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I am trying to calculate the cut off frequency (Wo) and quality factor (Q) from the transfer function I derived for a Chebyshev filter. I need these two values to further calculate the component value for the low pass filter. I calculated the transfer function of the Chebyshev filter as per this formula:

Tranfer function of Chebyshev filter I ended up with a second transfer function at the denominator like this:

enter image description here

I want to cascade the second order and first order filters using the above functions and make a circuit. For that I need to know how to calculate the Wo and Q factor from this. Any pointers or hints would be helpful thank you. I don't know how to proceed after this.

The specs I was given: enter image description here

I am to design the Chebyshev filter.

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  • \$\begingroup\$ What specifications did you use to calculate the transfer function? Normally the cut-off frequency would be one of those. \$\endgroup\$
    – Barry
    Jul 18, 2021 at 0:36
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    \$\begingroup\$ Deadpool, for the 2nd order factors in the denominator, just take the square root of the rightmost constant to get \$\omega_{_0}\$. With that in hand, just multiply that value by 2 and divide it by the middle constant to get \$Q\$. (\$Q\$ isn't relevant for 1st order.) \$\endgroup\$
    – jonk
    Jul 18, 2021 at 0:52
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    \$\begingroup\$ \$Q\$ is a property of a second-order "biquad" section of your filter. Each section has it's own \$Q\$. now you can define the highest \$Q\$ as your overall filter parameter, but that is just a definition. \$\endgroup\$ Jul 18, 2021 at 4:13
  • \$\begingroup\$ @robert bristow-johnson ,as given T(s) in question,there are many ways to combine (product) two factors in denominator to make it quadratic for ex -(s-P1)(s-P2) or (s-P1)(s-P3) and other combinations are also possible , so isn't that gives different Q values and hence highest Q value might also be different (depending on how we combine the factors)? \$\endgroup\$
    – user215805
    Jul 18, 2021 at 6:05
  • \$\begingroup\$ you cannot match a complex pole with a real pole nor with another complex pole except for the complex conjugate. if you have three or more real poles that are different value, then you can make three different second-order sections out of three different pairs and each SOS will have a different \$Q\$ but they will all be no greater than \$\frac12\$. \$\endgroup\$ Jul 18, 2021 at 6:16

3 Answers 3

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Your specs are, at last, finally clearer. Though none of your steps have been shown, as yet. So you should update your question with what steps you took.

  • Pass-band edge is \$15\:\text{kHz}\$. So \$\Omega_p= 94247.78\:\frac{\text{rad}}{\text{s}}\$.
  • Stop-band edge is \$20\:\text{kHz}\$. So \$\Omega_s= 125663.71\:\frac{\text{rad}}{\text{s}}\$.
  • Maximum pass-band attenuation is \$1\:\text{dB}\$. So \$\delta_p=10^{^\frac{-1}{20}}\approx 0.891251\$.
  • Minimum stop-band attenuation is \$23\:\text{dB}\$. So \$\delta_s=10^{^\frac{-23}{20}}\approx 0.0707946\$.
  • Voltage gain is \$A_v=8\$. (Not needed, so far.)

I'd like to see your calculations expressed in your question. I agree that the order should be at least \$N=6\$. And I agree that you may use a 7th stage to add the required voltage gain.

But you should present your steps in your question and how you arrived at your pole values.

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  • \$\begingroup\$ I used this link equation to find the N. I just substituted the values for N=1,2,3... so on and checked if it is greater than the Amin.I found N to be around 6 and Then I calculated the poles using this equation link. So When I got the to transfer function, I didn't know how to make the gain to be the total of 8 at the end. So, instead, I made N=7 and planned to add a first order active filter system with a gain to make the global gain to be 8 at the end. I can add the steps here, because of a scribble of handwriting I have, I decided not to \$\endgroup\$
    – Deadpool
    Jul 18, 2021 at 20:29
  • \$\begingroup\$ @Deadpool There is a way of adding equations that read beautifully in your question, you know. You've only to look at my answer here to see that it can be done. Look here to see some discussion about it's use here. But I also seem to agree with your calculation that N=6 in this case. And if you want voltage gain (you do), then perhaps another stage is appropriate. \$\endgroup\$
    – jonk
    Jul 18, 2021 at 20:44
  • \$\begingroup\$ Thanks for the sharing the link about equations! I will be sure use it. And Thanks for confirming my answer. In the end, I am guessing that what I did was correct. When it comes to the calculation part of the filter. \$\endgroup\$
    – Deadpool
    Jul 18, 2021 at 21:42
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Express each second-order section (SOS) in this form:

$$ H_k(s) = \frac{b_0 + b_1 \, s + b_2 \, s^2}{1 + \frac{1}{Q}\frac{s}{\omega_0} + \left(\frac{s}{\omega_0}\right)^2} $$

Put it in that form and you will get your \$Q\$ and your \$\omega_0\$ .

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I have captured the expression in a Mathcad sheet and factored the terms in a low-entropy form featuring a leading term \$H_0\$ which is the dc gain of this filter. From there you can extract the values for the different resonating points and quality factors:

enter image description here

Then a plot confirms the two expressions (the raw and re-organized ones) are similar in magnitude and phase:

enter image description here

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    \$\begingroup\$ This is exactly what i got! all the values are same. But when I moved on with designing the active filters, the values I needed to satisfy this were.. a bit impractical. \$\endgroup\$
    – Deadpool
    Jul 18, 2021 at 18:51
  • \$\begingroup\$ In the sense that the capacitance was way off than needed. also to comfirm with the first order transfer function, the Wo of the first order function in the denominator is 19.5 rad? \$\endgroup\$
    – Deadpool
    Jul 18, 2021 at 19:17

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