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

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

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

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.


1 Answer 1


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.

  • $\begingroup$ I have computed them with Matlab and checked a thousand times, but maybe I am misleading some theory, in fact! So you are telling me that I have plotted them with the imaginary axis as $\omega$ and not $j\omega$? $\endgroup$
    – Martín
    May 2, 2018 at 23:12

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