Performance evaluation of pre- and post-FFT beamforming methods in pilot-assisted SIMO-OFDM systems

Abstract

Spatial domain processing using adaptive antenna arrays has been widely used to suppress co-channel interferences and mitigate the channel effect in single-input multiple-output (SIMO) OFDM systems. Two main mechanisms are pre-FFT scheme based on time-domain beamforming and post-FFT scheme based on frequency-domain beamforming. Both schemes have been recently extended to multiple-input multiple-output (MIMO) OFDM systems. However, there is no study, neither in SIMO-OFDM systems nor in MIMO-OFDM systems, to compare these two methods in details and provide specific guidelines to the designers on which scheme to deploy. Since performance improvement due to beamforming in both SIMO- and MIMO-OFDM systems are based on the same principles and noting the complexity of analysis in MIMO-OFDM beamforming systems, in this article we compare these two schemes in the simpler structure of a pilot assisted SIMO-OFDM system. This comparison is performed both from analytical perspective and by simulation results. It is shown under what conditions the more complex post-FFT scheme can exhibit better results and under what conditions the less costly pre-FFT scheme can demonstrate better performance. This greatly helps the designers to choose the right scheme based on their system characteristics.

Appendices

Appendix 1

From (9), (33), and (36) and ignoring the noise term, the post-FFT weights $\vec {W}(k)$ are obtained as,

(50)

Using the eigenvalue decomposition of , (50) yields,

(51)

Note that for a normalized channel as assumed here we have $\sqrt{\xi _1}=1$. If we define, then (51) becomes,

$$\begin{aligned} {\vec {g}} \vec {g}^{H} \vec {W}(k)=\frac{\sigma _0^{2}}{\xi _1} {\vec {g}} \end{aligned}$$

(52)

Equality holds when ${\vec {g}}^{H} \vec {W}(k)= \sigma _0^{2}/\xi _1$. This means $\vec {W}(k) =\left( {\sigma _0^{2}/\xi _1} \right) {\vec {g}} /\left\| {{\vec {g}}} \right\| ^{2}$, which gives rise to (37).

Appendix 2

From (17), (33), and (36), the pre-FFT weights $\vec {V}$ are obtained as,

(53)

For the left-hand side of (53) we can write,

(54)

Since,

$$\begin{aligned}&\sum _{k=1}^N {{\vec {g}} (k)} {\vec {g}}^{H}(k)\nonumber \\& \quad =\frac{1}{\xi _1}\sum _{k=1}^N {\left\{ {\sum \limits _{i=1}^{P} {\alpha _i e^{-j\frac{2\pi k}{N}\tau _i}{\vec {S_{0,i}}}}} \right\} } \left\{ {\sum \limits _{p=1}^{P} {\alpha _p^{{*}} e^{j\frac{2\pi k}{N}\tau _p}{\vec {S_{0,p}}}^{H}}} \right\} \nonumber \\& \quad =\frac{N}{\xi _1}\left( {\sum \limits _{i=1}^{P} {\left| {\alpha _i} \right| }^{2}{\vec {S_{0,i}}} {\vec {S_{0,i}}}^{H}} \right) \end{aligned}$$

(55)

we get,

$$\begin{aligned} \left( {\sum \limits _{i=1}^{P} {\left| {\alpha _i} \right| }^{2}{\vec {S_{0,i}}} {\vec {S_{0,i}}}^{H}} \right) \vec {V}=\sigma _0^{2}\alpha _1 {\vec {S_{0,1}}} \end{aligned}$$

(56)

Appendix 3

Consider the set of symmetric positive matrices with maximum eigenvalues $\xi _{\max ,i}$ and minimum eigenvalues $\xi _{\min ,i}$. Define $\chi _i =\frac{\xi _{\max ,i}}{\xi _{\min ,i}}, i=1,\ldots ,N$. Now if we define and denote its maximum eigenvalue by $\xi _{\max }$ and its minimum eigenvalue by $\xi _{\min }$ then $\chi =\frac{\xi _{\max }}{\xi _{\min }}$. On the other hand, $\xi _{\max } = \mathop {\max }\limits _{\left\| {\vec {X}} \right\| =1} \left( {\vec {X}^{H}G\vec {X}} \right)$ and $\xi _{\min } = \mathop {\min }\limits _{\left\| {\vec {X}} \right\| =1} \left( {\vec {X}^{H}G\vec {X}} \right)$ then

Since both and are symmetric positive definite matrices we can write,

(57)

With the same line of reasoning we can conclude that, $\xi _{\min } = \sum \limits _{i=1}^N {\xi _{\min ,i}}$. Consequently,

$$\begin{aligned} \chi =\frac{\sum \limits _{i=1}^N {\xi _{\max ,i}}}{\sum \limits _{i=1}^N {\xi _{\min ,i}}} \end{aligned}$$

(58)

Now let’s assume $\chi _i , i=1,\ldots ,N$ are arranged in ascending order $\chi _{1} \le \chi _2 \le \cdots \le \chi _{N}$. We would like to show that $\chi _{1} \le \chi \le \chi _{N}$.

Since $\chi _i =\frac{\xi _{\max ,i}}{\xi _{\min ,i}}\le \frac{\xi _{\max ,N}}{\xi _{\min ,N}} ,i=1,\ldots ,N$ we get

$$\begin{aligned} \xi _{\max ,i} \quad \xi _{\min , N} \le \xi _{\min ,i} \quad \xi _{\max , N} , i=1,\ldots ,N \end{aligned}$$

(59)

This means,

$$\begin{aligned} \sum \limits _{i=1}^N {\xi _{\max ,i} \quad \xi _{\min , N}} \le \sum \limits _{i=1}^N {\xi _{\min ,i} \quad \xi _{\max , N}} \end{aligned}$$

(60)

Therefore,

$$\begin{aligned} \frac{\sum \limits _{i=1}^N {\xi _{\max ,i}}}{\sum \limits _{i=1}^N {\xi _{\min ,i}}}\le \frac{\xi _{\max ,N}}{\xi _{\min ,N}} \end{aligned}$$

(61)

This in turn means $\chi \le \chi _{N}$. With the same line of reasoning, we can show that $\chi _{1} \le \chi$. As a result we obtain $\chi _{1} \le \chi \le \chi _{N}$.