Re: Mystery FFT Code...

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Article #815

#815

From: John Vlontzos
Date: Mon, 11 Jun 1990 14:24

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Looking at the code:
nn is the log(N) (base 2) that is for 1024 points, nn=10
isign controls the sign of exp(2kp/N) (forward/inverse fft)
ifft=(1/N)fft(X*) where X*=complex conjugate.
I didn't look too hard but I think the routine doesn't
divide by N when computing the inverse fft. 

John Vlontzos
Siemens

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Path: arc1-ca.pugleaf.net!utzoo!attcan!uunet!wuarchive!udel!princeton!siemens!xydeco!vlo
From: vlo@xydeco.siemens.com (John Vlontzos)
Newsgroups: comp.dsp
Subject: Re: Mystery FFT Code...
Message-ID: <32040@siemens.siemens.com>
Date: 11 Jun 90 14:24:05 GMT
References: <30678@cup.portal.com>
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Reply-To: vlo@xydeco.siemens.com (John Vlontzos)
Organization: Siemens Corporate Research, Inc.
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