FFT录音机Android

Question

请!我需要你的帮助！我正在尝试做一个录音机，然后用这个声音做傅立叶快速变换他的图形。问题是我不知道怎么做图形，我找到了一个快速傅立叶变换函数，但它没有图形。对此有什么建议吗？

package proiektua.proiektua;



import android.media.MediaPlayer;
import android.net.Uri;
import android.os.Bundle;

import android.provider.MediaStore;
import android.app.Activity;
import android.content.Intent;
import android.view.Menu;
import android.view.View;


public class MainActivity extends Activity  {

        int peticion = 1;
        Uri url1;


        @Override
        protected void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.activity_main);
        }

        @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;
        }

        public void grabar(View v) {
            Intent intent = new Intent(MediaStore.Audio.Media.RECORD_SOUND_ACTION);
            startActivityForResult(intent, peticion);}




        public void reproducir(View v) {
            MediaPlayer mediaPlayer = MediaPlayer.create(this, url1);
            mediaPlayer.start();
        }

        protected void onActivityResult(int requestCode, int resultCode, Intent data) {
            if (resultCode == RESULT_OK && requestCode == peticion) {
                url1 = data.getData();
            }
        }

    }

这是FFT：

public class FFT {

    // compute the FFT of x[], assuming its length is a power of 2
    public static Complex[] fft(Complex[] x) {
        int N = x.length;

        // base case
        if (N == 1) return new Complex[] { x[0] };

        // radix 2 Cooley-Tukey FFT
        if (N % 2 != 0) { throw new RuntimeException("N is not a power of 2"); }

        // fft of even terms
        Complex[] even = new Complex[N/2];
        for (int k = 0; k < N/2; k++) {
            even[k] = x[2*k];
        }
        Complex[] q = fft(even);

        // fft of odd terms
        Complex[] odd  = even;  // reuse the array
        for (int k = 0; k < N/2; k++) {
            odd[k] = x[2*k + 1];
        }
        Complex[] r = fft(odd);

        // combine
        Complex[] y = new Complex[N];
        for (int k = 0; k < N/2; k++) {
            double kth = -2 * k * Math.PI / N;
            Complex wk = new Complex(Math.cos(kth), Math.sin(kth));
            y[k]       = q[k].plus(wk.times(r[k]));
            y[k + N/2] = q[k].minus(wk.times(r[k]));
        }
        return y;
    }

和他的复杂类：

public class Complex {
    private final double re;   // the real part
    private final double im;   // the imaginary part

    // create a new object with the given real and imaginary parts
    public Complex(double real, double imag) {
        re = real;
        im = imag;
    }

    // return a string representation of the invoking Complex object
    public String toString() {
        if (im == 0) return re + "";
        if (re == 0) return im + "i";
        if (im <  0) return re + " - " + (-im) + "i";
        return re + " + " + im + "i";
    }

    // return abs/modulus/magnitude and angle/phase/argument
    public double abs()   { return Math.hypot(re, im); }  // Math.sqrt(re*re + im*im)
    public double phase() { return Math.atan2(im, re); }  // between -pi and pi

    // return a new Complex object whose value is (this + b)
    public Complex plus(Complex b) {
        Complex a = this;             // invoking object
        double real = a.re + b.re;
        double imag = a.im + b.im;
        return new Complex(real, imag);
    }

    // return a new Complex object whose value is (this - b)
    public Complex minus(Complex b) {
        Complex a = this;
        double real = a.re - b.re;
        double imag = a.im - b.im;
        return new Complex(real, imag);
    }

    // return a new Complex object whose value is (this * b)
    public Complex times(Complex b) {
        Complex a = this;
        double real = a.re * b.re - a.im * b.im;
        double imag = a.re * b.im + a.im * b.re;
        return new Complex(real, imag);
    }

    // scalar multiplication
    // return a new object whose value is (this * alpha)
    public Complex times(double alpha) {
        return new Complex(alpha * re, alpha * im);
    }

    // return a new Complex object whose value is the conjugate of this
    public Complex conjugate() {  return new Complex(re, -im); }

    // return a new Complex object whose value is the reciprocal of this
    public Complex reciprocal() {
        double scale = re*re + im*im;
        return new Complex(re / scale, -im / scale);
    }

    // return the real or imaginary part
    public double re() { return re; }
    public double im() { return im; }

    // return a / b
    public Complex divides(Complex b) {
        Complex a = this;
        return a.times(b.reciprocal());
    }

    // return a new Complex object whose value is the complex exponential of this
    public Complex exp() {
        return new Complex(Math.exp(re) * Math.cos(im), Math.exp(re) * Math.sin(im));
    }

    // return a new Complex object whose value is the complex sine of this
    public Complex sin() {
        return new Complex(Math.sin(re) * Math.cosh(im), Math.cos(re) * Math.sinh(im));
    }

    // return a new Complex object whose value is the complex cosine of this
    public Complex cos() {
        return new Complex(Math.cos(re) * Math.cosh(im), -Math.sin(re) * Math.sinh(im));
    }

    // return a new Complex object whose value is the complex tangent of this
    public Complex tan() {
        return sin().divides(cos());
    }



    // a static version of plus
    public static Complex plus(Complex a, Complex b) {
        double real = a.re + b.re;
        double imag = a.im + b.im;
        Complex sum = new Complex(real, imag);
        return sum;
    }



    // sample client for testing
    public static void main(String[] args) {
        Complex a = new Complex(5.0, 6.0);
        Complex b = new Complex(-3.0, 4.0);

        System.out.println("a            = " + a);
        System.out.println("b            = " + b);
        System.out.println("Re(a)        = " + a.re());
        System.out.println("Im(a)        = " + a.im());
        System.out.println("b + a        = " + b.plus(a));
        System.out.println("a - b        = " + a.minus(b));
        System.out.println("a * b        = " + a.times(b));
        System.out.println("b * a        = " + b.times(a));
        System.out.println("a / b        = " + a.divides(b));
        System.out.println("(a / b) * b  = " + a.divides(b).times(b));
        System.out.println("conj(a)      = " + a.conjugate());
        System.out.println("|a|          = " + a.abs());
        System.out.println("tan(a)       = " + a.tan());
    }

}

感谢大家！

Answer 1

好吧，只是..。制作一个图形表示-所有你需要做的就是计算每个频率段的Magnitude (or Euclidian Norm)，然后在Canvas或诸如此类的东西上绘制相应的线条。

在Google Code上有一个名为Audalyzer的开源项目，可以在那里获得完整的源代码--你可以在那里获得一些指针，如果你重用了那里的代码，请务必遵守许可证:)