stwc bildblaetter ss2011.book

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UNIVERSITT STUTTGART

INSTITUT FR NACHRICHTENBERTRAGUNG

Prof. Dr.-Ing. J. Speidel

www.inue.uni-stuttgart.de

Supplementary Material

for the LectureSpace-Time WirelessCommunications

(to be enhanced during lecture)

S T


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INSTITUT FR NACHRICHTENBERTRAGUNG UNIVERSITTProf. Dr.-Ing. Joachim Speidel STUTTGART

Space-Time Wireless Communications

Overview

1.)Multiple Input Multiple Output (MIMO) channel

2.)Spatial multiplex, diversity, beamforming principles

3.)Linear flat fading and frequency selective fading wireless MIMO channel

4.)MIMO receiver: Zero Forcing, Minimum Mean Square Error (MMSE),Maximum Likelihood (ML)

5.)MIMO channel capacity

6.)Space-time coding methods

7.)Convolutional coding, Turbo coding

8.)Decoding principles, iterative receivers

9.)Applications


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Literature

[1] J. Speidel:Multiple Input Multiple Output (MIMO) - Drahtlose Nachrichtenbertragung hoherBitrate und Qualitt mit Mehrfachantennen. Telekommunikation Aktuell, vol. 59, issue 7-10/05,July-Oct. 2005, pp. 1-63.

[2] B. Vucetic et al.: Space-Time Coding. John Wiley Publisher, 2003.

[3] A. Paulraj et al.:Introduction to Space-Time Wireless Communications. Cambridge UniversityPress, 2003.

[4] E. Larsson; P. Stoica: Space-Time Block Coding for Wireless Communications. Cambridge Uni-

versity Press, 2003.

[5] S. Alamouti:A simple transmit diversity technique for wireless communications. IEEE Transac-tions on Selected Areas of Communications, vol. 16, Oct. 1998.

[6] V. Tarokh; N. Seshardi; A.R. Calderbank: Space-time codes for high data rates wireless commu-nications: Performance criterion and code construction. IEEE Transactions on InformationTheory, vol. 44, 1998.

[7] J. Proakis: Digital Communications. Mc Graw-Hill Book Company, 4th edn. 2008.

[8] E. Lee; D. Messerschmitt; J. Barry: Digital Communication. Springer, 2004.


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Reprint prohibited - all rights reserved 3

Block diagram wireless MIMO transmission (details)

rN k

bandpass lowpasse

j 0t

g

r t r k b n

detector

mapper

impulseshaper

Re

u t s k

ej0t

sM k

uM t

g

r1 t r1 k

t0 kT+

MIMO receiver

bit sequence

b n

bit ratevB

P

S

serial-parallelconverter

a t

symbol ratevs 1 T= rN t

s1 k

MIMO channel

u1 t

MIMO transmitter

equivalent analog MIMO lowpass channel h t


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complex time-variant impulse response, ( ; )

Block diagram of wireless MIMO transmission (model)

Definition

MIMO Multiple Input Multiple Output : ,

SISO Single Input Single Output :

t0 kT+

s1

k

s k

sM k

b n b n

r1 k

r k

rN k

h t r t

r1 t

rN t

equivalent analogMIMO lowpass channel

MIMO Tx MIMO Rx

equivalent discrete-timeMIMO lowpass channel h m k

h t 1 M = 1 N =

M 1 N 1

M 1= N 1=


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MIMO noise vector

a) Gaussian , real , zero mean , variance

pdf

b) complex Gaussian

with property (1)

statistically independent

variance of

of course, has zero mean

c) random vector

all as in b), i. e. zero mean, variance , pdf (2)

all , statistically independent

x 2

p x 1

2

---------------- e

1

2---

x2

2-------

=

n x jy+= n

x y

x y

pnn p x p y

1

2----------------

2 e

1

2---

x2

y2

+

2---------------------

1

22-------------- e

1

2---

n

n

2-------------

1

n2

--------- e

n

n

n2

-------------

= = = =

22 n2

= n

n

n n1nN T

=

n n2

n n

pn n pn n 1=

N

1 n

2-----------

N

e

nH

n

n2---------

1

n2

----------- N

e

n2

n2---------

= = =(2)

nH

n n2

=

(1)

(2)

(3)


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Minimum mean squared error (MMSE) receiver

Given:

tx signal with autocorrelation matrix (1)

noise with autocorrelation matrix (2)

and statistically independent

Error ; squared error (3a,b)

MMSE: (4a,b)

Result: (5)

Proof of (5):

Introduce error autocorrelation matrix (6)

Then (7)

With (3a):

MIMO channel MMSE rx

H W

s H s

n

r y W r=

H N M

s Rss E ssH =

n Rnn E nn

H

=

s n

e s y s W r= = e2

eH

e=

J E eH

e mi n W min Jarg= = =

W RssHH

H RssHH

Rnn+ 1

=

Ree E eeH =

J trRee trE eeH = = trA sum of main diagonal elements=

J tr E s W r sH rHWH tr E ssH srHWH W rsH W rrHWH+ = =


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Eigenvalue decomposition of :

(9)

is called "square root" of

diagonal matrix with eigenvalues of

According to Lemma 1, decomposition (9) is always possible

In (10), 2nd and 3rd term are artificially introduced and cancel out.

With from (10)

(11a)

and with Lemma 2

(11b)

Due to (4a) for all also for , for which . Thus, the first term in(11b) is also . Consequently, results for

=

=

= (10)

-

=

=

=

Rrr

Rrr UH U UH1 2 1 2 U 1 2 U

H1 2 U AHA= = = =

A Rrr

diag 1 N = Rrr

Rrr AH

A andA1A I into (8) yields=

J tr Rss RsrA1AW

H W A

HA

H 1Rrs W A

HAW

H+

tr Rss RsrA1

AH

1Rrs RsrA

1A

H 1Rrs+

RsrA1 A WH W AH AH 1 Rrs W AHA WH+

RsrH

Rrs=

J tr Rss RsrA1

AH

1Rrs W A

HRsrA

1 W AH RsrA

1

H+ =

J tr Rss RsrA1

AH

1Rrs tr W A

HRsrA

1 W AH RsrA

1

H +=

J 0 W W W AH RsrA1

0=0 J min=

tr W AH

RsrA1

W AH RsrA1

H

W AH RsrA1

F2

0= =

W AH RsrA1

0

W AH

RsrA1

W RsrA1

AH

1Rsr A

HA

1RsrRrr

1= =

(9)

1


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As and are statistically independent, they are also uncorrelated, is of zero mean.

Thus , (16)

(16) into (13): (17)

(16) into (15): (18)

(17), (18) into (12): (19)

and (5) is proven.

Important case: ,

From (5) follows after some manipulation

; (20)

Lemma 3: Matrix inversion

With Lemma 3 follows from (20)

(21)

Proof of (21):

s n n

Rsn 0= Rns 0=

Rrr H RssHH

Rnn+=

Rsr RssHH

=

W RssHH

H RssHH

Rnn+ 1

=

Rss EsIM= Rnn n2

IN=

W HH

H HH IN+

1= n

2Es=

A BC D+

1A

1A

1B C

1D A

1B+

1D A

1=

W HH

H IM+ 1

HH=

HH

H HH IN+

1H

H IN HIMHH

+ 1

= =

A B C D

H 1 1 H 1 1 H 1 H 1 H 1 H 1 1 H 1

Lemma 3


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Proof of some statements of linear algebra

Let be Hermiteian matrix with eigenvalues .

A) is a quadratic form.

for any (22)

Proof: q. e. d.

B) A matrix with property (22) is called positive semi-definite.

(all are real)

Proof:

Let be the eigenvector associated to , i. e.

(23)

We calculate

q. e. d.

; (24)

Q N N 1 N =

Q aaH

=

zH

Qz

zH

Qz 0 z 0

zH

Qz zH

aaH

z zH

a zH

a H

zH

a2

0= = =

Q

Q with property (22) 0 1 N =

u Q u u=

uH

Q u uH u u

Hu u

20= = =

(23) (22)

0

Rrr E rrH = Rrr

HE rr

H H

E rrH Rrr= = =

0


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(27)

As , , and we conclude q. e. d.

Note, that (25) holds also for non-normalized vectors, i. e.

If the eigenvectors are normalized, i. e. , then they are orthonormal.

Matrix of orthonormal eigenvectors

(28)

due to orthonormality of (29)

with property (29) is called an unitary matrix. As can be seen column vectors

of an unitary matrix are orthogonal and normalized. (30)

Then (31)

(32)

From (31): =

=

With (29) = (33)

Q u2 2 u2=

1 u1 H

u2 Q u1 H

= u2 u1H

QH

u2 u1H

Q u2= =

u1H

= u22(27)

1 0 2 0 1 2 u1H

u2 0=

1u1H 2u2 0= 1 2 0

u u 2 1=

U u1 uN =

U1

UH

= u

U

UH

QU =

diag 1 N =

UH

Q U 1

Q UH

1U 1

Q UUH

1

u1H

u2 2

u1H

u2=

(23)


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Singular Value Decomposition (SVD):

SVD Lemma:

Given the matrix

An eigenvalue decomposition of does not always exist. In this case an SVD is feasible.

(1)

is called SVD of , where

(2)

and , are unitary matrices, i. e.

, (3a,b)

are called singular values of . (4)

are the positive (non-zero) eigenvalues of the Hermiteian matrix

(5)

where *) (6)

and (7)

The remaining eigenvalues of are zero.

The eigenvalue decomposition of is

H CN M

H

H UD VH

=

H

D

1 00

0 P0 0

RN M=

U CN N V CM M

U1

UH

= V1

VH

=

1 P = H 1 P =

QN H HH

CN N=

rankH rankQN P= =

P min M N

P 1 N += QNQN

H


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One method to find is the eigenvalue decomposition of as follows

(10)

with (11)

and (12)

Note,

Thus, is the matrix of eigenvectors with respect to the eigenvalues of .

SVD (1) exists for all cases and . (13)

Proof of SVD Lemma:

a) Proof of(1) - (9):

(8) exists with (9), (6) and (7).

With and (5) follows from (8)

; (14a)

We decompose

(14b)

V QM

VH

QMV M=

QM HH

H CM M=

M1 0

P0

0 0

RM M=

rankQM rankQN P= =

V 1 M QM

M N M N

QN QNH

is Hermiteian =

V VH

IM=(3b)

UH

H V VH

HH

U N= A CN M

N =1 0

00

1 00

0 DDH

=

A= A

H


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Comparing (14a) and (14b) we conclude

(15)

From (15)

and (1) is proven.

Note, that the only requirement on is (unitary).

b) We now prove, that can be obtained by eigenvalue decomposition of as outlined

by (10) - (12).

From (1) follows using Hermiteian operation

(16)

Multiplying (1) and (16) results in

(17)

With and in (14b) follows

(18)

with in (12)

Comparing and we conclude that and have the same non-

zero eigenvalues . Of course, the remaining eigenvalues are zero. Note, that in general

the total number and of eigenvalues is different for and , respectively. Also the

associated matrices and of eigenvectors are different in general.

UH

HV D=

H UD VH

=

V CM M V 1 VH=

V QM HH

H=

HH

V DH

UH

=

HH

H V DH

UH

UD VH

V DH

DVH

= =

DH

D

DH

D M=

M

N M QN H HH= QM H

HH=

1 P N M QN QM

U V

IN=(3a)


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MIMO channel capacity

Derivation of

Given (1)

Eigenvalue decomposition of yields

; (2)

(2) --> (1) (3)

Matrix calculus:

, also (4)

(4) --> (3)

q. e. d.

Derivation of

C ld IN H HHES

n2

------+=

C ld IN NES

n2

------+=

H HH

UH

H HH

U NP 0

0 0N N = = P diag 1 P =

C ld IN UHH HHU ES

n2

------+=

IM AB+ IN BA+ A M N B N M = M N=

C = ld IN H HH

U UHES

n2

------+

= ld IN H HHES

n2

------+

C ld IM HH

HES

n2

------+=

IN


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Max. channel capacity in bit/channel use

Prerequisites: with full rank, no fading, . Total tx mean power .

MIMO

Remark

for large SNR

MIMO larger than forN=M

MIMO same asN=M

SISO

SIMOlarger than for SISOgain of for large SNR

MISO same as SISO

Cmax

NM

rxtx

M N= Nld 1Estot

n2

----------+ N

ldEstot

n2

----------

M N Mld 1 NM-----

Estot

n2

----------+

M N Nld 1 Estotn

2----------+

M N 1= = ld 1Estot

n2

----------+

M 1=

N 1ld 1 NE

stot

n2

----------+ ldN

N 1=

M 1ld 1

Estot

n2

----------+

H h 1= 1 N ; 1 M = = Estot


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Orthogonal space-time block code matrices

a) For real symbols , tx antennas, spatial code rate 1

b) For complex symbols , tx antennas, spatial code rate 1/2

c) For complex symbols , tx antennas, spatial code rate 1: Alamouti ST code

S

s k M 4=

S

s1 s2 s3 s4

s2

s1

s4

s3

s3 s4 s1 s2

s4 s3 s2 s1

=

s k M 3=

S

s1 s2 s3 s4 s1* s2* s3* s4*

s2 s1 s4 s3 s2* s1

* s4* s3

*

s3 s4 s1 s2 s3* s4

* s1* s2

*

=

s k M 2=

Ss1 s2

*

s2 s1*

=

space

time


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Space-Time Trellis Coding

Input-output relation given by state transition diagram or trellis diagram (which includes time axis)

Example: Transmit delay diversity with M=2 antennas

c k b n

s1

k c k =

4 PS K

Tc k 1

s2

k c k 1 =

state machine

c1c2

c3 c4

Im c k

Re c k c k c1 c4 =

state 1

state 2

state 3

state 4

Trellis Diagram

transitions areindicated by

c c

input outputof state machine

c1 c1

c4 c4

c2 c2

c3 c3 12

3 4

k

c2 c

1

c1 c4

c

4

c

1

c 1

c 4

c

4c

2

not all transitions

are plotted


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Vertical Encoding (VE)

V-BLAST Vertical-Bell Labs Layered Architecture for Space-Time Coding

c k b'' n' b' n'

bitsequence

temporalencoder

b n demux

bitsequence

spatialdemux

symbolsequence

bitsequence

1 :M

interleaver QAM-mapper

1

M

c k c k L-1+

vector symbolsequence

c k M 1+ c k L-1-M+1+

y1

mux

1

ZF/MMSE-

receiver

QAM-

demapperdecision

device

de-

interleaver

temporal

decoder

1

1

r1


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Horizontal Encoding (HE)

H-BLAST Horizontal Bell Labs Layered Architecture for Space-Time Coding

cM

k b''M n'' b'M n'' bM

n' M

c1 k b''1 n'' b'1 n'' b1 n'

bitsequence

temporalencoder

b n demux

bitsequence

interleaverQAM

mapper

symbolsequence

c1 k

cM

k

c1 k L-1+

cM

k L-1+

vector symbolsequence

antenna 1

antennaM

Transmitter

1 :M

1


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One solution for H-BLAST receiver

Note:rx is rather simple, because it can operate just in temporal direction ("horizontally"), except for ZF or MMSE part, which operates in space direction.

cM k cM k bM n'' b' M n'' bM n' M

1

1 1

MN

c1 k c1 k b'' 1 n'' b1' n'' b1 n'

ZF orMMSEreceiver

QAMdemapper

QAMdecisiondevice

de-interleaver

temporaldecoder mux

b n

estimates

M: 1

r1 k

rN k


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The Q-Function

, where variance of with

, where

(erf: error function; erfc: error function complement)

Q 12

---------- e

1

2--- u

2

du ; Q 1 Q =

=

Q 1 ; Q 0 12--- ; Q 0===

P X Q ---

= 2 X p x 12

--------------- e

1

2---

x

---

2

=

P A X B Q A---

Q B---

=

Q 1

2--- 1 erf

2-------

1

2---erfc

2-------

= = erf z

2

------- eu

2

du0

z

=

erfc z 1 erf z =

1

103

4

Q

1--------------- 1

1------

e1

2--- 2

1

2--- e

1

2--- 2

1

2 --------------- e

1

2--- 2

Chernov bound

0.1

0.01


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Bit error ratio (BER) as a function of SNR for MxN MIMO system, with ZF receiver, 4-PSK,

Gray mapping. Encoded with convolutional code or uncoded.

Bit error ratio (BER) as a function of SNR for 4x4 MIMO system with ZF receiver, 4-PSK,

Gray mapping. Parallel detection or sequential detection with interferencecancellation (SAL). Encoded with convolutional code or uncoded.

Bit error ratio (BER) as a function of SNR for 6x6 MIMO system. ZF or MMSE receiver, 16-QAM,

Gray mapping, parallel detection. Encoded with convolutional code or uncoded.

SAL (Symbolauslschung) meansordered successive interference cancellation (OSIC)


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Iterative V-BLAST receiver

Bit error ratio (BER) as a function of SNR for 6x6 V-BLAST MIMO system with regular

16-QAM, Gray mapping, convolutional encoding and iterative V-BLAST receiver.

Bit error ratio (BER) as a function of SNR for 6x6 V-BLAST MIMO system

with convolutional encoding, Anti-Gray mapping, and iterative V-BLAST MMSE receiver.A1 regular 16-QAM, A2 is non-regular 16-QAM.

Non-regular 16-QAM


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Ergodic MIMO channel capacity for a 4x4 MIMO system as a function of transmit

correlation coefficient, no receive correlation. Exponential correlation matrix model,

uncorrelated fading. Large transmit correlation coefficients have a catastrophic impact.

Ergodic capacity for a 4x4 MIMO system as a function of SNR and

various transmit correlation coefficients. Receive correlation 0,5.

Exponential correlation model, uncorrelated fading.

Relative water-filling gain for a 4x4 MIMO system as a function of SNR

and various transmit correlation coefficients, receive correlation 0.7.

Exponential correlation matrix model, uncorrelated fading.

Bit error ratio (BER) as a function of SNR for uncorrelated MIMO channel

with T=4 transmit and R receive antennas. Rank of channel matrix is L=4.

The slope of the asymptotics depend on R, which is clearly visible.

T - number or tx antennas

R - number of rx antennas

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