Ooura 逆 FFT 与 Matlab ifft 的区别
今天，在努力让我的 C++ 代码（使用 Ooura FFT 库）给出与 Matlab 相同的结果时，我终于找到了一个问题和解决方案。
在Matlab中，计算幅频响应网格的反向系数是通过以下方式完成的（Matlab内部fir2()函数的代码片段）：
%H contains 8192 points of AFR data
Hconj = [H conj(H(npt-1:-1:2))]; % Fourier transform of real series
ht = real(ifft(Hconj)); % Symmetric real series
结果我们得到16384个bin，其中后半部分可以扔掉，但前半部分稍后可以用作 FIR 系数。
但是，如果我使用 Real DFT rdft() 函数在 Ooura 中执行相同操作，我会得到在生成的 AFR 中创建镜像效果的系数，AFR 图上的所有频率都除以 2。
所以我想到了一个想法：在我的C++ 代码 我将 H 放大两倍（16384 点），并用频率数据填充它们，而不进行镜像。瞧！它成功了，现在我得到了 16384 点，扔掉 8192 点之后的所有内容，现在得到的 AFR 与 Matlab 的相匹配。
我确信所有标准 FFT 实现都需要这种镜像。这是否只是 Ooura 的一个怪癖，它不需要在输入中镜像数据，或者可能还有其他原因？
Today while struggling to make my C++ code (using Ooura FFT library) to give the same results as Matlab does, I finally found a problem and a solution.
In Matlab calculating reverse coefficients for a grid of amplitude-frequency response is done in a following way (code fragment from Matlab's internal fir2() function):
%H contains 8192 points of AFR data
Hconj = [H conj(H(npt-1:-1:2))]; % Fourier transform of real series
ht = real(ifft(Hconj)); % Symmetric real series
as a result we get back 16384 bins, and the second half of them can be thrown away, but the first half can later be used as FIR coefficients.
But if I do the same in Ooura using Real DFT rdft() function, I get coefficients which create a mirror effect in the resulting AFR, all the frequencies on the AFR plot are divided by 2.
So an idea came to me: in my C++ code I made my H twice as big (16384 points) and filled them all with frequency data without mirroring. And voila! it worked, now I got 16384 points, throw away everything after 8192 point and now the resulting AFR matches the Matlab's.
I was convinced that all standard FFT implementations need this mirroring. Is it just a quirk of Ooura that it DOES NOT need mirroring data in input, or maybe there is something else going on?
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“真正的”FFT 通常通过自动使用镜像值的输入的复共轭来进行内部镜像。标准FFT/IFFT 不会这样做，因为它的自由度是限制为实数输出的FFT/IFFT 实现的两倍（例如，从IFFT 产生复数输出，或者说，来自IFFT 的复数域滤波器）。
FFT 的频率仓步长由其长度控制。一半的长度产生一半的频率步长，就像您最初的试验一样。
A "real" FFT often does the mirroring internally by automatically using the complex conjugates of the input for the mirrored values. A standard FFT/IFFT doesn't do this because it has twice the degrees of freedom (e.g. to produce a complex output or, say, a complex domain filter from an IFFT) of an FFT/IFFT implementation constrained to real-only output.
The frequency bin step size of an FFT is controlled by its length. Half the length produces half the frequency step size, as in your original trial.