I'm trying to derive the transfer function for Chebyshev filter.


where $$T_n(x)=\cos(N\cos^{-1}(x)) \forall x \le 1$$

$$T_n(x)=\cosh(N\cosh^{-1}(x)) \forall x \ge 1$$


To calculate the poles I made the following substituion which looks like this:

$$\cos^{-1}(\frac{s}{j\Omega_c})=\alpha + j \beta$$


$$s=j\Omega_c(\cos\alpha \cosh\beta - j \sin\alpha \sinh\beta)$$ $$s=\Omega_c \sin\alpha \sinh\beta + j \Omega_c \cos\alpha \cosh\beta$$

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?


1 Answer 1


Note that the squared magnitude of the frequency response is given by


In the $s$-domain we have


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 $H(s)$.

  • $\begingroup$ I understand that @Matt L. , but in the derivation of the transfer function of the chebyshcev filter...how do I ensure that the poles are stable...? $\endgroup$
    – Orpheus
    Commented Apr 2, 2021 at 14:45
  • $\begingroup$ @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 $H(s)$. $\endgroup$
    – Matt L.
    Commented Apr 2, 2021 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.