lecture 6: modeling of mimo channels theoretical...

Lecture 6Modeling of MIMO

Channels

Lars KildehøjCommTh/EES/KTH

Lecture 6: Modeling of MIMO ChannelsTheoretical Foundations of Wireless Communications1


Wednesday, May 11, 20169:00-12:00, Conference Room SIP

1Textbook: D. Tse and P. Viswanath, Fundamentals of Wireless Communication1 / 1


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Overview

Lecture 5: Spatial Diversity, MIMO Capacity

• SIMO, MISO, MIMO

• Degrees of freedom

• MIMO capacity

Lecture 6: MIMO Channel Modeling

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Overview

Motivation:

• How does the multiplexing capability of MIMO channels depend onthe physical environment?

• When can we gain (much) from MIMO?

• How do we have to design the system?

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Physical Modeling– Line-of-Sight Channels: SIMO• Free space without scattering

and reflections.

• Antenna separation ∆rλc , withcarrier wavelength λc and thenormalized antenna separation∆r ; nr receive antennas.

• Distance between transmitterand i-th receive antenna: di

(D. Tse and P. Viswanath, Fundamentals of Wireless Communi-

cations.)

• Continuous-time impulse between transmitter and i-th receiveantenna:

hi (τ) = a · δ(τ − di/c)

• Base-band model (assuming di/c 1/W , signal BW W ):

hi = a · exp

(−j 2πfcdi

c

)= a · exp

(−j 2πdi

λc

)• SIMO model: y = h · x + w, with w ∼ CN (0,N0I)

→ h: signal direction, spatial signature.

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Physical Modeling– Line-of-Sight Channels: SIMO

• Paths are approx. parallel, i.e.,

di ≈ d + (i − 1)∆rλc cos(φ)

• Directional cosine

Ω = cos(φ) (D. Tse and P. Viswanath, Fundamentals of Wireless Communi-

cations.)

• Spatial signature can be expressed as

h = a · exp

(−j 2πd

λc

)

1exp(−j2π∆rΩ)

exp(−j2π2∆rΩ)...

exp(−j2π(nr − 1)∆rΩ)

→ Phased-array antenna.

• SIMO capacity (with MRC)

C = log

(1 +

P‖h‖2

N0

)= log

(1 +

Pa2nrN0

)→ Only power gain, no degree-of-freedom gain.

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Physical Modeling– Line-of-Sight Channels: MISO

• Similar to the SIMO case:∆t , λc , di , φ, Ω,...

• MISO channel model:

y = h∗x + w ,

with w ∼ CN (0,N0). (D. Tse and P. Viswanath, Fundamentals of Wireless Communi-

cations.)

• Channel vector

h = a · exp

(j

2πd

λc

)

1exp(−j2π∆tΩ)

exp(−j2π2∆tΩ)...

exp(−j2π(nt − 1)∆tΩ)

• Unit spatial signature in the directional cosine Ω:

e(Ω) = 1/√n · [1, exp(−j2π∆Ω), . . . , exp(−j2π(n − 1)∆Ω)]T

→ et(Ωt) and er(Ωr ) with nt ,∆t and nr ,∆r , respectively.

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Physical Modeling– Line-of-Sight Channels: MIMO

• Linear transmit and receive array with nt ,∆t and nr ,∆r .

• Gain between transmit antenna k and receive antenna i

hik = a · exp (−j2πdik/λc)

• Distance between transmit antenna k and receive antenna i

dik = d + (i − 1)∆rλc cos(φr )− (k − 1)∆tλc cos(φt)

• MIMO channel matrix (with Ωt = cos(φt) and Ωr = cos(φr ))

H = a√ntnr exp

(−j 2πd

λc

)er(Ωr )et(Ωt)

∗

→ H is a rank-1 matrix with singular value λ1 = a√ntnr

→ Compare with SVD decomposition in Lecture 5:

H =k∑

i=1

λiuiv∗i

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Physical Modeling– Line-of-Sight Channels: MIMO

• MIMO capacity

C = log

(1 +

Pa2nrntN0

)→ Only power gain, no degree-of-freedom gain.

• nt = 1: power gain equals nr → receive beamforming.

• nr = 1: power gain equals nt → transmit beamforming.

• General nt , nr : power gain equals nr · nt→ Transmit and receive beamforming.

• Conclusion: In LOS environment, MIMO provides only a power gainbut no degree-of-freedom gain.

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Physical Modeling– Geographically Separated Antennas at the Transmitter

Example/special case

(D. Tse and P. Viswanath, Fundamentals of

Wireless Communications.)

• 2 distributed transmit antennas,attenuations a1, a2, angles of incidenceφr 1, φr 2, negligible delay spread.

• Spatial signature (nr receive antennas)

hk = ak√nr exp

(−j 2πd1k

λc

)er(Ωrk)

• Channel matrix H = [h1, h2]

• H has independent columns as long as (Ωr i = cos(φr i ))

Ωr = Ωr 2 − Ωr 1 6= 0 mod1

∆r

→ Two non-zero singular values λ21, λ

22; i.e., two degrees of freedom.

→ But H can still be ill-conditioned!

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• Conditioning of H is determined by how the spatial signatures arealigned (with Lr = nr∆r ):

| cos(θ)| = |fr (Ωr2−Ωr1)| = | er(Ωr1)∗er(Ωr2)︸︷︷︸=fr (Ωr2−Ωr1)

| =

∣∣∣∣ sin(πLrΩr )

nr sin(πLrΩr/nr )

∣∣∣∣• Example (a1 = a2 = a)

λ21 = a2nr (1 + | cos(θ)|)λ2

2 = a2nr (1− | cos(θ)|) ⇒λ1

λ2=

√1 + | cos(θ)|1− | cos(θ)|

(D. Tse and P. Viswanath, Fundamentals of Wireless Commu-

nications.)

• fr (Ωr ) is periodic with nr/Lr .

• Maximum at Ωr = 0; fr (0) = 1.

• fr (Ωr ) = 0 at Ωr = k/Lr withk = 1, . . . , nr − 1.

• Resolvability 1/Lr ,if Ωr 1/Lr , then the signalsfrom the two antennas cannotbe resolved.

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Beamforming pattern

(D. Tse and P. Viswanath, Fundamentals of Wireless

Communications.)

• Assumption: signal arrives withangle φ0; receive beamformingvector er(cos(φ0)).

• A signal form any other direction φwill be attenuated by a factor

|er(cos(φ0))∗er(cos(φ))|= |fr (cos(φ)− cos(φ0))|

• Beamforming pattern

( φ, |fr (cos(φ)− cos(φ0))| )

• Main lobes around φ0 and any angle φ for which cos(φ) = cos(φ0).

→ In a similar way, separated receive antennas can be treated.

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Physical Modeling– LOS Plus One Reflected Path

• Direct path:φt1, Ωr1, d (1), and a1.

• Reflected path:φt2, Ωr2, d (2), and a2.

(D. Tse and P. Viswanath, Fundamentals of Wireless Communications.)

• Channel model follows from signal superposition

H = ab1er(Ωr1)et(Ωt1)∗ + ab2er(Ωr2)et(Ωt2)∗,

with

abi = ai√ntnr exp

(−j 2πd (i)

λc

).

→ H has rank 2 as long as

Ωt1 6= Ωt2 mod1

∆tand Ωr1 6= Ωr2 mod

1

∆r.

→ H is well conditioned if the angular separations |Ωt |, |Ωr | at thetransmit/receive array are of the same order or larger than 1/Lt,r .

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• Direct path:φt1, Ωr1, d (1), and a1.

• Reflected path:φt2, Ωr2, d (2), and a2.


• H can be rewritten as H = H′′H′, with

H′′ = [ab1er(Ωr1), ab2er(Ωr2)] and H′ =

[et∗(Ωt1)

et∗(Ωt2)

]→ Two imaginary receivers at points A and B (virtual relays).

• Since the points A and B are geographically widely separated, H′

and H′′ have rank 2 and hence H has rank 2 as well.

• Furthermore, if H′ and H′′ are well-conditioned, H will bewell-conditioned as well.

→ Multipath fading can be viewed as an advantage which can beexploited!

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• Significant angular separation is required at both the transmitterand the receiver to obtain a well-conditioned matrix H.

• If the reflectors are close to the receiver (downlink), we have a smallangular separation ⇒ not very well-conditioned matrix H.

• Similar, if the reflectors are close to the transmitter (uplink).

→ Size of an antenna array at a base station will have to be manywavelengths to be able to exploit the spatial multiplexing effect.

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Physical Modeling– Summary


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Modeling of MIMO Fading Channels– General Concept

• Antenna lengths Lt , Lr limit the resolvability2 of the transmit andreceive antenna in the angular domain.

→ Sample the angular domain at fixed angular spacings of 1/Lt at thetransmitter and 1/Lr at the receiver.

→ Represent the channel (the multiple paths) in terms of these inputand output coordinates.

• The (k, l)-th channel gainfollows as the aggregation of allpaths whose transmit andreceive directional cosines lie ina (1/Lt × 1/Lr ) bin around thepoint (l/Lt , k/Lr ).


2Note, if Ωr,t 1/Lr,t , the paths cannot be separated.16 / 1

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Modeling of MIMO Fading Channels– Angular Domain Representation (ADR)

• Orthonormal basis for the received signal space (nr basis vectors)

Sr =

er(0), er(

1

Lr), . . . , er(

nr − 1

Lr)

→ Orthogonality follows directly from the properties of fr (Ω).

• Orthonormal basis for the transmitted signal space (nt basis vectors)

St =

et(0), et(

1

Lt), . . . , et(

nt − 1

Lt)

→ Orthogonality follows directly from the properties of ft(Ω).

• Orthonormal bases provide a very simple (but approximate)decomposition of the total received/transmitted signal up to aresolution 1/Lr , 1/Lt .

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Modeling of MIMO Fading Channels– Angular Domain Representation

Examples: Receive beamform patterns of the angular basis vectors in Sr

• (a) Critically spaced (∆r = 1/2),each basis vector has a single pairof main lobes.

• (b) Sparsely spaced (∆r > 1/2),some of the basis vectors havemore than one pair of main lobes.

• (c) Densely spaced (∆r < 1/2),some of the basis vectors have nopair of main lobes.


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Modeling of MIMO Fading Channels– ADR of MIMO Channels

(Assumption: critically spaced antennas)

• Observation: The vectors in St and Sr form unitary matrices Ut andUr with dimensions (nt × nt) and (nr × nr ), respectively. (→ IDFTmatrices!)

• With3 xa = U∗t x and ya = U∗r y we get

ya = U∗r HUtxa + U∗r w = Haxa + wa, with wa ∼ CN (0,N0Inr ).

• Furthermore, with H =∑

i abi er(Ωri )et(Ωti )

∗, we get

hakl = er(k/Lr )∗Het(l/Lt)

=∑i

abi [er(k/Lr )∗er(Ωri )]︸︷︷︸(1)

· [et(Ωti )∗et(l/Lt)]︸︷︷︸(2)

• The terms (1) and (2) are significant for the i − th path if∣∣∣∣Ωri −k

Lr

∣∣∣∣ < 1

Lrand

∣∣∣∣Ωti −k

Lt

∣∣∣∣ < 1

Lt.

(→ Projections on the basis vectors in Sr , St .)

3The superscript “a” denotes angular domain quantities.19 / 1


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Modeling of MIMO Fading Channels– Statistical Modeling in the Angular Domain

• Let Tl and Rk be the sets of physical paths which have most energyin directions of et(l/Lt) and er(k/Lr ).

• hakl corresponds to the aggregated gains abi of paths which lie inRk ∩ Tl .

• Independence and time variation• Gains of the physical paths abi [m] are independent.

⇒ The path gains hakl [m] are independent across m.

• The angles φri [m]m and φti [m]m evolve slower than abi [m].

⇒ The physical paths do not move from one angular bin to another.

⇒ The path gains hkl [m] are independent across k and l .

• If there are many paths in an angular bin ⇒ Central Limit Theorem

⇒ hakl [m] can be modeled as complex circular symmetric Gaussian.

• If there are no paths in an angular bin ⇒ hakl [m] ≈ 0.

• Since Ut and Ur are unitary matrices, the matrix H has the samei.i.d. Gaussian distribution as Ha.

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Modeling of MIMO Fading Channels– Statistical Modeling in the Angular Domain

Example for Ha


• (a) Small angular spread at the transmitter.

• (b) Small angular spread at the receiver.

• (c) Small angular spread at both transmitter and receiver.

• (d) Full angular spread at both transmitter and receiver.

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Modeling of MIMO Fading Channels– Degrees of Freedom and Diversity

Degrees of freedom

• Based on the derived statistical model, we get the following result:with probability 1, the rank of the random matrix Ha is given by

rank(Ha) = min number of non-zero rows,

number of non-zero columns .

• The number of non-zero rows and columns depends on two factors:• Amount of scattering and reflection; the more scattering and

reflection, the larger the number of non-zero entries in Ha.

• Lengths Lr and Lt ; for small Lr , Lt many physical paths are mappedinto the same angular bin; with higher resolution, more paths can berepresented.

Diversity: The diversity is given by the number of non-zero entries in Ha.Example: Same number of degrees of freedom but different diversity.


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Modeling of MIMO Fading Channels– Antenna Spacing

So far: critically spaced antennas with ∆r = 1/2.

• One-to-one correspondence between the angular windows and theresolvable bins.

Setup 1: vary the number of antennas for a fixed array length Lr,t .• Sparsely spaced case (∆r > 0.5)

• Beamforming patterns of some basis vectors have multiple mainlobes.

• Different paths with different directions are mapped onto the samebasis vector.

→ Resolution of the antenna array, number of degrees of freedom, anddiversity are reduced.

• Densely spaced case (∆r < 0.5)• There are basis vectors with no main lobes which do not contribute

to the resolvability.

• Adds zero rows and columns to Ha and creates correlation in H.

Setup 2: vary the antenna separation for a fixed number of antennas.

• Rich scattering: number of non-zero rows in Ha is already nr ; i.e.,no improvement possible.

• Clustered scattering: scattered signal can be received in more bins;i.e., increasing number of degrees of freedom.

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lecture 6: modeling of mimo channels theoretical...

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