Fundamentals of MIMO W Wireless Communications Part III+R. Bose, Information Theory, Coding and...

Fundamentals of MIMO WPart III

Prof. Rakhesh Singh

Wireless CommunicationsPart III

Singh Kshetrimayum

Fundamentals of MIMO Wireless CommunicationsPart III

• It covers

• Chapter 7: Introduction to Space-time codes

• Chapter 8: Space-time block and trellis codes

2/8/2018Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless

Communications, Cambridge University Press, 2017

Fundamentals of MIMO Wireless Communications

time codes

time block and trellis codes

, Fundamentals of MIMO Wireless


Why do we need channel coding?

• It is done to

• improve transmission quality

• when the signal encounters

• noise• noise

• interference,

• Doppler shift,

• multipath propagation

• What is space time codes?



Why do we need channel coding?



Alamouti space-time codes

• Space-time coding: coding over both space and time

• Code-word matrix for Alamouti space

*s s −

= =1

s



1 2

*

2 1

s s

s s

−= =

2

sS

s

+ S. M. Alamouti, “A simple transmit diversity technique for wireless communications,” IEEE Journal on Selected Areas in Communicationspp. 1451-8.

time codes

time coding: coding over both space and time

space-time code



, “A simple transmit diversity technique for wireless IEEE Journal on Selected Areas in Communications, 16(8), 1998,

Why space-time codes?

• Space-time codes:

• Two types

• Space-time block codes which is an extension of block codes

• Space-time block codes give

• diversity gain



• diversity gain

• but no coding gain

• Space-time trellis codes which is an extension of

• space-time trellis codes give

• both coding and diversity gain

time block codes which is an extension of block codes

5, Fundamentals of MIMO Wireless


time trellis codes which is an extension of convolutional codes

both coding and diversity gain


• What do we mean by the terms diversity and coding gain?

• the approximate symbol error rate (SER) for spacesystems+

( ) dGc

eSG

cP ≈

• where S represents the SNR,

• c is some constant,

• Gc is the coding gain and

• Gd represents the diversity order of the system



( )cSG

+ R. Bose, Information Theory, Coding and Cryptography

What do we mean by the terms diversity and coding gain?

the approximate symbol error rate (SER) for space-time coded

represents the diversity order of the system



Information Theory, Coding and Cryptography, Tata McGraw Hill, 2008.


• If we take log to the base 2 of the above equation, we have,

• What was diversity order/gain?

( )2 2 2 2log log log loge d c d

P c G G G S≈ − −

• What was diversity order/gain?

• The diversity gain/order determines the negative slope of an

• error rate curve plotted vs SNR on a log



If we take log to the base 2 of the above equation, we have,

2 2 2 2log log log loge d c d

P c G G G S≈ − −

The diversity gain/order determines the negative slope of an

SNR on a log-log scale



Why space-time codes?• In other words, a space-time coded scheme with diversity order

• has an error probability at high SNR behaving as

• If there is some coding gain,

• then average probability of error will be of the form

( ) dG

eP S

−≈

• then average probability of error will be of the form



( )

1e G

c

PG S

≈

+ E. Biglieri, R. Calderbank, A. Constantinides, A. Goldsmith, A.

MIMO wireless communications, Cambridge University Press, 2007

.

time coded scheme with diversity order Gd

has an error probability at high SNR behaving as

then average probability of error will be of the formthen average probability of error will be of the form



dG

, A. Goldsmith, A. Paulraj and H. V. Poor,

Cambridge University Press, 2007


• If there were no array or power gain

• then the probability of error expression will be of the form

• The coding gain gives the horizontal shifting of the

( )

1d

e G

c

PG S

≈

• The coding gain gives the horizontal shifting of the error rate curve

• to the space time coded error rate curve plotted on a logobtained for the same diversity order



( )2 2 2 2log log log loge d c d

P c G G G S≈ − −

If there were no array or power gain

then the probability of error expression will be of the form

The coding gain gives the horizontal shifting of the uncoded system The coding gain gives the horizontal shifting of the uncoded system

to the space time coded error rate curve plotted on a log-log scale obtained for the same diversity order



2 2 2 2log log log loge d c d

P c G G G S≈ − −





Code design criteria

• Let us consider two of the simplest cases:

• First case:

• Consider we have two codewords{C1=0, C2= 1}

Pairwise error probability (PEP)1 2

• Pairwise error probability (PEP)

• You send codeword C1 and decode that it is codeword Creceiver

• If the two bits have equal probability then,

• PEP should be ½ for a binary symmetric channel with equal probabilities



Let us consider two of the simplest cases:

codewords which simply send 0 and 1 bits

error probability (PEP)error probability (PEP)

and decode that it is codeword C2 at the

If the two bits have equal probability then,

PEP should be ½ for a binary symmetric channel with equal



Code design criteria

• Second case:

• Consider we have two codewordsfor bit 0 and codeword C2 for bit 1

• {C1=00000, C2= 11111}

• We have added a lot of redundancy (coding)

• PEP in this case is reduced (½×½• PEP in this case is reduced (½×½

• Hamming distance between codewordsand second case is 5

• Code design criteria:

• Maximize the Hamming distance between two

• Minimize the error in decoding



codewords in which we send codeword C1

for bit 1

We have added a lot of redundancy (coding)

½×½×½×½)½×½×½×½)

codewords C1 and C2 for the first case is 1

Maximize the Hamming distance between two codewords



What are the design criteria for space

• three criteria for designing space-time codes

• Rank criterion

• Determinant criterion

• Trace criterion

• Note in space-time codes, codewordsspace-time codespace-time code

• Pairwise error probability is defined as

• the probability that the decoder decides in

• favor of codeword matrix C2

• instead of the codeword matrix



+ H. Jafarkhani, Space time coding: theory and practice

What are the design criteria for space-time codes?

time codes

codewords are matrices like in Alamouti

error probability is defined as

the probability that the decoder decides in

instead of the codeword matrix C1 which was transmitted



Space time coding: theory and practice, Cambridge University Press, 2005.


Example

Given that

Show that the diversity gain of space

where r is the rank of the codeword difference matrix

( )1 2

1

1 1 4

2 4 2

r

boundn

Pγλ µ

=

⇒ → ≤ =

∏C C

where r is the rank of the codeword difference matrix

which is basically the difference of two codeword matrices

Solution: Diversity gain is defined as



((

2

2

loglim

log

e

dSNR

PG

γ→∞= −


Show that the diversity gain of space-time codes is rNR

codeword difference matrix ( )21

1

1 1 4

2 4 2

RR

R

R

N rN

n m

Nr

rN

n

n

γλ µ

λ γ

−

=

→ ≤ =

∏

codeword difference matrix

which is basically the difference of two codeword matrices

Solution: Diversity gain is defined as



))

eP

γ

( )21,CCD


( )

2

1

1 4log

2

lim

R

R

R

rN

Nr

rN

n

nG

λ γ=

= −

∏



( )

( ) (

( )

{ }( )

2

2 2 2 2

1

2

2

2

limlog

log 2 log log 4 log

limlog

log glim

log

R

d

Nr

n R

n

R

R

G

rNrN

γ

γ

γ

γ

λ γ

γ

γ

γ

→∞

=

→∞

→∞

= −

+ − +

=

≈ =

∏


( )2 2 2log log log loge d c d

P c G G G≈ − −



( ) { }2 2 2 2log 2 log log 4 logRrN

n RrNλ γ+ − +

( )2 2 2log log log loge d c d

P c G G G≈ − −


• A sound design policy to assure full diversity is

• to ensure that for every possible codeword pairs codeword distance matrix

( ) ( ) ( ijiHji CDCCDCCA ,, =

• is full rank (RANK CRITERION)



( ) ( ) (CDCCDCCA ,, =


A sound design policy to assure full diversity is

to ensure that for every possible codeword pairs Ci and Cj, iǂj, the

)jC,



)C,


• Similarly the coding gain is defined as

• where the codeword distance matrix is • where the codeword distance matrix is



( ) ( ) ( 12121 ,,, CCDCCDCCA •= H


Similarly the coding gain is defined as

where the codeword distance matrix is

( )1

1

1 2

1

,r r

r

n

n

λ=

=

∏ A C C

where the codeword distance matrix is



) ( ) ( )21212 CCCCC −•−=H


• We can define coding gain distance (CGD) between two and

( ) (21 det, ACC =CGD

• Coding gain is defined as



( )[ ]rc CGDG

121 , CC=


We can define coding gain distance (CGD) between two codewords

( ))21 , CC



r

1


• To augment the coding gain of a full diversity code,

• further good design policy would be to determinant of the matrices(DETERMINANT CRITERION)

( iCCA ,

(DETERMINANT CRITERION)

Example

• Find the CGD and coding gain of two codeword matrices and for Space time trellis code given by




To augment the coding gain of a full diversity code,

further good design policy would be to maximize the minimum

)jC i j∀ ≠

Find the CGD and coding gain of two codeword matrices and for




• Also find the rank and trace of the codeword distance matrix.

• Solution:

=

= ,

11

11 21CC

• Solution:

• The codeword difference matrix is

• The codeword distance matrix is



( ) −=, 121CCCCD


Also find the rank and trace of the codeword distance matrix.

−

−=

11

11



=

02

202C


• The rank and trace of the codeword distance matrix is 2 and 8

( ) ( ) (

=

=

=

40

04

02

20

02

20

,, 2121CDCCDCCA

• The rank and trace of the codeword distance matrix is 2 and 8 obviously

• The coding gain distance



( ) (, == 121CACCCGD


The rank and trace of the codeword distance matrix is 2 and 8

),H21

CC

The rank and trace of the codeword distance matrix is 2 and 8



) 16, =21C


• The coding gain is

• The diversity order of the system is twice the number of receiving antennas

( )1

2C

G CGD= = =

• We can also express average PEP upper bound of spaceover flat fading iid Rayleigh channel as



( ) (1 2 1 21exp ,

4 4R

P N

→ ≤ −

C C D C C


The diversity order of the system is twice the number of receiving

16 4= = =

We can also express average PEP upper bound of space-time codes Rayleigh channel as



( )2

1 2 1 2exp ,

4 4F

γ

C C D C C


• Another good design policy is to maximize the minimum squared

Frobenius norm of codeword difference matrix(TRACE CRITERION)

( ) ( )(2

, , , ,i j H i j i j i jTrace Trace= • =D C C D C C D C C A C C

• If we maximize the minimum trace of codeword distance matrix between all pairs of codeword matrices

• then we can minimize the probability of error bound



( ) ( )(, , , ,i j H i j i j i j

FTrace Trace= • =D C C D C C D C C A C C


maximize the minimum squared

norm of codeword difference matrix among all possible iǂj

( )) ( )( ), , , ,i j H i j i j i jTrace Trace= • =D C C D C C D C C A C C

If we maximize the minimum trace of codeword distance matrix between all pairs of codeword matrices

then we can minimize the probability of error bound



( )) ( )( ), , , ,i j H i j i j i jTrace Trace= • =D C C D C C D C C A C C


• Coding over space and time

*

1 2

*

s s

s s

−= =

1

2

sS

s



*

2 1s s= =

2S

s

+ S. M. Alamouti, “A simple transmit diversity technique for wireless communications,” IEEE Journal on Selected Areas in Communicationspp. 1451-8.

time codes



( )1 2 1 2, 0H

= =s s s s

, “A simple transmit diversity technique for wireless IEEE Journal on Selected Areas in Communications, 16(8), 1998,


• If we allow the channel gain coefficients to remain equal for two successive symbol periods

( ) ( ) 1111 ehhTthth ==+=



( ) ( ) 222 hhTthth ==+=

time codes

If we allow the channel gain coefficients to remain equal for two

1θje



22

θjeh


• At the receiver the signals after passing through the additive white Gaussian noise (AWGN) channel may be expressed as

• Signal received in first time instant by the receiving antenna is

• Signal received in second time instant by the receiving antenna is

122111 nshshr ++=

• Signal received in second time instant by the receiving antenna is

• In matrix form, component-wise



2*12

*212 nshshr ++−=

*1 1 11 2

*2 2 22 1

T

r h ns s

r h ns s

− = +

time codes

At the receiver the signals after passing through the additive white Gaussian noise (AWGN) channel may be expressed as

Signal received in first time instant by the receiving antenna is

Signal received in second time instant by the receiving antenna is

1

Signal received in second time instant by the receiving antenna is



1 1 1

2 2 2

T

r h n

r h n

= +


• An equivalent form of the above equation,

−=

2

1

*1

*2

21

*2

1

s

s

hh

hh

r

r

• In compact matrix form, the received signal vector may be written as

• The combining operation at the receiver is

• where



= +r Hs n

H= +r H Hs n% %

nHn H=~

time codes

An equivalent form of the above equation,

+

*2

1

n

n

In compact matrix form, the received signal vector may be written as

The combining operation at the receiver is




• The 2×1 vector signal output of the combiner is given as:

=

−=

*

1

*

2*11

~

~

r

r

hh

hh

r

r

• Similarly the 2×1 vector of additive complex noise in the combiner output is



−

*21

*22

~ rhhr

−=

*2

1

1*2

2*1

2

1

~

~

n

n

hh

hh

n

n

time codes

1 vector signal output of the combiner is given as:

−

+**

*221

*1

rhrh

rhrh

1 vector of additive complex noise in the combiner



− *211

*2 rhrh

−

+=

*211

*2

*221

*1

nhnh

nhnh


• In matrix form, the output of the combiner can be written as

−

−=

*1

*2

21

1*2

2*1

2

1

~

~

hh

hh

hh

hh

r

r

• In matrix form, the output of the combiner can be written as



H= +r H Hs n% %

time codes

In matrix form, the output of the combiner can be written as

+

2

1

2

1

*1

~

~

n

n

s

s

In matrix form, the output of the combiner can be written as




• which can be simplified as

(=

+

+

+=

2

1

2

1

2

2

2

1

2

2

2

1

2

1

~

~

0

0~

~

hn

n

s

s

hh

hh

r

r

• In order to perform this operation, the combiner needs the channel state information (CSI)

• The above matrix equation can be written as



+ 22212 0 hh

( ) 211

2

2

2

11~;~~ rnshhr =++=

time codes

) ( )

+

+=

+

+

2

1

2

12

2

2

12

1

2

12

2

2

2

1 ~

~

~

~

n

n

s

shh

n

n

s

shh I

In order to perform this operation, the combiner needs the channel

The above matrix equation can be written as



2222

( ) 22

2

2

2

1~nshh ++=


• ML detection

• is one of the M-ary

( 2 2

1 1 1 2

argˆ

min{ }s r h h s

m= − +%

{ }Mss ∈• is one of the M-ary

• We may express above decision criterion as



{ }M

kkm ss1=∈

(1

1 2 2

1 2

argˆ

min{ }

rs s

m h h= −

+

%

time codes

ary symbols

)2 2

1 1 1 2 ms r h h s= − + Tt ≤≤0

ary symbols

We may express above decision criterion as



)2 2

1 2

ms sh h

= −


• ML detection

• is one of the M-ary{ }Mss ∈

(2 2 1 2

argˆ

min{ }s r h h s

m= − +%

• is one of the M-ary

• We may express above decision criterion as



{ }M

kkm ss1=∈

(2 2 2

1 2

argˆ

min{ }s s

m h h= −

time codes

ary symbols

TtT 2≤≤)2 2

2 2 1 2 ms r h h s= − +

ary symbols

We may express above decision criterion as



)2

2 2

1 2

m

rs s

h h= −

+

%

What is the equivalent MRC receiver diversity?

• 2×1 Alamouti space-time code have similar performance to 1system

• For SIMO system, the received signals are

1011 nshr +=

• and the signal after the Maximal rat



1011 nshr +=

2022 nshr +=

( 2

2

12*21

*10

~ hhrhrhr +=+=

+ M. Jankiraman, Space-time codes and MIMO systems


time code have similar performance to 1×2 SIMO

For SIMO system, the received signals are

l ratio combining operation is given by



) 2*21

*10

2

2 nhnhs ++

time codes and MIMO systems, Artech House, 2004


• Note that the MRC signal is equivalent to

• the resulting combined signals of the transmit diversity scheme (Alamouti space-time code) above,

• except for a phase difference in the noise components• except for a phase difference in the noise components

• Hence, the two schemes have the same effective SNR

• This shows that the diversity order from (with one receive antenna) is the

• same as that of the two branch MRC




Note that the MRC signal is equivalent to

the resulting combined signals of the transmit diversity scheme time code) above,

except for a phase difference in the noise componentsexcept for a phase difference in the noise components

Hence, the two schemes have the same effective SNR

This shows that the diversity order from Alamouti space-time code (with one receive antenna) is the

same as that of the two branch MRC




• In general,

• Alamouti space-time code with Nantennas

• has the same performance of a MRC with 2N• has the same performance of a MRC with 2N




time code with NT=2 and NR number of receiving

has the same performance of a MRC with 2NR receive antennashas the same performance of a MRC with 2NR receive antennas



Alamouti Space-Time Code

• Diversity gain, coding gain and code rate

• Consider two distinct Alamouti codeword matrices

* *

1 2 1 2

* *;A A B B

A B

s s s s

s s s s

− −= =

S S

• The code word difference matrix is given by



* *

2 1 2 1

A B

A A B Bs s s s

( )

−−

+−−=

*1

*122

**211,

BABA

BABABA

ssss

ssssSSD

Time Code

Diversity gain, coding gain and code rate

codeword matrices

* *

1 2 1 2

* *

A A B Bs s s s

s s s s

− −

( ) ( )2121 BBAA ssss ≠

The code word difference matrix is given by



* *

2 1 2 1A A B Bs s s s

1

*2B

2121 BBAA


• Distance matrices of every two uneqtwo

• Alamouti scheme provides full transmit diversity of two

• The code word distance matrix• The code word distance matrix



( ) ( )BABA DSSDSSA ,, =

( )

+−= 2

2

11

0, ABA

BA

sssSSA

Time Code

unequal code words have a full rank of

scheme provides full transmit diversity of two



( )BAH

SSD ,

−+−

−2

22

2

11

2

22 0

BABA

B

ssss

s


• Coding gain

( )

( )

2

2

11

0det, ABA

BA

sss

+−=SSA

• We can observe that the Gc is equal to that of the



( )22

22

2

11 BABA ssss −+−=

( )1 2

2 2 2 2

1 1 2 2 1 1 2 2r

A B A B A B A BCGD s s s s s s s s

= − + − = − + −

Time Code

2

22

2

11

2

2 0

BABA

B

ssss

s

−+−

−

is equal to that of the uncoded system



( )1/ 2

22 2 2 2

1 1 2 2 1 1 2 2A B A B A B A BCGD s s s s s s s s

= − + − = − + −


• Note that code rate is 1

• For SER analysis, there are two basic steps:

• a) First, find the conditional error probability (CEP) for the specific modulation scheme (refer to any wireless book for this)modulation scheme (refer to any wireless book for this)

• b) Second, average it over the pdfaverage symbol error rate (SER)



+ A. Goldsmith, Wireless Communications, Cambridge University Press, 2005.

Time Code

For SER analysis, there are two basic steps:

a) First, find the conditional error probability (CEP) for the specific modulation scheme (refer to any wireless book for this)modulation scheme (refer to any wireless book for this)

pdf of the received SNR to obtain



, Cambridge University Press, 2005.


• SER analysis for Alamouti space-time code over fading channels

• Since we know that symbol error rate (SER)

• is a function of the received SNR

• For Alamouti space-time code, the received signal for 0• For Alamouti space-time code, the received signal for 0



( ) *221

*11

2

2

2

11~ nhnhshhr +++=

Time Code

time code over fading channels

symbol error rate (SER)

is a function of the received SNR

time code, the received signal for 0≤t≤T istime code, the received signal for 0≤t≤T is




• Hence the received SNR can be expressed as

( )( )

( 2

1

22

2

2

1

22

2

2

1

σσ

+=

+

+=

s

Alamouti

h

hh

EhhSNR

• If channel state information (CSI) is not available at the transmitter,

• then we will allot equal power to each channel



( )21σσ+ hh

Time Code

Hence the received SNR can be expressed as

) ( ) ( )212

2

2

2

2

1

2

2

2γγ

σσσ+=+=

+ sss EhEhEh

If channel state information (CSI) is not available at the transmitter,

then we will allot equal power to each channel



σσσ


• Hence, we should modify the above equation as follows

• where T is the symbol period

( ) ( ) (22

2

2

2

2

2

1

σσ=+=

hE

hE

h

SNR

ss

Alamouti

• where T is the symbol period

• For Alamouti space-time code, condBPSK is given by



( ) ( )21 2,| γγ

== QSNRQEPb

Time Code

Hence, we should modify the above equation as follows

) ( )2

22 21

2

2

2

2

2

1 γγ

σσ

+=+

PTh

PTh

onditional error probability (CEP) for



( )2121

22 γγ

γγ+=

+Q


• Then average bit error rate (BER) of BPSK for code over a fading channel can be calculated as

( ) ( )∫ ∫∞ ∞

+= 21 1γγ γpQEPb

• For independent channel, we have,



( ) ( )∫ ∫ +=

0 0

21 1γγ γpQEPb

( ) ( ) (∫ ∫∞ ∞

+=

0 0

121 1γγγ γpQEPb

Time Code

Then average bit error rate (BER) of BPSK for Alamouti space-time code over a fading channel can be calculated as

( ) 2121, ,21

γγγγγγ dd

For independent channel, we have,



( ) 2121, ,21

γγγγγγ dd

) ( ) 2121 2γγγγ γ ddp


• Hence, for iid case

( )2 2

1 2 1 2

1 2 2 2 2

0 0

1 1exp exp exp

2sin 2sin 2sinQ d d

π π

γ γ γ γγ γ θ θ

π πθ θ θ

+ + = − = − −

∫ ∫Q

( )2

1 21exp expP E d p p d d

π

γ γ∞ ∞

= − − ∫ ∫ ∫



( ) 1 2

2 2

0 0 0

2

1 2

2 2

0 0 0

1exp exp

2sin 2sin

1exp exp

2sin 2sin

bP E d p p d d

π

γ γ

π θ θ

γ γ

π θ θ

∞ ∞

= − −

= − −

∫ ∫ ∫

∫ ∫ ∫

J. Craig, “A new, simple and exact result for calculating the probability of error for two

signal constellations”, in Proc. IEEE MILCOM, pp. 25.5.1

Time Code2 2

1 2 1 2

2 2 2

0 0

1 1exp exp exp

2sin 2sin 2sinQ d d

π π

γ γ γ γγ γ θ θ

π πθ θ θ

+ = − = − −

∫ ∫

( ) ( )1 2P E d p p d dγ γ

θ γ γ γ γ

= − −



( ) ( )

( ) ( )

1 2

1 2

1 2

1 2 1 22 2

1 2 1 22 2

2sin 2sinP E d p p d d

p p d d d

γ γ

γ γ

γ γθ γ γ γ γ

θ θ

γ γ γ γ θθ θ

= − −

J. Craig, “A new, simple and exact result for calculating the probability of error for two-dimensional

, pp. 25.5.1-25.5.5, Boston, MA, 1991.


• Expressing in terms of MGF, we have,

( )1 2

2

2 2

0

1 1 1

2sin 2sinb

P E d

π

γ γπ θ θ

= Μ − Μ −

∫

• For identical channels, we have the BER of BPSK for time code over any fading channel is given as



02sin 2sinπ θ θ

( )2

2

2

0

1 1

2sinb

P E d

π

γπ θ

= Μ −

∫

Time Code

Expressing in terms of MGF, we have,

2 2

1 1 1

2sin 2sinP E dθ

θ θ

( ) ( )[ ]sXEsM X exp=

( ) ( ) ( )expX XM s sx p x dx

∞

−∞

= ∫

For identical channels, we have the BER of BPSK for Alamouti space-time code over any fading channel is given as



2sin 2sinθ θ

2

P E dθ


• For Rayleigh fading case, the average BER for code employing BPSK is given by

( )221 2sin

π

θ = ∫



( )22

2

0

1 2sin

2sinb

P E dθ

π θ γ

=

+ ∫

( ) ( )γ

γγ

γ

γγγ dss

=

−=Μ

∞∞

∫∫ exp1

exp1

exp

00

Time Code

For Rayleigh fading case, the average BER for Alamouti space-time

2

θ



P E dθθ γ

γ

γ

γ

γγ

γγ

γ

γγ

ss

s

ds−

=

−

−

=

−

∞

1

1

1

exp1

0


• BER for DBPSK over Nakagami fading channel

• Note that in BPSK and MPSK information is carried in signal phase

• In differential modulation,

• we can utilize the previous symbol’s phase as phase reference for • we can utilize the previous symbol’s phase as phase reference for the current symbol

• Hence, we do not have the necessitat the receiver



Time Code

fading channel

Note that in BPSK and MPSK information is carried in signal phase

we can utilize the previous symbol’s phase as phase reference for we can utilize the previous symbol’s phase as phase reference for

ssity of the coherent phase reference

Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless



• For differential binary phase shift keying (DBPSK), the CEP for SISO case is given as

• For Alamouti space-time code,

( ) γγ −= eEPb2

1|


• we have average BER for DBPSK over any fading channel is given as



( ) ∫ ∫∞ ∞ +

−=

0 0

2

21

2

1γγ

eEPb

Time Code

For differential binary phase shift keying (DBPSK), the CEP for SISO

we have average BER for DBPSK over any fading channel is given as



( ) ( ) 2121 21

2

γγγγ γγ ddpp


• Expressing in terms of MGF,

( ) ∫ ∫∞ ∞

−−

=⇒0 0

22

21

2

1γ

γγ

peeEPb


• we have the average BER of DBPSK over any fading channel as



( )1 2 1

1 1 1 1 1

2 2 2 2 2bP E γ γ γ

= Μ − Μ − = Μ −

Time Code

( ) ( ) 2121 21γγγγ γγ ddp

we have the average BER of DBPSK over any fading channel as



1 2 1

2

1 1 1 1 1

2 2 2 2 2γ γ γ

= Μ − Μ − = Μ −


• Hence for Nakagami fading,

• the average BER for Alamouti spaceobtained as



( )m

bm

EP

2

21

2

1−

+=

γ

Time Code

space-time code employing DBPSK is

( )m

s−

−=Μγ



( )m

ss

−=Μ

γγ 1


• SER for MPSK over Hoyt fading channel

• For M-ary phase shift keying (M-PSK), CEP is given by

−Mπ

1



( )

−= ∫

gEP

MPSK

bθ

γ

πγ

π

0

2sinexp

1|

Time Code

SER for MPSK over Hoyt fading channel

PSK), CEP is given by



=

Mgd PSK

πθ

θ

γ 2sin;



• MPSK over fading channel, we have SER as

∞ ∞

−

+1

21 γγπgM

M

PSK



( ) ∫ ∫ ∫∞ ∞

−=

0 0 0

2

21

sin

2exp1

θπ

g

EP

M PSK

b

Time Code

MPSK over fading channel, we have SER as

2



( ) ( )

2121

2

21γγγγθ γγ ddppd



• we have the average SER of MPSK for

• over any fading channel as



( )π

π

γ

gEP

M

M

b ∫

−

−Μ=

1

0sin2

1

Time Code

we have the average SER of MPSK for Alamouti space-time code



θθ

dg PSK

2

2sin


• For Hoyt fading MGF is given by

( ) ( )(

(2

21

+−=Μ ss γγ



( ) ( )(

21

+−=Μ ss γγ

Time Code

)

( )

2

1

222

−

qsγ



( )221

+ q


• For Hoyt fading,

• the average SER for Alamouti space



( ) γθ

π

π

π

MEP

M

M

b

1

0

2

2

sin

sin

11

−

∫

+

+=

Time Code

space-time code employing M-PSK is

π1

2

2sin

−



( )θ

γθ

d

q

qM

22

2

2

2

1

sin

sin

+


• SER for M-QAM over Rice fading channel

• In MQAM, information is carried in both phase and amplitude of the signal

• For M-ary quadrature amplitude modulation (M• For M-ary quadrature amplitude modulation (MSISO is given as



+ A. Goldsmith, Wireless Communications, Cambridge University Press, 2005.

( ) ( ) 421

14| −

−= gQ

MEP QAMb γγ

Time Code

QAM over Rice fading channel

In MQAM, information is carried in both phase and amplitude of the

amplitude modulation (M-QAM), the CEP for amplitude modulation (M-QAM), the CEP for



, Cambridge University Press, 2005.

( )( )12

3;2

114 2

2

−=

−

MggQ

MQAMQAM γ



( )sin2

exp1

4

0

2

−= ∫

xxQ

π

π

Q


• M-QAM over fading channel, we have SER as



( ) ∫ ∫∞ ∞

−−

+

−=

0 0

21 142

21

14γγ

gQM

EP QAMb

Time Code

0;sin 2

2

≥

xd

xθ

θ

QAM over fading channel, we have SER as



( ) ( )

+

2121

212

2

2122

1γγγγ

γγγγ ddppgQ

MQAM


• Integrating w.r.t. γ1 and γ2 first, we have,

( )π

π

g

MEP

QAM

b

2

sinexp

1114 ∫ ∫ ∫

∞ ∞

−

−=



θ

γ

π

π

π

g

M

M

QAM4

0 0 0

2

12

0 0 0

sin

2exp11

14

sin

∫ ∫ ∫

∫ ∫ ∫

∞ ∞

+

−

−−

Time Code

first, we have,

( ) ( ) θγγγγθ

γγ

γγ dddppQAM

21212

21

21sin

2+



( ) ( ) θγγγγ

γ

θ

γγ dddpp 2121

2

2

21

21

2

sin

+


• Expressing in terms of MGF, we have,

( )θπ

π

γ

g

MEP

QAM

b

−Μ

−= ∫

2

2sin2

11

41



θπ

θπ

γ

π

γ

γ

g

M

M

QAM

b

Μ

−Μ

−−

∫

∫

4

0

2

2

0

2

sin2

11

4

sin2

21

1

Time Code

Expressing in terms of MGF, we have,

θθ

γ dgQAM

−Μ

2sin2

2



θθ

θγ

dgQAM

−

2

2

sin2

sin22


• For equal channels,

• we have the SER of M-QAM for Alamoutifading channel as

π



( ) θθπ

π

γ dg

MEP

QAM

b ∫

−Μ

−=

2

0

2

2sin2

11

4

Time Code

Alamouti space-time code over any

π



θθπ

θ

π

γ dg

M

QAM

∫

−Μ

−−

4

0

2

2

2

sin2

11

4


• For Rice fading,

• the SER of M-QAM for Alamouti space

π

K

+ 2114



( )

γθ

π

γθ

π

π

gK

K

M

gK

K

MEP

QAM

QAMb

∫

∫

++

+

−−

++

+

−=

4

02

2

2

02

1

exp

sin21

111

4

exp

sin21

111

4

Time Code

space-time code is given by

γθ

gK

QAM

−

2

2



θ

γθ

γθ

θ

γθ

γθ

dg

K

gK

dg

K

K

QAM

QAM

QAM

++

−

++

−

2

2

2

2

2

sin21

sin2

sin21

sin2exp

Space-time block codes

• The Alamouti space-time code was for two transmitting antennas only

• How do we generalize it for any number of transmitting antennas?

• Orthogonal space-time block codes (OSTBC) for any number of • Orthogonal space-time block codes (OSTBC) for any number of transmitting antennas

• based on the theory of orthogonal designs

• The orthogonal property makes the decoding complexity minimal

• and we can decouple the symbols at the decoder



time code was for two transmitting antennas

How do we generalize it for any number of transmitting antennas?

time block codes (OSTBC) for any number of time block codes (OSTBC) for any number of

based on the theory of orthogonal designs

The orthogonal property makes the decoding complexity minimal

and we can decouple the symbols at the decoder








• Much of the characteristics of spacespecified by the generator matrix

SS L1211



=

LL NN SS

SS

SS

L

OMM

L

L

21

2221

1211

G

Much of the characteristics of space-time block code (STBC) is

TNSL 1



TL

T

T

NN

N

N

S

S

S

L

MO

L

L

2

1

LN

kR =


• We have observed that Alamouti space

• the maximum diversity

• and can be detected easily using maximum likelihood (ML) detectordetector

• These two desirable characteristics were achieved

• as a result of the orthogonality charG for the Alamouti code



space-time code achieves

and can be detected easily using maximum likelihood (ML)

These two desirable characteristics were achieved

characteristic of the generator matrix




• The column vectors are orthogonal

−= *

2

11

s

ss

= *

1

22

s

ss

• orthogonality property of G



[ ] 2

2

2

2

1 IGG ssH +=

The column vectors are orthogonal

( ) 021 =• ssH



1 2

* *

2 1

s s

s s

= −

G


• The Alamouti space-time code provides full diversity of two

• without channel state information (CSI) at the transmitter and

• an uncomplicated ML decoding system at the receiver

• Such a system provides a guaranteed overall diversity gain• Such a system provides a guaranteed overall diversity gain

• Because of these, the scheme was extended to any number of transmit antennas

• by using the theory of orthogonal designs



time code provides full diversity of two

without channel state information (CSI) at the transmitter and

an uncomplicated ML decoding system at the receiver

Such a system provides a guaranteed overall diversity gainSuch a system provides a guaranteed overall diversity gain

Because of these, the scheme was extended to any number of

by using the theory of orthogonal designs




• The generalized schemes are known as codes

• These codes achieve full transmit diversity of

• while allowing an uncomplicated ML decoding algorithm• while allowing an uncomplicated ML decoding algorithm

• The entries of the G are chosen such that they are linear combinations of and their conjugates

• The matrix itself is constructed base



TNsss ,,, 21 L

HssGG

++=2

2

2

1

The generalized schemes are known as orthogonal space time block

These codes achieve full transmit diversity of

while allowing an uncomplicated ML decoding algorithmRT NN

while allowing an uncomplicated ML decoding algorithm

are chosen such that they are linear combinations of and their conjugates

based on orthogonal designs such that



TT NNs I+

2L


• OSTBC presumes that the channel coa period of symbols, i.e.,

• This block fading assumption is needed for uncomplicated linear

( )ij ij Th t h t N= =

• This block fading assumption is needed for uncomplicated linear decoding of OSTBC

• The orthogonality allows us to achieve full transmit diversity

• and enables receiver to decouple the signals transmitted from different antennas



el coefficients remains the same over

is needed for uncomplicated linear

; 1,2, ,ij ij T

h t h t N= = L

is needed for uncomplicated linear

allows us to achieve full transmit diversity

and enables receiver to decouple the signals transmitted from




• Dependent on the type of signal constellation used,

• space-time block codes can be divided into

• OSTBC with real signals

• or OSTBC with complex signals• or OSTBC with complex signals

• OSTBC for real signals

• Such matrices exist if the number of transmit antennas

• Real orthogonal generator matrix designs provide a diversity order of



8,4,2=TN

RT NN

Dependent on the type of signal constellation used,

time block codes can be divided into

Such matrices exist if the number of transmit antennas

Real orthogonal generator matrix designs provide a diversity order of




• each column of an orthogonal design is a permutation with sign change of the first column

• Just a simple check, see all the columns are orthogonal or not

s s s s− − −



1 2

2

2 1

s s

s s

− =

G

1 2 3 4

2 1 4 3

4

3 4 1 2

4 3 2 1

s s s s

s s s s

s s s s

s s s s

− − − = −

G

each column of an orthogonal design is a permutation with sign

Just a simple check, see all the columns are orthogonal or not

s s s s− − −



1 2 3 4

2 1 4 3

3 4 1 2

4 3 2 1

s s s s

s s s s

s s s s

s s s s

− − −

−

−


• The code rate for all these matrices is equal to one

1 2 3 4 5 6 7 8

2 1 4 3 6 5 8 7

s s s s s s s s

s s s s s s s s

s s s s s s s s

− − − − − − −

− − − − − −



3 4 1 2 7 8 5 6

4 3 2 1 8 7 6 5

8

5 6 7 8 1 2 3 4

6 5 8 7 2 1 4 3

7 8 5 6 3 4 1 2

8 7 6 5 4 3 2 1

s s s s s s s s

s s s s s s s s

s s s s s s s s

s s s s s s s s

s s s s s s s s

s s s s s s s s

− − −

− − − =

− − − − − −

− − −

G

The code rate for all these matrices is equal to one

1 2 3 4 5 6 7 8

2 1 4 3 6 5 8 7

s s s s s s s s

s s s s s s s s

s s s s s s s s

− − − − − − −

− − − − − −



3 4 1 2 7 8 5 6

4 3 2 1 8 7 6 5

5 6 7 8 1 2 3 4

6 5 8 7 2 1 4 3

7 8 5 6 3 4 1 2

8 7 6 5 4 3 2 1

s s s s s s s s

s s s s s s s s

s s s s s s s s

s s s s s s s s

s s s s s s s s

s s s s s s s s

− − −

− − − − − −

− − − − − −

− − −


• We can also find the diversity gain for the given generator matrices

• For instance, for 2×2 generator matrix of

−

=21

2ss

ssG

• For two different space-time codewords

• let us find the codeword difference matrix and

• codeword distance matrix



=12

2ss

G

We can also find the diversity gain for the given generator matrices

2 generator matrix of

codewords,

let us find the codeword difference matrix and




• codeword difference matrix of SA and

• The codeword distance matrix of S

( ) 1 1 2 2

2 2 1 1

, A B A B

A B

A B A B

s s s s

s s s s

− − +=

− − D S S

• The codeword distance matrix of S



( ) ( ) ( )( )( 1 1 2 2

, , ,T A B A B

A B A B A B

s s s s − + − = =

A S S D S S D S S

and SB is

S and S is

1 1 2 2

2 2 1 1

A B A B

A B A B

s s s s

s s s s

− − +

− −

SA and SB is



) ( )

( ) ( )

2 2

1 1 2 2

2 2

1 1 2 2

0

0

A B A B

A B A B

s s s s

s s s s

− + − − + −


• Rank of this matrix is 2,

• hence it achieves diversity gain of 2N

• Similarly, we can show that generator matrices of

• also achieve diversity gain of 4N• also achieve diversity gain of 4N

• But there is no coding gain

• which is the same as the squared Euclidean distance of the symbols for uncoded system



( ) ( )( ) (112,r

d A B A B A BG CGD s s s s= = = − + −A S S

hence it achieves diversity gain of 2NR

Similarly, we can show that generator matrices of G4 and G8

also achieve diversity gain of 4NR and 8NR respectivelyalso achieve diversity gain of 4NR and 8NR respectively

which is the same as the squared Euclidean distance of the symbols



( ) ( )( )2 2

1 1 2 2d A B A B A BG CGD s s s s= = = − + −


• Assuming a single receiving antenna (two transmitting antennas)

• and we can check the decoding of this real OSTBC

• For instance, let us consider generator matrix of

• For this case, the received signal is• For this case, the received signal is



[ ] [ ]210

212

hhN

Eyy s

=

Assuming a single receiving antenna (two transmitting antennas)

and we can check the decoding of this real OSTBC

For instance, let us consider generator matrix of G2

For this case, the received signal isFor this case, the received signal is



[ ]2112

21nn

ss

ss+

−


• Note that we have normalized the power

• by introducing a term of the form explicitly

• If this term is not included,

• then we have to consider this power division separately • then we have to consider this power division separately

• when we calculate the received SNR

• The above equation can be expressed in equivalent form as



Note that we have normalized the power

by introducing a term of the form explicitly

then we have to consider this power division separately

0NN

E

T

s

then we have to consider this power division separately

when we calculate the received SNR

The above equation can be expressed in equivalent form as




• Hence, the ML detection is

(

argˆ

min{ }

j

j m

rs s j

m E= − =

%

• where is one of the Mmodulation (MPAM) symbols



( 2 2

1 2

0

min{ }

2

Sm E

h hN

+

{ }M

kkm ss1=∈

); 1,2

j ms s j= − =

where is one of the M-ary pulse amplitude



)2 2

1 2h h


• OSTBC for Complex Signal Constellations

• The Alamouti scheme gives the full diversity of 2Nrate of 1

• It has been shown in the literature• It has been shown in the literaturewith code rate R=1 do not exist for N



+ H. Jafarkhani, Space time coding: theory and practice

2005.

V. Tarokh, H. Jafarkhani and A. R. Calderbank, “Space

communications: performance results,” IEEE Journal on Selected Areas in

Communications, 17(3), 1999, pp. 451-60

OSTBC for Complex Signal Constellations

scheme gives the full diversity of 2NR with a full code

It has been shown in the literature+ that orthogonal complex designs It has been shown in the literature+ that orthogonal complex designs with code rate R=1 do not exist for NT>2 transmit antennas



Space time coding: theory and practice, Cambridge University Press,

, “Space-time block coding for wireless

IEEE Journal on Selected Areas in


• However by reducing the code rate,

• it is possible to devise complex orthogonal designs for 2constellations

• For example, • For example,

• an orthogonal generator matrix for a STBC that transmits four

• complex-valued Phase shift keying (PSK) or

• Quadrature amplitude modulation (QAM) symbols

• on 4 transmit antennas is given below



However by reducing the code rate,

it is possible to devise complex orthogonal designs for 2-D signal

an orthogonal generator matrix for a STBC that transmits four

valued Phase shift keying (PSK) or

amplitude modulation (QAM) symbols

on 4 transmit antennas is given below




1 2 3 4

2 1 4 3

3 4 1 2

4 3 2 1

4 * * * *

1 2 3 4

s s s s

s s s s

s s s s

s s s s

s s s s

− − − − − − =

G

• For this code generator, the four complextransmitted in eight consecutive time slots (R=1/2)



1 2 3 4

* * * *

2 1 4 3

* * * *

3 4 1 2

* * * *

4 3 2 1

s s s s

s s s s

s s s s

s s s s

− − − − − −

1 2 3 4

2 1 4 3

3 4 1 2

4 3 2 1

* * * *

1 2 3 4

s s s s

s s s s

s s s s

s s s s

s s s s

− − − −

For this code generator, the four complex-valued symbols are transmitted in eight consecutive time slots (R=1/2)



1 2 3 4

* * * *

2 1 4 3

* * * *

3 4 1 2

* * * *

4 3 2 1

s s s s

s s s s

s s s s

s s s s

− − − −


• We also observe that

• where c is a constant

[ ]2

4

2

3

2

2

2

1 IGG sssscH +++=

• where c is a constant

• So this code provides

• fourth order diversity in the case of one receive antenna and

• 4NR diversity with NR receive antennas



4I

fourth order diversity in the case of one receive antenna and

receive antennas




• Differential modulation introduces memory in the transmitted data

• This memory can be used to estimate the transmitted data without knowing CSIR

• It is very useful for high speed wireless communication• It is very useful for high speed wireless communication

• Estimate of the channel fading is poor for fast fading channel

• Hence differential modulation can be used for fast fading channels



+ M. R. Bhatnagar, Differential Coding for MIMO and Cooperative Communication

thesis, University of Oslo, Norway, 2008.

Differential modulation introduces memory in the transmitted data

This memory can be used to estimate the transmitted data without

It is very useful for high speed wireless communicationIt is very useful for high speed wireless communication

Estimate of the channel fading is poor for fast fading channel

Hence differential modulation can be used for fast fading channels



Differential Coding for MIMO and Cooperative Communication, PhD


• Differential OSTBC

• For non-coherent STBC,

• we can apply differential STBC so that

• we can estimate the transmitted symbol without CSIR• we can estimate the transmitted symbol without CSIR

• Consider I-O model of a NT×NR MIMO system

• The received signal at time t can be expressed as



t t t= +Y HX N

we can apply differential STBC so that

we can estimate the transmitted symbol without CSIRwe can estimate the transmitted symbol without CSIR

MIMO system

The received signal at time t can be expressed as




• where

• Yt is an NR×NL matrix

• H is an NR×NT matrix (for frequency H are iid distributed as NC(0,1) )H are iid distributed as NC(0,1) )

• Xt is an NT×NL OSTBC encoded transmission matrix at time t

• Nt is NR×NL matrix with i.i.d. complex circular Gaussian random variables (RVs) with each components distributed as



ncy flat Rayleigh fading, elements of

OSTBC encoded transmission matrix at time t

. complex circular Gaussian random variables (RVs) with each components distributed as NC(0,σ2)




• Let S denote the symbol alphabet from a unitary constellation

• ( ), for instance, BPSK, QPSK, M

• Assume be a block of p symbols to be sent at a time t

2

1,j j

s S s∀ ∈ =

{ }p

s• Assume be a block of p symbols to be sent at a time t

• Then define



{ }1

jj

s=

(1

1 preal imag

t j j j jj

s i s ip =

= + = −∑U V W

denote the symbol alphabet from a unitary constellation

( ), for instance, BPSK, QPSK, M-PSK

Assume be a block of p symbols to be sent at a time tAssume be a block of p symbols to be sent at a time t



) 1,real imag

t j j j js i s i= + = −U V W


• where satisfy the amicable orthogonal designs

• Then we can show that

,j j

V W

• which means that Ut is a unitary matrix



T T

H

t t N N×=U U I

+ G. Ganesan and P. Stoica, “Differential Modulation Using Space

IEEE Signal Processing Letters, vol. 9, no. 2, Feb. 2002, pp. 57

where satisfy the amicable orthogonal designs+

is a unitary matrix



Differential Modulation Using Space–Time Block Codes,”

, vol. 9, no. 2, Feb. 2002, pp. 57-60.


• Some examples are

• (a) For two antennas and 2 symbols

• Let s1 and s2 be two symbols from a unitary symbol constellation



−=

*

12

*

212

2

1

ss

sstU

(a) For two antennas and 2 symbols

be two symbols from a unitary symbol constellation




• (b) For three antennas and 3 symbols

• (c)For 4 antennas and 3 symbols

−−

−

=

0

0

0

3

1

*

13

*

2

*

2

*

31

321

3

sss

sss

sss

tU

• (c)For 4 antennas and 3 symbols



−

−−

−

=

*

12

*

3

*

13

*

2

*

2

*

31

21

4

0

0

0

0

3

1

sss

sss

sss

sss

tU

(b) For three antennas and 3 symbols

3



*

1

*

2

3

0

s


• Let {Uk} be a set of unitary matrices to be transmitted

• Instead of transmitting {Uk} we encode the information differentially for forming a new set of matrices {

=X X U

• Since {Uk} are unitary and X0 is an identity matrix, {



t t t=X X U

0=X I

} be a set of unitary matrices to be transmitted

} we encode the information differentially for forming a new set of matrices {Xk}

X X U

is an identity matrix, {Xk} are also unitary



1t t t−X X U

T TN N×

=X I


• Differential modulation and detection:

• Assume we initially transmit

• At time t, we may transmit

0=X I

t t t=X X U

• The received signal matrix at time t can be expressed as



t t t

t t t t t t−= + = +Y HX N HX U N

Differential modulation and detection:

T TN N×

=X I

1t t t−=X X U

The received signal matrix at time t can be expressed as



1t t t−

1t t t t t t−= + = +Y HX N HX U N


• the received signal at the previous time

• Since is a Gaussian white noise

1 1 1t t t− − −= +Y HX N

N• Since is a Gaussian white noise

• can be taken as the ML estimate of

• It is assumed that channel H remains constant over the transmission period of Xt-1 and Xt



1t−N

1t−Y

the received signal at the previous time t-1 was

Since is a Gaussian white noise

1 1 1t t t− − −= +Y HX N

Since is a Gaussian white noise

can be taken as the ML estimate of

remains constant over the transmission



1t−HX


• Substituting this, we have,

• Hence the ML detection could be carried out without CSIR

1 1t t t t t t t− −= + ≅ +Y HX U N Y U N

• Hence the ML detection could be carried out without CSIR

• We can estimate Ut and correspondingly block of p symbols



{ } {{ } Sss

traces

jj

p

jj

∈

==

,

maxargˆ1

Hence the ML detection could be carried out without CSIR

1 1t t t t t t t− −= + ≅ +Y HX U N Y U N

Hence the ML detection could be carried out without CSIR

and correspondingly block of p symbols



( ) ( ){ }tttH

ttt −− −− 11 UYYUYY

Algebraic Space Time Codes

• Alamouti Space Time Codes is the only full rate (rate r=1) and full diversity

• OSTBC for complex signal constellations

• There are no OSTBC with full rate and full diversity for N• There are no OSTBC with full rate and full diversity for N

• for complex signal constellations

• Can we have a full rate and full diversity STBC for Nsystem?

• This is possible with some class of space time codes

• popularly known as Algebraic space time codes




Space Time Codes is the only full rate (rate r=1) and full

OSTBC for complex signal constellations

There are no OSTBC with full rate and full diversity for NT>2 There are no OSTBC with full rate and full diversity for NT>2

for complex signal constellations

Can we have a full rate and full diversity STBC for N×N MIMO

This is possible with some class of space time codes

Algebraic space time codes




• Diagonal Algebraic Space Time Codes

• full diversity and rate one code

• Threaded Algebraic Space Time Codes

• full diversity and rate r code where r• full diversity and rate r code where r

• Perfect space time codes

• full diversity and rate r code where r




Diagonal Algebraic Space Time Codes

full diversity and rate one code

Threaded Algebraic Space Time Codes

full diversity and rate r code where r≥1full diversity and rate r code where r≥1

full diversity and rate r code where r≥1




• Diagonal algebraic space time codes

• can be constructed for

• NT=N equal to 2n where n is a positive integer

• The code construction is a two step process:• The code construction is a two step process:

• find an optimal unitary matrix of dimension diversity

• use Hadamard matrix to multipleand time




Diagonal algebraic space time codes

where n is a positive integer

The code construction is a two step process:The code construction is a two step process:

of dimension N having the maximal

ltiplex information symbols over space




• Code construction:

• Unitary transformation of symbol vector

• Use Hadamard matrix for multiplexing symbol over space and time



= Hadm

N

N

DAST

N

x

x

x

L

MOMM

L

L

2

1

00

00

00

HG


Unitary transformation of symbol vector

matrix for multiplexing symbol over space and time



=

N

N

N

Hadm

s

s

s

x

x

x

MM

2

1

2

1

; U


• Unitary matrices:

NN

H

N IUU =

2012.0



−

−=

−=

7859.0

4857.0

3255.0

2012.0

;5257.08507.0

8507.05257.042 UU


−− 7859.04857.03255.02012



−

−

−−

2012.03255.04857.07859

3255.02012.07859.04857

4857.07859.02012.03255

7859.04857.03255.02012


• Unitary matrices:

• For N=2n dimension constructed on number fields

−× 114



[

−××

−×

−××= 12

14

124

4cos

2

N

NNN

M

πU

+ M. O. Damen, K. Abed-Meraim and J. C. Belfiore

Codes,” IEEE Trans. Inform. Theory, vol. 48, no. 3, March 2002, pp. 628


dimension constructed on number fields

NQ

8

2cos

π



]

−×−×− 121221 NL

Belfiore, “Diagonal Algebraic Space Time

, vol. 48, no. 3, March 2002, pp. 628-636.


• An N×N Hadamard matrix is a binary matrix -1} such that

• It can be constructed for N=2n where n is a positive integer

( ) ( Hadm

N

THadm

N

Hadm

N HHH =

• It can be constructed for N=2n where n is a positive integer

• That’s why we can construct DiagonN=2n only

• For example, a 2×2 Hadamard matrix is



−=

11

112

HadmH


is a binary matrix with elements {1,


) N

Hadm

N

THadmNIH =

Hadm

NH


gonal Algebraic Space Time Codes for

matrix is




• is the Hadamard matrix of N dimension

• which could be obtained from

Hadm

NH

2 2

N N

2 2/ /

/ /

Hadm Hadm

N N

−

H H

• For example



2 2

/ /

/ /

Hadm HadmHadmN NN

− =

H HH


matrix of N dimension

which could be obtained from Hadamard matrix as follows2 2

N N×



−−

−−

−−=

1111

1111

1111

1111

4

HadmH


• Diagonal Algebraic Space-time Code for N

11

11

0

0

0

0

2

1

2

2

1

2

−

=

=

x

x

x

xDAST HG

• The symbols after unitary transformation is



−=

=

2

1

2

1

2

2

1

5257.08507.0

8507.05257.0

s

s

s

s

x

xU


time Code for NT=2

;1

1

22

11

−=

xx

xx

The symbols after unitary transformation is



+−

+=

21

21

5257.08507.0

8507.05257.0

ss

ss


• Diagonal Algebraic Space-time Code for N

0

0

0

000

000

000

000 1

4

3

2

1

4

=

=

x

x

x

x

x

DAST HG



;

0000

4444

3333

2222

1111

4

−−

−−

−−=

xxxx

xxxx

xxxx

xxxx

x


time Code for NT=4

1111

1111

1111

1111

00

00

00

000

3

2

−−

−−

−−

x

x

x



111100 4

−−x


• The symbols after unitary transformation is

−=

=

3

2

1

4

3

2

1

7859.04857.0

2012.03255.0

3255.02012.0

s

s

s

x

x

x

U



−+−

++

++−

−+

=

⇒

−

21

21

21

21

4

3

2

1

4

3

4

3

3255.04857.07859.0

2012.07859.04857.0

7859.02012.03255.0

4857.03255.02012.0

4857.07859.0

ss

ss

ss

ss

x

x

x

x

sx


The symbols after unitary transformation is

−

−−

3

2

1

3255.02012.0

4857.07859.0

7859.04857.0

s

s

s



−

+

−

−

−

43

43

43

43

4

3

2012.03255

3255.02012

4857.07859

7859.04857

2012.03255.0

ss

ss

ss

ss

s


• Groups:

• A Group (G) is a set of elements with a binary operation that obeys

• the following four properties (or axioms) and

• an optional fifth property +• an optional fifth property +

• Closure:

• Associativity:



, ,a b G a b G∀ ∈ • ∈

( ), , ,a b c G a b c a b c∀ ∈ • • = • •+ B. A. Forouzan, Cryptography and Network Security


A Group (G) is a set of elements with a binary operation that obeys

the following four properties (or axioms) and



( )a b c G a b c a b c∀ ∈ • • = • •

Cryptography and Network Security, Tata Mc-Graw Hill, 2007.


• Existence of identity:

• Existence of an inverse: a G a a a a a∀ ∈ ∃ − − = − + =

a G e a e e a a∀ ∈ ∃ • = • =

• Commutativity (optional):

true only for Abelian (commutative) group



a G a a a a a∀ ∈ ∃ × = × =

, ,a b G a b b a∀ ∈ • = •


0, ,a G a a a a a∀ ∈ ∃ − − = − + =

, ,a G e a e e a a∀ ∈ ∃ • = • =

true only for Abelian (commutative) group



1 1 1 1, ,a G a a a a a− − −∀ ∈ ∃ × = × =

, ,a b G a b b a∀ ∈ • = •


• Ring:

• A ring is a set R along with

• two binary operations “+” and “

• obeying the following axioms+:• obeying the following axioms+:

• R is an Abelian group for the operation “+”

• satisfies all five axioms for “+” operation



+ R. B. J. T. Allenby, Rings, Fields and Groups, Edward Arnold, 1991.


two binary operations “+” and “×”

is an Abelian group for the operation “+”

satisfies all five axioms for “+” operation



, Edward Arnold, 1991.


• The operation “×” is associative and

• certainly closed also

• i.e, satisfies first two axioms closure and associativity only for the “×” operation“×” operation

• The operations satisfy the Distributive Laws

• the second operation is distributed over the first operation



( ), , , ,a b c R a b c a b a c a b c a c b c∀ ∈ × + = × + × + × = × + ×


” is associative and

, satisfies first two axioms closure and associativity only for the

Distributive Laws

the second operation is distributed over the first operation



( ), , , ,a b c R a b c a b a c a b c a c b c∀ ∈ × + = × + × + × = × + ×


• If the second operation “×” is commutative,

• we call R a commutative ring

• Sometimes the ring has a multiplicative identity and

• we say it a Ring with identity and • we say it a Ring with identity and

• the multiplicative identity is denoted by 1

• Example

• Assume R (set of real numbers) be a commutative ring with an identity




” is commutative,

the ring has a multiplicative identity and

and and

the multiplicative identity is denoted by 1

(set of real numbers) be a commutative ring with an




• Consider a polynomial of degree n

• whose coefficients in R with an indeterminate

• may be represented as

( )p α

• Addition and multiplication of polynomials are carried out as usual

• The ring of such polynomials is denoted by



( )0

0, ,n

i

i n ii

p a a a Rα α=

= ≠ ∈∑


polynomial of degree n

with an indeterminate α

Addition and multiplication of polynomials are carried out as usual

The ring of such polynomials is denoted by



, ,i n i

p a a a R= ≠ ∈

R α


• Fields:

• A field is a commutative ring with identity

• in which each non-zero element has a multiplicative inverse

• Let Z be the set of integers • Let Z be the set of integers



, , , , , , ,Z

= − −

L L

, ,a G a a a a a− − −∀ ∈ ∃ × = × =


is a commutative ring with identity

zero element has a multiplicative inverse

( )1 0≠



2 1 0 1 2 3, , , , , , ,

= − −

L L

1 1 1 1, ,a G a a a a a− − −∀ ∈ ∃ × = × =


• Q is the set of rational numbers

• R is the set of real numbers and

| , ,a

Q a b Z bb

= ∈ ≠

• R is the set of real numbers and

• C is the set of complex numbers

• The rings Q, R, C are fields



| , ,


0| , ,Q a b Z b

= ∈ ≠



| , ,


• Field extension of Q: Example 1:

• Consider an irreducible polynomial in α (it product of two non-constant polynomials over Q, but it can be factored over irrational number)

2 2 0 2α α− = ⇒ =

• α is not an element of Q (set of rational numbers)

• One may extend Q by adding to Q,

• denoted by which contains both Q and



2

( )2Q

+E. Viterbo and F. Oggier, Algebraic Number Theory and Code Design for Rayleigh Fading Channels

publishers, 2004


Consider an irreducible polynomial in α (it cann’t be factored into constant polynomials over Q, but it can be

α is not an element of Q (set of rational numbers)

One may extend Q by adding to Q,

denoted by which contains both Q and



2

Algebraic Number Theory and Code Design for Rayleigh Fading Channels, Now


• Any element in extended field can be expressed in polynomial form

(Q

2 , ,a b a b Q+ ∈

• The basis vector for

• and the dimension of is 2 over Q



( )2Q

1 2,

( )2Q


Any element in extended field can be expressed in )2

and the dimension of is 2 over Q



1 2


• Field extension of q-QAM: Example 2

• Consider a number field Q(i) (q-QAM constellations)

• Consider the minimum polynomial of

,K Q i eθ

= =

• Its root is called an algebraic number and

• we will consider algebra of



( ) 2p x x i

θ= −

4

i

i e

π

θ = =

θ


QAM: Example 2:

of degree two over base field

Consider the minimum polynomial of

4,i

K Q i e

π = =

θ

Its root is called an algebraic number and




• Its conjugate

• the integral basis of

4

i

e

π

θ = −

[ ]θ1=b

• and each element of canas



( ),K Q i θ=

, ,x a b a b Q iθ= + ∈


can be expressed in polynomial form



( )x a b a b Q i


• Consider a generator matrix as

• You may also find the value of

=

θ

θ

1

1G

θ =• By finding the value of ω which will minimize the PEP (trace

criterion)

• Or by minimizing the trace of codeword distance matrix

• The unitary matrix of dimension two can be constructed as



2

2=

GU

θ =


ωje=which will minimize the PEP (trace

Or by minimizing the trace of codeword distance matrix

The unitary matrix of dimension two can be constructed as



e=


• For two QAM information symbols

• Unitary transformation: we can rotate this input symbol vector by

=

2

1

s

ss

• Unitary transformation: we can rotate this input symbol vector by multiplying with the unitary matrix



−== 2

1

1

2

1

θ

θsUx


For two QAM information symbols

Unitary transformation: we can rotate this input symbol vector by Unitary transformation: we can rotate this input symbol vector by multiplying with the unitary matrix U as



−

+=

21

21

2

1

2

1

ss

ss

s

s

θ

θ

θ

θ


• This operation increase the algebraic dimension of constellation as

• is a vector space of dimension two over Q(

• An Space Time Block Code will not incur any information loss

• if the maximum instantaneous mutual information of the

( ),K Q iθ=

• if the maximum instantaneous mutual information of the equivalent MIMO channel that includes the Space Time Block Code codeword is equal

• to the maximum instantaneous mutual information of the MIMO channel without the Space Time Block Code codeword




This operation increase the algebraic dimension of constellation as

is a vector space of dimension two over Q(i)

An Space Time Block Code will not incur any information loss

if the maximum instantaneous mutual information of the if the maximum instantaneous mutual information of the equivalent MIMO channel that includes the Space Time Block

to the maximum instantaneous mutual information of the MIMO channel without the Space Time Block Code codeword




• Any unitary transform will preserve the mutual information

• while changing the diversity and coding gain

• The above codeword could be rewritten in a diagonal matrix form as

+ θ



+=

21

02

1

s

ss θX

+ B. S. Rajan, “Space-time block codes,” in Channel coding: Theory, Algorithms and Applications

Declerq, M. Fossorier and E. Biglieri, Eds., Oxford, UK: Elsevier Academic Press, 2014, pp. 451


Any unitary transform will preserve the mutual information

while changing the diversity and coding gain+

The above codeword could be rewritten in a diagonal matrix form as



− 21

0

ss θ

Channel coding: Theory, Algorithms and Applications, D.

, Eds., Oxford, UK: Elsevier Academic Press, 2014, pp. 451-495.


• Now applying the Hadamard transformation,

• we have the codeword matrix of Diafor N=2

+ −==

21111 ssHadmDAST θXHC



+

−==

2122

02

1

11

11 ssHadmDAST θXHC


transformation,

f Diagonal Algebraic Space Time code

( ) −−+=

212110 ssss θθ



( )

−+

−−+=

− 2121

2121

21 2

10

ssss

ssss

ss θθ

θθ

θ


• For dimension N, we can construct codes similarly

• Note that we need to construct an optimal unitary matrix in the

DAST Hadm Hadm

N N= • = • •C H X H U s

• Note that we need to construct an optimal unitary matrix in the above equation

• symbol vector



1

2

N

s

s

s

=

s M


, we can construct codes similarly

Note that we need to construct an optimal unitary matrix in the

( )DAST Hadm Hadm

N Ndiag= • = • •C H X H U s

Note that we need to construct an optimal unitary matrix in the




• The coding gain of N-dimension Diaare given by

2

22 4

5min

, ,

,

N

N

δ

=

=

• Minimum coding gain for N=2,4,8 are 0.8944, 0.6324 and 0.545 respectively



1

18

2

min

, ,

,N

N

N

δ

−

= ≥


Diagonal Algebraic Space Time codes

Minimum coding gain for N=2,4,8 are 0.8944, 0.6324 and 0.545




• Threaded Algebraic Space Time Codes

• are fully diverse and full rate (can achieve any rate rnumber of transmit antennas

• Note that the maximum value of r is min{N• Note that the maximum value of r is min{N

• The rate r TAST code for NR×NT MIMO system can be constructed as



(1

T T

rTAST

N k N kk=

=∑C D U x


Threaded Algebraic Space Time Codes

are fully diverse and full rate (can achieve any rate r≥1) for any

Note that the maximum value of r is min{NR,NT}Note that the maximum value of r is min{NR,NT}

MIMO system can be constructed as



)( )1

T T

k

N k N kγ

−

C D U x Π


• and matrix

• γ is a unit-magnitude complex number which makes the TAST full

1 2 1, , , ,

T T TN N N− −

=

Π e e e eL

• γ is a unit-magnitude complex number which makes the TAST full diverse and it is dependent on the QAM alphabet size and N

• is the diagonal matrix with diagonal elements consisting of a rotated version of the kth symbol



( ) ( )T T

k N k N kdiag=D U x U x


and ei is the ith column of a identity

magnitude complex number which makes the TAST full magnitude complex number which makes the TAST full diverse and it is dependent on the QAM alphabet size and NT

is the diagonal matrix with diagonal elements consisting of a rotated




• xk is the vector consisting of information symbols of the

0

1

,

,

k

k

x

x

• is the unitary rotation matrix



2

1

,

,

,

T

T

kk N

k N

x

x

−

−

=

xM

TN

UT T

N N×


is the vector consisting of information symbols of the kth thread

is the unitary rotation matrix




• For example for NT=2 and r=2, TAST code can be generated as follows:

( )( )1

2

2 1 2 1 2 2 2T

k

TAST

k N kγ γ

−

= = +∑C D U x Π D U x D U x

• where x1 and x2 are symbols sent for layer 1 and 2

• For this case,



( )( )2 1 2 1 2 2 21

Tk N k

k=

∑

4

42

1 0 1

1 1 01

2;

i

i

e

e

π

πγ γ

= = −

U Π


=2 and r=2, TAST code can be generated as

( ) ( )2 1 2 1 2 2 2γ γ= = +D U x D U x Π

are symbols sent for layer 1 and 2



( ) ( )2 1 2 1 2 2 2

1 0 1

1 0


• Hence

4 41 0 2 0

11 2 14 42

1 1 0

1 11 1

2 2

, ,

, ,

i i

i iTAST

e x e x

x xDiag Diage e

π π

π π

∴ = + − −

C



2 2, ,


4 41 0 2 0

11 2 14 4

1 1 0

1 1 01 1

2 2

, ,

, ,

i i

i i

e x e x

x xDiag Diage e

π π

π π

γ

γ

∴ = + − −



2 2, ,


• Diagonalize

4 41 0 11 2 0 2 1

0 0

1 1, , , ,

, ,

i i

iTAST

x e x x e x

π π

π

+ +

⇒ = +C



421 0 11

1 10

2 2

, , , ,

, ,

iTAST

x e x

π ⇒ = + −

C


4 41 0 11 2 0 2 1

0 0

1 1, , , ,

, , , ,

i i

i

x e x x e x

π π

π

γ

+ + = +



41 0 11 2 0 2 1

1 1

02 2

, , , ,

, , , ,

i

x e x x e x

π

γ

= + −


4 41 0 11 2 0 2 1

4 422 0 2 1 1 0 11

1

2

, , , ,

, , , ,

i i

i iTAST

x e x x e x

x e x x e x

π π

π π

γ

+ +

⇒ = − −

C

• where for for QPSK and for 16

• Note that ϒ is a unit-magnitude complex number that ensures a fulldiversity TAST code (rank criterion)



2 0 2 1 1 0 112 , , , ,x e x x e xγ − −

6

i

e

π

γ =


4 41 0 11 2 0 2 1

4 42 0 2 1 1 0 11

, , , ,

, , , ,

i i

i i

x e x x e x

x e x x e x

π π

π π

γ

+ + − −

QPSK and for 16-QAM constellations

magnitude complex number that ensures a full-(rank criterion)



2 0 2 1 1 0 11, , , ,x e x x e x− −

4

i

e

π

γ =


• Z ring of integers

• Q field of rational numbers

• R field of real numbers

• C field of complex numbers• C field of complex numbers

• Z(i) set of numbers a+ib, where a,b

• Q(i) set of numbers a+ib, where a,b

• R(i) is field of complex numbers C




a,b belongs to Z

a,b belongs to Q




• An algebraic number field K=Q(θ) is the

• set of all possible algebraic

• combinations of an algebraic number

• with the rational numbers of Q• with the rational numbers of Q

• θ is a root of an irreducible polynomial minimal polynomial

• Let n be the degree of this polynomial,

• it is also the degree of the algebraic number field




) is the

combinations of an algebraic number θ

irreducible polynomial mθ over Q which is called

Let n be the degree of this polynomial,

it is also the degree of the algebraic number field.




• mθ has n distinct roots in C called the

• (1, θ, θ2, …, θn-1) forms a basis of for K.

• K is called an extension of degree n over Q.

Example

• Let { }QbabaK ∈+== ,,,2 θθ• Let

• The minimal polynomial of θ is

• The conjugates of θ are

• The degree of the algebraic number field is 2

• (1, θ) forms a basis of for K=Q(θ).



{ }QbabaK ∈+== ,,,2 θθ

mθ

,2 21 == θθ


has n distinct roots in C called the conjugates of θ: θ1, θ2, …, θn.

) forms a basis of for K.

K is called an extension of degree n over Q.

The degree of the algebraic number field is 2



22 −= x

2−=


• Let F denote either the real field R or the complex field C.

• It is possible to uniquely represent each element of K by

• means of canonical embeddings (ring homomorphism in field theory) theory)

• with a point in Fn.

• The above mapping is additive homomorphism

• so it preserves the additive structure



( ) ( )(K∈ φσφσφ ,,, 21 La


Let F denote either the real field R or the complex field C.

It is possible to uniquely represent each element of K by

canonical embeddings (ring homomorphism in field

The above mapping is additive homomorphism

so it preserves the additive structure



( )) n

n F∈φσ,


Ring homomorphism:

• a transformation of one set into another that preserves in the second set the relations such between elements of the first

• i.e., it is a function f f:K �Fn such that

• f(a+b)=f(a)+f(b) for all a and b in K

• f(a×b)=f(a) × f(b) for all a and b in K• f(a×b)=f(a) × f(b) for all a and b in K

• f(1 in K)=1 in Fn

• We call the algebraic number field totally real if all the roots of the minimal polynomial are real.




a transformation of one set into another that preserves in the second set the relations such between elements of the first

such that

)=f(a)+f(b) for all a and b in K

f(b) for all a and b in Kf(b) for all a and b in K

We call the algebraic number field totally real if all the roots of the




• A set of algebraic integers w1,w2,…,K if every algebraic integers in K can be expressed in linear combinations of w1,w2,…,wn

• A set Let Fn=Rn and consider an integral basis (w

• The n-vectors vj=(σ1(wj), σ2(wj),…, σare linearly independentare linearly independent

• So they define a full rank lattice of F



( ) ( )( ) ( )

( ) ( )

=

nn ωσωσ

ωσωσ

ωσωσ

L

OMM

L

L

21

2221

1211

G


,…,wn in K is called an integral basis of K if every algebraic integers in K can be expressed in linear

and consider an integral basis (w1,w2,…,wn)

σn(wj)) belongs to Fn for j=1,2,…,n

So they define a full rank lattice of Fn with generator matrix.



( )( )

( )

nn

n

n

ωσ

ωσ

ωσ

L

MO

L

L

2

1


• Thus studying the properties of a lattice in Fthe algebraic number field K

• Example

• is a degree 4 over Q, the minimal polynomial is + 22Q• is a degree 4 over Q, the minimal polynomial is

• which has four distinct roots over R also called as conjugates of



+ 22Q

24 24 +−= xxmθ

{ ,22,22 21 +−=+= θθθ


Thus studying the properties of a lattice in Fn is equivalent to studying

is a degree 4 over Q, the minimal polynomial isis a degree 4 over Q, the minimal polynomial is

which has four distinct roots over R also called as conjugates of θ



}22,22 43 −−=−= θθ


• The four canonical embeddings are

θθσ

θθσ

θθσ

−

a

a

a

:

:

:

2

1

• One notices that {σ1, σ2, σ3, σ4} forms a Galois group function composition law



θθσ

θθσ

−a

a

:

:

4

3


The four canonical embeddings are

θ

} forms a Galois group w.r.t. the




• The basis {w1=1, w2=θ, w3=θ2, w4=θ

• The lattice of R4 defined by the following generator matrix has a full rank

11



−

−=

33

224

θθ

θθ

θθG


θ3} is integral.

defined by the following generator matrix has a full

11



−

−

33

22

θθ

θθ

θθ

Algebraic Space Time Codes• Q is the set of rational numbers,

• L = Q(i ) is the extended field of Q

• Choice of the extension

• In case of PSTBC, we shall then choose an extension cyclic base of degree nof degree n

• Golden codes

• [K=Q(i, θ) : L=Q(i)] = n (for Golden codes n=2)

• θ is root of minimal polynomial

• and θ is also called as Golden number

• The conjugates of θ are



11

+=θ


) is the extended field of Q

In case of PSTBC, we shall then choose an extension cyclic base K of L

n (for Golden codes n=2)

is root of minimal polynomial

also called as Golden number



2 1 0x x− − =

2

51,

2

52

−=

+θ


• Definition of the cyclic algebra

• Let K=Q(i,θ) be a cyclic extension of the base field Ldegree n with Galois group GK/L=<σ

• where the canonical embeddings are usually denoted as {σ, σ3= σ2, …, σn= σn-1}

• We can build a non-commutative algebra A based on K usually • We can build a non-commutative algebra A based on K usually denoted as A(K/L,σ,ϒ)

• A(K/L,σ,ϒ) is a cyclic algebra of degree n

• As a vector space A can be seen as a sum of n copies of the chosen number field K of degree n



1A K e K e K= • ⊕ • ⊕ ⊕L


) be a cyclic extension of the base field L (such as Q(i)) of σ>

where the canonical embeddings are usually denoted as {σ1=1, σ2=

commutative algebra A based on K usually commutative algebra A based on K usually

) is a cyclic algebra of degree n iff

As a vector space A can be seen as a sum of n copies of the chosen



1nA K e K e K

−= • ⊕ • ⊕ ⊕


• For example, A(K/L,σ,ϒ) is a cyclic algebra of degree 2

• where

• The PSTBC also known as Golden code is a finite subset of the Cyclic Algebra of degree n=2

2e γ=

• Galois group GK/L=<σ> with canonical embeddings {

• We can construct the cyclic algebra for any z

• where a1 = z1 + e z2 and a2 = z3 + e z



θθσ −a:


cyclic algebra of degree 2 iff

The PSTBC also known as Golden code is a finite subset of the Cyclic

KeKA •⊕•= 1

> with canonical embeddings {σ1=1, σ2= σ}

We can construct the cyclic algebra for any z1,z2 belongs to K

+ e z4



θ


• When we suffix e (non-commutative algebrabelongs to an extended field K, σ will operate on z, the outcome is

( ) ( )ze e z e a b e a bσ σ θ θ= = + = +

• There is a corresponding matrix representation for



( ) ( )ze e z e a b e a bσ σ θ θ= = + = +

( ) ( )1 2

2 1a

z z

z zA z z Kγσ σ

= = ∈

X


commutative algebra) to an element z which will operate on z, the outcome is

) ( )ze e z e a b e a bσ σ θ θ= = + = +

corresponding matrix representation for



) ( )ze e z e a b e a bσ σ θ θ= = + = +

1 2; ,A z z K

= = ∈

1 1 2a z ez A= + ∈


• How do we come up with this matrix representation?

• Let us find the multiplication of two elements

( ) ,ze e z z Kσ= ∀ ∈



( )( )( )( ) ( )(

1 2 1 2 3 4 1 3 1 4 2 3 2 4

1 3 1 4 2 3 2 4

1 3 2 4 1 4 2 3

a a z ez z ez z z z e z ez z e z e z

z z e z z ez z z z

z z z z e z z z z

σ γσ

γσ σ

= + + = + + +

= + + +

= + + +

( ) ,


How do we come up with this matrix representation?

Let us find the multiplication of two elements 1 2,a a A∈



( ) ( )( )

)

1 2 1 2 3 4 1 3 1 4 2 3 2 4

1 3 1 4 2 3 2 4

1 3 2 4 1 4 2 3

a a z ez z ez z z z e z ez z e z e z

z z e z z ez z z z

z z z z e z z z z

σ γσ

= + + = + + +

= + + +2

e γ=


• For the basis 1 e

( ) ( )(1 2 1 3 2 4 1 4 2 3a a z z z z e z z z zγσ σ= + + +

• we may express the above equation in matrix form as



( ) ( )(1 2 1 3 2 4 1 4 2 3a a z z z z e z z z zγσ σ= + + +

( )( )( )

=

4

3

12

21

21 1z

z

zz

zzeaa

σ

γσ


)1 2 1 3 2 4 1 4 2 3a a z z z z e z z z z= + + +

we may express the above equation in matrix form as



)1 2 1 3 2 4 1 4 2 3a a z z z z e z z z z= + + +


• Thus there is a corresponding matrix for every element

( )( )( )

1 2

2 11 1 2

T

z z

z za z ez A

γσ

σ

= + ∈ ↔ =

• A Space-time Block Code can be obtained by considering the matrix of left multiplication in the given basis (1, e)

• If for a1 = z1 + e z2 belongs to A and

• Then the corresponding multiplication matrix as given above




Thus there is a corresponding matrix for every element

( ) ( )1 2

2 1

z z

z zγσ σ

= + ∈ ↔ =

time Block Code can be obtained by considering the matrix of left multiplication in the given basis (1, e)

belongs to A and z1, z2 belongs to K=Q(i,θ)

Then the corresponding multiplication matrix as given above




• A(K/L,σ,ϒ) is a cyclic algebra of degree 4

• where

• Let K=Q(θ) be a cyclic extension of the base field L=Q of degree 4

KeKeKeKA •⊕•⊕•⊕•= 321

γ=4e

• Let K=Q(θ) be a cyclic extension of the base field L=Q of degree 4 with Galois group GK/L=<σ>

• where the canonical embeddings are usually denoted as {σ, σ3= σ2, σ4= σ3}




cyclic algebra of degree 4 iff

) be a cyclic extension of the base field L=Q of degree 4 ) be a cyclic extension of the base field L=Q of degree 4

where the canonical embeddings are usually denoted as {σ1=1, σ2=




• Q(θ)={x=a+b θ+c θ2+d θ3,a,b,c,d ϵ Q}

• It is of degree 4, that is of dimension 4 as a vector space over Q

• PAM symbols can be seen as elements of Q

• Q(θ) actually encodes 4 PAM symbols namely • Q(θ) actually encodes 4 PAM symbols namely

• The four canonical embeddings are



θθσσ

θθσσ

θθσσ

θθσ

−=

=

−=

=

a

a

a

a

:

:

:

:1

3

4

2

3

2

1


Q}

It is of degree 4, that is of dimension 4 as a vector space over Q

PAM symbols can be seen as elements of Q

) actually encodes 4 PAM symbols namely a,b,c,d) actually encodes 4 PAM symbols namely a,b,c,d

The four canonical embeddings are



θ

θ

θ

θ


• It yields a cyclic Galois group GK/L=<

• We can build a non-commutative algebra A on denote by K

• In our case n=4, KeKA ⊕•⊕•= 1• In our case n=4,

• where {1, e, e2, e3} forms a basis and

• ϒ:=e4 must be an element of the base field Q.

• Such a cyclic algebra is shortly denoted by



KeKA ⊕•⊕•= 1


=<σ>

commutative algebra A on GK/L=<σ> which we

KeKe •⊕•⊕ 32

} forms a basis and

must be an element of the base field Q.

Such a cyclic algebra is shortly denoted by A(K/L,σ,ϒ)



KeKe •⊕•⊕ 32


• A space-time block code can be obtained by

• considering the matrix of left multiplication

• in the given basis {1, e, e2, e3}

• If a1=z1+ez2+e2z3+e3z4 belongs to A, z• If a1=z1+ez2+e2z3+e3z4 belongs to A, z

• and a2=z5+ez6+e2z7+e3z8 belongs to A, z

• then the corresponding multiplication table can be obtained by

• considering multiplication of two elements a




time block code can be obtained by

considering the matrix of left multiplication

belongs to A, z1,z2,z3,z4 belongs to K=Q(θ), belongs to A, z1,z2,z3,z4 belongs to K=Q(θ),

belongs to A, z1,z2,z3,z4 belongs to K=Q(θ),

then the corresponding multiplication table can be obtained by

considering multiplication of two elements a1, a2 belongs to A




• Let us find the multiplication of two elements a

( )(( ) ( ) 3

71

22

6151

54

3

3

2

2121

ezzezzezz

ezzzezeezzaa

σσσ +++=

++++=



( ) ( )

( ) ( )

( ) ( )

( ) ( ) 74

2

6454

3

73

2

63

3

53

2

72

23

62

2

52

716151

zzezzzze

zzzzezze

zzezzezez

ezzezzezz

γσγσ

γσσ

σσ

σσσ

++++

++++

++++

+++=


Let us find the multiplication of two elements a1, a2 belongs to A

)( ) 81

3

8

3

7

2

6

zz

zezeez

σ

++



( )

( )

( )

( ) 84

32

83

3

82

3

81

zze

zze

zz

zz

γσ

σγ

γσ

σ

+


• which can be expressed as

( )

( ) ( )( ) ( )( ) ( )( ) ( )

2

1

2

23

4

2

12

3

2

41

321

zzz

zzz

zzz

zzz

eee

σσ

σσ

γσσ

γσγσ

• Transpose of the above matrix gives the space

• where the factor ϒ comes from e4=ϒ

• {1, σ ,σ2, σ3} are the elements of the Galois group,

• appearing due to the non-commutative multiplication defined in A

• by ze=e σ(z) for z belongs to K



( ) ( ) 2

2

34 zzz σσ


( )( )( )( )

7

6

5

3

4

3

3

3

2

3

z

z

z

z

z

z

z

z

σ

γσ

γσ

γσ

Transpose of the above matrix gives the space-time block code

ϒ and

} are the elements of the Galois group,

commutative multiplication defined in A



( )

81

3 zzσ


• Elements in a cyclic algebra A of degree n can be described by matrices of the form

( ) ( )

( ) ( )

0 1 1

1 0 2

1 1 1

n n

n n n

z z z

z z z

z z z

γσ σ σ

γσ γσ σ

− −

− − −

L

L

M O O M

L

• where

• zl belongs to K, l=0,1,2,…,n-1.

• the factor ϒ comes from and L* is the set of nonzero elements of L



( ) ( )1 1 1

1 2 0

n n nz z zγσ γσ σ− − −

L

, ne A e L∈ = ∈


Elements in a cyclic algebra A of degree n can be described by

( )

( )

0 1 1

1 0 2

1 1 1

n

n n

n n n

z z z

z z z

z z z

γσ σ σ

γσ γσ σ

−

− −

− − −

M O O M

comes from and L* is the set of non-



( )1 1 1

1 2 0

n n nz z zγσ γσ σ− − −

*,e A e Lγ∈ = ∈


• How to find ϒ?

• Example

• Let K=Q(i), L=Q and p be any rational prime integer of the form 4k+1

• Then any matrix of the form• Then any matrix of the form



=

2

1

xx

pxxX


), L=Q and p be any rational prime integer of the form 4k+1



*

1

*

2

x

px


• is an invertible matrix

• Because( ) ( ) (

*

2

*

11det

===⇒

−=

xx

xxpxxX

• From number theory, any prime of the form 4k+1 this way



*

*

22

*

11 , ===⇒ yyyxx

xxp


)*

2 0

−=

=x

From number theory, any prime of the form 4k+1 cann’t be factorized



1

21

−= xx


• If we restrict to take values of x1,x2 from Z(

• Then the design X will have a non-vanishing determinant

• Hence for

• p=5, ϒ=2+i

• p=13, ϒ=3+2i

• p=29, ϒ=5+2i

• p=37, ϒ=6+i

• p=53, ϒ=7+2i



+ P. Elias, K. R. Kumar, S. A. Pawar, P. V. Kumar and H.-F. Lu, “Explicit space

diversity-multiplexing gain tradeoff,” IEEE Trans. on Inform. Theory

3884.


If we restrict to take values of x1,x2 from Z(i) which is a subset of Q(i)

vanishing determinant



F. Lu, “Explicit space-time codes achieving the

IEEE Trans. on Inform. Theory, vol. 52, no. 9, Sept. 2006, pp. 3869-


• Another way: If ϒl for l=0,1,2,…,n-1, does not belong to norm Ni.e.,

• ( )∏−

=

≠1

0

n

i

il xσγ

• for any z belongs to K, then

• the cyclic algebra A is a division algebra

• every non-zero element in A has a multiplicative inverse

• The above condition imposed on ϒ

• A ϒ satisfying norm condition is said to be a



0i


1, does not belong to norm NK/L(K),

cyclic algebra A is a division algebra, i.e.,

zero element in A has a multiplicative inverse

ϒ is called norm condition.

satisfying norm condition is said to be a non-norm element.




• Finding Norm of field

Example:

• A basis for complex filed C=R(i) over real field R is {1,i} and with x=a+bi, we have,x=a+bi, we have,

• (a+bi)(1)=a+bi and (a+bi)(i)=-b+ai

• Therefore matrix representation of (



( )

−

b

ai1


) over real field R is {1,i} and with

Therefore matrix representation of (a+bi) is



−

a

b


• Hence the norm and trace of (a+bi

• N(a+bi)=a2+b2 and trace is 2a

Example

• Let be an element of the quadratic extensionmbax +=• Let be an element of the quadratic extension

• where m is a square-free integer

• Let us find the norm and trace of x

• For Galois group of the extension consists of the identity

• and the automorphism (one-to-one mapping)



mbax +=


a+bi) is

Let be an element of the quadratic extension ( ) QmQ /Let be an element of the quadratic extension

free integer

Let us find the norm and trace of x

For Galois group of the extension consists of the identity

one mapping)



( ) QmQ /

( ) mbamba −=+σ


• Norm is defined can be obtained as

• For our case, n=2, therefore,

• for canonical embeddings of identity ((σ =σ), we have,(σ2=σ), we have,

• and trace is T(x) =x+σ(x)=2a



( ) ( ) ( )( mbambaxxxN −+== σ


Norm is defined can be obtained as

for canonical embeddings of identity (σ1=1) and automorphism

( )∏=

n

i

i x1

σ



) 22mba −=


• For our case for code construction of algebraic space time code

• Note that norm of ϒ should be equal to 1

• in order to guarantee the same average transmitted energy

• from each antenna, • from each antenna,

• at each channel use

• This limits our choice to ϒ=+1,-1,+i,

• It means we need to show that norm for 2



+ F. Oggier, G. Rekaya, J,-C. Belfiore and E. Viterbo, “Perfect space

Theory, vol. 52, no. 9, Sept. 2006, pp. 3885-3902.

For our case for code construction of algebraic space time code

should be equal to 1

in order to guarantee the same average transmitted energy

1,+i,-i and usually ϒ=i is chosen

It means we need to show that norm for 2×2 case+



, “Perfect space-time block codes,” IEEE Trans. Inform.

( ) imbaxN ≠−= 22


• Let C be the codebook formed by codewords

• For it to be fully diverse,

• it is enough to have det{X’-X’’}≠0 for

• or equivalently det{X}≠0 for X ≠• or equivalently det{X}≠0 for X ≠

• it is true for codes designed from cyclic division algebra

• since every codeword matrix has an inverse




codewords X of the above form

’’}≠0 for X’ ≠X’’ in C

≠0 in C≠0 in C

it is true for codes designed from cyclic division algebra

since every codeword matrix has an inverse




• Perfect Space Time Codes:

• Exist for 2,3,4 and 6 transmit antennas

• Designed to meet the following four important criteria

• Full rate (minimum of N or N )• Full rate (minimum of NR or NT)

• Non-vanishing determinant (Fully diverse)

• Constellation shaping: Unitary transformation (No change in energy in transmitting symbols after the transformation)

• Uniform average transmitted energy (




Exist for 2,3,4 and 6 transmit antennas

Designed to meet the following four important criteria

))

vanishing determinant (Fully diverse)

Constellation shaping: Unitary transformation (No change in energy in transmitting symbols after the transformation)

Uniform average transmitted energy (ϒ should be unit magnitude)




• The rate r PSTBC for a MIMO system with Nbe constructed as

1 r

• where



1

1

T

rPSTBC k

N kkλ =

= ∑C D Γ

( ) ( )( 2, , , ,k k k k k

diag z z z zσ σ σ=D

+ P. Elia, B. A. Sethuraman and P. Vijay Kumar, “Perfect space

minimum delay for any number of transmit antennas,”

2007, pp. 3853-3868.


The rate r PSTBC for a MIMO system with NT transmit antennas+ can



1PSTBC k

N k

−Γ

( ))1, , , , T

N

k k k k kdiag z z z zσ σ σ −

L

and P. Vijay Kumar, “Perfect space-time codes with minimum and non-

minimum delay for any number of transmit antennas,” IEEE Trans. Inform. Theory, vol. 53, no. 11, Nov.


• λ is a suitable real-valued scalar desienergy constraint

• γ is a unit magnitude complex number makes the PSTBC fully diverse

( )1 2 2 1, , , , ,

N N Nγ

− −=Γ e e e e eL

• For rate r=2 and NT=2, λ=5, we have



( )1 2 2 1, , , , ,

T T TN N N

γ− −

=Γ e e e e eL

(1 2

1 22 1 2

0 0

1 1 0 0

5 5

PSTBC

z z

z zσ σ

= + = +

C D D Γ


designed so that the STBC meets the

is a unit magnitude complex number makes the PSTBC fully diverse

, we have



) ( )1 2

1 2

0 0 0 1

00 0

z z

z z γσ σ

= + = +


• For γ=i,

( ) (1 2 1 2

2 1 2 12

1 1

5 5

PSTBC

z z z z

z z i z zγσ σ σ σ

⇒ = =

C

• which is also called as Golden code




) ( ) ( )1 2 1 2

2 1 2 1

1 1

5 5

z z z z

z z i z zγσ σ σ σ

= =

which is also called as Golden code



Basics of convolutional codes

• In convolutional code+,

• the block of n code bits produced by the encoder in a particular moment is dependent on

• the block of k message bits in that particular moment and • the block of k message bits in that particular moment and

• the block of data bits for N-1 moments (N>1) before



+ B. P. Lathi, Modern digital and analog communication systems

E. Biglieri, Coding for wireless channels, Springer, 2005

S. G. Wilson, Digital Modulation and Coding, Pearson, 1996

T. Oberg, Modulation, Detection and Coding, John Wiley and Sons, 2001

codes

the block of n code bits produced by the encoder in a particular

the block of k message bits in that particular moment and the block of k message bits in that particular moment and

1 moments (N>1) before



Modern digital and analog communication systems, Oxford University Press, 2009 .

, Pearson, 1996

, John Wiley and Sons, 2001


• A convolutional code with constraint length N

• consists of an N-stage shift register (SR) and

• ν modulo-2 adders



codes

A convolutional code with constraint length N

stage shift register (SR) and






codes




• Fig. shows such a coder for the case N=3 and

• The output samples the ν modulo-

• once during each input-bit interval

ExampleExample

• Assume that the input digits are 0101

• Find the coded sequence output

• Initially, the SRs



0321 === sss

codes

Fig. shows such a coder for the case N=3 and ν=2

-2 adders in a sequence,

bit interval

Assume that the input digits are 0101



0


• Note that SR just shifts the input dainstant

• When the first message bit 1 enters the

,1 == sss

• Then

• The coder output is 11

• When the second message bit 0 enters the SR,



,1 321 == sss

1 21y y= =

codes

t data to the next SR in the next time

When the first message bit 1 enters the

0=

When the second message bit 0 enters the SR,



0=

0,1,0 321 === sss


• Then


• When the third message bit 1 enters the SR,

1 21 0,y y= =

• Then


• When the fourth message bit 0 enters the SR,



,1 21 = ss

1 20y y= =

,1,0 321 === sss

codes

When the third message bit 1 enters the SR,

When the fourth message bit 0 enters the SR,



1,0 3 == s

0=


• Then

• State diagram:

• When a message bit enters the SR (s

• the coder outputs are dependent on both the message bit in s

1 21 0,y y= =

• the coder outputs are dependent on both the message bit in sthe two past bits already in s3 and s

• There are four feasible combinations of the two past bits in s00, 01,10,11

• We will name these four states as a, b, c, d respectively



codes

When a message bit enters the SR (s1),

the coder outputs are dependent on both the message bit in s and the coder outputs are dependent on both the message bit in s1 and and s2

There are four feasible combinations of the two past bits in s3 and s2 :

We will name these four states as a, b, c, d respectively



Fig. 3-shift registers showing states a, b, c and d



shift registers showing states a, b, c and d , Fundamentals of MIMO Wireless



• we will adopt

• solid lines for 0 input bit, and

• dashed lines for 1 input bit

• State a goes to State a for 0 input and 00 output• State a goes to State a for 0 input and 00 output

• State a goes to State b for 1 input and 11 output

• State b goes to State c for 0 input and 00 output

• State b goes to State d for 1 input and 01 output



codes

State a goes to State a for 0 input and 00 outputState a goes to State a for 0 input and 00 output

State a goes to State b for 1 input and 11 output

State b goes to State c for 0 input and 00 output

State b goes to State d for 1 input and 01 output




• State c goes to State a for 0 input and 11 output

• State c goes to State b for 1 input and 00 output

• State d goes to State c for 0 input and 01 output

• State d goes to State d for 1 input and 01 output• State d goes to State d for 1 input and 01 output



codes

State c goes to State a for 0 input and 11 output

State c goes to State b for 1 input and 00 output

State d goes to State c for 0 input and 01 output

State d goes to State d for 1 input and 01 outputState d goes to State d for 1 input and 01 output







Fig. State diagram for the coder


• Trellis diagram:

• Trellis diagram may be easily drawn from the above state diagram

• It commences from entire 0s in the SR, i.e., state a and

• makes transitions depending on every input data digit• makes transitions depending on every input data digit

• Such changes are represented by

• a solid line (ensuing data digit 0) and

• by a dashed line (ensuing data digit 1)



codes

Trellis diagram may be easily drawn from the above state diagram

It commences from entire 0s in the SR, i.e., state a and

makes transitions depending on every input data digitmakes transitions depending on every input data digit

a solid line (ensuing data digit 0) and

by a dashed line (ensuing data digit 1)



Fig. Survivor paths after the 3rd branch of the Trellis diagram for received sequence 01 00 01



branch of the Trellis diagram for




• As usual, ML receiver selects a code word nearest to the received code word

• Since there are code words (k input data digits),

• the ML decoder needs to store 2• the ML decoder needs to store 2the received code word

• For large k, there is exponential incr

• Viterbi simplify such ML calculation by observing that

• every four nodes (a, b, c and d) has only two antecessors



codes

As usual, ML receiver selects a code word nearest to the received

Since there are code words (k input data digits),

the ML decoder needs to store 2k code words and compares with the ML decoder needs to store 2k code words and compares with

increase in complexity of the decoder

simplify such ML calculation by observing that

every four nodes (a, b, c and d) has only two antecessors




• only the path that matches most with the received sequence

• (the minimum distance path) requires to be kept for each node

• For a received bit stream,

• it is necessary to determine a path in the trellis diagram with • it is necessary to determine a path in the trellis diagram with

• the output digit stream which matches most with the received stream

Example

• Assume that the initial six received digits are 01 00 01

• Find the survivor paths (minimum-sequence)



codes

only the path that matches most with the received sequence

) requires to be kept for each node

it is necessary to determine a path in the trellis diagram with it is necessary to determine a path in the trellis diagram with

the output digit stream which matches most with the received

Assume that the initial six received digits are 01 00 01

-distance path with the received




• Survivor paths after the 3rd branch of the Trellis diagram for received sequence 01 00 01

• Note that every node can be checked in from two nodes only

• With four paths eliminated as illustrated in Table, the four survivor • With four paths eliminated as illustrated in Table, the four survivor paths are the only contestants

• To truncate the Viterbi algorithm and ultimately we need to resolve on single path rather than four



codes

branch of the Trellis diagram for received

Note that every node can be checked in from two nodes only

With four paths eliminated as illustrated in Table, the four survivor With four paths eliminated as illustrated in Table, the four survivor

algorithm and ultimately we need to resolve



After 3rd branches Paths Distance with received

Node a 00 00 00

11 10 11

Node b 00 00 11

11 10 0011 10 00

Node c 00 11 10

11 01 01

Node d 00 11 01

11 01 10



with received sequence Survivor?

2 Yes

3

2 Yes

33

5

2 Yes

3 Yes

4




• This is made possible for 00 given to the last two data digits

• When the first 0 is input to the SR,

• we look for the survivors at nodes a and c only

• When the second 0 enters the SR, we scrutinize survivor at node a• When the second 0 enters the SR, we scrutinize survivor at node a

• For Viterbi algorithm, storage and coconsiderably (proportional to 2N)

Table: Survivor paths after the 3rd branch of the Trellis diagram for received sequence 01 00 01



codes

This is made possible for 00 given to the last two data digits

When the first 0 is input to the SR,

we look for the survivors at nodes a and c only

When the second 0 enters the SR, we scrutinize survivor at node aWhen the second 0 enters the SR, we scrutinize survivor at node a

d computational complexity reduces

branch of the Trellis diagram for



Space-time trellis codes

• Space Time Trellis Code is

• an extension of conventional trellis codes to systems

• These codes may be designed to

• extract the diversity gain and coding • extract the diversity gain and coding

• The literal meaning of “trellis” is

• “a light frame made of long narrow pipes of wood that cross each other, used to support climbing plants

• Note the similarity of trellis with the trellis diagram



+ V. Kuhn, Wireless Communications over MIMO channels,

extension of conventional trellis codes to multi-antenna

the diversity gain and coding gain+the diversity gain and coding gain

“a light frame made of long narrow pipes of wood that cross each other, used to support climbing plants”

Note the similarity of trellis with the trellis diagram



Wireless Communications over MIMO channels, Wiley, 2006


Fig. Quadrature phase shift keying (QPSK) constellation diagram (Ξ(0)�e0=1; Ξ(1)�ejΠ/2=j; Ξ(2)�e2jΠ/2

ary mapping function)



phase shift keying (QPSK) constellation diagram /2=-1; Ξ(3)�e3jΠ/2=-j where Ξ is the M-




• The number of nodes in the trellis diagram

• corresponds to the number of states in the

• Fig. Trellis diagram of a QPSK, four-rate of 2bps/Hz rate of 2bps/Hz



The number of nodes in the trellis diagram

to the number of states in the trellis

-state trellis code for NT=2 with a








Input States 0

0 00

1 011 01

2 02

3 03



1 2 3

10 20 30

11 21 3111 21 31

12 22 32

13 23 33




• For example,

• when the encoder is in state St=0 and

• the input bits are 11, the symbol is a

• The Space Time Trellis Code outputs the pair of symbols (3,0),

• corresponding to the phases 3π/2 and 0 • corresponding to the phases 3π/2 and 0

• (corresponding signals Ξ(3)�e3j

• and Ξ(0)�e0=1 where Ξ is the M

• The signal -j is transmitted in the second antenna

• and 1 signal is transmitted on the first



and

input bits are 11, the symbol is a 3

outputs the pair of symbols (3,0),

/2 and 0 /2 and 0 3jΠ/2=-j

=1 where Ξ is the M-ary mapping function)

j is transmitted in the second antenna

1 signal is transmitted on the first antenna




• At this point encoder goes to state S

• If the next two input bits are 01,

• the encoder outputs the symbols (2,3) which are transmitted on the two antennasthe two antennas

• (second antenna transmits -1 and first antenna transmits

• Thus the encoder goes to state St=2 and the process goes on

• At the end of a block of input bits, say a frame of data,

• zeros are inserted in the data strstate St=0



state St=3

encoder outputs the symbols (2,3) which are transmitted on

1 and first antenna transmits -j)

=2 and the process goes on

the end of a block of input bits, say a frame of data,

stream to return the encoder to the




• The delay diversity scheme was started by A.

• We will denote the generator matrixCode by

( )TNSM ,,W

• where M signifies M-ary modulation scheme,

• S is the number of states in the trellis diagram and

• NT is the number of transmitting antennas



( )TNSM ,,W

The delay diversity scheme was started by A. Wittneben (1993)

atrix for the above Space Time Trellis

modulation scheme,

S is the number of states in the trellis diagram and

is the number of transmitting antennas




• The generator matrix will have

• NT columns and m+s rows

• (m=log2M and s is the number ofand it is equal to m=log2S where S is the number of states)and it is equal to m=log2S where S is the number of states)

• Each entry is being a number between 0 to M

• The generator matrix for Wittneben



( )4, 4, 2T

=

W

time trellis codes

er of memory elements in the encoder S where S is the number of states)S where S is the number of states)

Each entry is being a number between 0 to M-1

Wittneben Delay diversity is given by



1 0

2 0

0 1

0 2

=


• The transmit antennas send delayed version of the message bits

• We are considering QPSK, trellis with four states and

• number of transmitting antennas equal to 2 as depicted in Fig.



( ) ( ) ( ) ( )[

( ) ( )( ) ( )

2 1 2 1

2 1

2 1

Input bits State bits

d n d n q n q n

Output bitss n s n M ary mapping function

o n o n

Ξ

= Ξ = Ξ −

G

time trellis codes

The transmit antennas send delayed version of the message bits

We are considering QPSK, trellis with four states and

number of transmitting antennas equal to 2 as depicted in Fig.

1 0



] ( )mod

;

TM

s n s n M ary mapping function

= Ξ = Ξ −

G ( )

1 0

2 04, 4, 2

0 1

0 2

T

=

W


• The output bits will be mapped to Mmapping function Ξ

• which maps integer values to the M

• For instance for M-PSK constellation, • For instance for M-PSK constellation,

• Ξ (i) =exp(2πji/M)

• where j is the √(-1) and

• i is an integer between 0 and M



The output bits will be mapped to M-ary symbols using the M-ary

which maps integer values to the M-ary symbols

PSK constellation, PSK constellation,

is an integer between 0 and M-1








• Let us revisit the space time trellis code of Fig.

• Let us assume that the input bit stream for this code is 01110010

• Fig. below shows the trellis path corresponding to this input bits streamstream

• Note that we have to add 00 at the end to guarantee that the statemachine return to state zero

• Using the generator matrix ,

• find the set of transmitted symbols



time trellis codes

Let us revisit the space time trellis code of Fig.

Let us assume that the input bit stream for this code is 01110010

Fig. below shows the trellis path corresponding to this input bits

Note that we have to add 00 at the end to guarantee that the state-

find the set of transmitted symbols



1 0

2 0

0 1

0 2

T

=

G




time trellis codes




• First two inputs are 10 and state is 00 (State 0)

• Hence, the output can be obtained as

1 0



( ) (

1 0

2 01 0 0 0 1 0

0 1

0 2

=

time trellis codes

First two inputs are 10 and state is 00 (State 0)

Hence, the output can be obtained as



)1 0 0 0 1 0


• Therefore, Ξ(1)= ejΠ/2=j and Ξ(0)= ethe second and first antennas respectively

• Now the next state is 10 (State 1)

• Next two input bits are 00, hence, the outputs are• Next two input bits are 00, hence, the outputs are



( )

1 0

2 00 0 1 0 0 1

0 1

0 2

time trellis codes

=j and Ξ(0)= ej0Π/2=1 are sent at time t=1 from the second and first antennas respectively

Next two input bits are 00, hence, the outputs areNext two input bits are 00, hence, the outputs are



( )0 0 1 0 0 1

=


• Therefore, Ξ(0)=ej0Π/2=1 and Ξ(1)= ethe second and first antennas respectively





( )

1 0

2 01 1 0 0 3 0

0 1

0 2

time trellis codes

=1 and Ξ(1)= ej1Π/2=j are sent at time t=2 from the second and first antennas respectively




( )

1 0

2 01 1 0 0 3 0

0 1

0 2

=


• Therefore, Ξ(3)= ej3Π/2=-j and Ξ(0)= ethe second and first antennas respectively





( )

1 0

2 00 1 1 1 2 3

0 1

0 2

=

time trellis codes

j and Ξ(0)= ej0Π/2=1 are sent at time t=3 from the second and first antennas respectively




( )0 1 1 1 2 3

=


• Therefore, Ξ(2)= ej2Π/2=-1 and Ξ(3)= ethe second and first antennas respectively





( )

1 0

2 00 0 0 1 0 2

0 1

0 2

time trellis codes

1 and Ξ(3)= ej3Π/2=-j are sent at time t=4 from the second and first antennas respectively




( )

1 0

2 00 0 0 1 0 2

0 1

0 2

=


• Therefore, Ξ(0)= ej0Π/2=1 and Ξ(2)= ethe second and first antennas respectively

• Rank and coding gain distance (CGD) calculations

Example:Example:

• The first path stays in state zero during both transitions that is 000

• The second path goes to state 2 in the first transition and merges to state zero in the second transition that is 020

• Find the rank and CGD for these two trellis paths shown in next slide



time trellis codes

=1 and Ξ(2)= ej2Π/2=-1 are sent at time t=5 from the second and first antennas respectively

Rank and coding gain distance (CGD) calculations

The first path stays in state zero during both transitions that is 000

The second path goes to state 2 in the first transition and merges to state zero in the second transition that is 020

Find the rank and CGD for these two trellis paths shown in next slide






time trellis codes




• The corresponding codewords for thdoing M-ary mapping Ξ for QPSK shown in Fig.

( ) ( )0 00 1 0 1j je e Ξ = = Ξ = =

• Then codeword difference and distance matrix of can be calculated as



( ) ( )( ) ( )

0 0

1 2

0 0

0 1 0 1;

0 1 0 1

j j

j j

e e

e e

Ξ = = Ξ = == = Ξ = = Ξ = =

C C

time trellis codes

or the two different trellis paths after mapping Ξ for QPSK shown in Fig.

( ) ( )2

022 1 0 1j

je e

π Ξ = = − Ξ = =

Then codeword difference and distance matrix of C1 and C2 and CGD



( ) ( )

( ) ( )

021 2

20 2

2 1 0 1

0 1 2 1

j

jj

e e

e e

π

Ξ = = − Ξ = = = =

Ξ = = Ξ = = −

C C


( )1 2 1 2, = − =D C C C C

( ) ( ) ((

2/8/2018 Rakhesh Singh Kshetrimayum, of MIMO Wireless Communications,

Cambridge UniFundamentals

( ) ( ) ((1 2 1 2 1 2, , ,= = =A C C D C C D C C

( )1 2, 16CGD = =A C C (dG = = = =A C C A C C

time trellis codes

1 2 1 22 0

0 2

= − =

D C C C C

))2 0 2 0 4 0T

, of MIMO Wireless Communications,

UniFundamentals versity Press, 2017207

))1 2 1 2 1 22 0 2 0 4 0

, , ,0 2 0 2 0 4

T = = =

A C C D C C D C C

( ) ) ( )( )1 1

21 2 1 2, , 16 4r

= = = =A C C A C C


• Assume that for an Space Time Trellis Code with Ntransmitted codeword is 220313 and the decoder decides in favor of the codeword 330122

• The symbols transmitted are from QPSK scheme• The symbols transmitted are from QPSK scheme

• Find the coding gain and diversity gain for this case (Ξ(0)Ξ(1)�ejΠ/2=j; Ξ(2)�e2jΠ/2=-1; Ξ(3)�



1 21 ; 1

1 1

j j

j j

− − −

= − = − − − −

C C

time trellis codes

Assume that for an Space Time Trellis Code with NT=2, the transmitted codeword is 220313 and the decoder decides in favor of

The symbols transmitted are from QPSK schemeThe symbols transmitted are from QPSK scheme

Find the coding gain and diversity gain for this case (Ξ(0)�e0=1; �e3jΠ/2=-j )



1 2

1 1

1 ; 1j j

j j

− − −

= − = − − − −

C C


• Codeword difference matrix

( 2 1 2 1, 0 2D C C C C

• Codeword distance matrix



( ) ( ) ( )(2 1 2 1 2 1, , , 2 2 4 2 2= = − − +A C C D C C D C C

time trellis codes

)2 1 2 1

1 1

, 0 2

1 1

j j

j

j j

− − − +

= − = − + − +

D C C C C



1 1j j − + − +

))4 2 2 2 2

, , , 2 2 4 2 2

2 2 2 2 4

Hj j

j j

j j

+ − −

= = − − + − + − −


• The codeword distance matrix has only two 9.4641

• Therefore, rank of codeword distance matrix is 2 (diversity gain is and coding gain is and coding gain is



( ) ( )1/ 1/ 2 1/ 2r

dG CGD CGD= = = × =

time trellis codes

The codeword distance matrix has only two eigenvalues 2.5359 and

Therefore, rank of codeword distance matrix is 2 (diversity gain is 2NR)



( )1/ 1/ 2 1/ 2

2.5359 9.4641 24= = = × =


• Space Time Trellis Code employing QPSK for two transmit antennas designed based on rank and determinant criteria

• T(4,4,2)( )

=

0

12,4,4T

• M=4, S=4, NT=2 (rows = 2, columns = 4)

• Rank 2, Coding gain 2



( )

=0

2,4,4T

+ V. Tarokh, N. Seshadri and A. Calderbank, “Spacecommunication: performance criterion and codeTheory, 44(2), 1998, pp. 744-765.

time trellis codes

Space Time Trellis Code employing QPSK for two transmit antennas designed based on rank and determinant criteria

210

002

=2 (rows = 2, columns = 4)



210

“Space-time codes for high data rate wirelessconstruction”, IEEE Transactions on Information


• Space Time Trellis Code employing QPSK for two transmit antennas designed based on rank and determinant criteria

• Y(4,16,2)( )

=

22

202,16,4Y


• Rank 2, Coding gain



( )

=22

2,16,4Y

32

+ Q. Yan and R. S. Blum, “Optimum space-time convolutional

and Networking Conference, 2000, Chicago, USA.

time trellis codes

Space Time Trellis Code employing QPSK for two transmit antennas designed based on rank and determinant criteria

20212

02112




20212

convolutional codes”, in Proc. Wireless Communications


• B(4,16,2)

• M=4, S=16, N =2 (rows = 2, columns = 6)

( )

=

20

122,16,4B


• Rank 2, Coding gain



20

+ S. Baro, G. Baush and A. Hansmann, “Improved Codes

Communications Letters, 4(1), 2000, pp. 20-22.

time trellis codes


20122

02201




Codes for Space-time trellis coded modulation”, IEEE


• P(4,16,3)


( ) 120

• Full rank and coding gain is



( )

=

312

2123,16,4P

24

time trellis codes


022



212

230

Performance analysis of Spaceover separately correlated MIMO channel• Let us denote the codeword difference matrix between two

codewords C1 and C2 by

• Assume that we are sending the spacedecoder decides in favor of C2

1 2= −∆ C C

• Hence the conditional pairwise error probability (PEP)



( )1 2

02 2

sE

P Q QN

→ = =

C C H H

+ G. Taricco and E. Biglieri, “Exact pairwise error probability

Theor., 48(2), Feb. 2002, pp. 510-513.

Performance analysis of Space-time codes over separately correlated MIMO channel

Let us denote the codeword difference matrix between two

Assume that we are sending the space-time codeword C1 and the

error probability (PEP)+ is given as



2 2

2 2P Q Q

γ → = =

C C H H∆ H∆

probability of space-time codes”, IEEE Trans. On Inform.

Performance analysis of Spaceover separately correlated MIMO channel

• Using the alternate form of Q-function, we have,

( )2

1 2

0

1expP d

π

π

→ = −

∫C C H

• Hence the unconditional pairwise error probability (PEP) is given as



0π

( )2

2 21 2

2 2

0 0 0

1 1exp

4sin 4sinP p d d M d

π π

γ

π πθ θ

∞ → = − = −

∫ ∫ ∫H∆

H∆C C


function, we have,

2

2exp

4sinP d

γθ

θ

→ = −

H∆

error probability (PEP) is given as



4sin θ

( )2 2

2 2

2 2

0 0 0

1 1


π π

γζ ζ θ θ

π πθ θ

→ = − = −

∫ ∫ ∫∆ H∆


• In the above equation, assuming ,

• we can denote as the m

of

( )2M sH∆

2of



2H∆


In the above equation, assuming ,

e moment generating function (MGF)

24sins

γ

θ= −




• Note that

• For separately correlated MIMO channel, we have,

( ) ({2 H H H H Htrace vect vect= = ⊗H∆ H∆∆ H H I

• For separately correlated MIMO channel, we have,



( ) (H wvect vect=H R H


For separately correlated MIMO channel, we have,

( )} ( ) ( )R

HH H H H H

Ntrace vect vect= = ⊗H H I ∆∆ H

For separately correlated MIMO channel, we have,



)H wH R H


• Therefore,

( ){ } ( )2 / 2 1/ 2H HH H H

w H N H wvect vect= ⊗H∆ H R I

• where is the spatial correlation matrix

• Assume that A is a Hermitian matrix



X XR T= ⊗

HR R R

( ) (/ 2 1/ 2

R

H

H N H= ⊗A R I ∆∆


( )( ) ( )/ 2 1/ 2

R

H H H

w H N H wvect vect= ⊗H R I ∆∆ R H

where is the spatial correlation matrix

matrix



)( )/ 2 1/ 2H

H N H∆∆ R


• Rayleigh fading MIMO channel

• Assuming is a zero mean Gaussian ( ){ }H

H

wvect=v H

vector with covariance matrix as an identity matrix



=vR I


Assuming is a zero mean Gaussian ( )=vµ 0

vector with covariance matrix as an identity matrix



R TN N=R I


Theorem:

• Consider the random quadratic form of a complex Gaussian multivariate (row vector)



• 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 complex Gaussian multivariate (row vector) ( ),N

CN= v Vv µ R



{ } ( )1 H

v

v

s s

s

− −

−

v vµ A I R A µ

I R A


• Note that for , the expone

of equation will become 1

( )=vµ 0

(yM s

Hence, the MGF for random quadratic form of a correlated Rayleigh fading channel is



( ) = IsMTR NNy

(y


onential term in the MGF expression

){ } ( )

1exp

H

vs sM s

s

− − =

−

v vµ A I R A µ

I R A

Hence, the MGF for random quadratic form of a Hermitian matrix A for correlated Rayleigh fading channel is



1−− As

T

)vs−I R A

R TN N=vR I


• Then, after substituting A, the MGF of y becomes

( ) ( ) (/ 2 1/ 2

R T R

H

y N N H N HM s s= − ⊗I R I

• Since , we have,



BAIABI +=+

( ) (R T Ry N N N H

M s s= − ⊗I I


, the MGF of y becomes

)( )1

/ 2 1/ 2

R T R

H

y N N H N H

−

= − ⊗I R I ∆∆ R

Since , we have,



)1

R T R

H

y N N N H

−

= − ⊗I I ∆∆ R


• Since for Kronecker

• Since , we have,

X XH R T= ⊗R R R

( ) (R T R X Xy N N N R T

M s s= − ⊗ ⊗I I ∆∆

( )( ) BDACDCBA ⊗=⊗⊗• Since , we have,

• If λA is an eigenvalue of A and λB is an

• then λA λB is an eigenvalue of



( )( ) BDACDCBA ⊗=⊗⊗

⊗A B

( ) (R T X Xy N N R TM s s= − ⊗I R ∆∆


Kronecker model, we have,

Since , we have,

)( )1

R T R X X

H

y N N N R T

−

= − ⊗ ⊗∆∆ R R

Since , we have,

is an eigenvalue of B,



⊗A B

)1

R T X X

H

y N N R T

−

∆∆ R


• where

• r is the rank of

( ) (ˆ

1 1

1r r

y n m

n m

M s sλ µ= =

= −∏∏

H∆∆ R M s s• r is the rank of

• is the rank of

• is the eignvalue of

• is the eignvalue of



X

H

T∆∆ R

XRRr̂

X

H

T∆∆ R

XRR

nλ

mµ

M s s


)1

y n mλ µ−

( ) ( )1

HM s s

−

= − ⊗I R ∆∆ R



XT

( ) ( )R T X X

H

y N N R TM s s= − ⊗I R ∆∆ R


• In our case, we have taken,

• hence, the exact PEP becomes 4sin

s = −

( ) (2

2 21 2 1 1

expP p d d M d

π π

γ∞ → = − = −∫ ∫ ∫

H∆C C



( ) ∫∏∏= =

−

+=→

2

0 1

ˆ

1

1

2

21

sin41

1

π

θ

µγλ

π

r

n

r

m

mn dP CC

( ) (2 2

1 2

2 2

0 0 0

1 1exp


γ

π πθ θ

→ = − = −

∫ ∫ ∫H∆

H∆C C


24sin

γ

θ= −

( )2 21 1

P p d d M d

π π

γζ ζ θ θ

→ = − = −∫ ∫ ∫



θd

( )2 22 2

0 0 0

1 1


γζ ζ θ θ

π πθ θ

→ = − = −

∫ ∫ ∫ H∆

( ) ( )ˆ

1

1 1

1r r

y n m

n m

M s sλ µ−

= =

= −∏∏


• Find the PEP for the following two space

−

−

−

=

=

11

1

1

;

11

11

1121

CC

• Assume i.i.d. Rayleigh fading MIMO channel

• Note that Δ is the codeword difference

• Δ ΔH is the codeword distance matrix



−

1111


Find the PEP for the following two space-time codewords:

−

−

1

1

1

. Rayleigh fading MIMO channel.

is the codeword difference matrix

is the codeword distance matrix



1


• For i.i.d. Rayleigh fading MIMO channel,

X T X

H H

T N T= ⇒R I ∆∆ R



mNR RX=⇒= µIR

( ) ( )=

= ,;

02

20

20

, 2121CCACCD


fading MIMO channel,

X T X

H H

T N T=R ∆∆



RNr =ˆ,1

=

80

04


• Hence the eigenvalues of codeword distance matrix are

( ) ∫∏∏2 ˆ

1

π

r r



( ) ∫

+

+=→

2

0

2

2

2

221

8sin4

sin4

4sin4

sin41

π

γθ

θ

γθ

θ

πdP

RR NN

CC

( ) ∫∏∏= =

=→

0 1

ˆ

1

21 1

π

r

n

r

m

P CC


of codeword distance matrix are 4 and 8

∏−

1µγλ



∫

+

+=

2

0

2

2

2

2

2sin

sin

sin

sin1

π

θγθ

θ

γθ

θ

πθ dd

RR NN

∏

+

1

1

2sin41 θ

θ

µγλ mn d


• For Chernoff bound of PEP, put or

• for PEP, then

θ

( )ˆ

1 2 1 r r

P → ≤ +∏∏C C

• For high SNR case, PEP is bounded as



( )1 12 4bound

n m

P= =

→ ≤ +∏∏C C

( )1 2

1 1

1

2 4

r r

boundn m

P= =

→ ≤ ∏∏C C


bound of PEP, put or 2

πθ = ( )2sin 1θ =

1ˆ

1r r

n mγλ µ

−

→ ≤ + ∏∏

For high SNR case, PEP is bounded as



1 1

12 4= =

→ ≤ +

∏∏

1ˆ

1 12 4

r rn m

n m

γλ µ−

= =

∏∏


• For i.i.d. Rayleigh fading MIMO channel,

X

H H

T =∆∆ R ∆∆

Nr ==⇒= ˆ,1µIR (P



RmNR NrRX

==⇒= ˆ,1µIR

( )1 2

1

1 1 4

2 4 2

RNrn

boundn

Pγλ

−

=

⇒ → ≤ =

∏C C

(P


. Rayleigh fading MIMO channel,

( )1ˆ

1 2 1 r rn mP

γλ µ−

→ ≤ ∏∏C C



1

1 1 4

2 4 2

RR

R

R

rN

Nr

rN

n

n

λ γ=

→ ≤ = ∏

( )1 2

1 1

1

2 4

n m

boundn m

Pγλ µ

= =

→ ≤

∏∏C C


• Once we have the PEP between all

• we can find the union bound for error probability

( ) ∑ ∑≤1

PM

eP

• where M is the total number of codewords



( ) ∑ ∑≠

≤C CC ˆ

PM

eP


Once we have the PEP between all codewords,

we can find the union bound for error probability

( )→ CC ˆ

codewords



( )→ CC

Capacity approaching codes

• Capacity approaching codes

• Codes which can approach the Shannon’s capacity limit

• It can be obtained from the Shannon’s capacity for AWGN channel

• where Es is the energy per symbol and

• N0 is the variance of the noise



( ) =+= 2 log1log SNRη


Codes which can approach the Shannon’s capacity limit

It can be obtained from the Shannon’s capacity for AWGN channel

is the energy per symbol and



+

0

2 1logN

Es


• An error correcting code maps k input bits to n symbols

• Code rate r=k/n is a measure of spectral efficiency of code

• For reliable communication, code rate r should be less than or equal to η

• Note that Eb=Es/r is the energy per bit• Note that Eb=Es/r is the energy per bit

• Assume that we are transmitting at η

• From Shannon’s capacity, one may obtain



+ D. J. Costello, Jr. and G. D. Forney, Jr., “Channel coding

vol. 95, no. 6, June 2007, pp. 1150-1177.+ M. C. Valenti, “Turbo codes and iterative processing”,


An error correcting code maps k input bits to n symbols

Code rate r=k/n is a measure of spectral efficiency of code

For reliable communication, code rate r should be less than or equal

=Es/r is the energy per bit=Es/r is the energy per bit

g at the Shannon’s spectral efficiency

From Shannon’s capacity, one may obtain+,



( )121

0

−> η

ηN

Eb

coding: the road to channel capacity”, Proceedings of IEEE,

processing”, in Proc. IEEE New Zealand Wire. Symp., Nov. 1998.


• This is called the Shannon’s limit (lower bound) on the function of spectral efficiency η

• For example

• η=0, lower bound on Eb/N0 is ln2=0.693 and

• η=1/2, lower bound on E /N is 0.828 and • η=1/2, lower bound on Eb/N0 is 0.828 and

• η=2/3, lower bound on Eb/N0 is 0.88 and

• η=1, lower bound on Eb/N0 is 1 and 0 dB

• It can be also calculated for any higher values of



+ S. Benedetto, G. Montorsi and D. Divsalar, “Concatenated

Communications Magazine, Aug. 2003, pp. 102-109.


(lower bound) on the Eb/N0 as a

is ln2=0.693 and -1.59 dB

is 0.828 and -0.817 dB,( )

is 0.828 and -0.817 dB,

is 0.88 and -0.55 dB

is 1 and 0 dB

It can be also calculated for any higher values of η as well



“Concatenated Convolutional Codes with Interleavers”, IEEE

( )121

0

−> η

ηN

Eb


• Large coding gains for a given spectral efficiency by coding such as

• Turbo codes

• Low density parity check (LDPC) codes

• Turbo codes started by Claude Berrou• Turbo codes started by Claude Berrou

• Reported a turbo code of code rate ½

• very close to Shannon limit (0.5 dB) in 1993

• It is possible since

• it consists of two separated codes (concatenated code)

• which are combined to form a larger code




Large coding gains for a given spectral efficiency by coding such as

Low density parity check (LDPC) codes

BerrouBerrou

Reported a turbo code of code rate ½

very close to Shannon limit (0.5 dB) in 1993

it consists of two separated codes (concatenated code)

which are combined to form a larger code



Basics of turbo codes

• Encoder:

• Turbo codes are

• basically parallel concatenation of

• 2 systematic recursive convolutional• 2 systematic recursive convolutional

• Length k-message u encoded

• by the 1st encoder produces parity bits

• Interleaved u, i.e., Π(u) encoded

• by the 2nd encoder produces parity bits



+ S. J. Johnson, Iterative error corrections, Cambridge University

basically parallel concatenation of

convolutional codes (SRCC)convolutional codes (SRCC)

encoder produces parity bits p(1)

encoder produces parity bits p(2)



University Press, 2010.

Fig. Turbo code encoder



encoder




• Interleaver:

• It is represented by a permutation sequence

[ ΠΠΠ=∏ ,,, 21 L

• where the sequences

• is a permutation of the integers 1 to n



[ ΠΠ=∏ ,1

It is represented by a permutation sequence

]nΠ

is a permutation of the integers 1 to n



]nΠΠ ,,2 L


• Interleaver ensures that parity bits of encoder 2 is

• completely different than encoder 1

• If low-weight parity sequence for encoder 1 then

• high-weight parity sequence for encoder 2, • high-weight parity sequence for encoder 2,

• avoids low-weight turbo codewords

• Example

• Π=[4 2 5 3 1] acting on the input vector

• will produce



( ) [ 524 uuuu=Π u

ensures that parity bits of encoder 2 is

completely different than encoder 1

weight parity sequence for encoder 1 then

weight parity sequence for encoder 2, weight parity sequence for encoder 2,

codewords (improves error rate)

=[4 2 5 3 1] acting on the input vector



[ ]54321 uuuuu=u

]13 uu


• u=[1 0 1 1 0]� Π(u)=[1 0 0 1 1]

• Code rate:

knknk

k

−+−+ 21

• where code rates for encoder 1 and encoder 2 are chosen as

• and respectively

• Note that n1-k and n2-k are parity bits for SRCC1 and SRCC2



knknk −+−+ 21

1n

k

2n

k

( ) [ ]13524 uuuuu=Π u

where code rates for encoder 1 and encoder 2 are chosen as

are parity bits for SRCC1 and SRCC2




• Puncturing:

• To increase code rate,

• we may puncture the output of one or both

• For example, encoding message bits • For example, encoding message bits

• [0 1 0 1 0 0] with an encoder produces the two codeword bits

• Present code rate=6/12=1/2



[ ];001010 21 = CC

we may puncture the output of one or both convolutional codes

For example, encoding message bits For example, encoding message bits

[0 1 0 1 0 0] with an encoder produces the two codeword bits



[ ]0101102 =


• Puncturing every third bit in codeword 2 will produce

• where x indicates that the corresponding bit is not transmitted

[ ×= 0102C

• where x indicates that the corresponding bit is not transmitted

• That means, for every 6 message bits,

• only 10 codeword bits which means the

• code rate=6/10=3/5



Puncturing every third bit in codeword 2 will produce

where x indicates that the corresponding bit is not transmitted

]×1

where x indicates that the corresponding bit is not transmitted

That means, for every 6 message bits,

only 10 codeword bits which means the




• The puncturing pattern is specified by puncturing matrix P

• For encoder with n output bits the matrix P

• will have n rows one for each output stream

• The zero entry in the third column of the second row indicates • The zero entry in the third column of the second row indicates

• that every third bit in the output is to be punctured



=

011

111P

The puncturing pattern is specified by puncturing matrix P

For encoder with n output bits the matrix P

will have n rows one for each output stream

The zero entry in the third column of the second row indicates The zero entry in the third column of the second row indicates

that every third bit in the output is to be punctured




• Systematic recursive convolutional

• Note there is a feedback loop in encoder diagram

• unlike convolutional codes

• hence, the name recursive• hence, the name recursive

• we can modify the generator polynomial for SRCC accordingly

• First part of the code c(1) is the message bit itself

• so the name systematic



+ S. Haykin and M. Moher, Communication systems, John

convolutional code (SRCC)

Note there is a feedback loop in encoder diagram

we can modify the generator polynomial for SRCC accordingly

is the message bit itself



John Wiley & Sons, 2010.

Fig. A SRCC encoder



Fig. A SRCC encoder




• Consider SRCC shown in Fig.

(a) Find the generator matrix of this SRCC

+=

21 D

• 1 in the generator matrix means

• we are sending the first part of the codeword same as the message bit



++

+=

2

2

1

11

DD

DG

(a) Find the generator matrix of this SRCC

we are sending the first part of the codeword same as the message




(b) Consider an input message bit of 100, find the output of the code

Solution:

• Assume input bits are coming as 1, 0 and 0

• Note that subscript denote the time, at time t=1,• Note that subscript denote the time, at time t=1,

• First part of the code is equal to the input message bit at that time

• Second part of the code can be calculated as



( )11

11 == uc

( ) ( ) ( )( ) (1

21

111

21 ⊕⊕⊕= sssuc

(b) Consider an input message bit of 100, find the output of the code

Assume input bits are coming as 1, 0 and 0

Note that subscript denote the time, at time t=1,Note that subscript denote the time, at time t=1,

First part of the code is equal to the input message bit at that time

Second part of the code can be calculated as



( )1

21 =


• Note that subscript denote the time

• At time t=2,





( )02

12 == uc

( ) ( ) ( )( )22

122

22 ssuc ⊕⊕⊕=

Note that subscript denote the time





( )22s⊕


• We need to find the states of the SR

( ) ( ) (( 21

111

12 ⊕⊕= ssus

( ) ( )0

12 == ss

• Hence,



( ) ( )0

11

22 == ss

( ) ( ) ( )( )22

122

22 ⊕⊕⊕= ssuc

We need to find the states of the SR1 and SR2 first

) ) 12 =



( )1

22 =⊕ s


• At time t=3,


( )0

1 == uc




( )03

13 == uc

( ) ( ) ( )( ) ( )23

23

133

23 sssuc ⊕⊕⊕=






• We need to find the states of the SR

( ) ( ) ( )( ) 12

21

221

3 =⊕⊕= ssus

( ) ( )1

12 == ss

• Hence,



( ) ( )1

12

23 == ss

( ) ( ) ( )( )23

133

23 ⊕⊕⊕= ssuc

We need to find the states of the SR1 and SR2 first



( )1

23 =⊕ s


• The final code word for the input message bit 100 is

• ½ rate turbo code by Berrou, Glavieux

( ) ( ) ( ) ( ) ( )[ ;;1

32

21

22

11

1= cccccC

• ½ rate turbo code by Berrou, Glavieux

• It uses the same encoder 1 and 2: rate ½ SRCC shown in Fig.

• Hence it will produce rate 1/3 turbo code



+ C. Berrou, A. Glavieux and P. Thitimajshima, “Near Shannon

turbo-codes,” in Proc. International Conference on Communications

1993. {citations 10,721 as of 07-Feb-2018}

The final code word for the input message bit 100 is

Glavieux and Thitimajshima

) ( ) ] [ ];10;10;11;2

3 =c

Glavieux and Thitimajshima

It uses the same encoder 1 and 2: rate ½ SRCC shown in Fig.

Hence it will produce rate 1/3 turbo code



Shannon limit error-correcting coding and decoding:

Communications, Geneva, Switzerland, pp. 1064-1070,

Fig. Berrou, Glavieux and Thitimajshima2/8/2018



Thitimajshima turbo code encoder, Fundamentals of MIMO Wireless



• we can increase the code rate to 1/2 by encoders

=

=

10

00;

01

1121 PP

• The generator matrix for the SRCC depicted in Fig.



1001

+++

+=

21

11

DD

DG


we can increase the code rate to 1/2 by puncturing both the SRCC

1

0

The generator matrix for the SRCC depicted in Fig.



1

+ 43

4

DD

D


• Interleaver used is

• Assume a message bit

[ ]4,9,8,1,10,5,2,6,7,3=Π

• Assume a message bit

• u=[1 0 1 0 1 0 1 0 1 0] is entering the turbo code encoder

• Let us find out what is the output codeword

• The input bit to SRCC 1 is u



=[1 0 1 0 1 0 1 0 1 0] is entering the turbo code encoder

Let us find out what is the output codeword




• The bit interleaved input for the SRCC 2 is given by

• v=Π(u)=[1 1 0 0 1 0 1 0 1 0]

• At time t=1,

• First part of the code is equal to the input message bit at that time• First part of the code is equal to the input message bit at that time

• Second part of the code (for SRCC 1) can be calculated as



( )11

11 == uc

The bit interleaved input for the SRCC 2 is given by

First part of the code is equal to the input message bit at that timeFirst part of the code is equal to the input message bit at that time

Second part of the code (for SRCC 1) can be calculated as






( ) ( ) ( ) ( )( 21

111

11

21 ⊕⊕⊕== ssupc


• Also note that

• Third part of the code (SRCC 2) can be calculated as

• At time t=1, c1=111



11 =v

( ) ( ) ( ) (( 21

111

21

31 ⊕⊕== ssvpc



( ) ( ) ) ( )1

41

41

31 =⊕⊕ sss


Third part of the code (SRCC 2) can be calculated as



) ( ) ( ) ) ( )1

41

41

31

2 =⊕⊕⊕ sss


• At time t=2,


( )02

12 == uc




022 == uc

( ) ( ) ( ) ( )( 22

122

12

22 ssupc ⊕⊕==





) ( ) ( ) ) ( )42

42

32 sss ⊕++


• We need to find the SRCC 1 states of the SR1, SR2, SR3

( ) ( ) ( )( 21

111

12 ⊕⊕⊕= ssus

( ) ( )1

12 == ss

• Hence,



( ) ( )1

11

22 == ss

( ) ( )0

42

32 == ss

( ) ( )0

12

22 == pc

We need to find the SRCC 1 states of the SR1, SR2, SR3 and SR4 first

( ) ( ) ) 14

13

1 =⊕ ss




• Note that

• Third part of the code (SRCC 2) can be calculated as

12 =v

• We need to find the SRCC 1 states of the SR1, SR2, SR3



( ) ( ) ( ) ( )( 22

122

22

32 sssvpc +⊕⊕==

Third part of the code (SRCC 2) can be calculated as

We need to find the SRCC 1 states of the SR1, SR2, SR3 and SR4 first



( ) ( ) ) ( )42

42

32 sss ⊕+


( ) ( ) ( )( 21

111

12 ⊕⊕⊕= sssvs

( ) ( )1

11

22 == ss

( ) ( )0

43 == ss

• Hence,

• Therefore, at time t=2, c2=001



( ) ( )0

42

32 == ss

( ) ( )1

12

32 == pc

( ) ( ) ) 14

13

1 =⊕ ss




• Hence the output code from SRCC 1

• u=[1 0 1 0 1 0 1 0 1 0]

( ) [ ]L101 =p

• For puncturing matrix



( ) [ ]L101 =p

=

01

111P

Hence the output code from SRCC 1




• We will send u as it is, therefore, u

• We will send 1st, 3rd, 5th, 7th and 9th

therefore,( ) [ ]L11 =p

• Hence, the output code from SRCC 2

• v=Π(u)=[1 1 0 0 1 0 1 0 1 0]



[ ]L1=p

( ) [ ]L112 =p

u=[1 0 1 0 1 0 1 0 1 0]

th bits from parity bit matrix 1,

Hence, the output code from SRCC 2




• For puncturing matrix

• we have,

• We will not send any bit from v

=

0

02P

• We will not send any bit from v

• We will send 2nd, 4th, 6th, 8th and 10therefore,



( )21 = p L

1

0

and 10th bits from parity bit matrix 1,




• We can observe that for 2 input bits,

• we are sending the same 2 input bits and

• 2 parity bits

• The code rate is ½• The code rate is ½



We can observe that for 2 input bits,

we are sending the same 2 input bits and



Introduction to Space-time turbo encoders

• Note that the space-time turbo codes are

• extension of turbo codes for multiple antennas

• We will consider the case of two transmit antennas

• for illustration purpose• for illustration purpose

• Fig. depicts turbo space-time coded modulation scheme

• It comprises of two

• parallel concatenated systematic and recursive space time trellis codes (STTCs)



time turbo encoders

time turbo codes are

extension of turbo codes for multiple antennas

We will consider the case of two transmit antennas

time coded modulation scheme

parallel concatenated systematic and recursive space time trellis



Fig. Turbo space-time coded modulation scheme (STTC: Spacetime trellis code)



time coded modulation scheme (STTC: Space-




• One of the antennas is always linked to

• the systematic output of the systematic and recursive STTC 1,

• and the second antenna is joined to

• the parity symbols of the two systematic and recursive STTCs• the parity symbols of the two systematic and recursive STTCs

• With puncturing (full rate may be achieved),

• one parity symbol for each of the

• two systematic and recursive STTCs may be sent through the channel

• and ignore the other parity symbol



time turbo encoders

One of the antennas is always linked to

the systematic output of the systematic and recursive STTC 1,

and the second antenna is joined to

the parity symbols of the two systematic and recursive STTCsthe parity symbols of the two systematic and recursive STTCs

With puncturing (full rate may be achieved),

two systematic and recursive STTCs may be sent through the

and ignore the other parity symbol




• But then full diversity may not be achieved

• This space-time turbo codes is very similar to the binary turbo codes

• But there are some differences

• The interleavers operate on symbols rather than on bits• The interleavers operate on symbols rather than on bits

• There are interleavers for the systematic and recursive STTC 2

• and de-interleaving operation before sending over the channel

• It makes sure that the systematic symbols for

• both the systematic and recursive STTCs are equal



time turbo encoders

But then full diversity may not be achieved

time turbo codes is very similar to the binary turbo codes

operate on symbols rather than on bitsoperate on symbols rather than on bits

for the systematic and recursive STTC 2

interleaving operation before sending over the channel

It makes sure that the systematic symbols for

both the systematic and recursive STTCs are equal




• Its decoders are very similar to the binary turbo decoders

• except that trellis diagram will be

• One may also find out the Eucliddistance in Viterbi decoders distance in Viterbi decoders

• Iterative decoders are employed in



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

time turbo encoders

Its decoders are very similar to the binary turbo decoders

ill be used for symbols rather than bits

clidean distance instead of Hamming

in turbo codes



communication systems, John Wiley & Sons, 2007.

Coding theory

• We call a 0 and a 1 a digit

• A word is a sequence of digits

• The length of a word is the number of digits in the word

• For example, 11001 is a word of length five• For example, 11001 is a word of length five

• A binary code is a set C of codewords

• The code consisting of all words of length two is

• C={00,01,10,11}

• A block code is a code having all its word of the same length



The length of a word is the number of digits in the word

For example, 11001 is a word of length fiveFor example, 11001 is a word of length five

codewords.

The code consisting of all words of length two is

is a code having all its word of the same length



Coding theory

• This number is called the length of a code

• We shall denote the number of codewords

• Let K={0,1} and Kn be the set of all binary words of length n

• Multiplication is AND• Multiplication is AND

• Addition is XOR

• A code C is called a linear code if v+ww are in C

• For example



This number is called the length of a code

codewords in a code C by |C|

be the set of all binary words of length n

v+w is a word in C whenever v and



Coding theory

• C={000, 111} is a linear code

• A linear code C must contain the zero word

• But C1={000,001,101} is not a linear code

• But zero being in a code does no• But zero being in a code does no

• We can use tools from linear algebra for linear code

• Polynomial representation of codewords

• The polynomial of degree

• at most n-1 over K may be regarded as the word v=a

• of length n in Kn



( ) = axf

A linear code C must contain the zero word

But C1={000,001,101} is not a linear code

s not guarantee that the code is linears not guarantee that the code is linear

We can use tools from linear algebra for linear code

codewords

1 over K may be regarded as the word v=a0a1a2…an-1



1

1210

−−++++ n

n xaxaxaa L

Coding theory

• For example codeword and polynomial are

2

20111

11100

00000

xxx

x

+

++

+

• Let v be a codeword of length n, then



211010 x+

( ) 1001101110

0011111100

v

v

π

For example codeword and polynomial are

3x

Let v be a codeword of length n, then cyclic shift is denoted by π(v)



011111000010011

111100000100111

Coding theory

• A code C is a cyclic code

• if the cyclic shift of each codeword is also a codeword

Example:

• The code C={000,110, 101, 011} is a linear cyclic code

• π(000)=000, π(110)=011, π(101)=101,• π(000)=000, π(110)=011, π(101)=101,

• If we wish to construct a cyclic linear code then

• we pick a codeword v from a set S consisting of

• v and all of its cyclic shifts



( ) ( ) ( ){ } (L ;,,,, 212vvvvS

n ππππ= −

if the cyclic shift of each codeword is also a codeword

The code C={000,110, 101, 011} is a linear cyclic code

(101)=101, π(011)=110(101)=101, π(011)=110

If we wish to construct a cyclic linear code then

we pick a codeword v from a set S consisting of



( ) ( )( ) ( ) ( )( )( ) L,, 3vvvv ππππππ ==

Coding theory

• We say v is the generator for the cyclic code and

• the corresponding polynomial is called by g(x)

• For example

Let n=3 and v=100, • Let n=3 and v=100,

• then

• S={v, π(v), π(v)}={100,010,001}



for the cyclic code and

the corresponding polynomial is called generator polynomial denoted

( ) ( )vavavaw2ππ ++=



( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )vavavaw

vavavaw

vavavaw

2

102

3

2

2

10

2

210

πππ

ππππ

ππ

++=⇒

++=⇒

++=

Coding Theory

• The simplest generator matrix for linear cyclic codes

• the matrix in which the rows are the

• corresponding to the generator polynomial

• and its first k-1 cyclic shifts

• For example, • For example,

• for v=1010,

• n=4,

• k=2,

• g(x) has degree =n-k=2



( )

=

=

xxg

xG

generator matrix for linear cyclic codes is

the matrix in which the rows are the codewords

corresponding to the generator polynomial

( )( )

( )

=

−xgx

xxg

xg

k 1

MG



=

+

+

0101

101013

2

xx

x

Coding theory

• Galois field

• Galois field a special class of cyclic fields

• The order of a finite field is always a

• For each prime power, there exists• For each prime power, there exists

Example:

• Galois Field (2) is the smallest field and

• it has (1,0) as elements

• Addition is XOR and

• Multiplication is AND operation



Galois field a special class of cyclic fields

The order of a finite field is always a prime or a power of a prime

prime power, there exists exactly one finite field GF(pn)prime power, there exists exactly one finite field GF(pn)

Galois Field (2) is the smallest field and

Multiplication is AND operation



Coding theory• Consider the construction of GF(23) using the primitive polynomial

h(x)=1+x+x3

001

010

100

modxwordi

• Every non-zero codeword in Kn can be represented by some power of x



6

5

4

3

101

1111

011

110

x

x

x

x

≡

≡

≡

) using the primitive polynomial

( )

2

1

mod

x

x

xh

can be represented by some power of x



2

2

2

1

1

1

x

xx

xx

x

+≡

++

+≡

+≡

Coding theory

• Galois Field (GF) construction

• A 4-tuple (a,b,c,d) expressing a number in GF(2

2 3a b c dα α α+ + +

• In other words,

• The elements of GF(24) can be represented asdegree less than 4

• A polynomial p(α) that cannot be factored into polynomials of lower degree is referred to as irreducible



a b c dα α α+ + +

) expressing a number in GF(24) can be expressed as

2 3a b c dα α α+ + +

represented as polynomials with

) that cannot be factored into polynomials of lower degree is referred to as irreducible



a b c dα α α+ + +

Coding theory

• Any irreducible polynomial of degree

• Consider the irreducible polynomial over GF(2

( ) 4 41 1p α α α α α= + + ⇒ = +

• Hence we may simplify all higher order polynomials to a polynomial with degree less than 3



5 4 2 6 5 2 3 7 6 3 4 3, ,α αα α α α αα α α α αα α α α α= = + = = + = = + = + +

of degree n yields the field

Consider the irreducible polynomial over GF(24)

1 1α α α α α= +

Hence we may simplify all higher order polynomials to a polynomial with degree less than 3



5 4 2 6 5 2 3 7 6 3 4 3 1, ,α αα α α α αα α α α αα α α α α= = + = = + = = + = + +

Coding theory

• A subgroup H of a group G is a group constructed from a subset of

elements in a group with the same operation “

• If a subgroup of a group can be produced using the power of an element σ (generator),

• the subgroup is known as a cyclic subgroup, denoted by• the subgroup is known as a cyclic subgroup, denoted by

• For example, set of even numbers is <2>

• Elements in a finite cyclic subgroup can be written as

• where and is the identity element

• The order of an element σ in a grou



nσ ζ=

nσ ζ=

ζ

A subgroup H of a group G is a group constructed from a subset of

elements in a group with the same operation “.”

If a subgroup of a group can be produced using the power of an

the subgroup is known as a cyclic subgroup, denoted by σthe subgroup is known as a cyclic subgroup, denoted by

For example, set of even numbers is <2>

Elements in a finite cyclic subgroup can be written as

where and is the identity element

roup is the smallest integer for which



σ

{ }2 1, , , , nζ σ σ σ −L

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