我需要开发一个应用程序，使用手机的麦克风实时录制频率，然后显示它们（文本中）。我在这里发布我的代码。 FFT和复杂类已经从http://introcs.cs.princeton.edu/java/97data/FFT.java.html和http://introcs.cs.princeton.edu/java/97data/Complex.java.html使用。问题是当我在模拟器上运行时，频率从一些随机值开始并持续增加到7996.然后重复整个处理。有人可以帮助我吗？
public class Main extends Activity {
TextView disp;
private static int[] sampleRate = new int[] { 44100, 22050, 11025, 8000 };
short audioData[];
double finalData[];
int bufferSize,srate;
String TAG;
public boolean recording;
AudioRecord recorder;
Complex[] fftArray;
float freq;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
disp = (TextView) findViewById(R.id.display);
Thread t1 = new Thread(new Runnable(){
public void run() {
Log.i(TAG,"Setting up recording");
for (int rate : sampleRate) {
try{
Log.d(TAG, "Attempting rate " + rate);
bufferSize=AudioRecord.getMinBufferSize(rate,AudioFormat.CHANNEL_CONFIGURATION_MONO,
AudioFormat.ENCODING_PCM_16BIT)*3; //get the buffer size to use with this audio record
if (bufferSize != AudioRecord.ERROR_BAD_VALUE) {
recorder = new AudioRecord (MediaRecorder.AudioSource.MIC,rate,AudioFormat.CHANNEL_CONFIGURATION_MONO,
AudioFormat.ENCODING_PCM_16BIT,2048); //instantiate the AudioRecorder
Log.d(TAG, "BufferSize " +bufferSize);
srate = rate;
}
} catch (Exception e) {
Log.e(TAG, rate + "Exception, keep trying.",e);
}
}
bufferSize=2048;
recording=true; //variable to use start or stop recording
audioData = new short [bufferSize]; //short array that pcm data is put into.
Log.i(TAG,"Got buffer size =" + bufferSize);
while (recording) { //loop while recording is needed
Log.i(TAG,"in while 1");
if (recorder.getState()==android.media.AudioRecord.STATE_INITIALIZED) // check to see if the recorder has initialized yet.
if (recorder.getRecordingState()==android.media.AudioRecord.RECORDSTATE_STOPPED)
recorder.startRecording(); //check to see if the Recorder has stopped or is not recording, and make it record.
else {
Log.i(TAG,"in else");
// audiorecord();
finalData=convert_to_double(audioData);
Findfft();
for(int k=0;k<fftArray.length;k++)
{
freq = ((float)srate/(float) fftArray.length) *(float)k;
runOnUiThread(new Runnable(){
public void run()
{
disp.setText("The frequency is " + freq);
if(freq>=15000)
recording = false;
}
});
}
}//else recorder started
} //while recording
if (recorder.getState()==android.media.AudioRecord.RECORDSTATE_RECORDING)
recorder.stop(); //stop the recorder before ending the thread
recorder.release(); //release the recorders resources
recorder=null; //set the recorder to be garbage collected.
}//run
});
t1.start();
}
private void Findfft() {
// TODO Auto-generated method stub
Complex[] fftTempArray = new Complex[bufferSize];
for (int i=0; i<bufferSize; i++)
{
fftTempArray[i] = new Complex(finalData[i], 0);
}
fftArray = FFT.fft(fftTempArray);
}
private double[] convert_to_double(short data[]) {
// TODO Auto-generated method stub
double[] transformed = new double[data.length];
for (int j=0;j<data.length;j++) {
transformed[j] = (double)data[j];
}
return transformed;
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
// Inflate the menu; this adds items to the action bar if it is present.
getMenuInflater().inflate(R.menu.main, menu);
return true;
}
}
答案 0 :(得分：1)
你的问题就在这里：
Findfft();
for(int k=0;k<fftArray.length;k++) {
freq = ((float)srate/(float) fftArray.length) *(float)k;
runOnUiThread(new Runnable() {
public void run() {
disp.setText("The frequency is " + freq);
if(freq>=15000) recording = false;
}
});
}
所有这些for循环确实是通过你的FFT值数组，将数组索引转换为以Hz为单位的频率，然后打印它。
如果你想输出你正在录制的频率，你至少应该看一下阵列中的数据 - 最粗略的方法是计算平方实际幅度并找到最大的频率仓。
除此之外，我不认为您使用的FFT算法会进行任何预先计算 - 还有其他一些预先计算，并且当您正在为移动设备开发时，您可能希望获得CPU使用率和电力使用考虑在内。
JTransforms是一个使用预先计算来降低CPU负载的库，其文档非常完整。
你也可以在Wikipedia找到有关如何解释从FFT返回的数据的有用信息 - 没有冒犯，但看起来你不确定你在做什么，所以我给了指针。
最后，如果您希望将此应用用于音符，我似乎记得有很多人说FFT不是最好的方法，但我不记得是什么。也许其他人可以添加那个位？
答案 1 :(得分：1)
然而，为了进一步实现目标并完成循环，您的问题已经得到了简明扼要的回答......
是的，在有限的CPU上进行音调/频率识别时，FFT不是最佳选择。更优化的方法是YIN描述here。您可以在Tarsos找到实施方案。 您将面临的问题是ADK中缺少javax.sound.sampled，因此将SMSRecord中的短路/字节转换为引用实现所需的浮点数。
答案 2 :(得分：0)
我在几天之后找到了这个解决方案 - 在Hrz中获得频率最佳：
同时下载Jtransforms和this Jar - Jtransforms需要它。
然后我使用这个任务：
public class MyRecorder extends AsyncTask<Void, short[], Void> {
int blockSize = 2048;// = 256;
private static final int RECORDER_SAMPLERATE = 8000;
private static final int RECORDER_CHANNELS = AudioFormat.CHANNEL_IN_MONO;
private static final int RECORDER_AUDIO_ENCODING = AudioFormat.ENCODING_PCM_16BIT;
int BufferElements2Rec = 1024; // want to play 2048 (2K) since 2 bytes we use only 1024
int BytesPerElement = 2;
@Override
protected Void doInBackground(Void... params) {
try {
final AudioRecord audioRecord = new AudioRecord(MediaRecorder.AudioSource.MIC,
RECORDER_SAMPLERATE, RECORDER_CHANNELS,
RECORDER_AUDIO_ENCODING, BufferElements2Rec * BytesPerElement);
if (audioRecord == null) {
return null;
}
final short[] buffer = new short[blockSize];
final double[] toTransform = new double[blockSize];
audioRecord.startRecording();
while (started) {
Thread.sleep(100);
final int bufferReadResult = audioRecord.read(buffer, 0, blockSize);
publishProgress(buffer);
}
audioRecord.stop();
audioRecord.release();
} catch (Throwable t) {
Log.e("AudioRecord", "Recording Failed");
}
return null;
}
@Override
protected void onProgressUpdate(short[]... buffer) {
super.onProgressUpdate(buffer);
float freq = calculate(RECORDER_SAMPLERATE, buffer[0]);
}
public static float calculate(int sampleRate, short [] audioData)
{
int numSamples = audioData.length;
int numCrossing = 0;
for (int p = 0; p < numSamples-1; p++)
{
if ((audioData[p] > 0 && audioData[p + 1] <= 0) ||
(audioData[p] < 0 && audioData[p + 1] >= 0))
{
numCrossing++;
}
}
float numSecondsRecorded = (float)numSamples/(float)sampleRate;
float numCycles = numCrossing/2;
float frequency = numCycles/numSecondsRecorded;
return frequency;
}