Lecture 6: MIMO Channel and Spatial Multiplexing I-Hsiang - - PowerPoint PPT Presentation

Lecture ¡6: ¡MIMO ¡Channel ¡and ¡ Spatial ¡Multiplexing

I-Hsiang Wang ihwang@ntu.edu.tw 5/1, 2014

Mutliple ¡Antennas

• Multi-Antennas so far:
• Provide diversity gain and increase reliability
• Provide power gain via beamforming (Rx, Tx, opportunistic)
• But no degrees of freedom (DoF) gain
• because at high SNR the capacity curves have the same slope
• DoF gain is more significant in the high SNR regime
• MIMO channels have a potential to provide DoF gain by

spatially multiplexing multiple data streams

• Key questions:
• How the spatial multiplexing capability depends on the physical

environment?

• How to establish statistical models that capture the properties

succinctly?

2

Plot

• First study the spatial multiplexing capability of MIMO:
• Convert a MIMO channel to parallel channel via SVD
• Identify key factors for DoF gain: rank and condition number
• Then explore physical modeling of MIMO with examples:
• Angular resolvability
• Multipath provides DoF gain
• Finally study statistical modeling of MIMO channels:
• Spatial domain vs. angular domain
• Analogy with time-frequency channel modeling (Lecture 1)

3

Outline

• Spatial multiplexing capability of MIMO systems
• Physical modeling of MIMO channels
• Statistical modeling of MIMO channels

4

5

Spatial ¡Multiplexing ¡in ¡ MIMO ¡Systems

MIMO ¡AWGN ¡Channel

• MIMO AWGN channel (no fading):
• nt := # of Tx antennas; nr := # of Rx antennas
• Tx power constraint P
• Singular value decomposition (SVD) of matrix H:
• Unitary
• Rectangular
• with zero off-diagonal elements and

diagonal elements

• These λ’s are the singular values of matrix H

6

y[m] = Hx[m] + w[m]

y[m] ∈ Cnr, x[m] ∈ Cnt, H ∈ Cnr×nt, w ∼ CN (0, Inr)

H = UΛV∗

U ∈ Cnr×nr, V ∈ Cnt×nt (UU∗ = U∗U = I) Λ ∈ Cnr×nt λ1 ≥ λ2 ≥ · · · ≥ λmin(nt,nr) ≥ 0

MIMO ¡Capacity ¡via ¡SVD

• Change of coordinate:
• Let
• , get an equivalent channel
• Power of x and w are preserved since U and V are unitary
• Parallel channel: since the off-diagonal entries of Λ are

all zero, the above vector channel consists of nmin := min{nt,nr} parallel channels:

• Capacity can be found via water-filling

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e y := U∗y, e x := V∗x, e w := U∗w y = Hx + w = UΛV∗x + w ⇐ ⇒ U∗y = ΛV∗x + U∗w

e y = Λe x + e w e yi = λie xi + e wi, i = 1, 2, . . . , nmin

Spatially ¡Parallel ¡Channels

8

V* ...

λ1 λnmin e wnmin e w1

U x y V U* e y e x x y H w H = UΛV∗

Multiplexing ¡over ¡Parallel ¡Channels

9 + AWGN coder AWGN coder {x1[m]} ~ {y1 [m]} ~ {xnmin[m]} ~ {ynmin[m]} ~ . . . . . . . . . n min information streams {0} {0} {w[m]} U* H V Decoder Decoder

P ∗

i =

✓ ν − σ2 λ2

i

◆+ , ν satisfies

nmin

X

i=1

P ∗

i = P

CMIMO =

nmin

X

i=1

log ✓ 1 + λ2

i P ∗ i

σ2 ◆

Rank ¡= ¡# ¡of ¡Multiplexing ¡Channels

10

V* ...

λ1 λnmin e wnmin e w1

U x y V U* e y e x

• If λi = 0 ⟹ the i-th channel contributes 0 to the capacity
• Rank of H = # of non-zero singular values

CMIMO =

k

X

i=1

log ✓ 1 + λ2

i P ∗ i

σ2 ◆ , k := rank (H)

Rank ¡= ¡# ¡of ¡Multiplexing ¡Channels

• DoF gain is more significant at high SNR
• At high SNR, uniform power allocation is near-optimal:
• Rank of H determines how many data streams can be

multiplexed over the channel ⟹ k := multiplexing gain

• Full rank matrix is the best (∵k ≤ nmin)

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CMIMO ≈

k

X

i=1

log ✓ 1 + λ2

i P

kσ2 ◆ ≈

k

X

i=1

log ✓λ2

i P

kσ2 ◆ = k log SNR +

k

X

i=1

log ✓λ2

i

k ◆

Condition ¡Number

• Full rank is not enough:
• If the some λi < 1, then log(λi2/k) will be negative
• How to maximize the second term?
• By Jensen’s inequality:
• For a family of full-rank channel matrices with fixed
• ,

since

• , maximum is attained

when all λ’s are equal ⟺ λmax = λmin

• Well-conditioned (smaller condition number λmax/λmin)
• nes attain higher capacity

12

CMIMO ≈ k log SNR +

k

X

i=1

log ✓λ2

i

k ◆ P

i,j |hi,j|2

P

i,j |hi,j|2 = Tr (HH∗) = Pk i=1 λ2 i

1 k

k

X

i=1

log ✓λ2

i

k ◆ ≤ log Pk

i=1 λ2 i

k2 !

Key ¡Channel ¡Parameters ¡for ¡MIMO

• Rank of channel matrix H
• Rank of H determines how many data streams can be

multiplexed over the channel

• Condition number of channel matrix H
• An ill-conditioned full-rank channel can have smaller capacity

than that of a rank-deficient channel

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14

Physical ¡Modeling ¡of ¡ MIMO ¡Channels

15

Line-­‑of-­‑Sight ¡SIMO ¡Channel

...

φ d ∆rλc

• If distance d ≫ antenna distance spread, then
• Phase difference between consecutive antennas is
• ⟹
• Channel vector h lies along the direction er(Ω), where

Rx antenna i

...

di 2π∆r cos φ carrier wavelength: λc antenna spacing: ∆rλc hi = ae−j2π fcdi

c

= ae−j2π di

λc

channel to i-th antenna: di = d + (i − 1)∆rλc cos φ, ∀ i = 1, 2, . . . , nr h = ae−j2π d

λc ⇥

1 e−j2π∆r cos φ · · · e−j2π(nr−1)∆r cos φ⇤T Ω := cos φ, er (Ω) := 1 √nr ⇥ 1 e−j2π∆rΩ · · · e−j2π(nr−1)∆rΩ⇤T directional cosine

y = hx + w

φ

• If distance d ≫ antenna distance spread, then
• Phase difference between consecutive antennas is
• ⟹
• Channel vector h lies along the direction et(Ω), where

di = d − (i − 1)∆tλc cos φ, ∀ i = 1, 2, . . . , nt

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Line-­‑of-­‑Sight ¡MISO ¡Channel

φ d Tx antenna i

... ...

di directional cosine ∆tλc

y = h∗x + w

−2π∆t cos φ h = aej2π d

λc ⇥

1 e−j2π∆t cos φ · · · e−j2π(nt−1)∆t cos φ⇤T Ω := cos φ, et (Ω) := 1 √nt ⇥ 1 e−j2π∆tΩ · · · e−j2π(nt−1)∆tΩ⇤T φ

Line-­‑of-­‑Sight ¡SIMO ¡and ¡MISO

• Line-of-sight SIMO:
• y = hx+w, h is along the receive spatial signature er(Ω), where
• nr -fold power gain, no DoF gain
• Line-of-sight MISO:
• y = h*x+w, h is along the transmit spatial signature et(Ω), where
• nt -fold power gain, no DoF gain

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er (Ω) := 1 √nr ⇥ 1 e−j2π∆rΩ · · · e−j2π(nr−1)∆rΩ⇤T et (Ω) := 1 √nt ⇥ 1 e−j2π∆tΩ · · · e−j2π(nt−1)∆tΩ⇤T

...

18

Line-­‑of-­‑Sight ¡MIMO ¡Channel

d

... y = Hx + w

Tx k Rx i

di,k

= ⇒ hi,k = ae−j2π d

λc e−j2π(i−1)∆rΩrej2π(k−1)∆tΩt

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

• Rank of H = 1 ⟹ no spatial multiplexing gain!
• In line-of-sight MIMO, still power gain (nt×nr -fold) only

= ⇒ H = ae−j2π d

λc √ntnrer (Ωr) et (Ωt)∗

The ¡Need ¡of ¡Multi-­‑Paths

• Line-of-sight environment: only power gain, no DoF gain
• Reason: there is only single path
• Because Tx/Rx antennas are co-located
• Multi-paths are needed in order to get DoF gain
• Multi-paths are common due to reflections

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Single ¡ReWlector, ¡Two-­‑Paths ¡MIMO ¡(1)

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φr1 φr2 Rx antenna 1

• Two paths:
• Path 1:
• Path 2:
• By the linear superposition principle, we get the channel matrix
• rank(H) = 2 ⟺ er(Ωr1) ∦ er(Ωr2) and et(Ωt1) ∦ et(Ωt2):
• Ωr := Ωr2 – Ωr1 ≠ 0

mod 1/Δr

• Ωt := Ωt2 – Ωt1 ≠ 0

mod 1/Δt

φt1 φt2 Tx antenna 1

y = Hx + w H1 = a1e−j2π d1

λc √ntnrer (Ωr1) et (Ωt1)∗

H2 = a2e−j2π d2

λc √ntnrer (Ωr2) et (Ωt2)∗

H = ab

1er (Ωr1) et (Ωt1)∗ + ab 2er (Ωr2) et (Ωt2)∗

ab

i := aie−j2π di

λc √ntnr

Single ¡ReWlector, ¡Two-­‑Paths ¡MIMO ¡(2)

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φr2 φr1 Rx antenna 1

• Question: what affects the condition number of H?
• To understand better, let us place two virtual antennas at

A and B, and break down the system into two stages:

• Tx antenna array to {A,B} and {A,B} to Rx antenna array
• H = HrHt, where
• Note: {A,B} form a geographically separated virtual antenna array

φt1 φt2 Tx antenna 1

y = Hx + w H = ab

1er (Ωr1) et (Ωt1)∗ + ab 2er (Ωr2) et (Ωt2)∗

A B

Hr = ⇥ ab

1er (Ωr1)

ab

2er (Ωr2)

⇤ , Ht = et (Ωt1)∗ et (Ωt2)∗

{A,B} ¡to ¡Rx ¡Antenna ¡Array

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φr2 φr1 Rx antenna 1

• Observation:
• Hr has two columns along the directions er(Ωr1) and er(Ωr2)
• The more aligned er(Ωr1) & er(Ωr2) are, the worse the conditioning
• Hr’s conditioning depends on the angular resolvability:

the angle θ between er(Ωr1) and er(Ωr2), where

A B Hr = ⇥ ab

1er (Ωr1)

ab

2er (Ωr2)

⇤

ab

1er (Ωr1)

ab

2er (Ωr2)

y = Hr xB xA

• + w

| cos θ| = |er (Ωr1)∗ er (Ωr2) |

Computation ¡of ¡|cos ¡θ|

• Let Ωr := Ωr2 – Ωr1: the inner product is a function of Ωr
• Since
• , we have
• Let Lr := nrΔr (length of antenna array in the unit of carrier wavelength), the

above expression becomes

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er (Ωr1)∗ er (Ωr2) = 1 nr

nr

X

i=1

e−j2π(i−1)∆rΩr = 1 nr 1 − e−j2πnr∆rΩr 1 − e−j2π∆rΩr |1 − e−j2θ| = 2| sin θ| |er (Ωr1)∗ er (Ωr2) | =

• sin (πLrΩr)

nr sin (πLrΩr/nr)

• |er (Ωr1)∗ er (Ωr2) | = 1

nr |1 − e−j2πnr∆rΩr| |1 − e−j2π∆rΩr| =

• sin (πnr∆rΩr)

nr sin (π∆rΩr)

Properties ¡of ¡|cos ¡θ|

24

| cos θ| = |er (Ωr1)∗ er (Ωr2) | =

• sin (πLrΩr)

nr sin (πLrΩr/nr)

• Ωr = Ωr2 − Ωr1 = cos φr2 − cos φr1
• Its peak is 1 and it peaks at Ωr = 0
• It is equal to 0 at Ωr = k/Lr, k = 1, 2, …, nr–1
• It is periodic with period nr/Lr = 1/Δr
• Hence the channel matrix Hr is ill conditioned whenever
• 1/Lr: resolvability in the angular domain
• Ωr m

∆r

• ⌧ 1

Lr , for some integer m

Angular ¡Resolvability

• If Ωr := Ωr2 – Ωr1 ≠ m/Δr for some m, then based on the

previous discussion the rank of H = 2

• If Ωr is close to some m/Δr within distance ≪ 1/Lr, the

two signals from A and B cannot be resolved

• Angular resolvability depends only on the length (Lr) of

the (linear) antenna array

• The smaller 1/Lr is, the better the angular resolvability
• Packing more Rx antennas won’t help
• Its condition number depends on Lr×minm{|Ωr – m/Δr|}:
• If its value ≪ 1, then Hr is ill-conditioned
• If its value ≥ 1, then Hr is well-conditioned

25

Single ¡ReWlector, ¡Two-­‑Paths ¡MIMO ¡(3)

26

φr2 φr1 Rx antenna 1

• Recall: we broke down the system into two stages:
• H = HrHt, where
• If both Hr and Ht are well conditioned, then so is H:
• We only need both Lr×Ωr and Lt×Ωt ≥ 1 (for Δr, Δt ≤ 1/2)
• Reflectors and scatters should be placed such that both Tx

angular separation and Rx angular separation are large enough

φt1 φt2 Tx antenna 1

y = Hx + w H = ab

1er (Ωr1) et (Ωt1)∗ + ab 2er (Ωr2) et (Ωt2)∗

A B

Hr = ⇥ ab

1er (Ωr1)

ab

2er (Ωr2)

⇤ , Ht = et (Ωt1)∗ et (Ωt2)∗

The ¡# ¡of ¡Resolvable ¡Paths ¡Matters

• Assume that Δr, Δt ≤ 1/2 in the previous example
• rank(H) = 2 ⟺ Ωr, Ωt ≠ 0
• Two paths emitting and injecting at different angles suffice
• H is well-conditioned if Ωr ≥ 1/Lr, Ωt ≥ 1/Lt
• Different angles are not enough!
• Two paths have to be resolvable both by the Tx and the Rx array
• Resolvability depends on the size of the antenna arrays
• The larger the array is, the better the resolvability
• Physical model that counts the # of paths could be

misleading and may NOT be the right level of abstraction for the design and analysis of communication systems

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28

Statistical ¡Modeling ¡of ¡ MIMO ¡Channels

Modeling ¡Approach

• Recall how we modeled multi-path channels in Lecture 1:
• Start with a deterministic continuous-time model
• Sampling to get a discrete-time model with delay taps
• Physical paths are grouped into delay bins of width 1/W, one for

each tap

• Each tap gain hl is an aggregation of several physical paths and

can be modeled as Gaussian

• We follow the same approach for MIMO:
• Paths are grouped into angular bins of size (1/Lt)×(1/Lr)
• Results in an channel matrix Ha in the angular domain
• Each entry of Ha is an aggregation of several physical paths and

can be modeled as Gaussian

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4 4 5 5 2 2 2 2 3 1 1 1 1 3 3 3 +1 +1 –1 –1 path B 1 / Lr 1 / Lt path A path B path A Resolvable bins Ωt Ωr

Tx Antenna Array Rx Antenna Array

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Spatial-­‑Domain ¡MIMO ¡Channel

y = Hx + w

er (Ω) := 1 √nr      1 e−j2π∆rΩ . . . e−j2π(nr−1)∆rΩ      , et (Ω) := 1 √nt      1 e−j2π∆tΩ . . . e−j2π(nt−1)∆tΩ      H = X

i

ab

ier (Ωri) et (Ωti)∗ ,

ab

i := ai

√ntnre−j2π di

λc

• Parameters:
• di: distance between Tx antenna 1 and Rx antenna 1 along path i
• λc: carrier wavelength
• Δt, Δr: antenna spacing in the unit of λc
• er(Ω), et(Ω): unit spatial signatures along the direction Ω
• Ω := cosϕ: directional cosine

Angular-­‑Domain ¡Bases

32

• The above two matrices are both unitary
• Because
• ,
• whenever Ωr := Ωr2 – Ωr1 = k/Lr, k = 1, 2, …, nr–1
• Hence Ur is unitary; similarly Ut is also unitary
• Hence, columns of Ur & Ut form orthonormal bases for

the nr-dim. Rx signal space and the nt-dim. Tx signal space respectively!

|er (Ωr1)∗ er (Ωr2) | =

• sin (πLrΩr)

nr sin (πLrΩr/nr)

• = 0

Ur := h er (0) er ⇣

1 Lr

⌘ · · · er ⇣

nr−1 Lr

⌘ i Ut := h et (0) et ⇣

1 Lt

⌘ · · · et ⇣

nt−1 Lt

⌘ i

Spatial-­‑Angular-­‑Domain ¡Transformation

• Change of coordinate:
• Expand Tx signal x over the columns of Ut: x = Ut xa
• Expand Rx signal y over the columns of Ur: y = Ur ya
• Plug it back we get:
• Note: entries of Ha are independent from one another

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Ur ya = HUt xa + w ⟹ ¡ya = (Ur*HUt) xa + (Ur*w) ⟹ ya = Haxa + wa

Ha := U∗

rHUt

wa := U∗

rw ∼ CN

• 0, σ2Inr
• Here

= ⇒ ha

k,l = er (k/Lr)∗ H et (l/Lt)

= X

i

ab

i

• er (k/Lr)∗ er (Ωri)

×

• et (Ωti)∗ et (l/Lt)

Beamforming ¡Pattern

34

• Rx beamforming patterns: for k = 0, 1, …, nr –1
• Tx beamforming patterns: for l = 0, 1, …, nt –1
• Beamforming pattern gives the antenna gain in that

given Tx/Rx direction

Br,k (Ωr) := |er (k/Lr)∗ er (Ωr) | =

• sin (π (LrΩr − k))

nr sin (π (LrΩr − k) /nr)

• Bt,l (Ωt) := |et (l/Lt)∗ et (Ωt) | =
• sin (π (LtΩt − l))

nt sin (π (LtΩt − l) /nt)

35

0.5 0.5 0.5 1 30 210 60 240 90 270 120 300 150 330 180 0.5 1 30 210 60 240 90 270 120 300 150 330 180 1 30 210 60 240 90 270 120 300 150 330 180 1 30 210 60 240 90 270 120 300 150 330 180

Polar plots of (ϕ, Br,k(cosϕ)) for Lr = 2, nr = 4

k = 0 k = 1 k = 2 k = 3

Critically ¡Spaced ¡Antennas: ¡∆ = 1/2

• When ∆ = L/n = 1/2, in the polar plot there will be one

pair of main lobes for all k

• When ∆ = L/n > 1/2, in the polar plot there will be

more than a pair of main lobes for some k

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Polar plots of (ϕ, Br,k(cosϕ)) for Lr = 2, nr = 2

0.5 0.5 1 30 210 60 240 90 270 120 300 150 330 180 1 30 210 60 240 90 270 120 300 150 330 180

k = 0 k = 1

Sparsely ¡Spaced ¡Antennas: ¡∆ > 1/2

• In the polar plot, some basis will have no main lobes

37

Polar plots of (ϕ, Br,k(cosϕ)) for Lr = 2, nr = 8

0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 1 30 210 60 240 90 270 120 300 150 330 180 1 30 210 60 240 90 270 120 300 150 330 180 1 30 210 60 240 90 270 120 300 150 330 180 1 30 210 60 240 90 270 120 300 150 330 180 1 30 210 60 240 90 270 120 300 150 330 180 1 30 210 60 240 90 270 120 300 150 330 180 1 30 210 60 240 90 270 120 300 150 330 180 1 30 210 60 240 90 270 120 300 150 330 180

k = 0 k = 1 k = 2 k = 3 k = 4 k = 5 k = 6 k = 7

Densely ¡Spaced ¡Antennas: ¡∆ < 1/2

Angular ¡Bins

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ha

k,l =

X

i

ab

i

• er (k/Lr)∗ er (Ωri)

×

• et (Ωti)∗ et (l/Lt)
• Path i contributes to the (k,l)-th entry of Ha iff
• ϕri falls inside the main lobe of the Rx beamforming pattern Br,k
• ϕti falls inside the main lobe of the Tx beamforming pattern Bt,l
• Width of the main lobes are 1/Lr and 1/Lt respectively
• Paths within the bin are unresolvable and aggregate to

the effective channel matrix entry

Dependency ¡on ¡Antenna ¡Spacing ¡∆

• Let us fix the array size (length) Lt and Lr
• nt, nr ↑ ⟹ ∆t, ∆r ↓
• Focus on the Rx side and fix Lr = 3:
• 2Lr = 6 angular windows of width 1/Lr = 1/3

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3 2 4 5 1 1 5 4 2 3

Critically ¡Spaced ¡Antennas: ¡∆ = 1/2

• ∆r = 1/2 ⟹ in total nr = Lr/∆r = 6

windows

• Each of the 6 basis vectors has a

pair of main lobes

• Hence, 1-to-1 correspondence

between angular windows and resolvable bins!

40

L r = 3, n r = 6 2 4 1 2 3 4 5 k 5 1 1 5 4 2 3 3 Bins

Sparsely ¡Spaced ¡Antennas: ¡∆ > 1/2

41

• ∆r > 1/2 ⟹ in total nr = Lr/∆r < 6 windows
• Each basis vector has more than a pair of main lobes
• Hence, several angular windows will be lumped into a

resolvable bin

• Effectively reduces the number of resolvable bins

42

Bins 1 1 1 1 1 1 k 1 L r = 3, n r = 2 Bins 2 3 1 4 2 3 2 1 4 3 k 1 2 3 4 L r = 3, n r = 5

Densely ¡Spaced ¡Antennas: ¡∆ < 1/2

• ∆r < 1/2 ⟹ in total nr = Lr/∆r > 6 windows
• Some basis vectors have no main lobes
• Hence, the bins corresponding to these basis vectors are

empty and effectively useless

• # of non-empty resolvable bins = 6, the same as the

critically spaced case

43

44

7 8 9 1 2 3 2 1 9 8 k 0 1 9 8 7 6 5 4 3 2 Empty bins L r = 3, n r = 10

10 20 30 40 50 5 10 15 20 25 30 35 40 45 50 1 2 3 4 5 L = 16, n = 50 |hkl|

a

l – Transmitter bins K–Receiver bins

Examples: ¡Limited ¡Angular ¡Separation

45

~ ~ ~ ~ Tx antenna array Rx antenna array Very small angular separation Large angular separation (a) ~ ~ ~ ~ Tx antenna array Rx antenna array (b) 5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 20 25 30 k – Receiver bins l – Transmitter bins

k – Receiver bins 5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 20 25 k – Receiver bins l – Transmitter bins |hkl|

a

46

5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 20 25 30 k – Receiver bins

(a) 60° spread at transmitter, 360° spread at receiver (c) 60° spread at transmitter, 60° spread at receiver

l – Transmitter bins 5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 20 25 k – Receiver bins

(b) 360° spread at transmitter, 60° spread at receiver (d) 360° spread at transmitter, 360° spread at receiver

l – Transmitter bins 5 10 15 20 25 30 5 10 15 20 25 30 10 20 30 40 50 k – Receiver bins l – Transmitter bins 5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 k – Receiver bins l – Transmitter bins |hkl|

a

|hkl|

a

|hkl|

a

|hkl|

a

Degrees ¡of ¡Freedom

• Fact: rank(H) = rank(Ha)
• rank(Ha) = min{# of non-zero rows, # of non-zero columns},

for a random matrix Ha (with each entry independent from one another and

has a continuous distribution)

• # of non-zero rows and columns depends on
• The amount of scattering and reflection (environment)
• The sizes (length) Lt and Lr of the antenna arrays (device)
• More scatterers and reflectors ⟹ more non-zero entries

⟹ larger DoF

• Larger Lt and Lr ⟹ better angular resolvability ⟹ more

non-zero entries ⟹ larger the DoF

47

48

Clustered ¡Model

Cluster of scatterers Receive array Transmit array

φ t φ r Θ t,1 Θ t,2 Θ r,1 Θ r,2

• # of DoF = min{Lt Ωt,total , Lr Ωr,total}
• # of non-zero resolvable Tx bins = Ωt,total / (1/Lt) = Lt Ωt,total
• # of non-zero resolvable Rx bins = Ωr,total / (1/Lr) = Lr Ωr,total

Ωt,total := X

k

|Θt,k| Ωr,total := X

k

|Θr,k|

49

Examples

5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 20 5 10 15 20 25 30 5 10 15 20 25 30 5 15 10 120° –175° –20° 40° Tx Rx 10° 5° 15° 10° 70° –175° –120° –60° Tx Rx 10° 5° 15°

10°

(a) (b)

|hkl|

a

|hkl|

a

l – Transmitter bins K – Receiver bins l – Transmitter bins K – Receiver bins

Dependency ¡on ¡Arrary ¡Size

50

Cluster of scatterers (a) Array length of L1 (b) Array length of L2 > L1 Cluster of scatterers Receive array Receive array 1/L1 1/L1 1/L2 1/L2 Transmit array Transmit array

Better angular resolvability when L is larger

Dependency ¡on ¡Carrier ¡Frequency

51

2 3 4 5 6 7 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 2 3 4 5 6 7 1 2 3 4 5 6 7 Frequency (GHz) Frequency (GHz) (b) (a) Ω total in townhouse Ω total Ω total /λ c (m−1) 1/λ (m-1) 1/λ c Ω total in office Office Townhouse 5 10 15 20 25 8 8

• Normalized array size L increases with fc when the

actual physical size is fixed

• However Ωtotal decreases as fc increases since
• signals at higher freq. attenuates faster ⟹ # of effective paths ↓
• scattering is more specular ⟹ angular spread ↓

Diversity

• Diversity order = # of non-zero entries in Ha
• In the clustered model, Div ≤ Lt Ωt,total×Lr Ωr,total
• Degrees of freedom = rank(Ha)
• Channels that have the same DoF can have very

different amount of diversity

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(a) nt nr nr nr nt nt (b) (c)

Div = 4 Div = 8 Div = 16

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5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 20 5 10 15 20 25 30 5 10 15 20 25 30 5 15 10 120° –175° –20° 40° Tx Rx 10° 5° 15° 10° 70° –175° –120° –60° Tx Rx 10° 5° 15°

10°

(a) (b)

|hkl|

a

|hkl|

a

l – Transmitter bins K – Receiver bins l – Transmitter bins K – Receiver bins

Amount of diversity ↑ when there are more bounces

i.i.d. ¡Rayleigh ¡Model

• Scatterers at all angles from Tx and Rx (rich scattering)
• A lot of multi-paths in each of the resolvable angular bin
• Entries of Ha : i.i.d. circular symmetric complex Gaussian
• Since Ha = Ur*HUt , entries of H also i.i.d. Gaussian
• Angular spread Ωt = Ωr = 2
• ⟹ # of DoF = min{2Lt , 2Lr} = min{nt , nr} when the

antennas are critically spaced

54

k – Receiver bins 5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 k – Receiver bins l – Transmitter bins

Correlated ¡Fading

• When scattering only comes from certain angles, Ha has

zero entries

• Corresponding spatial H has correlated entries
• Same happens when antenna separation ∆ is less than

1/2 (can be reduced to a lower-dimensional i.i.d. matrix)

• Angular domain model provides a physical explanation of

correlation

55

10 20 30 40 50 5 10 15 20 25 30 35 40 45 50 1 2 3 4 5

kl

l – Transmitter bins K–Receiver bins

Spatial-­‑Angular ¡vs. ¡Time-­‑Frequency

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Time-Frequency Spatial-Angular

Domains Time Angular Frequency Spatial Resources Resolution of Multi-paths signal duration T bandwidth W angular spreads Ωt,Ωr bandwidth W delay bins of 1/W angular bins of 1/Lt × 1/Lt DoF WT min{Lt Ωt, Lr Ωr} Diversity # of non-zero delay bins # of non-zero angular bins