I am looking to replicate in Swift what the FFT function does in Matlab. Essentially, it take an arbitrary length signal (not necessarily a multiple of \$2^n\$) and gives real and complex DFT coefficients.

Since the FFT described in Accelerate can only handle sample sizes that are multiples of \$2^n\$, I wrote a brute force algorithm in Swift that produces exactly the same results as Matlab FFT function for arbitrary sample size.

The problem: When my sample size > 15,000 sample (say), this algorithm takes about 20 s to complete. Could this be sped up?

import Foundation

public func fft(x: [Double]) -> ([Double],[Double]) {

    let N = x.count
    var Xre: [Double] = Array(repeating:0, count:N)
    var Xim: [Double] = Array(repeating:0, count:N)

    for k in 0..<N
    {
        Xre[k] = 0
        Xim[k] = 0
        for n in 0..<N {
            let q = (Double(n)*Double(k)*2.0*M_PI)/Double(N)
            Xre[k] += x[n]*cos(q) // Real part of X[k]
            Xim[k] -= x[n]*sin(q) // Imag part of X[k]
        }
    }
    return (Xre, Xim)
}

// Call FFT
let x: [Double] = [1, 2, 3, 4, 5, 6] // works rapidly
// let x = Array(stride(from: 0, through: 15000, by: (1.0))) // Will   choke it
let (fr, fi) = fft (x: x)
print("Real:", fr)
print(" ")
print("Imag:", fi)
// Call FFT

I am looking to replicate in Swift what the FFT function does in Matlab. Essentially, it takes an arbitrary length signal (not necessarily a multiple of \$2^n\$) and gives real and complex DFT coefficients.

Since the FFT described in Accelerate can only handle sample sizes that are multiples of \$2^n\$, I wrote a brute force algorithm in Swift that produces exactly the same results as the Matlab FFT function for arbitrary sample size.

The problem: When my sample size > 15,000 sample (say), this algorithm takes about 20 s to complete. Could this be sped up?

import Foundation

public func fft(x: [Double]) -> ([Double],[Double]) {

    let N = x.count
    var Xre: [Double] = Array(repeating:0, count:N)
    var Xim: [Double] = Array(repeating:0, count:N)

    for k in 0..<N
    {
        Xre[k] = 0
        Xim[k] = 0
        for n in 0..<N {
            let q = (Double(n)*Double(k)*2.0*M_PI)/Double(N)
            Xre[k] += x[n]*cos(q) // Real part of X[k]
            Xim[k] -= x[n]*sin(q) // Imag part of X[k]
        }
    }
    return (Xre, Xim)
}

// Call FFT
let x: [Double] = [1, 2, 3, 4, 5, 6] // works rapidly
// let x = Array(stride(from: 0, through: 15000, by: (1.0))) // Will   choke it
let (fr, fi) = fft (x: x)
print("Real:", fr)
print(" ")
print("Imag:", fi)
// Call FFT