

with C2 = ld(

1 + |h2|2 P2

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 130 / 140

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 constraint

P1 ≤ Px1

α, α 6= 0

Transmitter Tx2

k ∈ K2 = κ + 1, . . . ,K

1K

K∑k=κ+1

P2 ≤ Px2

average power constraint

P2 ≤ Px2

1− α , α 6= 1Lehrstuhl fürDigitale Kommunikationssysteme



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 133 / 140

Successive Interference Cancelation

Upper Bounds

1 Tx1 transmits only

R1 ≤ C1 , with C1 = ld(

1 + |h1|2 P1

σ2z

)


R2 ≤ C2 , with C2 = ld(

1 + |h2|2 P2

σ2z

)

3 Tx1, Tx2 are cooperating (MISO)

R2 ≤ −R1 + C , with C = ld(

1 + |h1|2 P1

σ2z+ |h2|2 P2

σ2z

)

Are the upper bounds achievable?





Successive Interference Cancelation Rate Region

0 R12 C10

R21

C2Tx1 transmits only

Tx1 , Tx

2 are cooperating

Tx2

transmits

only

R1

R2

Upper bounds achievable?






0 R12 C10

R21


Tx1 , Tx

2 are cooperating

Tx2

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 P1

σ2z+ |h2|2 P2

σ2z

)− R2

Achievability

(R1,R2) = (R12,C2)





Known upper bound

3 R1 ≤ ld(

1 + |h1|2 P1

σ2z+ |h2|2 P2

σ2z

)− R2

Achievability

(R1,R2) = (R12,C2)

R12 ≤ ld(

1 + |h2|2 P2

σ2z+ |h1|2 P1

σ2z

)− ld

(1 + |h2|2 P2

σ2z

)





Known upper bound

3 R1 ≤ ld(

1 + |h1|2 P1

σ2z+ |h2|2 P2

σ2z

)− R2

Achievability

(R1,R2) = (R12,C2)

R12 ≤ ld(

1 +|h1|2P1

σ2z + |h2|2P2

)





Known upper bound

3 R1 ≤ ld(

1 + |h1|2 P1

σ2z+ |h2|2 P2

σ2z

)− R2

Achievability

(R1,R2) = (R12,C2)

R12 ≤ ld(

1 +|h1|2P1

σ2z + |h2|2P2

)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!






0 R12 C10

R21


Tx1 , Tx

2 are cooperating

Tx2

transmits

only

TDMA

Time-Sharing

(0, C2)

(C1, 0)

(R12, C2)

(C1, R21)

R1

R2





Known upper bound

3 R2 ≤ ld(

1 + |h1|2 P1

σ2z+ |h2|2 P2

σ2z

)− R1

Achievability

(R1,R2) = (C1,R21)

R21 ≤ ld(

1 +|h2|2P2

σ2z + |h1|2P1

)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!






0 R12 C10

R21


Tx1 , Tx

2 are cooperating

Tx2

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) = (C1,R21)

4 receiver successively cancels interference

(R1,R2) = (R12,C2)

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 - Lecture slides · Wireless Communications Contents 1 Motivation 2 Linear...

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