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

Spatial ¡Multiplexing ¡in ¡ MIMO ¡Systems 5

MIMO ¡AWGN ¡Channel • MIMO AWGN channel (no fading): y [ m ] = Hx [ m ] + w [ m ] - y [ m ] ∈ C n r , x [ m ] ∈ C n t , H ∈ C n r × n t , w ∼ CN (0 , I n r ) - n t := # of Tx antennas; n r := # of Rx antennas - Tx power constraint P • Singular value decomposition (SVD) of matrix H : H = U Λ V ∗ - U ∈ C n r × n r , V ∈ C n t × n t ( UU ∗ = U ∗ U = I ) Unitary - Rectangular � � � � with zero off-diagonal elements and Λ ∈ C n r × n t diagonal elements λ 1 ≥ λ 2 ≥ · · · ≥ λ min( n t ,n r ) ≥ 0 - These λ ’s are the singular values of matrix H 6

MIMO ¡Capacity ¡via ¡SVD • Change of coordinate: y = Hx + w = U Λ V ∗ x + w ⇐ ⇒ U ∗ y = Λ V ∗ x + U ∗ w - Let � � y := U ∗ y , e � � x := V ∗ x , e � � � w := U ∗ w � � � , get an equivalent channel e y = Λ e e x + e w - 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 n min := min{ n t , n r } parallel channels: y i = λ i e e x i + e w i , i = 1 , 2 , . . . , n min - Capacity can be found via water-filling 7

Spatially ¡Parallel ¡Channels w H y x H = U Λ V ∗ λ 1 e w 1 y ... x V * U * V U e e x y λ n min e w n min 8

Multiplexing ¡over ¡Parallel ¡Channels ~ ~ { x 1 [ m ]} { y 1 [ m ]} AWGN Decoder coder . . n min . { w [ m ]} . information . . streams ~ ~ { x n min [ m ]} { y n min [ m ]} AWGN U * H V + Decoder coder {0} . . . {0} n min 1 + λ 2 ◆ + n min ✓ ◆ i P ∗ ν − σ 2 ✓ X i X C MIMO = log P ∗ P ∗ i = ν satisfies i = P , σ 2 λ 2 i =1 i i =1 9

Rank ¡= ¡# ¡of ¡Multiplexing ¡Channels λ 1 e w 1 y ... x V * U * V U e e x y λ n min e w n min • If λ i = 0 ⟹ the i -th channel contributes 0 to the capacity • Rank of H = # of non-zero singular values k 1 + λ 2 ✓ ◆ i P ∗ X i C MIMO = log k := rank ( H ) , σ 2 i =1 10

Rank ¡= ¡# ¡of ¡Multiplexing ¡Channels • DoF gain is more significant at high SNR • At high SNR, uniform power allocation is near-optimal: k k 1 + λ 2 ✓ λ 2 ✓ ◆ ◆ i P i P X X C MIMO ≈ log log ≈ k σ 2 k σ 2 i =1 i =1 k ✓ λ 2 ◆ X i = k log SNR + log k i =1 • Rank of H determines how many data streams can be multiplexed over the channel ⟹ k := multiplexing gain • Full rank matrix is the best ( ∵ k ≤ n min ) 11

Condition ¡Number • Full rank is not enough: k ✓ λ 2 ◆ X i C MIMO ≈ k log SNR + log k i =1 - If the some λ i < 1 , then log( λ i 2 / k ) will be negative - How to maximize the second term? P k ! k i =1 λ 2 ✓ λ 2 ◆ 1 • By Jensen’s inequality: X i i log ≤ log k 2 k k i =1 - i,j | h i,j | 2 For a family of full-rank channel matrices with fixed � P � � , i,j | h i,j | 2 = Tr ( HH ∗ ) = P k i =1 λ 2 since �� � � � � � � � � � , maximum is attained P i when all λ ’s are equal ⟺ λ max = λ min • Well-conditioned (smaller condition number λ max / λ min ) ones attain higher capacity 12

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 13

Physical ¡Modeling ¡of ¡ MIMO ¡Channels 14

Line-­‑of-­‑Sight ¡SIMO ¡Channel d φ ∆ r λ c carrier wavelength: λ c d i ... φ antenna spacing: ∆ r λ c Rx antenna i channel to i -th antenna: y = h x + w ... = ae − j 2 π di h i = ae − j 2 π fcdi λ c c - If distance d ≫ antenna distance spread, then d i = d + ( i − 1) ∆ r λ c cos φ , ∀ i = 1 , 2 , . . . , n r - Phase difference between consecutive antennas is 2 π ∆ r cos φ - e − j 2 π ( n r − 1) ∆ r cos φ ⇤ T h = ae − j 2 π d ⟹ e − j 2 π ∆ r cos φ λ c ⇥ 1 · · · - Channel vector h lies along the direction e r ( Ω ) , where 1 e − j 2 π ( n r − 1) ∆ r Ω ⇤ T e − j 2 π ∆ r Ω ⇥ Ω := cos φ , e r ( Ω ) := 1 · · · √ n r directional cosine 15

Line-­‑of-­‑Sight ¡MISO ¡Channel ... d i Tx antenna i φ ∆ t λ c y = h ∗ x + w d ... φ - If distance d ≫ antenna distance spread, then d i = d − ( i − 1) ∆ t λ c cos φ , ∀ i = 1 , 2 , . . . , n t - Phase difference between consecutive antennas is − 2 π ∆ t cos φ - e − j 2 π ( n t − 1) ∆ t cos φ ⇤ T h = ae j 2 π d ⟹ e − j 2 π ∆ t cos φ λ c ⇥ 1 · · · - Channel vector h lies along the direction e t ( Ω ) , where 1 e − j 2 π ( n t − 1) ∆ t Ω ⇤ T e − j 2 π ∆ t Ω ⇥ Ω := cos φ , e t ( Ω ) := 1 · · · √ n t directional cosine 16

Line-­‑of-­‑Sight ¡SIMO ¡and ¡MISO • Line-of-sight SIMO: - y = h x + w , h is along the receive spatial signature e r ( Ω ) , where 1 e − j 2 π ( n r − 1) ∆ r Ω ⇤ T e − j 2 π ∆ r Ω ⇥ e r ( Ω ) := 1 · · · √ n r - n r -fold power gain, no DoF gain • Line-of-sight MISO: - y = h * x + w , h is along the transmit spatial signature e t ( Ω ) , where 1 e − j 2 π ( n t − 1) ∆ t Ω ⇤ T e − j 2 π ∆ t Ω ⇥ e t ( Ω ) := 1 · · · √ n t - n t -fold power gain, no DoF gain 17

Line-­‑of-­‑Sight ¡MIMO ¡Channel d Tx k y = Hx + w ... ... d i,k Rx i d i,k = d + ( i − 1) ∆ r λ c cos φ r − ( k − 1) ∆ t λ c cos φ t ⇒ h i,k = ae − j 2 π d λ c e − j 2 π ( i − 1) ∆ r Ω r e j 2 π ( k − 1) ∆ t Ω t = λ c √ n t n r e r ( Ω r ) e t ( Ω t ) ∗ H = ae − j 2 π d = ⇒ • Rank of H = 1 ⟹ no spatial multiplexing gain! • In line-of-sight MIMO, still power gain ( n t × n r -fold) only 18

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 19

Single ¡ReTlector, ¡Two-­‑Paths ¡MIMO y = Hx + w Tx antenna 1 φ t 1 φ t 2 Rx antenna 1 φ r 2 φ r 1 • Two paths: - H 1 = a 1 e − j 2 π d 1 λ c √ n t n r e r ( Ω r 1 ) e t ( Ω t 1 ) ∗ Path 1: - H 2 = a 2 e − j 2 π d 2 Path 2: λ c √ n t n r e r ( Ω r 2 ) e t ( Ω t 2 ) ∗ - By the linear superposition principle, we get the channel matrix 1 e r ( Ω r 1 ) e t ( Ω t 1 ) ∗ + a b H = a b i := a i e − j 2 π di 2 e r ( Ω r 2 ) e t ( Ω t 2 ) ∗ a b λ c √ n t n r • rank( H ) = 2 ⟺ e r ( Ω r 1 ) ∦ e r ( Ω r 2 ) and e t ( Ω t 1 ) ∦ e t ( Ω t 2 ) : - Ω r 2 – Ω r 1 ≠ 0 � mod 1/ Δ r - Ω t 2 – Ω t 1 ≠ 0 � � mod 1/ Δ t 20

Single ¡ReTlector, ¡Two-­‑Paths ¡MIMO y = Hx + w A Tx antenna 1 φ t 1 φ t 2 B Rx antenna 1 φ r 2 φ r 1 1 e r ( Ω r 1 ) e t ( Ω t 1 ) ∗ + a b H = a b 2 e r ( Ω r 2 ) e t ( Ω t 2 ) ∗ • 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 -  e t ( Ω t 1 ) ∗ � H = H r H t , where a b a b ⇥ ⇤ H r = 1 e r ( Ω r 1 ) 2 e r ( Ω r 2 ) H t = , e t ( Ω t 2 ) ∗ - Note: {A,B} form a geographically separated virtual antenna array 21

{A,B} ¡to ¡Rx ¡Antenna ¡Array A a b 2 e r ( Ω r 2 )  x B � y = H r + w x A B a b 1 e r ( Ω r 1 ) Rx antenna 1 φ r 2 φ r 1 ⇥ a b a b ⇤ H r = 1 e r ( Ω r 1 ) 2 e r ( Ω r 2 ) • Observation: - H r has two columns along the directions e r ( Ω r 1 ) and e r ( Ω r 2 ) - The more aligned e r ( Ω r 1 ) and e r ( Ω r 2 ) are, the worse the conditioning of H r • The conditioning of H r depends on the angle θ between | cos θ | = | e r ( Ω r 1 ) ∗ e r ( Ω r 2 ) | e r ( Ω r 1 ) and e r ( Ω r 2 ) ; 22