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Fundamentals of MIMO Wireless Fundamentals of MIMO Wireless Fundamentals of MIMO Wireless Fundamentals of MIMO Wireless CommunicationsCommunicationsCommunicationsCommunications

Part IPart IPart IPart I

Prof. Rakhesh Singh

Fundamentals of MIMO Wireless Fundamentals of MIMO Wireless Fundamentals of MIMO Wireless Fundamentals of MIMO Wireless CommunicationsCommunicationsCommunicationsCommunications

Part IPart IPart IPart I

Singh Kshetrimayum

Fundamentals of MIMO Wireless CommunicationsPart I

It covers

Chapter 1: Introduction to MIMO systems

Chapter 2: Classical and generalized fading distributions

Chapter 3: Analytical MIMO channel modelsChapter 3: Analytical MIMO channel models

Chapter 4: Power allocation in MIMO systems

Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless

Communications, Cambridge University Press, 2017

Fundamentals of MIMO Wireless Communications

Chapter 1: Introduction to MIMO systems

Chapter 2: Classical and generalized fading distributions

Chapter 3: Analytical MIMO channel modelsChapter 3: Analytical MIMO channel models

Chapter 4: Power allocation in MIMO systems



SISO Systems

Fig. 1 Single-input, single-output (SISO) system (N



output (SISO) system (NT =NR =1)



SIMO systems

Fig. 2 Single-input, multiple-output (SIMO) system Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


output (SIMO) system (NT =1 and NR≥2)Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


Receiver diversity techniques

• Receiver diversity techniques (

• Equal gain combining (EGC)

• co-phases signals on each branch

• and then combines them with equal weight• and then combines them with equal weight

• Selection Combining (SC)

• selects the signal branch with the highest signal(SNR)




Receiver diversity techniques (combat multipath fading)

Equal gain combining (EGC)

phases signals on each branch

and then combines them with equal weightand then combines them with equal weight

selects the signal branch with the highest signal-to-noise ratio




• Maximal ratio combining (MRC)

• MRC outputs the weighted sum of all the branches

• Weights = complex conjugate of the channel gain coefficients

• MRC is optimal in terms of SNR• MRC is optimal in terms of SNR

• but complex to implement




Maximal ratio combining (MRC)

MRC outputs the weighted sum of all the branches

Weights = complex conjugate of the channel gain coefficients

MRC is optimal in terms of SNRMRC is optimal in terms of SNR



Single-user MIMO

Fig. 3 Multiple-input, Single-output (MISO) system Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


Single User: Point

output (MISO) system (NT≥2 and NR =1)Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


User 1

Single User: Point-to-point MIMO communicatio

Multi-user MIMO

In cellular communication:

• Multiple users with

• Single antenna

• Base station with multiple antennas• Base station with multiple antennas

+H. Huang, C. B. Papadias and S. Venkatesan, MIMO communication for cellular

networks, Springer, 2012.Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


Base station with multiple antennasBase station with multiple antennas

MIMO communication for cellular



Multi-user MIMO

Fig. 4 Multi-user MIMO (1 BS with NRakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


user MIMO (1 BS with NT antennas & K users)Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


MIMO systems

Fig. 5 Point-to-point NT × NR multiple-input multiple

(NT transmitting antennas and NR receiving antennas)



input multiple-output (MIMO) system

receiving antennas)



SISO system capacity

CSISO=BW log2 (1+SNR)+

In order to increase data rate either• BW

• Signal to noise ratio (SNR)

should increaseshould increase

BW is precious, almost always fixed for different applications

Signal power increases• Device’s battery life time decreases

• causes higher interference

• needs expensive RF amplifiers

+ T. M. Cover and J. A. Thomas, Elements of Information Theory



In order to increase data rate either

BW is precious, almost always fixed for different applications

Elements of Information Theory, Wiley, 1999.



MIMO capacity

MIMO (pronounced “My-Moe”)+

• capacity boosters for wireless channels

• without penalty in bandwidth and power

• In a rich Rayleigh scattering environment • In a rich Rayleigh scattering environment

• capacity increases linearly with the minimum N

• CMIMO= m CSISO

+J. G. Andrews, A. Ghosh and R. Muhamed, Fundamentals of WIMAX

2007.Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


capacity boosters for wireless channels

without penalty in bandwidth and power

In a rich Rayleigh scattering environment In a rich Rayleigh scattering environment

capacity increases linearly with the minimum NT or NR=m

Fundamentals of WIMAX, Prentice Hall,



MIMO advantages over SISO

Basically, two gains of MIMO over SISO systems

• Multiplexing (rate) gain

• Diversity gain

• For example, for 3×3 MIMO system• Rate gain =3

• Diversity gain= 9




Basically, two gains of MIMO over SISO systems

3 MIMO system



Diversity gain in MIMO systems

Fig. 6 3×3 MIMO system

1. E. Biglieri et. al, MIMO wireless communications



Diversity gain in MIMO systems

3 MIMO system1

MIMO wireless communications, Cambridge University Press, 2007



Multiplexing gain in MIMO systems

001 1

0


0

0



Multiplexing gain in MIMO systems

1

0

001

3 MIMO system

0

0



Advantages of Diversity gain in MIMO systems

0

0

0


0

0



Advantages of Diversity gain in MIMO systems

0

0

0

3 MIMO system

0

0




• Spatially multiplexed MIMO systems• 3 times data rate than that of SISO system for a 3

• Different message bits are sent in para

• Increases the capacity linearly with the number of antennas • Increases the capacity linearly with the number of antennas

• MIMO for diversity gain• Same message bits are sent from all the 3 transmitting antennas

• If any link is broken or down, receiver can decode message bit from the remaining working links

• It minimizes the probability of error in detection




Spatially multiplexed MIMO systems3 times data rate than that of SISO system for a 3×3 MIMO system

parallel from the 3 transmitting antennas

Increases the capacity linearly with the number of antennas Increases the capacity linearly with the number of antennas

Same message bits are sent from all the 3 transmitting antennas

If any link is broken or down, receiver can decode message bit from the

It minimizes the probability of error in detection



Diversity Multiplexing trade-

• Instead of using all the antennas for rate gain or diversity gain

• We may employ some antennas for rate and diversity gain

• If we use more antennas for div• If we use more antennas for divmay be used rate gain, hence, a trade

• Diversity multiplexing trade-off

• doptimal=(NR-r) (NT-r); 0 ≤ r ≤ min

• Implies d increase, r decreases +L. Zheng and D. N. Tse, “Diversity and multiplexing: A fundamental trade

antenna channels,” IEEE Trans. Information Theory



-off

Instead of using all the antennas for rate gain or diversity

We may employ some antennas for rate and diversity gain

r diversity gain then less antennas r diversity gain then less antennas may be used rate gain, hence, a trade-off

off

≤ r ≤ minNR, NT

Implies d increase, r decreases , “Diversity and multiplexing: A fundamental trade-off in multiple

IEEE Trans. Information Theory, pp. 1073-96, May 2003.



Diversity Multiplexing trade-off: Case study I

Fig. 8 3×5 MIMO system (r = minN



off: Case study I

5 MIMO system (r = minNR,NT=3)



Diversity Multiplexing trade-off: Case study II

Fig. 9 5×5 MIMO system (r=2, Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


off: Case study II

5 MIMO system (r=2, doptimal=9)Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


Diversity Multiplexing trade-off: Case study III

Fig. 10 5×5 MIMO system (r=3, Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


off: Case study III

5 MIMO system (r=3, doptimal=4)Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


Applications

3G, 4G LTE, one of the proponents for 5G

IEEE 802.11n

IEEE 802.16m

WiMAXWiMAX

WiFi



3G, 4G LTE, one of the proponents for 5G



Review questions

Review question 1.1: What is coherence bandwidth of the channel?

Review question 1.2: What is coherence time of the channel?

Review question 1.7: What are non



Review question 1.1: What is coherence bandwidth of the channel?

Review question 1.2: What is coherence time of the channel?

Review question 1.7: What are non-coherent and coherent systems?

, Fundamentals of MIMO Wireless


Example 1.1

Assume that the multiplexing gain (diversity-multiplexing trade-off

( )(NrNd Topt −=

for SNR∞.

Assume MIMO system with an SNR of 10dB, one needs a spectral efficiency of R=16 bps per Hertz.

Find the supreme diversity gain such MIMO system can achieve.



Assume that the multiplexing gain (r) and diversity gain (d) satisfy the

)rNR −

Assume MIMO system with an SNR of 10dB, one needs a spectral

Find the supreme diversity gain such MIMO system can achieve.



Example 1.1

Given SNR =10 dB, R=16bps

r=4.8165

Therefore five antennas may be use

( ) RSNRr =2log

Therefore five antennas may be use(7-2) two antennas may used for diversity

The maximum diversity gain can be calculated as



( )(−= NrNd Topt

used for multiplexing and remaining used for multiplexing and remaining 2) two antennas may used for diversity

The maximum diversity gain can be calculated as



) ( )( ) 45757 =−−=− rN R

Exercises

Exercise 1.3

What are close loop, open loop and blind MIMO systems?

Exercise 1.4

Which diversity was left aside for many years? Why?Which diversity was left aside for many years? Why?

Exercise 1.5

What are frequency flat, frequency selective, fast and slow fading channels?



What are close loop, open loop and blind MIMO systems?

Which diversity was left aside for many years? Why?Which diversity was left aside for many years? Why?

What are frequency flat, frequency selective, fast and slow fading



Course webpage & quick revision of SISO fading channel

http://www.iitg.ac.in/engfac/krs/public_html/lectures/ee634/

A quick review of SISO wireless systems over fading channels

What is a wireless fading channel?

A brief mention on large-scale fading (PL, shadowing) and smallA brief mention on large-scale fading (PL, shadowing) and smallfading (multipath)

How do we model it?

It is modelled as a multiplicative term to the transmitted signal

What are its performance metrics of wireless fading channels?



Course webpage & quick revision of SISO fading channel

http://www.iitg.ac.in/engfac/krs/public_html/lectures/ee634/

A quick review of SISO wireless systems over fading channels

What is a wireless fading channel?

scale fading (PL, shadowing) and small-scale scale fading (PL, shadowing) and small-scale

It is modelled as a multiplicative term to the transmitted signal

What are its performance metrics of wireless fading channels?



Random SISO channel

SISO fading channel:

For a SISO channel, the I-O relationship can be expressed as

where y is the received signal,

nhxy +=

where y is the received signal,

x is the transmitted signal and

n is the AWGN



O relationship can be expressed as



SISO fading channel

We will consider two performance metrics for random channel

• Average or Ergodic capacity

• Outage probability



We will consider two performance metrics for random channel h



Capacity of random SISO channel

Average or Ergodic capacity for SISO fading channel

• Average of instantaneous capacity



( ) 0

2 log1log WSNRWEC =+= ∫∞

α


Average or Ergodic capacity for SISO fading channel

Average of instantaneous capacity



( ) ( ) 2

2 ;1log hdpSNR =+ αααα α


The outage probability denoted as P

• probability that the channel capacity C drops below a certain threshold information rate R

• the probability that the rate R is • the probability that the rate R is C(h)



( ) ( ) (=<= WobRCobRPout logPrPr


denoted as Pout is the

probability that the channel capacity C drops below a certain

is greater than the channel capacity is greater than the channel capacity



( ) )

−

<=<+SNR

obRSNRW

R

12Pr1log 2 αα


Average or Ergodic capacity is the average of the instantaneous capacity of the random channel

It is found out by taking the expectaover the probability density function (PDF) of over the probability density function (PDF) of

where h is the random channel gain coefficient



h=α


is the average of the instantaneous

ectation of the instantaneous capacity over the probability density function (PDF) of over the probability density function (PDF) of

is the random channel gain coefficient



2h


Outage probability can be obtained function (CDF) of the random variable

( ) ( ) αα dpxP

x

∫=



( ) ( ) αααα dpxP ∫=

0

SNR

ex

W

R

12ln −=


ned from the cumulative distribution function (CDF) of the random variable α




For example

Rayleigh fading

is exponential i.e., it has PDF2

h=α is exponential i.e., it has PDF

where α0 is the mean value of RV α

u(α) is the unit step function



h=α

( ) ( )αα

α

ααα up

−=

00

exp1


is exponential i.e., it has PDFis exponential i.e., it has PDF

α and




The average or ergodic capacity is given by

( )( )2

0 00

1log 1 expC W SNR d

αα α

α α

∞ = + −

∫

The outage probability can be obtained from



0 00

( ) ( ) 01α

αα αα

xx

edpxP−

∞−

−== ∫ P


capacity is given by

0 0

C W SNR dα

α αα α

= + −

The outage probability can be obtained from



0 0

( ) 0

2ln 1

1αSNR

e

out

W

R

eRP

−−

−=


• there cannot be any reliable transzero outage probability

• regardless of the value of the bandwidth (BW) and

• transmit power for a Rayleigh fading channeltransmit power for a Rayleigh fading channel

• From outage probability,

• we may express data rate in terms of

• SNR and

• Outage probability as



(WSNR

⇒

−

2ln

1 α


ransmission at any rate guaranteeing a

regardless of the value of the bandwidth (BW) and

transmit power for a Rayleigh fading channel2ln 1eW

R

−transmit power for a Rayleigh fading channel

we may express data rate in terms of



( )( )) ( )( )( ) RRPSNR

eRP

out

W

R

out

=−−

=−

1ln1ln2

1ln

0

2ln0

α

α

( ) 0

2ln 1

1αSNR

e

out

W

eRP

−−

−=


We may express spectral efficiency in bps per Hertz as

If we want to have a zero outage probability,

( SNRW

R−=⇒ 1log 02 α


• Pout=0

• we obtain data rate, R=0+

Hence zero outage probability is an impossibility even for Rayleigh fading channel, the mostly widely us



+ S. Barbarossa, Multiantenna Wireless Communication Systems


We may express spectral efficiency in bps per Hertz as


( )( ))RPout−1ln0


Hence zero outage probability is an impossibility even for Rayleigh ly used wireless fading channel model



Wireless Communication Systems, Artech House, 2003.

SER of various modulation schemes for SISO system over various fading channels

One may employ different modulation schemes

• BPSK, QPSK, QAM, PSK, etc at thebits to symbols to transmit it over the channel

Besides AWGN (additive term), Besides AWGN (additive term),

• we also have different wireless fading channel models (multiplicative term) like Rayleigh, channel.



SER of various modulation schemes for SISO system over

modulation schemes like

t the transmitter side to convert the bits to symbols to transmit it over the channel

we also have different wireless fading channel models ) like Rayleigh, Rician, Nakagami, etc for the




Assume Maximum likelihood decoding

• (Nearest neighbourhood rule) at the receiver side

How do we find the bit error rate or symbol error rate the receiver side?side?




Maximum likelihood decoding

(Nearest neighbourhood rule) at the receiver side

How do we find the bit error rate or symbol error rate the receiver




Monte Carlo simulation

• Generate a stream of bits, convert it to symbols, simulate the channel, add AWGN

• Find the number of bits in error a• Find the number of bits in error abits sent which gives the bit error rate (BER)

Analytical

• Closed form formula of the BER

• Helps in designing the system

• Benchmark is the simulation results




Generate a stream of bits, convert it to symbols, simulate the

ror and divide by the total number of ror and divide by the total number of bits sent which gives the bit error rate (BER)

Closed form formula of the BER

Benchmark is the simulation results




For single antenna case,

• total transmit power is P

Hence,



( ) ( )σσ

==2

2

2

2PThEh

SNRs

SISO




γ=PT


For SER analysis, there are two basic steps:

a) First, find the conditional error probability (CEP) for the specific modulation scheme modulation scheme

• error probability over AWGN channel (y=

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




For SER analysis, there are two basic steps:

a) First, find the conditional error probability (CEP) for the specific

error probability over AWGN channel (y=x+n)

pdf of the received SNR to obtain




In the moment generating function (MGF) based approach

we may express the SER as function of the MGF of the particular fading channel

For example,For example,

BER for BPSK over Rayleigh fading channel

For single antenna case, conditionalgiven by



( ) (γ 2| SNRQEPb =

+ M. K. Simon and M.-S. Alouini, Digital communications over fading channels

2005.


moment generating function (MGF) based approach+

we may express the SER as function of the MGF of the particular

BER for BPSK over Rayleigh fading channel

onal error probability (CEP) for BPSK is



) ( )γ2QSNR =

Digital communications over fading channels, Wiley,


Q function is the tail probability of the standard normal distribution

Then average bit error rate (BER) cathe pdf of received SNR γ

( ) ( )[ ] ( γγ QEPEEP bb ∫∞

== 2|

where E is the expectation operator



( ) ( )[ ] ( γγ QEPEEP bb ∫==0

2|

( ) ;sin2

exp1

2

0

2

2

−= ∫ d

xxQ θ

θπ

π

Q

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

constellations”, in Proc. IEEE MILCOM, pp. 25.5.1-25.5.5, Boston, MA, 1991. (


Q function is the tail probability of the standard normal distribution

) can be obtained by averaging over

) ( ) γγγ dp

where E is the expectation operator



) ( ) γγγ dp

0; ≥x

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

25.5.5, Boston, MA, 1991. (826 citations as of July 26, 2019)


we can obtain the average BER as

( )θ

γ

π

π

EPb ∫ ∫∞

−=

2

2sin2

2exp

1

Integrating w.r.t. γ first, we have, the average BER of BPSK for SISO over any fading channel as



θπ∫ ∫ 0 0

sin2

( ) (γθ

γ

π

π

γpEPb

−= ∫ ∫

∞2

0 0

2sinexp

1


( ) γγθθ

γ dpd

. γ first, we have, the average BER of BPSK for SISO



θ

) θθπ

θγγ

π

γ ddd

−Μ=

∫ 2

2

0sin

11


where Mϒ is the moment generating function (MGF) of SNR γ

What are advantages?

• Converted two indefinite integrations to a definite integration

For example,For example,

For Rayleigh fading, the SNR (γ) is exponentially distributed

Hence, MGF of γ is given by



( ) ( )γ

γγ

γ

γγγ dss =

−=Μ

∞∞

∫∫ exp1

exp1

exp

00

(sM X


is the moment generating function (MGF) of SNR γ

Converted two indefinite integrations to a definite integration

For Rayleigh fading, the SNR (γ) is exponentially distributed



γ

γ

γ

γγ

γγ

γ

γγ

ss

s

ds−

=

−

−

=

−

∞

1

1

1

exp1

exp

0

) ( )[ ]sXEs exp=


where

Hence, the BER of BPSK for SISO casgiven by

Hence, the BER of BPSK for SISO cas

( )γγ E=

Hence, the BER of BPSK for SISO casgiven by



( )π

θ

γθ

π

ππ

dEPb ∫∫ =

+

=

2

0

2

2

02

sin

sin1

sin

11

11


case over Rayleigh fading channel is

case over Rayleigh fading channel is case over Rayleigh fading channel is



θγθ

θd

+2

2sin

SISO fading channels

Performance metrics for wireless communications

• Average capacity

• Expectation of instantaneous capacity over

• Outage probability

• Can be obtained from CDF of

• BER/SER

• Can be obtained from MGF of • Can be obtained from MGF of

• CEP for various modulation schemes

For any SISO fading channel, in order to obtain the above three performance metrics,

• MGF, pdf and cdf of ϒ



Kulkarni, L. Choudhary, B. Kumbhani and R. S. KshetrimayumTAS/MRC and OSTBC in Equicorrelated Rayleigh Fading

2014, pp. 1850-1858.

Performance metrics for wireless communications

Expectation of instantaneous capacity over pdf of α

Can be obtained from CDF of α

Can be obtained from MGF of ϒ

( ) ( )γ

σσ===

2

2

2

2PThEh

SNRs

SISO

Can be obtained from MGF of ϒ

CEP for various modulation schemes+

For any SISO fading channel, in order to obtain the above three



Kshetrimayum, Performance Analysis Comparison oFading MIMO Channels, IET Communications, vol. 8, Is

3 important parameters for a random channel

For any random variable X

PDF:

CDF: The cdf of a RV X is defined as

MGF: The mgf of a RV X is defined as

( )Xp x

MGF: The mgf of a RV X is defined as

From mgf we can obtain characteristic function (



( ) ( )[ ]sXEsM X exp= M s sx p x dx

+A. Papoulis and S. U. Pillai, Probability, Random Variables and Stochastic Processes

Tata McGraw Hill, 2002.

3 important parameters for a random channel

is defined as

is defined as

( ) [ ] ( )x

X UP x P X x p u du

−∞

= ≤ = ∫is defined as

we can obtain characteristic function (cf) by putting s=jω



−∞

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

∞

−∞

= ∫

Probability, Random Variables and Stochastic Processes,

Generalized fading distributions

These are three recent generalized fading distributions viz.,

k-μ,

α-μ and

η-μ fading distributionsη-μ fading distributions

In these fading distributions,

fading is generally considered as composed of many clusters of multipaths or rays



+M. D. Yacoub, “The k-μ distribution and the η-μ

Mag., vol. 49, no. 1, pp. 68-81, Feb. 2007.


These are three recent generalized fading distributions viz.,

fading is generally considered as composed of many clusters of



distribution,” IEEE Antennas Propagat.


Fig. 10 Typical Power Delay Profile (illustration of clusters and rays)




Fig. 10 Typical Power Delay Profile (illustration of clusters and rays)




Note that multipath waves within any one cluster the phases of scattered waves are random and

• have similar delay times with

delay-time spreads of different clusters being relatively largedelay-time spreads of different clusters being relatively large




Note that multipath waves within any one cluster the phases of

time spreads of different clusters being relatively largetime spreads of different clusters being relatively large




How are different clusters formed?

Along the path of signals from the transmitter to receiver,

• Let us assume there are different group of disturbances

• Each group of disturbances will form different clusters• Each group of disturbances will form different clusters

Every multipath will have a different amplitude and phase

• Hence it is a complex RV




How are different clusters formed?

Along the path of signals from the transmitter to receiver,

Let us assume there are different group of disturbances

Each group of disturbances will form different clustersEach group of disturbances will form different clusters

Every multipath will have a different amplitude and phase




How do I find its magnitude?

Usually take square root of the

• real part (in-phase) and

• imaginary part (quadrature component)• imaginary part (quadrature component)

Generally, all RV are Gaussian in nature

Both in-phase and quadrature component may assumed Gaussian distributed




component)component)

Generally, all RV are Gaussian in nature

component may assumed Gaussian



k-μ fading distributions

In such a model,

where several clusters (say n) of many

the representation of envelope, X of the fading signal is



( )2 2

2

1

n

i i i ii

X I p Q q=

= + + + ∑

where several clusters (say n) of many multipaths or rays are formed

the representation of envelope, X of the fading signal is



( )2 2

i i i iX I p Q q

= + + +


where n is the number of clusters in the received signal (usually denoted by µ in the literature)

(Ii+pi) and (Qi+qi) are respectively the component of the resultant signal of component of the resultant signal of

Both Ii and Qi are mutually independent and Gaussian distributed with

• zero mean, i.e. E(Ii) = E(Qi) = 0 and

• equal variance, i.e.



( ) (2 2 2

i iE I E Q= =

is the number of clusters in the received signal (usually

) are respectively the in-phase and quadrature phase of the resultant signal of ith clusterof the resultant signal of i cluster

are mutually independent and Gaussian distributed

) = 0 and



)2 2 2

i iE I E Q σ= =


pi and qi are the respective means

• in-phase and quadrature components of

ith cluster in the received signal

The non zero mean of in-phase and The non zero mean of in-phase and reveal the presence of a

• dominant component in the clusters of the received signal



of

components of

phase and quadrature phase components phase and quadrature phase components

dominant component in the clusters of the received signal




The pdf of k-μ distributed random variable (RV)

( )( )+

−+

ek αµ µµ

12 2

1



( )( )

Ω

+=

+−−

lk

ll

ek

ekp

k

αµα

µµ

µ

µ

α µ

12

22

1

2

distributed random variable (RV) αl is given by

( )

( )

+

Ω

+−

k

kkl

l

αµ

1

1 2



( )

Ω

+−+

Ω

l

l

kkI

l

αµµ1

21

2

1


In the above equation, k > 0 and μ > 0 distribution,

That’s why the name k-μ fading distributions

• k is the ratio of the total power dthe total power due to scattered waves and

• μ represents the number of clusters• μ represents the number of clusters

By varying these two parameters, one can obtain various fading channels

Iυ(・) represents the υth order modified Bessel function of the first kind (MATLAB function is besseli(nu,Z



( )2ll E α=Ω

k > 0 and μ > 0 are the main parameters of the

fading distributions

er due to dominant components to the total power due to scattered waves and

μ represents the number of clustersμ represents the number of clusters

By varying these two parameters, one can obtain various fading

order modified Bessel function of the first nu,Z), where nu is order)




Special cases:

Rice fading distribution

Limit the number of clusters in the received signal to 1 which represents μ = 1represents μ = 1

Rice (μ = 1 and k = K),



( ) (2 2

X I p Q q= + + +

( ) ( +=

== l

Kp

Kkαα µ

121,

Limit the number of clusters in the received signal to 1 which



)2 2

X I p Q q= + + +

)( )

( )

Ω

+

Ω

Ω

+−

−

lll

K

lK

KKI

eeKl

l

αα

α

120

1 2


Nakagami-m fading distribution

consider that the received signal arrives in clusters but the clusters

does not have any dominant component, i.e. put p

with each Ii and Qi being mutually independent and Gaussian distributed with zero mean and equal variance



( )2 2

2

1

n

i ii

X I Q=

= + ∑

consider that the received signal arrives in clusters but the clusters

does not have any dominant component, i.e. put pi = qi = 0

being mutually independent and Gaussian distributed with zero mean and equal variance



( )2 2

i iX I Q

= +


Nakagami-m (k →0 and μ = m),

For small arguments,

µ

−1y

( )(12 +µ k



( )( )µ

µΓ

≅−1

2yI

( )(

2

1

12−

+=

− µα

µα

µ l

k

kp

k

( ) ( )e

km

k

Limp

ml

mk

ml

mm

lmk ΓΩ

+

→=⇒

−

=→ ,0

212

0α

αα

µ

)( )

( )1

1

2

1

1

2−Ω

+−+

+

µαµ

µµ

α

k

kkekl

l



)

( )

( )

2

11

2 1+

Ω

Ω

+

ΓΩ

µµ

µ

αµ

µ

αl

l

lk

l kk

e

ek l

( )

( ) ( )m

em

m

eml

m

ml

m

kml

ll

l

ΓΩ=

Γ

Ω−

−Ω

+−

−

22

12

1

1 2αα

α


Rayleigh fading distribution

For μ = 1, we consider that there is n= 0, it reduces to

( ) ( )2 2

X I Q= +

For m=μ=1, in the above pdf we have the Rayleigh distribution



( ) ( )X I Q= +

( ) 21,0

αα

α

αΩ

=Ω

−

=→l

llmk

ep

e is no dominant component, i.e. p = q

2 2

we have the Rayleigh distribution



2

2

2

2;

2

22

σσ

α σ

αα

=Ω=

−Ω

ll

l

l

l

e


For the k-µ distribution, the pdf of instantaneous SNR is

( ) ( )( )

γ

µµµ

µµ

γ

µµµ

γ µI

ek

exkxp

k

xk

k

+=

+−

+−−

+

−

1

2

1

2

1

1

2

1

2

1

The mgf of instantaneous SNR for k



γµek

k 22

( ) ( ) ( )ssdpeeEsM

kk γγγ

γ γµµ

∞

−− == ∫ −−

0

of instantaneous SNR is

( ) ( )γγγ

µ Exkk

=

+− ;

121

k -μ fading distribution is given by



( )( )

( )( )

ksk

kk

esk

kd

µγµ

µµ

γµ

µγ

−++

+−

++

+= 1

12

1

1


function x=kappa_mu_channel(kappa,mu,Nr,Nt

m = sqrt( kappa/((kappa+1))) ;

s = sqrt( 1/(2(kappa+1)) );

ni=0;

nq=0;nq=0;

for j=1:2*mu

• if mod(j,2)==1

• norm1=m+s*randn(Nr,Nt);



+ B. Kumbhani and R. S. Kshetrimayum, MIMO Wireless Communications over

Generalized Fading Channels, CRC Press, 2017

kappa,mu,Nr,Nt);



MIMO Wireless Communications over


• ni=ni+norm1.ˆ2;

• else

• norm2=s*randn(Nr,Nt);

• nq=nq+norm2.ˆ2;• nq=nq+norm2.ˆ2;

• end

end

h_abs=sqrt(ni+nq)/sqrt(mu);

theta = 2*pi*rand(Nr,Nt);

x=h_abs.*cos(theta)+sqrt(−1)*h_abs



h_abs.*sin(theta);



Weibull fading distributions

Weibull fading distribution (one cluster)

Weibull fading statistics is best fit foin 800/900 MHzin 800/900 MHz

The envelope of the fading signal R is obtained as phase and quadrature components



( )2 2R X Yα= +

fading distribution (one cluster)

it for mobile radio systems operating

The envelope of the fading signal R is obtained as αth root of the in-components



α-μ fading distributions

Weibull distribution for μ = 1 (one cluster)

( )α

α

αβα α

α

er

rrf r

r

Rˆ

ˆ1

==−−

How about extending this classical ffading distribution?



αr

( ) ( )ˆ ˆ;r E R E R rα α αα= Ω = =

(one cluster)

αβα βαβ

α

rer

r

ˆ

1;1 =−−

cal fading distribution to generalized



αr

ˆ ˆr E R E R rα α α= Ω = =


α-μ distribution can be used to model fading channels in the environment characterized by

• non-homogeneous obstacles that may be nonlinear in nature

Suppose that such a non-linearity isSuppose that such a non-linearity isα > 0 thereby the emerging envelope R is

It is more general case as the usual cthis



( )∑=

+=n

i

ii YXR

1

22α

μ distribution can be used to model fading channels in the

homogeneous obstacles that may be nonlinear in nature

ty is expressed by a power parameter ty is expressed by a power parameter thereby the emerging envelope R is

ual case of α =2 is a particular case of



)


The probability density function of the follows

( )( )

α

α

µ

αµ

αµµ

µ

αµr

r

R er

rrf ˆ

1

ˆ

−−

Γ=

Special cases

Weibull fading distribution (one cluster)



( )µr Γ

( )2 2R X Yα= +

The probability density function of the α-μ signal is obtained as

fading distribution (one cluster)




Weibull distribution for μ = 1 (one cluster)

( )α

α

αβα α

α

er

rrf r

r

Rˆ

ˆ1

==−−

Weibull fading statistics is best fit foin 800/900 MHz



αr

( ) ( )ˆ ˆ;r E R E R rα α αα= Ω = =

(one cluster)

αβα βαβ

α

rer

r

ˆ

1;1 =−−

it for mobile radio systems operating



αr

ˆ ˆr E R E R rα α α= Ω = =


exponential fading distribution

considering α = 1 and μ = 1 in the expression of physical model described by

2 2R X Y= +

This represents exponential distribution



( ) (r

erf

r

r

R expˆ

ˆ

−==

−

χ

2 2R X Y= +

in the expression of physical model

This represents exponential distribution



)r

rˆ

1; =χχ



for α = 2 (note that n below is usually denoted by

( )2 2n

= +∑



( )2 2

2

1

n

i ii

X I Q=

= + ∑

( )( )

ˆ2

12

ˆ

2e

r

rrf r

r

RΓ

=−− µ

µ

µµ

µ

µ

α = 2 (note that n below is usually denoted by µ)

( )2 2 = +



( )2 2

i iX I Q

= +

( )2

12ˆ ˆ;

22

2

2

rer

r

r

r

=ΩΓΩ

= Ω−− µ

µ

µµ

µ

µ


Relation of α-μ fading distribution and

In MATLAB, Gamma RV can be generated using the function

( )2 2 2

1

n

i i Nakagamii

R X Y Xα

=

= + =∑

In MATLAB, Gamma RV can be generated using the function

shape parameter, k, and scale parameter,

k and θ are related to the Nakagami

k=m and respectively



m

γθ =

μ fading distribution and Nakagami-m distribution

In MATLAB, Gamma RV can be generated using the function gamrnd

2 2 2

i i NakagamiR X Y X

In MATLAB, Gamma RV can be generated using the function gamrnd

shape parameter, k, and scale parameter, θ, as input arguments

Nakagami-m fading parameter m as




MATLAB code to generate α-μ channel matrix

• n=gamrnd(mu,a_SNR/mu,Nr,Nt

• a_inv=1/alpha;

• phi=2*pi*rand(Nr,Nt);• phi=2*pi*rand(Nr,Nt);

• H=(n.ˆa_inv).*exp(j*phi);





μ channel matrix

mu,Nr,Nt);





Rayleigh fading distribution

considering α = 2 and μ = 1 in the expression of physical model described by

( ) ( )2 2

X I Q= +

Note that in this case, Nakagami-m distribution with Rayleigh distribution



( ) ( )X I Q= +

( )ˆ

2 ˆ2

2

2

er

rrf r

r

R =−

in the expression of physical model

2 2

m distribution with μ = 1 is like



2

ˆ;

222

2

2

2

2

2

re

rr

==−

γγ

γ

η-μ fading distributions

First case, k-μ fading distribution, extension of Rice fading channel

• consideration of μ clusters instead of single clusters

Second case, we have considered the case of

• where instead of taking square root of the inquadrature component of the fading signal

• we have taken α-th root, extension of • we have taken α-th root, extension of

One more extension is possible

• where we assume the variance of the incomponents are different

• It is the third case of η-µ fading distributionNakagami-m fading channel



extension of Rice fading channel

clusters instead of single clusters

Second case, we have considered the case of α-μ fading distribution

where instead of taking square root of the in-phase and component of the fading signal

extension of Weibull fading channelextension of Weibull fading channel

where we assume the variance of the in-phase and quadrature

fading distribution, extension of




μ is number of clusters

η is defined as the

• ratio of power of the in-phase component to power of phase component of the receivephase component of the receivephase and quadrature components are uncorrelated)

• Correlation coefficient of the incomponents in format II (assuming incomponents are correlated)



phase component to power of quadratureeived signals in format I (assuming in-eived signals in format I (assuming in-

components are uncorrelated)

Correlation coefficient of the in-phase and quadraturecomponents in format II (assuming in-phase and quadrature




η-µ fading distribution is similar to

• it is assumed that the multi-pathare in the form of several clusters and

the clusters does not have any domithe clusters does not have any domidistribution

However, the parameter η makes it different from as mentioned before



( )2 2

2

1

n

i ii

X I Q=

= + ∑

similar to Nakagami-m fading model,

path components in received signal are in the form of several clusters and

omina`ng or LOS component in η − μ omina`ng or LOS component in η − μ

However, the parameter η makes it different from Nakagami-m fading



( )2 2

i iX I Q

= +


In η −μ fading, the varia`on from Nakagami

that the variance which is same as the power content of Idifferent

( ) ( )2 2 2 2;i I i Q

E I E Qσ σ= =

The probability density function (pdf



( ) ( );i i

i I i QE I E Qσ σ= =

( )

( ) ΩΓ

=−

+

−

2

1

22

1

4 ll

H

hp

µ

αµπα

µ

µµµ

α µη

Nakagami-m fading is

that the variance which is same as the power content of Ii and Qi is

2 2 2 2

i I i Qσ σ= =

pdf) of η-μ distributed RV is given by



i ii I i Q

σ σ= =

ΩΩ

−+

Ω−

2

2

1

2

1

2

2

2

ll

l

h

HI

el

l

αµ

µµ

αµ


H and h are the functions of fading parameter η

• which is the fading parameter defined in two ways and

• thus there are two formats for η − μ

Format IFormat I

In this format, the in-phase component and component of the resultant signal in each cluster are

• assumed to be independent and with different powers



are the functions of fading parameter η

defined in two ways and

η − μ fading channels

phase component and quadrature phase component of the resultant signal in each cluster are

assumed to be independent and with different powers




η is defined as the ratio of power of in phase component to the power of quadrature component, i.e.

∈The parameter η ∈ (0, ∞) is also assfor all the clusters in the received signal



( )η

η

η

η −=

+=

4

1;

4

12

Hh

η is defined as the ratio of power of in phase component to the component, i.e. ( )

( )

2 2

22

i

i

i I

Qi

E I

E Q

ση

σ= =

assumed that this ratio is constant for all the clusters in the received signal



( )ii

η

η

η

η

+

−=

1

1;

2

h

H


Since

The pdf has Iʋ which is function of H and it is symmetric for H=0

( ) ( ) ( )1I z I zν

ν ν− = − ( )I zν

=

(−

p αα µη

Hence the distribution is symmetric around

• therefore power distribution maythe regions

It can be shown that the values of h= 1, i.e.

• the values of H and h for 0 < η ≤ 1



which is function of H and it is symmetric for H=0

( )

2

0

1

21!

i

i

z

i i

ν

ν

+∞

=

=

Γ + + ∑

)

( )

ΩΩΓ

=−+−

Ω−+

2

2

1

2

1

2

1

2

22

1

24

2

ll

l

h

ll

HI

H

ehl

l

αµ

µ

αµπα

µµµ

αµ

µµµ

Hence the distribution is symmetric around η=1 since for η=1, H=0,

may be considered only within one of

of h and H are symmetrical around η

0 < η ≤ 1 are same for the range 1 ≤ η < ∞




Format II

assumption made is that the in-phase and components of MPCs within each cluster are

• correlated and

∈

• correlated and

• have same powers

The parameter η ∈ (-1, 1) is the corrcomponents



( )( )

(2 2

, ,i i i i

i i

E I Q E I Q

E Q E Iη = =

phase and quadrature phase components of MPCs within each cluster are

correlation coefficient between these



)( )2 2

, ,i i i i

i i

E I Q E I Q

E Q E I


It is also assumed that the

• correlation coefficient between in phase component and the quadrature component is same

• for all the clusters in the received signal• for all the clusters in the received signal

It can be shown that the values of h= 0 i.e. H=0

• (0 ≤ η < 1 or −1 < η ≤ 0 needs to be considered)



21

1

η−=h

21 η

η

−=H

H

hη=

correlation coefficient between in phase component and the component is same

for all the clusters in the received signalfor all the clusters in the received signal

of h and H are symmetrical around η

needs to be considered)




Relation between format I and format II can be obtained by

equating the ratio H/h of both the formats

1 ηη

−=

Special cases

η−μ distribution for format I, differs from in only one parameter

the different variance of in-phase components and components of resultant of each cluster in the received signal



1formatI

ηη

=+

Relation between format I and format II can be obtained by

equating the ratio H/h of both the formats

formatIIη−

distribution for format I, differs from Nakagami-m fading model

phase components and quadrature phase components of resultant of each cluster in the received signal



formatII

formatIIη+


μ = n/2 but it constraints the values of

• μ to be discrete on the account of discrete values of n

For η=1 and μ=m/2, it gives the Nakagami

( )

μ=0.5 and η=1 gives Rayleigh distribution



( )( )

2 2

22

i

i

i I

Qi

E I

E Q

ση

σ= = =

( ) ( )2 2

X I Q= +

n/2 but it constraints the values of

μ to be discrete on the account of discrete values of n


=1 gives Rayleigh distribution



1= = =

)2 2

X I Q


Hoyt fading distribution (η= q2 and fading

( ) (2 2

X I qQ= +

The pdf of the instantaneous signalη-µ distribution is



( ) (

( )

( )µ

µπ

µ

µµ

γ µη

H

hxp

Γ

=

+

−

2 2

1

and μ = 0.5) also called as Nakagami-q

)2 2

X I qQ

of the instantaneous signal-to-noise ratio (SNR) of the



)2lαγ =

( )γγγ

µ

γµµµ

γ

µµ

µ

EHx

Iex

xh

=

−+−

−−

;2

2

1

2

1

2

1

2

2

1


The mgf of instantaneous SNR for η

Special cases

( ) ( ) ( )γγγ

γ γγµηµη

=== ∫∞

−−

−−dpeeEsM

ss

0

Special cases

Rayleigh fading



( )5.0,1 == µη

( )γµηγ

ssM

+=

== 1

15.0,1

+N. Ermolova, “Moment generating functions of the generalized

their applications to performance evaluations of communication systems,”

Letters, vol. 12, no. 7, pp. 502-504, 2008.

η -μ fading distribution+ is given by

( )( ) ( )( )

µ

γµγµ

µ

+++−=

sHhsHh

h

22

42



, “Moment generating functions of the generalized η-μ and k-μ distributions and

their applications to performance evaluations of communication systems,” IEEE Communications


Nakagami-m

• For format 1, η 1, implies

==

2,1

mµη

( )

• For format 2, η 0 implies



( )4

1;1

4

122

−==

+=

η

η

η

ηHh

01

;11

122

=−

==−

=η

η

ηHh



0=

0


Nakagami-q

( )m

sm

msM

m

+=

== γµηγ

2,1

( )5.0,2 == µη qNakagami-q



( )5.0,2 == µη q

( )( )( ) ( )(5.0,

+++−=

= γµηγHhsHh

hsM



)

5.0

γs


For format 1,

( ) ( ) 2

2

222

4

1;

4

1

4

1H

q

qh =

−=

+=

+=

η

η

η

η



( )

( )

2 0 5

22

2

2 2

2

1

4

1 1

2 2

, .q

q

qM s

q qs s

q

η µγ

γ γ= =

+

= = + +

+ +

2

22

2

4

2

1;

2

1;

4

1

q

qHh

qHh

q

q +=+

+=−

−=



( )( )( )

0 5

02

2

2 2 2

1

1 2 1 2

.

q

q s q q ss s

γ γγ γ

+ = = + + + + + +


It is more suitable for NLOS signal propagation

Hoyt fading (or Nakagami-q) is best fit for satellite links subject to strong atmospheric scintillation



( )(15.0,2

+

=∴== µη

γ

q

sMq

Scintillation effects are because of arbitrary refraction

small-scale fluctuations in air density

gradients

It is more suitable for NLOS signal propagation

q) is best fit for satellite links subject to



) ( )5.0

22222

2

421

1

++

+

γγ sqsq

q

Scintillation effects are because of arbitrary refraction caused by

air density due to temperature


function x=eta_mu_channel(eta,mu,Nr,Nt

coeff=sqrt(eta);

ni=0;

nq=0;nq=0;

for j=1:2*mu

• norm1=randn(Nr,Nt);





eta,mu,Nr,Nt);

1,m

Nakagami η µ

= =




12

,m

Nakagami η µ

= =


• ni=ni+norm1.ˆ2;

• norm2=coeff*randn(Nr,Nt);

• nq=nq+norm2.ˆ2;

end

h_abs=(sqrt(ni+nq))/sqrt((2*mu*(1+eta)));

theta = 2*pi*rand(Nr,Nt);

x=h_abs.*cos(theta)+sqrt(−1)*h_abs



( )( )

2 2

22

i

i

i I

Qi

E I

E Q

ση

σ= =

((2*mu*(1+eta)));

h_abs.*sin(theta);



MIMO I-O system model

Fig. 11 2×2 MIMO systemRakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


2 MIMO system, Fundamentals of MIMO Wireless



Received signal at antenna 1

y1=h11x1+h12x2+n1

Received signal at antenna 2

y =h x +h x +ny1=h21x1+h22x2+n2

In matrix form,1 11 12 1

2 21 22 2

y h h x

y h h x

=



1 11 12 1

2 21 22 2

y h h x

y h h x




For NT ×NR MIMO systems, I-O model is given by

y=Hx+n

where

y n h h h x 1 1 11 12 1 1

2 2 21 22 2 2

1 2

; ; ;

R R R R R T TN N N N N N N

y n h h h x

y n h h h x

y n h h h x

= = = =

y n H xM M M O N M M



O model is given by

y n h h h x K1 1 11 12 1 1

2 2 21 22 2 2

1 2

; ; ;

T

T

R R R R R T T

N

N

N N N N N N N

y n h h h x

y n h h h x

y n h h h x

= = = =

y n H x

K

L

M M M O N M M

L




Notation:

• hij, i is the receiver antenna indexindex

• Bold face small letters are used for representing vectors• Bold face small letters are used for representing vectors

• Bold face capital letters are used for representing matrices



ndex and j is the transmitter antenna

Bold face small letters are used for representing vectorsBold face small letters are used for representing vectors

Bold face capital letters are used for representing matrices



Analytical MIMO channel models

Some of the analytical MIMO channel models are:

• iid (uncorrelated) MIMO channel model

• fully correlated MIMO channel model

• separately correlated MIMO channel model• separately correlated MIMO channel model

• Uncorrelated keyhole MIMO channel model



Analytical MIMO channel models

Some of the analytical MIMO channel models are:

(uncorrelated) MIMO channel model

fully correlated MIMO channel model

separately correlated MIMO channel modelseparately correlated MIMO channel model

Uncorrelated keyhole MIMO channel model



iid MIMO channel model

In MIMO communication, the channel

• whose elements are hij, i=1,2,..,N

A complex RV Z=X+jY, a pair of real RVs X and Y

The pdf of a complex RV, the joint pdfThe pdf of a complex RV, the joint pdf

pdf of a complex normal RV

A complex normal RV (Z=X+jY) is a complex RV

• whose real (X) and imaginary (Y) parts are mean and variance ½



In MIMO communication, the channel H is a matrix

=1,2,..,NR, j=1,2,…,NT complex RVs

, a pair of real RVs X and Y

pdf of its real and complex partspdf of its real and complex parts

) is a complex RV

whose real (X) and imaginary (Y) parts are i.i.d. Gaussian with zero




From probability theory:

Independent and non-identically distributed (

RVs X1,X2,…,XN are iid if for all x1,x2,…,

( ), , 1 1 2, ,X X N X X X Np x x p x p x p x=L L L

Independent and identically distributed (

RVs X1,X2,…,XN are iid if for all x1,x2,…,

( )1 , , 1 1 2, ,

NX X N X X X Np x x p x p x p x=L L L

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

N NX X N X X X Np x x p x p x p x=L L L

+A. Papoulis and S. U. Pillai, Probability, Random Variables and Stochastic Processes

Tata McGraw Hill, 2002.



identically distributed (i.i.n.d.) RVs:

,…,xN

( ) ( ) ( ), , 1 1 2X X N X X X Np x x p x p x p xL L

Independent and identically distributed (i.i.d.) RVs+:

,…,xN

( ) ( ) ( ), , 1 1 2X X N X X X Np x x p x p x p xL L

( ) ( ) ( )1 1 2, , 1 1 2N NX X N X X X Np x x p x p x p xL L

Probability, Random Variables and Stochastic Processes,




For example, hij, i=1,2,..,NR, j=1,2,…,N

( ) 1

/

/

; ~ ,real imag real imag

ij ij ij ij

real imag

h h jh h N

p h e

= +

=

Rayleigh distributed

( ) 1

12

2

/real imag

ijp h e

π

=



( real imag

ij ij ijh h h= +

, j=1,2,…,NT are complex normal RV

( )2

12

2

10

2

/

/; ~ ,

real imagij

real imag real imag

h

h h jh h N

p h e

−

22p h e



) ( )2 2

real imag

ij ij ijh h h= +


( )0 1~ ,ij C

h N

complex normal RV hij has pdf as

( ) ( ) ( )

( )( ) ( )

2 2

1 1

2 2

1 1real imagij ij

real imag

ij ij ij

h h

ij

p h p h p h e e

p h e e

π π

π π

− +

= =

= =



( ) ( )2 2

real imagij ij

h h( ) ( )

2

1 12 2

2 21 1

1 12 2

2 2

ij ij

ij

h h

h

p h p h p h e e

p h e e

π π

− −

−




• A random matrix is a matrix whose elements are RVs

• If the elements of random matrix are complex RVs, then it is a complex random matrix

• A random matrix can have joint pdf

• For iid MIMO channel model, all elements of the channel matrix i=1,2,..,NR, j=1,2,…,NT) are iid complex normal RVs also called as Rayleigh MIMO fading channel

• For example, a complex normal matrixelements are complex normal RVs

+T. W. Anderson, An introduction to multivariate statistical analysis

3rd edition, 2003.Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


A random matrix is a matrix whose elements are RVs

If the elements of random matrix are complex RVs, then it is a

pdf of its elements+

MIMO channel model, all elements of the channel matrix H (hij

complex normal RVs also called as iid

complex normal matrix H is a random matrix whose elements are complex normal RVs

An introduction to multivariate statistical analysis, John Wiley & Sons,




For H an iid normal matrix, the pdfcomplex normal RVs, hij, i=1,2,..,NR

( )2 2

1 1 1, ,

R T R TN N N N

h h− −= = =∏ ∏

Note that trace for a square matrix is equal to the sum total of its diagonal elements

HHH is a square matrix whose trace is

( )1 1

1 1 1, ,

, ,

R T R T

ij ij

R T R T

h h

N N N Ni j i j

p e e eπ π π

− −

= =

= = =∏ ∏HH



pdf is the multiplication of pdfs of

R, j=1,2,…,NT

22 2

1 1 1

,

,

, ,

N NR T

R T R T ijN N N N h

h h−

− − ∑= = =∏ ∏

Note that trace for a square matrix is equal to the sum total of its

is a square matrix whose trace is

1

1 1

1 1 1

,

,

, ,

, ,

R T R T ijij ij i j

R T R T

hh h

N N N Ni j i j

p e e eπ π

=

−− −

= =

∑= = =∏ ∏




( )

11 12 1 11 21 1

21 22 2 12 22 2

1 2 1 2

22 2

T R

T R

R R R T T T R T

H

N N

N N

N N N N N N N N

trace

h h h h h h

h h h h h htrace

h h h h h h

=

+ + +

HH

L L

L L

M O O M M O O M

L L

22 2

11 12 1

2 2

21 22 2

TNh h h

h h htrace

+ + +

+ + +=

L L L L

L L L L

M O O M

L L L

,2

,

, 1

R TN N

i j

i j

h

=

= ∑Rakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


*

11 12 1 11 21 1

21 22 2 12 22 2

T R

T R

R R R T T T R T

N N

N N

N N N N N N N N

h h h h h h

h h h h h h

h h h h h h

L L

L L

M O O M M O O M

L L

22 2

21 22 2

2

1

T

R

N

N N

h h h

h h

+ + +

+

L L L L

L L L L

M O O M

L L L2 2

2R R TN Nh

+ +

L




Hence the pdf of the normal matrix

( ) ((1exp

R TN Np Trace

π= −H H HH

Short hand notation of exponential (trace) =

( ) (1

R TN Np etri

π= −H H HH



of the normal matrix H is

( ))Hp TraceH HH

Short hand notation of exponential (trace) =etri

)H= −H HH




All the components of H, hij are assu

Assume that the path gains are identically distributed RV

From probability theory

The covariance and correlation of two RVs X and Y is defined asThe covariance and correlation of two RVs X and Y is defined as

( ) ( )(, X YCov X Y E X Yµ µ = − −

( ) [ ],Cor X Y E XY=

+R. D. Yates and D. J. Goodman, Probability and Stochastic Processes

Sons, 2005.



assumed independent (uncorrelated)

Assume that the path gains are identically distributed RV

The covariance and correlation of two RVs X and Y is defined as+The covariance and correlation of two RVs X and Y is defined as+

)X Yµ µ = − −

( ) ( ), , X YCov X Y Cor X Y µ µ= −

Probability and Stochastic Processes, John Wiley and




For zero mean RVs X and Y,

The two RVs X and Y are

orthogonal if

( ) (, ,Cov X Y Cor X Y=

orthogonal if

uncorrelated if

Note that uncorrelated means covariance is zero

( ), 0Cor X Y =

( ), 0Cov X Y =



), ,Cov X Y Cor X Y

Note that uncorrelated means covariance is zero




If X and Y are independent, then

which implies that

Cor X Y

( ), , 0Cov X Y Cor X Y= − =

A Gaussian random vector X has independent components covariance matrix is diagonal matrix

A normal Gaussian random vector covariance matrix is an identity matrix



( ), X YCor X Y µ µ=

( ), , 0X YCov X Y Cor X Y µ µ= − =

has independent components iffcovariance matrix is diagonal matrix

A normal Gaussian random vector X has independent components iffcovariance matrix is an identity matrix




For iid MIMO channel matrix H, the covariance matrix matrix

Example,

uncorrelated (i.i.d.) fading MIMO channel model for a 2uncorrelated (i.i.d.) fading MIMO channel model for a 2system

Show that the covariance matrix is

Let us consider a 2×2 MIMO systemchannel matrix is

=

2221

1211

hh

hhH



, the covariance matrix RH is a diagonal

.) fading MIMO channel model for a 2×2 MIMO .) fading MIMO channel model for a 2×2 MIMO

Show that the covariance matrix is I4.

tem (for illustration purpose) whose

22

12h




Assume that

we can find the covariance matrix

0; , 1,2ijE h i j= =

“vect” (vectorization) stacks all the cvector

“Unvec” converts back a vectorizedmatrix



we can find the covariance matrix RH as ; ( )HH E vect= =R hh h H

the columns of a matrix into a column

vectorized matrix into its corresponding




We can “vect” (vectorize) the above system and find the covariance matrix

h

( ) ( ) [ ]H

H

h

hE vect vect E h h h h

h

h

= =

R H H



) the above 2×2 H matrix for 2×2 MIMO system and find the covariance matrix RH as follows

h

[ ]

11

*21

11 21 12 22

12

22

h

hE vect vect E h h h h

h

h




2

11 11 21 11 12 11 22

* * *

21 11 21 21 12 21 22

[ ]H

E h E h h E h h E h h

E h h E h E h h E h h

E

= =R hh

* * *

12 11 12 21 12 12 22

* * *

22 11 22 21 22 12 22

[ ]H

HE

E h h E h h E h E h h

E h h E h h E h h E h

= =

R hh



* * *

11 11 21 11 12 11 22

2* * *

21 11 21 21 12 21 22

E h E h h E h h E h h

E h h E h E h h E h h

2* * *

12 11 12 21 12 12 22

2* * *

22 11 22 21 22 12 22

E h h E h h E h E h h

E h h E h h E h h E h




Note that h11, h12, h21 and h22 are all mutually independent and identically distributed (uncorrelated) RVs with zero mean

2

11E h

( ) ( )

11

0 0 0

[ ]

0 0 0

0 0 0

H

E h

E vect vect

=

H H



are all mutually independent and identically distributed (uncorrelated) RVs with zero mean

0 0 0

2

21

2

12

2

22

0 0 0

0 0 0

0 0 0

0 0 0

E h

E h

E h




If we choose

then

2 2 2 2

11 12 21 22E h E h E h E h = = = =

( ) ( )

1 0 0 0

0 1 0 0[ ]

H

E vect vect

=H H

This analysis can be easily done for any arbitrary Nand show that

( ) ( ) 0 1 0 0

[ ]0 0 1 0

0 0 0 1

H

E vect vect =

H H

T RH N N=R I



2 2 2 2

11 12 21 22 1E h E h E h E h = = = =

1 0 0 0

0 1 0 0

This analysis can be easily done for any arbitrary NT×NR MIMO system

0 1 0 0

0 0 1 0

0 0 0 1




An important matrix we will frequenis the random complex Wishart matrix which is defined as

≥

<=

H

RH

NN

NN

,

,

HH

HHQ

where H is the random MIMO channel matrix

From spectral theorem

Q is a random matrix

Λ is also a random matrix

≥=

RH

NN,HHQ

=Q U Λ



quently use in MIMO capacity analysis matrix which is defined as

T

N

N

is the random MIMO channel matrixTN

H

Λ U




1

2

0 0

0 0

0 0m

λ

λ

λ

= =

Λ

L

L

M M O M

L

Marginal distribution of an eigenvaluematrix+

( ) ( ) ( )( )∑∑∑

−

= = =− +−

−=

1

0 0

2

021

!!!2

!211m

i

i

j

j

lli

l

jmnlj

j

mp λ

+E. Biglieri, Coding for Wireless Channels, Springer, 2005.



; min ,R T

m N N

= =

M M O M

,max TR NNn =

eigenvalue (λ1) of complex Wishart

( ) −−+

−

−+

−

−1

1

2

22222 mnle

lj

mnj

ji

ji λλ

; min ,

, Springer, 2005.



,max TR NNn =


Fig. 12 NT×NR MIMO systemRakhesh Singh Kshetrimayum, Fundamentals of MIMO Wireless


MIMO system, Fundamentals of MIMO Wireless


Fully correlated MIMO channel model

In practice, MIMO channel is never uncorrelated

In fully correlated MIMO channel model

we need to consider all the

• co- and cross-correlations between all the channel coefficients • co- and cross-correlations between all the channel coefficients

for various channel paths between the transmitting antennas and receiving antennas

For a given H matrix, the correlation matrix



+ N. Costa and S. Haykin, Multiple-input multiple

Sons, 2010.

; ( )HH E vect= =R hh h H


In practice, MIMO channel is never uncorrelated

In fully correlated MIMO channel model

correlations between all the channel coefficients correlations between all the channel coefficients

for various channel paths between the transmitting antennas and

matrix, the correlation matrix+ can be defined as



input multiple-output channel models, John Wiley &

; ( )E vect= =R hh h H


11

1

12 * * * * *11 1 12 2

2

* * * * *

NR

H NR N N NR R T

NR

N NR T

h

h

hE h h h h h

h

h

=

R

M

L L LM

M



* * * * *11 11 11 1 11 11 1112 2

* * *1 11 1 1 1 112

NR N N N

N N N N NR R R R R N

h h h h h h h h h h

h h h h h h h h

E=

L L L

M O M M O M O M

L L

* * * * *12 11 12 1 12 12 1212 2

* * * * *2 11 2 1 2 2 212 2

* * * * *11 1 12 2

NR N N N

N N N N N NR R R R R N R N N

N N N N N N N N N N NR T R T R R T R T N R T N N

h h h h h h h h h h

h h h h h h h h h h

h h h h h h h h h h

L L L

M O M M O M O M

L L L

M O M M O M O M

L L L


* * * * *

12 2

* * * * *

R N N NR R TE h h h h h

L L L



* * * * *11 11 11 1 11 11 1112 2

1 11 1 1 1 1

R N N NR R T

N N N N NR R R R R N

h h h h h h h h h h

h h h h h h h h

L L L

M O M M O M O M

* *12

* * * * *12 11 12 1 12 12 1212 2

* * * * *2 11 2 1 2 2 212 2

* * * * *

12 2

NR N NR R T

R N N NR R T

N N N N N NR R R R R N R N NR R T

N N N N N N N N N N NR T R T R R T R T N R T N NR R T

h h

h h h h h h h h h h

h h h h h h h h h h

h h h h h h h h h h

L

L L L

M O M M O M O M

L L L

M O M M O M O M

L L L


mean vector m and covariance matrix

How do you calculate them?

they can be estimated as

This covariance matrix size and consof the covariance matrix become prohibitively large

• as the number of transmitting and receiving antennas increase



( )( )∑=

=−−=N

i

H

iiN 1

ˆ;ˆˆ1ˆ mmxmxΦ


and covariance matrix Φ are unknown

consequently the number of elements of the covariance matrix become prohibitively large

as the number of transmitting and receiving antennas increase



∑=

N

i

iN 1

1x


Let us take an example to find out this

Consider a 2×3 MIMO system as follows:

Vectorize it 11

h

h



21

31

12

22

32

h

h

h

h

h

=

h


Let us take an example to find out this

3 MIMO system as follows: 11 12

21 22

31 32

h h

h h

h h

=

H



31 32


Covariance matrix

( )

11

21

31

12H

h

h

h

hE h h h h h h

h

=

hh

How many components we need tomatrix? 6×6



( )22

32

h

h


11 21 31 12 22 32

*

E h h h h h h

d to calculate for finding covariance



11 21 31 12 22 32


For NT×NR MIMO system,

Obviously the above correlation matrix will have

For example, you consider a slightly larger MIMO system

10×10 MIMO system, we will have 10,00010×10 MIMO system, we will have 10,000

Not manageable

How do one reduce this number of components calculation in the covariance matrix?




Obviously the above correlation matrix will have (NRNT)2 components

For example, you consider a slightly larger MIMO system

10 MIMO system, we will have 10,000 components10 MIMO system, we will have 10,000 components

How do one reduce this number of components calculation in the



Separately correlated (Kronecker

Finite correlation exist between ante

• But transmitter and Receiver are at very far distance

• We may assume receiver antenntransmitter antenna correlation and vice versatransmitter antenna correlation and vice versa

We can separate the transmitter andwrite



X X

H T

T RE vect= = ⊗ =

HR hh R R h H

Kronecker) MIMO channel model

antennas because of limited spacing

But transmitter and Receiver are at very far distance

enna correlation is independent of transmitter antenna correlation and vice versatransmitter antenna correlation and vice versa

r and receiver antenna correlation and



; ( )X XT R

E vect= = ⊗ =R hh R R h H


Kronecker product:

Let A be an N×M matrix and B be an L

The Kronecker product of A and Bmatrixmatrix

It can be obtained as



=⊗

B

B

BA

NA

A

L

OM

L

1

11


be an L×K matrix

represented by is an NL×MK ⊗A B



B

B

NM

M

A

A

M

1


11 12

21 22

a a

a a

=

A

11 12

21 22

b b

b b

=

B

a b a b a b a b

Example:



11 11 11 12 12 11 12 12

11 12 11 21 11 22 12 21 12 22

21 22 21 11 21 12 22 11 22 12

21 21 21 22 22 21 22 22

a b a b a b a b

a a a b a b a b a b

a a a b a b a b a b

a b a b a b a b

⊗ = =

B B

B BA B


11 12

21 22

b b

b b

a b a b a b a b



11 11 11 12 12 11 12 12

11 12 11 21 11 22 12 21 12 22

21 22 21 11 21 12 22 11 22 12

21 21 21 22 22 21 22 22

a b a b a b a b

a a a b a b a b a b

a a a b a b a b a b

a b a b a b a b


How to calculate receiver and transmitter correlation matrices?

The correlation matrices at the transmitter and the receiver are calculated as

( ) ( )T

Example:

Let us consider a 2×2 MIMO system for illustration purpose whose channel matrix is



( ) ( )* ;T

H T HE E E

= = = X XT RR H H H H R HH

=

2221

1211

hh

hhH


How to calculate receiver and transmitter correlation matrices?

The correlation matrices at the transmitter and the receiver are

( )

2 MIMO system for illustration purpose whose



( );H T HE E E= = =

X XT RR H H H H R HH


The transmitter correlation matrix (of H) is given by

*

11 21 11 12[ ]

T

H T h h h hE E

h h h h

= = = XTR H H



12 22 21 22

2 2 * *11 21 12 11 22 21

*11 12

h h h h

E h h E h h h h

E h h h

+ + =

+

XT

2 2*21 22 12 22h E h h

+


The transmitter correlation matrix (correlation between the columns

2 2 * *11 21 11 12 21 22

2 2* *

TT

E h h E h h h h

E h h h h E h h

+ + = = = + +



2 2* *12 11 22 21 12 22E h h h h E h h

+ +


The receiver correlation matrix (correlation between the rows of given by

11 12 11 21[ ]H h h h h

E Eh h h h

= = = XRR HH



21 22 12 22

[ ]E Eh h h h

= = =

XRR HH


correlation between the rows of H) is

2 2 * **

11 12 11 21 12 22E h h E h h h h + + = = =



2 2* *21 11 22 12 21 22E h h h h E h h

= = = + +


In total in order to find

we require to find elements only

For example, 10×10 MIMO system, we will have 200

How to introduce correlation in an otherwise

X X

T

T R= ⊗HR R R

( )2 2R TN N+

How to introduce correlation in an otherwise channel matrix Hw?

Define

What happens to covariance matrix



( H wunvec=H R h


In total in order to find

elements only

10 MIMO system, we will have 200 components only

How to introduce correlation in an otherwise iid or spatially white

X XT RR R R

How to introduce correlation in an otherwise iid or spatially white

trix when we multiply to ?



)H wH R h

HR w

h


Hermitian square root for the square root of the matrix last operation

R T

H H

H H w w H

H

H N N H H H H

E E = =

= = =

R hh R h h R

R I R R R R

last operation

Correlation matrix of iid hw has identity matrix

But h now has covariance matrix as

We have introduced correlation matrix




square root for the square root of the matrix RH due to the

( )1/2 /2

HH H

H H w w H

H

H N N H H H H

E E

= = =

R hh R h h R

R I R R R R

has identity matrix

now has covariance matrix as RH

We have introduced correlation matrix RH into h from hw




We have introduced the correlation in an otherwise white channel matrix Hw

It means that the correlation matrix for It means that the correlation matrix for matrix for h becomes RH

What happens to channel matrix H

• For the correlation matrix RH above,

• let us find the expression for correlated channel matrix




We have introduced the correlation in an otherwise iid or spatially

It means that the correlation matrix for hw is I but the correlation It means that the correlation matrix for hw is I but the correlation

H?

above,

let us find the expression for correlated channel matrix H




in terms of

• receiver correlation,

• transmitter correlation and

• spatially white channel matrix spatially white channel matrix

Using the identity



( ) == wH unvecunvec hRH

(( 2 1 2/ /

x x

T

T R wunvec vect= ⊗H R R H

( ) ( )vect vect⊗ =A B C BCA




⊗ wR

TT Xx

hRR

))T R wH R R H

( )Tvect vect⊗ =A B C BCA


Take

channel matrix H for Kronecker channel

( 1 2 1 2 1 2 1 2/ / / /

x x x xR w T R w T

unvec vect= =H R H R R H R

/ 2 1/2, ,x x

T

T R w= = =A R B R C H

channel matrix H for Kronecker channel

where ()1/2 denotes the Hermitian



12/1

X TR RHRH w=

+ J. P. Kermoal, L. Schumacher, K. I. Pederson, P. E.

Stochastic MIMO Radio Channel Model With Experimental Validation,”

Selected Areas in Communications, vol. 20, no. 6, Aug. 2002, pp. 1211


channel+ model can be written as

)1 2 1 2 1 2 1 2/ / / /

x x x xR w T R w T

= =H R H R R H R

channel+ model can be written as

square root of matrix



2/1

XT

, L. Schumacher, K. I. Pederson, P. E. Mogensen and F. Frederiksen, “A

Stochastic MIMO Radio Channel Model With Experimental Validation,” IEEE Journal on

vol. 20, no. 6, Aug. 2002, pp. 1211-26.

Separately correlated (Kronecker) MIMO channel model

How do we obtain Kronecker or sepmodel from iid channel model?

• In the iid MIMO channel model

• Premultiply by Hermitian square• Premultiply by Hermitian square

• Postmultiply by Hermitian square root of correlation at the transmitter

How to find the pdf of such matrices?



) MIMO channel model

separately correlated MIMO channel

MIMO channel model

are root of correlation at the receiverare root of correlation at the receiver

square root of correlation at the

of such matrices?




All matrices can be vectorized, better find the

A complex Gaussian random vector its mean (m=E(z)) and covariance matrix [

Once we have mean and covariance matrices, we can write its Once we have mean and covariance matrices, we can write its

When x is real we write and its



+ H. Wymeersch, Iterative receiver design, Cambridge University Press, 2007*G. A. F. Seber and A. J. Lee, Linear Regression Analysis

(~ ,nRNx m

( )( ) ( )

(1 1

exp2

2 detn

p

π

= − − −

x x m

Φ


, better find the pdf for random vectors

A complex Gaussian random vector z+ is completely characterized by )) and covariance matrix [Φ=(E(z-m)(z-m)H)]

Once we have mean and covariance matrices, we can write its pdfOnce we have mean and covariance matrices, we can write its pdf

is real we write and its pdf* is given by



, Cambridge University Press, 2007.

Linear Regression Analysis, John Wiley & Sons, 2003

)~ ,x m Φ

) ( )1T − = − − −

x x m Φ x m


For example

For n=1, mean=0 and variance =1

For a complex n-multivariate Gaussian distribution with mean For a complex n-multivariate Gaussian distribution with mean and covariance matrix is denoted by

• Note that the subscript c denote

• superscript n means that it is an n

• N means it is normal or Gaussian distribution



nnC

×∈Φ


multivariate Gaussian distribution with mean

( )2

1exp

22

xp x

π

= −

n∈mmultivariate Gaussian distribution with mean and covariance matrix is denoted by

otes that it is a complex distribution

superscript n means that it is an n-multivariate distribution and

N means it is normal or Gaussian distribution



nC∈m

( )~ ,n

CNz m Φ


It is basically a complex z

• with independent imaginary andmatrix ½ Φ

( )1

Its pdf+ is given by

For example

For n=1, mean=0 and variance=1



+ A. van den Bos, “The Multivariate Complex Normal Distribution

IEEE Trans. Inform. Theory, vol. 41, no. 2, Mar. 1995, pp. 537

( )( ) (

1

detn

pπ

= − − −z z mΦ


and real parts with same covariance

( ) ( )( )1H −



, “The Multivariate Complex Normal Distribution- A Generalization,”

, vol. 41, no. 2, Mar. 1995, pp. 537-539.

)( ) ( )( )1exp

H −= − − −z z m Φ z mΦ

( ) ( )21expp z z

π= −


For NR×NT MIMO wireless channel, when we

It gives a vector with NR×NT components, , its

( )( ) ( )

1

detT RN N

H

pπ

×= −h h R h

R

For example, NR=NT=2, iid case RH=I



11

11 12 21

21 22 1211 21 12 22

22

* * * *; ; H

h

h h h

h h h h h h h

h

= = =

H h h


MIMO wireless channel, when we vectorize it

components, , its pdf is( )~ 0,R TN N

C HN×

h R

( ) ( )( )1expH

H−= −h h R h

=I4



11 21 12 22

* * * *h h h h


Kronecker MIMO channel model+ used for

( )( )

( ) ( )( ) (4 4

1 1exp exp

Hp h h h h

ππ= − = − − − −h h h

Kronecker MIMO channel model+ used for

• IEEE 802.11n and

• IEEE 802.20 (Mobile Broadband Wireless Access)




used for

)2 2 2 2

11 21 12 22exp expp h h h h= − = − − − −

used for

IEEE 802.20 (Mobile Broadband Wireless Access)




Based on the antenna array geomettypes

• Constant

• Circular• Circular

• Exponential



+ K.-L. Du and M. N. S. Swamy, Wireless Communications

2010.


metry, correlation could be of various



Wireless Communications, Cambridge University Press,


Constant correlation model is the worst case scenario

• suitable for antenna array of three antennas placed on an equilateral triangle or

• for closely spaced antennas other than linear arrays• for closely spaced antennas other than linear arrays



1

1

1tancons t

x x

x x

x x

= < <

R

L

L

M M M M

L


Constant correlation model is the worst case scenario

suitable for antenna array of three antennas placed on an

for closely spaced antennas other than linear arraysfor closely spaced antennas other than linear arrays



0 1,

x x

x x

x

= < <

M M M M

1 0 3 0 3 0 3

0 3 1 0 3 0 3

0 3 0 3 1 0 3

0 3 0 3 0 3 1

. . .

. . .

. . .

. . .X

T

=

R


Circular correlation model is appropriate for

• antennas lying on a circle, or

• four antennas placed on a square

1 L xx



=−−

−

1

1

1

*2

*1

2*

1

11

L

MMMM

L

L

xx

xx

xx

nn

n

circularR


Circular correlation model is appropriate for

four antennas placed on a square

1 0 1 0 2 0 3. . .

. . .

. . .



2

1 0 3 1 0 1 0 2

0 2 0 3 1 0 1

0 1 0 2 0 3 1

. . .

. . .

. . .

. . .X

R

=

R


Exponential correlation model is suitable model for

• equally spaced linear antenna array

−1 1

Ln

xx



( ) ( )

=

−−

−

1

1

1

2*1*

2*

exp

L

MMMM

L

L

nn

n

onential

xx

xx

xx

R


Exponential correlation model is suitable model for

equally spaced linear antenna array

2 3

2

1 0 2 0 2 0 2

0 2 1 0 2 0 2

. . .

. . .

. . .



2

2

3 2

0 2 1 0 2 0 2

0 2 0 2 1 0 2

0 2 0 2 0 2 1

. . .

. . .

. . .

. . .X

R

=

R

Keyhole MIMO channel model

Such model is appropriate for indoor wireless communication through

• corridor or

• underpass or • underpass or

• subway

Cooperative communication

• employing the amplify-and-forward protocol

may be considered as keyhole channels



Such model is appropriate for indoor wireless communication

forward protocol

may be considered as keyhole channels






Fig. 13 3×3 keyhole MIMO channel, Fundamentals of MIMO Wireless


3 keyhole MIMO channel


Let us assume that the transmitted signal vector is

= 2

1

x

x

x

x

The signal incident at the keyhole is given by



3x

[ ]1

1 2 3 2

3

left

x

y h h h x

x

= =

h x

Let us assume that the transmitted signal vector is

The signal incident at the keyhole is given by



1

1 2 3 2

3

x

y h h h x

x


The signal at the other side of the keyhole is given by

The signal vector at the receive antennas on the right side of the

1y yα=

The signal vector at the receive antennas on the right side of the keyhole is given by



1right right lefty α= =r h h h x

The signal at the other side of the keyhole is given by

The signal vector at the receive antennas on the right side of the The signal vector at the receive antennas on the right side of the



right right leftα= =r h h h x


The equivalent channel matrix can be represented as

4 1 4 2 4 3 1 4 1

5 1 5 2 5 3 2 5 1 2 3 2

6 1 6 2 6 3 3 6 3

h h h h h h x h x

h h h h h h x h h h h x

h h h h h h x h x

α α

= = =

r Hx

The rank of this channel matrix is onmultiplexing gain (multiplexing gain is equal to rank of channel



4 1 4 2 4 3

5 1 5 2 5 3

6 1 6 2 6 3

h h h h h h

h h h h h h

h h h h h h

α

=

H

The equivalent channel matrix can be represented as

[ ]4 1 4 2 4 3 1 4 1

5 1 5 2 5 3 2 5 1 2 3 2

6 1 6 2 6 3 3 6 3

h h h h h h x h x

h h h h h h x h h h h x

h h h h h h x h x

is one which implies that there is no multiplexing gain (multiplexing gain is equal to rank of channel H)




Let us do the analysis for NR×NT keyhole MIMO channel

Assume α=hleft is for the equivalent (1

1 2 Nα α α L

Assume β=hright is for the equivalent (



1 2 Nα α α L

keyhole MIMO channel

is for the equivalent (1×NT) MISO channel

β Nα α α

is for the equivalent (NR×1) SIMO channel



1

2

RN

β

β

β

M

TNα α α


The channel matrix H for keyhole M

== T

T

N

N

Tβαβαβα

βαβαβα

MOMM

L

L

22221

11211

βαH

For example for 2×2 keyhole MIMO channel



==

TRR NNNN βαβαβα L

MOMM

21

βαH

1 1 2 1

1 2 2 2

α β α β

α β α β

=

H

le MIMO channels can be obtained as

2

1

2 keyhole MIMO channel



RN


Note that α is a row vector with NT

NC(0,1)

Hence α row vector is distributed as Hence α row vector is distributed as

and similarly β is an equivalent SIMO channel (column vector with elements which are distributed as N

Hence β is distributed as



(RNCNβ ,0~

T elements which are distributed as

row vector is distributed as ( )T

NN

N Iα ,0~row vector is distributed as

is an equivalent SIMO channel (column vector with NR

elements which are distributed as NC(0,1))



( )T

T

NN

CN Iα ,0~

)RNI,


Now we can calculate the Z as

where

( )2 2

*H

H T T T HZ UV= = = = =HH βα βα βα α β α β

where



22 2 2

1 2T

NU α α α= = + + +α L

22 2 2

1 2R

NV β β β= = + + +β L

2 2

Z UV= = = = =βα βα βα α β α β




What is the distribution of U and V?

• Distribution of

• is exponential (square of Rayleigh distribution)

2

1 2, , , ,i T

i Nα = L

• is exponential (square of Rayleigh distribution)

• U and V are the sums of NT and Nrespectively

• hence they are central Chi-square distributed with 2 Ndegrees of freedom



What is the distribution of U and V?

is exponential (square of Rayleigh distribution)

, , , ,i T

i N2

1 2, , , ,j R

j Nβ = L

is exponential (square of Rayleigh distribution)

and NR independent exponential RVs

square distributed with 2 NT and 2NR




Note that pdf of Chi-square RV

• which is the sum of squares of i.i.dcommon variance σ2

with 2N degrees of freedom is given bywith 2N degrees of freedom is given by



( )( )

21 21

; 01 !

Np e

N

χ

σχ χ χ χ

−−= >

−

+ A. Paulraj, R. Nabar and D. Gore, Introduction to Space


i.i.d. zero mean Gaussian RVs with

degrees of freedom is given bydegrees of freedom is given by



; 0= >

Introduction to Space-Time Wireless

, Cambridge University Press, 2003.


Hence, for our case σ2=1/2 , N=NT

have,( )

( )11

1 !TN

U

T

p u u e uN

−= >−

1



( )( )

11

1 !RN

V

R

p v v eN

−= >−

for RV U and N=NR for RV V, we

1; 0u

p u u e u− −= >



1; 0v

p v v e ν−= >


let us consider the transformation of functions of two RVs

as Z=XY and W=Y,

then Jacobian for this transformation is

( )



( )

( )1 1

,

,

xy xyz z

x y x yz w w w y yJ y wx y x y x y

∂ ∂∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= = = = = ∂ ∂ ∂ ∂

let us consider the transformation of functions of two RVs X and Y

for this transformation is

( )



( )

( ) 0 1

xy xyy x

x y x yy yJ y w

x y

∂ ∂

∂ ∂ ∂ ∂∂ ∂= = = = =

∂ ∂


The joint pdf of Z and W after transfX and Y

( ), ,

,

, ,

,,

,

X Y X Y

Z W

z zp w p w

w wp z w

z w

= =

For the marginal density w.r.t. z, we have,



1 1

,

, ,

,,

,

z wJ

x y

( )w

zp

wzp YXZ ∫

∞

∞−

=

1,

ansformation of functions of two RVs

, ,, ,

X Y X Y

z zp w p w

w w

w

Z=XY and W=Y

, we have,



, ,

w

dwww

,


For our case, the variables are Y=U, X=V, Z=UV

The transformed variables are w=y=u

Hence, the pdf of Z when U and V are independent RVs is given

( ) dwww

zp

wzp YXZ ∫

∞

= ,

1,



( ) ( )1

Z U V

zp z p u p du

u u

∞

−∞

=

∫+ H. Shin and J. H. Lee, “Effect of keyholes on the symbol error rate of space

codes,” IEEE Comm. Lett., vol. 7, pp. 27-29, Jan. 2003.

( ) dwww

pw

zp YXZ ∫∞−

= ,,

Y=U, X=V, Z=UV

w=y=u and z=xy=uv

are independent RVs is given

dw



zp z p u p du

u u

H. Shin and J. H. Lee, “Effect of keyholes on the symbol error rate of space-time block

29, Jan. 2003.

dw


Putting the pdf of U and V, we have,

( )( )

1

0

1 1 1

1 ! 1 !TN u

Z

T R

p z u e e duu N N u

∞

− −=− −∫

Expressing the terms of z in terms of



( )

( ) ( )( )

0

1

0

1 1

R T

T R

N NT R

z e duN N u

∞

− +=

Γ Γ ∫

of U and V, we have,

( )

11 1 1

1 ! 1 !

R zNu u

T R

zp z u e e du

u N N u

−−

−

− −

Expressing the terms of z in terms of



( )

)1R

T R

zuN uz e du

− −−

z


If we assume that

1 1∞

( )( ) ( )

( )zuNN

zpN

NN

RT

z

R

TR

2

0

1

11 −∞

+−∫ΓΓ=

zx

ut ==2

,



( )( ) ( ) ( )0

1 1

R TZ N N

T R

p z e dtN N t

∞

− +=

Γ Γ ∫

( )( ) ( )

2 1 1

2 2 2

R T T R TN N N N N

Z

T R

x xp z e dt

N N

− + − −

⇒ = Γ Γ

2 ( ) 22 2

1 1 R T T xN N Ntx

− + −− −

( )due u

zu

2

2 −−



( )2 2

41

1 1

2

R T T

R T

xN N Nt

tN N

xp z e dt

− + −− −

− +

( )

22 2

41

0

2 1 1

2 2 2

R T T R T

R T

xN N N N Nt

tN N

x xp z e dt

t

∞− + − −− −

− +

∫


The nth order modified Hankel function is expressed as

MATLAB command is besselh(nu,Z)

( )2

1

0

1 1exp ; arg , Re 0

2 2 4 2

n

n n

x xK x t dt x x

tt

∞

+

= − − < >

∫

MATLAB command is besselh(nu,Z)

For our case, n=NR-NT and

Hence, the pdf of Z is



( )( ) ( )

(1

222 ; 0

T R

R T

N N

Z N N

T R

zp z K z z

N N

+−

−= ≥Γ Γ

x z=

function is expressed as

), where nu is order

( )2exp ; arg , Re 02 2 4 2

K x t dt x xπ

= − − < >

), where nu is order



)2 ; 0p z K z z= ≥

2x z=

MIMO channel parallel decomposition



Fig. 14 Parallel decomposition of a MIMO channel using

shaping




Fig. 14 Parallel decomposition of a MIMO channel using precoding and


What is precoding?

• In precoding, the input x to the ainto the input vector

; =x = Vx x V x% %

What is receiver shaping?

• In receiver shaping, we multiply the channel output



; =x = Vx x V x% %

UyUy == H~


he antennas is linearly transformed

H=x = Vx x V x

In receiver shaping, we multiply the channel output y by UH



=x = Vx x V x

( )nHxU +H


From Singular Value Decomposition (SVD) of the channel matrix have,

Hence,

HVUH ∑=

Hence,

Since U and V are unitary matrices (



( VΣUUy =⇒ ~ H

nxΣy ~~~ +=⇒


From Singular Value Decomposition (SVD) of the channel matrix H, we

are unitary matrices (UHU=VHV=I)



)nxVV +~H


y

y

000

000~

~

2

1

2

1

σ

σ

It may be good to write the above matrices component

in order to interpret clearly



=

+

H

R

H

H R

N

R

R

y

y

y

y

0000

0000

000

000

000

~

~

~

~2

1

2

OMMMM

O

M

M

σ

σ


n

n

x

x~

~

~

~

00

00

2

1

2

1

It may be good to write the above matrices component-wise



+

++

R

H

H

T

H

H

N

R

R

N

R

R

n

n

n

n

x

x

x

x

~

~

~

~

~

~

~

~

00

00

00

00

00

1

2

1

2

M

M

M

M

MO


HHHH RRRR nxy

nxy

nxy

~~

~~~

~~~

~~~

2222

1111

+=

+=

+=

MOM

σ

σ

σ



RR

HH

NN

RR

ny

ny

~~

~~11

=

= ++

MOM

We can use RH parallel Gaussian channels, R

channel matrix




parallel Gaussian channels, RH is the rank of the


Hence in order to get U and V matrices,

• we need the channel state informshaping) and transmitter (for transmitter

• Such MIMO systems are called Closed loop MIMO system • Such MIMO systems are called Closed loop MIMO system

It may be noted that the product of vector does not modify the noise distribution




matrices,

formation at the receiver (for receiver shaping) and transmitter (for transmitter precoding)

Closed loop MIMO system Closed loop MIMO system

t of a unitary matrix with the noise vector does not modify the noise distribution




How to find the singular matrix of H

• find the eigenvalues λi of HHH

• take the square root of the eigenvalues

• Put those singular values in descending order in a diagonal matrix• Put those singular values in descending order in a diagonal matrixwhich gives singular matrix Σ=diag

How to find the U and V matrices?

• Columns of U are the eigenvectors of

• Columns of V are the eigenvectors of




H?

eigenvalues gives the singular values σiPut those singular values in descending order in a diagonal matrixPut those singular values in descending order in a diagonal matrix

diag(σi)

are the eigenvectors of HHH

are the eigenvectors of HHH




Find the SVD of a MIMO channel given as

MATLAB command “[V D]=eig(H)” will give the

++

++=

ii

ii

5443

3221H

MATLAB command “[V D]=eig(H)” will give the

• diagonal matrix D with eigenvalues

• V matrix whose columns are the eigenvectors.



−−

+=

8844.00283.05899.0

0482.04642.08070.0

i

iV D


the SVD of a MIMO channel given as

(H)” will give the (H)” will give the

eigenvalues and

matrix whose columns are the eigenvectors.



+

−−=

i

i

2631.73567.50

02631.03567.0


We can see that the

• eigenvalues and eigenvectors

of a complex H matrix may be also complex

For our case, eigenvalues of HHH can be obtained as For our case, eigenvalues of HHH can be obtained as



=

0.88710.4615-

0.0539i + 0.45830.1037i + 0.8811V


matrix may be also complex

can be obtained as can be obtained as



0.0539i

=

83.80910

00.1909D


Taking the square root and keeping in descending order of eigenvalues, we get,

=09.1547

Σ

We can also find the SVD decomposition directly



=0.43690

Σ

HVUH ∑=


Taking the square root and keeping in descending order of

We can also find the SVD decomposition directly




The SVD of H for our example is

=

9.1547

0.2469i - 0.3899-0.7160i - 0.5238-

0.5589i + 0.68890.4017i - 0.2271-H

where U and V are unitary matrices



0.2469i - 0.3899-0.7160i - 0.5238-


0.0298i - 0.59630.8022-

0.0401i + 0.8012-0.5971-

0.43690

09.1547

are unitary matrices



0.0298i - 0.59630.8022-0.43690

Power allocation in MIMO systems

SISO

• we allocate all the power to the single transmit antenna

MIMO

• We have numerous antennas at the transmitter • We have numerous antennas at the transmitter

• The fundamental question is howtransmit antennas

• Note that power allocation plays a significant role in deciding MIMO capacity (this will be discussed




we allocate all the power to the single transmit antenna

We have numerous antennas at the transmitter We have numerous antennas at the transmitter

how much power we allocate to each

Note that power allocation plays a significant role in deciding MIMO capacity (this will be discussed in later)




Open loop MIMO system:

• CSI is available at the receiver but not at the transmitter

• Uniform (equal) power allocation is employed

Closed loop MIMO system: Closed loop MIMO system:

• CSI is available at the transmitter as well as at the receiver

• we may allocate more power to better channels than the bad channels

• adaptive power allocation




CSI is available at the receiver but not at the transmitter

Uniform (equal) power allocation is employed

CSI is available at the transmitter as well as at the receiver

we may allocate more power to better channels than the bad



Uniform power allocation

From parallel decomposition of MIMO channels, we have,

Using the Shannon capacity formula fo

Hiiii Rinxy ,,2,1;~~~L=+= σ

Using the Shannon capacity formula fochannel capacity for equal power allocation is

where W is the bandwidth of the channel, P is the total power, each antenna will receive P/NT power for equal power allocation



∑∑==

+=

+=

HH

i

R

i

R

i

rW

PWC

1

2

122 1log1log

σ

From parallel decomposition of MIMO channels, we have,

la for parallel Gaussian channels, the la for parallel Gaussian channels, the channel capacity for equal power allocation is

where W is the bandwidth of the channel, P is the total power, each power for equal power allocation



∏=

+=

HR

i T

i

T

i

N

PW

N

P

1222

1logσ

λ

σ

λ


Since λi are eigenvalues of Q matrix

where λ are the eigenvectors for Q

Hiii Ri ,,2,1; L== xQx λ

where λi are the eigenvectors for Q

eigenvalues (rank of Q matrix is RH



i

T

i

T N

P

N

P22

=

xxQ λ

σσ

matrix

Q and Q has R non-zero

≥

<=

TRH

TRH

NN

NN

,

,

HH

HHQ

Q and Q has RH non-zero

H)



Hi Ri ,,2,1; L=x


Since the identity matrix has all its

T

i

T

RN

P

N

PH

12

+=

+ xQI

σσ

We also know that determinant of aof its eigenvalues



TT NN σσ

+=

+∏

=12

det1σ

λ

T

R

R

i T

i

N

P

N

P

H

H

I

Since the identity matrix has all its eigenvalues equal as 1

Hii Ri ,,2,1;2

L=

xλ

σ

of a matrix equals the multiplication



σ

2σT

PQ


Therefore the capacity formula+

+= detlogWC I



+= 2 detlog RWC

HI

+ B. Vucetic and J. Yuan, Space-time coding, John Wiley and Sons, 2003.

+PQ

2 21

1logH

R

i

i T

PC W

N

λ

σ=

= +

∏



+

2σTN

PQ

, John Wiley and Sons, 2003.

Adaptive power allocation

Usually the channel state information is available at the receiver (CSIR) using pilot signals

• If the receiver sends the CSI to thchannel, channel,

• then, the channel state information is also available at the transmitter (CSIT)

we may distribute power adaptivelyboost the spectral efficiency



Usually the channel state information is available at the receiver

to the transmitter through a feedback

then, the channel state information is also available at the

ively to individual transmit antenna to




the channel capacity may be expressed as

∑=

+=

HR

i

ii PWC

122 1log

σ

λ

where Pi is the transmit power at the

We need to maximize C by choosing P

Water-filling algorithm can be utilizethe ensuing power constraint



= i 1 σ

∑=

TN

i 1

the channel capacity may be expressed as

is the transmit power at the ith transmit antenna

We need to maximize C by choosing Pi properly

tilized in obtaining the capacity under



=i PP


Hence the capacity can be written as

Using the method of Lagrange multipliers

2 22 21 1

1 1log log ;H H

R R

i i i i i

i i

P P P PC W W

P

λ γ λ

σ σ= =

= + = + =

∑ ∑

Using the method of Lagrange multipliersor objective function as



log log ;

+

+=∑

=

HR

i

iiP

P

PF

1

2 1log ζγ

+ G. B. Arfken and H. J. Weber, Mathematical methods for physicists

2005.

Hence the capacity can be written as

Using the method of Lagrange multipliers+, let us introduce the cost

2 22 21 1

1 1log log ;H H

R R

i i i i i

ii i

P P P PC W W

P

λ γ λγ

σ σ= =

= + = + =

∑ ∑

Using the method of Lagrange multipliers+, let us introduce the cost



log log ;

−∑

=

HR

i

iP

1

Mathematical methods for physicists, Academic Press,


where is the Lagrange multiplier

The unknown transmit power Pi are determined

• by setting the partial derivative oto zero

ζ

to zero



0=idP

dF

1log 2

+

⇒i

ii

dP

P

Pd

γ

where is the Lagrange multiplier

are determined

ive of the cost or objective function F



0=

−

iPζ


Change the log2 to natural loge

( )

11

2

log

ln

i i

e i

Pd P

P

dP

γ + −

⇒



( )1

1

2ln

1=−

+

⇒ ζγ

γ P

P

P

i

ii

( ) 02ln1

=−

+

⇒ ζ

γi

i

PP

( )2

log

ln idP

0

i i

e i

Pd P

Pζ

+ −

=



0=

0


( )1 1

2ln

iP

P P γζ⇒ = −

(12ln

i

i

PP

ζ

γ

⇒ =

+

Since power allocated should be greater than or equal to zero ( Pwe have,



( )2lnP P γζ

−=⇒ i

P

P

γ

1

0

1 1

iγ

= −i

i

P

P

γγ

11

0

−=⇒

)2ln( )

1

2lni

i

PP

γ ζ⇒ + =

Since power allocated should be greater than or equal to zero ( Pi ≥ 0),



iγ

iP γγ 0

+

iγ

1


where the notation

The MIMO channel capacity may be rewritten as follows

[ ]

≤

>=

+

00

0,

k

kkk




∑∑==

=

+=

HH R

i

R

i

ii WP

PWC

1

2

1

2 1log1logγ


0

2

0:

logH

i

R

i

i

C Wγ γ

γ

γ

+

≥

=

∑




∑≥

++

=

−+

H

i

R

i

i

i

i W

0: 02

0

log11

1

γγγ

γ

γγγ


Find the spectral efficiency and optimal power distribution for the MIMO channel

++

++=

ii

ii

5443

3221H

assuming and BW=1 Hz

Solution:

The SVD of is given by



++

=ii 5443

H

dBP

52

==σ

γ

HVUH ∑=

Find the spectral efficiency and optimal power distribution for the

and BW=1 Hz

The SVD of is given by




The singular values of the channel are

=

9.1547

0.2469i - 0.3899-0.7160i - 0.5238-

0.5589i + 0.68890.4017i - 0.2271-H


Hence,



1547.91 =λ 4369.02 =λ

iiii

Pλλ

σγλγ 1623.3

2=== γ


0.0298i - 0.59630.8022-

0.0401i + 0.8012-0.5971-

0.43690

09.1547




0276.2651 =γ 6037.02 =γ


Considering that power is distributepower constraint becomes

2.66021

12

111

22

∑∑ =+=⇒=

−

the second channel is not allocated any power



2.66021

12

111

101 0∑∑

==

=+=⇒=

−i ii i γγγγ

uted to the two parallel channels, the

2.6602

the second channel is not allocated any power



2.6602

751.002 =< γγ


Then the power constraint yields

1.00381

11

111

1010

=+=⇒=−γγγγ

The capacity is given by



99624.001 => γγ

0.99624

265.0276loglog 2

0

12

=

=

γ

γC

1.0038



zbits/sec/H8.05540.99624

265.0276=

Interpretation on log2 (1+SNR) curve

Interpretation on log2 (1+SNR) curve

Low SNR regions

( )2 2 21 2 7183 1 4427log logSNR SNR e+ ≈ ≈ ≈

High SNR regions



( )2 2 21 2 7183 1 4427log logSNR SNR e+ ≈ ≈ ≈

( ) SNRSNR 22 log1log ≈+

(1+SNR) curve

(1+SNR) curve

( )2 2 21 2 7183 1 4427log log . . ( )e SNR SNR+ ≈ ≈ ≈



( )2 2 21 2 7183 1 4427log log . . ( )e SNR SNR+ ≈ ≈ ≈

Near optimal power allocation

High SNR

• noise power level is much lower than the threshold

• it is advantageous to distribute equal power to all subwith the non-zero eigenvalues

How many non-zero eigenvalues?How many non-zero eigenvalues?

• rank decides this

condition number of the channel matrix also decides the performance



( ) max

min

condσ

σ=H

noise power level is much lower than the threshold

it is advantageous to distribute equal power to all sub-channel

condition number of the channel matrix also decides the




Capacity for RH parallel Gaussian channels

In adaptive power allocation

C W W= + = +



i

i

P

P

γγ

11

0

−=⇒i

Pγ

σ=

2

0

1i

i

P

P P

σ

γ λ⇒ = −

iP⇒ = −

0

;i i threshold i

PP N P N

γ λ⇒ + = = =

parallel Gaussian channels

2 22 21 1

1 1log logH H

R R

i i i

i i

i

P PC W W

λ

σ σ

λ

= =

= + = +

∑ ∑



2

iPλ

σ2

0 i

P σ

γ λ= −

2

;i i threshold i

i

P N P Nσ

γ λ+ = = =

log log

iλ



Fig. 15 Waterfilling algorithm (adaptive power allocation)

most power into less noisy channels to make equal power +

noise in each channel



algorithm (adaptive power allocation) put

most power into less noisy channels to make equal power +



Fig. 16 Waterfilling algorithm:

and recalculate



algorithm: If Pthreshold < N4 then set P4=0



Fig. 17 Near optimal power allocation for high SNR (usually

signal power is much higher than effective noise power)



Near optimal power allocation for high SNR (usually

signal power is much higher than effective noise power)


∑∑==

≈

+=

HH R

i

R

i

iiW

PWC

122 1log

σ

λ

Equal power allocation



∑=

=

≈⇒

HR

i

H

H

i WRR

PWC

122 loglog

σ

λ

Equal power allocation

∑=

H

ii P

122log

σ

λ



∑=

+

HR

i H

i

RW

P

1

222 loglogλ

σ


Low SNR

most noise power level is high and the threshold

it is advantageous to supply power to the strongest it is advantageous to supply power to the strongest exclusively

We need to fill water of the deepest vessel (communication)

rank and condition number does not influence the performance



and will be equal to or greater than

it is advantageous to supply power to the strongest eigenmodeit is advantageous to supply power to the strongest eigenmode

We need to fill water of the deepest vessel (opportunistic

rank and condition number does not influence the performance





Fig. 17 Near optimal power allocation for low SNR

(N3,N4>Pthreshold, N1=Pthreshold)



Near optimal power allocation for low SNR

)


2 22 21 1

log 1 logH HR R

i i i i

i i

P PC W W e

λ λ

σ σ= =

= + ≈

∑ ∑



( )max

22log

PC W e

λ

σ⇒ ≈

( )2 22 21 1

log 1 logH HR R

i i i i

i i

P PC W W e

λ λ

σ σ= =

∑ ∑



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