Well yes, but I thought one can convert one to the other, say
by H(highpass) = 1 - H(lowpass). I also thought maybe that by
substituting 1/s in the transfer function one switches from lowpass
to highpass. Either way, I can't reconcile them.
Thanks, I've bookmarked it and I'm studying it now.
-Alex
Reply by Tim Wescott●August 12, 20042004-08-12
axlq wrote:
> [posted and emailed]
>
> Peter, I apologize for reviving this part of a past discussion, but
> there's a point I'm not clear on, and it has mostly to do with my
> ignorance about where to start from when constructing a filter.
>
> In article <XvydnQv1cPdCKZDcRVn-qg@comcast.com>,
> Peter Nachtwey <peter@deltacompsys.com> wrote:
>
>
>>ftp://ftp.deltacompsys.com/public/mcd/Bessel%20High%20Pass.htm
>>A high pass Bessel filter. It doesn't look much different from
>>the Butterworth filter. Beware, the corner frequencies are really
>>higher than the value used in the calculations. The filters aren't
>>hard to make if you have Mathcad.
>
>
> Your normalized Bessel function (is this what's also called the
> "transfer function"?), about halfway down your page, is
> s^2 / (s^2 + 3s + 3).
>
> Here's a page full of Bessel filter constants:
> http://www.crbond.com/papers/bsf.pdf
>
> ...which shows (equation 6) the transfer function as
> 3 / (s^2 + 3s + 3).
>
> How are the two reconciled? Your page describes clearly the steps
> in constructing a filter (assuming I have something like MathCad
> available), but it seems to start from a different filter function
> than what I can find published elsewhere for Bessel filters.
>
> Thanks.
> -Alex
>
> [posted and emailed - not sure if you're still following comp.dsp]
[posted and emailed]
Peter, I apologize for reviving this part of a past discussion, but
there's a point I'm not clear on, and it has mostly to do with my
ignorance about where to start from when constructing a filter.
In article <XvydnQv1cPdCKZDcRVn-qg@comcast.com>,
Peter Nachtwey <peter@deltacompsys.com> wrote:
>ftp://ftp.deltacompsys.com/public/mcd/Bessel%20High%20Pass.htm
>A high pass Bessel filter. It doesn't look much different from
>the Butterworth filter. Beware, the corner frequencies are really
>higher than the value used in the calculations. The filters aren't
>hard to make if you have Mathcad.
Your normalized Bessel function (is this what's also called the
"transfer function"?), about halfway down your page, is
s^2 / (s^2 + 3s + 3).
Here's a page full of Bessel filter constants:
http://www.crbond.com/papers/bsf.pdf
...which shows (equation 6) the transfer function as
3 / (s^2 + 3s + 3).
How are the two reconciled? Your page describes clearly the steps
in constructing a filter (assuming I have something like MathCad
available), but it seems to start from a different filter function
than what I can find published elsewhere for Bessel filters.
Thanks.
-Alex
[posted and emailed - not sure if you're still following comp.dsp]
Reply by glen herrmannsfeldt●August 8, 20042004-08-08
> The merit of this receiving instrument is, that
> it indicates with extreme sensibility all the variations of the current
> in the cable, so that, instead of having to wait until each signal wave
> sent into the cable has travelled to the receiving end before sending
> another, a series of waves may be sent after each other in rapid
> succession. These waves, encroaching upon each other, will coalesce at
> their bases; but if the crests remain separate, the delicate decipherer
> at the other end will take cognisance of them and make them known to the
> eye as the distinct signals of the message.
> Mirror galvanometer
(snip)
And note that Nyquist was not working one sampling of analog
signals, but on how fast pulses could be sent down a cable
with a limited bandwidth. Different problems with the same
solution.
-- glen
Reply by Jerry Avins●August 7, 20042004-08-07
axlq wrote:
> In article <41144572$0$5908$61fed72c@news.rcn.com>,
> Jerry Avins <jya@ieee.org> wrote:
>
>>But as Alex correctly pointed out, the response of a 1st-order analog
>>high pass to a step is 1 - exp(-t/T), which has monotonic decay without
>>undershoot. That single counterexample disproves my argument.
>
>
> Not really -- your argument holds for a filter order higher than
> 1. My empirical tests indicate that the number of zero-crossings of
> a critically-damped high-pass filter is proportional to the order,
> as n-1 crossings for order n. If there's no way to "fix" that
> situation, then I have to live with it.
>
> -Alex
The conclusion is probably true, but the "proof" itself is invalid.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by axlq●August 7, 20042004-08-07
In article <41144572$0$5908$61fed72c@news.rcn.com>,
Jerry Avins <jya@ieee.org> wrote:
>But as Alex correctly pointed out, the response of a 1st-order analog
>high pass to a step is 1 - exp(-t/T), which has monotonic decay without
>undershoot. That single counterexample disproves my argument.
Not really -- your argument holds for a filter order higher than
1. My empirical tests indicate that the number of zero-crossings of
a critically-damped high-pass filter is proportional to the order,
as n-1 crossings for order n. If there's no way to "fix" that
situation, then I have to live with it.
-Alex
Reply by Jerry Avins●August 6, 20042004-08-06
Jon Harris wrote:
> "axlq" <axlq@spamcop.net> wrote in message news:cf0in5$iif$1@blue.rahul.net...
...
>>My expectation of a HP critically damped step response is that the
>>filter output immediately jumps, then approaches zero as rapidly as
>>possible without dipping below.
>
>
> As far as I know, this isn't possible. Jerry outlined a proof in another post.
But as Alex correctly pointed out, the response of a 1st-order analog
high pass to a step is 1 - exp(-t/T), which has monotonic decay without
undershoot. That single counterexample disproves my argument.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by axlq●August 6, 20042004-08-06
In article <2nhtgcF14t3fU1@uni-berlin.de>,
Jon Harris <goldentully@hotmail.com> wrote:
>> > On my page http://unicorn.us.com/alex/buttercrit.html is there
>> > a way to get Q from the information there? Or do I need a
>> > different approach?
>
>I took another look, Q is related to your damping d. So you should
>just be able to pick any value of d you like to get a smoother time
>response (and the corresponding less sharp frequency response).
I tried that. It shifts the cutoff frequency, and when I adjust the
cutoff to shift it back, I end up with something that looks like a
2nd-order filter again, at least in the time domain -- I haven't
checked the frequency response yet.
>> I haven't looked at your page, but try
>> http://www.harmony-central.com/Computer/Programming/Audio-EQ-Cook
>> book.txt. There are simple equations for the biquad coefs given
>> cutoff frequency and Q. Pick any value of Q you like to obtain
>> the desired time-domain response.
Thanks! I think you or someone else posted that link before in this
thread, and I have it bookmarked.
-Alex
Reply by axlq●August 6, 20042004-08-06
In article <xxpsmb0116p.fsf@usrts005.corpusers.net>,
Randy Yates <randy.yates@sonyericsson.com> wrote:
>> H(s) = S^2 / (S^2 + 3S + 3) where S = s/omega
>>
>> Opening my Schaum's Mathematical Handbook and turning to the chapter
>> on Bessel functions, I can find nothing that looks like the function
>> above.
>
>All I can say is, don't feel bad. The relationship between the two seems
>to come from something called "umbral calculus." This document seems
>to describe it well (not that I've read the entire thing)
>
> http://www.combinatorics.org/Surveys/ds3.pdf
Ugh. I thought there was a simple answer. Thanks for the info.
-Alex
Reply by axlq●August 6, 20042004-08-06
In article <4113d391$0$5922$61fed72c@news.rcn.com>,
Jerry Avins <jya@ieee.org> wrote:
> Proof that All High-Pass Filters Ring when Presented with Step Inputs
> ���������������������������������������������������������������������
>1) DC can't get through a high-pass filter, so the integral of its
> response to any signal must be zero.
>2) The rising edge of a step input is all high frequency. It passes
> the filter unchanged in sign and magnitude (scaling excepted).
>3) In order that the integral of a curve be zero, it must have as much
> negative area as positive. Therefore, a high-pass-filiered positive
> step must have both positive and negative regions. That's ringing.
>
>Does the argument break down?
Only in the case of a 1st order filter, which doesn't ring at all.
But based on this, and other messages you and others have posted, I'm
convinced that what I seek is a futile search, unless there's some trick
to combine filters to cancel out the ringing. I could do this with
low-pass filters, but never managed with a high pass filter.
-Alex