Fast Computation of the Discrete Fourier Transform Square Index Coefficients

TL;DR

This paper introduces an efficient algorithm for computing specific DFT coefficients, called square index coefficients, reducing computational complexity by leveraging elementary mathematical properties, especially for power-of-two sizes.

Contribution

The authors present a novel algorithm that computes SICs of the DFT with fewer points and no multiplications, and combines it with FFT for faster computation when N is a power of two.

Findings

Reduces SIC computation from N to √N points

No multiplications needed for SIC calculation

Achieves O(√N log √N) complexity when combined with FFT

Abstract

The $N$-point discrete Fourier transform (DFT) is a cornerstone for several signal processing applications. Many of these applications operate in real-time, making the computational complexity of the DFT a critical performance indicator to be optimized. Unfortunately, whether the $\mathcal{O}(N\log_2 N)$ time complexity of the fast Fourier transform (FFT) can be outperformed remains an unresolved question in the theory of computation. However, in many applications of the DFT -- such as compressive sensing, image processing, and wideband spectral analysis -- only a small fraction of the output signal needs to be computed because the signal is sparse. This motivates the development of algorithms that compute specific DFT coefficients more efficiently than the FFT algorithm. In this article, we show that the number of points of some DFT coefficients can be dramatically reduced by means of elementary mathematical properties. We present an algorithm that compacts the square index coefficients (SICs) of DFT (i.e., $X_{k\sqrt{N}}$, $k=0,1,\cdots, \sqrt{N}-1$, for a square number $N$) from $N$ to $\sqrt{N}$ points at the expense of $N-1$ complex sums and no multiplication. Based on this, any regular DFT algorithm can be straightforwardly applied to compute the SICs with a reduced number of complex multiplications. If $N$ is a power of two, one can combine our algorithm with the FFT to calculate all SICs in $\mathcal{O}(\sqrt{N}\log_2\sqrt{N})$ time complexity.