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)

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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

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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

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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

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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

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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 well-conditioned rank-deficient channel

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Physical ¡Modeling ¡of ¡ MIMO ¡Channels

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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

...

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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)∗

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 need in order to get DoF gain
• Multi-paths are common due to reflections

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Single ¡ReTlector, ¡Two-­‑Paths ¡MIMO

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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):
• Ωr2 – Ωr1 ≠ 0

mod 1/Δr

• 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 ¡ReTlector, ¡Two-­‑Paths ¡MIMO

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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) and er(Ωr2) are, the worse the

conditioning of Hr

• The conditioning of Hr depends on the angle θ between

er(Ωr1) and er(Ωr2);

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|

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| 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 H is ill conditioned whenever
• 1/Lr: resolvability in the angular domain
• Ωr m

∆r

• ⌧ 1

Lr , for some integer m