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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?

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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 $H(s)$.

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  • $\begingroup$ 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 Apr 2 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. Apr 2 at 14:57

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