Java声音生成会产生噪音

Question

我正在使用javax.sound来制作声音,但是当你演奏时它们会在背景中产生某种噪音,如果你一次演奏几个音符,它甚至可以克服声音.这是代码:

public final static double notes[] = new double[] {130.81, 138.59, 146.83, 155.56, 164.81, 174.61, 185,
        196, 207.65, 220, 233.08, 246.94, 261.63, 277.18, 293.66,
        311.13, 329.63, 349.23, 369.99, 392, 415.3, 440, 466.16,
        493.88, 523.25, 554.37};



public static void playSound(int note, int type) throws LineUnavailableException {          //type 0 = sin, type 1 = square

    Thread t = new Thread() {
        public void run() {
            try {

                int sound = (int) (notes[note] * 100);


                byte[] buf = new byte[1];
                AudioFormat af = new AudioFormat((float) sound, 8, 1, true,
                        false);
                SourceDataLine sdl;

                sdl = AudioSystem.getSourceDataLine(af);

                sdl = AudioSystem.getSourceDataLine(af);
                sdl.open(af);
                sdl.start();
                int maxi = (int) (1000 * (float) sound / 1000);
                for (int i = 0; i < maxi; i++) {
                    double angle = i / ((float) 44100 / 440) * 2.0
                            * Math.PI;
                    double val = 0;
                    if (type == 0) val = Math.sin(angle)*100;
                    if (type == 1) val = square(angle)*50;

                    buf[0] = (byte) (val * (maxi - i) / maxi);
                    sdl.write(buf, 0, 1);
                }
                sdl.drain();
                sdl.stop();
                sdl.close();
            } catch (LineUnavailableException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        };
    };

    t.start();
}

public static double square (double angle){
    angle = angle % (Math.PI*2);
    if (angle > Math.PI) return 1;
    else return 0;
}

此代码来自此处:https://stackoverflow.com/a/1932537/3787777

Answer 1

在这个答案中,我将参考1)你的代码,2)更好的方法(恕我直言:)和3)在同一时间播放两个音符.

你的代码

首先,采样率应该不依赖于音符频率.因此尝试:

AudioFormat(44100,...

接下来,使用16位采样(听起来更好!).这是你的代码,在没有噪音的情况下播放简单的音调 - 但我会稍微使用它(见后面).请查看评论:

Thread t = new Thread() {
    public void run() {
        try {

            int sound = (440 * 100);        // play A
            AudioFormat af = new AudioFormat(44100, 16, 1, true, false);
            SourceDataLine sdl;
            sdl = AudioSystem.getSourceDataLine(af);
            sdl.open(af, 4096 * 2);
            sdl.start();

            int maxi = (int) (1000 * (float) sound / 1000); // should not depend on notes frequency!
            byte[] buf = new byte[maxi * 2];   // try to find better len!
            int i = 0;
            while (i < maxi * 2) {
                // formula is changed to be simple sine!!
                double val = Math.sin(Math.PI * i * 440 / 44100);
                short s = (short) (Short.MAX_VALUE * val);
                buf[i++] = (byte) s;
                buf[i++] = (byte) (s >> 8);   // little endian
            }
            sdl.write(buf, 0, maxi);
            sdl.drain();
            sdl.stop();
            sdl.close();
        } catch (LineUnavailableException e) {
            e.printStackTrace();
        }
    }
};

t.start();

建议更好的代码

以下是代码的简化版本,可以播放一些没有噪音的音符(频率).我更喜欢它,因为我们首先创建doubles的数组,这是通用值.这些值可以组合在一起,或者存储或进一步修改.然后我们将它们转换为(8位或16位)样本值.

private static byte[] buffer = new byte[4096 * 2 / 3];
private static int bufferSize = 0;

// plays a sample in range (-1, +1).
public static void play(SourceDataLine line, double in) {
    if (in < -1.0) in = -1.0;  // just sanity checks
    if (in > +1.0) in = +1.0;

    // convert to bytes - need 2 bytes for 16 bit sample
    short s = (short) (Short.MAX_VALUE * in);
    buffer[bufferSize++] = (byte) s;
    buffer[bufferSize++] = (byte) (s >> 8);   // little Endian

    // send to line when buffer is full
    if (bufferSize >= buffer.length) {
        line.write(buffer, 0, buffer.length);
        bufferSize = 0;
    }
    // todo: be sure that whole buffer is sent to line!
}

// prepares array of doubles, not related with the sampling value!
private static double[] tone(double hz, double duration) {
    double amplitude = 1.0;
    int N = (int) (44100 * duration);
    double[] a = new double[N + 1];
    for (int i = 0; i <= N; i++) {
        a[i] = amplitude * Math.sin(2 * Math.PI * i * hz / 44100);
    }
    return a;
}

// finally:
public static void main(String[] args) throws LineUnavailableException {
    AudioFormat af = new AudioFormat(44100, 16, 1, true, false);
    SourceDataLine sdl = AudioSystem.getSourceDataLine(af);

    sdl.open(af, 4096 * 2);
    sdl.start();

    double[] tones = tone(440, 2.0);    // play A for 2 seconds
    for (double t : tones) {
        play(sdl, t);
    }

    sdl.drain();
    sdl.stop();
    sdl.close();
}

听起来不错 ;)

同时播放两个音符

只需结合两个笔记:

double[] a = tone(440, 1.0);        // note A
double[] b = tone(523.25, 1.0);     // note C (i hope:)
for (int i = 0; i < a.length; i++) {
    a[i] = (a[i] + b[i]) / 2;
}
for (double t : a) {
    play(sdl, t);
}

请记住,使用双阵列,您可以组合和操纵您的音调 - 即组合同时播放的音调.当然,如果添加3个音调,则需要通过除以3来规范化该值,依此类推.

叮咚 :)