|
[Date Index]
[Thread Index]
[Author Index]
Re: Fourier Question
- To: mathgroup at smc.vnet.net
- Subject: [mg25653] Re: Fourier Question
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Wed, 18 Oct 2000 02:52:25 -0400 (EDT)
- Organization: Universitaet Leipzig
- References: <8seaq8$6s8@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
Fourier[] make a discrete Fourier transform of equal sampled data
on the interval [0,2Pi). It make a comples to complex transform
i.e. with basis functions Exp[i k t] and not Cos[] or Sin[] only.
A nice introduction text about discret Fourier transform may be
at:
http://www.ulib.org/webRoot/Books/Numerical_Recipes/bookcpdf.html
chapter 12.
Mathematica is clearly much smarter than the NR functions but the
text is very clear.
Regards
Jens
xxx at uunet.uu.net wrote:
>
> Hello. I am reading a few books on Fourier Analysis, and I am stuck on
> understanding a few concepts.
> I was hoping someone who uses this function in Mathematica could point me in
> the right direction.
>
> To make it simple, suppose I have 4 sample points:
>
> data = {1, 3, 2, 4}
>
> Fourier[data, FourierParameters -> {-1, 1}] // Chop
>
> This returns:
>
> {2.5, -0.25 - 0.25 \[ImaginaryI], -1., -0.25 + 0.25 \[ImaginaryI]}
>
> I am not sure how to interpret the results.
> The best I understand is:
>
> results = 2.5 + (-.25-.25 I ) Cos[1 t] -1 Sin [ 1 t] + (-.25 + .25 I Cos
> 2 t]...
>
> As you can tell, I am sure this is not correct.
> Could someone point me in the right direction. I am sure I will figure it
> out from there once I understand how the results are used.
>
> I am using FourierParameters -> {-1, 1} because I am mostly doing data
> analysis.
> Thank you in advance for any insight. Dana.
Prev by Date:
Re: Simple equation checking..
Next by Date:
RE: Simple equation checking..
Previous by thread:
Fourier Question
Next by thread:
Re: Fourier Question
|