Poles and zeros in time domain of analogs filters I am currently studying two Butterworth and Chebyshev low-pass filters of order $n =3$ and $n=2$ respectively, whcih are in fact two prototypes to make a bandpass filter. The transfer function that I have is the following one:
$$H(j\omega) = 2\sqrt{\frac{R_G}{R_L}}\frac{1}{R_G(C+\frac{D}{R_L}) + A + \frac{B}{R_L}}$$
where $R_G$ is the impedance in series with the generator and $R_L$ is the load impedance

For the Butterworth filter the data is the following:
$$\begin{align}R_G&amp;=R_L=50\;\Omega\\
A &amp;=1 -0.708\omega^2\\
B &amp;=j59\omega \\
C &amp;=j(0.024\omega -0.008496\omega^3)\\
D &amp;= 1-0.708\omega^2\end{align}$$

and for the Chebyshev filter the data is the following:
$$\begin{align}R_G&amp;=50\Omega\quad R_L=18.8\;\Omega\\
A &amp;=1\\
B &amp;=j34.25\omega \\
C &amp;=j0.036\omega\\
D &amp;=1-1.233\omega^2\end{align}$$
and this is the poles diagram computed with the data above of such circuits:


where:
$$\begin{align}
\omega_{pB_1} &amp;= j1.66\\
\omega_{pB_2} &amp;= -1.45+j0.83\\
\omega_{pB_3} &amp;= 1.45+j0.83\\
\omega_{pC_1} &amp;= -0.9 + j0.55\\
\omega_{pC_2} &amp;= 0.9 + j0.55
\end{align}$$

It is clear that both systems are going to be unstable since they have poles in the right-half plane, but how will these poles look in the time domain? I have read articles like this one or this one and some questions on this website, and I know that if they were complex conjugated poles the response in time domain would be a sinusoid type signal.

I am asking this because the majority of the examples have two conjugated poles, which in this case they are not, and hence the time domain expressions won't be sinusoids, or I am wrong? Could someone answer with the inverse Laplace transform in Matlab of my transfer function?

Note that I am not asking for my job to be done, I have already made it, but some more information about those poles will be appreciated.

Thank you.
I think you calculated the poles incorrectly, namely for $\omega$ instead of the Laplace variable $s=j\,\omega$. You can find the appropriate poles in terms of $s$ by multiplying your poles by $j$. This will give you conjugate poles, with negative real parts.