Block-Circulant Inverse Orthogonal Jacket Matrices and Its Applications to the Kronecker MIMO Channel

Abstract

This paper presents a note on the block-circulant generalized Hadamard matrices, which is called inverse orthogonal Jacket matrices of orders $N=2p, 4p, 4^kp, np$, where k is a positive integer for the Kronecker MIMO channel. The class of block Toeplitz circulant Jacket matrices not only have many properties of the circulant Hadamard conjecture but also have the construction of block-circulant, which can be easily applied to fast algorithms for decomposition. The matrix decomposition is with the form of the products of block identity $I_2$ matrix and block Hadamard $H_2$ matrix. In this paper, a block fading channel model is used, where the channel is constant during a transmission block and varies independently between transmission blocks. The proposed block-circulant Jacket matrices can also achieve about 3db gain in high SNR regime with MIMO channel. This algorithm for realizing these transforms can be applied to the Kronecker MIMO channel.

Appendix

1.1 The Proof of Eq. (1) is as follows:

To analyze the fundamentals of the Jacket patterns and their related properties, first the basic format is investigated: Symmetric Jacket Matrix of Rank 2. Some important properties and patterns are investigated.

Given an two-by-two symmetric pattern, such as

$$\begin{aligned} {[}A]_2=\left( \begin{array}{cc} a &{} b \\ b &{} -c \end{array}\right) . \end{aligned}$$

(42)

where a, b, c can be arbitrary.

Theorem A1

A two-by-two symmetric Jacket matrix pattern is

$$\begin{aligned} {[}J]_2=\left( \begin{array}{cc} a &{} \sqrt{ac} \\ \sqrt{ac} &{} -c \end{array}\right) \ \ or \ \ [J]_2=\left( \begin{array}{cc} a &{} -\sqrt{ac} \\ -\sqrt{ac} &{} -c \end{array}\right) , \end{aligned}$$

(43)

where the entries of this Jacket are nonzero.

Proof

If the two-by-two symmetric matrix pattern $[S]_2$ should have Eq. (44).

$$\begin{aligned} {[}A]_2^{-1}=\frac{1}{2}\left( \begin{array}{cc} a^{-1} &{} b^{-1} \\ b^{-1} &{} c^{-1} \end{array}\right) . \end{aligned}$$

(44)

then

$$\begin{aligned} {[}A]_2 \cdot [A]_2^{-1}=[I]_2, \left( \begin{array}{cc} a &{} b \\ b &{} -c \end{array}\right) \cdot \frac{1}{2}\left( \begin{array}{cc} a^{-1} &{} b^{-1} \\ b^{-1} &{} c^{-1} \end{array}\right) =[I]_2, \end{aligned}$$

(45)

thus it can be obtained

$$\begin{aligned} \frac{1}{2}\left( a\cdot \frac{1}{a} +b \cdot \frac{1}{b}\right) =1, \nonumber \\ \frac{1}{2}\left( a\cdot \frac{1}{b} -b \cdot \frac{1}{c}\right) =0, \nonumber \\ \frac{1}{2}\left( b\cdot \frac{1}{a} -c \cdot \frac{1}{b}\right) =0, \nonumber \\ \frac{1}{2}\left( b\cdot \frac{1}{b} +c \cdot \frac{1}{c}\right) =1, \nonumber \\ \end{aligned}$$

(46)

clearly, it should be forced that

$$\begin{aligned} \frac{a}{c}-\frac{b}{c}=0 \ \ \mathrm{and} \ \ \frac{b}{a}-\frac{c}{b}=0, \end{aligned}$$

(47)

thus

$$\begin{aligned} \frac{ac-b^2}{bc}=0 \ \ \mathrm{and} \ \ \frac{b^2-ac}{ab}=0, \end{aligned}$$

(48)

then it can be obtained

$$\begin{aligned} b=\pm \sqrt{ac}. \end{aligned}$$

(49)

$\square$

Lemma A1

An orthogonal symmetric two-by-two Jacket matrix considers always with Hadamard matrix, if the entries of the matrix are nonzero.

Proof

The orthogonal property has

$$\begin{aligned} {[}J]_2 \cdot [J]_2^T=2[I]_2, \end{aligned}$$

(50)

thus it can be obtained

$$\begin{aligned} {[}J]_2^T=2[J]_2^{-1}. \end{aligned}$$

(51)

If $[J]_2=\left( \begin{array}{cc} a &{} \sqrt{ac} \\ \sqrt{ac} &{} -c \end{array}\right)$, then it can be obtained

$$\begin{aligned}{}[J]_2[J]_2^T=\left( \begin{array}{cc} a &{} \sqrt{ac} \\ \sqrt{ac} &{} -c \end{array}\right) \left( \begin{array}{cc} a^{-1} &{} \sqrt{ac}^{-1} \\ \sqrt{ac}^{-1} &{} -c^{-1} \end{array}\right) =2[I]_2. \end{aligned}$$

(52)

If $[J]_2=\left( \begin{array}{cc} a &{} -\sqrt{ac} \\ -\sqrt{ac} &{} -c \end{array}\right)$, then it can be obtained

$$\begin{aligned}{}[J]_2[J]_2^T=\left( \begin{array}{cc} a &{} -\sqrt{ac} \\ -\sqrt{ac} &{} -c \end{array}\right) \left( \begin{array}{cc} a^{-1} &{} -\sqrt{ac}^{-1} \\ -\sqrt{ac}^{-1} &{} -c^{-1} \end{array}\right) =2[I]_2. \end{aligned}$$

(53)

Clearly, the solution is $a=c=1$. Thus, the Jacket pattern can be written by

$$\begin{aligned} {[}J]_2=\left( \begin{array}{cc} a &{} -\sqrt{ac} \\ -\sqrt{ac} &{} -c \end{array}\right) =\left( \begin{array}{cc} 1 &{} -1 \\ -1 &{} -1 \end{array}\right) , \end{aligned}$$

(54)

also

$$\begin{aligned} {[}J]_2=\left( \begin{array}{cc} a &{} \sqrt{ac} \\ \sqrt{ac} &{} c \end{array}\right) =\left( \begin{array}{cc} 1 &{} 1 \\ 1 &{} -1 \end{array}\right) , \end{aligned}$$

(55)

$\square$