Android 更改块大小会导致FFT分析失败
我正在试着录制音频并获取频率。我可以用44100的采样率和2048的块大小成功地做到这一点。我相信垃圾箱的大小大约是20个。但是,如果我尝试将块大小增加到4096,那么我不会得到准确的频率,而是得到相同的不准确频率,没有幅度/分贝 我的录音任务如下:Android 更改块大小会导致FFT分析失败,android,audio,fft,Android,Audio,Fft,我正在试着录制音频并获取频率。我可以用44100的采样率和2048的块大小成功地做到这一点。我相信垃圾箱的大小大约是20个。但是,如果我尝试将块大小增加到4096,那么我不会得到准确的频率,而是得到相同的不准确频率,没有幅度/分贝 我的录音任务如下: private class RecordAudio extends AsyncTask<Void, double[], Boolean> { @Override protected Boolean doInBackg
private class RecordAudio extends AsyncTask<Void, double[], Boolean> {
@Override
protected Boolean doInBackground(Void... params) {
int bufferSize = AudioRecord.getMinBufferSize(frequency,
channelConfiguration, audioEncoding);
audioRecord = new AudioRecord(
MediaRecorder.AudioSource.DEFAULT, frequency,
channelConfiguration, audioEncoding, bufferSize);
int bufferReadResult;
short[] buffer = new short[blockSize];
double[] toTransform = new double[blockSize];
try {
audioRecord.startRecording();
} catch (IllegalStateException e) {
Log.e("Recording failed", e.toString());
}
while (started) {
if (isCancelled() || (CANCELLED_FLAG == true)) {
started = false;
//publishProgress(cancelledResult);
Log.d("doInBackground", "Cancelling the RecordTask");
break;
} else {
bufferReadResult = audioRecord.read(buffer, 0, blockSize);
for (int i = 0; i < blockSize && i < bufferReadResult; i++) {
toTransform[i] = (double) buffer[i] / 32768.0; // signed 16 bit
}
transformer.ft(toTransform);
publishProgress(toTransform);
}
}
return true;
}
@Override
protected void onProgressUpdate(double[]...progress) {
int mPeakPos = 0;
double mMaxFFTSample = 150.0;
for (int i = 100; i < progress[0].length; i++) {
int x = i;
int downy = (int) (150 - (progress[0][i] * 10));
int upy = 150;
//Log.i("SETTT", "X: " + i + " downy: " + downy + " upy: " + upy);
if(downy < mMaxFFTSample)
{
mMaxFFTSample = downy;
mMag = mMaxFFTSample;
mPeakPos = i;
}
}
mFreq = (((1.0 * frequency) / (1.0 * blockSize)) * mPeakPos)/2;
//Log.i("SSS", "F: " + mFreq + " / " + "M: " + mMag);
Log.i("SETTT", "FREQ: " + mFreq + " MAG: " + mMaxFFTSample);
}
@Override
protected void onPostExecute(Boolean result) {
super.onPostExecute(result);
try{
audioRecord.stop();
}
catch(IllegalStateException e){
Log.e("Stop failed", e.toString());
}
}
}
私有类RecordAudio扩展异步任务{
@凌驾
受保护的布尔doInBackground(Void…params){
int bufferSize=AudioRecord.getMinBufferSize(频率,
信道配置、音频编码);
音频记录=新的音频记录(
MediaRecorder.AudioSource.DEFAULT,频率,
信道配置、音频编码、缓冲区大小);
int缓冲读取结果;
short[]buffer=新的short[blockSize];
double[]toTransform=新的double[blockSize];
试一试{
录音。开始录制();
}捕获(非法状态){
Log.e(“记录失败”,例如toString());
}
while(启动){
如果(isCancelled()||(CANCELLED|FLAG==true)){
开始=错误;
//出版进度(取消结果);
Log.d(“doInBackground”,“取消RecordTask”);
打破
}否则{
bufferReadResult=audioRecord.read(缓冲区,0,块大小);
对于(int i=0;i
希望有一个快速修复我错过了。谢谢。您需要仔细查看
realdublefft.ft
函数的文档。进入函数的值是实数,但出来的值是复数FFT系数,toTransform[0]
是第一个系数的实部,toTransform[1]
是第一个系数的虚部,依此类推。最终的数组大小是相同的,但由于每个复数占2倍,因此总共有N/2个系数,其中最后一个系数是sampleRate/2的系数
接下来,因为您对数量感兴趣,所以需要计算复数的数量。对于复数x=a+bj
,幅值|x |=sqrt(a*a+b*b)
double maxMag=0;
int-peakIndex=0;
对于(int i=0;i
transformer是如何初始化的,其类型是什么,以便我们至少可以参考其文档?还有,你所说的“没有大小”是什么意思?一切都是零吗?我使用的是RealDoubleFFT,可以在这里找到:
double maxMag = 0;
int peakIndex = 0;
for (int i = 0; i < progress[0].length/2; i++)
{
double re = progress[i*2];
double im = progress[i*2+1];
double mag = Math.sqrt(re*re + im*im);
if (mag > maxMag)
{
peakIndex = i;
maxMag = mag;
}
}
double peakFreq = sampleRate/fftLen * i/2; // might need a bit of tweaking.
double magInDb = 20*Math.log10(mag);