BIBO stability

On Sat, 09 Jun 2007 20:59:19 -0400, Randy Yates wrote:

> p.kootsookos@remove.ieee.org (Peter K.) writes:
> 
>> Tim Wescott <tim@seemywebsite.com> writes:
>>
>>> Why do you say that the ideal low-pass filters are not BIBO stable?
>>> The impulse response you give has a finite amount of energy in it, and
>>> it goes to zero over time -- that says "BIBO stable" to me.
>>
>> For BIBO stability, the impulse response has to be absolutely summable
>> (http://en.wikipedia.org/wiki/BIBO_stability).
>>
>> The sinc function is not absolutely summable (or integrable in
>> continuous time http://en.wikipedia.org/wiki/Sinc_function), so the
>> ideal low pass filter is not BIBO stable. :-)
> 
> Ayup. Here's a copy of the post (from Google Groups) on a proof I made a few years back:
> 
-- snip --

Well, I'll be dipped.

For what evil (yet bounded) signals does a filter with this impulse
response respond with infinite strength?  My intuition tells me that it
would be sin(wt).  But then, my intuition told me earlier today that the
system was stable, so I'm not sure it's trustworthy.

-- 
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

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"Applied Control Theory for Embedded Systems" by Tim Wescott
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Tim Wescott <tim@seemywebsite.com> writes:

> On Sat, 09 Jun 2007 20:59:19 -0400, Randy Yates wrote:
>
>> p.kootsookos@remove.ieee.org (Peter K.) writes:
>> 
>>> Tim Wescott <tim@seemywebsite.com> writes:
>>>
>>>> Why do you say that the ideal low-pass filters are not BIBO stable?
>>>> The impulse response you give has a finite amount of energy in it, and
>>>> it goes to zero over time -- that says "BIBO stable" to me.
>>>
>>> For BIBO stability, the impulse response has to be absolutely summable
>>> (http://en.wikipedia.org/wiki/BIBO_stability).
>>>
>>> The sinc function is not absolutely summable (or integrable in
>>> continuous time http://en.wikipedia.org/wiki/Sinc_function), so the
>>> ideal low pass filter is not BIBO stable. :-)
>> 
>> Ayup. Here's a copy of the post (from Google Groups) on a proof I made a few years back:
>> 
> -- snip --
>
> Well, I'll be dipped.
>
> For what evil (yet bounded) signals does a filter with this impulse
> response respond with infinite strength?  

It was embedded in the proof:

  Now assume x[L*m] = sgn(h[1-L*m]).

> My intuition tells me that it would be sin(wt).  But then, my
> intuition told me earlier today that the system was stable, so I'm
> not sure it's trustworthy.

When I first realized this I had a hard time swallowing it, too. My
encounter came about when I was trying to predict what the bound of
the output of interpolating filter in my delta-sigma DAC would be.
-- 
%  Randy Yates                  % "Though you ride on the wheels of tomorrow,
%% Fuquay-Varina, NC            %  you still wander the fields of your
%%% 919-577-9882                %  sorrow."
%%%% <yates@ieee.org>           % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr

On Jun 9, 9:22 pm, Tim Wescott <t...@seemywebsite.com> wrote:
> On Sat, 09 Jun 2007 20:59:19 -0400, Randy Yates wrote:
> > p.kootsoo...@remove.ieee.org (Peter K.) writes:
>
> >> Tim Wescott <t...@seemywebsite.com> writes:
>
> >>> Why do you say that the ideal low-pass filters are not BIBO stable?
> >>> The impulse response you give has a finite amount of energy in it, and
> >>> it goes to zero over time -- that says "BIBO stable" to me.
>
> >> For BIBO stability, the impulse response has to be absolutely summable
> >> (http://en.wikipedia.org/wiki/BIBO_stability).
>
> >> The sinc function is not absolutely summable (or integrable in
> >> continuous timehttp://en.wikipedia.org/wiki/Sinc_function), so the
> >> ideal low pass filter is not BIBO stable. :-)
>
> > Ayup. Here's a copy of the post (from Google Groups) on a proof I made a few years back:
>
> -- snip --
>
> Well, I'll be dipped.
>
> For what evil (yet bounded) signals does a filter with this impulse
> response respond with infinite strength?  My intuition tells me that it
> would be sin(wt).  But then, my intuition told me earlier today that the
> system was stable, so I'm not sure it's trustworthy.

A sinc has bounded energy because, even though the positive
peaks only decrease roughly harmonically, the negative lobes
cancel out most of the previous positive lobe, and the difference
between the positive and negative lobes decreases faster than
harmonically (more like an alternating harmonic series, which
does converge).

However, say you have a filter input which is positive for
the positive lobes of a sinc filter function, but zero or
negative for all the negative lobes of a sinc.  Then it looks
like the filter output might not converge for an arbitrarily
long filter and filter input.

I think (sin(t-t0) * sgn(t-t0)) or similar input would do it.


IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M