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my problem is perhaps very simple but I just can not find the answer, even though this is used (though not explained) in several books:

What is $\frac{\partial G (z, \theta)}{\partial \theta}$ when $G$ is some transfer function, and $\theta$ the coefficients? (at least I call them coefficients). I tried using "regular" calculus on the following, but the result is very strange.

Example: $$G(z, \theta) = \frac{b_1 z^{-1} + b_2 z^{-2}}{1 + a_1 z^{-1} + a_2 z^{-2} + a_3 z^{-3}} = \frac{B}{A}$$

$$ \theta = [b_1, b_2, a_1, a_2, a_3] $$ Using the chain rule, and treating this as a regular rational function in $z$, I get: $$ \frac{\partial G (z, \theta)}{\partial \theta} = [\frac{\partial G (z, \theta)}{\partial b_1}, ..., \frac{\partial G (z, \theta)}{\partial a_3}] \\[19pt] = [\frac{z^{-1}}{A}, \frac{z^{-2}}{A}, (-1)z^{-1}\frac{ B}{A^2}, (-1)z^{-2}\frac{ B}{A^2 }, (-1)z^{-3}\frac{ B}{A^2}] \\[19pt] $$

I agree, this looks weird.

EDIT: What I am looking for is the gradient, $\nabla_\theta G$.

Perheps there are some rules of differentiation in this domain that I do not know. The problem is related to parameter estimation.

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  • $\begingroup$ Well, the question here is "what kind of derivative do you need?" We've got different ones to pick from: What you describe is closest to what's called total derivative and noted $\mathrm d G$, but it's possible you actually need the gradient, $\nabla_{\hspace{-0.25ex}\theta} G$. You say "related to parameter estimation", but that's not really clarifying what you're trying to do with that derivative. $\endgroup$ Feb 5 at 12:41
  • $\begingroup$ Ok, thanks for feedback. The last comment was just for some context, since this might be unusual (?). It is indeed the gradient that I need, I will edit the post. $\endgroup$
    – haavbj
    Feb 5 at 12:49
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    $\begingroup$ well, then what you're doing looks right! $\endgroup$ Feb 5 at 12:51
  • $\begingroup$ This is right. Take a look at this answer where I've written down those derivatives. $\endgroup$
    – Matt L.
    Feb 5 at 18:14

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