Hi,
Given a filter in cascaded form. So I have the coefficients b_{0,1,2} and a_{1,2} for each biquad.
Ho to check if the filter is stable? I mean I can multiply it out to get the direct form, but are there easier ways?
Thank you in advance!
Cheers Marcus
Proof stability of filter in cascaded form
Started by ●November 24, 2013
Reply by ●November 26, 20132013-11-26
On 11/24/13 3:23 AM, Marcus Jenkins wrote:> Hi, > > Given a filter in cascaded form. So I have the coefficients b_{0,1,2} and a_{1,2} for each biquad. > Ho to check if the filter is stable? I mean I can multiply it out to get the direct form, but are there easier ways? >if every biquad section is stable, the filter is stable. if any biquad section is unstable, the filter is unstable. the fact that you are given the filter in cascaded form means that 95% of the work (the factorization of the denominator) has been done for you. don't undo it by "multiplying it out". -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●November 27, 20132013-11-27
On Tuesday, November 26, 2013 3:57:59 PM UTC-5, robert bristow-johnson wrote:> On 11/24/13 3:23 AM, Marcus Jenkins wrote: > > > Hi, > > > > > > Given a filter in cascaded form. So I have the coefficients b_{0,1,2} and a_{1,2} for each biquad. > > > Ho to check if the filter is stable? I mean I can multiply it out to get the direct form, but are there easier ways? > > > > > > > if every biquad section is stable, the filter is stable. if any biquad > > section is unstable, the filter is unstable. > > > > the fact that you are given the filter in cascaded form means that 95% > > of the work (the factorization of the denominator) has been done for > > you. don't undo it by "multiplying it out". > > > > > > -- > > > > r b-j rbj@audioimagination.com > > > > "Imagination is more important than knowledge."A "mean or tricky" professor, can certainly put a zero in one biquad that cancels out a pole in another biquad. So for mathematical completeness one would factor everything possible with real coefs and cancel out common terms. Clay
Reply by ●November 27, 20132013-11-27
>> A "mean or tricky" professor, can certainly put a zero in one biquadthat cancels out a pole in another biquad. So for mathematical completeness one would factor everything possible with real coefs and cancel out common terms. No problem on paper (see CIC filter as a special case). Still, I don't think it's a good idea if I think of a "biquad" as some physical structure, with finite precision arithmetics. http://www.ee.usyd.edu.au/tutorials_online/matlab/extras/PZ.html _____________________________ Posted through www.DSPRelated.com
Reply by ●November 27, 20132013-11-27
On Wed, 27 Nov 2013 11:51:57 -0800, clay wrote:> On Tuesday, November 26, 2013 3:57:59 PM UTC-5, robert bristow-johnson > wrote: >> On 11/24/13 3:23 AM, Marcus Jenkins wrote: >> >> > Hi, >> >> >> > >> > Given a filter in cascaded form. So I have the coefficients b_{0,1,2} >> > and a_{1,2} for each biquad. >> >> > Ho to check if the filter is stable? I mean I can multiply it out to >> > get the direct form, but are there easier ways? >> >> >> > >> >> >> if every biquad section is stable, the filter is stable. if any biquad >> >> section is unstable, the filter is unstable. >> >> >> >> the fact that you are given the filter in cascaded form means that 95% >> >> of the work (the factorization of the denominator) has been done for >> >> you. don't undo it by "multiplying it out". >> >> >> >> >> >> -- >> >> >> >> r b-j rbj@audioimagination.com >> >> >> >> "Imagination is more important than knowledge." > > A "mean or tricky" professor, can certainly put a zero in one biquad > that cancels out a pole in another biquad. So for mathematical > completeness one would factor everything possible with real coefs and > cancel out common terms.Thereby intentionally missing the case where the system has an unstable pole that's masked by a zero? If I were that "mean" or "tricky" prof, and I did that, and you said that such a system was stable, you'd lose points! -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●November 30, 20132013-11-30
On Wednesday, November 27, 2013 5:33:42 PM UTC-5, Tim Wescott wrote:> On Wed, 27 Nov 2013 11:51:57 -0800, clay wrote: > > > > > On Tuesday, November 26, 2013 3:57:59 PM UTC-5, robert bristow-johnson > > > wrote: > > >> On 11/24/13 3:23 AM, Marcus Jenkins wrote: > > >> > > >> > Hi, > > >> > > >> > > >> > > > >> > Given a filter in cascaded form. So I have the coefficients b_{0,1,2} > > >> > and a_{1,2} for each biquad. > > >> > > >> > Ho to check if the filter is stable? I mean I can multiply it out to > > >> > get the direct form, but are there easier ways? > > >> > > >> > > >> > > > >> > > >> > > >> if every biquad section is stable, the filter is stable. if any biquad > > >> > > >> section is unstable, the filter is unstable. > > >> > > >> > > >> > > >> the fact that you are given the filter in cascaded form means that 95% > > >> > > >> of the work (the factorization of the denominator) has been done for > > >> > > >> you. don't undo it by "multiplying it out". > > >> > > >> > > >> > > >> > > >> > > >> -- > > >> > > >> > > >> > > >> r b-j rbj@audioimagination.com > > >> > > >> > > >> > > >> "Imagination is more important than knowledge." > > > > > > A "mean or tricky" professor, can certainly put a zero in one biquad > > > that cancels out a pole in another biquad. So for mathematical > > > completeness one would factor everything possible with real coefs and > > > cancel out common terms. > > > > Thereby intentionally missing the case where the system has an unstable > > pole that's masked by a zero? > > > > If I were that "mean" or "tricky" prof, and I did that, and you said that > > such a system was stable, you'd lose points! > > > > -- > > > > Tim Wescott > > Wescott Design Services > > http://www.wescottdesign.comIt depends on how the question is asked. I would think a failure to mention that the polynomial describing the transfer equation is reducible is worthy of losing a few points. Just because the equation is factored one way doesn't mean the physical system is actually implemented that way. Plus who said there is actually a physical system? It could just be an academic problem where only a partially factored equation is given. Clay
Reply by ●November 30, 20132013-11-30
On Fri, 29 Nov 2013 22:18:13 -0800, clay wrote:> On Wednesday, November 27, 2013 5:33:42 PM UTC-5, Tim Wescott wrote: >> On Wed, 27 Nov 2013 11:51:57 -0800, clay wrote: >> >> >> >> > On Tuesday, November 26, 2013 3:57:59 PM UTC-5, robert >> > bristow-johnson >> >> > wrote: >> >> >> On 11/24/13 3:23 AM, Marcus Jenkins wrote: >> >> >> >> >> >> > Hi, >> >> >> >> >> >> >> >> >> >> > >> >> > Given a filter in cascaded form. So I have the coefficients >> >> > b_{0,1,2} >> >> >> > and a_{1,2} for each biquad. >> >> >> >> >> >> > Ho to check if the filter is stable? I mean I can multiply it out >> >> > to >> >> >> > get the direct form, but are there easier ways? >> >> >> >> >> >> >> >> >> >> > >> >> >> >> >> >> >> >> if every biquad section is stable, the filter is stable. if any >> >> biquad >> >> >> >> >> >> section is unstable, the filter is unstable. >> >> >> >> >> >> >> >> >> >> >> >> the fact that you are given the filter in cascaded form means that >> >> 95% >> >> >> >> >> >> of the work (the factorization of the denominator) has been done for >> >> >> >> >> >> you. don't undo it by "multiplying it out". >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> -- >> >> >> >> >> >> >> >> >> >> >> >> r b-j rbj@audioimagination.com >> >> >> >> >> >> >> >> >> >> >> >> "Imagination is more important than knowledge." >> >> >> > >> > A "mean or tricky" professor, can certainly put a zero in one biquad >> >> > that cancels out a pole in another biquad. So for mathematical >> >> > completeness one would factor everything possible with real coefs and >> >> > cancel out common terms. >> >> >> >> Thereby intentionally missing the case where the system has an unstable >> >> pole that's masked by a zero? >> >> >> >> If I were that "mean" or "tricky" prof, and I did that, and you said >> that >> >> such a system was stable, you'd lose points! >> >> >> >> -- >> >> >> >> Tim Wescott >> >> Wescott Design Services >> >> http://www.wescottdesign.com > > > It depends on how the question is asked.Hmph. I suppose so.> I would think a failure to > mention that the polynomial describing the transfer equation is > reducible is worthy of losing a few points. Just because the equation is > factored one way doesn't mean the physical system is actually > implemented that way. Plus who said there is actually a physical system? > It could just be an academic problem where only a partially factored > equation is given.I certainly wouldn't accept a set of cascaded filters with an unstable pole canceled by zeros. That's not asking for trouble -- that's insisting on disaster. While I've had occasion to implement systems where a _stable_ pole is canceled by a zero, even that is something that should be approached with care, and often great care. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
Reply by ●November 30, 20132013-11-30
clay@claysturner.com writes:> [...] > It depends on how the question is asked. I would think a failure to > mention that the polynomial describing the transfer equation is > reducible is worthy of losing a few points. Just because the equation > is factored one way doesn't mean the physical system is actually > implemented that way. Plus who said there is actually a physical > system? It could just be an academic problem where only a partially > factored equation is given.I was going to say the same thing but didn't get a round tuit. If the point was to "synthesize" a transfer function, then of course the pole/zero pair could be removed. If it represents an actual physical system, well then no it can't. -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
Reply by ●November 30, 20132013-11-30
On Sat, 30 Nov 2013 20:18:32 -0500, Randy Yates <yates@digitalsignallabs.com> wrote:>clay@claysturner.com writes: >> [...] >> It depends on how the question is asked. I would think a failure to >> mention that the polynomial describing the transfer equation is >> reducible is worthy of losing a few points. Just because the equation >> is factored one way doesn't mean the physical system is actually >> implemented that way. Plus who said there is actually a physical >> system? It could just be an academic problem where only a partially >> factored equation is given. > >I was going to say the same thing but didn't get a round tuit.Here's one for ya: http://www.brooksgroup.com/blog/wp-content/uploads/2013/08/Round-tuit1.jpg Eons ago when I worked in a lumber yard they sometimes had wooden tuits like these to give out.>If the point was to "synthesize" a transfer function, then of course the >pole/zero pair could be removed. If it represents an actual physical >system, well then no it can't. >-- >Randy Yates >Digital Signal Labs >http://www.digitalsignallabs.comEric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●December 1, 20132013-12-01
eric.jacobsen@ieee.org (Eric Jacobsen) writes:> On Sat, 30 Nov 2013 20:18:32 -0500, Randy Yates > <yates@digitalsignallabs.com> wrote: > >>clay@claysturner.com writes: >>> [...] >>> It depends on how the question is asked. I would think a failure to >>> mention that the polynomial describing the transfer equation is >>> reducible is worthy of losing a few points. Just because the equation >>> is factored one way doesn't mean the physical system is actually >>> implemented that way. Plus who said there is actually a physical >>> system? It could just be an academic problem where only a partially >>> factored equation is given. >> >>I was going to say the same thing but didn't get a round tuit. > > Here's one for ya: > > http://www.brooksgroup.com/blog/wp-content/uploads/2013/08/Round-tuit1.jpg > > Eons ago when I worked in a lumber yard they sometimes had wooden > tuits like these to give out.Can you get me a few dozen Eric? I seem to need them frequently. --Randy> >>If the point was to "synthesize" a transfer function, then of course the >>pole/zero pair could be removed. If it represents an actual physical >>system, well then no it can't. >>-- >>Randy Yates >>Digital Signal Labs >>http://www.digitalsignallabs.com > > Eric Jacobsen > Anchor Hill Communications > http://www.anchorhill.com-- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
