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Discussion Groups | Comp.DSP | ARMA models

There are 5 messages in this thread.

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ARMA models - 2012-10-29 00:04:00

Suppose you have a system B/A and you
estimate using an AR method g/A2 where g is a constant, the rest ie B,A,A2 are polynomials in z^-1. Here I am assuming that you cannot get A directly using an AR method since it is estimating an approximation of the zeros in the poles.So instead we call it A2.

Then you estimate B/A = C using a MA method.(where C is another polynomial).

Can you combine C,A2 and g in any way to get the "true" model B/A?


Re: ARMA models - Tim Wescott - 2012-10-29 12:46:00

On Sun, 28 Oct 2012 21:04:22 -0700, gyansorova wrote:

> Suppose you have a system B/A and you estimate using an AR method g/A2
> where g is a constant, the rest ie B,A,A2 are polynomials in z^-1. Here
> I am assuming that you cannot get A directly using an AR method since it
> is estimating an approximation of the zeros in the poles.So instead we
> call it A2.
> 
> Then you estimate B/A = C using a MA method.(where C is another
> polynomial).
> 
> Can you combine C,A2 and g in any way to get the "true" model B/A?

Good question.  I dunno.  There's probably enough information in there.

But why not just use ARMA to estimate B/A?

-- 
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com


Re: ARMA models - 2012-10-29 13:57:00

On Tuesday, October 30, 2012 5:46:59 AM UTC+13, Tim Wescott wrote:
> On Sun, 28 Oct 2012 21:04:22 -0700, gyansorova wrote:
> 
> 
> 
> > Suppose you have a system B/A and you estimate using an AR method g/A2
> 
> > where g is a constant, the rest ie B,A,A2 are polynomials in z^-1. Here
> 
> > I am assuming that you cannot get A directly using an AR method since it
> 
> > is estimating an approximation of the zeros in the poles.So instead we
> 
> > call it A2.
> 
> > 
> 
> > Then you estimate B/A = C using a MA method.(where C is another
> 
> > polynomial).
> 
> > 
> 
> > Can you combine C,A2 and g in any way to get the "true" model B/A?
> 
> 
> 
> Good question.  I dunno.  There's probably enough information in there.
> 
> 
> 
> But why not just use ARMA to estimate B/A?
> 
> 
> 
> -- 
> 
> My liberal friends think I'm a conservative kook.
> 
> My conservative friends think I'm a liberal kook.
> 
> Why am I not happy that they have found common ground?
> 
> 
> 
> Tim Wescott, Communications, Control, Circuits & Software
> 
> http://www.wescottdesign.com

I could tell you but I would have to shoot you afterwards.


Re: ARMA models - Vladimir Vassilevsky - 2012-10-29 15:05:00

<g...@gmail.com> wrote:
>Suppose you have a system B/A and you
>estimate using an AR method g/A2 where g is a constant, the rest ie B,A,A2 
>are polynomials in z^-1. Here I am assuming that you >cannot get A directly 
>using an AR method since it is estimating an approximation of the zeros in 
>the poles.So instead we call it A2.
>Then you estimate B/A = C using a MA method.(where C is another 
>polynomial).
>Can you combine C,A2 and g in any way to get the "true" model B/A?

No.

It is not possible to recreate entire data from two solution vectors. 
Therefore, you can compute _a_ solution in the form of A/B, but this 
solution is generally not going to be the same as true ARMA solution; unless 
for special cases.

Vladimir Vassilevsky
DSP and Mixed Signal Consultant
www.abvolt.com


Re: ARMA models - 2012-10-29 20:36:00

On Monday, October 29, 2012 5:04:22 PM UTC+13, gyans...@gmail.com wrote:
> Suppose you have a system B/A and you
> 
> estimate using an AR method g/A2 where g is a constant, the rest ie B,A,A2 are polynomials in z^-1. Here I am assuming that you cannot get A directly using an AR method since it is estimating an approximation of the zeros in the poles.So instead we call it A2.
> 
> 
> 
> Then you estimate B/A = C using a MA method.(where C is another polynomial).
> 
> 
> 
> Can you combine C,A2 and g in any way to get the "true" model B/A?

I beginning to believe you.