COMM 1003: Information

Theory

Parallel Channels MIMO

MIMO: Multiple Input – Multiple Output

WHY MIMO?

Frequency and Time are limited resources for increasing data rates

We have to find new avenues for parallel channels

Space processing itself is interesting because it does not increase

bandwidth (Think smart! More throughput in the same bandwidth)

2

𝟒 × 𝟐 MIMO Example

1. Can we Achieve Multiple Parallel Channels? 2. How Many Parallel Channels? 3. What are characteristics of these Parallel Channels ? 4. How can we use such Parallel Channels?

First Some Linear Algebra

Single Value Decomposition:

A rectangular matrix H can be broken into the product of three matrices:

An Orthogonal Matrix 𝑈

A Diagonal Matrix 𝐷

The Hermitian of an Orthogonal Matrix 𝑉

Hermitian?

𝑉 =

𝑣11 𝑣12 … 𝑣1𝑛𝑣21 𝑣22 … 𝑣2𝑛⋮ ⋮ ⋱ ⋮𝑣𝑛1 𝑣𝑛2 … 𝑣𝑛𝑛

⇒ 𝑉′ =

𝑣11∗ 𝑣21

∗ … 𝑣𝑛1∗

𝑣12∗ 𝑣22

∗ … 𝑣𝑛2∗

⋮ ⋮ ⋱ ⋮𝑣1𝑛∗ 𝑣2𝑛

∗ … 𝑣𝑛𝑛∗

3

Hermitian is the Conjugate Transpose Matrix (𝑣𝑖𝑗∗ is the complex conjugate of 𝑣𝑖𝑗)

First Some Linear Algebra

Single Value Decomposition:

A rectangular matrix H can be broken into the product of three matrices:

An Orthogonal Matrix U

A Diagonal Matrix D

The Hermitian of an Orthogonal Matrix V

Orthogonal?

𝑉′ × 𝑉 =

1 0 … 0 0 1 … 0⋮ ⋮ ⋱ ⋮0 0 … 1

= 𝐼𝑛

Example: The Constellation Diagram Coordinate System

Real-Axis

Imag-Axis

𝑉 =1 00 𝑗

⇒ 𝑉′ × 𝑉 =1 00 −𝑗

×1 00 𝑗

=1 00 1

= 𝐼2

4

Single Value Decomposition

Let 𝐻 be an 𝑚 × 𝑛 complex matrix. Then there exists

orthogonal matrices 𝑈 of size 𝑚 ×𝑚 and 𝑉 of size 𝑛 × 𝑛 such

that

𝑯 = 𝑼𝑫𝑽′ where

𝐷 is an 𝑚 × 𝑛 matrix with non-diagonal entries of all zeros and

𝑑11 ≥ 𝑑22 ≥ … ≥ 𝑑𝑝𝑝 ≥ 0, 𝑝 = min 𝑚, 𝑛 .

𝐷 =

𝑑11 0 … 0 0 … 00 𝑑22 … 0 0 … 0⋮ ⋮ ⋱ ⋮ ⋮ ⋱ ⋮0 0 … 𝑑𝑚𝑚 0 … 0

𝐷 =

𝑑11 0 … 00 𝑑22 … 0⋮ ⋮ ⋱ ⋮0 0 … 𝑑𝑛𝑛0 0 … 0⋮ ⋮ ⋱ ⋮0 0 … 0

5

𝒎 < 𝒏

𝒏 < 𝒎

Single Value Decomposition

𝑯 = 𝑼𝑫𝑽′

𝑈′𝑈 = 𝐼𝑚

𝑉′𝑉 = 𝐼𝑛

𝑈 is constituted from the 𝑚 eigenvectors of 𝐻×𝐻′

𝑉 is constituted from the 𝑛 eigenvectors of 𝐻′×𝐻

where 𝐻 is an 𝑚 × 𝑛 complex matrix.

𝐷 is a diagonal matrix containing the square roots

of eigenvalues from 𝑈 or 𝑉 in descending order.

6

SVD Example

Let 𝐻 = ℎ11 ℎ12

𝑈 = 𝑈1 is constituted from the 1 eigenvector of 𝐻×𝐻′

Eigenvector-Eigenvalue Problem for 𝑼:

𝐻 × 𝐻′ 𝑈1 = 𝜆1𝑈1

⇒ ℎ11 ℎ12 ×ℎ11∗

ℎ12∗ 𝑈1 = 𝜆1𝑈1

⇒ ℎ11ℎ11∗ + ℎ11ℎ12

∗ 𝑈1 = 𝜆1𝑈1

⇒ ℎ112 + ℎ12

2 𝑈1 = 𝜆1𝑈1

⇒ 𝜆1 = ℎ112 + ℎ12

2

⇒ 𝑈1 = 1 and 𝑈 = 1

7

SVD Example

Let 𝐻 = ℎ11 ℎ12

𝑉 is constituted from the 2 eigenvectors of 𝐻′×𝐻

𝑉 = 𝑉1 𝑉2 =𝑣11 𝑣21𝑣12 𝑣22

Eigenvector-Eigenvalue Problem for 𝑽: 𝐻′ × 𝐻 𝑉𝑖 = 𝜆𝑖 , 𝑉𝑖 𝑎𝑛𝑑 𝑖 = 1,2

⇒ ℎ11∗

ℎ12∗ × ℎ11 ℎ12 𝑉𝑖 = 𝜆𝑖𝑉𝑖

⇒ ℎ112 ℎ11

∗ ℎ12ℎ12∗ ℎ11 ℎ12

2 𝑉𝑖 = 𝜆𝑖𝑉𝑖

⇒ ℎ112 ℎ11

∗ ℎ12ℎ12∗ ℎ11 ℎ12

2 − 𝜆𝑖𝐼2 𝑉𝑖 = 0

⇒ ℎ112 ℎ11

∗ ℎ12ℎ12∗ ℎ11 ℎ12

2 −𝜆𝑖 00 𝜆𝑖

𝑣𝑖1𝑣𝑖2= 0

8

Matrix (𝐻′ × 𝐻) Multiplied by a Vector = Constant (𝜆, i.e., eigenvalue) Multiplied by a Vector

SVD Example

Let 𝐻 = ℎ11 ℎ12

𝑉 is constituted from the 2 eigenvectors of 𝐻′×𝐻

𝑉 = 𝑉1 𝑉2 =𝑣11 𝑣21𝑣12 𝑣22

Eigenvector-Eigenvalue Problem for 𝑽:

⇒ ℎ112 − 𝜆𝑖 ℎ11

∗ ℎ12ℎ12∗ ℎ11 ℎ12

2 − 𝜆𝑖

𝑣𝑖1𝑣𝑖2= 0

⇒ ℎ112 − 𝜆𝑖 𝑣𝑖1 + ℎ11

∗ ℎ12 𝑣𝑖2 = 0 → 𝐼

⇒ ℎ12∗ ℎ11 𝑣𝑖1 + ℎ12

2 − 𝜆𝑖 𝑣𝑖2 = 0 → 𝐼𝐼

𝐅𝐫𝐨𝐦 𝑰:

𝑣𝑖1 = −ℎ11∗ ℎ12ℎ112 − 𝜆𝑖

𝑣𝑖2

𝐒𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐞 𝐢𝐧 𝑰𝑰:

− ℎ12∗ ℎ11

ℎ11∗ ℎ12ℎ112 − 𝜆𝑖

𝑣𝑖2 + ℎ122 − 𝜆𝑖 𝑣𝑖2 = 0

9

SVD Example

Let 𝐻 = ℎ11 ℎ12

𝑉 is constituted from the 2 eigenvectors of 𝐻′×𝐻

𝑉 = 𝑉1 𝑉2 =𝑣11 𝑣21𝑣12 𝑣22

Eigenvector-Eigenvalue Problem for 𝑽:

⇒ − ℎ12∗ ℎ11

ℎ11∗ ℎ12ℎ112 − 𝜆𝑖

+ ℎ122 − 𝜆𝑖 𝑣𝑖2 = 0

⇒ − ℎ112 ℎ12

2 + ℎ112 − 𝜆𝑖 ℎ12

2 − 𝜆𝑖 = 0 ⇒ − ℎ11

2 ℎ122 + 𝜆𝑖

2 − ℎ112 + ℎ12

2 𝜆𝑖 + ℎ112 ℎ12

2 = 0

⇒ 𝜆𝑖2 − ℎ11

2 + ℎ122 𝜆𝑖 = 0

⇒ 𝜆𝑖 𝜆𝑖 − ℎ112 + ℎ12

2 = 0

⇒ 𝜆1 = ℎ112 + ℎ12

2 and 𝜆2 = 0

10

SVD Example

11

Let 𝐻 = ℎ11 ℎ12

𝑉 is constituted from the 2 eigenvectors of 𝐻′×𝐻

𝑉 = 𝑉1 𝑉2 =𝑣11 𝑣21𝑣12 𝑣22

Eigenvector-Eigenvalue Problem for 𝑽:

Substituting for 𝜆1 = ℎ112 + ℎ12

2 back in (I )

(𝑖. 𝑒., 𝜆𝑖 − ℎ112 𝑣𝑖1 + ℎ11

∗ ℎ12 𝑣𝑖2 = 0)

⇒ 𝑣𝑖1 = −ℎ11∗ ℎ12ℎ112 − 𝜆𝑖

𝑣𝑖2

⇒ 𝑣11 =ℎ11∗ ℎ12ℎ122 𝑣12

⇒ 𝑣11 =ℎ11∗ ℎ12ℎ12ℎ12

∗ 𝑣12

⇒ 𝑣11 =ℎ11∗

ℎ12∗ 𝑣12

SVD Example

12

Let 𝐻 = ℎ11 ℎ12

𝑉 is constituted from the 2 eigenvectors of 𝐻′×𝐻

𝑉 = 𝑉1 𝑉2 =𝑣11 𝑣21𝑣12 𝑣22

Eigenvector-Eigenvalue Problem for 𝑽:

⇒Since 𝑉1 =𝑣11𝑣12

is a unit vector

⇒ 𝑣112 + 𝑣12

2 = 1

⇒ℎ11∗

ℎ12∗ 𝑣12 ×

ℎ11ℎ12𝑣12∗ + 𝑣12

2 = 1

⇒ 𝑣122 =

1

1 +ℎ112

ℎ122

⇒ 𝑣122 =

ℎ122

ℎ112 + ℎ12

2

⇒ 𝑣12 =ℎ12∗

ℎ112 + ℎ12

2, 𝑣11 =

ℎ11∗

ℎ112 + ℎ12

2

SVD Example

Let 𝐻 = ℎ11 ℎ12

𝑉 is constituted from the 2 eigenvectors of 𝐻′×𝐻

𝑉 = 𝑉1 𝑉2 =𝑣11 𝑣21𝑣12 𝑣22

Eigenvector-Eigenvalue Problem for 𝑽:

Substituting for 𝜆2 = 0 back in (I )

(𝑖. 𝑒., 𝜆𝑖 − ℎ112 𝑣𝑖1 + ℎ11

∗ ℎ12 𝑣𝑖2 = 0)

⇒ 𝑣21 = −ℎ11∗ ℎ12ℎ112 − 𝜆2

𝑣22 ⇒ 𝑣21 = −ℎ11∗ ℎ12ℎ112 𝑣22

⇒ 𝑣21 = −ℎ11∗ ℎ12ℎ11ℎ11

∗ 𝑣22

⇒ 𝑣21 = −ℎ12ℎ11𝑣22

13

SVD Example

Let 𝐻 = ℎ11 ℎ12

𝑉 is constituted from the 2 eigenvectors of 𝐻′×𝐻

𝑉 = 𝑉1 𝑉2 =𝑣11 𝑣21𝑣12 𝑣22

Eigenvector-Eigenvalue Problem for 𝑽:

⇒Since 𝑉2 =𝑣21𝑣22

is a unit vector

⇒ 𝑣212 + 𝑣22

2 = 1

⇒ −ℎ12ℎ11𝑣22 × −

ℎ12∗

ℎ11∗ 𝑣22∗ + 𝑣22

2 = 1

⇒ 𝑣222 =

1

1 +ℎ122

ℎ112

⇒ 𝑣222 =

ℎ112

ℎ112 + ℎ12

2

⇒ 𝑣22 =−ℎ11

ℎ112 + ℎ12

2, 𝑣21 =

ℎ12

ℎ112 + ℎ12

2

14

SVD Example

Let 𝐻 = ℎ11 ℎ12

𝑉 is constituted from the 2 eigenvectors of 𝐻′×𝐻

𝑉 = 𝑉1 𝑉2 =𝑣11 𝑣21𝑣12 𝑣22

Eigenvector-Eigenvalue Problem for 𝑽:

𝑉 = 𝑉1 𝑉2 =

ℎ11∗

ℎ112 + ℎ12

2

ℎ12

ℎ112 + ℎ12

2

ℎ12∗

ℎ112 + ℎ12

2

−ℎ11

ℎ112 + ℎ12

2

𝑉′ × 𝑉 =1

ℎ11 2+ ℎ12 2

ℎ11 ℎ12ℎ12∗ −ℎ11

∗ × ℎ11∗ ℎ12ℎ12∗ −ℎ11

=1 00 1

= 𝐼2

15

SVD Example

𝐻 = ℎ11 ℎ12

Lets Now See, Does SVD Work? 𝐻 = 𝑈𝐷𝑉′ 𝑈 = 1

𝐷 = 𝜆1 0 = ℎ112 + ℎ12

2 0

𝑉 =1

ℎ112+ ℎ12

2 ℎ11∗ ℎ12ℎ12∗ −ℎ11

𝑈𝐷𝑉′ = 1 × ℎ112 + ℎ12

2 01

ℎ112 + ℎ12

2 ℎ11 ℎ12ℎ12∗ −ℎ11

∗

𝑼𝑫𝑽′ = 𝒉𝟏𝟏 𝒉𝟏𝟐 = 𝑯

16

MIMO Channel Model

17

TX RX

𝒉𝟏𝟏

𝒉𝟏𝟐 𝒉𝟏𝒏

𝑛 - Rx Antenna

𝒉𝟐𝟏

𝒉𝟐𝟐

𝒉𝟐𝒏

𝒉𝒎𝟏 𝒉𝒎𝟐

𝒉𝒎𝒏

𝒙𝟏

𝒙𝟐

𝒙𝒎

𝒚𝟏

𝒚𝟐

𝒚𝒏

𝑚 - Tx Antenna

𝑦1 = ℎ11𝑥1 + ℎ21𝑥2 +⋯+ ℎ𝑚1𝑥𝑚 + 𝑧1

𝑦2 = ℎ12𝑥1 + ℎ22𝑥2 +⋯+ ℎ𝑚2𝑥𝑚 + 𝑧2

𝑦𝑛 = ℎ1𝑛𝑥1 + ℎ2𝑛𝑥2 +⋯+ ℎ𝑚𝑛𝑥𝑚 + 𝑧𝑛

𝑤ℎ𝑒𝑟𝑒 𝑧1, 𝑧2, … , 𝑧𝑛 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑁𝑜𝑖𝑠𝑒 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑓𝑜𝑟 𝑅𝑥 − 𝑎𝑛𝑒𝑡𝑛𝑛𝑎 1, 2, …𝑛

Define

𝑋 = 𝑥1 𝑥2 … 𝑥𝑚

𝑌 = 𝑦1 𝑦2 … 𝑦𝑛

𝑍 = 𝑧1 𝑧2 … 𝑧𝑛

𝐻 =

ℎ11 ℎ12 … ℎ1𝑛ℎ21 ℎ22 … ℎ2𝑛⋮ ⋮ ⋱ ⋮ℎ𝑚1 ℎ𝑚2 … ℎ𝑚𝑛

𝒀 = 𝑿𝑯+ 𝒁 = 𝑿 𝑼𝑫𝑽′ + 𝒁

MIMO Channel Model

18

𝒙𝟐

𝒙𝒎

𝑦1 = ℎ11𝑥1 + ℎ21𝑥2 +⋯+ ℎ𝑚1𝑥𝑚 + 𝑧1

𝑦2 = ℎ12𝑥1 + ℎ22𝑥2 +⋯+ ℎ𝑚2𝑥𝑚 + 𝑧2

𝑦𝑛 = ℎ1𝑛𝑥1 + ℎ2𝑛𝑥2 +⋯+ ℎ𝑚𝑛𝑥𝑚 + 𝑧𝑛

𝑤ℎ𝑒𝑟𝑒 𝑧1, 𝑧2, … , 𝑧𝑛 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑁𝑜𝑖𝑠𝑒 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑓𝑜𝑟 𝑅𝑥 − 𝑎𝑛𝑒𝑡𝑛𝑛𝑎 1, 2, …𝑛

Define

𝑋 = 𝑥1 𝑥2 … 𝑥𝑚

𝑌 = 𝑦1 𝑦2 … 𝑦𝑛

𝑍 = 𝑧1 𝑧2 … 𝑧𝑛

𝐻 =

ℎ11 ℎ12 … ℎ1𝑛ℎ21 ℎ22 … ℎ2𝑛⋮ ⋮ ⋱ ⋮ℎ𝑚1 ℎ𝑚2 … ℎ𝑚𝑛

𝒀 = 𝑿𝑯+ 𝒁 = 𝑿 𝑼𝑫𝑽′ + 𝒁

𝒙𝟏

𝑈 V’

𝜆1

𝜆2

𝜆𝑘

𝐍𝐨𝐭𝐞: 𝑘 = min 𝑚, 𝑛

𝒚𝟏

𝒚𝟐

𝒛𝟏

𝒛𝟐

𝒛𝒏

𝒚𝒏

𝑯

MIMO Channel Model

19

𝒙𝟐

𝒙𝒎

𝑌 = 𝑋𝐻 + 𝑍 = 𝑋 𝑈𝐷𝑉′ + 𝑍

𝒙𝟏

𝑈 V’

𝜆1

𝜆2

𝜆𝑘

𝐍𝐨𝐭𝐞: 𝑘 = min 𝑚, 𝑛

𝒚𝟏

𝒚𝟐

𝒛𝟏

𝒛𝟐

𝒛𝒏

𝒚𝒏

𝑯

𝑈′

Pre-Processing (Precoding)

𝒔𝟐

𝒔𝒎

𝒔𝟏

Define

𝑆 ⇒ 𝑋 = 𝑆𝑈′ 𝑆 = 𝑠1 𝑠2 … 𝑠𝑚

𝑌 = 𝑆𝑈′ 𝑈𝐷𝑉′ + 𝑍 = 𝑆𝐷𝑉′ + 𝑍

𝑉

𝒓𝟐

𝒓𝒏

𝒓𝟏

𝑅 ⇒ 𝑅 = 𝑌𝑉 𝑅 = 𝑟1 𝑟2 … 𝑟𝑛

Post-Processing

𝑅 = 𝑌𝑉 = 𝑆𝐷𝑉′ 𝑉 + 𝑍𝑉 = 𝑆𝐷 + 𝑍𝑉

𝑅 = 𝑆𝐷 +𝑊

𝑊 = 𝑍𝑉

Important Note: 𝑊 is just a rotation of 𝑊 and has the same statistical properties.

Equivalent MIMO Channel Model

20

𝜆1

𝜆2

𝜆𝑘

𝐍𝐨𝐭𝐞: 𝑘 = min 𝑚, 𝑛

𝒘𝟏

𝒘𝒌

𝒔𝟐

𝒔𝒌

𝒔𝟏

where

𝒓𝟐

𝒓𝒌

𝒓𝟏

𝑅 = 𝑆𝐷 +𝑊

𝒘𝟐

𝑆 = 𝑠1 𝑥2 … 𝑠𝑚

𝑅 = 𝑟1 𝑟2 … 𝑟𝑛

𝑊 = 𝑤1 𝑤2 … 𝑤𝑛

𝐷 =

𝜆1 0 … 0 0 … 0

0 𝜆2 … 0 0 … 0

⋮ ⋮ ⋱ ⋮ ⋮ ⋱ ⋮

0 0 … 𝜆𝑘 0 … 0

𝐷 =

𝜆1 0 … 0

0 𝜆2 … 0

⋮ ⋮ ⋱ ⋮

0 0 … 𝜆𝑘0 0 … 0⋮ ⋮ ⋱ ⋮0 0 … 0

𝑘 = 𝑚

𝑘 = 𝑛

Modes for MIMO Operation

21

𝜆1

𝜆2

𝜆𝑘 𝒘𝒌

𝒔𝟐

𝒔𝒌

𝒔𝟏

𝒓𝟐

𝒓𝒌

𝒓𝟏

𝒘𝟐

𝒘𝟏

Multiplexing Diversity

𝜆1

𝜆2

𝜆𝑘 𝒘𝒌

𝒔

𝒔

𝒔

𝒓

𝒓

𝒓

𝒘𝟐

𝒘𝟏

Sending a Different Symbol on each Orthogonal Channel

Main target is increasing data rate

Given 𝑝𝑖 = 𝐸 𝑠𝑖2 is power

allocated on channel 𝑖:

Sending the same Symbol on all Orthogonal Channels

Main target is enhancing SNR (and in turn data rate as well)

Given 𝑝𝑖 is power allocated on channel 𝑖 for each symbol 𝑠:

𝐶 = log 1 +𝑝𝑖𝜆𝑖𝜎2

𝑘

𝑖=1

𝐶 = log 1 + 𝑝𝑖𝜆𝑖𝜎2

𝑘

𝑖=1

Capacities of MIMO Multiplexing

With Channel Knowledge

22

𝜆1

𝜆2

𝑝1

𝜆3

𝜆4

𝑝2

𝑝3

𝑝4

V’

Pre-Processing(Precoding)

Post-Processing

It is possible to implement water filling solution:

𝑃𝑇 = 𝑝𝑖

𝑘

𝑖=1

𝑝𝑖 = max 0, 𝛽 −𝜎2

𝜆𝑖

𝐶 = log 1 +𝑝𝑖𝜆𝑖𝜎2

𝑘

𝑖=1

Capacities of MIMO Multiplexing

Without Channel Knowledge

23

𝜆1

𝜆2

𝑃𝑇/𝑚

𝜆3

𝜆4

We are forced to transmit equal power from ALL Tx Antenna. We know that some of Tx power is wasted

𝑝𝑖 =𝑃𝑇𝑚

𝐶 = log 1 +𝑃𝑇𝜆𝑖𝑚𝜎2

𝑘

𝑖=1

𝑃𝑇/𝑚

𝑃𝑇/𝑚

𝑃𝑇/𝑚

V’

𝑃𝑇/𝑚

𝑃𝑇/𝑚

𝑃𝑇/𝑚

Capacities of MIMO Diversity

With Channel Knowledge

24

V’

Pre-Processing(Precoding)

Post-Processing

We can direct all power to strongest channel

𝐶 = log 1 +𝑃𝑇𝜆𝑚𝑎𝑥𝜎2

Capacities of MIMO Diversity

Without Channel Knowledge

25

𝜆1

𝜆2

𝑃𝑇/𝑚

𝜆3

𝜆4

We are forced to transmit equal power from ALL Tx Antenna.

𝑃𝑇/𝑚

𝑃𝑇/𝑚

𝑃𝑇/𝑚

V’

𝑃𝑇/𝑚

𝑃𝑇/𝑚

𝑃𝑇/𝑚

𝐶 = log 1 +𝑃𝑇𝑚𝜎2 𝜆𝑖

𝑘

𝑖=1

Classical Receive (Rx) Diversity (SIMO)

26

ℎ11

𝐻 = ℎ11 ℎ12

𝐻 = 𝑈𝐷𝑉′ = 1 × ℎ112 + ℎ12

2 01

ℎ112 + ℎ12

2 ℎ11 ℎ12ℎ12∗ −ℎ11

∗

MIMO Model: 𝑟 = ℎ112 + ℎ12

2𝑠 + 𝑤

𝐶𝑅𝑥−𝐷𝑖𝑣𝑒𝑟𝑠𝑖𝑡𝑦 = log 1 +𝑃𝑇 × ℎ11

2 + ℎ122

𝜎2

𝜆1 = ℎ11 2 + ℎ12 2

ℎ12 𝜆2 = 0

𝟎

𝒔

𝒓

𝒓

𝒘𝟐

𝒘𝟏

Equivalent MIMO Model

Classical Receive (Rx) Diversity (SIMO)

27

ℎ11

In MRC: 𝑟 = ℎ11𝑠 + 𝑧1 × ℎ11

∗ + ℎ12𝑠 + 𝑧1 × ℎ12∗

𝑟 = ℎ112 + ℎ12

2 𝑠 + ℎ11∗ 𝑧1 + ℎ12

∗ 𝑧2 𝑟 = ℎ11

2 + ℎ122 𝑠 + 𝑤

where 𝑤 is AWGN with Noise Power ℎ112 + ℎ12

2 𝜎2

Therefore SNR in MRC ⇒ 𝑆𝑁𝑅𝑀𝑅𝐶 = 𝑃𝑇 ℎ11

2+ ℎ122 2

ℎ11 2+ ℎ12 2 𝜎2 =𝑃𝑇 ℎ11

2+ ℎ122

𝜎2

Same as MIMO Model!!

ℎ12

𝒓

𝒛𝟏

Implementation Remember MRC in Wireless

ℎ11

𝒛𝟐 ℎ12

ℎ11∗

ℎ12∗

𝒔

Constellation Representation of MRC

28

ℎ11

ℎ12

𝒓

𝒛𝟏

Implementation Remember MRC in Wireless

ℎ11

𝒛𝟐 ℎ12

ℎ11∗

ℎ12∗

𝒔

Assume BPSK at Tx

At Rx #2

At Rx #1

𝒉𝟏𝟏

𝒉𝟏𝟐

𝟏

After MRC

𝒉𝟏𝟏𝒉𝟏𝟏∗ 𝒉𝟏𝟐𝒉𝟏𝟐

∗

Transmit Diversity

29

ℎ11

ℎ21

𝜆1 = ℎ11 2 + ℎ21 2

𝜆2 = 0

𝟎

𝒔

𝒓

𝒓

𝒘𝟐

𝒘𝟏

Equivalent MIMO Model with Channel Knowledge at TX (i.e., Beam Forming - BF)

𝐶𝑇𝑥−𝐷𝑖𝑣𝑒𝑟𝑠𝑖𝑡𝑦−𝐵𝐹 = log 1 +𝑃𝑇 × ℎ11

2 + ℎ212

𝜎2= 𝐶𝑅𝑥−𝐷𝑖𝑣𝑒𝑟𝑠𝑖𝑡𝑦

Without Channel Knowledge:

𝐶𝑇𝑥−𝐷𝑖𝑣𝑒𝑟𝑠𝑖𝑡𝑦 = log 1 +𝑃𝑇2×ℎ112 + ℎ21

2

𝜎2< 𝐶𝑅𝑥−𝐷𝑖𝑣𝑒𝑟𝑠𝑖𝑡𝑦

Transmit Diversity versus Receive Diversity

30

𝐶𝑇𝑥−𝑑𝑖𝑣𝑒𝑟𝑠𝑖𝑡𝑦−𝐵𝐹 = 𝑙𝑜𝑔2 1 + ℎ𝑖1

2𝑵𝒕𝒊=𝟏

𝜎2𝑃𝑇

𝐶𝑅𝑥−𝑑𝑖𝑣𝑒𝑟𝑠𝑖𝑡𝑦 = 𝑙𝑜𝑔2 1 + ℎ1𝑖

2𝑵𝒓𝒊=𝟏

𝜎2𝑃𝑇

𝐶𝑇𝑥−𝑑𝑖𝑣𝑒𝑟𝑠𝑖𝑡𝑦 = 𝑙𝑜𝑔2 1 + ℎ𝑖1

2𝑵𝒕𝒊=𝟏

𝑁𝑡𝜎2𝑃𝑇

Channel Knowledge at Rx

No Channel Knowledge at Tx

Channel Knowledge at Tx

Simulation of MIMO

Set 𝑚 and 𝑛 number of transmit and receive antenna

Generate 𝑁𝐶 (e.g., 100,000) realizations of the channel matrix 𝐻 (𝑚 × 𝑛)

Each element ℎ𝑖𝑗 = ℎ𝑖𝑗 𝑒𝑗𝜃𝑖𝑗

ℎ𝑖𝑗 is Rayleigh 𝑓 ℎ =ℎ

𝜎2 𝑒− ℎ 2𝜎

2 with 2𝜎2 = 1 (unit power on

each channel)

𝜃𝑖𝑗 is uniform in the interval 0,2𝜋

For each channel realization compute SVD to derive diagonal matrix 𝐷

Compute capacity based on Water Filling solution

𝑝𝑖 = max 0, 𝛽 −𝜎2

𝜆𝑖

𝑃𝑇 = 𝑝𝑖𝑘𝑖=1

𝐶 = log 1 +𝑝𝑖𝜆𝑖

𝜎2𝑘𝑖=1

Average the capacity over all 𝑁𝐶 channel realizations

31

Example for Multiplexing Capacity Curves

32