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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

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

2 Answers 2

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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: -

enter image description here

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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.

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