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Topic: derivative of discrete fourier transform interpolation
Replies: 8   Last Post: Jun 6, 2013 7:24 AM

 Messages: [ Previous | Next ]
 Peter Spellucci Posts: 2,760 Registered: 12/7/04
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(N-1), we get a known real sequence x0,x1,x2,...x(N-1).
>
>Now we make an interpolation of the original function. Define X0,X1,X2....X(N-1) as the discrete fourier transform of x0,x1...x(N-1). The Xj's are typically complex numbers. Now, I believe we can interpolate x(s) as
>
>x(s)= (1/N)*sum {n=0...(N-1)} 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...(N-1)} 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

Date Subject Author
9/30/05 Gareth Davies
9/30/05 Rusty
9/30/05 Rusty
9/30/05 Rusty
9/30/05 Rusty
9/30/05 Peter Spellucci
9/30/05 Steven G. Johnson
10/1/05 Rusty
6/6/13 Wesley