问题描述
我在android中做声音分析器应用。我可以使用AudioTrack生成18 khz到20 khz的超声波Api.i可以使用AudioRecord Api录制音频。但是我不知道如何计算频率形式的麦克风输入数据。我看到了多个问题 并它没有给出正确的频率。请帮助我。接受我的沟通。
这是我的频率计算代码
int bufferSizeInBytes = 1024;
short[] buffer = new short[bufferSizeInBytes];
class Recording extends Thread {
@Override
public void run() {
while (true) {
bufferReadResult = audioInput.read(buffer, 0, bufferSizeInBytes); // record data from mic into buffer
if(bufferReadResult > 0){
calculate();
}
}
}
public void calculate() {
DoubleFFT_1D fft1d = new DoubleFFT_1D(bufferSizeInBytes);//using JTransforms lib
double[] fftBuffer = new double[bufferSizeInBytes * 2];
double[] magnitude = new double[bufferSizeInBytes / 2];
// copy real input data to complex FFT buffer
for (int i = 0; i < bufferSizeInBytes - 1; ++i) {
fftBuffer[2 * i] = buffer[i];
fftBuffer[2 * i + 1] = 0;
}
//perform FFT on fft[] buffer
fft1d.realForward(fftBuffer);
// calculate power spectrum (magnitude) values from fft[]
for (int i = 0; i < (bufferSizeInBytes / 2) - 1; ++i) {
double real = fftBuffer[2 * i];
double imaginary = fftBuffer[2 * i + 1];
magnitude[i] = Math.sqrt(real * real + imaginary * imaginary);
}
// find largest peak in power spectrum
double max_magnitude = magnitude[0];
int max_index = 0;
for (int i = 0; i < magnitude.length; ++i) {
if (magnitude[i] > max_magnitude) {
max_magnitude = (int) magnitude[i];
max_index = i;
}
}
double freq = max_index * 44100 / bufferSizeInBytes;
Log.e("AudioBEacon", "" + freq);
}
}
这是我的输出。请让我知道我错在哪里。
02-10 12:33:04.450 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21662.0
02-10 12:33:04.451 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21317.0
02-10 12:33:04.453 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21791.0
02-10 12:33:04.471 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21748.0
02-10 12:33:04.472 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21533.0
02-10 12:33:04.474 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21834.0
02-10 12:33:04.491 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21533.0
02-10 12:33:04.493 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21705.0
02-10 12:33:04.511 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21533.0
02-10 12:33:04.512 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21447.0
02-10 12:33:04.513 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21490.0
02-10 12:33:04.531 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21576.0
02-10 12:33:04.551 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21619.0
02-10 12:33:04.591 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21877.0
02-10 12:33:04.613 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21576.0
02-10 12:33:04.633 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21920.0
02-10 12:33:04.653 17013-17063/com.org.sohamsaa.audiobeacontransmitter E/AudioBEacon: 21791.0
我的频率范围是18 khz到20 khz。但我没有得到我的频率。我可以过滤我的频率。谢谢你。
1楼
看我的示例项目。 ：D 。
基于Google开发人员开发的Spectrum Analyzer应用程序，这种检索频率具有高精度和高速度。
2楼
最后我找到了答案。我的代码中只应用FFT而不是JTransforms lib。这段代码对我有用。
int bufferSizeInBytes = 1024;
short[] buffer = new short[bufferSizeInBytes];
class Recording extends Thread {
@Override
public void run() {
while () {
if (true) {
int bufferReadResult = audioInput.read(buffer, 0, bufferSizeInBytes); // record data from mic into buffer
if (bufferReadResult > 0) {
calculate();
}
}
}
}
}
public void calculate() {
double[] magnitude = new double[bufferSizeInBytes / 2];
//Create Complex array for use in FFT
Complex[] fftTempArray = new Complex[bufferSizeInBytes];
for (int i = 0; i < bufferSizeInBytes; i++) {
fftTempArray[i] = new Complex(buffer[i], 0);
}
//Obtain array of FFT data
final Complex[] fftArray = FFT.fft(fftTempArray);
// calculate power spectrum (magnitude) values from fft[]
for (int i = 0; i < (bufferSizeInBytes / 2) - 1; ++i) {
double real = fftArray[i].re();
double imaginary = fftArray[i].im();
magnitude[i] = Math.sqrt(real * real + imaginary * imaginary);
}
// find largest peak in power spectrum
double max_magnitude = magnitude[0];
int max_index = 0;
for (int i = 0; i < magnitude.length; ++i) {
if (magnitude[i] > max_magnitude) {
max_magnitude = (int) magnitude[i];
max_index = i;
}
}
double freq = 44100 * max_index / bufferSizeInBytes;//here will get frequency in hz like(17000,18000..etc)
}