# Chebyshev filter with tow thomas

I'm designing a 4th order chebyshev filter. I decided to implement it with two tow thomas filter. I've calculate the poles, but I don't know how to calculate Q and w0 for each tow thomas.

How do I calculate Q and w0 for a 4th order chebyshev filter?

Problem solved. Thanks to everybody

• Please write a specific question instead of "I need help", you'll get better answers. Commented Jul 31, 2017 at 15:48
• Please also note that Chebyshev, Tow and Thomas are proper names and should be capitalised. Commented Aug 6, 2017 at 9:15

## 2 Answers

i've calculate the poles, but i don't know how to calculate Q and w0 for each tow thomas.

If you have calculated the poles then this means you have calculated their position in terms of $\sigma$ and $j\omega$ and therefore you know the natural resonant frequency $\omega_n$ because it is the distance from the origin to each pole.

The sigma position is $-\zeta\omega_n$ where $\zeta$ = 1/2Q

This might help: -

Your transfer function is in the form of a numerator $N(s)$ divided by a denominator $D(s)$. For a second-order expression, the denominator can be written as $D(s)=1+b_1s+b_2s^2$. You can factor it under the well-known polynomial form $D(s)=1+\frac{s}{Q\omega_0}+\left(\frac{s}{\omega_{0}}\right)^2$ in which $Q=\frac{\sqrt{b_2}}{b_1}$ and $\omega_0=\frac{1}{\sqrt{b_2}}$. From the expression you have derived featuring the capacitors and resistances values, rearrange the denominator as in the above (starting with 1 + ...) and identify the $s$ factors to determine $Q$ and $\omega_0$.

Now, if you have a fourth-order denominator, you have to factor it into two second-order polynomial forms. If your denominator follows the form $D(s)=1+b_1s+b_2s^2+b_3s^3+b_4s^4$ and assuming it features two well-separated resonances, it can be approximated as the following product: $D(s)\approx \left(1+\frac{s}{Q_1\omega_{01}}+\left(\frac{s}{\omega_{01}}\right)^2\right) \left(1+\frac{s}{Q_2\omega_{02}}+\left(\frac{s}{\omega_{02}}\right)^2\right)$ in which: $\omega_{01}=\frac{1}{\sqrt{b_2}}$, $Q_1=\frac{1}{b_1\omega_{01}}$, $\omega_{02}=\frac{1}{\sqrt{b_4}\;\omega_{01}}$ and $Q_2=\frac{\omega_{02}}{\frac{b_3}{b_4}-\frac{\omega_{01}}{Q_1}}$.

I have found the following link interesting in a previous Stackexchange discussion and it describes the realization of a Tow-Thomas filter.