Given a signal $x[n] = [1, 2, 3, 4, 5]$. How many transfer functions can be found with Padé approximation which have a causal impulse response and start with those five samples. How many of them are stable?

Found this question and I'm not sure if I understood it right. I'd say none of them is stable since Padé can't guarantee stable solutions. The first five samples will be perfectly recreated in the impulse response. Therefor $h(z)$ between $0-4$ looks like $x[n]$. But what about how many? Are there multiple solutions? Is this question badly formulated?

A further question is then: Find the Padé model with exaclty one pole. One pole? Isn't that an all pole model? Or is there some kind of rule like always the same amount of poles and zeros? Can't solve it with the Padé euqations.

  • $\begingroup$ what's p and q ? their sum p+q+1 = 5 , but what's the particular pair ? Then given the order p, you find the coefficeints a[k] (and b[k]) and look for possible transfer functions... $\endgroup$ – Fat32 Feb 6 '19 at 19:53
  • $\begingroup$ p and q are not given in the first question. I assumed its known from the fact that 5 samples need to be perfectly correct. For the second question: So I know q from 5-1-p? That would be 3 zeros and 1 pole? Doesn't this extra lags? I thought p and q should only differ one to be stable and causal. $\endgroup$ – Mr.Sh4nnon Feb 6 '19 at 19:57

The following might help. Given the data of $5$ samples,

$$ x[n] = [1, 2, 3, 4, 5]$$

You will have possible choices for the orders $(p,q)$ of $a[k]$ and $b[k]$, such as $\{(4,0),(3,1),(2,2),(1,3),(0,4)\}$.

You can solve or each of the possibilities and check whether they yield stable systems or not, by looking at the roots of $a[k]$... afaik they should all be causal as this is Pade modeling assumption (unless equations modified to handle otherwise).

Note that for certain data sets (including this one) some tail coefficients might turn to be $0$ and the actual order can be less than that of indicated by $p$ or $q$.

Also note that a model with one pole is not an all-pole model. All-pole model requires that $q=0$, and does not depend on $p$.

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    $\begingroup$ In this context of exact mathcing for modeling, (4,0) is an all-pole Padé model with 4 poles and no zeros. $\endgroup$ – Fat32 Feb 6 '19 at 20:27
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    $\begingroup$ All-pole means that there are no zeros. Since for Padé modeling $p+q+1 = N$ (N data number) then all pole means $q=0$ and $p = N-1$... $\endgroup$ – Fat32 Feb 6 '19 at 20:28
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    $\begingroup$ it's 1 pole (p =1) and then N-p-1 = q = 5 - 1 - 1= 3 zeros... $\endgroup$ – Fat32 Feb 6 '19 at 20:37
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    $\begingroup$ No probably you misunderstood. Let me show an example. $$H_1(z) = \frac{ 1 + 0.4 z^{-1} + 0.79 z^{-2} + 0.26 z^{-3} }{ 1 - 0.79z^{-1} } $$ is a causal system as the numerator powers are all less than zero, however the following $$H_2(z) = \frac{ 1 + 0.4 z^{1} + 0.79 z^{2} + 0.26 z^{-3} }{ 1 - 0.79z^{-1} } $$ is non-causal as numerator has positive power of $z$. $\endgroup$ – Fat32 Feb 6 '19 at 22:01
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    $\begingroup$ Ou^^ That easy. Obviously you can see it on the highest "z" power. Knew that negativ powers are causal, but thought it might depends on the number of poles and zeros too. Silly me. $\endgroup$ – Mr.Sh4nnon Feb 6 '19 at 22:09

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