我在这个网站上看到过与此相似的类似问题，但我的问题有点不同。我用来捕获音频的代码是this。我想简单地获取捕获的音频，然后用256点应用FFT。
我意识到这个line count = line.read(buffer, 0, buffer.length);将音频分解为“块”。
我可以找到我正在使用的FFT here。
我的问题是：
答案 0 :(得分：6)
所有javax.sound.sampled包都读取文件中的原始字节并将它们写入输出。因此，您需要做的是“介于两者之间”，即自行转换样本。
以下显示了如何对PCM执行此操作（带注释），取自我的代码示例WaveformDemo：
public static float[] unpack(
byte[] bytes,
long[] transfer,
float[] samples,
int bvalid,
AudioFormat fmt
) {
if(fmt.getEncoding() != AudioFormat.Encoding.PCM_SIGNED
&& fmt.getEncoding() != AudioFormat.Encoding.PCM_UNSIGNED) {
return samples;
}
final int bitsPerSample = fmt.getSampleSizeInBits();
final int bytesPerSample = bitsPerSample / 8;
final int normalBytes = normalBytesFromBits(bitsPerSample);
/*
* not the most DRY way to do this but it's a bit more efficient.
* otherwise there would either have to be 4 separate methods for
* each combination of endianness/signedness or do it all in one
* loop and check the format for each sample.
*
* a helper array (transfer) allows the logic to be split up
* but without being too repetetive.
*
* here there are two loops converting bytes to raw long samples.
* integral primitives in Java get sign extended when they are
* promoted to a larger type so the & 0xffL mask keeps them intact.
*
*/
if(fmt.isBigEndian()) {
for(int i = 0, k = 0, b; i < bvalid; i += normalBytes, k++) {
transfer[k] = 0L;
int least = i + normalBytes - 1;
for(b = 0; b < normalBytes; b++) {
transfer[k] |= (bytes[least - b] & 0xffL) << (8 * b);
}
}
} else {
for(int i = 0, k = 0, b; i < bvalid; i += normalBytes, k++) {
transfer[k] = 0L;
for(b = 0; b < normalBytes; b++) {
transfer[k] |= (bytes[i + b] & 0xffL) << (8 * b);
}
}
}
final long fullScale = (long)Math.pow(2.0, bitsPerSample - 1);
/*
* the OR is not quite enough to convert,
* the signage needs to be corrected.
*
*/
if(fmt.getEncoding() == AudioFormat.Encoding.PCM_SIGNED) {
/*
* if the samples were signed, they must be
* extended to the 64-bit long.
*
* so first check if the sign bit was set
* and if so, extend it.
*
* as an example, imagining these were 4-bit samples originally
* and the destination is 8-bit, a mask can be constructed
* with -1 (all bits 1) and a left shift:
*
* 11111111
* << (4 - 1)
* ===========
* 11111000
*
* (except the destination is 64-bit and the original
* bit depth from the file could be anything.)
*
* then supposing we have a hypothetical sample -5
* that ought to be negative, an AND can be used to check it:
*
* 00001011
* & 11111000
* ==========
* 00001000
*
* and an OR can be used to extend it:
*
* 00001011
* | 11111000
* ==========
* 11111011
*
*/
final long signMask = -1L << bitsPerSample - 1L;
for(int i = 0; i < transfer.length; i++) {
if((transfer[i] & signMask) != 0L) {
transfer[i] |= signMask;
}
}
} else {
/*
* unsigned samples are easier since they
* will be read correctly in to the long.
*
* so just sign them:
* subtract 2^(bits - 1) so the center is 0.
*
*/
for(int i = 0; i < transfer.length; i++) {
transfer[i] -= fullScale;
}
}
/* finally normalize to range of -1.0f to 1.0f */
for(int i = 0; i < transfer.length; i++) {
samples[i] = (float)transfer[i] / (float)fullScale;
}
return samples;
}
public static int normalBytesFromBits(int bitsPerSample) {
/*
* some formats allow for bit depths in non-multiples of 8.
* they will, however, typically pad so the samples are stored
* that way. AIFF is one of these formats.
*
* so the expression:
*
* bitsPerSample + 7 >> 3
*
* computes a division of 8 rounding up (for positive numbers).
*
* this is basically equivalent to:
*
* (int)Math.ceil(bitsPerSample / 8.0)
*
*/
return bitsPerSample + 7 >> 3;
}
这段代码假定为float[]而您的FFT需要double[]，但这是一个相当简单的更改。 transfer和samples是长度等于bytes.length * normalBytes的数组，bvalid是read的返回值。我的代码示例假设AudioInputStream，但相同的转换应适用于TargetDataLine。我不确定你是否可以直接复制和粘贴它，但这只是一个例子。
关于你的两个问题：
但是当完成FFT时，还有一些事情要做，我没有看到链接的类在做什么：
编辑，相关：