论文标题
FFT足够快以超过5G的通信吗？
Is FFT Fast Enough for Beyond-5G Communications?
论文作者
论文摘要
在本文中，我们研究了计算复杂性对{\ color {black}快速傅立叶变换（fft）}算法的吞吐量极限的影响。基于光谱计算{\ color {\ corcorcao}复杂性}（sc）分析，{\ color {\ corcorcorecao}，我们验证$ n $ point fft的复杂性是否比$ ofdm符号中的位比$ n $ n of n of n of n of n of fft null goption。傅立叶变换（DFT）问题验证为$ω（n）$，这仍然是理论计算机科学中的“迷人”开放问题。另外，由于FFT要求$ n $成为两个$ 2^i $（$ i> 0 $）的功率，因此频谱扩大会导致$ i $的指数复杂性，即$ o（2^ii）$。为了克服这些限制，{\ color {\ corcorrecao}我们考虑了向量ofdm（v-ofdm）的替代频率时间变换公式，其中$ n $ point fft被$ n/l $ n/l $（$ l $ l $ l $ u> $ $> $ $ 0 $ 0 $ 0 $）较小{ Ofdm。在此基础上，我们用直接的DFT算法代替了FFT，以释放V-OFFDM参数从增长为两个的功能，并从灵活的命理学中受益（例如，$ L = 3 $，$ n = 156 $）。此外，通过将$ l $设置为$θ（1）$，生成的解决方案可以在$ n $上线性运行（而不是在$ i $上呈指数级），同时将非零吞吐量保持在$ n $增长。 }
In this paper, we study the impact of computational complexity on the throughput limits of the {\color{black}fast Fourier transform (FFT)} algorithm for {\color{black}orthogonal frequency division multiplexing(OFDM)} waveforms. Based on the spectro-computational {\color{\corcorrecao}complexity} (SC) analysis, {\color{\corcorrecao} we verify that the complexity of an $N$-point FFT grows faster than the number of bits in the OFDM symbol.} Thus, we show that FFT nullifies the OFDM throughput on $N$ unless the $N$-point discrete Fourier transform (DFT) problem verifies as $Ω(N)$, which remains a "fascinating" open question in theoretical computer science. Also, because FFT demands $N$ to be a power of two $2^i$ ($i>0$), the spectrum widening leads to an exponential complexity on $i$, i.e. $O(2^ii)$. To overcome these limitations, {\color{\corcorrecao} we consider the alternative frequency-time transform formulation of vector OFDM (V-OFDM), in which an $N$-point FFT is replaced by $N/L$ ($L$$>$$0$) smaller {\color{\corcorrecao}$L$-point} FFTs to mitigate the cyclic prefix overhead of OFDM. Building on that, we replace FFT by the straightforward DFT algorithm to release the V-OFDM parameters from growing as powers of two and to benefit from flexible numerology (e.g., $L=3$, $N=156$). Besides, by setting $L$ to $Θ(1)$, the resulting solution can run linearly on $N$ (rather than exponentially on $i$) while sustaining a non null throughput as $N$ grows. }
