
Re: derivative of discrete fourier transform interpolation
Posted:
Sep 30, 2005 1:07 PM


In article <24659772.1128056066767.JavaMail.jakarta@nitrogen.mathforum.org>, Gareth Davies <gd340@uow.edu.au> writes: >Hi everyone >I'm trying to understand this problem which arises in a paper I've been reading. I'm quite new to fourier analyses, and I haven't been able to find anything in books/web which answers it. > > >If x(s) is an (unknown) real function which is sampled at evenly spaced intervals denoted s0,s1,s2,s3...s(N1), we get a known real sequence x0,x1,x2,...x(N1). > >Now we make an interpolation of the original function. Define X0,X1,X2....X(N1) as the discrete fourier transform of x0,x1...x(N1). The Xj's are typically complex numbers. Now, I believe we can interpolate x(s) as > >x(s)= (1/N)*sum {n=0...(N1)} Xn*exp(2i*pi*n*s/N) > >This is a real valued function thanks to the Xn's occurring in suitable conjugate pairs. > >My problem arises when trying to calculate dx/ds using this interpolation. Differentiating term by term, we find > >dx/ds(s)= (1/N^2)*2i*pi*sum {n=0...(N1)} n*Xn*exp(2i*pi*n*s/N) > >Now, it seems to me that dx/ds is typically not real valued, even though x(s) is real valued. > >From this I have 2 questions > >1. Is what I've said above correct?? >2. What is a better approach to calculating the derivatives. > >Thanks, >Gareth Davies the FFT gives a real sin/cos series only for special N and in this case your sum is running from N/2 to N/2 . the real sin/cos (Fourier ) sum can then be differentiated in the usual manner with no trouble. hth peter

