# 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?

$$\big|H(j\omega)\big|^2=\frac{1}{1+\epsilon^2T^2_N\left(\frac{\omega}{\omega_c}\right)}\tag{1}$$
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 $$H(s)$$.
• @Orpheus: You get both poles in the left and in the right half-plane, and you're the one who decides which poles are assigned to $H(s)$. Apr 2 at 14:57