在 Java 中交替播放 2 个不同的频率

Question

我是 Java Sounds 的新手。我想在指定的时间内循环播放 2 个不同的频率，每个频率播放 1 秒。比如，如果我有 2 个频率 440hz 和 16000hz，并且时间段是 10 秒，那么每 'even' 秒播放 440hz，每 'odd' 秒播放 16000hz，即交替播放 5 秒。

我通过一些示例学到了一些东西，并且我还制作了一个程序，该程序针对单个用户指定的频率运行一段时间，该频率也是用户在这些示例的帮助下给出的。

如果有人能帮我解决这个问题，我将不胜感激。谢谢

我也附上那个单频代码供参考。

  import java.nio.ByteBuffer;
  import java.util.Scanner;
  import javax.sound.sampled.*;

  public class Audio {

   public static void main(String[] args) throws InterruptedException, LineUnavailableException {
    final int SAMPLING_RATE = 44100;            // Audio sampling rate
    final int SAMPLE_SIZE = 2;                  // Audio sample size in bytes

    Scanner in = new Scanner(System.in);
    int time = in.nextInt();                      //Time specified by user in seconds
    SourceDataLine line;
    double fFreq = in.nextInt();                         // Frequency of sine wave in hz

    //Position through the sine wave as a percentage (i.e. 0 to 1 is 0 to 2*PI)
    double fCyclePosition = 0;

    //Open up audio output, using 44100hz sampling rate, 16 bit samples, mono, and big 
    // endian byte ordering
    AudioFormat format = new AudioFormat(SAMPLING_RATE, 16, 1, true, true);
    DataLine.Info info = new DataLine.Info(SourceDataLine.class, format);

    if (!AudioSystem.isLineSupported(info)) {
        System.out.println("Line matching " + info + " is not supported.");
        throw new LineUnavailableException();
    }

    line = (SourceDataLine) AudioSystem.getLine(info);
    line.open(format);
    line.start();

    // Make our buffer size match audio system's buffer
    ByteBuffer cBuf = ByteBuffer.allocate(line.getBufferSize());

    int ctSamplesTotal = SAMPLING_RATE * time;         // Output for roughly user specified time in seconds

    //On each pass main loop fills the available free space in the audio buffer
    //Main loop creates audio samples for sine wave, runs until we tell the thread to exit
    //Each sample is spaced 1/SAMPLING_RATE apart in time
    while (ctSamplesTotal > 0) {
        double fCycleInc = fFreq / SAMPLING_RATE;  // Fraction of cycle between samples

        cBuf.clear();                            // Discard samples from previous pass

        // Figure out how many samples we can add
        int ctSamplesThisPass = line.available() / SAMPLE_SIZE;
        for (int i = 0; i < ctSamplesThisPass; i++) {
            cBuf.putShort((short) (Short.MAX_VALUE * Math.sin(2 * Math.PI * fCyclePosition)));

            fCyclePosition += fCycleInc;
            if (fCyclePosition > 1) {
                fCyclePosition -= 1;
            }
        }

        //Write sine samples to the line buffer.  If the audio buffer is full, this will 
        // block until there is room (we never write more samples than buffer will hold)
        line.write(cBuf.array(), 0, cBuf.position());
        ctSamplesTotal -= ctSamplesThisPass;     // Update total number of samples written 

        //Wait until the buffer is at least half empty  before we add more
        while (line.getBufferSize() / 2 < line.available()) {
            Thread.sleep(1);
        }
    }

    //Done playing the whole waveform, now wait until the queued samples finish 
    //playing, then clean up and exit
    line.drain();
    line.close();
}

}

Answer 1

您最好的选择可能是创建 Clips，如下面的示例代码所示。也就是说，MHz 范围通常是听不见的——看起来您的问题中有错字。如果没有错字，您将运行遇到 Mr. Nyquist.

的问题

另一个提示：没有人在 Java 中使用 Hungarian Notation。

import javax.sound.sampled.*;
import java.nio.ByteBuffer;
import java.nio.ShortBuffer;

public class AlternatingTones {

    public static void main(final String[] args) throws LineUnavailableException, InterruptedException {

        final Clip clip0 = createOneSecondClip(440f);
        final Clip clip1 = createOneSecondClip(16000f);

        clip0.addLineListener(event -> {
            if (event.getType() == LineEvent.Type.STOP) {
                clip1.setFramePosition(0);
                clip1.start();
            }
        });
        clip1.addLineListener(event -> {
            if (event.getType() == LineEvent.Type.STOP) {
                clip0.setFramePosition(0);
                clip0.start();
            }
        });
        clip0.start();

        // prevent JVM from exiting
        Thread.sleep(10000000);
    }

    private static Clip createOneSecondClip(final float frequency) throws LineUnavailableException {
        final Clip clip = AudioSystem.getClip();
        final AudioFormat format = new AudioFormat(AudioFormat.Encoding.PCM_SIGNED, 44100f, 16, 1, 2, 44100, true);
        final ByteBuffer buffer = ByteBuffer.allocate(44100 * format.getFrameSize());
        final ShortBuffer shortBuffer = buffer.asShortBuffer();
        final float cycleInc = frequency / format.getFrameRate();
        float cyclePosition = 0f;
        while (shortBuffer.hasRemaining()) {
            shortBuffer.put((short) (Short.MAX_VALUE * Math.sin(2 * Math.PI * cyclePosition)));
            cyclePosition += cycleInc;
            if (cyclePosition > 1) {
                cyclePosition -= 1;
            }
        }
        clip.open(format, buffer.array(), 0, buffer.capacity());
        return clip;
    }
}

Answer 2

我将使用的方法是在输出到 SourceDataLine 时计算帧数。当你写完一秒钟的帧时，切换频率。这将比尝试 fiddle 使用 Clips 提供更好的计时准确性。

我不清楚您显示的代码是您编写的还是复制粘贴的。如果您对它如何不起作用有疑问，如果您展示您的尝试以及产生了哪些错误或异常，我很乐意提供帮助。

当输出到 SourceDataLine 时，必须有一个步骤，根据您拥有的音频格式中指定的 16 位编码，将短值 (-32768..+32767) 转换为两个字节。我看不到这是在您的代码中完成的。 [编辑：可以看到 putShort() 方法在哪里执行此操作，尽管它仅适用于 BigEndian，而不适用于更常见的 LittleEndian。]

您看过 Java 教程了吗 Sound Trail?

在 Java 中交替播放 2 个不同的频率

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