用SLM方法降低PAPR的OFDM信道容量研究

Yaron Handali, Itamar Nizan, D. Wulich
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引用次数: 13

摘要

OFDM的一个主要缺点是峰值平均功率比(PAPR)大。OFDM是一种线性调制,因此大的PAPR要求发射器(TX)和接收器(RX)路径的大线性动态范围。TX-RX链的弱点是功率放大器(PA)。众所周知,能够放大具有大PAPR的信号的线性PA必须具有大的动态范围,从而导致低功率效率,因此通过降低PAPR可以提高PA的效率。研究了一种基于选择性映射(SLM)的OFDM中PAPR减小方法。众所周知,SLM需要传输侧信息(SI)。增加SLM阶数会降低PAPR,但会增加SI的大小。为了传输SI,需要一些功率和带宽,这些都是从总功率和带宽预算中扣除的,从而降低了容量。另一方面,减小的PAPR提高了PA效率,从而增加了信道容量。因此,期望信道容量的最大值以及最优的SLM顺序。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
On channel capacity of OFDM with SLM method for PAPR reduction
One major drawback of OFDM is the large Peak to Average Power Ratio (PAPR). OFDM is a linear modulation, therefore large PAPR requires large linear dynamic range of the transmitter (TX) as well as the receiver (RX) paths. The weak point in the TX-RX chain is the Power Amplifier (PA). As known, a linear PA which is able to amplify signals with large PAPR has to have a large dynamic range that results with low power efficiency, thus by reducing PAPR one may improve the efficiency of PA. A PAPR reducing method in OFDM, based on SeLective Mapping (SLM) is considered. As known, SLM requires transmitting a Side Information (SI). Increasing the SLM order reduces the PAPR but increases the size of SI. In order to transmit the SI some power and bandwidth are needed, those are taken from the total power and bandwidth budget and thus lowering the capacity. On the other hand the reduced PAPR improves the PA efficiency, thus increases the channel capacity. Accordingly maximal value of channel capacity is expected as well as an optimal SLM order.
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