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SoundPlayer+LibDSP: Move the FFT implementation to LibDSP

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LibDSP can greatly benefit from this nice FFT implementation, so let's
move it into the fitting library :^)

Note that this now requires linking SoundPlayer against LibDSP. That's
not an issue (LibDSP is rather small currently anyways), as we can
probably make great use of it in the future anyways.
This commit is contained in:
kleines Filmröllchen 2021-11-20 17:44:12 +01:00 • committed by Linus Groh
parent 4cbf7f24be
commit a099a77e82
5 changed files with 14 additions and 6 deletions
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58
Userland/Applications/SoundPlayer/AudioAlgorithms.cpp
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@ -1,58 +0,0 @@
/*
* Copyright (c) 2021, Cesar Torres <shortanemoia@protonmail.com>
*
* SPDX-License-Identifier: BSD-2-Clause
*/
#include "AudioAlgorithms.h"
#include <AK/Complex.h>
#include <AK/Math.h>
// This function uses the input vector as output too. therefore, if you wish to
// leave it intact, pass a copy to this function
//
// The sampling frequency must be more than twice the frequency to resolve.
// The sample window must be at least large enough to reflect the periodicity
// of the smallest frequency to be resolved.
//
// For example, to resolve a 10 KHz and a 2 Hz sine waves we need at least
// a samplerate of 20 KHz and a window of 0.5 seconds
//
// If invert is true, this function computes the inverse discrete fourier transform.
//
// The data vector must be a power of 2
// Adapted from https://cp-algorithms.com/algebra/fft.html
void fft(Vector<Complex<double>>& sample_data, bool invert)
{
int n = sample_data.size();
auto data = sample_data.data();
for (int i = 1, j = 0; i < n; i++) {
int bit = n >> 1;
for (; j & bit; bit >>= 1)
j ^= bit;
j ^= bit;
if (i < j)
swap(data[i], data[j]);
}
for (int len = 2; len <= n; len <<= 1) {
double ang = 2 * AK::Pi<double> / len * (invert ? -1 : 1);
Complex<double> wlen(AK::cos(ang), AK::sin(ang));
for (int i = 0; i < n; i += len) {
Complex<double> w = { 1., 0. };
for (int j = 0; j < len / 2; j++) {
Complex<double> u = data[i + j], v = data[i + j + len / 2] * w;
data[i + j] = u + v;
data[i + j + len / 2] = u - v;
w *= wlen;
}
}
}
if (invert) {
for (int i = 0; i < n; i++)
data[i] /= n;
}
}