Fourier question# Computation - 科学计算
p*f
1 楼
Let's say we have
f(x)=sin(x), and thus f"(x)=-sin(x)
and let's use Fourier as
F(k)=\int f(x) e^{-i*2*pi*k*x}dx
and thus DFT as
F(n/NT)=\sum_{j=0}^{j=N-1} f(jT)e^{-i*2*pi*n*j/N}, n=0,...N-1
As a result, we should have
F[f"(x)]=-k^2*F[f(x)]
But if I calculate F[f(x)] use FFT, and then calculate F[f"(x)] according to
F[f"(x)]=-k^2*F[f(x)]=-(n/2pi)^2*F(f(x)), n=0,...N-1, NT=2pi
I cannot recover F[f"(x)]=F[-sin(x)]
what is the problem? I am really confused now. Thanks in advance.
f(x)=sin(x), and thus f"(x)=-sin(x)
and let's use Fourier as
F(k)=\int f(x) e^{-i*2*pi*k*x}dx
and thus DFT as
F(n/NT)=\sum_{j=0}^{j=N-1} f(jT)e^{-i*2*pi*n*j/N}, n=0,...N-1
As a result, we should have
F[f"(x)]=-k^2*F[f(x)]
But if I calculate F[f(x)] use FFT, and then calculate F[f"(x)] according to
F[f"(x)]=-k^2*F[f(x)]=-(n/2pi)^2*F(f(x)), n=0,...N-1, NT=2pi
I cannot recover F[f"(x)]=F[-sin(x)]
what is the problem? I am really confused now. Thanks in advance.
