1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.