Thank you to everyone who responded to my question. Now I clearly
understand how to solve this problem.

> 
> Hi 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-]

Hi Rick,

Thanks very much for that. Yes indeed, I am very glad the whole
process is over. Because the explanations are for potential readers
and not for the author, small details become important and assumptions
cannot be made. In the end, I often spent days or even weeks trying to
figure out why certian things are true, having in the past just
accepted them. This is a problem I am sure you have faced yourself
with your publications.

Patrick

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%20015

Hi 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-]
 

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

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? Thanks

Bobby,

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

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.
�����������������������������������������������������������������������

"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 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
>
>
It will work but the phase does not match.

Tom

"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 

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!

"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