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Re: Butterworth Band-Pass Filter [message #73913 is a reply to message #73912]

Sun, 12 December 2010 09:02

David Fanning
Messages: 11724
Registered: August 2001

Senior Member

burton449 writes:

> According to your documentation, a high-pass butterworth filter will
> be defined like that:
> filter = 1 / [1 + C(Ro/R)^2n]
>
> So in IDL, for a cuttoff of 15, and freqImage = FFT(image, -1), the
> filter will be defined like that:
> filter = 1.0 / ( 1.0d + (15.0/freqImage)^2 )
>
> Is it right?

For what it's worth, I got confused by all this when
I was reading this article, too, in preparation for
including this information in my new book. I think I
explained it better in the book, and I'll probably go
back and fix this article, too. But not now. I can see
the end of this book from where I am standing and I am
at the point where bathing is completely extraneous to
my purpose! :-)

Cheers,

David

--
David Fanning, Ph.D.
Fanning Software Consulting, Inc.
Coyote's Guide to IDL Programming: http://www.dfanning.com/
Sepore ma de ni thui. ("Perhaps thou speakest truth.")

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[Message index]

		Re: Butterworth Band-Pass Filter By: David Fanning on Sun, 12 December 2010 09:24
		Re: Butterworth Band-Pass Filter By: David Fanning on Sun, 12 December 2010 09:19
		Re: Butterworth Band-Pass Filter By: burton449 on Sun, 12 December 2010 09:07
		Re: Butterworth Band-Pass Filter By: David Fanning on Sun, 12 December 2010 09:02
		Re: Butterworth Band-Pass Filter By: David Fanning on Sun, 12 December 2010 08:56
		Re: Butterworth Band-Pass Filter By: burton449 on Sun, 12 December 2010 08:29
		Re: Butterworth Band-Pass Filter By: David Fanning on Sat, 11 December 2010 20:37
		Re: Butterworth Band-Pass Filter By: burton449 on Sun, 12 December 2010 09:28
		Re: Butterworth Band-Pass Filter By: burton449 on Sun, 12 December 2010 09:30

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