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 ≤ k log k 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 ) 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 ... 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 ... - 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 SIMO: - 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 r ) ∗ H = ae − j 2 π d = ⇒ • Rank of H = 1 ⟹ no spatial multiplexing gain! • In line-of-sight MIMO, still power gain only 18