FFTW versus NRC FFT

Hello,

I have a problem with the FFTW algorithm, special the discrete sine
transform. Before exchanging the FFT routine known from the NRC
(numerical recipes) book by that given from FFTW, I performed some
tests. I started with the function exp(-t) which can be analytically
transformed. Applying both codes for that function the FFTW algorithm
limits into a function value of one not zero as one would think and
what the FFT NRC code does. What is the problem here? The FFT NRC code
is checked by using it before on many problems.

Sincerely,

Volker

The code I used:

#include<fftw3.h>
#include<math.h>
#include<stdio.h>
#include<stdlib.h>

#define NR_END 1
#define FREE_ARG char*

int main(void)
{
double *vector(unsigned long start, unsigned long end);
void free_vector(double *vec, unsigned long start, unsigned long end);
void Transform_sine(double *function, unsigned long N);

unsigned int i, N=2048;

double dr = 0.1;

double *in_nrc, *out_nrc;

in_nrc=vector(1, N);
out_nrc=vector(1, N);

double *in, *out;
fftw_plan p;

in=(double*)fftw_malloc(sizeof(double)*N);
out=(double*)fftw_malloc(sizeof(double)*N);
p=fftw_plan_r2r_1d(N,in,out,FFTW_RODFT10,FFTW_ESTIMATE);

for(i=0;i<N;i++) {
in[i]=exp(-dr*i);
in_nrc[i+1]=in[i];
}

Transform_sine(in_nrc, N);

fftw_execute(p);

for(i=0;i<N;i++) {
printf("%e %e %e\n", in[i], out[i], in_nrc[i+1]);
}

fftw_free(in); fftw_free(out);

return(0);
}

The NRC routines you will find free online-readable at: www.nr.com

.