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Blind Detection Design for AF Two-Way Relaying Over Frequency Selective Channels

Abstract

This paper considers an amplify-and-forward two-way relay system with frequency selective fading channels. The knowledge of channel state information is assumed not available for all nodes. An efficient transmission and signaling scheme is proposed to identify the transmitted symbols and multi-tap channel coefficients using the M-ary phase shift keying transmitted symbols. A blind receiver is derived using the least square error criterion. Simulation results demonstrate that the proposed transmission and signaling scheme outperforms an existing conventional training scheme where pilots are applied to estimate the channel coefficients.

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Appendices

Appendix

Proof of Proposition 1

Some Basic Definitions and Results in Galois Theory and Algebraic Number Theory

We first briefly introduce some definitions and properties in [23, 24], which play an important role in proving the proposition.

Galois Theory The main idea of Galois theory is to associate any polynomial f(x) with a group of permutations of the roots of f(x).

Definition 1

\({\mathbb {F}}\) is a field if \({\mathbb {F}}\) is an Abelian group under addition, \({\mathbb {F}}-\{0\}\) is an Abelian group under multiplication and multiplication distributes over addition.

If \({\mathbb {F}}\subseteq {\mathbb {K}}\) are fields, then, \({\mathbb {K}}\) is called a field extension of \({\mathbb {F}}\), which is denoted by \({\mathbb {K}}/{\mathbb {F}}\). Given a field extension \({\mathbb {K}}/{\mathbb {F}}\), then, \({\mathbb {K}}\) can be considered as a vector space over \({\mathbb {F}}\).

Cyclotomic Extension Given a polynomial over \({\mathbb {F}}\), consider a minimal field extension of \({\mathbb {F}}\) that contains all the roots of the polynomial. Given a field extension \({\mathbb {L}}/{\mathbb {K}}\) and an element \(\omega \) of \({\mathbb {L}}\), \({\mathbb {K}}(\omega )\) denotes the smallest subfield of \({\mathbb {L}}\) which contains \({\mathbb {K}}\) and \(\omega \). A field extension of the form \({\mathbb {L}}={\mathbb {K}}(\omega )\) is called a simple extension and \(\omega \) is called a primitive element of the extension and \(\omega \) is called a primitive element of the extension.

Definition 2

An n-th root of of the unity is said to be a primitive n-th root of the unity if its order is n, i.e. z is a primitive n-th root of unity if \(z^k\ne 1\) for \(k=1,2,\ldots ,n-1\).

Next, we consider the field extension \({\mathbb {F}}(\omega )/{\mathbb {F}}\), where \(\omega \) is an n-th root of unity.

Definition 3

If \(\omega \) is an n-th root in \({\mathbb {C}}\), then, the field extension \({\mathbb {F}}(\omega )/{\mathbb {F}}\) is called a cyclotomic extension. Particularly, if \(\omega \) is a primitive n-th root of unity in \({\mathbb {C}}\), then, \({\mathbb {Q}}(\omega )/{\mathbb {Q}}\) is called cyclotomic field, denoted by \({\mathbb {Q}}_n\).

Definition 4

An algebraic integer is a complex number that is a root of some monic polynomial (leading coefficient 1) with coefficients in \({\mathbb {Z}}\).

Specially for a primitive n-th root of unity \(\zeta _n\), we have the following properties:

Property 1

\({\mathbb {Z}}[\zeta _n]\) is the ring of all algebraic integers of the cyclotomic field \({\mathbb {Q}}_n={\mathbb {Q}}(\zeta _n)\) [23, 24].

If \(\epsilon \) is an algebraic integer of \({\mathbb {Q}}(\zeta _n)\), the norm of \(\epsilon \) has the following properties:

Property 2

If \(\epsilon \in {\mathbb {Z}}[\zeta _n]\), then, \(N_{\frac{{\mathbb {Q}}_n}{{\mathbb {Q}}}}(\epsilon )\) is a rational integer [23, 24].

Property 3

Let \(d=\gcd (m,n)\) denote the greatest common factor of m and n. Then, we have

$$\begin{aligned} {\mathbb {Z}}[\zeta _m]\cap {\mathbb {Z}}[\zeta _n]={\mathbb {Z}}[\zeta _d]. \end{aligned}$$
(19)

A Fact

To prove Proposition 1, we first state the following fact:

Fact 1

Let \(a_i, y_i\) be \(p_i\)-PSK with symbol “0” removed, all \(p_i \) are pairwise coprime, \(i=1, \ldots , N\). If \(\sum _{i=1}^N a_i= \sum _{i=1}^N y_i\), then \(a_i=y_i, i=1, \ldots , N\).

Firstly, we justify for \(N=2\), i.e., if \((a_1+a_2)=(y_1+y_2)\), then \(a_i=y_i, i=1,2 \).

Denote \(a_i=\exp (\frac{j 2\pi m_i}{p_i}), y_i=\exp (\frac{j 2\pi k_i}{p_i}), p_1-1\ge m_1,k_1\ge 1, p_2-1\ge m_2,k_2\ge 1\). We know \(e^{ja}+e^{jb}= e^{\frac{j(a+b)}{2}}( e^{\frac{j(a-b)}{2}}+e^{\frac{-j(a-b)}{2}}) =2\cos (\frac{a-b}{2})e^{\frac{j(a+b)}{2}}\), and \(e^{ja}-e^{jb}= e^{\frac{j(a+b)}{2}}( e^{\frac{j(a-b)}{2}}-e^{\frac{-j(a-b)}{2}}) =2j\sin (\frac{a-b}{2})e^{\frac{j(a+b)}{2}}\). From \(a_1+a_2=(y_1+y_2)\), we have

$$\begin{aligned}&\cos \left( \frac{2 \pi m_1}{2p_1}-\frac{2 \pi m_2}{2p_2} \right) e^{j\frac{2 \pi m_1}{2p_1}+j\frac{2 \pi m_2}{2p_2}} \\&\quad =\cos \left( \frac{2 \pi k_1}{2p_1}-\frac{2 \pi k_2}{2p_2} \right) e^{j\frac{2 \pi k_1}{2p_1}+j\frac{2 \pi k_2}{2p_2}}, \\&\text {i.e.,}\; \cos \left( \frac{\pi m_1}{p_1}-\frac{ \pi m_2}{p_2} \right) e^{j\frac{ \pi m_1}{p_1}+j\frac{ \pi m_2}{p_2}} \\&\quad = \cos \left( \frac{ \pi k_1}{p_1}-\frac{ \pi k_2}{p_2} \right) e^{j\frac{ \pi k_1}{p_1}+j\frac{ \pi k_2}{p_2}} \end{aligned}$$

We obtain \( |\cos (\frac{ \pi m_1}{p_1}-\frac{ \pi m_2}{p_2} )|=|\cos (\frac{\pi k_1}{p_1}-\frac{ \pi k_2}{p_2} )|\), i.e. \((\frac{ \pi m_1}{p_1}-\frac{ \pi m_2}{p_2} )= (\frac{\pi k_1}{p_1}-\frac{ \pi k_2}{p_2} )+K_1 \pi , K_1\) is an integer. The imaginary part in \(e^{j\frac{\pi m_1}{p_1}+j\frac{ \pi m_2}{p_2} -j\frac{\pi k_1}{p_1}-j\frac{ \pi k_2}{p_2}}\) has to equal to 0. We have

$$\begin{aligned}&\frac{ (m_1-k_1)}{p_1}- \frac{ (m_2-k_2)}{p_2}=K_1\nonumber \\&\frac{ (m_1-k_1)}{p_1}+ \frac{ (m_2-k_2)}{p_2}=K_2\nonumber \\&\frac{ 2(m_1-k_1)}{p_1}=K_1+K_2 \end{aligned}$$
(20)

Since \(m_1, k_1 \in (1, p_1-1)\), then \((m_1-k_1)\in (-(p_1-2), p_1-2)\). The fraction \(\frac{ (m_1-k_1)}{p_1}\) cannot be an integer, \(p_1\) is prime, \(\frac{ 2(m_1-k_1)}{p_1}\) cannot be an integer either. Therefore, \(K_1+K_2=0\), we have \(m_1=k_1\).

Another proof: from \(a_1+a_2=(y_1+y_2)\), we have \(a_1-y_1=y_2-a_2\). Since \(a_1-y_1 \in {{\mathbb {Z}}}[\zeta _{ p_1}]\), and \(y_2-a_2 \in {{\mathbb {Z}}}[\zeta _{ p_2}]\), therefore, \(a_1-y_1 \in {{\mathbb {Z}}}\) and \(a_2-y_2 \in {{\mathbb {Z}}}\). The imaginary parts in \(a_i-y_i,i=1,2\) must equal to zero.

$$\begin{aligned}&2j\sin \left( \frac{ \pi (m_1-k_1)}{p_1}\right) \exp \left( \frac{ \pi (m_1+k_1)}{p_1}\right) \in {{\mathbb {Z}}}\nonumber \\&\text {Suppose }\; \sin \left( \frac{ \pi (m_1-k_1)}{p_1}\right) \ne 0,\; \text {then},\; \cos \left( \frac{ \pi (m_1+k_1)}{p_1}\right) =0\nonumber \\&\text {We have }\; \frac{ \pi (m_1+k_1)}{p_1}=\frac{(2K+1)\pi }{2}, K\; \text {integer}\nonumber \\&\frac{ (m_1+k_1)}{p_1}=\frac{2K+1}{2}\nonumber \\&2(m_1+k_1)=p_1 (2K+1) \end{aligned}$$
(21)

Since \(2(m_1+k_1)\) is even, \(p_1 (2K+1)\) is odd, therefore, there is no solution to make \(\cos (\frac{ \pi (m_1+k_1)}{p_1})=0\). To make \(2j\sin (\frac{ \pi (m_1-k_1)}{p_1}) \exp ( \frac{ \pi (m_1+k_1)}{p_1}) \in {{\mathbb {Z}}}\), one must have \(\sin (\frac{ \pi (m_1-k_1)}{p_1})= 0\), then \(m_1=k_1\). Similarly, we have \(m_2=k_2\).

For \(n \ge 3\), if \(\sum _{i=1}^{n-1} a_i+a_n= \sum _{i=1}^{n-1} y_i+y_n\), we have \(\sum _{i=1}^{n-1} a_i- \sum _{i=1}^{n-1} y_i=y_n-a_n\). The left side \(\in {{\mathbb {Z}}}[\zeta _{ \prod _i^{n-1} p_i}]\), the right side \(\in {{\mathbb {Z}}}[\zeta _{ p_n}]\). Since all \(p_i\) are pairwise coprime, \(y_n-a_n \in {{\mathbb {Z}}}\). Following the steps in the case for \(a_1+a_2=y_1+y_2\), one can achieve \(m_n=k_n\).

Proof of Proposition 1

Denote \(a_i=\exp (\frac{j 2\pi m_i}{p_i}), b_i=\exp (\frac{j 2\pi n_i}{p_i}), p_i-1\ge m_i,n_i\ge 1, x_i=\exp (\frac{j 2\pi \ell _i}{q_i}), y_i=\exp (\frac{j 2\pi k_i}{q_i}), q_i-1\ge \ell _i,k _i\ge 1, i=1, \ldots , N \).

We use deduction for the proof.

  • For \(M=1\), given \(\sum _{i=1}^N a_i x_1=\sum _{i=1}^N b_i y_1\), we only need to prove \(x_1=y_1\). Then, one can achieve \(a_i=b_i, i=1, \ldots , N\) based on Fact 1.

    Rewriting \(\sum _{i=1}^N a_i x_1=\sum _{i=1}^N b_i y_1\), we have

    $$\begin{aligned} \frac{\sum _{i=1}^{N} a_i}{ \sum _{i=1}^{N}b_i }= \frac{y_1 }{x_1} \end{aligned}$$
    (22)

    The left side \(\in {{\mathbb {Q}}}[\zeta _{ \prod _i^{N} p_i}]\), the right side \(\in {{\mathbb {Q}}}[\zeta _{q_1}]\). Since all \(p_i,q_1\) are pairwise coprime, we have \(\frac{y_1 }{x_1}\in {{\mathbb {Q}}}\). Let \(y_1=v x_1\), where v is real. Since \(x_1\) and \(y_1\) are both complex with unit magnitude, given \(p_i\) is prime, v must equal to one. Therefore, \(x_1=y_1\).

  • Suppose the following statement is true: if \(\sum _{i=1}^N a_i \sum _{j=1}^{M-1} x_j=\sum _{i=1}^N b_i \sum _{i=j}^{M-1} y_j\) holds, then \(a_i=b_i, i=1, \ldots , N\) and \(x_j=y_j, j=1, \ldots , M-1\)

  • We need to prove if \(\sum _{i=1}^N a_i \sum _{j=1}^{M} x_j=\sum _{i=1}^N b_i \sum _{j=1}^{M} y_j\) holds, then \(a_i=b_i, i=1, \ldots , N\), and \(x_j=y_j, j=1, \ldots , M\).

    Rewriting \(\sum _{i=1}^N a_i \sum _{j=1}^{M} x_j=\sum _{i=1}^N b_i \sum _{j=1}^{M} y_j\) as

    $$\begin{aligned} \frac{\sum _{i=1}^{N} a_i}{ \sum _{i=1}^{N}b_i }= \frac{\sum _{i=1}^{M-1}y_i+y_M }{\sum _{i=1}^{M-1} x_i+x_M}, \end{aligned}$$
    (23)

    we have \( \sum _{i=1}^{N} a_i \sum _{i=1}^{M-1} x_i+ \sum _{i=1}^{N} a_i x_M= \sum _{i=1}^{N}b_i \sum _{i=1}^{M-1}y_i+ \sum _{i=1}^{N}b_i y_M \). Based on the statement for the case of \(M-1\), we obtain \(\sum _{i=1}^{N} a_i x_M= \sum _{i=1}^{N}b_i y_M \). Following the same procedure to prove \(x_1=y_1\), we have \(x_M=y_M\). From Fact 1, we have \(a_i=b_i, i=1, \ldots , N\).

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Ding, Y., Li, L. Blind Detection Design for AF Two-Way Relaying Over Frequency Selective Channels. Int J Wireless Inf Networks (2021). https://doi.org/10.1007/s10776-021-00536-5

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Keywords

  • Two-way relay
  • Unique identification
  • Blind detection
  • Frequency selective fading
  • Least square error