Many high quality FFT libraries support fast implementations for small primes. (2,3,5,7). I’ll use CUDA fft here as an example.

When you have say an array of 18000 samples d_data, you might not want to do this:

nfft =32768; cudafftPlan1d(&m_plan, nfft, CUFFT_C2C, 1); cufftExecC2C(m_plan, d_data, d_data, CUFFT_FORWARD);

Especially if you are doing many, many iterations of this.

Instead, you can use a method to find the closest fft number to 18000, that is still made of of small primes (2,3,5,7).

With this findNFFT method, nfft = **18432**, which is

2^11 * 3 * 3

In this way, you process less samples, which saves on time. Any gain in speed by using a true radix-2 (32768) number is offset by the number of samples being processed, which is now much less. In one of my projects, this helped to cut the processing time by half.

Here’s the code for the findNFFT method – not the prettiest code but it does the trick.

/** * @brief Returns a vector containing the prime factors of n * * @param [in] The number to find the prime factors for * @return */ std::vector<int> primeFactors(int n) { std::vector<int> vec; while (n % 2 == 0) { vec.push_back(2); n /= 2; } for (int i = 3; i <= sqrt(n); i += 2) { while (n % i == 0) { vec.push_back(i); n /= i; } } if (n > 2) vec.push_back(n); // std::cout << "Prime factors:" << std::endl; // for (int j=0; j < vec.size(); j++) // { // printf("%d ", vec[j]); // } // printf("\n"); return vec; } /** * @brief Used to find the appropriate fft integer for the input n * This uses the "formula" (N + D - 1)/D * D * Criteria: Output nfft should be a factor of 2,3,5 * * @param [in] Integer to find nfft for */ int findNFFT(int n) { std::vector<int> ansPrimes; std::vector<int> firstPrimes; int d = 0; do { if (n > 2048) d = 512; else if (n > 1024) d = 256; else if (n > 128) d = 64; else if (n > 32) d = 32; else if (n > 8) d = 8; else d = 2; int fn = (n + d - 1) / d * d; firstPrimes = primeFactors(fn); for (int i = 0; i < firstPrimes.size(); i++) { if (firstPrimes[i] == 2 || firstPrimes[i] == 3 || firstPrimes[i] == 5) { ansPrimes.push_back(firstPrimes[i]); firstPrimes.erase(firstPrimes.begin() + i); i -= 1; } } int newN = 1; if (firstPrimes.size() > 0) { for (int i = 0; i < firstPrimes.size(); i++) newN *= firstPrimes[i]; } n = newN; firstPrimes = {}; } while (n != 1); // if n == 1 means that firstPrimes int ans = 1; for (int i = 0; i < ansPrimes.size(); i++) ans *= ansPrimes[i]; return ans; }