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Lecture 6: MIMO Channel and Spatial Multiplexing I-Hsiang Wang [email protected] 5/1, 2014

Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang [email protected] 5/1,

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Page 1: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

Lecture 6: MIMO Channel and Spatial Multiplexing

I-Hsiang [email protected]

5/1, 2014

Page 2: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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

Page 3: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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

Page 4: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

Outline• Spatial multiplexing capability of MIMO systems

• Physical modeling of MIMO channels

• Statistical modeling of MIMO channels

4

Page 5: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

5

Spatial Multiplexing in MIMO Systems

Page 6: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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] 2 Cnr , x[m] 2 Cnt , H 2 Cnrnt , w CN (0, Inr )

H = UV

U 2 Cnrnr , V 2 Cntnt (UU = UU = I)

2 Cnrnt

1 2 · · · min(nt,nr) 0

Page 7: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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 := minnt,nr parallel channels:

- Capacity can be found via water-filling

7

ey := U

y, e

x := V

x, e

w := U

w

y = Hx+w = UVx+w () U

y = V

x+U

w

ey =

ex+ e

w

eyi = iexi + ewi, i = 1, 2, . . . , nmin

Page 8: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

Spatially Parallel Channels

8

V* ...

1

nmin ewnmin

ew1

Ux

yV U* eye

x

x

yH

w

H = UV

Page 9: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

Multiplexing over Parallel Channels

9

294 MIMO I: spatial multiplexing and channel modeling

+

AWGNcoder

AWGNcoder

x1[m]~ y1 [m]~

xnmin[m]~ ynmin[m]~

.

.

.

.

.

.

.

.

.

n min information

streams

0

0

w[m]

U*HV

Decoder

Decoder

There is a clear analogy between this architecture and the OFDM systemFigure 7.2 The SVD architecturefor MIMO communication. introduced in Chapter 3. In both cases, a transformation is applied to convert a

matrix channel into a set of parallel independent sub-channels. In the OFDMsetting, the matrix channel is given by the circulant matrix C in (3.139),defined by the ISI channel together with the cyclic prefix added onto theinput symbols. In fact, the decomposition C=Q−1!Q in (3.143) is the SVDdecomposition of a circulant matrix C, with U = Q−1 and V∗ = Q. Theimportant difference between the ISI channel and the MIMO channel is that,for the former, the U and V matrices (DFTs) do not depend on the specificrealization of the ISI channel, while for the latter, they do depend on thespecific realization of the MIMO channel.

7.1.2 Rank and condition number

What are the key parameters that determine performance? It is simpler tofocus separately on the high and the low SNR regimes. At high SNR, thewater level is deep and the policy of allocating equal amounts of power onthe non-zero eigenmodes is asymptotically optimal (cf. Figure 5.24(a)):

C ≈k!

i=1

log"1+ P"2

i

kN0

#≈ k log SNR+

k!

i=1

log""2i

k

#bits/s/Hz# (7.12)

where k is the number of non-zero "2i , i.e., the rank of H, and SNR $= P/N0.

The parameter k is the number of spatial degrees of freedom per second perhertz. It represents the dimension of the transmitted signal as modified bythe MIMO channel, i.e., the dimension of the image of H. This is equal tothe rank of the matrix H and with full rank, we see that a MIMO channelprovides nmin spatial degrees of freedom.

P i =

2

2i

+

, satisfiesnminX

i=1

P i = PCMIMO =

nminX

i=1

log

1 +

2iP

i

2

Page 10: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

Rank = # of Multiplexing Channels

10

V* ...

1

nmin ewnmin

ew1

Ux

yV U* eye

x

• If λi = 0 ⟹ the i-th channel contributes 0 to the capacity

• Rank of H = # of non-zero singular values

CMIMO =

kX

i=1

log

1 +

2iP

i

2

, k := rank (H)

Page 11: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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)

11

CMIMO kX

i=1

log

1 +

2iP

k2

kX

i=1

log

2iP

k2

= k log SNR+

kX

i=1

log

2i

k

Page 12: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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) ones attain higher capacity

12

CMIMO k log SNR+

kX

i=1

log

2i

k

Pi,j |hi,j |2P

i,j |hi,j |2 = Tr (HH) =Pk

i=1 2i

1

k

kX

i=1

log

2i

k

log

Pki=1

2i

k2

!

Page 13: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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

13

Page 14: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

14

Physical Modeling of MIMO Channels

Page 15: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

15

Line-­‐of-­‐Sight SIMO Channel

...

d

rc

- 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

2r cos

carrier wavelength: c

antenna spacing: rc

hi = aej2fcdi

c = aej2dic

channel to i-th antenna:

di = d+ (i 1)rc cos, 8 i = 1, 2, . . . , nr

h = aej2 dc

1 ej2r cos · · · ej2(nr1)r cos

T

:= cos, er () :=1

pnr

1 ej2r · · · ej2(nr1)r

T

directional cosine

y = hx+w

Page 16: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

- 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)tc cos, 8 i = 1, 2, . . . , nt

16

Line-­‐of-­‐Sight MISO Channel

d

Tx antenna i

...

... di

directional cosine

tc y = h

x+ w

2t cos

h = aej2dc

1 ej2t cos · · · ej2(nt1)t cos

T

:= cos, et () :=1

pnt

1 ej2t · · · ej2(nt1)t

T

Page 17: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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

17

er () :=1

pnr

1 ej2r · · · ej2(nr1)r

T

et () :=1

pnt

1 ej2t · · · ej2(nt1)t

T

Page 18: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

...

18

Line-­‐of-­‐Sight MIMO Channeld

...

y = Hx+wTx k

Rx idi,k

=) hi,k = aej2 dc ej2(i1)rrej2(k1)tt

di,k = d+ (i 1)rc cosr (k 1)tc cost

• Rank of H = 1 ⟹ no spatial multiplexing gain!

• In line-of-sight MIMO, still power gain (nt×nr -fold) only

=) H = aej2 dcpntnrer (r) et (t)

Page 19: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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

19

Page 20: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

Single ReWlector, Two-­‐Paths MIMO (1)

20

r1r2Rx 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

t1 t2

Tx antenna 1

y = Hx+w

H1 = a1ej2

d1cpntnrer (r1) et (t1)

H2 = a2ej2

d2cpntnrer (r2) et (t2)

H = ab1er (r1) et (t1) + ab2er (r2) et (t2)

abi := aie

j2dicpntnr

Page 21: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

Single ReWlector, Two-­‐Paths MIMO (2)

21

r2 r1Rx 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

t1 t2

Tx antenna 1

y = Hx+w

H = ab1er (r1) et (t1) + ab2er (r2) et (t2)

A

B

Hr =ab1er (r1) ab2er (r2)

, Ht =

et (t1)

et (t2)

Page 22: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

A,B to Rx Antenna Array

22

r2 r1Rx 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 =ab1er (r1) ab2er (r2)

ab1er (r1)

ab2er (r2)y = Hr

xB

xA

+w

| cos | = |er (r1) er (r2) |

Page 23: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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

23

er (r1) er (r2) =

1

nr

nrX

i=1

ej2(i1)rr =1

nr

1 ej2nrrr

1 ej2rr

|1 ej2| = 2| sin |

|er (r1) er (r2) | =

sin (Lrr)

nr sin (Lrr/nr)

|er (r1) er (r2) | =

1

nr

|1 ej2nrrr ||1 ej2rr | =

sin (nrrr)

nr sin (rr)

Page 24: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

Properties of |cos θ|

24

| cos | = |er (r1) er (r2) | =

sin (Lrr)

nr sin (Lrr/nr)

r = r2 r1 = cosr2 cosr1

• 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

r m

r

1

Lr, for some integer m

Page 25: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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

Page 26: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

Single ReWlector, Two-­‐Paths MIMO (3)

26

r2 r1Rx 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

t1 t2

Tx antenna 1

y = Hx+w

H = ab1er (r1) et (t1) + ab2er (r2) et (t2)

A

B

Hr =ab1er (r1) ab2er (r2)

, Ht =

et (t1)

et (t2)

Page 27: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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

27

Page 28: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

28

Statistical Modeling of MIMO Channels

Page 29: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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

29

Page 30: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

30

310 MIMO I: spatial multiplexing and channel modeling

of as a (time-)resolvable path, consisting of an aggregation of individualphysical paths. The bandwidth of the system dictates how finely or coarselythe physical paths are grouped into resolvable paths. From the point of viewof communication, it is the behavior of the resolvable paths that matters,not that of the individual paths. Modeling the taps directly rather than theindividual paths has the additional advantage that the aggregation makesstatistical modeling more reliable.Using the analogy between the finite time-resolution of a band-limited

system and the finite angular-resolution of an array-size-limited system, wecan follow the approach of Section 2.2.3 in modeling MIMO channels. Thetransmit and receive antenna array lengths Lt and Lr dictate the degree ofresolvability in the angular domain: paths whose transmit directional cosinesdiffer by less than 1/Lt and receive directional cosines by less than 1/Lr

are not resolvable by the arrays. This suggests that we should “sample” theangular domain at fixed angular spacings of 1/Lt at the transmitter and atfixed angular spacings of 1/Lr at the receiver, and represent the channel interms of these new input and output coordinates. The !k" l#th channel gain inthese angular coordinates is then roughly the aggregation of all paths whosetransmit directional cosine is within an angular window of width 1/Lt aroundl/Lt and whose receive directional cosine is within an angular window ofwidth 1/Lr around k/Lr . See Figure 7.11 for an illustration of the lineartransmit and receive antenna array with the corresponding angular windows.In the following subsections, we will develop this approach explicitly foruniform linear arrays.

Figure 7.11 A representationof the MIMO channel in theangular domain. Due to thelimited resolvability of theantenna arrays, the physicalpaths are partitioned intoresolvable bins of angularwidths 1/Lr by 1/Lt . Herethere are four receiveantennas (Lr = 2) and sixtransmit antennas (Lr = 3).

4

45

5

0

0

0

0

2

2

22

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 binsΩt

Ωr

Tx Antenna Array

Rx Antenna Array

Page 31: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

31

Spatial-­‐Domain MIMO Channely = Hx+w

er () :=1

pnr

2

6664

1ej2r

...ej2(nr1)r

3

7775, et () :=

1pnt

2

6664

1ej2t

...ej2(nt1)t

3

7775

H =X

i

abier (ri) et (ti) , abi := ai

pntnre

j2dic

• 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

Page 32: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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 (r1) er (r2) | =

sin (Lrr)

nr sin (Lrr/nr)

= 0

Ur :=her (0) er

1Lr

· · · er

nr1Lr

i

Ut :=het (0) et

1Lt

· · · et

nt1Lt

i

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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

33

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

Ha := UrHUt wa := U

rw CN0,2Inr

Here=) ha

k,l = er (k/Lr)Het (l/Lt)

=X

i

abier (k/Lr)

er (ri)

et (ti)

et (l/Lt)

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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 (r) := |er (k/Lr) er (r) | =

sin ( (Lrr k))

nr sin ( (Lrr k) /nr)

Bt,l (t) := |et (l/Lt) et (t) | =

sin ( (Ltt l))

nt sin ( (Ltt l) /nt)

Page 35: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

35

313 7.3 Modeling of MIMO fading channels

Figure 7.12 Receivebeamforming patterns of theangular basis vectors.Independent of the antennaspacing, the beamformingpatterns all have the samebeam widths for the mainlobe, but the number of mainlobes depends on the spacing.(a) Critically spaced case; (b)Sparsely spaced case. (c)Densely spaced case.

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(b) L r = 2, n r = 2

(c) L r = 2, n r = 8

decomposition of the overall transmitted signal into the components transmit-ted along the different physical directions, up to a resolution of 1/Lt .

Examples of angular basesExamples of angular bases, represented by their beamforming patterns, areshown in Figure 7.12. Three cases are distinguished:

• Antennas are critically spaced at half the wavelength (!r = 1/2). In thiscase, each basis vector er"k/Lr# has a single pair of main lobes around theangles ± arccos"k/Lr#.

• Antennas are sparsely spaced (!r > 1/2). In this case, some of the basisvectors have more than one pair of main lobes.

• Antennas are densely spaced (!r < 1/2). In this case, some of the basisvectors have no main lobes.

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

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• When ∆ = L/n > 1/2, in the polar plot there will be more than a pair of main lobes for some k

36

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

313 7.3 Modeling of MIMO fading channels

Figure 7.12 Receivebeamforming patterns of theangular basis vectors.Independent of the antennaspacing, the beamformingpatterns all have the samebeam widths for the mainlobe, but the number of mainlobes depends on the spacing.(a) Critically spaced case; (b)Sparsely spaced case. (c)Densely spaced case.

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(b) L r = 2, n r = 2

(c) L r = 2, n r = 8

decomposition of the overall transmitted signal into the components transmit-ted along the different physical directions, up to a resolution of 1/Lt .

Examples of angular basesExamples of angular bases, represented by their beamforming patterns, areshown in Figure 7.12. Three cases are distinguished:

• Antennas are critically spaced at half the wavelength (!r = 1/2). In thiscase, each basis vector er"k/Lr# has a single pair of main lobes around theangles ± arccos"k/Lr#.

• Antennas are sparsely spaced (!r > 1/2). In this case, some of the basisvectors have more than one pair of main lobes.

• Antennas are densely spaced (!r < 1/2). In this case, some of the basisvectors have no main lobes.

k = 0 k = 1

Sparsely Spaced Antennas: ∆ > 1/2

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

37

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

313 7.3 Modeling of MIMO fading channels

Figure 7.12 Receivebeamforming patterns of theangular basis vectors.Independent of the antennaspacing, the beamformingpatterns all have the samebeam widths for the mainlobe, but the number of mainlobes depends on the spacing.(a) Critically spaced case; (b)Sparsely spaced case. (c)Densely spaced case.

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decomposition of the overall transmitted signal into the components transmit-ted along the different physical directions, up to a resolution of 1/Lt .

Examples of angular basesExamples of angular bases, represented by their beamforming patterns, areshown in Figure 7.12. Three cases are distinguished:

• Antennas are critically spaced at half the wavelength (!r = 1/2). In thiscase, each basis vector er"k/Lr# has a single pair of main lobes around theangles ± arccos"k/Lr#.

• Antennas are sparsely spaced (!r > 1/2). In this case, some of the basisvectors have more than one pair of main lobes.

• Antennas are densely spaced (!r < 1/2). In this case, some of the basisvectors have no main lobes.

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

k = 4 k = 5 k = 6 k = 7

Densely Spaced Antennas: ∆ < 1/2

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Angular Bins

38

hak,l =

X

i

abier (k/Lr)

er (ri) et (ti)

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

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Dependency on Antenna Spacing ƥ 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

39

324 MIMO I: spatial multiplexing and channel modeling

Figure 7.20 An antenna arrayof length Lr partitions thereceive directions into 2Lrangular windows. Here, Lr = 3and there are six angularwindows. Note that because ofsymmetry across the 0! −180!

axis, each angular windowcomes as a mirror image pair,and each pair is only countedas one angular window.

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0

Figure 7.21 Antennas arecritically spaced at half thewavelength. Each resolvablebin corresponds to exactly oneangular window. Here, thereare six angular windows andsix bins.

L r = 3, n r = 6

24

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the sparsely spaced case (!r > 1/2), the beamforming patterns of some of thebasis vectors have multiple main lobes. Thus, paths arriving in the differentangular windows corresponding to these lobes are all lumped into one binand cannot be resolved by the array (Figure 7.22). In the densely spaced case(!r < 1/2), the beamforming patterns of 2Lr of the basis vectors have a singlemain lobe; they can be used to resolve among the 2Lr angular windows. Thebeamforming patterns of the remaining nr −2Lr basis vectors have no mainlobe and do not correspond to any angular window. There is little receivedenergy along these basis vectors and they do not participate significantly inthe communication process. See Figure 7.23.The key conclusion from the above analysis is that, given the antenna

array lengths Lr and Lt , the maximum achievable angular resolution canbe achieved by placing antenna elements half a wavelength apart. Placingantennas more sparsely reduces the resolution of the antenna array and can

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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

324 MIMO I: spatial multiplexing and channel modeling

Figure 7.20 An antenna arrayof length Lr partitions thereceive directions into 2Lrangular windows. Here, Lr = 3and there are six angularwindows. Note that because ofsymmetry across the 0! −180!

axis, each angular windowcomes as a mirror image pair,and each pair is only countedas one angular window.

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Figure 7.21 Antennas arecritically spaced at half thewavelength. Each resolvablebin corresponds to exactly oneangular window. Here, thereare six angular windows andsix bins.

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the sparsely spaced case (!r > 1/2), the beamforming patterns of some of thebasis vectors have multiple main lobes. Thus, paths arriving in the differentangular windows corresponding to these lobes are all lumped into one binand cannot be resolved by the array (Figure 7.22). In the densely spaced case(!r < 1/2), the beamforming patterns of 2Lr of the basis vectors have a singlemain lobe; they can be used to resolve among the 2Lr angular windows. Thebeamforming patterns of the remaining nr −2Lr basis vectors have no mainlobe and do not correspond to any angular window. There is little receivedenergy along these basis vectors and they do not participate significantly inthe communication process. See Figure 7.23.

The key conclusion from the above analysis is that, given the antennaarray lengths Lr and Lt , the maximum achievable angular resolution canbe achieved by placing antenna elements half a wavelength apart. Placingantennas more sparsely reduces the resolution of the antenna array and can

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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

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42

325 7.3 Modeling of MIMO fading channels

(b)

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reduce the number of degrees of freedom and the diversity of the channel.Figure 7.22 (a) Antennas aresparsely spaced. Some of thebins contain paths frommultiple angular windows.(b) The antennas are verysparsely spaced. All binscontain several angularwindows of paths.

Placing the antennas more densely adds spurious basis vectors which do notcorrespond to any physical directions, and does not add resolvability. In termsof the angular channel matrix Ha, this has the effect of adding zero rows andcolumns; in terms of the spatial channel matrixH, this has the effect of makingthe entries more correlated. In fact, the angular domain representation makesit apparent that one can reduce the densely spaced system to an equivalent2Lt ×2Lr critically spaced system by just focusing on the basis vectors thatdo correspond to physical directions (Figure 7.24).Increasing the antenna separation within a given array length Lr does not

increase the number of degrees of freedom in the channel. What about increas-ing the antenna separation while keeping the number of antenna elements nr

the same? This question makes sense if the system is hardware-limited ratherthan limited by the amount of space to put the antenna array in. Increasingthe antenna separation this way reduces the beam width of the nr angularbasis beamforming patterns but also increases the number of main lobes ineach (Figure 7.25). If the scattering environment is rich enough such that thereceived signal arrives from all directions, the number of non-zero rows ofthe channel matrix Ha is already nr , the largest possible, and increasing thespacing does not increase the number of degrees of freedom in the channel.On the other hand, if the scattering is clustered to within certain directions,increasing the separation makes it possible for the scattered signal to be

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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

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44

326 MIMO I: spatial multiplexing and channel modeling

Figure 7.23 Antennas aredensely spaced. Some binscontain no physical paths.

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L r = 3, n r = 10

Figure 7.24 A typical Ha

when the antennas aredensely spaced.

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received in more bins, thus increasing the number of degrees of freedom(Figure 7.25). In terms of the spatial channel matrix H, this has the effect ofmaking the entries look more random and independent. At a base-station ona high tower with few local scatterers, the angular spread of the multipaths issmall and therefore one has to put the antennas many wavelengths apart todecorrelate the channel gains.

Sampling interpretationOne can give a sampling interpretation to the above results. First, think ofthe discrete antenna array as a sampling of an underlying continuous array!−Lr/2"Lr/2#. On this array, the received signal x$s% is a function of the

326 MIMO I: spatial multiplexing and channel modeling

Figure 7.23 Antennas aredensely spaced. Some binscontain no physical paths.

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Figure 7.24 A typical Ha

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received in more bins, thus increasing the number of degrees of freedom(Figure 7.25). In terms of the spatial channel matrix H, this has the effect ofmaking the entries look more random and independent. At a base-station ona high tower with few local scatterers, the angular spread of the multipaths issmall and therefore one has to put the antennas many wavelengths apart todecorrelate the channel gains.

Sampling interpretationOne can give a sampling interpretation to the above results. First, think ofthe discrete antenna array as a sampling of an underlying continuous array!−Lr/2"Lr/2#. On this array, the received signal x$s% is a function of the

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Examples: Limited Angular Separation

45

308 MIMO I: spatial multiplexing and channel modeling

one can interpret the second matrix H′′ as the matrix channel from twoimaginary transmitters at A and B to the receive antenna array. This matrixhas rank 2 as well; its conditioning depends on the parameter Lr!r . If bothmatrices are well-conditioned, then the overall channel matrix H is also well-conditioned.

The MIMO channel with two multipaths is essentially a concatenation of thent by 2 channel in Figure 7.8 and the 2 by nr channel in Figure 7.4. Althoughboth the transmit antennas and the receive antennas are close together, mul-tipaths in effect provide virtual “relays”, which are geographically far apart.The channel from the transmit array to the relays as well as the channel fromthe relays to the receive array both have two degrees of freedom, and sodoes the overall channel. Spatial multiplexing is now possible. In this con-text, multipath fading can be viewed as providing an advantage that can beexploited.

It is important to note in this example that significant angular separationof the two paths at both the transmit and the receive antenna arrays is crucialfor the well-conditionedness of H. This may not hold in some environments.For example, if the reflector is local around the receiver and is much closerto the receiver than to the transmitter, then the angular separation !t at thetransmitter is small. Similarly, if the reflector is local around the transmitterand is much closer to the transmitter than to the receiver, then the angularseparation !r at the receiver is small. In either case H would not be verywell-conditioned (Figure 7.10). In a cellular system this suggests that if thebase-station is high on top of a tower with most of the scatterers and reflectorslocally around the mobile, then the size of the antenna array at the base-station

Figure 7.10 (a) The reflectorsand scatterers are in a ringlocally around the receiver;their angular separation at thetransmitter is small. (b) Thereflectors and scatterers are ina ring locally around thetransmitter; their angularseparation at the receiver issmall.

~~

~~

~~

~~

Tx antenna array

Tx antenna array

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Rx antennaarray

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Large angularseparation

(a)

(b)

308 MIMO I: spatial multiplexing and channel modeling

one can interpret the second matrix H′′ as the matrix channel from twoimaginary transmitters at A and B to the receive antenna array. This matrixhas rank 2 as well; its conditioning depends on the parameter Lr!r . If bothmatrices are well-conditioned, then the overall channel matrix H is also well-conditioned.

The MIMO channel with two multipaths is essentially a concatenation of thent by 2 channel in Figure 7.8 and the 2 by nr channel in Figure 7.4. Althoughboth the transmit antennas and the receive antennas are close together, mul-tipaths in effect provide virtual “relays”, which are geographically far apart.The channel from the transmit array to the relays as well as the channel fromthe relays to the receive array both have two degrees of freedom, and sodoes the overall channel. Spatial multiplexing is now possible. In this con-text, multipath fading can be viewed as providing an advantage that can beexploited.

It is important to note in this example that significant angular separationof the two paths at both the transmit and the receive antenna arrays is crucialfor the well-conditionedness of H. This may not hold in some environments.For example, if the reflector is local around the receiver and is much closerto the receiver than to the transmitter, then the angular separation !t at thetransmitter is small. Similarly, if the reflector is local around the transmitterand is much closer to the transmitter than to the receiver, then the angularseparation !r at the receiver is small. In either case H would not be verywell-conditioned (Figure 7.10). In a cellular system this suggests that if thebase-station is high on top of a tower with most of the scatterers and reflectorslocally around the mobile, then the size of the antenna array at the base-station

Figure 7.10 (a) The reflectorsand scatterers are in a ringlocally around the receiver;their angular separation at thetransmitter is small. (b) Thereflectors and scatterers are ina ring locally around thetransmitter; their angularseparation at the receiver issmall.

~~

~~

~~

~~

Tx antenna array

Tx antenna array

Rx antennaarray

Rx antennaarray

Very smallangular separation

Large angularseparation

(a)

(b)

317 7.3 Modeling of MIMO fading channels

7.3.5 Statistical modeling in the angular domain

The basis for the statistical modeling of MIMO fading channels is the approxi-mation that the physical paths are partitioned into angularly resolvable bins andaggregated to form resolvable pathswhose gains are ha

kl!m". Assuming that thegains ab

i !m" of the physical paths are independent, we can model the resolvablepathgainsha

kl!m" as independent.Moreover, the angles #$ri!m"%m and #$ti!m"%mtypically evolve at a much slower time-scale than the gains #ab

i !m"%m; there-fore, within the time-scale of interest it is reasonable to assume that paths donot move from one angular bin to another, and the processes #ha

kl!m"%m can bemodeled as independent acrossk and l (seeTable 2.1 inSection 2.3 for the analo-gous situation for frequency-selective channels). In an angular bin &k' l(, wherethere are many physical paths, one can invoke the Central Limit Theorem andapproximate the aggregate gain ha

kl!m" as a complex circular symmetric Gaus-sian process. On the other hand, in an angular bin &k' l( that contains no paths,the entries ha

kl!m" can be approximated as 0. For a channel with limited angularspread at the receiver and/or the transmitter,many entries ofHa!m"maybe zero.Some examples are shown in Figures 7.14 and 7.15.

Figure 7.14 Some examples ofHa . (a) Small angular spread atthe transmitter, such as thechannel in Figure 7.10(a). (b)Small angular spread at thereceiver, such as the channel inFigure 7.10(b). (c) Smallangular spreads at both thetransmitter and the receiver. (d)Full angular spreads at both thetransmitter and the receiver.

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

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|a

317 7.3 Modeling of MIMO fading channels

7.3.5 Statistical modeling in the angular domain

The basis for the statistical modeling of MIMO fading channels is the approxi-mation that the physical paths are partitioned into angularly resolvable bins andaggregated to form resolvable pathswhose gains are ha

kl!m". Assuming that thegains ab

i !m" of the physical paths are independent, we can model the resolvablepathgainsha

kl!m" as independent.Moreover, the angles #$ri!m"%m and #$ti!m"%mtypically evolve at a much slower time-scale than the gains #ab

i !m"%m; there-fore, within the time-scale of interest it is reasonable to assume that paths donot move from one angular bin to another, and the processes #ha

kl!m"%m can bemodeled as independent acrossk and l (seeTable 2.1 inSection 2.3 for the analo-gous situation for frequency-selective channels). In an angular bin &k' l(, wherethere are many physical paths, one can invoke the Central Limit Theorem andapproximate the aggregate gain ha

kl!m" as a complex circular symmetric Gaus-sian process. On the other hand, in an angular bin &k' l( that contains no paths,the entries ha

kl!m" can be approximated as 0. For a channel with limited angularspread at the receiver and/or the transmitter,many entries ofHa!m"maybe zero.Some examples are shown in Figures 7.14 and 7.15.

Figure 7.14 Some examples ofHa . (a) Small angular spread atthe transmitter, such as thechannel in Figure 7.10(a). (b)Small angular spread at thereceiver, such as the channel inFigure 7.10(b). (c) Smallangular spreads at both thetransmitter and the receiver. (d)Full angular spreads at both thetransmitter and the receiver.

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46

317 7.3 Modeling of MIMO fading channels

7.3.5 Statistical modeling in the angular domain

The basis for the statistical modeling of MIMO fading channels is the approxi-mation that the physical paths are partitioned into angularly resolvable bins andaggregated to form resolvable pathswhose gains are ha

kl!m". Assuming that thegains ab

i !m" of the physical paths are independent, we can model the resolvablepathgainsha

kl!m" as independent.Moreover, the angles #$ri!m"%m and #$ti!m"%mtypically evolve at a much slower time-scale than the gains #ab

i !m"%m; there-fore, within the time-scale of interest it is reasonable to assume that paths donot move from one angular bin to another, and the processes #ha

kl!m"%m can bemodeled as independent acrossk and l (seeTable 2.1 inSection 2.3 for the analo-gous situation for frequency-selective channels). In an angular bin &k' l(, wherethere are many physical paths, one can invoke the Central Limit Theorem andapproximate the aggregate gain ha

kl!m" as a complex circular symmetric Gaus-sian process. On the other hand, in an angular bin &k' l( that contains no paths,the entries ha

kl!m" can be approximated as 0. For a channel with limited angularspread at the receiver and/or the transmitter,many entries ofHa!m"maybe zero.Some examples are shown in Figures 7.14 and 7.15.

Figure 7.14 Some examples ofHa . (a) Small angular spread atthe transmitter, such as thechannel in Figure 7.10(a). (b)Small angular spread at thereceiver, such as the channel inFigure 7.10(b). (c) Smallangular spreads at both thetransmitter and the receiver. (d)Full angular spreads at both thetransmitter and the receiver.

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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

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48

Clustered Model

320 MIMO I: spatial multiplexing and channel modeling

Table 7.1 Examples of some indoor channel measurements. The Intelmeasurements span a very wide bandwidth and the number of clusters andangular spread measured are frequency dependent. This set of data is furtherelaborated in Figure 7.18.

Frequency (GHz) No. of clusters Total angular spread (!)

USC UWB [27] 0–3 2–5 37Intel UWB [91] 2–8 1–4 11–17Spencer [112] 6.75–7.25 3–5 25.5COST 259 [58] 24 3–5 18.5

Cluster of scatterers

Receivearray

Transmitarray

φ t φ rΘ t,1

Θ t,2

Θ r,1

Θ r,2

Figure 7.16 The clustered response model for the multipath environment. Each cluster bouncesoff a continuum of paths.

In such a model, the directional cosines !r along which paths arriveare partitioned into several disjoint intervals: !r = ∪k!rk. Similarly, onthe transmit side, !t = ∪k!tk. The number of degrees of freedom in thechannel is

min

!"

k

#Lt$!tk$%""

k

#Lr$!tk$%#

(7.77)

For Lt and Lr large, the number of degrees of freedom is approximately

min#Lt$t"total"Lr$r"total%" (7.78)

where

$t"total &="

k

$!tk$ and $r"total &="

k

$!rk$ (7.79)

• # of DoF = minLt 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

t,total :=X

k

|t,k| r,total :=X

k

|r,k|

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49

Examples

318 MIMO I: spatial multiplexing and channel modeling

7.3.6 Degrees of freedom and diversity

Degrees of freedomGiven the statistical model, one can quantify the spatial multiplexing capa-bility of a MIMO channel. With probability 1, the rank of the random matrixHa is given by

rank!Ha"=min#number of non-zero rows, number of non-zero columns$

(7.74)

(Exercise 7.6). This yields the number of degrees of freedom available in theMIMO channel.

The number of non-zero rows and columns depends in turn on two separatefactors:

• The amount of scattering and reflection in the multipath environment. The

Figure 7.15 Some examples ofHa . (a) Two clusters ofscatterers, with all paths goingthrough a single bounce.(b) Paths scattered via multiplebounces.

more scatterers and reflectors there are, the larger the number of non-zeroentries in the random matrix Ha, and the larger the number of degrees offreedom.

• The lengths Lt and Lr of the transmit and receive antenna arrays. With smallantenna array lengths, many distinct multipaths may all be lumped into asingle resolvable path. Increasing the array apertures allows the resolution

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Dependency on Arrary Size

50

321 7.3 Modeling of MIMO fading channels

are the total angular spreads of the clusters at the transmitter and at thereceiver, respectively. This formula shows explicitly the separate effectsof the antenna array and of the multipath environment on the number ofdegrees of freedom. The larger the angular spreads the more degrees offreedom there are. For fixed angular spreads, increasing the antenna arraylengths allows zooming into and resolving the paths from each cluster,thus increasing the available degrees of freedom (Figure 7.17).One can draw an analogy between the formula (7.78) and the classic

fact that signals with bandwidth W and duration T have approximately2WT degrees of freedom (cf. Discussion 2.1). Here, the antenna arraylengths Lt and Lr play the role of the bandwidth W , and the total angularspreads !t"total and !r"total play the role of the signal duration T .

Effect of carrier frequencyAs an application of the formula (7.78), consider the question of howthe available number of degrees of freedom in a MIMO channel dependson the carrier frequency used. Recall that the array lengths Lt and Lr

are quantities normalized to the carrier wavelength. Hence, for a fixedphysical length of the antenna arrays, the normalized lengths Lt and Lr

increase with the carrier frequency. Viewed in isolation, this fact wouldsuggest an increase in the number of degrees of freedom with the carrierfrequency; this is consistent with the intuition that, at higher carrier fre-quencies, one can pack more antenna elements in a given amount of areaon the device. On the other hand, the angular spread of the environment

Cluster of scatterers

(a) Array length of L1

(b) Array length of L2 > L1

Cluster of scatterers

Receivearray

Receivearray

1/L1 1/L1

1/L21/L2

Transmitarray

Transmitarray

Figure 7.17 Increasing the antenna array apertures increases path resolvability in the angulardomain and the degrees of freedom.

Better angular resolvability when L is larger

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Dependency on Carrier Frequency

51

322 MIMO I: spatial multiplexing and channel modeling

typically decreases with the carrier frequency. The reasons aretwo-fold:• signals at higher frequency attenuate more after passing through orbouncing off channel objects, thus reducing the number of effectiveclusters;

• at higher frequency the wavelength is small relative to the feature sizeof typical channel objects, so scattering appears to be more specular innature and results in smaller angular spread.

These factors combine to reduce !t"total and !r"total as the carrier frequencyincreases. Thus the impact of carrier frequency on the overall degrees offreedom is not necessarily monotonic. A set of indoor measurements isshown in Figure 7.18. The number of degrees of freedom increases andthen decreases with the carrier frequency, and there is in fact an optimalfrequency at which the number of degrees of freedom is maximized. Thisexample shows the importance of taking into account both the physicalenvironment as well as the antenna arrays in determining the availabledegrees of freedom in a MIMO channel.

2 3 4 5 6 70

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

2 3 4 5 6 70

1

2

3

4

5

6

7

Frequency (GHz) Frequency (GHz)

(b)(a)

Ω total in townhouse

Ω to

tal

Ω to

tal /

λ c (m

−1)

1/λ (

m-1

)

1/λ c

Ω total in officeOfficeTownhouse

0

5

10

15

20

25

8 8

Figure 7.18 (a) The total angular spread !total of the scattering environment (assumed equal atthe transmitter side and at the receiver side) decreases with the carrier frequency; the normalizedarray length increases proportional to 1/"c . (b) The number of degrees of freedom of the MIMOchannel, proportional to !total/"c , first increases and then decreases with the carrier frequency.The data are taken from [91].

DiversityIn this chapter, we have focused on the phenomenon of spatial multiplexingand the key parameter is the number of degrees of freedom. In a slow fadingenvironment, another important parameter is the amount of diversity in thechannel. This is the number of independent channel gains that have to be ina deep fade for the entire channel to be in deep fade. In the angular domainMIMO model, the amount of diversity is simply the number of non-zero

• 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 ↓

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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

52

323 7.3 Modeling of MIMO fading channels

Figure 7.19 Angular domainrepresentation of three MIMOchannels. They all have fourdegrees of freedom but theyhave diversity 4, 8 and 16respectively. They modelchannels with increasingamounts of bounces in thepaths (cf. Figure 7.15).

(a)

nt

n r n r n r

nt nt

(b) (c)

entries in Ha. Some examples are shown in Figure 7.19. Note that channelsthat have the same degrees of freedom can have very different amounts ofdiversity. The number of degrees of freedom depends primarily on the angularspreads of the scatters/reflectors at the transmitter and at the receiver, whilethe amount of diversity depends also on the degree of connectivity betweenthe transmit and receive angles. In a channel with multiple-bounced paths,signals sent along one transmit angle can arrive at several receive angles(see Figure 7.15). Such a channel would have more diversity than one withsingle-bounced paths with signal sent along one transmit angle received at aunique angle, even though the angular spreads may be the same.

7.3.7 Dependency on antenna spacing

So far we have been primarily focusing on the case of critically spacedantennas (i.e., antenna separations !t and !r are half the carrier wavelength).What is the impact of changing the antenna separation on the channel statisticsand the key channel parameters such as the number of degrees of freedom?To answer this question, we fix the antenna array lengths Lt and Lr and vary

the antenna separation, or equivalently the number of antenna elements. Letus just focus on the receiver side; the transmitter side is analogous. Given theantenna array length Lr , the beamforming patterns associated with the basisvectors "er#k/Lr$%k all have beam widths of 2/Lr (Figure 7.12). This dictatesthe maximum possible resolution of the antenna array: paths that arrive withinan angular window of width 1/Lr cannot be resolved no matter how manyantenna elements there are. There are 2Lr such angular windows, partitioningall the receive directions (Figure 7.20). Whether or not this maximum reso-lution can actually be achieved depends on the number of antenna elements.Recall that the bins !k can be interpreted as the set of all physical

paths which have most of their energy along the basis vector et#k/Lr$. Thebins dictate the resolvability of the antenna array. In the critically spaced case#!r = 1/2), the beamforming patterns of all the basis vectors have a singlemain lobe (together with its mirror image). There is a one-to-one correspon-dence between the angular windows and the resolvable bins !k, and pathsarriving in different windows can be resolved by the array (Figure 7.21). In

Div = 4 Div = 8 Div = 16

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53

318 MIMO I: spatial multiplexing and channel modeling

7.3.6 Degrees of freedom and diversity

Degrees of freedomGiven the statistical model, one can quantify the spatial multiplexing capa-bility of a MIMO channel. With probability 1, the rank of the random matrixHa is given by

rank!Ha"=min#number of non-zero rows, number of non-zero columns$

(7.74)

(Exercise 7.6). This yields the number of degrees of freedom available in theMIMO channel.

The number of non-zero rows and columns depends in turn on two separatefactors:

• The amount of scattering and reflection in the multipath environment. The

Figure 7.15 Some examples ofHa . (a) Two clusters ofscatterers, with all paths goingthrough a single bounce.(b) Paths scattered via multiplebounces.

more scatterers and reflectors there are, the larger the number of non-zeroentries in the random matrix Ha, and the larger the number of degrees offreedom.

• The lengths Lt and Lr of the transmit and receive antenna arrays. With smallantenna array lengths, many distinct multipaths may all be lumped into asingle resolvable path. Increasing the array apertures allows the resolution

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Amount of diversity ↑ when there are more bounces

Page 54: Lecture6:MIMOChanneland SpatialMultiplexinghomepage.ntu.edu.tw/~ihwang/Teaching/Sp14/Slides/Lecture...Lecture6:MIMOChanneland SpatialMultiplexing I-Hsiang Wang ihwang@ntu.edu.tw 5/1,

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 = min2Lt , 2Lr = minnt , nr when the antennas are critically spaced

54

317 7.3 Modeling of MIMO fading channels

7.3.5 Statistical modeling in the angular domain

The basis for the statistical modeling of MIMO fading channels is the approxi-mation that the physical paths are partitioned into angularly resolvable bins andaggregated to form resolvable pathswhose gains are ha

kl!m". Assuming that thegains ab

i !m" of the physical paths are independent, we can model the resolvablepathgainsha

kl!m" as independent.Moreover, the angles #$ri!m"%m and #$ti!m"%mtypically evolve at a much slower time-scale than the gains #ab

i !m"%m; there-fore, within the time-scale of interest it is reasonable to assume that paths donot move from one angular bin to another, and the processes #ha

kl!m"%m can bemodeled as independent acrossk and l (seeTable 2.1 inSection 2.3 for the analo-gous situation for frequency-selective channels). In an angular bin &k' l(, wherethere are many physical paths, one can invoke the Central Limit Theorem andapproximate the aggregate gain ha

kl!m" as a complex circular symmetric Gaus-sian process. On the other hand, in an angular bin &k' l( that contains no paths,the entries ha

kl!m" can be approximated as 0. For a channel with limited angularspread at the receiver and/or the transmitter,many entries ofHa!m"maybe zero.Some examples are shown in Figures 7.14 and 7.15.

Figure 7.14 Some examples ofHa . (a) Small angular spread atthe transmitter, such as thechannel in Figure 7.10(a). (b)Small angular spread at thereceiver, such as the channel inFigure 7.10(b). (c) Smallangular spreads at both thetransmitter and the receiver. (d)Full angular spreads at both thetransmitter and the receiver.

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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

326 MIMO I: spatial multiplexing and channel modeling

Figure 7.23 Antennas aredensely spaced. Some binscontain no physical paths.

0

0

7

8

9 1

2

3

219

8

k0 1 98765432

Empty bins

L r = 3, n r = 10

Figure 7.24 A typical Ha

when the antennas aredensely spaced.

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1

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L = 16, n = 50

|hkl

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received in more bins, thus increasing the number of degrees of freedom(Figure 7.25). In terms of the spatial channel matrix H, this has the effect ofmaking the entries look more random and independent. At a base-station ona high tower with few local scatterers, the angular spread of the multipaths issmall and therefore one has to put the antennas many wavelengths apart todecorrelate the channel gains.

Sampling interpretationOne can give a sampling interpretation to the above results. First, think ofthe discrete antenna array as a sampling of an underlying continuous array!−Lr/2"Lr/2#. On this array, the received signal x$s% is a function of the

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Spatial-­‐Angular vs. Time-­‐Frequency

56

Time-Frequency Spatial-Angular

DomainsTime 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 minLt Ωt, Lr Ωr

Diversity # of non-zero delay bins

# of non-zero angular bins