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251 lines
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251 lines
7.3 KiB
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original version by: Nikos Drakos, CBLU, University of Leeds
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<H2><A ID="SECTION001233000000000000000"></A>
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<A ID="sect8.shelving"></A>
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<BR>
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Shelving filter
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</H2>
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<P>
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Generalizing the one-zero, one-pole filter above, suppose we place the
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zero at a point <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img592.png"
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ALT="$q$">, a real number close to, but less than, one. The
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pole, at the point <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img57.png"
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ALT="$p$">, is similarly situated, and might be either
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greater than or less than <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img592.png"
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ALT="$q$">, i.e., to the right or left, respectively,
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but with both <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img592.png"
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ALT="$q$"> and <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img57.png"
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ALT="$p$"> within the unit circle. This situation is
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diagrammed in Figure <A HREF="#fig08.14">8.14</A>.
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<P>
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<DIV ALIGN="CENTER"><A ID="fig08.14"></A><A ID="10346"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 8.14:</STRONG>
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One-pole, one-zero shelving filter: (a) pole-zero diagram; (b)
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frequency response.</CAPTION>
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<TR><TD><IMG
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WIDTH="540" HEIGHT="380" BORDER="0"
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SRC="img914.png"
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ALT="\begin{figure}\psfig{file=figs/fig08.14.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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At points of the circle far from <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img57.png"
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ALT="$p$"> and <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img592.png"
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ALT="$q$">, the effects of the pole and the
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zero are nearly inverse (the distances to them are nearly equal), so the filter
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passes those frequencies nearly unaltered. In the neighborhood of <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img57.png"
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ALT="$p$"> and
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<IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img592.png"
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ALT="$q$">, on the other hand, the filter will have a gain greater or less than one
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depending on which of <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img57.png"
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ALT="$p$"> or <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img592.png"
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ALT="$q$"> is closer to the circle. This configuration
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therefore acts as a low-frequency shelving filter. (To make a high-frequency
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shelving filter we do the same thing, only placing <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img57.png"
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ALT="$p$"> and <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img592.png"
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ALT="$q$"> close to -1
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instead of 1.)
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<P>
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To find the parameters of a shelving filter given a desired
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transition frequency <IMG
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WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img27.png"
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ALT="$\omega $"> (in angular units) and low-frequency
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gain <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img29.png"
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ALT="$g$">, first we choose an average distance <IMG
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WIDTH="11" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img28.png"
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ALT="$d$">, as pictured in the figure,
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from the pole and the zero to the edge of the circle. For small values of <IMG
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WIDTH="11" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img28.png"
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ALT="$d$">,
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the region of influence is about <IMG
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WIDTH="11" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img28.png"
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ALT="$d$"> radians, so
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simply set <IMG
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WIDTH="43" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img915.png"
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ALT="$d = \omega$"> to get the desired transition frequency.
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<P>
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Then put the pole at <!-- MATH
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$p = 1 - d / \sqrt{g}$
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-->
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<IMG
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WIDTH="97" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img916.png"
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ALT="$p = 1 - d / \sqrt{g}$"> and the zero at
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<!-- MATH
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$q = 1 - d \sqrt{g}$
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-->
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<IMG
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WIDTH="89" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
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SRC="img917.png"
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ALT="$q = 1 - d \sqrt{g}$">. The gain at zero frequency is then
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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\begin{displaymath}
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{{1-q} \over {1-p}} = g
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\end{displaymath}
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-->
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<IMG
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WIDTH="67" HEIGHT="41" BORDER="0"
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SRC="img918.png"
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ALT="\begin{displaymath}
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{{1-q} \over {1-p}} = g
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\end{displaymath}">
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</DIV>
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<BR CLEAR="ALL">
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<P></P>
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as desired. For example, in the figure, <IMG
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WIDTH="11" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img28.png"
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ALT="$d$"> is 0.25 radians and <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img29.png"
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ALT="$g$"> is 2.
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HREF="node143.html">Band-pass filter</A>
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<B> Up:</B> <A ID="tex2html2697"
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HREF="node139.html">Designing filters</A>
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<B> Previous:</B> <A ID="tex2html2691"
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HREF="node141.html">One-pole, one-zero high-pass filter</A>
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HREF="node4.html">Contents</A></B>
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<ADDRESS>
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Miller Puckette
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2006-12-30
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