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swift - 试图了解 AudioKit 中 AKFFTTap 的输出

转载 作者:可可西里 更新时间:2023-11-01 00:54:32 25 4
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我正在尝试使用 AudioKit 构建一个应用程序来分析麦克风的输入并将传入的声音分成 3 个频率范围(低、中、高)及其振幅。

这是我的代码:

class ViewController: UIViewController {

var mic: AKMicrophone!
var amplitude: AKAmplitudeTracker!
var fftTap: AKFFTTap?
var timer: Timer!

override func viewDidLoad() {
super.viewDidLoad()
// Do any additional setup after loading the view, typically from a nib.

mic = AKMicrophone()
fftTap = AKFFTTap.init(mic)

}

override func viewDidAppear(_ animated: Bool) {
super.viewDidAppear(animated)

do {
try AudioKit.start()
} catch {
AKLog("AudioKit did not start!")
}

mic.start()

timer = Timer.scheduledTimer(withTimeInterval: 0.01, repeats: true, block: { (timer) in

for i in 0...256 {
print(Double(self.fftTap?.fftData[i] ?? 0.0))
}

})
}

}

但现在我不知道输出的实际含义是什么?

如何获得特定频率范围内的最大幅度?我同时需要所有三个范围,所以我认为仅仅 Frequency-Tracker 是做不到的。

通过阅读有关 FFT 的文档,我了解到前 256 个 bin 表示某个频率的幅度。但我只找到了将这些值转换为绘图的 Matlab 绘图示例(这对我来说真的没有意义)。

最佳答案

我在 Google 上找到了帮助我解决问题的代码片段:

https://groups.google.com/forum/#!topic/comp.dsp/cZsS1ftN5oI

特别是这部分:

/* do FFT (taken from NR [http://www.nr.com] but uses array of doubles) */
four1(fftBuffer-1, FFT_SIZE, 1);

/* display 15 bins around the frequency of interest */
for (long k = 80; k < 110; k += 2) {

/* real */
double re = fftBuffer[k];

/* imaginary */
double im = fftBuffer[k+1];

/* get normalized bin magnitude */
double normBinMag = 2.*sqrt(re*re + im*im) / FFT_SIZE;

/* convert to dB value */
double amplitude = 20. * log10( normBinMag );

/* and display */
printf("bin: %d,\tfreq: %f [Hz],\tmag: %f,\t ampl.: %f [dB]\n", \
k/2, sampleRate*.5*(double)k/FFT_SIZE, normBinMag, amplitude);
}
}

/* Program output:

bin: 40, freq: 861.328125 [Hz], mag: 0.000000, ampl.: -182.347994 [dB]
bin: 41, freq: 882.861328 [Hz], mag: 0.000000, ampl.: -180.895076 [dB]
bin: 42, freq: 904.394531 [Hz], mag: 0.000000, ampl.: -179.201401 [dB]
bin: 43, freq: 925.927734 [Hz], mag: 0.000000, ampl.: -177.156879 [dB]
bin: 44, freq: 947.460938 [Hz], mag: 0.000000, ampl.: -174.555312 [dB]
bin: 45, freq: 968.994141 [Hz], mag: 0.000000, ampl.: -170.934049 [dB]
bin: 46, freq: 990.527344 [Hz], mag: 0.000000, ampl.: -164.817195 [dB]
bin: 47, freq: 1012.060547 [Hz], mag: 1.000000, ampl.: 0.000000 [dB]
bin: 48, freq: 1033.593750 [Hz], mag: 0.000000, ampl.: -164.633624 [dB]
bin: 49, freq: 1055.126953 [Hz], mag: 0.000000, ampl.: -170.566625 [dB]
bin: 50, freq: 1076.660156 [Hz], mag: 0.000000, ampl.: -174.003468 [dB]
bin: 51, freq: 1098.193359 [Hz], mag: 0.000000, ampl.: -176.419757 [dB]
bin: 52, freq: 1119.726562 [Hz], mag: 0.000000, ampl.: -178.277857 [dB]
bin: 53, freq: 1141.259766 [Hz], mag: 0.000000, ampl.: -179.783660 [dB]
bin: 54, freq: 1162.792969 [Hz], mag: 0.000000, ampl.: -181.046952 [dB]

*/

[编辑]

根据要求,这是 Swift 代码:

//
// ViewController.swift
//

import AudioKit
import UIKit

class ViewController: UIViewController {

var mic: AKMicrophone!
var fftTap: AKFFTTap?
var timer: Timer!
let FFT_SIZE = 512
let sampleRate:double_t = 44100

override func viewDidLoad() {
super.viewDidLoad()

mic = AKMicrophone()

fftTap = AKFFTTap.init(mic)

}

override func viewDidAppear(_ animated: Bool) {
super.viewDidAppear(animated)

do {
try AudioKit.start()
} catch {
AKLog("AudioKit did not start!")
}

mic.start()

timer = Timer.scheduledTimer(withTimeInterval: 0.1, repeats: true, block: { (timer) in

for i in 0...510 {

let re = self.fftTap!.fftData[i]
let im = self.fftTap!.fftData[i + 1]
let normBinMag = 2.0 * sqrt(re * re + im * im)/self.FFT_SIZE
let amplitude = ((20.0 * log10(normBinMag))

print("bin: \(i/2) \t freq: \(frequency)\t ampl.: \(amplitude)")
}

// Now do anything you like with the data
// Be aware, though, that the amplitude is a negative number
// the lower, the less input it represents
// in my tests, the lowest number was around -260
// Read more on Google about converting the negative
// number to a positive

})
}

}

关于swift - 试图了解 AudioKit 中 AKFFTTap 的输出,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52687711/

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