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Hi All,

Hoping someone can help me out.  I'm working with Butterworth Filters
and have some great equations for calculating the coefficients as a
function of cut-off frequency:

http://www.planetanalog.com/showArticle.jhtml?articleID=12802683&pgno=3

However I need a function which gives me the phase delay of the filter
as a function of cut-off frequencies.  For example, a second order
Butterworth Filter (Low Pass) with a cut-off frequency of 0.05 has a
lag (phase shift) of ??? radians.

Hope someone has a formula for this.  If it was genereric for N-Order
that would be best.

Regards,

Stewart

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Stewart wrote:

> Hi All,
> 
> Hoping someone can help me out.  I'm working with Butterworth Filters
> and have some great equations for calculating the coefficients as a
> function of cut-off frequency:
> 
> http://www.planetanalog.com/showArticle.jhtml?articleID=12802683&pgno=3
> 
> However I need a function which gives me the phase delay of the filter
> as a function of cut-off frequencies.  For example, a second order
> Butterworth Filter (Low Pass) with a cut-off frequency of 0.05 has a
> lag (phase shift) of ??? radians.
> 
> Hope someone has a formula for this.  If it was genereric for N-Order
> that would be best.
> 
> Regards,
> 
> Stewart

The phase performance of a digital Butterworth filter varies not only 
with the particular frequency in the passband, but also with the ratio 
of cut-off to sampling frequencies. It's not just one number.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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"Jerry Avins" <j...@ieee.org> wrote in message
news:406d8bf0$0$3057$6...@news.rcn.com...
> Stewart wrote:
>
> > Hi All,
> >
> > Hoping someone can help me out.  I'm working with Butterworth Filters
> > and have some great equations for calculating the coefficients as a
> > function of cut-off frequency:
> >
> > http://www.planetanalog.com/showArticle.jhtml?articleID=12802683&pgno=3
> >
> > However I need a function which gives me the phase delay of the filter
> > as a function of cut-off frequencies.  For example, a second order
> > Butterworth Filter (Low Pass) with a cut-off frequency of 0.05 has a
> > lag (phase shift) of ??? radians.
> >
> > Hope someone has a formula for this.  If it was genereric for N-Order
> > that would be best.
> >
> > Regards,
> >
> > Stewart
>
> The phase performance of a digital Butterworth filter varies not only
> with the particular frequency in the passband, but also with the ratio
> of cut-off to sampling frequencies. It's not just one number.

When the OP mentioned sample rate ("0.05"), since no units were given, I assumed
this was normalized to the sample rate.  If that is indeed the case, then the
order and normalized cut-off would be enough to calculate the phase shift vs.
frequency.

However, it is not a simple calculation, especially for higher orders.  In the
past, when I've needed this, I've started with the s-domain transfer function,
and just worked it out with brute-force algebra (substitute s = jw, separate
real/imaginary parts, take the arctan of imaginary/real).  I also found I had to
do some pre-warping correction for improved accuracy near the Nyquist frequency.
It was messy and ugly and what I came up with wasn't that general--only did 1st
and 2nd order filters.  However, if you have that, you can work out the higher
order filters by first decomposing them to 1st/2nd order sections (factoring)
and then simply summing the phase responses.

Probably there is a better way to do this directly in the z-domain, but I don't
know how.  That's the way Matlab does it.  If you don't have Matlab, then you'd
probably at least want to use Excel or some other tool for crunching the
numbers.

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Jon Harris wrote:

   ...

> When the OP mentioned sample rate ("0.05"), since no units were given, I assumed
> this was normalized to the sample rate.  If that is indeed the case, then the
> order and normalized cut-off would be enough to calculate the phase shift vs.
> frequency.

I assumed that too, and .05 makes agreement between the digital filter
and its analog prototype very good. The OP seemed to want a single
number, though, not a function of frequency. If that's the case, the
rest of any discussion is beside the point. Let's wait to hear from him.

   ...

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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