请！我需要帮助！我正在尝试做一个录音机，后来用这个声音用他的图形做傅立叶快速变换。问题是我不知道如何做图形，我找到了一个FFT函数，但它没有图形。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();
}
}
}
这就是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;
}
他复杂的课程：
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());
}
}
谢谢大家！
答案 0 :(得分：1)
嗯，只是...做一个图形表示 - 你需要做的就是计算每个频率仓的Magnitude (or Euclidian Norm)，然后在Canvas或某些事情上绘制相应的行。
有一个名为Audalyzer的开源项目，可在Google Code获得完整的源代码 - 您可以在那里获得一些指示，如果您从那里重用代码，请确保遵守许可证：）