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理解FFT办法
原标题:Understanding FFT in aurioTouch2
  • 时间:2012-01-13 03:29:05
  •  标签:
  • ios
  • audio
  • fft
最佳回答

这对我来说不是微不足道的,而是我如何理解。 如果超过简化,这纯粹是为了我的利益,就意味着赞助。

根据Nyquist频率计算结果:

I m going to suppose we are computing the forward FFT of 1024 input samples. At 44100hz input this is usually true in my case (but isn t what AurioTouch is doing, which i find a bit weird, but i m no expert). It s easier for me to understand with specific values.

鉴于1024(n)输入样本,按需要排列(7个指数,先是奇指数{[0],载于[4],......,载于1 ,载于[3],载于[5], ......} (使用代码>vPDS_ctoz(oz)(<)/rp)。

FFT 1024 (n) 输入样本的输出是513 ((n/2)+1) 复杂值.ie 513 reallement and 513 imaginarylement, a total of 1026 Value。

然而,imaginary [0] and imaginary [512] (n/2), 当然,zero。 因此,将real[512](Nyquistcies bin的真正组成部分)放在inginary [0],并插入ginary[512]——这一数值始终为零,可以推算,结果被包装为1024(n) 长度的缓冲。

So, for the returned results to be valid you must at least set imaginary[0] back to zero. If you require all 513 ((n/2)+1) frequency bins you need to append another complex value to the result and set it thus..

unpackedVal = imaginary[0]
real[512]=unpackedVal, imaginary[512]=0
imaginary[0] = 0

在Ausrio Touchi, 总是说他们只是 t。 n/2 成果显然更方便地与视力者合作,而且你很难从视力者那里了解:“Ohook,在Nyquist频率<<>/em>上遗漏了一层。

The UsingFourier Transforms docs interpret the Packing

www.un.org/Depts/DGACM/index_spanish.htm NB 具体数值1024、513、512等,不是Ausrio Touch的N、(n/2)+1、N/2的实际数值。

产出价值的范围并不大,即与投入样本数量有关,因此必须加以规范。 比额表为1.0/(2*inNumberFrames)。

扩大范围后为-1.0->+1.0。 随之而来的是复杂病媒的规模(这个阶段被忽视),每个频率在0 和1.0之间给出了一定数值。

然后将这一数值解释为-128和0之间的折中值。

绘画......+80 /64. ......*120......i m不确定。 我可能是完全错误的,也可能是......艺术许可证?

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