A Quadrature Signals Tutorial: Complex, But Not Complicated

Understanding the 'Phasing Method' of Single Sideband Demodulation

Complex Digital Signal Processing in Telecommunications

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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?

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

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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.

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<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

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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.