I have unstable transfer function H(z). One of the poles located outside of unit circle. I need to find stable transfer function G(z), which has the same Magnitude frquency response as H(z). Any ideas how can I do that? Thanks
How to find stable transfer function?
Started by ●October 29, 2004
Reply by ●October 29, 20042004-10-29
waystark@yahoo.com (Bobby) writes:> I have unstable transfer function H(z). > One of the poles located outside of unit circle. > I need to find stable transfer function G(z), > which has the same Magnitude frquency response as H(z). > Any ideas how can I do that? ThanksTwo hints: 1. Think of the geometric view of a transfer function's poles and zeros (re: Signals and Systems, Oppenheim et al.) 2. Read your textbook. -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
Reply by ●October 29, 20042004-10-29
"Bobby" <waystark@yahoo.com> wrote in message news:10b25cdd.0410290657.45afddda@posting.google.com...>I have unstable transfer function H(z). > One of the poles located outside of unit circle. > I need to find stable transfer function G(z), > which has the same Magnitude frquency response as H(z). > Any ideas how can I do that? ThanksBobby, Offhand, I don't know but I wonder what happens if you replace the poles outside the unit circle with poles that are polar reciprocals, thus inside the unit circle. I know that works for zeros so I imagine the math may work out similarly. So, if a pole is at |1.2| L 45 degrees then move it to |1/1.2| L 45 degrees, etc. (where "L" means "at an angle of" - polar notation). It's worth a try. Then you can see if the math works out or get an insight into what might work. Fred
Reply by ●November 2, 20042004-11-02
If I am not wrong, you will have an H(z) of infinity at least a one frequency, otherwise the transfer function would be stable. So, at least at these frequencies, you cannot generate a stable filter with equivalent response. Fred Marshall wrote:> "Bobby" <waystark@yahoo.com> wrote in message > news:10b25cdd.0410290657.45afddda@posting.google.com... > >>I have unstable transfer function H(z). >>One of the poles located outside of unit circle. >>I need to find stable transfer function G(z), >>which has the same Magnitude frquency response as H(z). >>Any ideas how can I do that? Thanks > > > Bobby, > > Offhand, I don't know but I wonder what happens if you replace the poles > outside the unit circle with poles that are polar reciprocals, thus inside > the unit circle. I know that works for zeros so I imagine the math may work > out similarly. > > So, if a pole is at |1.2| L 45 degrees then move it to |1/1.2| L 45 degrees, > etc. (where "L" means "at an angle of" - polar notation). > > It's worth a try. Then you can see if the math works out or get an insight > into what might work. > > Fred > >-- Please change no_spam to a.lodwig when replying via email!
Reply by ●November 2, 20042004-11-02
"Andre" <no_spam@fischer-zoth.de> wrote in message news:cm877u$o7j$00$1@news.t-online.com...> If I am not wrong, you will have an H(z) of infinity at least a one > frequency, otherwise the transfer function would be stable. > So, at least at these frequencies, you cannot generate a stable filter > with equivalent response. >Andre, What about poles inside the unit circle? A pole is "an H(z) of infinity" and would be part of a stable IIR response as long as all the poles are inside the unit circle. Fred
Reply by ●November 3, 20042004-11-03
"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:N5adnXw6PJCh5x_cRVn-sQ@centurytel.net...> > "Bobby" <waystark@yahoo.com> wrote in message > news:10b25cdd.0410290657.45afddda@posting.google.com... > >I have unstable transfer function H(z). > > One of the poles located outside of unit circle. > > I need to find stable transfer function G(z), > > which has the same Magnitude frquency response as H(z). > > Any ideas how can I do that? Thanks > > Bobby, > > Offhand, I don't know but I wonder what happens if you replace the poles > outside the unit circle with poles that are polar reciprocals, thus inside > the unit circle. I know that works for zeros so I imagine the math maywork> out similarly. > > So, if a pole is at |1.2| L 45 degrees then move it to |1/1.2| L 45degrees,> etc. (where "L" means "at an angle of" - polar notation). > > It's worth a try. Then you can see if the math works out or get aninsight> into what might work. > > Fred > >It will work but the phase does not match. Tom
Reply by ●November 3, 20042004-11-03
Number 6 wrote: ...> It will work but the phase does not match.Something needs to change. What would create stability if nothing did? There are only two important characteristics: magnitude and phase. Both change with many stabilization techniques. If one may not change, the other must. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 3, 20042004-11-03
waystark@yahoo.com (Bobby) wrote in message news:<10b25cdd.0410290657.45afddda@posting.google.com>...> I have unstable transfer function H(z). > One of the poles located outside of unit circle. > I need to find stable transfer function G(z), > which has the same Magnitude frquency response as H(z). > Any ideas how can I do that? ThanksBobby, Clearly if one of the poles lies outside of the unit circle then the transfer function will be unstable. In order to control the response, and to see how it changes with order complexity, it is a good idea to cascade the filter in biquad sections. The math for this is quite straightforward. Since I am new to this user group,I should mention that we in my group have produced a hardware and software platform that allows you to design and run many kinds of filter in real time, including IIR types. In addition to the standtard Butterworth etc, you can enter your poles and zeros and see the response instantly displayed. Please take a look at: http://www2.umist.ac.uk/dias/pag/signalwizard.htm Also, I should mention that I have just completed a book on DSP which is suitable for a wide audience, from beginner to expert; it starts with simple maths and goes on to explain quite sophisticated filters, such as adaptive etc, and also explains how to build real-time hardware. A CD accompanies the book, containing 30 windows-based programs complete with source code. The book is published by the IEE and will be available from November 12. In case you are wondering, it also explains how to break down high-order IIRs using the biquad method. Take a look at: http://www.iee.org/Publish/Books/Cds/index.cfm?book=CS%20015
Reply by ●November 3, 20042004-11-03
On 2004-11-03 17:13:08 +0100, patrick.gaydecki@manchester.ac.uk (Patrick Gaydecki) said:> ...a lot of things...All well and good, but that filter design book/program ad doesn't help him get his problem solved. Bobby, reflecting the poles back into the unit circle is the best bet I have. A pole at d becomes a pole at 1/d. As Jerry and Tom have pointed out, this is equivalent to multiplying the original transfer function with an allpass filter. If your magnitude needs to be preserved multiply by 1/d for each reflected pole - IIRC. -- Stephan M. Bernsee http://www.dspdimension.com
Reply by ●November 3, 20042004-11-03
On 3 Nov 2004 08:13:08 -0800, patrick.gaydecki@manchester.ac.uk (Patrick Gaydecki) wrote: (snipped)> >Also, I should mention that I have just completed a book on DSP which >is suitable for a wide audience, from beginner to expert; it starts >with simple maths and goes on to explain quite sophisticated filters, >such as adaptive etc, and also explains how to build real-time >hardware. A CD accompanies the book, containing 30 windows-based >programs complete with source code. The book is published by the IEE >and will be available from November 12. In case you are wondering, it >also explains how to break down high-order IIRs using the biquad >method. Take a look at: > >http://www.iee.org/Publish/Books/Cds/index.cfm?book=CS%20015Hi Patrick, Congratulations on the publication of your book! (I'll bet you're glad to have the entire publication process finished, huh?) Good Luck, [-Rick-]