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In a paper I am interested in, the following equation is given to determine the cut-off frequency $\omega$ of a filter:

$$t^{2N}+\left[-2\left(1-\frac{1}{2Q^2}\right)\right]t^N +\left[-\left(\frac{|H(\omega_{CF})|}{\gamma \, \omega_0^{2N-1}}\right)^{-\frac{2}{N}}\right] t +\omega_0^4 =0$$ Where $t=\omega^{\frac{2}{N}}$, $Q$ is the quality factor, $N$ is the filter order, $\omega_0$ is the natural frequency, $|H(\omega_{CF})|$ is the filter peak-gain and $\gamma=\sqrt{2}$ for the response at -3dB.

Since only $\omega$ is unknown, I can see that the equation has the form $x^{2N}a + x^{N}b + xc +d$. However, I am failing to understand how $\omega$ can be evaluated. In addition, is it possible to find the value of $Q$ for any given $\omega$.

Thank you for helping with this.

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  • $\begingroup$ Numeric methods? $\endgroup$ Commented May 22, 2022 at 18:48
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    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Commented May 22, 2022 at 19:03
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    $\begingroup$ That's an awkward equation. Not sure what a "typical" filter order is, but you end up with $2N+1$ solutions, most of which will probably be complex. $\endgroup$
    – Hilmar
    Commented May 22, 2022 at 19:24
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    $\begingroup$ @Hilmar In that case only one will be real, so maybe that's the one? But if $t=\omega^{2/N}$ then $t^{2N}=\omega^{2N\cdot 2/N}=\omega^4$, all the time, irrespective of $N$, similar for $t^N$, while $t=\omega^{2/N}$. If $N>2$ it ends up fractional. The plot thickens. $\endgroup$ Commented May 23, 2022 at 9:59

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Since only $\omega$ is unknown, I can see that the equation has the form $x^{2N}a + x^{N}b + xc +d$.

Exactly, you're looking for the roots (i.e., points where the value is 0) of a polynomial in $t$. The Fundamental Theorem of Algebra tells us that this polynomial has $2N$ roots, some of which might have multiplicity > 1 (so potentially, there's fewer different roots).

Although I think it's possible to find an elegant factorization of this polynomial by guessing one root, numerical methods for such problems abound. Your favourite math tool (Python, Julia, MATLAB or Mathematica, or whatever!) certainly comes with a root-finding toolbox as well.

However, I am failing to understand how $\omega$ can be evaluated.

  1. Find all $t_k$ solving the equation
  2. For each of these $t_k$, find all $\omega_{k,\ell}$ such that $t_k=\omega_{k,\ell}^{\frac{2}{N}}$. (Note: of course, $\omega_{·,·}=\sqrt[N]{t_k}^2$ if it exists is one solution, but so might $-\sqrt[N]{t_k}^2$ etc.)

I'm guessing that all possible $\omega$ necessarily are real, so that you can throw away a lot of potential solutions here. This might mean that you might already only be interested in real $t_k$, so that numerically searching for real roots might be a good approach if $N$ is very large. (Otherwise, your computer will spend most of its time computing irrelevant imaginary roots.)

In addition, is it possible to find the value of $Q$ for any given $\omega$.

Sure, just rearranging your original equations so that $Q^2$ stands alone, but there's going to be two solutions.


Personally: sit down, write down the polynomial in $t$ as polynomial in $\omega$, then implement it for a "reasonable example" selection of $Q$, $N$ and $\omega_0$, and plot it in real, imaginary part and absolute value. Might give you a bit of an intuition. It's not a very intuitive formula, as unconstrained as shown, for something as real as a cutoff frequency.

I'd also look for what the authors of your paper do with that equation. Maybe they implicitly give a hint.

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  • $\begingroup$ Yes, the solution should be real. I have been trying with mathlab but it didn't find any root. So I'm basically trying to solve a degree 3 polynomial equation here? what puzzles me is what should I do with the power $^{2N}$ and $^N$. How can I put it in the form $ax^3+bx^2+cx+d$? $\endgroup$
    – papaya
    Commented May 23, 2022 at 15:13
  • $\begingroup$ no, you're trying to solve a degree $2N$ polynomial equation. As said (in breadth!), any such equation has $2N$ solutions, no exceptions. However, as said, not all of these solution must be different; ther is not any guarantee there's a real solution, even. $\endgroup$ Commented May 23, 2022 at 15:17
  • $\begingroup$ for example, the equation $x^{2N}+1 = 0$ definitely has $2N$ solutions, but definitely no real one. $\endgroup$ Commented May 23, 2022 at 15:19

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