I am interested in how to add resonance (Q) to the magnitude response of a Butterworth low pass when it is expressed in the form:

$$ G^2(\omega)=\frac {1}{1+\left(\frac{\omega}{\omega_c}\right)^{2n}} $$

I know how to add resonance $Q > 0$ for $n >= 2$ i.e. include it in one (and only one) of the second second order terms (if the equation is expanded as a product of second order sections):

$$ H(s) = \frac{ \omega_0^2 }{ s^2 + \frac{ \omega_0 }{Q} + \omega_0^2 } $$

However, I wonder whether there is a general formula for resonance $Q$ that is a function of $\omega$ that can be used for fractional $n$ such that it can be added to the transfer function $G$ or $G^2$ to yield a new magnitude response?

  • $\begingroup$ How would you implement the filter if $n$ is fractional? $\endgroup$ – Olli Niemitalo Jan 30 '16 at 5:42
  • $\begingroup$ I know how to do that (it will be digital not an analogue circuit). $\endgroup$ – keith Jan 30 '16 at 8:32
  • $\begingroup$ I think a multiplicative modification would be better than an additive one because additive modifications of $G$ and $G^2$ can't be equivalent. $\endgroup$ – Olli Niemitalo Jan 30 '16 at 8:57
  • $\begingroup$ I agree, a multiplicative solution would work better. $\endgroup$ – keith Jan 30 '16 at 9:36
  • $\begingroup$ One idea is to have in series with a non-resonant Butterwoth lowpass filter an additional (multiplicative) pole-zero pair such that when $Q = 1$ the two would coincide with each other, and with the pole of your $H(s)$ above (with $Q=1$) in case of integer $n$. When changing $Q$, the zero would stay in place, canceling a pole from the unmodified filter in case of integer $N$, and the pole would move as in your $H(s)$ above. $\endgroup$ – Olli Niemitalo Jan 30 '16 at 11:22

Given the formula for second order sections:

$$ s_k = \omega_c e^{\frac{j(2k+n-1)\pi}{2n}}\qquad k = 1,2,3,\ldots, n \qquad s = i\omega $$

$$ H(s)=\frac{G_0}{\prod_{k=1}^n (s-s_k)/\omega_c} $$

Use k = 1 for the resonance section, add $Q$ in the normal way i.e. calculate $H_{k=1}^2 = H_{k=1}(s) \overline{H_{k=1}(s)}$ where $s = i\omega$ and the denominator will be of the form $w^4 + cw^2 + 1$ where $c$ is a constant calculated from $s_k$. We add $Q$ to the first order term (note that this is a quadratic in $\omega^2$). Using any positive real $n$ in $s_k$ gives valid values for $c$.

Let $R^2$ be a modified $H_{k=1}^2$ with $Q$ added in to the first order term of the quadratic in $w^2$ then to add resonance to the original equation in the question multiply it by:

$$ \frac {R^2}{H_{k=1}^2} $$

This effectively removes the $k = 1$ second order section and adds a resonant section back in. It turns out once you do this, $n$ can be any positive real value when used in $c$ but note this wouldn't work for fractional roots of the sections (you would have to use fractional calculus if you wanted the roots of the fractional equation).

Without resonance:

$$ c = 2 \cos^2(\frac{\pi (1+n) } {2n}) - 2 $$

With resonance:

$$ c = \frac{2}{Q^2}\cos^2(\frac{\pi (1+n) } {2n}) - 2 $$

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