Android audio FFT to display fundamental frequency

2019-02-07 10:10发布

问题:

I have been working on an Android project for awhile that displays the fundamental frequency of an input signal (to act as a tuner). I have successfully implemented the AudioRecord class and am getting data from it. However, I am having a hard time performing an FFT on this data to get the fundamental frequency of the input signal. I have been looking at the post here, and am using FFT in Java and Complex class to go with it.

I have successfully used the FFT function found in FFT in Java, but I am not sure if I am obtaining the correct results. For the magnitude of the FFT (sqrt[rere+imim]) I am getting values that start high, around 15000 Hz, and then slowly diminish to about 300 Hz. Doesn't seem right.

Also, as far as the raw data from the mic goes, the data seems fine, except that the first 50 values or so are always the number 3, unless I hit the tuning button again while still in the application and then I only get about 15. Is that normal?

Here is a bit of my code.

First of all, I convert the short data (obtained from the microphone) to a double using the following code which is from the post I have been looking at. This snippet of code I do not completely understand, but I think it works.

//Conversion from short to double
double[] micBufferData = new double[bufferSizeInBytes];//size may need to change
final int bytesPerSample = 2; // As it is 16bit PCM
final double amplification = 1.0; // choose a number as you like
for (int index = 0, floatIndex = 0; index < bufferSizeInBytes - bytesPerSample + 1; index += bytesPerSample, floatIndex++) {
    double sample = 0;
    for (int b = 0; b < bytesPerSample; b++) {
        int v = audioData[index + b];
        if (b < bytesPerSample - 1 || bytesPerSample == 1) {
            v &= 0xFF;
        }
        sample += v << (b * 8);
    }
    double sample32 = amplification * (sample / 32768.0);
    micBufferData[floatIndex] = sample32;
}

The code then continues as follows:

//Create Complex array for use in FFT
Complex[] fftTempArray = new Complex[bufferSizeInBytes];
for (int i=0; i<bufferSizeInBytes; i++)
{
    fftTempArray[i] = new Complex(micBufferData[i], 0);
}

//Obtain array of FFT data
final Complex[] fftArray = FFT.fft(fftTempArray);
final Complex[] fftInverse = FFT.ifft(fftTempArray);

//Create an array of magnitude of fftArray
double[] magnitude = new double[fftArray.length];
for (int i=0; i<fftArray.length; i++){
    magnitude[i]= fftArray[i].abs();
}


fft.setTextColor(Color.GREEN);
fft.setText("fftArray is "+ fftArray[500] +" and fftTempArray is "+fftTempArray[500] + " and fftInverse is "+fftInverse[500]+" and audioData is "+audioData[500]+ " and magnitude is "+ magnitude[1] + ", "+magnitude[500]+", "+magnitude[1000]+" Good job!");
for(int i = 2; i < samples; i++){
    fft.append(" " + magnitude[i] + " Hz");
}

That last bit is just to check what values I am getting (and to keep me sane!). In the post referred to above, it talks about needing the sampling frequency and gives this code:

private double ComputeFrequency(int arrayIndex) {
    return ((1.0 * sampleRate) / (1.0 * fftOutWindowSize)) * arrayIndex;
}

How do I implement this code? I don't realy understand where fftOutWindowSize and arrayIndex comes from?

Any help is greatly appreciated!

Dustin

回答1:

Recently I'm working on a project which requires almost the same. Probably you don't need any help anymore but I will give my thoughts anyway. Maybe someone need this in the future.

  1. I'm not sure whether the short to double function works, I don't understand that snippet of code neither. It is wrote for byte to double conversion.
  2. In the code: "double[] micBufferData = new double[bufferSizeInBytes];" I think the size of micBufferData should be "bufferSizeInBytes / 2", since every sample takes two bytes and the size of micBufferData should be the sample number.
  3. FFT algorithms do require a FFT window size, and it has to be a number which is the power of 2. However many algorithms can receive an arbitrary of number as input and it will do the rest. In the document of those algorithms should have the requirements of input. In your case, the size of the Complex array can be the input of FFT algorithms. And I don't really know the detail of the FFT algorithm but I think the inverse one is not needed.
  4. To use the code you gave at last, you should firstly find the peak index in the sample array. I used double array as input instead of Complex, so in my case it is something like: double maxVal = -1;int maxIndex = -1;

    for( int j=0; j < mFftSize / 2; ++j ) {
        double v = fftResult[2*j] * fftResult[2*j] + fftResult[2*j+1] * fftResult[2*j+1];
        if( v > maxVal ) {
            maxVal = v;
            maxIndex = j;
        }
    }
    

    2*j is the real part and 2*j+1 is the imaginary part. maxIndex is the index of the peak magnitude you want (More detail here), and use it as input to the ComputeFrequency function. The return value is the frequency of the sample array you want.

Hopefully it can help someone.



回答2:

You should pick an FFT window size depending on your time versus frequency resolution requirements, and not just use the audio buffer size when creating your FFT temp array.

The array index is your int i, as used in your magnitude[i] print statement.

The fundamental pitch frequency for music is often different from FFT peak magnitude, so you may want to research some pitch estimation algorithms.



回答3:

I suspect that the strange results you're getting are because you might need to unpack the FFT. How this is done will depend on the library that you're using (see here for docs on how it's packed in GSL, for example). The packing may mean that the real and imaginary components are not in the positions in the array that you expect.

For your other questions about window size and resolution, if you're creating a tuner then I'd suggest trying a window size of about 20ms (eg 1024 samples at 44.1kHz). For a tuner you need quite high resolution, so you could try zero-padding by a factor of 8 or 16 which will give you a resolution of 3-6Hz.