我有一个键盘踏板(如在钢琴键盘中),我已插入麦克风插孔。我希望能够将输入翻译成击键(例如,当我踏上踏板时,按下" V"在我的COMPUTER键盘上)。我知道我可以使用Java的AWT机器人来执行击键,但是我在处理麦克风输入时遇到了麻烦(我没有音频处理经验)。
这是我踩踏板并释放踏板的信号:
对于有经验的用户来说,这似乎是一项非常简单的任务,那里的任何人都知道我该怎么做吗?
到目前为止,我基本上都是从地方复制粘贴代码来获取此代码:
我使用javax.sound API将麦克风输入读取为字节。当我踏上踏板时,我试图检测振幅的跳跃......字节转换为短路,并将值与任意高的数字进行比较。它似乎工作但是在一次踏板踩下之后,机器人只是一直按下键(值> 32000)。
while (true) {
// Read the next chunk of data from the TargetDataLine.
numBytesRead = microphone.read(data, 0, data.length);
// Save this chunk of data.
out.write(data, 0, numBytesRead);
byte[] bytes = out.toByteArray();
short[] shorts = new short[bytes.length/2];
// to turn bytes to shorts as either big endian or little endian.
ByteBuffer.wrap(bytes).order(ByteOrder.LITTLE_ENDIAN).asShortBuffer().get(shorts);
for (short s : shorts) {
int value = Math.abs(s);
if (value > 32000)
{
robot.keyPress(KeyEvent.VK_V);
robot.keyRelease(KeyEvent.VK_V);
break;
} else {
//robot.keyRelease(KeyEvent.VK_V);
}
}
}
编辑:
它一直在继续,因为我没有清除ByteArrayOutputStream缓冲区,所以随着时间的推移,我不断地读取相同的字节集和新字节。 out.reset()为我解决了这个问题。
我现在的问题是关于我从踏板上读取的输入,如果我在短时间内按下并释放踏板,我无法正确解释按下或释放。
红色圆圈表示踩下(并保持)踏板时的状态,踏板松开时显示黑色矩形。
正如你所看到的,当它被压低时,它会下降然后迅速增加,然后再逐渐回到0。当它被释放时,它会向上射击,然后迅速下降,然后再次回到0。
我现在用来区分两者的方法是仅在两帧/间隔之间存在较大差异时才记录按下/释放。根据图表,当它是负值时我按下它,当它是正值时我发布它。
我遇到的问题是,当信号处于上升状态时(或当信号处于释放的下降状态时我踩下踏板)时,我松开踏板时,会赢得“won”两个帧之间的差异足以让我使用这种方法。
我不知道如何通过强大的方式来检测新闻/发布。
这是我输入输入的固定代码,如果有人有兴趣尝试同样的事情(顺便说一下我使用这个踏板:http://www.amazon.co.uk/Cherub-WTB-004-Keyboard-Sustain-Pedal/dp/B000UDVV6E)
import java.io.ByteArrayOutputStream;
import java.nio.ByteBuffer;
import java.nio.ByteOrder;
import javax.sound.sampled.*;
import java.awt.AWTException;
import java.awt.Robot;
import java.awt.event.KeyEvent;
public class PedalToKeyboard {
public static void main(String[] args) {
AudioFormat format = new AudioFormat(8000.0f, 16, 1, true, true);
Robot robot = null;
try {
robot = new Robot();
} catch (AWTException e) {
e.printStackTrace();
}
try {
TargetDataLine microphone = AudioSystem.getTargetDataLine(format);
System.out.println(microphone);
microphone.open(format);
ByteArrayOutputStream out = new ByteArrayOutputStream();
int numBytesRead;
byte[] data = new byte[microphone.getBufferSize()/5];
// Begin audio capture.
microphone.start();
boolean keep_going = true;
boolean keyPressed = false;
short previousShort = 0;
while (keep_going) {
// Read the next chunk of data from the TargetDataLine.
numBytesRead = microphone.read(data, 0, data.length);
// Reset the buffer (get rid of previous shit)
out.reset();
// Save this chunk of data.
out.write(data, 0, numBytesRead);
byte[] bytes = out.toByteArray();
short[] shorts = new short[bytes.length/2];
// to turn bytes to shorts as either big endian or little endian.
ByteBuffer.wrap(bytes).order(ByteOrder.BIG_ENDIAN).asShortBuffer().get(shorts);
for (short s : shorts) {
// Check if descending or ascending (pedal press is descending, release is ascending)
if (s < 0) { // descending
// make sure drop is large instantaneous drop
if (Math.abs(Math.abs(previousShort) - Math.abs(s)) > 10000 && s < -32700) {
if (!keyPressed) {
keyPressed = true;
//robot.keyPress(KeyEvent.VK_V);
System.out.println("Pressed: " + s);
break;
}
}
} else if (s > 0) { // ascending
// make sure increase is large instantaneous increase
if (Math.abs(Math.abs(previousShort) - Math.abs(s)) > 10000 && s > 32700) {
if (keyPressed) {
keyPressed = false;
//robot.keyRelease(KeyEvent.VK_V);
System.out.println("Released: " + s + "\n");
break;
}
}
}
previousShort = s;
}
}
} catch (LineUnavailableException e) {
e.printStackTrace();
}
}
}
答案 0 :(得分:1)
好吧我实际上调整了一些值并修正了一些逻辑(我在比较差异时错误地应用了Math.abs)。
这对我来说效果很好,任何有MIDI踏板的人都可以尝试调整参数供你自己使用。
package pedal2keyboard;
import java.io.ByteArrayOutputStream;
import java.nio.ByteBuffer;
import java.nio.ByteOrder;
import javax.sound.sampled.*;
import java.awt.AWTException;
import java.awt.Robot;
import java.awt.event.KeyEvent;
/***
* Author: Dois Koh
* Date: 27th October 2015
*
* Gets your microphone signal and you can go do whatever you want with it.
* Right now, it takes signals from my Cherub WTB-004 Keyboard Sustain Pedal, plugged into
* my microphone jack, and converts it into key presses (holds down V when depressed,
* releases V when released)
*/
public class PedalToKeyboard {
// Robot for performing keyboard actions (pressing V)
public static Robot robot = null;
// Currently 8KHz, 16 bit signal (2 bytes), single channel, signed (+ and -) and BIG ENDIAN format
public static AudioFormat format = new AudioFormat(8000.0f, 16, 1, true, true);
public static TargetDataLine microphone = null;
public static boolean pedalPressed = false;
public static void main(String[] args) {
try {
// Initialize robot for later use
robot = new Robot();
// Retrieve the line to from which to read in the audio signal
microphone = AudioSystem.getTargetDataLine(format);
// Open the line in the specified format -
// Currently 8KHz, 16 bit signal (2 bytes), single channel, signed (+ and -) and BIG ENDIAN format
microphone.open(new AudioFormat(8000.0f, 16, 1, true, true));
ByteArrayOutputStream out = new ByteArrayOutputStream();
byte[] data = new byte[microphone.getBufferSize()/8];
// Begin audio capture.
microphone.start();
int numBytesRead = 0;
short previousShort = 0;
// Continue until program is manually terminated
while (true) {
// Read the next chunk of data from the TargetDataLine.
numBytesRead = microphone.read(data, 0, data.length);
// Reset the buffer (get rid of previous data)
out.reset();
// Save this chunk of data.
out.write(data, 0, numBytesRead);
byte[] bytes = out.toByteArray();
short[] shorts = new short[bytes.length/2];
// to turn bytes to shorts as either big endian or little endian.
ByteBuffer.wrap(bytes).order(ByteOrder.BIG_ENDIAN).asShortBuffer().get(shorts);
// Iterate through retrieved 16 bit data (shorts)
for (short s : shorts) {
// Check if descending or ascending (pedal press is descending, release is ascending)
if (s < 0) { // descending
// make sure drop is large instantaneous drop
if (Math.abs(previousShort - s) > 200 && s < -32700) {
if (!pedalPressed) {
PedalPressedAction();
previousShort = s;
break;
}
}
} else if (s > 0) { // ascending
// make sure increase is large instantaneous increase
if (Math.abs(previousShort - s) > 200 && s > 32700) {
if (pedalPressed) {
PedalReleasedAction();
previousShort = s;
break;
}
}
}
previousShort = s;
}
}
} catch (LineUnavailableException | AWTException e) {
e.printStackTrace();
} finally {
if (microphone != null)
microphone.close();
}
}
/***
* The action to perform when the pedal is depressed
*/
public static void PedalPressedAction() {
pedalPressed = true;
robot.keyPress(KeyEvent.VK_V);
}
/***
* The action to perform when the pedal is released
*/
public static void PedalReleasedAction(){
pedalPressed = false;
robot.keyRelease(KeyEvent.VK_V);
}
}
答案 1 :(得分:0)
如果您查看附加的图表,您可以看到该值在相当长的一段时间内保持在32000以上 - 可能大约100ms左右。根据您的采样率,这可以转化为相当多的按键。例如,在44.1kHz,这将是4410。您最初只能超过阈值时才需要模拟按键。为此,您需要跟踪某些状态。你可以使用PRESS_THRESHOLD和RELEASE_THRESHOLD的值来玩,但是从32000和31000开始。不设置它们的原因是为了防止错误按键,如果信号在边缘附近有毛刺。
for (short s : shorts) {
int value = Math.abs(s);
switch (state)
{
case State.pressed:
if (value < RELEASE_THRESHOLD)
{
state = State.released;
}
break;
case State.released:
if (value > PRESS_THRESHOLD)
{
robot.keyPress(KeyEvent.VK_V);
robot.keyRelease(KeyEvent.VK_V);
state = State.pressed;
}
break;
}
}