Performance Analysis of Alamouti Space-Time Block Code Over Time-Selective Fading Channels

Abstract

Alamouti orthogonal space-time block code (Alamouti in IEEE J Sel Areas Commun 16(8):1451–1458, 1998) has been applied widely in wireless communication, e.g., IEEE 802.16e-2005 standard. In this paper, theoretical analysis of symbol error rate performance for Alamouti orthogonal space-time block code (AOSTBC) over time-selective fading channels with a zero-forcing linear receiver is derived. Firstly, a closed-form expression (i.e., not in integral form) is derived for the average symbol pair-wise error probability (SPEP) in time-selective frequency-nonselective independent identically distributed Rayleigh fading channels. Then, the SPEP is used to derive a tight upper bound (UB) for the symbol-error rate (SER) of AOSTBC via establishing algorithmic Bonferroni-type upper bound. Extensive simulation results show that the curves for the UB coincide with the simulated SER curves for various antenna configurations even at very low signal-to-noise ratio regimes. The UB thus can be used to accurately predict the performance of AOSTBC code over time-selective fading channels when a zero-forcing receiver is used.

Appendix

Assume that $\rho ^{2}<1$, let us introduce the random variables $\varepsilon _{1k}$ and $\varepsilon _{2k}$, $k = 1,\ldots , N$

$$\begin{aligned} \xi _{1k} =\frac{h_{1k} \text{(1) }-\rho h_{1k} \text{(2) }}{\sqrt{1-\rho ^{2}}}; \xi _{2k} =\frac{h_{2k} \text{(2) }-\rho h_{2k} \text{(1) }}{\sqrt{1-\rho ^{2}}} \end{aligned}$$

(29)

By construction, $\varepsilon _{1k}$ and $\varepsilon _{2k}$ are independent and identically distributed with the same PDF as $h_{1k}$(2) and $h_{2k}$(1), and furthermore, $\varepsilon _{1k}$ is independent of $h_{1k}$(2), $\varepsilon _{2k}$ and is independent of $h_{2k}$(1). Plugging (29) into (10) leads to

$$\begin{aligned} \omega =\left| {\rho \sqrt{\alpha _1 }+\sqrt{1-\rho ^{2}}\frac{\sum _{k=1}^N {\left( {h_{1k}^*\left( 2 \right) \xi _{1k} +h_{2k} \left( 1 \right) \xi _{2k}^*} \right) } }{\sqrt{\alpha _1 }}} \right| ^{2}\frac{SNR}{2}\Delta _s^2 \end{aligned}$$

(30)

To simplify this expression further, observe that the fraction in (30) can be expressed as an inner product

$$\begin{aligned} z_1 +\text{ j }z_2 =\frac{\sum _{k=1}^N {\left( {h_{1k}^*\left( 2 \right) \xi _{1k} +h_{2k} \left( 1 \right) \xi _{2k}^*} \right) } }{\sqrt{\alpha _1 }}=\left\langle {\frac{{\varvec{h}}}{\left\| {{\varvec{h}}} \right\| },{{\varvec{e}}}} \right\rangle \end{aligned}$$

(31)

where $z_{1}, z_{2}$ are real variables, ${{\varvec{h}}}=\left[ {h_{11} \left( 2 \right) ,h_{21}^*\left( 1 \right) ,\ldots ,h_{1N} \left( 2 \right) ,h_{2N}^*\left( 1 \right) } \right] ^{\text{ T }}$ and ${{\varvec{e}}}=\left[ {\xi _{11}^*,\xi _{21} ,\ldots ,\xi _{1N}^*,\xi _{2N} } \right] ^{\text{ T }}$. Since ${{\varvec{e}}}$ is symmetric, the distribution of $z_{1} + \text{ j }z_{2}$ reduces to the distribution of $h_{1k}$(1), independent of ${{\varvec{h}}}$ (and, thus, also independent of $\left\| {{\varvec{h}}} \right\| )$. Thus, (30) simplifies to

$$\begin{aligned} \omega&= \left| {\rho \sqrt{\alpha _1 }+\sqrt{1-\rho ^{2}}\left( {z_1 +\text{ j }z_2 } \right) } \right| ^{2}\frac{SNR}{2}\Delta _s^2\end{aligned}$$

(32)

$$\begin{aligned} \omega&= \left( {A+X_1 } \right) ^{2}+X_2^2 \end{aligned}$$

(33)

where $A=\sqrt{\alpha _1 \frac{\rho ^{2}SNR}{2}\Delta _s^2 }$ and $X_i =\sqrt{\left( {1-\rho ^{2}} \right) \frac{SNR}{2}\Delta _s^2 }z_i$. Therefore, given $A, \omega$ has a noncentral chi-square distribution with two degrees of freedom, and PDF

$$\begin{aligned} p_{W\left| A \right. } \left( {\omega \left| a \right. } \right) =\frac{1}{\left( {1-\rho ^{2}} \right) c}\text{ exp }\left( {-\frac{\omega +a^{2}}{\left( {1-\rho ^{2}} \right) c}} \right) J_0 \left( {\frac{2\text{ j }a}{\left( {1-\rho ^{2}} \right) c}\sqrt{\omega }} \right) \end{aligned}$$

(34)

where $J_{0}(\cdot )$ is the zeroth-order Bessel function of the first kind, $c=\frac{SNR}{2}\Delta _s^2$ and $\text{ j }=\sqrt{-1}$.

However, $A=\sqrt{\alpha _1 \frac{\rho ^{2}SNR}{2}\Delta _s^2 }=\sqrt{\alpha _1 \rho ^{2}c}$ is Rayleigh distributed with $2N$ degrees of freedom, and PDF

$$\begin{aligned} p_A \left( a \right) =\frac{2}{\rho ^{4N}c^{2N}\Gamma \left( {2N} \right) }a^{4N-1}\text{ exp }\left( {-\frac{a^{2}}{\rho ^{2}c}} \right) \end{aligned}$$

(35)

Integrating the product of (34) and (35) over the variable from 0 to $\infty$ leads to the following density probability for $\omega$ as

$$\begin{aligned} p_W \left( \omega \right)&= \int \limits _0^\infty {p_{W\left| A \right. } \left( {\omega \left| a \right. } \right) } p_A \left( a \right) da\end{aligned}$$

(36)

$$\begin{aligned} p_W \left( \omega \right)&= \frac{2}{\left( {1-\rho ^{2}} \right) \rho ^{4N}c^{2N+1}\Gamma \left( {2N} \right) }\text{ exp }\left( {-\frac{\omega }{\left( {1-\rho ^{2}} \right) c}} \right) \nonumber \\&\times \int \limits _0^\infty {a^{4N-1}\text{ exp }\left( {-\frac{a^{2}}{\left( {1-\rho ^{2}} \right) \rho ^{2}c}} \right) J_0 \left( {\frac{2\text{ j }\sqrt{\omega }}{\left( {1-\rho ^{2}} \right) c}a} \right) } da\nonumber \\ \end{aligned}$$

(37)

From (37), by using [10, Eq. (6.631.1)] we obtain

$$\begin{aligned} p_W \left( \omega \right) =\frac{\left( {1-\rho ^{2}} \right) ^{2N-1}}{c}\text{ exp }\left( {-\frac{\omega }{\left( {1-\rho ^{2}} \right) c}} \right) {{}_{1} F_1} \left( {2N,1;\frac{\rho ^{2}}{\left( {1-\rho ^{2}} \right) c}\omega } \right) \end{aligned}$$

(38)

where $_1 F_1 \left( {\alpha ,\beta ;z} \right)$ is Kummer confluent hypergeometric function [10]. By applying the transform equations given in [12, 13] we have

$$\begin{aligned} _1 F_1 \left( {2N,1;\frac{\rho ^{2}}{\left( {1-\rho ^{2}} \right) c}\omega } \right) =\text{ exp }\left( {\frac{\rho ^{2}\omega }{\left( {1-\rho ^{2}} \right) c}} \right) \sum _{k=0}^{2N-1} {\left( {{\begin{array}{l} {2N-1} \\ {2N-1-k} \\ \end{array} }} \right) } \frac{\rho ^{2k}}{k\text{! }\left( {1-\rho ^{2}} \right) ^{k}c^{k}}\omega ^{k}\nonumber \\ \end{aligned}$$

(39)

Plugging (39) into (38) leads to

$$\begin{aligned} p_W \left( \omega \right) =\sum _{k=0}^{2N-1} {\left( {{\begin{array}{l} {2N-1} \\ {2N-1-k} \\ \end{array} }} \right) } \frac{\rho ^{2k}\left( {1-\rho ^{2}} \right) ^{2N-1-k}}{k\text{! }c^{k+1}}\omega ^{k}\text{ exp }\left( {-\frac{\omega }{c}} \right) \end{aligned}$$

(40)

It is very interesting that the calculation process from (29) to (40) is performed under assumption $\rho ^{2}<1$. However, the final result (40) is general and can apply any value of $\rho$ (which includes $\rho ^{2} = 1$). The proof is not difficult, so it is omitted here for brevity.