Chebyshev Filter Transfer Function I'm trying to derive the transfer function for Chebyshev filter.
$$|H(\Omega)|^2=\frac{1}{\sqrt{(1+\epsilon^2T_n^2(\frac{\Omega}{\Omega_c})}}$$
where   $$T_n(x)=\cos(N\cos^{-1}(x)) \forall x \le 1$$
$$T_n(x)=\cosh(N\cosh^{-1}(x)) \forall x \ge 1$$
$$H(s)=\frac{1}{\sqrt{(1+\epsilon^2T_n^2(\frac{s}{j\Omega_c}))}}$$
To calculate the poles I made the following substituion which looks like this:
$$\cos^{-1}(\frac{s}{j\Omega_c})=\alpha + j \beta$$
So
$$s=j\Omega_c(\cos\alpha \cosh\beta - j \sin\alpha \sinh\beta)$$
$$s=\Omega_c \sin\alpha \sinh\beta + j \Omega_c \cos\alpha \cosh\beta$$
where $\alpha=\frac{(2k-1)\pi}{2N}$ and $\beta = \frac{1}{N} \sinh^{-1}{\frac{1}{\epsilon}}$
But with this substitution my poles yield to be stability or produce an ambiguity on the stability end.
But if I make the substitution as $$\cos^{-1}(\frac{s}{j\Omega_c})=\alpha - j \beta$$
poles seem stable.
Can someone help me with this. is my substitution wrong?
Note that the squared magnitude of the frequency response is given by
$$\big|H(j\omega)\big|^2=\frac{1}{1+\epsilon^2T^2_N\left(\frac{\omega}{\omega_c}\right)}\tag{1}$$
In the $s$-domain we have
$$H(s)H(-s)=\frac{1}{1+\epsilon^2T^2_N\left(\frac{s}{j\omega_c}\right)}\tag{2}$$
Computing the zeros of $(2)$ does not only result in the zeros of the filter's transfer function $H(s)$, but also in the zeros of $H(-s)$. The zeros of $(2)$ lie on an ellipse, and for each zero $s_0$ in the left half-plane, there's also a zero $-s_0$ in the right half-plane. Since we want a stable filter, we assign all zeros in the left half-plane to $H(s)$.