Fundamentals of MIMO W Wireless Communications Part IV Fundamentals of MIMO Wireless Communications 11. Introduction to MIMO detection • In general, ... Wireless Communication Systems

Fundamentals of MIMO WPart IV

Prof. Rakhesh Singh

Wireless CommunicationsPart IV

Singh Kshetrimayum

Introduction to MIMO detection

• In MIMO detection, we need to detect signals jointly

• since many signals are transmitted from

• the transmitter to the receiver

• For instance, consider a 2×1 MIMO system with • For instance, consider a 2×1 MIMO system with

• two transmit antennas and one single receive antenna

• Two antennas are transmitting two signals at the same time,

• hence the receiving antenna receives both signals

• Hence, we need to detect both the signals jointly

4/18/2017 Fundamentals of MIMO Wireless Communications


In MIMO detection, we need to detect signals jointly

since many signals are transmitted from

1 MIMO system with 1 MIMO system with

two transmit antennas and one single receive antenna

Two antennas are transmitting two signals at the same time,

hence the receiving antenna receives both signals

Hence, we need to detect both the signals jointly

Fundamentals of MIMO Wireless Communications 2










• Maximum likelihood (ML) detector

• Let us consider a MIMO system whose I

• at any symbol time t for frequency flat fading is given by

• r =H s +n• rt=Htst+nt

• where symbol time slot t=1,2,…N

• NL may be considered as frame or packet length



Maximum likelihood (ML) detector

Let us consider a MIMO system whose I-O relation (matrix form)

at any symbol time t for frequency flat fading is given by

where symbol time slot t=1,2,…NL and

may be considered as frame or packet length



• In component form, it can be expressed as

tt

tt

t

t

hhh

hhh

r

r

2,22,21

1,12,11

,2

,1

L

L


=

NtNtN

tt

tN

t

RRRhhh

hhh

r

r

,2,1

2,22,21

,

,2

L

OMM

L

M


In component form, it can be expressed as

t

t

t

t

tN

tNT

n

n

s

sh

,2

,1

,2

,1

,2

,1


+

tN

t

tN

t

tNN

tN

RTTR

T

n

n

s

s

,

,2

,

,2

,

,2

MMM


• ML detection outputs the vector

• which minimizes the Euclidean distance

• between the received vector and

• all possible combinations of the transmitted symbol vectors• all possible combinations of the transmitted symbol vectors


2min

ˆ arg= −s r Hss


which minimizes the Euclidean distance

between the received vector and

all possible combinations of the transmitted symbol vectorsall possible combinations of the transmitted symbol vectors



• Example:

• Explain the ML detection for a 2 × 2 MIMO system

• Consider a 2 × 2 MIMO system at time instant t

• We have the received signal, channel matrix, transmitted signal and • We have the received signal, channel matrix, transmitted signal and noise vector as follows


=

=

2221

1211

2

1;

hh

hh

r

rHr


2 MIMO system

2 MIMO system at time instant t

We have the received signal, channel matrix, transmitted signal and We have the received signal, channel matrix, transmitted signal and


=

=

2

1

2

1;;

n

n

s

sns


• Now we can write the received signal vector

• for frequency flat fading as follows

+= nHsr


=

⇒

+=

2221

1211

2

1

hh

hh

r

r

nHsr

12121111 ;rnshshr ++=


Now we can write the received signal vector

for frequency flat fading as follows


+

2

1

2

1

22 n

n

s

s

22221212 nshshr ++=


• At the detector, we want to detect santenna 1) and s2 (symbol transmitted from antenna 2) at time t,

• but there exist interference of these two signals

• for both the receiving antennas• for both the receiving antennas

• Assume that are modulated in M-ary

• We need to find the minimum metric of the Euclidean distance


{ }1 2, , ,

k Ms s s s∈ L

({ }

1 11 12 2 21 22

1 2

min

, , , ,

i j i jr h s h s r h s h s

i j M

− + + − +

∈ L


At the detector, we want to detect s1 (symbol transmitted from (symbol transmitted from antenna 2) at time t,

but there exist interference of these two signals

for both the receiving antennasfor both the receiving antennas

ary constellation

find the minimum metric of the Euclidean distance


}, , ,

) ( )2 2

1 11 12 2 21 22i j i jr h s h s r h s h s

− + + − +


• For instance,

• 16-QAM, (s1,s2) are (1 of 16 symbols, 1 of 16 symbols)

• implies 16×16 pairs

• Metric calculations of 256 are required• Metric calculations of 256 are required

• For 3×3 MIMO system, (s1,s2, s3) are (1 of 16 symbols, 1 of 16 symbols, 1 of 16 symbols)

• 163=4096 metric calculations of are required

• For 5×5 MIMO system, (s1,s2, s3, s4, s5) are 1 of 16 symbols each

• 165=10,48,576 metric calculations are required

• which is obviously impractical



QAM, (s1,s2) are (1 of 16 symbols, 1 of 16 symbols)

Metric calculations of 256 are requiredMetric calculations of 256 are required

3 MIMO system, (s1,s2, s3) are (1 of 16 symbols, 1 of 16

=4096 metric calculations of are required

5 MIMO system, (s1,s2, s3, s4, s5) are 1 of 16 symbols each

=10,48,576 metric calculations are required



• In general, the decoding complexity increases exponentially

• where NT is the number of transmit antennas and M is the signal

TN

S M=

• where NT is the number of transmit antennas and M is the signal constellation size

• Performance analysis

• Let us try to find the PEP for detecting

• when the signal vector transmitted was


( ) (21 Pr Hsyss −=→ obP


In general, the decoding complexity increases exponentially

is the number of transmit antennas and M is the signal

TN

S M=

is the number of transmit antennas and M is the signal

Let us try to find the PEP for detecting s2

when the signal vector transmitted was s1


)2

1

2

2 HsyHs −≤


• Define (it is like your codeword difference matrix)

• Note that

21 ssd −=

hhn 12111


=

=

RRR NNN hh

hh

hh

n

n

n

MMM

1

2221

1211

2

1

; Hn


Define (it is like your codeword difference matrix)

TN dhL 1112


=

TTRR

T

T

NNN

N

N

d

d

d

h

h

h

M

L

MO

L

L

2

1

2

222

112

;d


• The PEP can be calculated as

( )

=→0

2

212N

QPHd

ss

• Using Chernoff’s bound, PEP is bounded as


02N

( )

−≤→ 21 expP

Hdss


2

bound, PEP is bounded as


0

2

4N

Hd


• When we assume that the first matrix is

• Hd is already a vector, we are trying to find an alternate form of

( ) (CABCvecT ⊗=Q

( ) ( )IABAB vecvec =

• Hd is already a vector, we are trying to find an alternate form of representation

• which will be useful in calculating the average PEP from

• the MGF of a random quadratic form of a complex Gaussian multivariate v

• Using the above identity


(vec vec∴ = ⊗Hd d I H


When we assume that the first matrix is I, we have

is already a vector, we are trying to find an alternate form of

) ( )BA vec⊗

( ) ( )AIB vecT ⊗=

is already a vector, we are trying to find an alternate form of

which will be useful in calculating the average PEP from

random quadratic form of a Hermitian matrix A in


) ( ) ( )R

T

Nvec vec∴ = ⊗Hd d I H


• Therefore, the average PEP with respect to

( )

−≤

>→<

2

21

4exp

NE

P

Hd

ss


( )

( )( ) ( ) (

⊗

−=

−≤

−≤

0

0

0

4exp

4exp

4exp

N

vec

E

NE

NE

R

TH

NTH

H

dIdH

HdHd Q


Therefore, the average PEP with respect to h is given by


) ( )

⊗ vec

RNT

HI

( ) ( ) ( )R

T

Nvec vec= ⊗ =Hd d I H HdQ


• Using the identity on the Kronecker

( )( )

( )⊗∴

=⊗⊗

*

R NH

NT

IIdd

ACDCBAQ


( )( )

( )( )(

−≤

>→<

⊗∴

21

expvec

E

P

R NN

H

ss

IIdd


Kronecker product, we have,

⊗=

⊗

*

RR NT

Idd

BDAC


) ( ) ( ))

⊗

⊗=

0

*

4N

vecR

RR

NTH

N

HIdd

Idd


• Theorem:

• Consider the random quadratic form of a complex Gaussian multivariate =v


• The MGF of the y is given as

( ) Hy Quad= =

Av vAv

( ){exp

y

s sM s

=

v vµ A I R A


Consider the random quadratic form of a Hermitian matrix A in

( ),N

CN=

v Vv µ R


{ } ( )1 H

v

v

s s

s

− −

−

v vµ A I R A µ

I R A


• We can show that for a symmetric and positive semi

• and note that h=vect

• In the mgf, if we put s=-1 and µv=0

( )( ) (

( )hRh ,0~ cN

• For and iid Rayleigh fading


( )( ) (detexp =− IAhhH

E

0

*

4N

RNT Idd

A⊗

=

( )21 det≤→∴ P ss

+ J. Choi, Optimal Combining & Detection, Cambridge University Press, 2010


We can show that for a symmetric and positive semi-definite matrix A

vect(H) and µv=0

0 in the previous theorem,

)

Rayleigh fading Rh=I


) 1−+ hARI

( ) 1

0

*

4det

−

⊗+

N

RNT

IddI

, Cambridge University Press, 2010.


( )21 det

≤→∴ P ssIII =⊗Q

( )P ≤→∴ ss

• Diversity gain

• From the above equation on the upper bound on PEP

• we can say that the diversity gain of the ML detection is N


( )P ≤→∴ 21 ss

Introduction to MIMO detection1

0

*

4

−

⊗

+

RN

T

NI

ddI

RNT

−

+≤*

detdd

I

From the above equation on the upper bound on PEP

we can say that the diversity gain of the ML detection is NR


N

+≤

04det I


• Another alternative MIMO detection technique

• Employ simpler and easy to implement linear detectors

• but they have poorer performance

• Linear sub-optimal detectors• Linear sub-optimal detectors

• In linear detector,

• a linear preprocessor (W) is first applied to the received signal vector

• the estimated symbol is given by


rWsH=ˆ


Another alternative MIMO detection technique

Employ simpler and easy to implement linear detectors

but they have poorer performance

) is first applied to the received signal vector



• Then each element of estimate ( )

• is considered as the received signal

• in the absence of other signals and

• from which the associated signal is independently detected

s

• from which the associated signal is independently detected

• ZF detector

• In ZF detector,

• the linear preprocessor suppress the other signals completely



Then each element of estimate ( )

is considered as the received signal

in the absence of other signals and

from which the associated signal is independently detected

s

from which the associated signal is independently detected

the linear preprocessor suppress the other signals completely



• The preprocessor output is given by

• where is the Moore Penrose pseudo

srHrWs+ === H

ZFˆ

( ) HHH 1−+ ==• where is the Moore Penrose pseudoH

• Example

• Show that for ZF


( ) HHHZF HHHHW

1−+ ==

( )HHZF HHHW

1−+ ==


The preprocessor output is given by

where is the Moore Penrose pseudo-inverse of

nHs++

where is the Moore Penrose pseudo-inverse of


HH

1−


• Note that the ZF searches for unconstrained vector

• (not constrained to alphabet S) that

• minimizes the squared Euclidean to the received vector

• This can be done by taking partial derivative

• w.r.t. and setting to 0 as follows


2min

arg

Hsr

s

−

∈ TNC


Note that the ZF searches for unconstrained vector

(not constrained to alphabet S) that

minimizes the squared Euclidean to the received vector r as

TNC∈s

This can be done by taking partial derivative


2Hsr −


• Wirtinger Calculus

• Complex derivative of a complex function f(z)

• For a function f(z) of a complex variable z=x=

• its derivative w.r.t. z and z* are defined as• its derivative w.r.t. z and z* are defined as


( ) ( ) ( )1 1

2 2;

f z f z f z f z f z f zj j

z x y x y

∂ ∂ ∂ ∂ ∂ ∂ = − = + ∂ ∂ ∂ ∂ ∂∂

+ K. L. Du and M. N. S. Swamy, Wireless Communication Systems From RF Subsystems to

4G Enabling Technologies, Cambridge University Press, 2010.


Complex derivative of a complex function f(z)

For a function f(z) of a complex variable z=x=jy Є C, x,y Є R,

. z and z* are defined as. z and z* are defined as


( ) ( ) ( )1 1

2 2*;

f z f z f z f z f z f zj j

z x y x yz

∂ ∂ ∂ ∂ ∂ ∂ = − = + ∂ ∂ ∂ ∂ ∂∂

Wireless Communication Systems From RF Subsystems to



• For example,

• For multiple complex variable system

( )( ) (

( )( )

* *

, , ; , , ;

, ,

f z f z f z f zf z az a f z az a

z zz zf z f z

f z zz z zz

∂ ∂ ∂ ∂= = = = = =

∂ ∂∂ ∂∂ ∂

= = =∂

• For multiple complex variable system

• The gradient can be defined as


( )1 2, , ,

Tn

nz z z C= ∈z L


For multiple complex variable system

( )( )

( ) ( )

( )0 0*

* *

*

, , ; , , ;

, ,

f z f z f z f zf z az a f z az a

z zz zf z f z

f z zz z zz

∂ ∂ ∂ ∂= = = = = =

∂ ∂∂ ∂∂ ∂

= = =∂

1

*

f f

z z

∂ ∂

∂ ∂ For multiple complex variable system


1 1

2 2

*

*

*

*

,

n n

z z

f f

z zf f

f f

z z

∂ ∂

∂ ∂ ∂ ∂∂ ∂ = =

∂ ∂ ∂ ∂ ∂ ∂

z zM M


• Similarly,( )

( )

( )( )

*

, ,

, ,

T T

T H

f ff

f ff

∂ ∂= = = =

∂

∂ ∂= = = =

z zz c z z c c 0

z

z zz c z z c 0 c


( )( )

( )

( )1 2

*

* *

, ,

, , ;

, , ,

T H

H T T

Tn

n

f ff

f

c c c C

∂ ∂= = = =

∂

∂ ∂= = = =

= ∈

z zz c z z c 0 c

z

z z Mz z Mz M z Mz

c L


( )

) ( )

*

*

, ,

, ,

f f

f f

∂ ∂= = = =

∂

∂ ∂= = = =

z zz c z z c c 0

z

z zz c z z c 0 c


) ( )

( ) ( )

*

* *

*

, ,

, , ;H T T

f f

f f

∂ ∂= = = =

∂

∂ ∂= = = =

∂ ∂

z zz c z z c 0 c

z z

z zz z Mz z Mz M z Mz

z z


• Hence,( ) ( )

( HsHsrrrs

HsrHsrs

HHH

H

H

H

−−∂

∂=

−−∂

∂

• Then we obtain


(

HsHrH

sHH

H

+−=

∂

( )(H

ZF

H

HH

HHW

HHHs

rHHsH

1

+

−

==∴

=⇒

=


)HsHsrHHHH +


)

) HH

H

HHH

r

1−


• We could also obtain the same relation by taking gradient

• What happens to noise power for ZF?

• Let us denote noise after ZF as

( )−+ HH 1

• The error performance of MIMO depower of or

• Using the SVD , the post


( ) nHHHnH ==−+ HH 1

nH+ 2

2nH

+

( )HVΣUH =


We could also obtain the same relation by taking gradient w.r.t. s

What happens to noise power for ZF?

detection is directly related with the

Using the SVD , the post-detected noise power is


z=


( ) (

22

22

22

2

12

2

HHH

HH

HH

===

∑=∑∑=

∑==

−

−

xxxQxQxQx

VnUVVV

VnHHHz

Q


{ } {

( ){ } {

2min

2

12

2

11

2

2

12

2

2

2

2

2

σ

σ

σ

σ n

N

i i

n

HH

H

HHH

T

trEtr

trEEE

≈=

∑=∑∑=

=

∑=∴

===

∑=

−−

−

UUnn

nUz

xxxQxQxQxQ


)

2

21

2

2

1

H

HH

∑

∑

−

−

nU

nUVV


( )}

} { }2222

11

2

2

σ nn

HH

tr

tr

∑=∑

∑∑

−−

−−

σ

UnnU


• Looking at the above equation, for not well behaved channel matrix,

• is very small and hence will be a large number

• Main hurdle of linear detector:

2minσ 2

min

2

σ

σn

• Main hurdle of linear detector:

• noise power is getting amplified due to

• application of the linear preprocessor (

• for ill behaved channel matrix

• Possible solution: Employ techniques like lattice reduction (LR)



Looking at the above equation, for not well behaved channel matrix,

is very small and hence will be a large number

noise power is getting amplified due to

linear preprocessor (W)

Possible solution: Employ techniques like lattice reduction (LR)



• SINR for ZF

• Post-detected noise :

• is a zero mean circular symmetric complex Gaussian with covariance matrix given by

( HHnH =+ H

matrix given by


( ) ( ) (

( ) ( ) ( )

1

1 1 1 12 2

H H H H H H H H H

zz

H H

H H H H H

n n

E E E

σ σ

−

− − − −

= = = = = =

R zz H H H nn H H H H H H nn H H H

H H H H H H H H H H


is a zero mean circular symmetric complex Gaussian with covariance

) znHH =− H1


) ( ) ( ) ( )

( )

1 1 1

1 1 1 12 2

H H

H H H H H H H H H

H H

H H H H H

n n

E E E

σ σ

− − −

− − − −

= = = = = =

R zz H H H nn H H H H H H nn H H H

H H H H H H H H H H


• Then we can obtain the kth diagonal element of

( ) ( )Hnkk

=

−12,

HHRzz σ

• Consider the received signal in the i


[ iiii hhhr = 2,1, L


diagonal element of aszzR

kk

Consider the received signal in the ith antenna given by


] iNi nhT

+s,


If we assume that kth stream is the desired signal,

• then, we can express the above received signal

∑+=TN

jikkii hshr ,,

• Instantaneous signal to interference noise ratio (SINR) for the received symbol as


∑≠= kjj

jikkii

,1

,,

( )kk

kZFZF

ESINR ==

,R zz

ργ


stream is the desired signal,

then, we can express the above received signal

+ ij ns

Instantaneous signal to interference noise ratio (SINR) for the kth


ij

( )kk

Hn

kE

=−12

HHσ


• where is the mean SNR

• SINR of ZF has been shown+ to be a Chi

• is distributed with degrees

ρ

γ χ• is distributed with degrees


ZFγ

( )2 1R T

N Nχ

− +

+ M. Rupp, C. Mecklenbrauker and G. Gritsch, “High diversity with simple space

block-codes and linear receivers,” in Proc. IEEE GLOBECOM


to be a Chi-square RV

is distributed with degrees-of-freedom( )− +is distributed with degrees-of-freedom


( )2 1R T

N N− +

, “High diversity with simple space-time

IEEE GLOBECOM, 2003, pp. 302-306.


• Example

• Find the outage probability of ZF

• Consider the separate spatial encoding case

• the data is demultiplexed (DMUX) to several sub• the data is demultiplexed (DMUX) to several sub

• each one of them separately encoded and

• feed to the corresponding transmitting antenna and

• sent through the channel


+ A. Hedayat and A. Nostrania, “Outage and diversity of linear receivers in flat

MIMO channels,” IEEE Trans. Signal Processing, vol. 55, no. 12, Dec. 2007, pp. 5868

5873.


Consider the separate spatial encoding case+,

(DMUX) to several sub-streams, (DMUX) to several sub-streams,

each one of them separately encoded and

feed to the corresponding transmitting antenna and


, “Outage and diversity of linear receivers in flat-fading

, vol. 55, no. 12, Dec. 2007, pp. 5868-


• If any one of the data sub-stream is for each sub-streams),

• the whole MIMO system is in outage

• The mutual information between • The mutual information between

• the kth transmitted symbol vector and

• kth estimated symbol vector at the output of the ZF detector



ks

( ) (kk SINRI += 1logˆ; 2ss


is in outage (assume equal data rate

the whole MIMO system is in outage

transmitted symbol vector and

estimated symbol vector at the output of the ZF detector+



ks

) ( )ZFZFSINR ργ+= 1log2


• outage probability for a target data rate of R

(

−= ITN

out

Iob

P

ˆ;Pr1 ss


(

(

−=

−=

=

=

I

I

TN

k

k

kk

ob

Iob

1

2

1

1logPr1

ˆ;Pr1 ss


outage probability for a target data rate of R

)

≥R


)

)

≥+

≥

T

ZF

T

k

N

R

N

ργ

Introduction to MIMO detection• Assume independent and equal sub

( ZF

out

ob

P

+−= ργ1logPr1 2

• For outage probabilities for sub-channels are small, we have,


( )

(

+=

<+≈

ZFT

ZF

out

obN

ob

P

ργ

ργ

1logPr

1logPr

2

2

Introduction to MIMO detectionAssume independent and equal sub-channel outage probabilities

)TN

T

ZFN

R

≥

channels are small, we have,


T

)

<

<

T

ZF

N

R

N

R

N

RT

Introduction to MIMO detection• Since is distributed , outage probability from the CDF ZF

γ ( )2 1R T

N Nχ

− +

−

<≈12

PrN

R

ZFT

out

obN

P

T

ργ


−= ∑=

−

− 1+N-N

1

12

TR

1

i

T eN

TN

R

ρ

Introduction to MIMO detectionSince is distributed , outage probability from the CDF

− 1iR


( )

−

−

− 1

!1

12

i

N

R

i

T

ρ


• Example: Show that the outage probability for ZF MIMO detection decays as

• Solution

1

1+− TR NNρ

• Solution

• Let i goes from 0 to


R TN N−

=outP


Example: Show that the outage probability for ZF MIMO detection

iR


( )

−

− ∑=

−

− N-NTR

0

12

!

12

1

i

N

R

Ti

eN

T

TN

R

ρ

ρ


• Using the infinite series expansion of exponential function, we get,

−

− 212 TN

R

TN

R


−=

−

−

−

212

1Tout eeNP

TNTN

ρρ


Using the infinite series expansion of exponential function, we get,

−

−1

12

i

N

R

T

ρ


( )

− ∑

∞

+=

−

N-NTR

1

1

!i

i

ρ


−12

2N

TN

R

• Pout


= ∑

∞

+=

−

1N-N TRi

T eN ρ


−1

i

N

R

T

ρ


( )

!i

ρ


• For high SNR case ( ), we have, ∞→ρ

2

N

R

T


=

∞→ −

2

NNTout NPLim

TR

T

ρρ


For high SNR case ( ), we have,

−

+−

1

1NNR TR

T


( )

+−

−

+!1

1

1TR NNT

T


• Hence the diversity gain is

• Performance analysis

• The post-detection SINR of ZF detector is given by

R TN N− +

• Assume hi is the ith row vector of H

• hi has complex multivariate normal distribution


( )Hn

kZFZF

ESINR

12

==−

HHσργ


detection SINR of ZF detector is given by

1R T

N N− +

H, then,

has complex multivariate normal distribution


T

kk

Nk ,,2,1;

,

1L=

( )iiNCi

TN ∑,~ µh


• Suppose all the row vectors hi have

• the complex multivariate normal distribution with

• the same covariance matrix Σ

• Then follows a complex H=Z H H• Then follows a complex

• where


H=Z H H

( )ΣMZ ,,~ RN

C NW T

[ ]TNRµµµM ,,, 21 L=


have

the complex multivariate normal distribution with

Then follows a complex Wishart distribution denoted byThen follows a complex Wishart distribution denoted by



• For M=0, we have central complex

• M≠0, then we have non-central complex

• One can also convert

• non-central complex Wishart to central complex • non-central complex Wishart to central complex distribution

• The non-central complex Wishart distribution can be approximated

• by central complex Wishart distribution as


( )ΣZ RN

C NW T ;ˆ,~


, we have central complex Wishart distribution and

central complex Wishart distribution

to central complex Wishartto central complex Wishart

distribution can be approximated

distribution as


) MMΣΣH

RN

1ˆ; +=


• The pdf of post-detected SINR (γk) for distribution is given by +

γ

ZFSINR =


( )( )

( ) ( ) (

k

TRkk

kk

k

k

NN

p

ˆ1

ˆ

ˆ

exp

11

1

+−Γ

−

=

−−

−

ΣΣ

Σ

ρ

γ

ρ

ρ

γ

γ

+ D. Gore, R. W. Heath and A. Paulraj, “On performance of the zero forcing receiver in presence

of transmit correlation,” in Proc. IEEE Int. Symp. on Information Theory

2002, pp. 159.


) for Z following complex Wishart

( )T

kk

Hn

kZF Nk

E,,2,1;

,

12

L=

=−

HHσργ


)T

NN

kk

Nk

TR

,,2,1;

1

L=

−

, “On performance of the zero forcing receiver in presence

. on Information Theory, Lausanne, Switzerland,


• Hence CDF is given by

,1

+− TR NN

ργ

• where is the mean SNR


( )( −Γ

=TR

kNN

P

ρ

γ

2n

kE

σρ =


( )

k

ρ

γ


( ))1

ˆ 1

+

−

T

kkΣ

ρ


• The average BER for kth symbol is given by

( )

( ) ( )

e kP

∞


( ) ( )

( ) ( )( )k

TRkk

kkk

Q

NN

dpQ

γρ

γγγ

∞

−

∞

+−Γ

=

=

∫

∫

01

0

exp2

1ˆ

1

2

Σ


symbol is given by


( ) ( )k

NN

kk

k

kk

kd

TR

γρ

γ

ρ

γ

−

−−

−

11 ˆˆ ΣΣ


• Let , then( )

= −kk

k 1Σ

ργγ

( )( ) (

ργQNN

kPe

∞

+−Γ= ∫ ˆ

21

1

Σ


( ) (NN TR

e

+−Γ ∫ ˆ1

0Σ

( )( ) ( )1

0

, ,m

q mqI p q m Q p e d

m

γγ γ γ∞

− −=Γ

∫

2

=p


) ( )( ) γγγ dTR NN −− −

exp1Σ


)kk−

1

Σ

1q mI p q m Q p e d

γγ γ γ− −

( ) 1,1,ˆ 1 +−==

− TRkk

NNmqΣ

ρ


• The above integration can be further simplified (for positive integer values of m) to

• where

( )

−=1 12

1,, mqpI

• where


(qp

p

2

2

2

=+

=ρ

ρ

ζ

2


The above integration can be further simplified (for positive integer

−

∑

−

=

1

0

2

4

12m

k

k

k

k ζζ


( )

( )( )kk

kk

kk

1

1

1

ˆ2ˆ

ˆ

−

−

−

+=

+

Σ

Σ

Σ

ρ

ρ

ρ

=04

kk


• Therefore, average BER for symbol k is simply

• Hence we need to find the

( ) ( )

= − ,1,ˆ

2 11 Rkk

e NIkPΣ

ρ

( )1ˆ −• Hence we need to find the

• Let us consider i.i.d. Rayleigh fading MIMO channel


( ) 1ˆ

ˆ

0

1 =⇒

==∴

=

−kk

NTI

Σ

ΣΣ

MQ

( )kk1ˆ −

Σ


Therefore, average BER for symbol k is simply

+− 1TR N

. Rayleigh fading MIMO channel



• Therefore, average BER for symbol k is

• MMSE detector

( ) (12 1 1, ,

e R TP k I N Nρ= − +

• MMSE detector

• As we have seen for ZF, noise was getting enhanced

• even if the spatial interference was removed

• MMSE detector minimizes the mean

• of the spatial interference plus noise



Therefore, average BER for symbol k is

)2 1 1, ,e R T

P k I N N= − +

As we have seen for ZF, noise was getting enhanced

even if the spatial interference was removed

MMSE detector minimizes the mean-square value

of the spatial interference plus noise



• MMSE preprocessor

HHrHIHHrWs HH

s

HHMMSE

E

N1

0ˆ

−

+=

+==

• MMSE pre-processor matrix is similar to ZF preexcept

• for an extra term which will reduce the noise enhancement


sE

ISE

N0


( ) ( )nHIHHHsHI H

s

HH

s E

N

E

N1

0

1

0

−−

++

+

processor matrix is similar to ZF pre-processor matrix

for an extra term which will reduce the noise enhancement


ss EE


• Show that

• Solution

• In MMSE detector, one tries to minimize

( 22 += Hs

Hs

HMMSE HHHW σσσ

• In MMSE detector, one tries to minimize

• the mean square error between the


min

arg

rW

s ∈ ×

H

NN

E

C RT


In MMSE detector, one tries to minimize

) 12 −

TNn Iσ

In MMSE detector, one tries to minimize

the actual signal and detected signal


2

sr −


• This can be done by taking partial derivative

• w.r.t. W and setting to 0

( )(WsrW

−

∂ HHtrE


( )(

({[(

(

srrr

rr

RRW

RWWRWW

WWrrWW

WsrWW

−=

−∂

∂=

−∂

∂=

−

∂

H

HH

HH

HH

trE

trE


This can be done by taking partial derivative 2

srW −HE

)sr −

HH


)

)}])

)sssrrs RWRR

ssWsrrsW

sr

+−

+−

−

HHHH

H


• Hence,

• Assuming noise vector and signal vector are independent

1−= rrsrRRWHMMSE

Hs

HR HHRHHR nnssrr

2σ +=+=

• Therefore,


Hs

HR HHR sssr

2σ==

( 222 +=TNn

Hs

Hs

HMMSE IHHHW σσσ


Assuming noise vector and signal vector are independent

RTT NnNsNn IRIRI nnss222 ,; σσσ ==+


RTT

) 1−

T


• Example

• What happens to noise power for MMSE?

• Let us denote noise after MMSE as

• Using SVD of H+ as , we have,


HIHH

+

−

s

H

E

N1

0

HVΣUH =


What happens to noise power for MMSE?

Let us denote noise after MMSE as

as , we have,


( ) znH =H


2 21 1

20 0

2

2 2

H H H H

s s

N N

E E

− −

= + = +

z H H I H n V

z2

2∴


( ) ∑=−−

VVΣ11 H

Q

V

V

z

s

H

E

N 0

2

+∑=

+∑=

∴


2 21 1

20 0

2 2

H H H H

s s

N N

E E

− −

= + = +

z H H I H n VΣ V I VΣU n


( )

( )nUΣ

nUVΣ

H

s

HH

sE

N

1

10

1

10

−

−

−

−


• Since the multiplication of a unitary matrix do not change the Frobenius norm, we have

z2

2

• We also know that,


( )nUΣH

sE

N1

10

−

−

+∑=

( ) ( ) BBBBB == HHTrTr


Since the multiplication of a unitary matrix do not change the


)

2

2


( )1

210

2

1 1

1 10 0

s

H H

s s

NE E

E

N NE tr

E E

−

−

− −

− −

= +

= + +

z Σ Σ U n

Σ Σ U nn U


(1 1

1 10 0

2

10

0

s s

H H

s s

s

E E

N Ntr E

E E

Ntr N

E

− −

− −

−

−

= + +

= +

Σ Σ U nn U

Σ Σ

Introduction to MIMO detection2

2

1 1

1 10 0

H

H H

s s

N N

E E

− −

− −

= + +

U n

U nn U Σ Σ


)1 1

1 10 0

s s

H H

s s

E E

N N

E E

− −

− −

= + +

U nn U Σ Σ


( )

(

2

20 0

0 021 1

2

0 0 0

T T

T T

N N

ii is i s i

N N

s i s i s

N E NE N N

E E

E E EN N N

σσ σ

σ σ σ

−

= =

= + =

= = ≈

∑ ∑

∑ ∑

z

• Hence, unlike ZF where ,

• the noise enhancement in MMSE is ZF detector


(0 0 02 2 2

1 10i is i

N N NE N E N E Nσ= =

= = ≈ + ∑ ∑

( )2

2min

NE

σ=z

+ Y. S. Cho, J. Kim, W. Y. Yang and C.-G. Kang, MIMO

MATLAB, Wiley, 2010.


) ( )

22

0 0

0 01 1

2 2 2 2

0 0 0

min

min

T TN N

s i

i is i s i

s i s i s

N E NE N N

E E

E E EN N N

σ

σ σ

σ σ σ

−

= =

+= + =

= = ≈

∑ ∑

Hence, unlike ZF where ,

E is less pronounced than that of the


) ( )0 0 02 2 2

2 2

0 0

min

mins i s

N N N

E N E Nσ σ= = ≈

+ +

0

2

min

N

σ

MIMO-OFDM wireless communications using

Introduction to MIMO detection• Performance analysis

• In linear detector, a linear preprocessor (received signal vector

rWs H=ˆ

• Then we can do individual detection of

• Without loss of generality, let us assume that we are detecting


rWs H=ˆ

1ˆ ,k

s k =


In linear detector, a linear preprocessor ( W) is first applied to the

( ) 1 2

H

H

N

=

W w w wL

Then we can do individual detection of

Without loss of generality, let us assume that we are detecting


s

( ) 1 2T

N =

W w w wL


• Then

• One may show that

nwhwrwHHH

ss 111111 +==1 1 11 12 1

= =h w

• One may show that

• The conditional error probability (CEP) for sub


( )21 ,0~ nCH

N σnw

= 12 hw

HQCEP


11

21

1 1 11 12 1

1

;R

H

N

N

h

h

w w w

h

= =

h wM L

The conditional error probability (CEP) for sub-channel 1 is given by


1h

1

;

RN

h


• For sub-channel 1, the corresponding weight vector is proportional to

1

1

0111

ˆˆ hIHHw

−

+∝

s

H

E

N

• We can partition the channel matrix

• where h1 is the first column vector for the desired sub

• and is the matrix after removing the first column


sE

H


channel 1, the corresponding weight vector is proportional to

1

1

0

1

0

H H H

s

H

N

EN

−

−

= +

⇒ = +

W H H I H

W H HH I

Q

We can partition the channel matrix H as

is the first column vector for the desired sub-channel 1

and is the matrix after removing the first column


[ ]HhH ˆ1=

0

sE

⇒ = +

W H HH I


• Using eigen-decomposition of

= 11

ˆˆ2 HHhH

QCEP

H• Using eigen-decomposition of

Assuming


1H

s

H

E

NλUIHH

=+ 0

11ˆˆ

1hUxH=


decomposition of , we have,

+

−

1

1

01 hIH

s

H

E

N

HHdecomposition of , we have,


H11H

H

sE

NUIλ

+ 0


• Note that the rank of is

1

01

1

0111

ˆˆN

i s

i

s

HH

E

N

E

N R

=

−

∑

+=

+ λhIHHh

H N −1• Note that the rank of is

• hence, eigenvalues of are zero


1H TN −1

1+− TR NN

∑−

=

−

=

+

TR NN

i

s

s

HH

N

E

E

N

101

1

0111

ˆˆ hIHHh


2

1

0i

s

x

−

N≤1

1hUx H=

HH

E

N

E

NUIλUIHH

+=+ 00

11ˆˆ

of are zero


RN≤1

H11

ˆˆ HH

∑∑+−=

−+

++

R

TR

T N

NNi

i

s

ii xE

Nx

2

21

0

12

λ

ss EE

11


• Hence

• Since

−≤ ∑

+−

=

TR NN

i

is

xN

ECEP

1

1

2

0

expexp

( )2H σ=• Since

• we know that all are independent of each other


( )IhUx2

1 ,0~ hcH

N σ=

−≤ ∑

+−

=

TR NN

i

is Ex

N

EEBER

1

1

2

0

exp


+− ∑

+−=

−R

TR

N

NNi

i

s

i xE

N

2

21

0exp λ

we know that all are independent of each other


+− ∑

+−=

−R

TR

N

NNi

i

s

i xE

N

2

21

0exp λ


• which can be approximated as

11

1N

BER

RNN TRγ

+

≅

+−


11

1

N

BER

R

γ

γ

++

+≅

+ J. Choi, Optimal Combining & Detection, Cambridge University Press,


2

1

;

1

E hb

NT

σγ

γ=

−


0

;1 N

E hbσγ

γ

γ=



• The diversity gain for ML detection was N

• whereas ZF and MMSE detectors have diversity gain of

• MMSE has slightly higher diversity than ZF

• we will discuss this in conservation theorem• we will discuss this in conservation theorem

• Sphere Decoding

• the complexity of ML detection grows exponentially

• Is there way to reduce this complexity without compromising the performance?

• That’s what sphere decoding (SD) exactly does



The diversity gain for ML detection was NR

whereas ZF and MMSE detectors have diversity gain of

MMSE has slightly higher diversity than ZF

we will discuss this in conservation theorem

1R T

N N− +

we will discuss this in conservation theorem

the complexity of ML detection grows exponentially

Is there way to reduce this complexity without compromising the

That’s what sphere decoding (SD) exactly does



• How does SD achieve this?

• It tries to find the ML solution vector within a sphere

• instead of all possible transmitted signal vectors (ML detection)

• But there may be • But there may be

• no vector at all or

• numerous vectors

• inside the chosen sphere

• How to handle such situations?



It tries to find the ML solution vector within a sphere

instead of all possible transmitted signal vectors (ML detection)



• In the first case,

• one may increase the radius of the sphere

• In the second case,

• one may decrease the radius of the sphere • one may decrease the radius of the sphere

• so that only one vector exists inside the sphere

• which will give us the ML solution

• Hence, SD is an iterative decoding

• which converges to the ML solution

• when the number of iterations is unbounded



one may increase the radius of the sphere

one may decrease the radius of the sphere one may decrease the radius of the sphere

so that only one vector exists inside the sphere

which will give us the ML solution

Hence, SD is an iterative decoding

which converges to the ML solution

when the number of iterations is unbounded



• First step:

• converting the complex I-O MIMO system model

• into an equivalent real system model


real real real real

equi equi equi equi⇒ = +y H x n

( )( )

( ) ( )( ) ( )

Re Re Im Re Re

Im Im Re Im Im

− ⇒ = +

y H H x n

y H H x n


O MIMO system model

into an equivalent real system model

= +y Hx n


( )( )

( )( )

Re Re Im Re Re

Im Im Re Im Im

= +

y H H x n

y H H x n


• Example

• Convert a complex MIMO I-O model to an equivalent real system model

• Solution:• Solution:

• For a MIMO system,


1 11 12 1 1

2 21 22 2 2

y h h x n

y h h x n

= +


O model to an equivalent real system


1 11 12 1 1

2 21 22 2 2

y h h x n

y h h x n

= +


• Separating the imaginary and real parts, we have,

1 1 11 11 12 12 1 1 1 1

2 2 21 21 22 22 2 2 2 2

real imag real imag real imag real imag real imag


y jy h jh h jh x jx n jn


+ + + + +

+ + + + + ⇒ = +

• Hence the real equivalent model is


2 2 21 21 22 22 2 2 2 2y jy h jh h jh x jx n jn+ + + + + ⇒ = +


Separating the imaginary and real parts, we have,

1 1 11 11 12 12 1 1 1 1

2 2 21 21 22 22 2 2 2 2





+ + + + +

+ + + + + = +

Hence the real equivalent model is


2 2 21 21 22 22 2 2 2 2y jy h jh h jh x jx n jn+ + + + + = +


1 11 12 11 12 1

2 21 22 21 22 2

1 11 12 11 12 1

real real real imag imag real


imag imag imag real real imag

imag imag imag real real ima

y h h h h x

y h h h h x

y h h h h x

y h h h h x

− −

− −

=

• MLD for the real equivalent system

• can be expressed as


2 21 22 21 22 2y h h h h x


1 11 12 11 12 1

2 21 22 21 22 2

1 11 12 11 12 1



imag imag imag real real imag

imag imag imag real real ima

y h h h h x

y h h h h x

y h h h h x

y h h h h x

− −

− −

1

2

1

real

real

imag

g imag

n

n

n

n

+

MLD for the real equivalent system


2 21 22 21 22 2y h h h h x

2n

2

arg min real real real

equi equi equi

real real

equi equi

−

∈

y H x

x χ


• MLD search for ML solution over the symbol alphabet

• But for SD, we will search the solution over a sphere of radius

• Hence

(2

real real realr− ≤y H x

• Let us consider the QR decomposition of the real equivalent channel matrix (R is upper triangular matrix)


(real real real

equi equi equi SDr− ≤y H x

( )2 2 2 2 2 21 2R T T R T T

realN N N N N Nequi − × − ×

= =

R R

0 0H Q Q Q


MLD search for ML solution over the symbol alphabet

But for SD, we will search the solution over a sphere of radius rSD only

real

equiχ

)2

r

Let us consider the QR decomposition of the real equivalent channel matrix (R is upper triangular matrix)


)equi equi equi SDr

( )2 2 2 2 2 21 2R T T R T TN N N N N N− × − ×

R R

0 0H Q Q Q


• Note that is a matrix

• Multiplying by and

real

equiH 2 2

R TN N×

1

H

H

Q

• Multiplying by and

• using the unitary property of the Q


2

HH =

QQ


Note that is a matrix

Q matrix, we have,



• Therefore,

(

1

2 2 22R T T

H

H real realN N Nequi equi SD− ×

− ≤

Q R

0Q y x


2 2

1 2

H real real H real

equi equi SD equi⇒ − ≤ −Q y Rx Q y

+ F. A. Monteiro, I. J. Wassell and N. Souto, “MIMO Detection Methods,” in

4G and beyond, M. M. da Silva and F. A. Monteiro


) ( )

2

2

2 2 2R T T

real realN N Nequi equi SD

r− ×

− ≤

Q R

y x


( )2 22

1 2

H real real H real

equi equi SD equir− ≤ −Q y Rx Q y

, “MIMO Detection Methods,” in MIMO Processing for

Monteiro, Eds., Boca Raton: CRC Press, 2014, pp. 47-117.


• Substituting the new

1

n H real

equi=y Q y ( )

22

2

H real

n SD equir r= − Q y

• Hence,

• Since R is upper triangular matrix,

• we can write the above inequality in component form as


( )2 2

n r

nr− ≤y Rx


2r real

equi=x x

is upper triangular matrix,

we can write the above inequality in component form as



• Example

22 2

1 1

T TN N

n r

i ij j ni j

y R x r= =

− ≤

∑ ∑

• Example

• Find the above SD metric for a 2×2

• Solution



( ) 2

i ij j ny R x r− ≤

2 MIMO system



• SD metric for a 2×2 MIMO system

1 111 12 13 14

2 22 23 24 20

0 0

n r

n r

n r

y xR R R R

y R R R x

R R


33 343 3

4 4

0 0

0 0 0

n r

n r

R Ry x

y x

− ≤

1 11 1 12 2 13 3 14 4 2 22 2 23 3 24 4

2 2

3 33 3 34 4 4 44 4

n r r r r n r r r

n r r n r

y R x R x R x R x y R x R x R x

y R x R x y R x r

⇒ − − − − + − − −

+ − − + − ≤


( )

2

1 111 12 13 14

2 22 23 24 2

n r

n r

n r

y xR R R R

y R R R x

R R


( ) 233 343 3

444 40 0 0

n r

nn r

R Ry x r

Ry x

− ≤

( )

2 2

1 11 1 12 2 13 3 14 4 2 22 2 23 3 24 4

2 2 2

3 33 3 34 4 4 44 4

n r r r r n r r r

n r r n r

n

y R x R x R x R x y R x R x R x

y R x R x y R x r

− − − − + − − −

+ − − + − ≤


• Reordering the terms in the LHS, we have,

2 2 2

4 44 4 3 34 4 33 3 2 24 4 23 3 22 2

n r n r r n r r r

n r r r r

y R x y R x R x y R x R x R x

y R x R x R x R x r

− + − − + − − −

+ − − − − ≤

• Similarly, expanding SD metric for a MIMO system, we have,


1 14 4 13 3 12 2 11 1y R x R x R x R x r+ − − − − ≤

2 2

2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 1

1 1 2 2 1 2 1 2 1 11 1

, , ,

, , ,

T T T T T T T T T T T

T T T T

n r n r r

N N N N N N N N N N N

n r r r

N N N N n

y R x y R x R x

y R x R x R x r

− − − − −

− −

− + − −

+ + − − − − ≤L L


Reordering the terms in the LHS, we have,

( )

2 2 2

4 44 4 3 34 4 33 3 2 24 4 23 3 22 2

2 2

n r n r r n r r r

n r r r r

y R x y R x R x y R x R x R x

y R x R x R x R x r

− + − − + − − −

+ − − − − ≤

Similarly, expanding SD metric for a MIMO system, we have,


( )1 14 4 13 3 12 2 11 1 ny R x R x R x R x r+ − − − − ≤

( )

2 2

2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 1

2 2

1 1 2 2 1 2 1 2 1 11 1

, , ,

, , ,


n r n r r


n r r r

N N N N n

y R x y R x R x

y R x R x R x r

− − − − −

− −

− + − −

+ + − − − − ≤L L


• Note that the first term is dependent only on ,

• therefore, we can have a necessary condition as follows

( )2 2

2 2 2 2,

n r

N N N N ny R x r− ≤

• In other words, we can look for in the interval


( )2 2 2 2,T T T T

N N N N ny R x r− ≤

2

rx

2 2

2 2 2

2 2 2 2, ,

T T

T T T

T T T T

n n

n N n Nr

N N N

N N N N

r y r yLB x UB

R R

− + + = ≤ ≤ =


Note that the first term is dependent only on ,

therefore, we can have a necessary condition as follows

2T

r

Nx

In other words, we can look for in the interval


2T

r

Nx

2 2

2 2 2

2 2 2 2, ,

T T

T T T

T T T T

n n

n N n N

N N N

N N N N

r y r yLB x UB

R R

− + + = ≤ ≤ =


• where is the lower bound for ,

• is the upper bound for ,

• is the smallest integer greater than a and

• is the greatest integer smaller than a

2T

NLB

2T

NUB 2

r

Nx

a • is the greatest integer smaller than a

• The second term depends only on and


a

2 2

2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 1

1 1 2 2 1 2 1 2 1 11 1

, , ,

, , ,


T T T T

n r n r r


n r r r

N N N N n

y R x y R x R x

y R x R x R x r

− − − − −

− −

− + − −

+ + − − − − ≤L L


where is the lower bound for ,

is the upper bound for ,

is the smallest integer greater than a and

is the greatest integer smaller than a

2T

r

Nx

2T

r

Nx

is the greatest integer smaller than a

The second term depends only on and


2T

r

Nx 2 1

T

r

Nx

−

( )

2 2

2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 1

2 2

1 1 2 2 1 2 1 2 1 11 1

, , ,

, , ,


n r n r r


n r r r

N N N N n

y R x y R x R x

y R x R x R x r

− − − − −− + − −

+ + − − − − ≤L L


• We can have second condition from the

• first and second term of the SD metric inequality as follows

2 2

2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 1, , ,

n r n r r

N N N N N N N N N N N ny R x y R x R x r

− − − − −− + − − ≤

• Therefore we can look for in the interval


2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 1, , ,T T T T T T T T T T T


− − − − −− + − − ≤

2 1T

r

Nx

−

2 1 2 1

2 12 2 12

2 1 2 1 2 1

2 1 2 1 2 1 2 1

| |

, ,

T T

T T T T

T T T

T T T T

N Nn n

n N N n N Nr

N N N

N N N N

r y r yLB x UB

R R

− −

− −

− − −

− − − −

− + + = ≤ ≤ =


We can have second condition from the

first and second term of the SD metric inequality as follows

( )2 2 2

2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 1, , ,

n r n r r


− − − − −− + − − ≤

Therefore we can look for in the interval


( )2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 1, , ,T T T T T T T T T T T


− − − − −− + − − ≤

2 1 2 1

2 12 2 12

2 1 2 1 2 1

2 1 2 1 2 1 2 1

| |

, ,

T T

T T T T

T T T

T T T T

N Nn n

n N N n N N

N N N

N N N N

r y r yLB x UB

R R

− −

− −

− − −

− − − −

− + + = ≤ ≤ =


• Where2 12 2 1 2 1 2 2| ,

T T T T T T

n n r

N N N N N Ny y R x

− − −= −

( ) ( )2 22 1

2 2 2 2,T

T T T T

N n r

n n N N N Nr r y R x

−= − −

• Following the same procedure,

• we can find the interval in which one can look for


( ) ( ) 2 2 2 2,T T T T

n n N N N N

2 2 2 3 1, , ,

T T

r r r

N Nx x x

− −L


2 12 2 1 2 1 2 2| ,T T T T T T

n n r

N N N N N Ny y R x

2

2 2 2 2,T T T T

n r

n n N N N Nr r y R x

we can find the interval in which one can look for


2 2 2 2,T T T T

n n N N N N


• In SD, the multidimensional search of MLD is transformed

• to multiple searches in one dimension

• Example

• Write the SD pseudo code for a simple 2• Write the SD pseudo code for a simple 2

• Solution:

• Step 1:

• Find the QR factorization of



In SD, the multidimensional search of MLD is transformed

to multiple searches in one dimension

Write the SD pseudo code for a simple 2×2 MIMO system.Write the SD pseudo code for a simple 2×2 MIMO system.



(2 2 21 2

real

equi

=

0H Q Q

• and

• Step 2:

• Set k=4,


1

n H real

equi=y Q y

( )22

2

H real

n SD equir r= − Q y y y


)2 2 2R T T

N N N− ×

R

0


45 4|

n ny y=


• Step 3:

• Set the bounds

1 1| |

, ,

k n k n

n k k n k krr y r y

LB x UB+ +

− + + = ≤ ≤ =

• Step 4:

• Increase


| |

, ,

k k k

k k k k

LB x UBR R

= ≤ ≤ =

1r

k kx LB= −

1k k

x x= +


1 1| |

, ,

k n k n

n k k n k kr y r y

LB x UB+ +

− + + = ≤ ≤ =


| |

, ,

k k k

k k k k

LB x UBR R

= ≤ ≤ =


• Decision 1: ?

• If no then

• Step 6:

• k=k+1

r

k kx UB≤

• k=k+1

• Decision 2: k=5?

• If yes then

• stop.

• If no then go to step 4





• If yes then

• Decision 3: k=1?

• If no then

• Step 5: Decrease k=k-1• Step 5: Decrease k=k-1


2

11

| ,

TN

n n r

k k k k j jj k

y y R x+

= +

= − ∑ ( ) (2 2 2

k k n r

n n k k k k kr r y R x

+= − −



) ( )2 2 2

1

1 2 1 1 1| ,

k k n r


+

+ + + + += − −

Introduction to MIMO detection• If yes then

• Step 7: Save xr and find its distance from

• Go to Step 4: Increase 1k k

x x= +



and find its distance from real

equiy



• Let us summarize

• Find the QR factorization of

• Set for k=4

real

equi=H Q Q

• Set for k=4


1

n H real

equi=y Q y

( )22

2

H real

n SD equir r= − Q y

45 4|

n ny y=


( )2 2 21 2 R T T

realN N Nequi − ×

=

R

0H Q Q



• Choose a candidate point from the following range

• If there exists not candidate point in the range,

1 1| |

, ,

k n k n

n k k n k kr

k k k

k k k k

r y r yLB x UB

R R

+ + − + + = ≤ ≤ =

rx 4

• If there exists not candidate point in the range,

• the radius needs to be increased

• If the a candidate point has been chosen successfully,

• then we proceed to find a candidate point in the range for k=3



MATLAB, Wiley, 2010.


Choose a candidate point from the following range

If there exists not candidate point in the range,

1 1| |

, ,

k n k n

n k k n k kr

k k k

k k k k

r y r yLB x UB

R R

+ + − + + = ≤ ≤ =

If there exists not candidate point in the range,

If the a candidate point has been chosen successfully,

then we proceed to find a candidate point in the range for k=3


rx 3

MIMO-OFDM wireless communications using


• where

1 1| |

, ,

k n k n

n k k n k kr

k k k

k k k k

r y r yLB x UB

R R

+ + − + + = ≤ ≤ =

• where


2

11

| ,

TN

n n r

k k k k j jj k

y y R x+

= +

= − ∑ ( ) (2 2 2

k k n r

n n k k k k kr r y R x= − −


1 1| |

, ,

k n k n

n k k n k kr

k k k

k k k k

r y r yLB x UB

R R

+ + − + + = ≤ ≤ =


) ( )2 2 2

1

1 2 1 1 1| ,

k k n r


+

+ + + + += − −


• If a candidate value for does not exist,

• then go back to the previous step and choose another value of

• Then search for that meets the bound for that new value of

• In case no candidate exist for for

rx 3

rx 3

rx• In case no candidate exist for for

• Increase the radius of the sphere

• Assume that and are the final chosen candidate points

• Given and , a candidate fointervals


rx 3

rx 4

rx 3

rx 4

rx 3


If a candidate value for does not exist,

then go back to the previous step and choose another value of

Then search for that meets the bound for that new value of

for all possible values of

rx 4

rx 4

rxfor all possible values of

Assume that and are the final chosen candidate points

te for is chosen for k=2 from the


rx 4

rx 2


• where

1 1| |

, ,

k n k n

n k k n k kr

k k k

k k k k

r y r yLB x UB

R R

+ + − + + = ≤ ≤ =

• where


2

11

| ,

TN

n n r

k k k k j jj k

y y R x+

= +

= − ∑ ( ) (2 2 2

k k n r

n n k k k k kr r y R x= − −


1 1| |

, ,

k n k n

n k k n k kr

k k k

k k k k

r y r yLB x UB

R R

+ + − + + = ≤ ≤ =


) ( )2 2 2

1

1 2 1 1 1| ,

k k n r


+

+ + + + += − −


• If a candidate value for does not exist,

• then go back to the previous step and choose another value of

• Then search for that meets the bound for that new value of

• In case no candidate exist for for

rx 2

rx 2

rx• In case no candidate exist for for

• Then we go back the step before previous,

• and choose another value of

• Assume that and are the final chosen candidate points


rx 2

rx 4

rx 2

rx 3

rx 4


If a candidate value for does not exist,

then go back to the previous step and choose another value of

Then search for that meets the bound for that new value of

for all possible values of

rx 3

rx 3

rxfor all possible values of

Then we go back the step before previous,

Assume that and are the final chosen candidate


rx 3

r

2

Introduction to MIMO detection• Similarly, we can find

• and are the final chosen candidate points

• And it turns out to be a single point

• It is declared as the ML solution vector and search stops

• How do we decide the radius of the sphere?

rx1

rx 4

rx 3

rx 2

rx1

• How do we decide the radius of the sphere?

• We can choose where is the noise variance

• Hence for low SNR the radius of the sphere is large and SD is less efficient

• SD is highly efficient for high SNR since the sphere radius is small


( ) 22nRSD Nr σ∝

+ T. Kailath, H. Vikalo and B. Hassibi, “MIMO Receive

From Array Processing to MIMO Communications,

van der Veen, Eds., Cambridge, UK: Cambridge University Press, 2006, pp. 302


and are the final chosen candidate points

oint within the sphere with that radius

It is declared as the ML solution vector and search stops

How do we decide the radius of the sphere? How do we decide the radius of the sphere?

We can choose where is the noise variance

Hence for low SNR the radius of the sphere is large and SD is less

SD is highly efficient for high SNR since the sphere radius is small


2nσ

, “MIMO Receive Algorithms,”in Space-Time Wireless Systems:

From Array Processing to MIMO Communications, H. Bolcskei, D. Gesbert, C. Papadias, and A. J.

, Eds., Cambridge, UK: Cambridge University Press, 2006, pp. 302-321

Advanced MIMO detection techniques

Fig. Vertical Layered Space Time Transmission



Fig. Vertical Layered Space Time Transmission



• Vertical Bell Laboratories Layered Space Time Transmission

• At the transmitter the data is passed through a

• serial-to-parallel converter (S/P converter) and

• transformed into sub-streams, • transformed into sub-streams,

• where each sub-stream is sent through a different transmit antenna



Vertical Bell Laboratories Layered Space Time Transmission

At the transmitter the data is passed through a

parallel converter (S/P converter) and

streams, streams,

stream is sent through a different transmit antenna



• As usual in any communication system,

• after the S/P converter,

• all sub-streams will be modulated and

• may be interleaved and • may be interleaved and

• sent through the transmitting antenna



As usual in any communication system,

streams will be modulated and

sent through the transmitting antenna



• The transmission matrix (X) for V-BLAST can be represented as

11 2 1 3 1

1 2 2 2 3 2

, , ,

, , ,

x x x

x x x

• where in , t is the time index and j is the antenna index


1 2 2 2 3 2

1 2 3

, , ,

, , ,

, , ,T T T

N N Nx x x

=

X M M M M

,t jx


BLAST can be represented as

11 2 1 3 1

1 2 2 2 3 2

, , ,

, , ,

x x x

x x x

L

L

, t is the time index and j is the antenna index


1 2 2 2 3 2

1 2 3

, , ,

, , ,

, , ,T T T

N N Nx x x

M M M M

L


• First row of matrix is transmitted from the first antenna for time t=1,2,3,…

• Second row of matrix is transmitted from the second antenna for time t=1,2,3,… time t=1,2,3,…

• and so on

• Example

• For NT=5, write down the transmission matrix for Vtransmission.

• Assume that there are 35 sub-streams.



matrix is transmitted from the first antenna for time

matrix is transmitted from the second antenna for

=5, write down the transmission matrix for V-BLAST

streams.



• The V-BLAST demultiplex the data stream into subto as layers

• and sent one sub-stream over one transmit antenna

161161


=

2015105

191494

181383

171272

161161

X


the data stream into sub-streams referred

stream over one transmit antenna

312621


353025

342924

332823

322722

312621


• It can be seen that the original stream is

• mapped vertically into the columns of the transmission matrix,

• hence the name Vertical BLAST

• This transmission method will yield inter• This transmission method will yield inter

• For instance,

• consider the first column of the transmission matrix

• All the antennas are transmittingt=1



It can be seen that the original stream is

mapped vertically into the columns of the transmission matrix,

This transmission method will yield inter-stream interferenceThis transmission method will yield inter-stream interference

consider the first column of the transmission matrix X

itting simultaneously at the time index



• Hence any antenna at the receiver will receive

• all the signals streams from transmitting antennas 1 to N

• V-BLAST detection is done with

• zero forcing successive interference cancellation (ZF• zero forcing successive interference cancellation (ZF

• minimum mean square error successive interference cancellation (MMSE-SIC)

• which will cancel the interference from the previously detected signals



Hence any antenna at the receiver will receive

all the signals streams from transmitting antennas 1 to NT

zero forcing successive interference cancellation (ZF-SIC)zero forcing successive interference cancellation (ZF-SIC)

minimum mean square error successive interference cancellation

which will cancel the interference from the previously detected



Fig. Horizontal Layered Space Time Transmission4/18/2017 Fundamentals of MIMO Wireless Communications


Fig. Horizontal Layered Space Time TransmissionFundamentals of MIMO Wireless Communications 110

Advanced MIMO detection techniques• If we introduce channel coding for each data sub

• before modulation in the V-BLAST

• then we have horizontal BLAST

• V-BLAST architecture can also include

• optional encoder in series after the message source• optional encoder in series after the message source

• But the main difference between

• V-BLAST and H-BLAST

• is that in V-BLAST


+ D.-S. Shiu and M. Kahn, “Layered space-time codes for wireless communications using multiple

transmit antennas,” IEEE International Conference on Communications

440

Advanced MIMO detection techniquesIf we introduce channel coding for each data sub-streams

BLAST

BLAST architecture can also include

optional encoder in series after the message sourceoptional encoder in series after the message source


time codes for wireless communications using multiple

IEEE International Conference on Communications, June 1999, vol. 1, pp. 436 -


• the channel coding can be done over time

• whereas in H-BLAST the channel coding

• can be done over space and time

• We can still employ the • We can still employ the

• ZF-SIC or MMSE-SIC for detection

• The only difference now will be to inreceiver



the channel coding can be done over time

BLAST the channel coding

can be done over space and time

SIC for detection

to introduce a channel decoder at the



• Diagonal Bell Labs Layered Space Time Transmission

• The information stream is DMUX into sub

• and each data sub-stream is transmitted by a different antenna

• through a diagonal interleaving scheme• through a diagonal interleaving scheme

• Table in next slide shows four different data subfor instance

• The sub-streams are cyclically shifted

• before sending it over the NT antennas



Diagonal Bell Labs Layered Space Time Transmission

The information stream is DMUX into sub-streams

stream is transmitted by a different antenna

through a diagonal interleaving schemethrough a diagonal interleaving scheme

Table in next slide shows four different data sub-streams a, b, c and d

streams are cyclically shifted

antennas



Transmitting antenna 1 a1,1 b2,1

Transmitting antenna 2 a2,2

Table: Diagonal Bell Labs Layered Space Time Transmission (D

Transmitting antenna 3

Transmitting antenna 4

Time slots 1 2



c3,1 d4,1

b3,2 c4,2 d5,2

Table: Diagonal Bell Labs Layered Space Time Transmission (D-BLAST)

a3,3 b4,3 c5,3 d6,3

a4,4 b5,4 c6,4 d7,4

3 4 5 6 7



• Note that a1,1, a2,2, a3,3 and a4,4 are referring to the same sub

• which has been transmitted from the 1transmitting antenna

• at time slots 1, 2, 3 and 4 respectively• at time slots 1, 2, 3 and 4 respectively

• It will ensure higher diversity order than the H

• since same sub-streams are transmitted from different antennas

• This results in diagonally layered signal in space and time



are referring to the same sub-stream

which has been transmitted from the 1st, 2nd, 3rd and 4th

at time slots 1, 2, 3 and 4 respectivelyat time slots 1, 2, 3 and 4 respectively

It will ensure higher diversity order than the H-BLAST

streams are transmitted from different antennas

This results in diagonally layered signal in space and time



• As we can see from Table,

• for NT=4, there are four layers and

• each codeword is divided into four blocks

• number of blocks should be equal to N• number of blocks should be equal to N

• The decoder decode layer (sub-stream) by layer

• The first layer is detected without any error

• since it is transmitted alone ( refer to Table)



=4, there are four layers and

each codeword is divided into four blocks

number of blocks should be equal to NTnumber of blocks should be equal to NT

stream) by layer

The first layer is detected without any error

since it is transmitted alone ( refer to Table)



• After that, the second layer is demodulated and detected

• and it has only one interferer from the first layer

• But the first layer is already decoded,

• it can be subtracted• it can be subtracted

• The third will face two interferers

• But the first and second layers are already detected and

• they can be subtracted



After that, the second layer is demodulated and detected

and it has only one interferer from the first layer

But the first layer is already decoded,

But the first and second layers are already detected and



• The process goes on

• Note that the decoding in the previous layers should be error free,

• otherwise the whole process would suffer from error propagation

• ZF-SIC and MMSE-SIC algorithms could be employed• ZF-SIC and MMSE-SIC algorithms could be employed

• There are many unused time slots

• some of the transmitting antennas are sitting idle in D

• threaded D-BLAST can be employed to increase transmission rate


+ T. M. Duman & A. Ghrayeb, Coding for MIMO communication systems


Note that the decoding in the previous layers should be error free,

otherwise the whole process would suffer from error propagation

SIC algorithms could be employedSIC algorithms could be employed

some of the transmitting antennas are sitting idle in D-BLAST,

BLAST can be employed to increase transmission rate


Coding for MIMO communication systems, John Wiley & Sons, 2007.


• Example

• For NT=5, write down the transmission matrix for Dtransmission

3451


=

2345

1234

5123

4512

3451

X


=5, write down the transmission matrix for D-BLAST

5123


4512

3451

2345

1234

5123


• There are as many sub-streams or layers

• as the number of transmit antennas

• But the sub-streams are not transmitted

• as it is unlike V-BLAST• as it is unlike V-BLAST

• The sub-streams are cyclically reordered

• and are transmitted repeatedly



streams or layers

as the number of transmit antennas

streams are not transmitted

streams are cyclically reordered

and are transmitted repeatedly



• Successive interference cancellation

• For V-BLAST detection

• Zero forcing successive interference cancellation (ZF

• Aim is to detect and decode these streams one by one• Aim is to detect and decode these streams one by one

• When the receiver want to decode a stream from one transmitting antenna,

• all other streams from the remaining transmitting antennas are acting as an interferer

• Is there any way of removing these inter



Successive interference cancellation detection

Zero forcing successive interference cancellation (ZF-SIC)

Aim is to detect and decode these streams one by oneAim is to detect and decode these streams one by one

When the receiver want to decode a stream from one transmitting

all other streams from the remaining transmitting antennas are

Is there any way of removing these inter-stream interferers?



• Assume any channel matrix

• which can be decomposed as H=QR

• where Q is a matrix

• with its orthonormal columns being the ZF

R TN N×

R TN N×

• with its orthonormal columns being the ZF


=⇒

==

MMMM

L

MMMM

N

HH

qqqQ

IQQQQ

21


Assume any channel matrix H where

QR

columns being the ZF nulling vectors

T RN N≤

columns being the ZF nulling vectors


TN


• R is upper triangular matrixT T

N N×

−

−

T

T

N

N

RR

RRR

01

1

221

11211

L

L


=

−−

−

TT

T

NN

N

R

RR

RR

000

00

0

11

1221

L

L

MMMM

L

R


is upper triangular matrix

T

T

N

N

R

R

2

1


−

TT

TT

T

NN

NN

N

R

R

R

1

2

M


• We can calculate the y vector by pre

• with , which is virtually the nullingHQ

( ) (QRxQnHxQrQy =+== HHH

• z is another Gaussian noise vector with same mean and variance as


zRxnQRx +=+= H


vector by pre-multiplying the r vector

nulling step, as

)nQRx+

is another Gaussian noise vector with same mean and variance as n



• The above equation can be written in element wise format of the matrix as

NRRR

y11211

1

L


=

−

−

T

T

T

N

N

N

NR

RR

y

y

y

1

2211

000

00

0

L

L

MMM

LM


The above equation can be written in element wise format of the

− TT NNzx

R1

111


+

−−

−−−

−

−

T

T

T

T

TT

TTT

TT

TT

N

N

N

N

NN

NNN

NN

z

z

z

x

x

x

R

R

R

11

111

1

1

112

0

MMMM


• which is basically

+

+++

=

xR

RxRxR

y

y 2121111 L

M

• Note that


+=

−−−−

TT

TTT

T

T

NN

NNN

N

N

R

xR

y

y1111

i

N

j

jiji izxRyT

,,2,1;

1

L=+=∑=


++

++−− TTTT NNNN

zxR

zxRxR11 111

M


+

++−

TT

TTTT

NN

NNNN

zx

zxR1

TN,L


• After the QR decomposition of the channel matrix as described above,

• we can do the MIMO detection in the following ways:

• Detect for , then estimate ( ) using nearest neighborhood i N=• Detect for , then estimate ( ) using nearest neighborhood rule

• Then cancel estimate ( ) from to detect


Ti N=

TN

x

1 1 1 1 1 1

1 1 1 1 1 1

T T T T T T T T

T T T T T T T T

N N N N N N N N

N N N N N N N N

y R x R x z

y R x R x z

− − − − − −

− − − − − −

= + +

⇒ − = +


After the QR decomposition of the channel matrix as described

we can do the MIMO detection in the following ways:

Detect for , then estimate ( ) using nearest neighborhood xDetect for , then estimate ( ) using nearest neighborhood

Then cancel estimate ( ) from to detect


TN

x

1T

Ny

− 1T

Nx

−

1 1 1 1 1 1

1 1 1 1 1 1

T T T T T T T T

T T T T T T T T

N N N N N N N N

N N N N N N N N

y R x R x z

y R x R x z

− − − − − −

− − − − − −

= + +

− = +


• If we have estimated correctly that means , thenT

Nx

1 1

1ˆ T T T T

T

N N N N

N

y R xx g

R

− −

−

− =

• where g is the slicing function

• Note that can be expressed as


1

1 1

ˆT

T T

N

N NR

−

− −

iy

N

ij

jijiiii xRxRyT

+= ∑+= 1


If we have estimated correctly that means , thenˆT T

N Nx x=

1 1ˆ

T T T TN N N N

y R x− −

Note that can be expressed as


1 1

ˆ

T T− −

ij z+


• where xi is the current detected signal

• yi contains a lower level of interference than the received signal

• as the interference from xl for l<available estimates which are already detected available estimates which are already detected

• Hence the current signal xi can be estimated as


1ˆ

TN

i ij jj i

i

ii

y R x

x gR

= +

−

=

∑


is the current detected signal

contains a lower level of interference than the received signal r

for l<i can be cancelled from the available estimates which are already detected available estimates which are already detected

can be estimated as


ˆi ij j

y R x

• Fig.

Illustration of Successive interference cancellation (SIC)

4/18/2017 Fundamentals of MIMO Wireless CommunicationsFundamentals of MIMO Wireless Communications 130


• The basic idea of successive interference cancellation (SIC) is to

• cancel the interference from the previous detected symbols

• as depicted in Fig.

• This will reduce the interference and • This will reduce the interference and

• hence increases effective SINR



The basic idea of successive interference cancellation (SIC) is to

cancel the interference from the previous detected symbols

This will reduce the interference and This will reduce the interference and


• BER performance comparison of conventional detectors in 2 ×2 MIMO system using 64-QAM over iid Rayleigh iid Rayleigh fading MIMO channel



• Fig. depicts the performance of various conventional detectors viz.

• ZF,

• MMSE,

• ML and • ML and

• ZF-SIC

• over iid Rayleigh fading MIMO channel

• As expected,

• ML has the best BER performance,

• then ZF-SIC, MMSE and ZF



Fig. depicts the performance of various conventional detectors viz.

Rayleigh fading MIMO channel

ML has the best BER performance,



• Example 10.3

• Explain ZF-SIC for 3×3 MIMO system

• Solution:

• For N =N =3, we have• For NT=NR=3, we have



3 MIMO system

=⇒

+=

++=

+++=

3

33333

13231222

13132121111

ˆy

gx

zxRy

zxRxRy

zxRxRxRy


−−=

−=

=⇒

11

32321211

22

32322

33

33

ˆˆˆ

ˆˆ

ˆ

R

xRxRygx

R

xRygx

Rgx


• Minimum-mean squared successive interference cancellationdetector

• We can see the equivalent ZF detection of MMSE detection

• by defining an extended channel matrix and received vector as • by defining an extended channel matrix and received vector as follows


=

=0

yy

I

H

H ext

s

ext

E

N0 ;

+ R. Bohnke, D. Wubben, V. Kuhn and K. D. Kammeyer

architectures,” in Proc. IEEE Proceedings of Global Conference on Telecommunications

CA, 2003.


mean squared successive interference cancellation

We can see the equivalent ZF detection of MMSE detection

by defining an extended channel matrix and received vector as by defining an extended channel matrix and received vector as


−=

I

n

n

s

ext

E

N0;

Kammeyer, “Reduced complexity MMSE detection for BLAST

IEEE Proceedings of Global Conference on Telecommunications, San Francisco,


• where is the signal-to-noise ratio (SNR)0

SE

N

HHH N

N0ˆ

==


( )yHIHH

IIHrWs

H

s

H

ss

HHZF

E

N

E

NE

N

1

0

00ˆ

−

+=

==


noise ratio (SNR)

yH N0

1−


0

yIH

Is

H

E

N0


• This is exactly what we do in the MMSE detector

• We can do the QR factorization of this new extended ZF detector and

• follow the same procedure of sequential detection

• which will behave like MMSE-SIC detector• which will behave like MMSE-SIC detector

• Conservation theorem

• sum of diversity gain (d) plus number of interferers ( ) equals the number of receive antennas (N


erNd + int+ J. R Barry, E. A. Lee and D. G. Messerschmitt, Digital Communications


This is exactly what we do in the MMSE detector

We can do the QR factorization of this new extended ZF detector and

follow the same procedure of sequential detection

SIC detectorSIC detector

) plus number of interferers ( ) equals NR)


erNint

RN=Digital Communications, Kluwer Publications, 2010.


• Example

• Explain that ZF-SIC has higher diversity order than ZF using conservation theorem

• Note that ZF-SIC involves • Note that ZF-SIC involves

• nulling and

• cancelling operation

• simultaneously

• First nulling vector must null the N



SIC has higher diversity order than ZF using

vector must null the NT-1 interferers



• From conservation theorem,

• the diversity order for first detection of 1

• Similarly, he diversity order for second detection of symbol is

+−

• Therefore, for detecting the symbol, the diversity order is


2+− TR NN

NN TR +−


the diversity order for first detection of 1st symbol is

Similarly, he diversity order for second detection of symbol is

1+− TR NN

Therefore, for detecting the symbol, the diversity order is


2

k


• So the last symbol detected will have full diversity order

• Note that if any symbols are not detected correctly,

• this diversity order decreases

• To sum up, • To sum up,

• there is diversity order gain for ZF



So the last symbol detected will have full diversity order

Note that if any symbols are not detected correctly,

there is diversity order gain for ZF-SIC over ZF MIMO detectors



• Example

• Explain that MMSE detector has higher diversity order than ZF using conservation theorem

• MMSE detectors ignores those interferers whose strength is below • MMSE detectors ignores those interferers whose strength is below noise floor level, thereby,

• the diversity order for MMSE detector can be expressed as

• where is the number of significant interferers


RN

effTN


Explain that MMSE detector has higher diversity order than ZF using

MMSE detectors ignores those interferers whose strength is below MMSE detectors ignores those interferers whose strength is below

the diversity order for MMSE detector can be expressed as

where is the number of significant interferers


1+−effTN


• Hence the effective diversity order of MMSE detectors

• could be higher than the ZF detector

• Ordered successive interference cancellation detector

• In SIC, if the detected stream in one step is incorrect, • In SIC, if the detected stream in one step is incorrect,

• its subtraction from the received vector will increase the interference and

• results in performance degradation

• This is also known as error propagation



Hence the effective diversity order of MMSE detectors

could be higher than the ZF detector

Ordered successive interference cancellation detector

In SIC, if the detected stream in one step is incorrect, In SIC, if the detected stream in one step is incorrect,

its subtraction from the received vector will increase the

results in performance degradation

This is also known as error propagation



• Hence the critical issue in ordering the detection of each stream

• so that error propagation is minimized

• There are techniques which combine

• ordered successive interference cancellation (OSIC) and • ordered successive interference cancellation (OSIC) and

• linear detection techniques like

• ZF-OSIC and

• MMSE-OSIC



Hence the critical issue in ordering the detection of each stream

so that error propagation is minimized

There are techniques which combine

ordered successive interference cancellation (OSIC) and ordered successive interference cancellation (OSIC) and

linear detection techniques like



• The idea is to detect the signal with minimal error first

• so that the error propagation may be minimized

• In the process, we may decide the order in which

• we detect the signals by various criteria listed below:

• (a) Signal to noise interference ratio (SINR)• (a) Signal to noise interference ratio (SINR)

• Signals with the higher SINR are detected earlier than the other signals

• (b) Signal to noise ratio (SNR)

• Signals with the higher SNR are detected earlier than the other signals



The idea is to detect the signal with minimal error first

so that the error propagation may be minimized

In the process, we may decide the order in which

we detect the signals by various criteria listed below:

(a) Signal to noise interference ratio (SINR)(a) Signal to noise interference ratio (SINR)

Signals with the higher SINR are detected earlier than the other

Signals with the higher SNR are detected earlier than the other



• (c) Log-likelihood ratio (LLR)

• The ordering is based on the LLR

• Example

• What is LLR?• What is LLR?

• For a given observation x, the likelihood function is defined as


( ) ( ),/ == iHxfxf ii

+ S. W. Kim, “Log-likelihood ratio based detection ordering for the V

2003, vol. 1, pp. 292-296.


For a given observation x, the likelihood function is defined as


1,0=

likelihood ratio based detection ordering for the V-BLAST,” in Proc. GLOBECOM, Dec.


• The ML decision is to choose the hypothesis (either H

• that maximizes the likelihood function

( )00 HH


( ) ( )( )( )

1

1

0

1

01

1

0

0

H

xf

xfxf

H

xf<

>⇒

<

>


The ML decision is to choose the hypothesis (either H0 or H1)

that maximizes the likelihood function

( )0H


( )( )( )

0ln1

1

0

1

0

H

xf

xfxLLR

<

>

=⇒


• Likelihood ratio

• Log likelihood ratio

( )( )xf

xf

1

0

( )( )

xf

xf

1

0ln

• let us consider a simple binary alphabet of

• So when we send 1 and -1, the received vectors are


( ) xf1

( ) rnhr =++= ;1


let us consider a simple binary alphabet of

1, the received vectors are


{ }1 1,S = − +

( ) nh +−1


• Then ML decision for signal s using the LR is given by

( )( )

((((r

r

−

−=

−=

+==

exp

exp

1|

1|

sf

sfLR


( ) ((( ) ( )( ){ hrRhr

r

−−−=

−−=

−1exp

exp1|

nH

sf

( ) ( ) xaaxax 422

=−−+Q



Then ML decision for signal s using the LR is given by

( ) ( ))( ) ( ))hrRhr

hrRhr

++

−−−

−

1

1

nH

nH


( ) ( ))) ( ) ( )( )}hrRhr

hrRhr

+++

++

−1n

H

n



• The sign of LLR is like ML detection (hard decision) and

• the absolute value of LLR will give an idea of how reliable is the

( ) {Re4ln == LRLLR

• the absolute value of LLR will give an idea of how reliable is the decision

• As we know that for V-BLAST detection (after have,


nw iii xy +=


Wiley, 2010.


The sign of LLR is like ML detection (hard decision) and

the absolute value of LLR will give an idea of how reliable is the

{ }rRh1Re −

nH

the absolute value of LLR will give an idea of how reliable is the

BLAST detection (after nulling operation), we


MIMO-OFDM wireless communications using MATLAB,


• The LLR λi for xi (assume equiprobable

(( |

|ln

si

si

iyExP

yExP

−=

+==λ

• It can be shown that the bit error probability and LLR is related by


( |si yExP −=

( )xxPP iiie =≠=1

ˆ,


equiprobable BPSK symbols) is given by

))

( )2

0

Re4

i

is

i

i y

N

E

y

y

w=

It can be shown that the bit error probability and LLR is related by


) 0 iiNy w

ieλ

+1

1


• Since the bit error probability decreases with increasing LLR,

• the detection ordering is to detect the component of

• that provides the largest firstiλ



Since the bit error probability decreases with increasing LLR,

the detection ordering is to detect the component of x



• Fig. V-BLAST MIMO system



BLAST MIMO system



• In the performance analysis,

• it is quite involved to find the exact performance analysis

• Hence, we will derive the

• upper and lower bound • upper and lower bound

• of the ZF-SIC detector’s performance

• Consider the I-O model of a V-BLAST MIMO system depicted in Fig.

• The received signal vector r for a narrowband MIMO channel be obtained as


= +r Hs n


it is quite involved to find the exact performance analysis

SIC detector’s performance

BLAST MIMO system depicted in Fig.

for a narrowband MIMO channel H can



• We can also express channel matrix

= 1 2 N

H h h h

h

• where for


1

2

R

k

k

k

N k

h

h

h

=

h M 1 2, , ,T

k N= L


We can also express channel matrix H in terms of column vectors

T

1 2 NH h h hL


T

, , ,T

k N


• Hence,1 1

2 2

s n

s n

s n

= + = +

T1 2 N

r h h h h nM ML


T RN N

s n

T1 2 N


1 1

2 2T

N

k k

s n

s n

s

s n =

= + = +

∑r h h h h nM M


1

T R

k kk

N Ns n =

∑


• We will assume SINR based ordering which is optimal, i.e.,

• only the layer with highest SINR is detected in each recursion

• which gives the lowest SER in overall

• We will also assume that perfect SIC • We will also assume that perfect SIC

• PSIC which means that the cancelled interference are accurate

• Hence optimal ordered perfect ZF-

• will give the lower bound of ZF-



We will assume SINR based ordering which is optimal, i.e.,

only the layer with highest SINR is detected in each recursion

which gives the lowest SER in overall

We will also assume that perfect SIC We will also assume that perfect SIC

PSIC which means that the cancelled interference are accurate

-SIC detector (ZF-OOPSIC)

-OSIC



• Let denotes the received signal

• for the jth recursive step after interference cancellation

• For PSIC, we can write

( )jr

• where is the hard decision of the estimated value of


( ) (hhr += ∑∑−

==

1

1

j

a

k

N

ja

kkj

a

T

aas

( )aky sQ ˆ


Let denotes the received signal

recursive step after interference cancellation

is the hard decision of the estimated value of


( )( ) n+− ˆkyk aa

sQs

aks


• In the above equation,

• the first summation term is for undetected symbols, and

• the second summation is the interfedetected symbolsdetected symbols

• For PSIC,

• the detected symbols are exactly

• what we have transmitted



the first summation term is for undetected symbols, and

terference cancellation of the already



• hence the second summation term should be zero for ideal case

( )hr ∑

=

+=T

aa

N

ja

kkj

s

• If is the nulling vector

• this is the lth row vector of Moore-Penrose pseudo


= ja

( )j

lw

( ) ( ) ( )[ jjj21 ,= hhH

+ J. Han, Q.-M. Cui, X.-F. Tao and P. Zhang, “SER bound for ordered ZF

system,” Journal of China Universities of Posts and Telecommunications


hence the second summation term should be zero for ideal case

n+

Penrose pseudo-inverse of


( ) ]j

jNR 1,,

+−hL

F. Tao and P. Zhang, “SER bound for ordered ZF-SIC receiver in M-QAM MIMO

Journal of China Universities of Posts and Telecommunications, Feb. 2010, pp. 51-55.


• for the lth undetected symbol in the

• We have

,,2,1 −= Nl TL

= pl1• We have

• the estimated symbols derived in the could be expressed as


( ) ( )

≠

==

pl

pljp

j

l 0

1hw

1+− jNT


undetected symbol in the jth recursive step where

1+− j

the estimated symbols derived in the jth recursive step



• Therefore, the SINR of the lth undetected layer in the

( ) ( ) ( ) ( ) ( )wnhwrw

j

l

N

ja

kkj

ljj

l

j

lss

T

aa=

+== ∑

=

ˆ

• Therefore, the SINR of the lth undetected layer in the step is given by


( ) ( )( )

0

2

2

j

l

lj

l

NE

sESINR

nw

=

=


undetected layer in the jth recursive

) ( ) ( ) ( ) ( )nwnwhwh

j

llj

l

N

lkja

kkj

llj

lsss

T

a

aa+=++ ∑

≠= ,

undetected layer in the jth recursive


( ) ( ) 2

0

2j

l

j

l

P

ww

γ=


• Note that follows Chi-square distribution with

• degrees of freedom and variance ½ and

• therefore

( ) 2

1

j

lw

( )

( ),2,1;

12

== lxj

j

l

w

• for any l will have generalized Rayleigh distribution with

• degrees of freedom and variance ½


( )j

lw

( )jNN TR +−2

+ J. G. Proakis and M. Salehi, Digital Communications


square distribution with

degrees of freedom and variance ½ and

( )jNN TR +−2

1, +− jNTL

for any l will have generalized Rayleigh distribution with

degrees of freedom and variance ½


Digital Communications, McGraw-Hill, 2008.


• Its pdf and CDF (for even ) are given by( NN R −2

( )( )( )

((!1

2 j

TR

jx

jNNxp

−+−=


( )!1TR jNN −+−

( )( ) ( )( ) (∑

−+−

=

−−=

1

0

12

jNN

k

xjTR

j

exP


and CDF (for even ) are given by)jNT +

) ) ( ) ( )( )212 jTR xjNN

e−−+−


( )( )2

!

kj

k

x


• For optimal ordering using order statistics,

• the maximum SINR in each recursive step should be selected from continuous population of pdf and CDF

• To select the lth undetected layer with the maximum SINR in the • To select the lth undetected layer with the maximum SINR in the recursive step,

• s are rearranged in ascending order of SINR say,


( )jlx

( ) ( ) ( )( jjjxxx 321 ≤≤≤ L

+ G. Casella and R. L. Berger, Statistical interference


For optimal ordering using order statistics,

the maximum SINR in each recursive step should be selected from and CDF

undetected layer with the maximum SINR in the jthundetected layer with the maximum SINR in the jth

s are rearranged in ascending order of SINR say,


( ) ( ) )jjN

jjN TT

xx1+−− ≤≤L

Statistical interference, Thomson learning, 2002.


• the rth highest value of SINR is selected

• According to order statistics, the PDF of is given by

( )( ) ( )({ jjjxPxp =

1

• where is the beta function and for positive arguments


( )( )( )

( )({ j

xT

jj

xxP

rjNrBxp

lr −+−=

2,

1

( ).,.B

( ) ( )(

!1,

+

−=

nm

mnmB

+ H. A. David and H. N. Nagaraja, Order Statistics, 3rd ed. New York: Wiley


highest value of SINR is selected

According to order statistics, the PDF of is given by

( ))} ( ) ( )( ){ } ( ) ( )( )jjrjNjjrjxpxP

T −+−−−

111

where is the beta function and for positive arguments


( ))} ( ) ( )( ){ } ( ) ( )( )jj

x

rjNjj

x

rjxpxP

l

T

l

−+−−−

111

( ))!1

!1!

−

−

n

n

, 3rd ed. New York: Wiley Interscience, 2003.


• Therefore the pdf of maximum SINR i.e., (or minimal

( ) ( )(

12,

−=−+−

rrjNrB T

• Therefore the pdf of maximum SINR i.e., (or minimal interference) for our case is given as


( ) ( )( )

( ) ( ){ } (pxP

jNBxp

j

N

jNj

jNT

j

jN T

T

TT 11 1,1

1 −

+−+− +−=


of maximum SINR i.e., (or minimal

) ( )( )!1

!1!1

+−

−+−

jN

rjN

T

T

( )jxof maximum SINR i.e., (or minimal

interference) for our case is given as


( )j

jNTx

1+−

) ( ) ( )( )

( ) ( ){ } ( ) ( )xpxPjN

jNx

j

jN

jNj

jNT

Tj T

T

TT 111 !

!1+−

−

+−+− −

+−=


• The pdf and cdf of SINR of any undetected layer for our case are

( )( )( 1

2

TR

j

jNNxp

−+−=


( 1TR jNN −+−

( )( ) ( )( ) ∑−

=

−−=0

12

NN

k

xjTR

j

exP


of SINR of any undetected layer for our case are( )j

x

)( )( ) ( ) ( )( )212

!1

jTR xjNNj

ex−−+−


)!1

( )( )∑

−+ 1

0

2

!

j kj

k

x


• Hence, the pdf of maximum SINR substituting is

• The CEP of M-QAM could be expressed as

x

( ) ( )1

1!

12 TjjN

eq

jNxp

T +−

−

+−=

• The CEP of M-QAM could be expressed as


( ) ( )

( )

−−

−

−=

−

−=

11

12

3112

1421

14|

2

2

erfcMM

SINRerfc

M

SINRgQM

SINREP QAMb


of maximum SINR substituting is

QAM could be expressed as

( )jx qjNN TR =−+− 1

22 1

0

2

!

xq

jNq

k

kx

exk

xe

T

−−

−

=

−

∑

QAM could be expressed as


( )( )

( )

−

−=

−

12

3

12

3;2

11 2

2

M

SINR

MgSINRgQ

MQAMQAM


• where erfc is the complementary error function defined as

( ) ∫∞

− ==

x

ydyexerfc

2 2

π


xπ

( ) = ∫∞

−

2

1

0

2

2

dyexQ

y

π

+ G. L. Stuber, Principles of Mobile Communication,


is the complementary error function defined as

( )−= xerf1


=

22

1 xerfc

, Kluwer Academic Publishers, 2001.


• There are NT layers to be detected

• we need to sum and average the probability of error for all layers

• The SER for M-QAM could be obtained as


( )

( )(EPN

PN

EP

T

T

N

j

jOOPSICZFb

T

N

jT

OOPSICZFb

∑∫

∑

=

∞

−

=

−

=

=

1 0

1

1

1


we need to sum and average the probability of error for all layers

QAM could be obtained as


( )( )

) ( ) ( )dxxpSINRE

EP

T

j

jN

jOOPSICZFb

+−

−

1/


• This is the lower bound of ZF-OSIC

( )

∑∫∞

−

+−

=TN

T

OOPSICZF

erfcq

jNA

P0

!

14

1

• where


( )∑

∫=

∞

−

+−=

jT

T

OOPSICZF

b

erfcq

jNA

q

NP

1

0

2

0

!

12

!1

( 12

3;

11 0

−=

−=

MB

MA

γ


( ) ∑ −−

−

=

−

−

−

T

xq

jNq

k

kx dxex

k

xeBxerfc 1

0

222

!1


( ) ∑ −−

−

=

−

=

−

T

xq

jNq

k

kx

k

dxexk

xeBxerfc

k

1

0

22

0

22

!1

!

)1


• For upper bound we can calculate the SER of ZF without SIC

• since in ZF we do not do the interference cancellation

• probability of error will be higher for this case than the ZF

• We can write SINR for ZF as• We can write SINR for ZF as


( )

( ) 2

0

j

jSINR

w

γ=


For upper bound we can calculate the SER of ZF without SIC

since in ZF we do not do the interference cancellation

probability of error will be higher for this case than the ZF-OSIC



• Note that follows Chi-square distribution with

• degrees of freedom and variance ½ and

• therefore

( ) 2

1

j

lw

( )

( ),2,1;

12

== lxj

j

l

w

• for any l will have generalized Rayleigh distribution with

• degrees of freedom and variance ½


( )j

lw

( )jNN TR +−2


square distribution with

degrees of freedom and variance ½ and

( )jNN TR +−2

1, +− jNTL

for any l will have generalized Rayleigh distribution with

degrees of freedom and variance ½



• Its pdf and CDF (for even ) are given by( NN R −2

( )( )( )

((!1

2 j

TR

jx

jNNxp

−+−=


( )!1TR jNN −+−

( )( ) ( )( ) (∑

−+−

=

−−=

1

0

12

jNN

k

xjTR

j

exP


and CDF (for even ) are given by)jNT +

) ) ( ) ( )( )212 jTR xjNN

e−−+−


( )( )2

!

kj

k

x


• Therefore, for M-QAM, the SER for ZF for Nby

( ) ( ) ( ) ( ) ( )EPEPN

EPjZF

b

NjZF

bT

ZFb

T1∑ ==


( ) ( )

( )( )( ) ( ) ( )

dueuBuerfcNN

A

duupSINREP

N

uNN

TR

ZFb

b

j

bT

TR212

0

0

1

!

4

/

−+−∞

∞

=

∫

∫

∑

−−

=

=


QAM, the SER for ZF for NT layer detection is given


( )( )( ) ( ) ( )

dueuBuerfcNN

A uNN

TR

TR212

0

22

!

2 −+−∞

∫−−


• SER for ZF-OSIC is bounded as

• Lattice reduction based detector

( ) PEPOSICZF

bOOPSICZF

b ≤ −−

• Lattice reduction based detector

• One of the major issues with linear detectors

• we have considered so far was the noise

• This effect becomes more pronounced

• when the channel matrix is not well



( ) ( )EPEZF

bOSIC ≤

One of the major issues with linear detectors

have considered so far was the noise enhancement

effect becomes more pronounced

the channel matrix is not well behaved



• Lattice reduction algorithm could be employed

• to reduce the condition number of channel

• bring it closer to 1

• The condition number of a matrix is defined as • The condition number of a matrix is defined as

• the ratio of the largest and smallest singular value of the

• The L2-norm of a matrix equals its largest singular value

• the L2-norm of inverse of a matrix equals the reciprocal of the smallest singular value



Lattice reduction algorithm could be employed

reduce the condition number of channel matrix

The condition number of a matrix is defined as The condition number of a matrix is defined as

ratio of the largest and smallest singular value of the matrix

norm of a matrix equals its largest singular value

norm of inverse of a matrix equals the reciprocal of the



• the condition number of matrix

• Hence for real orthogonal matrices

( )2

1

2min

max ≥== −HHH

σ

σcond

• Hence for real orthogonal matrices

• the condition number is one

• and no noise amplification for linear detectors

• Hence the matched filter and ZF are equivalent since


( ) HHHHHHH =

−1

+ V. Kuhn, Wireless Communications over MIMO channels


Hence for real orthogonal matrices with

1

T=−1Hence for real orthogonal matrices with

no noise amplification for linear detectors

Hence the matched filter and ZF are equivalent since


THH =−1

H

Wireless Communications over MIMO channels, John Wiley & Sons, 2006.


• MMSE also has an equivalent ZF by defining extended

• Hence matched filter and MMSE will also be

• Hence for well conditioned orthogonal matrices,

• the performance of linear detectors is • the performance of linear detectors is

• Therefore, it is desirable to have a roughly orthogonal matrix

• with condition number close to 1



MMSE also has an equivalent ZF by defining extended matrices

matched filter and MMSE will also be identical

Hence for well conditioned orthogonal matrices,

performance of linear detectors is goodperformance of linear detectors is good

, it is desirable to have a roughly orthogonal matrix

condition number close to 1



• Lattice is regular arrangements of points in Euclidean space

• Using the basis vectors

• we can generate the lattice points in 2

( )11 0,b ( )2

0 1,b

• we can generate the lattice points in 2

• They generate all the intersection points of the grid also known as lattice points

• Similarly, using new basis vectors


( ),1,121'1 bbbb +=


of points in Euclidean space

can generate the lattice points in 2-D space as depicted in Fig. (a) can generate the lattice points in 2-D space as depicted in Fig. (a)

They generate all the intersection points of the grid also known as


( )1,22 21'2 bbb +=

(a) Basis vectors in R2, (b) another equivalent basis vectors in R2 and (c) R and (c) not suitable basis vectors in R2



• we can also generate the lattice points in 2(b)

• But we can’t generate a lattice from the following two

• and they are not basis vectors

• because the basic parallelepiped generated from these two vectors

• contains the lattice point (1,0) and (2,1)


( ),1,1''221

''1 bbbb =+=


we can also generate the lattice points in 2-D space as shown in Fig.

But we can’t generate a lattice from the following two vectors

the basic parallelepiped generated from these two vectors

the lattice point (1,0) and (2,1)


( )0,22 1b


• Example

• What is a lattice?

• A lattice L (regularly arranged arrays of points) is a set of vectors

• that are obtained by the integer linear combination of • that are obtained by the integer linear combination of

• a set of linearly independent vectors known as basis vectors


1 2, , ,

k = B b b bL


arranged arrays of points) is a set of vectors

are obtained by the integer linear combination of are obtained by the integer linear combination of

set of linearly independent vectors known as basis vectors



• Mathematically,

• where Z is the set of integers and

== ∑=

K

k

k

1

| bBuBL

• where Z is the set of integers and

• Rank of lattice is K and dimension is N, Full rank N=M

• Example of lattice in communication theory is QAM constellation

• which is a finite subset of the complex integer

• Example: What is generator matrix of a lattice?



∈kkk Zuu ,

N ≥∈

Rank of lattice is K and dimension is N, Full rank N=M

of lattice in communication theory is QAM constellation

is a finite subset of the complex integer lattice

is generator matrix of a lattice?


KNRN

k ≥∈ ,b


• A vector v in lattice L be expressed

• where is the generator matrix

=v Bu u

, , , =• where is the generator matrix

• whose columns are basis/generator vectors


1 2, , ,

k = B b b bL

k b


be expressed as

the generator matrix

KZ∈u

the generator matrix

basis/generator vectors



• What is a unimodular matrix?

• A square matrix

• whose determinant is ±1 is called as

• An integer matrix U (whose elements are integer) • An integer matrix U (whose elements are integer)

• whose determinant is ±1 is called as integer

• It is quite possible that for the same lattice

• there could be two different generator matrices

• In that case,


21= B UB


1 is called as unimodular

(whose elements are integer) (whose elements are integer)

1 is called as integer unimodular

It is quite possible that for the same lattice L,

there could be two different generator matrices B1 and B2


Advanced MIMO detection techniques• Change of basis

• Multiply a basis by an invertible matrix, gives a new basis and a new lattice

• Multiply a basis by an unimodularand a same latticeand a same lattice

• Hence if U is unimodular, inverse of matrix

• For example, the following matrices are


( ) ( ) ( )detdet1det11 == −−

UUUU

=

=

= ;

10

01;

10

21111 UUU


Multiply a basis by an invertible matrix, gives a new basis and a new

matrix, gives an equivalent basis

, inverse of matrix U is also unimodular

For example, the following matrices are unimodular


) ( ) ( ) 1detdet1 ±==⇒ −

UU

− 11

01


• Example

• Explain in few words the lattice reduction based MIMO detection

• Lattice reduction (LR) basically assumes that

• the channel matrix (H) is a generator matrix for a lattice • the channel matrix (H) is a generator matrix for a lattice

• since H contains many column vectors

• Using efficient algorithms like

• Lenstra, Lenstra & Lovasz (LLL)

• it will find an equivalent generator matrix of the lattice

• which is well behaved



Explain in few words the lattice reduction based MIMO detection

Lattice reduction (LR) basically assumes that

) is a generator matrix for a lattice ) is a generator matrix for a lattice

contains many column vectors

it will find an equivalent generator matrix of the lattice



• We will apply the MIMO detection techniques to

• the new well behaved equivalent

• Sub-optimal MIMO detectors like

• ZF, ZF-SIC are surprisingly efficient • ZF, ZF-SIC are surprisingly efficient

• when they are employed on a reduced basis

• since the equivalent channel G

• hence the noise enhancement will be minimized



We will apply the MIMO detection techniques to

lent generator matrix of the channel

SIC are surprisingly efficient SIC are surprisingly efficient

when they are employed on a reduced basis

will be better behaved and

hence the noise enhancement will be minimized



• Diversity order of MIMO detection:

• ZF�MMSE�SIC�ML

• Diversity order of LR based ZF/MMSE linear detector is the same as

• achieved by ML detector which is N for N• achieved by ML detector which is N for N


+ X. Ma and W. Zhang, “Performance analysis of MIMO systems with lattice

equalization,” IEEE Trans. Comm., vol. 56, no. 2, Feb. 2008, pp. 309


Diversity order of MIMO detection:

Diversity order of LR based ZF/MMSE linear detector is the same as

achieved by ML detector which is N for N×N MIMO systemsachieved by ML detector which is N for N×N MIMO systems


X. Ma and W. Zhang, “Performance analysis of MIMO systems with lattice-reduction aided linear

, vol. 56, no. 2, Feb. 2008, pp. 309-318.


• Fig. Lattice reduction aided MIMO detection



Fig. Lattice reduction aided MIMO detection



• LR-ML detection:

• From lattice theory, since H and G

• they are related by a unimodular matrix

=G HU

• Therefore, the received signal vector can be rewritten as

• where


=G HU

1−= + = + = +r Hx n GU x n Gc n

1−=c U x


generate the same lattice,

matrix U as follows

Therefore, the received signal vector can be rewritten as


= + = + = +r Hx n GU x n Gc n


• We can apply the ML detection and find an estimate for

min

arg

ˆ Gcrc −

∈

=K

• From estimate of c, we can find the estimate of


c∈ KZ

ˆˆ =x Uc


We can apply the ML detection and find an estimate for c as follows

2Gc

we can find the estimate of x as follows



• LR-ZF detection:

• Similarly, we can apply the ZF over the new matrix

• which is more well behaved and

• will minimize the noise enhancement• will minimize the noise enhancement

• Note that G+ is the Moore-Penrose inverse of


( ) ( 1+ + + − + += + = + = + = +G r G Hx n G GU x n G Gc n c G n


Similarly, we can apply the ZF over the new matrix G instead of H

which is more well behaved and

will minimize the noise enhancementwill minimize the noise enhancement

Penrose inverse of G


) ( )1+ + + − + += + = + = + = +G r G Hx n G GU x n G Gc n c G n


• Hence we can estimate of x as integer closest to the estimate of

• BER performance comparison between ZF and LR

• for 2 × 2 MIMO system employing BPSK modulation scheme

• over iid Rayleigh fading MIMO channel • over iid Rayleigh fading MIMO channel

• is depicted in Fig.

• It can be observed that LR-ZF has superior BER performance than ZF



integer closest to the estimate of c

BER performance comparison between ZF and LR-ZF

2 MIMO system employing BPSK modulation scheme

Rayleigh fading MIMO channel Rayleigh fading MIMO channel

ZF has superior BER performance than ZF


• BER performance comparison between ZF and LR-ZF for 2 × 2 MIMO system employing BPSK employing BPSK modulation over iidRayeighfading MIMO channel



• LR-ZF-SIC detection:

• For SIC detection, do QR decomposition of

• Then multiply by to the received signal vectorHQ

( )

• where

• We can proceed with the same SIC detection techniques

• with this new equivalent system


( )1H H H H−= + = + = +Q r Q GU x n Q QRc n Rc Q n

1−=c U x


For SIC detection, do QR decomposition of

Then multiply by to the received signal vector

=G QR

( )

We can proceed with the same SIC detection techniques

with this new equivalent system


( )H H H H= + = + = +Q r Q GU x n Q QRc n Rc Q n


• Lattice reduction algorithms

• A lattice is generated as

• the integer linear combination of

• some set of linearly independent vectors• some set of linearly independent vectors

• A lattice in the n-D Euclidean space


( ) i

n

i

ii iZuuLL bB 1,;

1

=∈== ∑=


the integer linear combination of

some set of linearly independent vectorssome set of linearly independent vectors

D Euclidean space Rn is a set of the form


[ ]nn bbB ,,;,,1 1 LL =


• A lattice L can be generated by different bases for n≥2 and

• hence there is no unique basis

• We can obtain an equivalent basis from another basis

• by multiplying an integer unimodular• by multiplying an integer unimodular

• Basically this is the result of three operations

• (a) exchanging two columns

• (b) multiplying any column by -1 and

• (c) adding an integer multiple of one column to another



can be generated by different bases for n≥2 and

basis from another basis

unimodular matrixunimodular matrix

Basically this is the result of three operations (illustrated in Fig.):

1 and

(c) adding an integer multiple of one column to another


Fig. (a) basis vectors b1

and b2 for the integer lattice Z2 (b) exchanging two columns b1

and b2 (c) and b2 (c) multiplying both columns by -1 (d) adding twice of column b1 to column b2



• Lattice reduction algorithms are developed for real valued lattices

• for complex MIMO system

• Use equivalent real channel model as follows

= +y Hx n

• Use equivalent real channel model as follows

• Now the dimension will be doubled (say and )


( )( )

( )( )

=

H

H

y

y

Im

Re

Im

Re


Lattice reduction algorithms are developed for real valued lattices

Use equivalent real channel model as followsUse equivalent real channel model as follows

Now the dimension will be doubled (say and )


( )( )

( )( )

( )( )

+

−

n

n

x

x

H

H

Im

Re

Im

Re

Re

Im

2T

m N= 2R

n N=


• LLL reduced lattice:

• A basis with QL decomposition is

• LLL reduced with parameter δ (usually taken ¾), ,

• if the following two conditions hold true

redH

• if the following two conditions hold true

• (a) Size reduction


mlkllkl ≤<≤≤ 1;2

1,, LL


A basis with QL decomposition is

LLL reduced with parameter δ (usually taken ¾), ,

if the following two conditions hold true

red red red=H Q L

11

4δ≤ ≤

if the following two conditions hold true


4

m


• Note that column l is to the right of column k

• The condition (a) says the diagonal components of the

• are at least double as the off-diagonal components of the same rowrow

• Lred is a lower triangular matrix

• This is called the size-reduction condition and

• this ensures that there is no significant projection of one column on another



Note that column l is to the right of column k

The condition (a) says the diagonal components of the Lred

diagonal components of the same

reduction condition and

this ensures that there is no significant projection of one column on



• If it is not satisfied for (l,k) pair,

• we deduct an integer multiple of the column

• so that this condition is satisfied• so that this condition is satisfied

• The size reduction is carried out by

• subtracting integer multiples of the right column with index

• from the left column with index

• In other words,

• it makes sure that basis vectors are as orthogonal as possible



we deduct an integer multiple of the lth column from the kth

The size reduction is carried out by

subtracting integer multiples of the right column with index l

from the left column with index k

it makes sure that basis vectors are as orthogonal as possible



• This condition does not guarantee a minimal basis,

• there is another condition called as

• which will ensure the correct sorting of the columns

• Swapping of columns improve the basis• Swapping of columns improve the basis

• The last column of the reduced lattice can be thought of as the shortest vector

• (b) Sorting


[ ] [ ],1,122

≤++ kkkk redred LLδ


This condition does not guarantee a minimal basis,

there is another condition called as Lovasz condition

which will ensure the correct sorting of the columns

Swapping of columns improve the basisSwapping of columns improve the basis

The last column of the reduced lattice can be thought of as the


[ ] 11;,122

−≤≤++ mkkkredL


• The condition (b) ensures proper sorting

• since the lengths of the columns are only compared

• on the basis of a little 2×2 submatrix

• If the above condition is not satisfied, • If the above condition is not satisfied,

• we will interchange the columns

• Why are we considering 2×2 submatrices

• This will reduce the computational complexity

• at the price of lower performance especially for big channel matrices



The condition (b) ensures proper sorting

since the lengths of the columns are only compared

submatrix

If the above condition is not satisfied, If the above condition is not satisfied,

we will interchange the columns

submatrices?

This will reduce the computational complexity

at the price of lower performance especially for big channel



• The columns have to be ordered according to their lengths,

• shortest columns right

• and largest on the left

• Example • Example

• Explain the above condition (b) with the help of an example

• Solution:



The columns have to be ordered according to their lengths,

Explain the above condition (b) with the help of an example


;

0

00

000

0000

5554535251

44434241

333231

2221

11

=

LLLLL

LLLL

LLL

LL

L

L


• For instance for 5×5 lower triangular matrix given above

• we will consider the 2×2 sub-matrices

1,1L

2,2L

• If this condition is not satisfied,

• columns will be exchanged

• This condition is also known as Lovasz


2,21,2

1,1

LL

L

3,32,3

2,2

LL

L


5 lower triangular matrix given above

matrices

3,3L

4,4L

Lovasz condition


4,43,4

3,3

LL

5,54,5 LL


• Small value of δ leads to fast convergence,

• whereas large value of δ leads to better basis

• Usual choice of 3

4δ =

• Lovasz condition for submatrix


4

21,2

21,1

22,2

4

3LLL +≤


leads to fast convergence,

leads to better basis


2,21,2

1,1

LL

L


• What does this means for choice of

• Lovasz condition for submatrix

21,2

21,1

22,2

3LLL +≤

• It means left column has larger length than right column

• It also means we have shortest column on the right


1,21,12,24

LLL +≤


choice of 1δ =

1,1

LL

L

It means left column has larger length than right column

It also means we have shortest column on the right


2,21,2 LL


• Note that finding a nearly orthogonal basis vector is

• equivalent to finding minimal length vectors

• In Fig. (b), the basis vectors are short and orthogonal,

• it will have no noise enhancement• it will have no noise enhancement

• If none of the vectors have projection on the other then

• they have a smaller length and are almost orthogonal



Note that finding a nearly orthogonal basis vector is

equivalent to finding minimal length vectors

In Fig. (b), the basis vectors are short and orthogonal,

it will have no noise enhancementit will have no noise enhancement

If none of the vectors have projection on the other then

they have a smaller length and are almost orthogonal



• Fig. (a) Long and non-orthogonal babasis vectors



l basis vectors (b) Short and orthogonal



• LLL Algorithm for lattice reduction

• The inputs to the LLL algorithm are

00011 LL


= ;0

00

000

321

333231

2221

11

L

MOMMM

L

L

L

UL

mmmmm LLLL

LLL

LL

L

+ A. K. Lenstra, J. H. W. Lenstra and L. Lovasz, “Factorizing polynomials with rational coefficients,”

Ann., 216(4), 1982, pp. 515-534.


LLL Algorithm for lattice reduction

The inputs to the LLL algorithm are Q, L and U

0001 L


=

1000

0100

0010

0001

L

MOMMM

L

L

L

U

, “Factorizing polynomials with rational coefficients,” Math.


• The outputs are reduced matrices viz.

• LLL LR algorithm

• Initialization:

, ,red red m

= = =Q Q L L U I

• (%initial inputs, U is a unimodular matrix)

• (% k is the column under consideration and start from the last but second column, note that m=2NT)


, ,red red m

= = =Q Q L L U I

1k m= −


reduced matrices viz. Qred, Lred and Ured

matrix)

(% k is the column under consideration and start from the last but



• while (% for all columns of the matrix from the 1

• for (% l is larger than k, column)

•

1k ≥

1, ,l k m= + L

( ),,

,

redl k

l kµ =L•

• (% ratio of the off-diagonal and diagonal element in the same row)

• if

• (% off-diagonal element is larger than diagonal element)


( )( ),

,,

red

red

l kl k

l lµ =

L

L

0,l kµ ≠


while (% for all columns of the matrix from the 1st to the last)

for (% l is larger than k, lth is in the right side of kth

diagonal and diagonal element in the same row)

diagonal element is larger than diagonal element)



• (% subtract integer multiple of lth column from which has only l:m elements for lth

: , : , : ,l m k l m k l m lµ = − L L L

which has only l:m elements for lth

• (% subtract integer multiple of lth column from


:, :, :,k k lµ = − U U U


column from kth column of Lredth column)

: , : , : ,l m k l m k l m l L L L

th column)

column from kth column of U)


:, :, :,k k l U U U


• end

• if

• Exchange columns k and k+1 in L

[ ] [21,1 kkk redred ≥++ LLδ

• Exchange columns k and k+1 in Lred

• (% when we interchange k and k+1 column, now the new matrix will no longer be lower triangular, we need to force zero the element zero)



and U

] [ ] 22,1, kkkk red ++ L

red and U

% when we interchange k and k+1 column, now the new matrix will no longer be lower triangular, we need to force zero the L(k,k+1)



[[ :

1;

+

+=

−=

kk

k

red

red

L

LΘ α

αβ

βα

• (%Calculate Givens rotation matrix such that elementbecome zero)


[ ]1:1,1: =++ kkkred ΘLL

[ ] [redred kk QQ :,1::, =+


]]

[ ][ ]1,1:

1,

1,1

1,1

++

+=

++

+

kkk

kk

k

k

red

red

L

Lβ

(%Calculate Givens rotation matrix such that element L(k,k+1)


[ ]1:1,1: ++ kkkredΘL

Θ

] Tkk Θ1: +


• (% consider all columns, but only the k and k+1 rows only, Givens

• rotation operate on columns only)

{ }1 1min ,k k m= + −

• else

• k=k-1

• end

• end


{ }1 1min ,k k m= + −


(% consider all columns, but only the k and k+1 rows only, Givens



• Example

• What is Givens rotation?

• Solution:

• Consider a matrix which is an N( )θ,, kiΘ• Consider a matrix which is an N

• except for the elements

• It gives a rotation of θ in the N-D vector space


( )θ,, kiΘ

αθ === cos,*, kkii ΘΘ =− *

,ki ΘΘ

+ V. Kuhn, Wireless Communications over MIMO channels


Consider a matrix which is an N×N identity matrix Consider a matrix which is an N×N identity matrix

D vector space


βθ == sin,ikΘ

Wireless Communications over MIMO channels, John Wiley & Sons, 2006.


• What is the rotation matrix?

1

O


( )

=,,

O

θkiΘ



−

1

**

O

L

M

L

αβ

βα


• Define

( )

0 0 0 0

0 0 0

0 0 0 0

cos sin

, ,sin cos

i k

θ θ

θ

Θ =

I


( )0 0 0 0

0 0 0

0 0 0 0

cos sin

, ,sin cos

i k θ θ θ Θ =


0 0 0 0

0 0 0

0 0 0 0

cos sin

sin cos

θ θ

−

I


0 0 0 0

0 0 0

0 0 0 0

cos sin

sin cosθ θ

I

I


• where i is the row that contains

• and k is the row that contains

θα cos= θβ sin−=−

θβ sin= θα cos=

• If θ is chosen properly

• it can force the ith element of a colubelow

• It also force the kth element as


θβ sin= θα cos=

x


column vector equal to zero as shown


22ki xx +


• Assume xi and xk are the ith and kth

then

22cos k

xx

x

+

== θα


22

ki xx +

22sin

ki

i

xx

x

+

== θβ ( ,, kiΘ


element of the column vector,

1

O


)

−

=

1

,**

O

L

M

L

O

αβ

βα

θ


( )

−=−=

=

kx

ki

L

O

1

,,

βα

θ xΘy


=

+

=

−=−

+

=

⇒

ki

i

ki

k

xx

x

xx

L

M

L

*

22

**

22

αβ

βα


i

xxx

MM

11


+

=

+

+

N

ki

N

k

i

ki

k

ki

i

x

xx

x

x

x

xx

x

xx

M

M

M

M

M

M

O

22

22

*

220

1


• Example

• Explain LLL algorithm to find for a simple matrix, ,red red red

Q R U

=21

H

• Solution:

• We can find the condition number of cond(H)=14.9330


=

43H


Explain LLL algorithm to find for a simple matrix, ,red red red

Q R U

We can find the condition number of H by using MATLAB command



• Condition number of H should be closer to 1

• Apply LLL algorithm, the inputs are:

• The input to the LLL algorithm above is

• In MATLAB, one can find • In MATLAB, one can find

• [Q R]=qr(fliplr(H)); % flip columns

• L=fliplr(flipud(R)); % flip rows and then flip columns

• Q=fliplr(Q); % flip columns



should be closer to 1

Apply LLL algorithm, the inputs are:

The input to the LLL algorithm above is Q, L and U=I

(R)); % flip rows and then flip columns



• For example

• A = 1 2 3 4 5 6 7 8 9 10

• B = fliplr(A)%returns A with its columns flipped in the leftdirectiondirection

• B = 10 9 8 7 6 5 4 3 2 1

• A = [1 2 3 4 5 6 7 8 9 10]T

• B = flipud(A) % returns A with its rows flipped in the updirection

• B = [10 9 8 7 6 5 4 3 2 1 ]T



with its columns flipped in the left-right

with its rows flipped in the up-down



• Size reduction:

0 8944 0 4472 0 4472 0 1 0

0 4472 0 8944 3 1305 4 4721 0 1

. . .

. . . .; ;

− −

− − −= = =

Q L U

• Size reduction:

• Since the nearest integer of µ21 is 1, hence,


7039.04472.4

1305.3

22

2121 ===

L

Lµ


0 8944 0 4472 0 4472 0 1 0

0 4472 0 8944 3 1305 4 4721 0 1

. . .

. . . .; ;

− − −= = =

Q L U

is 1, hence,



• Substract column 1 from column 2 of

0 4472 0 1 0

1 3167 4 4721 1 0

.

. . ;red

− −= =

L U

• Sorting:


.

. . ;red

[ ] [ ]1,19998.142,24

3 22+≥= redred LL


column 1 from column 2 of Qred and Ured

red

0 4472 0 1 0

1 3167 4 4721 1 0;

− −= =

L U


red;

[ ] 9337.11,22

=+ redL


• Column 1 and 2 should be interchanged for

red

0 0 4472 0 1

4 4721 1 3167 1 1

.

. . ;red

− −= =

L U

• Now Lred is no more lower triangular

• We need to force zero the (1,2) element of

• apply Given rotation


.

. . ;


Column 1 and 2 should be interchanged for Qred and Ured

0 0 4472 0 1

4 4721 1 3167 1 1

− −= =

is no more lower triangular

We need to force zero the (1,2) element of Lred



• Hence,

+=

−=

3216.04472.03167.1

3167.1

cossin

sincos2,2,1

θθ

θθθG

0 8944 0 4472 0 7031 0 7111

0 4472 0 8944 0 7111 0 7031

. . . .

. . . .

− − − −

− − = = Q G


1 2

0 4472 0 8944 0 7111 0 7031

0 0 4472 1 4382 0

4 4721 1 3167 4 2346 1 3906, ,

. . . .

. . . .

. .

. . . .

red

red θ

− − = =

− −= =

Q G

L G


−=

+−=

9469.03216

3216.04472.03167.1

4472.09469.0

4472 222

0 8944 0 4472 0 7031 0 7111

0 4472 0 8944 0 7111 0 7031, ,

. . . .

. . . . ;T

− − − −

− − = = Q G


1 20 4472 0 8944 0 7111 0 7031

0 0 4472 1 4382 0

4 4721 1 3167 4 2346 1 3906

, ,

. . . .

. . . . ;

. .

. . . .

T

θ

− − = =

− −= =

Q G


• Therefore,2 0 9888 2 1

4 0 9777 4 1

.

.red red red red

− −

− −= = = =

H Q L HU

• Now the cond(Hred)=10.8753

• It is closer to 1


.


2 0 9888 2 1

4 0 9777 4 1red red red red

− −

− −= = = =

H Q L HU


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