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Re: How to perform the 1-D signal filter? [message #58429 is a reply to message #58425] Fri, 01 February 2008 08:57 Go to previous messageGo to previous message
Wox is currently offline  Wox
Messages: 184
Registered: August 2006
Senior Member
On Fri, 1 Feb 2008 06:54:42 -0800 (PST), "duxiyu@gmail.com"
<duxiyu@gmail.com> wrote:


> In the example, you set A=50 and Nterms=40 and I do not know where
> their values come from and how I get them.

The Nterms=40 determines the width of the filter: width=2.Nterms+1

The A=50 has something to do with the Gibbs phenomena, i.e. the
wiggles you get when you Fourier expand a function with a finite
number of terms (as opposed to infinite). The wiggles are visible at
discontinuities.

I don't know anything about the Kaiser-Bessel filter (I think that's
the one used here). So maybe someone else can explain this.

> Seondly, I want to know the relation between the filter in time and
> frequency domain.
> if the following command is excuted, I think newfsignal is exactly
> equal to fft(newsignal), isn't it?
> "newsignal=convol(signal,timefilter)
> fsignal=fft(signal)
> newfsignal=fsignal*freqfilter"

You are absolutly right. That's why in the code comments I wrote: ";
Frequency domain filter (instead of time domain filter)". This means
you can choose in which domain you are filtering, the time or
frequency(fourier) domain. You could even have a filter in the time
domain, take the fourier transform and use that to apply the filter in
the frequency domain. Keep in mind however that every fft introduces
some errors.

> And I also do not konw why you set steep=20.

If you plot the lowpass filter (in the fourier domain):

steep=20.
freqfilter= 1./(1.+(freq/f_high)^steep)
plot,freq,freqfilter

Do this for different "steep" and you will see that the edge becomes
sharper with higher "steep".
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