[编辑1]加数字来显示原始数据和所获得的数据分解在FORTRAN77
[编辑2]我发现我的错误,我用来代替fftw_estimate fftw_measure在的呼叫dfftw_plan_many_dft
中的代码(用dfftw_execute_dft_r2c U2D替换U)我试图执行使用多个1D FFT阵列的2D FFT而不是使用
[编辑3]修正笔误2D fft函数已经存在于fftw库中。随后,我需要执行逆2D fft。 这样做的原因是(将来)我的数组将会太大而无法一次加载来执行2D fft。
我的代码的第一个草案看上去或多或少像这样的时刻:
double precision u2d(nx,ny),u2d2(nx,ny)
double complex qhat2d(nx/2+1,ny),qhat2d2(nx/2+1,ny)
integer N(1)
integer howmany, idist, odist, istride, ostride
integer inembed(2), onembed(2)
integer rank
! some function to read the data into u2d
! perform x-fft
N(1) = NX
howmany = NY
inembed(1) = NX
inembed(2) = NY
istride = 1
idist = NX
ostride = 1
odist = (NX/2+1)
onembed(1) = (NX/2+1)
onembed(2) = NY
rank = 1
write(*,*) 'u', u2d(1,1)
CALL dfftw_plan_many_dft_r2c(PLAN,rank,N(1),howmany,
& u2d,inembed,
& istride,idist,
& qhat2d,onembed,
& ostride,odist,FFTW_ESTIMATE) !
CALL dfftw_execute_dft_r2c(PLAN,u2d,qhat2d) ! x-fft
CALL dfftw_destroy_plan(PLAN)
! perform y-fft
N(1) = NY
howmany = (NX/2+1)
inembed(1) = (NX/2+1)
inembed(2) = NY
istride = (NX/2+1)
idist = 1
ostride = (NX/2+1)
odist = 1
onembed(1) = (NX/2+1)
onembed(2) = NY
rank = 1
CALL dfftw_plan_many_dft(PLAN,rank,N(1),howmany,
& qhat2d,inembed,
& istride,idist,
& qhat2d2,onembed,
& ostride,odist,FFTW_FORWARD,
& FFTW_MEASURE) !
CALL dfftw_execute_dft(PLAN,qhat2d,qhat2d2) ! y-fft
CALL dfftw_destroy_plan(PLAN)
! normally here, perform some filtering operation
! but at the moment, I do nothing
! perform inv-y-fft
N(1) = NY
howmany = (NX/2+1)
inembed(1) = (NX/2+1)
inembed(2) = NY
istride = (NX/2+1)
idist = 1
ostride = (NX/2+1)
odist = 1
onembed(1) = (NX/2+1)
onembed(2) = NY
rank = 1
CALL dfftw_plan_many_dft(PLAN,rank,N(1),howmany,
& qhat2d2,inembed,
& istride,idist,
& qhat2d,onembed,
& ostride,odist,FFTW_BACKWARD,
& FFTW_MEASURE) !
CALL dfftw_execute_dft(PLAN,qhat2d2,qhat2d) ! inv-y-fft
CALL dfftw_destroy_plan(PLAN)
! perform inv-x-fft
N(1) = NX ! I'm not too sure about this value here
howmany = NY
inembed(1) = (NX/2+1)
inembed(2) = NY
istride = 1
idist = (NX/2+1)
ostride = 1
odist = NX
onembed(1) = NX
onembed(2) = NY
rank = 1
CALL dfftw_plan_many_dft_c2r(PLAN,rank,N(1),howmany,
& qhat2d,inembed,
& istride,idist,
& u2d2,onembed,
& ostride,odist,FFTW_ESTIMATE) !
CALL dfftw_execute_dft_c2r(PLAN,qhat2d,u2d2) ! x-fft
CALL dfftw_destroy_plan(PLAN)
write(*,*) 'u2d2', u2d2(1,1)
do i=1,nx
do j=1,ny
u2d2(i,j) = u2d2(i,j)/(nx*ny)
enddo
enddo
write(*,*) 'u2d2', u2d2(1,1) ! here the values u2d2(1,1) is different from u2d(1,1)
! some action to write u2d2 to file
end
我期待U2D和u2d2是相同的,但我获得相对不同的值。我在某个地方犯了错吗?
原文和结果如下所示。形状看起来相似,但数值相对不同(例如最小值和最大值)。
Field obtained after fft and i-fft
FFTW确实[没有标准化值](http://www.fftw.org/faq/section3.html#whyscaled)。因此,如果您在相同数据上进行前后变换,则会有数组长度的因素。 –
请参阅https://stackoverflow.com/questions/3721125/fftw-inverse-of-forward-fft-not-equal-to-original-function/7871634#7871634 –
或https://stackoverflow.com/ questions/4855958/normalizing-fft-data-fftw –