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When I've a transfer function that looks like:

$$\text{H}\left(\text{s}\right)=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{\text{K}}{\left(\frac{\text{s}}{\omega_0}\right)^2+2\cdot\beta\cdot\frac{\text{s}}{\omega_0}+1}\tag1$$

The final value theorem says:

$$\lim_{t\to\infty}\text{f}\left(t\right)=\lim_{\text{s}\to0}\text{s}\cdot\text{F}\left(\text{s}\right)\tag2$$

If all poles of $\text{s}\cdot\text{F}\left(\text{s}\right)$ are in the left half-plane.

Now, when:

$$\text{x}\left(t\right)=\delta\left(t\right)\space\space\space\to\space\space\space\text{X}\left(\text{s}\right)=1\tag3$$

We have:

$$\text{H}\left(\text{s}\right)=\frac{\text{Y}\left(\text{s}\right)}{1}=\text{Y}\left(\text{s}\right)=\frac{\text{K}}{\left(\frac{\text{s}}{\omega_0}\right)^2+2\cdot\beta\cdot\frac{\text{s}}{\omega_0}+1}\tag4$$

Question: Can we say that, in order to use the final value theorem, that the poles of $\text{s}\cdot\text{H}\left(\text{s}\right)$ have to lie in the left half of the plane (for $\Re\left(\text{s}\right)<0$)?

And the poles of $\text{s}\cdot\text{H}\left(\text{s}\right)$ are given by:

$$\left(\frac{\text{s}}{\omega_0}\right)^2+2\cdot\beta\cdot\frac{\text{s}}{\omega_0}+1=0\space\Longleftrightarrow\space\color{red}{\text{s}=\pm\sqrt{\omega_0^2\cdot\left(\beta-1\right)}-\beta\cdot\omega_0}\tag5$$

And so we must have:

$$\Re\left(\text{s}\right)=\Re\left(\pm\sqrt{\omega_0^2\cdot\left(\beta-1\right)}-\beta\cdot\omega_0\right)<0\tag6$$

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Yes, this is correct. If your poles had positive real part, the systems output $y(t)$ would grow unboundedly with time.

Note that the respective roots should be $s=[-\beta\pm \sqrt{\beta^\mathbf{2}-1}]\omega_0$. Thus you will have all left poles if $\beta>0$ and $\omega_0>0$.

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