[編輯1]加數字來顯示原始數據和所獲得的數據分解在FORTRAN77
[編輯2]我發現我的錯誤,我用來代替fftw_estimate fftw_measure在的呼叫dfftw_plan_many_dft
中的代碼(用dfftw_execute_dft_r2c U2D替換U)我試圖執行使用多個1D FFT陣列的2D FFT而不是使用
[編輯3]修正筆誤2D fft函數已經存在於fftw庫中。隨後,我需要執行逆2D fft。 這樣做的原因是(將來)我的數組將會太大而無法一次加載來執行2D fft。
我的代碼的第一個草案看上去或多或少像這樣的時刻:
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,但我獲得相對不同的值。我在某個地方犯了錯嗎?
原文和結果如下所示。形狀看起來相似,但數值相對不同(例如最小值和最大值)。
Field obtained after fft and i-fft
FFTW確實[沒有標準化值](http://www.fftw.org/faq/section3.html#whyscaled)。因此,如果您在相同數據上進行前後變換,則會有數組長度的因素。 –
請參閱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 –