fftw

Why is my data in the frequency domain “mirrored” when performing (2d) IDFT into DFT using FFTW?

由 狂风中的少年 提交于 2021-02-20 00:49:55
问题 I am manually initializing a state in the 2d frequency domain by setting the real components of certain modes in a 16x16 data set. I then perform a 2d IDFT to acquire the real domain data. This all works as expected. I then perform a DFT on the real domain data to get back (what should be) identical frequency modes to those that I manually initialized. However, they come back with their amplitudes halfed, and their vertical frequencies "mirrored". To illustrate: Input modes: k[1, 0]: 32 + 0i

calling fftw in multi thread program

由 Deadly 提交于 2021-02-08 05:47:23
问题 I want to use fftw3 in threads. But the code pasted at http://codepad.org/lIjdGF5z causes "double free or corruption" error. How to call fftw3 routines in threads properly. Thanks! You can compile the code through command "g++ test.cpp -lfftw3_threads -lfftw3 -lboost_thread" 回答1: I believe the call to fftw_plan_dft_2d is not reentrant, meaning that it can't be called in multiple threads simultaneously, even if you are creating different plans. The only fftw functions that are thread-safe are

FFT Multiple 1d transforms using FFTW

由 自作多情 提交于 2021-01-28 19:41:35
问题 I have a 3-dimensional array U(z,y,x) and I want to perform a complex Fourier transform in z for all values of y and x. I am planning to use the FFTW library. I figured out from the FFTW manual that there is a way to perform multiple 1d transforms at once(mentioned below). CALL dfftw_plan_many_dft(PLAN, rank, n, howmany, in, inembed, istride, idist, out, onembed, ostride, odist, FFTW_MEASURE) I don't clearly understand what inembed and outembed means. Could you provide more insight into this

FFT快速傅里叶变换算法

由 ◇◆丶佛笑我妖孽 提交于 2020-11-16 04:57:42
1、FFT算法概要： FFT（Fast Fourier Transformation）是离散傅氏变换（DFT）的快速算法。即为快速傅氏变换。它是根据离散傅氏变换的奇、偶、虚、实等特性，对离散傅立叶变换的算法进行改进获得的。 2、FFT算法原理： 离散傅里叶变换DFT公式： FFT算法(Butterfly算法) 设x(n)为N项的 复数 序列，由DFT变换，任一X（m）的计算都需要N次复数乘法和N-1次复数加法，而一次复数乘法等于四次 实数 乘法和两次实数加法，一次复数加法等于两次实数加法，即使把一次复数乘法和一次复数加法定义成一次“运算”（四次实数乘法和四次实数加法），那么求出N项复数序列的X（m）,即N点DFT变换大约就需要N^2次运算。当N=1024点甚至更多的时候，需要N 2=1048576次运算，在FFT中，利用WN的周期性和对称性，把一个N项序列（设N=2k,k为正整数），分为两个N/2项的子序列，每个N/2点DFT变换需要（N/2） 2次运算，再用N次运算把两个N/2点的DFT变换组合成一个N点的DFT变换。这样变换以后，总的运算次数就变成N+2*（N/2)^2=N+（N^2）/2。继续上面的例子，N=1024时，总的运算次数就变成了525312次，节省了大约50%的运算量。而如果我们将这种“一分为二”的思想不断进行下去，直到分成两两一组的DFT运算单元
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