another FFTW problem

Hi, I am using FFTW to try another test, f(t)=exp(-x), x>=0, the real part
of the Fourier transforms should be 1/(1+x^2). I use xmax=1000, and xmin=0,
let N=8192. I found the results seems shifted upper, every output value is
larger than the analytical solution. 
My code is like this:
  program test

  implicit none

  include "fftw3.f"

  integer N
  parameter(N=8192)
  integer pi
  parameter(pi=3.14159265357)

  double complex in
  double complex out
  double precision f
  dimension out(N)
  dimension in(N)
  dimension f(N)

  integer i
  real*8 :: x,y,xmax,xmin,dx
  integer*8  plan

  OPEN(UNIT=7,FILE='fftin')
  OPEN(UNIT=8,FILE='fftout')

 ! write(*,*) 'Input array:'

  xmax=1000.
  xmin=0.
  dx=(xmax-xmin)/N

  write(6,*)
'XMAX',xmax,'xmin',xmin,'dx',dx,'N',N,'omega',1.0/(2.0*dx)*2*pi

   
   do i = 1,N/2+1
      x=xmin+dx*(i-1)
      in(i)=exp(-x)*(-1)**(i+1)
      if(i.gt.N/2+1) in(i)=0.0
   enddo
   do i=1,N
      x=xmin+dx*(i-1)
      write(7,*) i,x,real(in(i)), imag(in(i))
   enddo
  close(7)


   !call dfftw_plan_r2r_1d (plan, N, in, out, fftw_r2hc, FFTW_ESTIMATE)
  !call dfftw_plan_dft_r2c_1d(plan,N,in,out,FFTW_ESTIMATE)
   call dfftw_plan_dft_1d (plan, N, in, out, FFTW_FORWARD, FFTW_ESTIMATE)

  call dfftw_execute(plan)

  !write(*,*) 'Output array:'

  do i = 1,N
     f(i)=(-N/2+i-1)/(N*dx)
     write(8,*)
i,f(i)*2*pi,real(out(i))*dx,imag(out(i))*dx,1./(1.+(f(i)*2.*pi)**2),-f(i)*2.*pi/(1.+(f(i)*2.*pi)**2)
  enddo
  close(8)
  STOP

can any one help me about it?
Thanks

jv_ju wrote:
> Hi, I am using FFTW to try another test, f(t)=exp(-x), x>=0, the real part
> of the Fourier transforms should be 1/(1+x^2). I use xmax=1000, and xmin=0,
> let N=8192. I found the results seems shifted upper, every output value is
> larger than the analytical solution. 
> My code is like this:
>   program test
> 
>   implicit none
> 
>   include "fftw3.f"
> 
>   integer N
>   parameter(N=8192)
>   integer pi
>   parameter(pi=3.14159265357)
> 
>   double complex in
>   double complex out
>   double precision f
>   dimension out(N)
>   dimension in(N)
>   dimension f(N)
> 
>   integer i
>   real*8 :: x,y,xmax,xmin,dx
>   integer*8  plan
> 
>   OPEN(UNIT=7,FILE='fftin')
>   OPEN(UNIT=8,FILE='fftout')
> 
>  ! write(*,*) 'Input array:'
> 
>   xmax=1000.
>   xmin=0.
>   dx=(xmax-xmin)/N
> 
>   write(6,*)
> 'XMAX',xmax,'xmin',xmin,'dx',dx,'N',N,'omega',1.0/(2.0*dx)*2*pi
> 
>    
>    do i = 1,N/2+1
>       x=xmin+dx*(i-1)
>       in(i)=exp(-x)*(-1)**(i+1)
>       if(i.gt.N/2+1) in(i)=0.0
>    enddo
>    do i=1,N
>       x=xmin+dx*(i-1)
>       write(7,*) i,x,real(in(i)), imag(in(i))
>    enddo
>   close(7)
> 
> 
>    !call dfftw_plan_r2r_1d (plan, N, in, out, fftw_r2hc, FFTW_ESTIMATE)
>   !call dfftw_plan_dft_r2c_1d(plan,N,in,out,FFTW_ESTIMATE)
>    call dfftw_plan_dft_1d (plan, N, in, out, FFTW_FORWARD, FFTW_ESTIMATE)
> 
>   call dfftw_execute(plan)
> 
>   !write(*,*) 'Output array:'
> 
>   do i = 1,N
>      f(i)=(-N/2+i-1)/(N*dx)
>      write(8,*)
> i,f(i)*2*pi,real(out(i))*dx,imag(out(i))*dx,1./(1.+(f(i)*2.*pi)**2),-f(i)*2.*pi/(1.+(f(i)*2.*pi)**2)
>   enddo
>   close(8)
>   STOP
> 
> can any one help me about it?

An FFT doesn't compute a continuous transform of a continuous function. 
Because what is being transformed is necessarily discrete in time, the 
result of the transform is necessarily periodic and discrete in 
frequency. 1/(1+x^2) meets neither of those constraints. Your tool can't 
produce it.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

>
>
>An FFT doesn't compute a continuous transform of a continuous function. 
>Because what is being transformed is necessarily discrete in time, the 
>result of the transform is necessarily periodic and discrete in 
>frequency. 1/(1+x^2) meets neither of those constraints. Your tool can't

>produce it.
>
>Jerry
>-- 
>Engineering is the art of making what you want from things you can get.
>�����������������������������������������������������������������������
>
How about you consider it as a periodic function with the period of 1000?

>
>
>An FFT doesn't compute a continuous transform of a continuous function. 
>Because what is being transformed is necessarily discrete in time, the 
>result of the transform is necessarily periodic and discrete in 
>frequency. 1/(1+x^2) meets neither of those constraints. Your tool can't

>produce it.
>
>Jerry
>-- 
>Engineering is the art of making what you want from things you can get.
>�����������������������������������������������������������������������
>
How about you consider it as a periodic function with the period of 1000?

jv_ju wrote:
>>
>> An FFT doesn't compute a continuous transform of a continuous function. 
>> Because what is being transformed is necessarily discrete in time, the 
>> result of the transform is necessarily periodic and discrete in 
>> frequency. 1/(1+x^2) meets neither of those constraints. Your tool can't>> produce it.

   ...

> How about you consider it as a periodic function with the period of 1000?

Will wishing make it so?

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������