我应该从 getFft 看到什么样的输出？

好吧，我正在努力创建一个 Android 音频可视化应用程序。问题是，我从 getFft() 方法中得到的结果与谷歌所说的它应该产生的结果不一致。我一直追溯到 C++ 源代码，但我对 C++ 或 FFT 不够熟悉，无法真正理解正在发生的事情。

我将尝试包含这里所需的一切：

(Java) Visualizer.getFft(byte[] fft) http://www.google.com/codesearch/p?hl=en#uX1GffpyOZk/media/java/android/media/audiofx/Visualizer.java

 /**
 * 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) http://www.google.com/codesearch/p?hl=en#uX1GffpyOZk/media/libmedia/Visualizer.cpp

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 *波形) http://www.google.com/codesearch/p?hl=en#uX1GffpyOZk/media/libmedia/Visualizer.cpp

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) http://www.google.com/codesearch/p?hl=en#uX1GffpyOZk/media/libmedia/fixedfft.cpp

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) http://www.google.com/codesearch/p?hl=en#uX1GffpyOZk/media/libmedia/fixedfft.cpp

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;
        }
    }
}
}

如果你能完成这一切，那你就太棒了！所以我的问题是，当我调用 java 方法 getFft() 时，我最终会得到负值，如果返回的数组旨在表示大小，则该负值不应该存在。所以我的问题是，我需要做什么才能使数组代表大小？

EDIT:看来我的数据实际上可能是傅里叶系数。我在网上浏览发现this http://www.mathcs.org/java/programs/FFT/。小程序“Start Function FFT”显示系数的图形表示，它是我绘制 getFft() 中的数据时所发生情况的真实图像。所以新问题：这就是我的数据吗？如果是这样，我如何从系数进行频谱分析？

FFT 不仅产生幅度，还产生幅度。它也产生相位（每个样本的输出是一个复数）。如果您想要幅度，那么您需要为每个输出样本显式计算它，如下所示re*re + im*im, where re and im分别是每个复数的实部和虚部。

Unfortunately, I can't see anywhere in your code where you're working with complex numbers, so perhaps some rewrite is required.

UPDATE

如果我必须猜测（浏览代码后），我会说实数分量位于偶数索引，奇数分量位于奇数索引。因此，要获得幅度，您需要执行以下操作：

uint32_t mag[N/2];
for (int i = 0; i < N/2; i++)
{
    mag[i] = fft[2*i]*fft[2*i] + fft[2*i+1]*fft[2*i+1];
}