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好的，所以我正在创建一个Android音频可视化应用程序。问题是，我从getFft（）方法得到的结果并没有与Google说它应该产生的结果有关。我将源代码追溯到C++，但我对C++或FFT不够熟悉，无法真正了解发生了什么。我应该从getFft看到什么样的输出？
我会努力的，包括在这里所需要的一切：
(Java) Visualizer.getFft(byte[] fft)
/**
* Returns a frequency capture of currently playing audio content. The capture is a 8-bit
* magnitude FFT. Note that the size of the FFT is half of the specified capture size but both
* sides of the spectrum are returned yielding in a number of bytes equal to the capture size.
* {@see #getCaptureSize()}.
* <p>This method must be called when the Visualizer is enabled.
* @param fft array of bytes where the FFT should be returned
* @return {@link #SUCCESS} in case of success,
* {@link #ERROR_NO_MEMORY}, {@link #ERROR_INVALID_OPERATION} or {@link #ERROR_DEAD_OBJECT}
* in case of failure.
* @throws IllegalStateException
*/
public int getFft(byte[] fft)
throws IllegalStateException {
synchronized (mStateLock) {
if (mState != STATE_ENABLED) {
throw(new IllegalStateException("getFft() called in wrong state: "+mState));
}
return native_getFft(fft);
}
}
(C++) Visualizer.getFft(uint8_t *fft)
status_t Visualizer::getFft(uint8_t *fft)
{
if (fft == NULL) {
return BAD_VALUE;
}
if (mCaptureSize == 0) {
return NO_INIT;
}
status_t status = NO_ERROR;
if (mEnabled) {
uint8_t buf[mCaptureSize];
status = getWaveForm(buf);
if (status == NO_ERROR) {
status = doFft(fft, buf);
}
} else {
memset(fft, 0, mCaptureSize);
}
return status;
}
(C++) Visualizer.doFft(uint8_t *fft, uint8_t *waveform)
status_t Visualizer::doFft(uint8_t *fft, uint8_t *waveform)
{
int32_t workspace[mCaptureSize >> 1];
int32_t nonzero = 0;
for (uint32_t i = 0; i < mCaptureSize; i += 2) {
workspace[i >> 1] = (waveform[i]^0x80) << 23;
workspace[i >> 1] |= (waveform[i + 1]^0x80) << 7;
nonzero |= workspace[i >> 1];
}
if (nonzero) {
fixed_fft_real(mCaptureSize >> 1, workspace);
}
for (uint32_t i = 0; i < mCaptureSize; i += 2) {
fft[i] = workspace[i >> 1] >> 23;
fft[i + 1] = workspace[i >> 1] >> 7;
}
return NO_ERROR;
}
(C++) fixedfft.fixed_fft_real(int n, int32_t *v)
void fixed_fft_real(int n, int32_t *v)
{
int scale = LOG_FFT_SIZE, m = n >> 1, i;
fixed_fft(n, v);
for (i = 1; i <= n; i <<= 1, --scale);
v[0] = mult(~v[0], 0x80008000);
v[m] = half(v[m]);
for (i = 1; i <n>> 1; ++i) {
int32_t x = half(v[i]);
int32_t z = half(v[n - i]);
int32_t y = z - (x^0xFFFF);
x = half(x + (z^0xFFFF));
y = mult(y, twiddle[i << scale]);
v[i] = x - y;
v[n - i] = (x + y)^0xFFFF;
}
}
(C++) fixedfft.fixed_fft(int n, int32_t *v)
void fixed_fft(int n, int32_t *v)
{
int scale = LOG_FFT_SIZE, i, p, r;
for (r = 0, i = 1; i < n; ++i) {
for (p = n; !(p & r); p >>= 1, r ^= p);
if (i < r) {
int32_t t = v[i];
v[i] = v[r];
v[r] = t;
}
}
for (p = 1; p < n; p <<= 1) {
--scale;
for (i = 0; i < n; i += p << 1) {
int32_t x = half(v[i]);
int32_t y = half(v[i + p]);
v[i] = x + y;
v[i + p] = x - y;
}
for (r = 1; r < p; ++r) {
int32_t w = MAX_FFT_SIZE/4 - (r << scale);
i = w >> 31;
w = twiddle[(w^i) - i]^(i << 16);
for (i = r; i < n; i += p << 1) {
int32_t x = half(v[i]);
int32_t y = mult(w, v[i + p]);
v[i] = x - y;
v[i + p] = x + y;
}
}
}
}
如果你通过所有的成功了，你真棒！所以我的问题是，当我调用java方法getFft（）时，我最终得到的是负值，如果返回的数组意味着代表幅度，则该值不应存在。所以我的问题是，我需要做些什么来使阵列代表幅度？
编辑：看来我的数据实际上可能是傅立叶系数。我在网上搜索，发现this。小程序“启动函数FFT”显示系数的图形表示，它是当我绘制来自getFft（）的数据时发生的事情的随机图像。所以新的问题：这是我的数据是什么？如果是的话，我怎样才能从系数到频谱分析呢？
你不是已经得到了答案，这在这里？ http://stackoverflow.com/questions/4720512/android-2-3-visualizer-trouble-understanding-getfft – 2011-01-24 22:22:39
这是一个不同的问题。原来是这样的：首先，“频谱的两边”是什么意思？这个输出与标准FFT有何不同？这次我有相关的源代码和一个更具体的问题。 – ebolyen 2011-01-24 22:26:11
@Evan，我面临同样的问题。当我在画布上绘制fft数据时，它看上去有线。你有这个解决方案吗？这是我在SO上发布的问题。 http://stackoverflow.com/questions/7024187/fft-data-displayed-on-line-graph-not-showing-smoothly – 2011-08-13 10:49:38