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Shelving filter

Generalizing the one-zero, one-pole filter above, supose we place the zero at a point $q$, a real number close to, but less than, one. The pole, at the point $p$, is similarly situated, and might be either greater than or less than $q$, i.e., to the right or left, respectively, but with both $q$ and $p$ within the unit circle. This situation is diagrammed in Figure 8.14.

Figure 8.14: One-pole, one-zero shelving filter: (a) pole-zero diagram; (b) frequency response.
\begin{figure}\psfig{file=figs/fig08.14.ps}\end{figure}

At points of the circle far from $p$ and $q$, the effects of the pole and the zero are nearly inverse (the distances to themare nearly equal), so the filter largely passes those frequencies unaltered. In the neighborhood of $p$ and $q$, on the other hand, the filter will have a gain greater or less than one depending on which of $p$ or $q$ is closer to the circle. This configuration therefore acts as a low-frequency shelving filter. (To make a high-frequency shelving filter we do the same thing, only placing $p$ and $q$ close to -1 instead of 1.)

To find the parameters of a desired shelving filter, start with a desired transition frequency $\omega $ (in angular units) and a desired low-frequency gain $g$. First we choose an average distance $d$, as pictured in the figure, from the pole and the zero to theedge of the circle. For small values of $d$, the region of influence (the crossover frequency) is about $d$ radians.

Then put the pole at $p = 1 - d \sqrt{g}$ and the zero at $q = 1 - d / \sqrt{g}$. The gain at zero frequency is then

\begin{displaymath}
{{1-q} \over {1-p}} = g
\end{displaymath}

as desired. For example, in the figure, $d$ is 0.25 radians and $g$ is 2.


next up previous contents index
Next: Band-pass filter Up: Designing filters Previous: One-pole, one-zero high-pass filter   Contents   Index
Miller Puckette 2005-02-21