Significance of poles in a Transfer Function

Question

Sorry for asking this basic question, but I am new to signal processing and have this doubt for a long time.

I have been studying signal modelling and have $$H(z) = B_q(z)/A_p(z)$$ where $A_p(z)$ represents the poles of the Filter's Transfer function.
When we increase poles, $A_p(z)$ increases in terms and the signal approximation becomes more accurate to the desired signal (I've understood this while doing Pade's approximation on a discrete signal in MATLAB, I increased the poles and the approximation error reduced).

However, when they teach us the basic definitions of poles and zeroes of a Transfer function, I've been told that poles are those values where the system is unstable.

Now, how can adding poles to the filter system make its approximation more accurate, while at the same time increasing its unstable region? Is this a tradeoff that design engineers have to take note of?

Please forgive if this question sounds too much basic, it's just a doubt I have and my professor doesn't intend to explain because it's too rudimentary for his calibre.

Answer 1

However, when they teach us the basic definitions of poles and zeroes of a Transfer function, I've been told that poles are those values where the system is unstable.

That is correct, if you have poles (that are not compensated by a zero at the same value), then the system (in Laplace domain) diverges at that point. But: for example, often you model your signal as periodic, and thus composable from complex sinusoids, whose s-domain loci are all on the unit circle. Poles are not necessarily all on the unit circle, so it's not clear whether your system would inherently be unstable. There's more criteria to check.

Now, how can adding poles to the filter system make its approximation more accurate, while at the same time increasing its unstable region?

You're not necessarily actually increasing the "unstable region" (decreasing the region of convergence, ROC, if you want something to google). Anyways:

Why shouldn't it? Making a model less stable doesn't mean it gets less accurate an approximation for the system that effected any given signal. I honestly don't see the connection there. Maybe that system you're modelling has exactly these poles! Even if it doesn't, your new approximation might be better at e.g. approximating the regions where the modelled system has high gain.