375 lines
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375 lines
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original version by: Nikos Drakos, CBLU, University of Leeds
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* revised and updated by: Marcus Hennecke, Ross Moore, Herb Swan
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* with significant contributions from:
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Jens Lippmann, Marek Rouchal, Martin Wilck and others -->
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<TITLE>Butterworth filters</TITLE>
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<B> Next:</B> <A NAME="tex2html2745"
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HREF="node146.html">Stretching the unit circle</A>
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HREF="node144.html">Peaking and stop-band filter</A>
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HREF="node201.html">Index</A></B>
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<BR>
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<!--End of Navigation Panel-->
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<H2><A NAME="SECTION001236000000000000000">
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Butterworth filters</A>
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</H2>
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<P>
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A filter with one real pole and one real zero can be configured as a shelving
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filter, as a high-pass filter (putting the zero at the point <IMG
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WIDTH="11" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img262.png"
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ALT="$1$">) or as a
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low-pass filter (putting the zero at <IMG
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WIDTH="23" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img401.png"
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ALT="$-1$">). The frequency responses of these
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filters are quite blunt; in other words, the transition regions are wide. It
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is often desirable to get a sharper filter, either shelving, low- or high-pass,
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whose two bands are flatter and separated by a narrower transition region.
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<P>
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A procedure borrowed from the analog filtering world transforms real,
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one-pole, one-zero filters to corresponding
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<A NAME="10375"></A>
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<I>Butterworth filters</I>,
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which have narrower transition regions. This procedure is described clearly
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and elegantly in the last chapter of [<A
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HREF="node202.html#r-steiglitz96">Ste96</A>]. The derivation
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uses more mathematics background than we have developed here, and we will simply
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present the result without deriving it.
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<P>
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To make a Butterworth filter out of a high-pass, low-pass, or shelving
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filter, suppose that either the pole or the
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zero is given by the expression
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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 - {r^2}} \over {{(1 + r)}^2}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="55" HEIGHT="48" BORDER="0"
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SRC="img928.png"
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ALT="\begin{displaymath}
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{{1 - {r^2}} \over {{(1 + r)}^2}}
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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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where <IMG
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WIDTH="11" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img467.png"
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ALT="$r$"> is a parameter ranging from 1 to <IMG
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WIDTH="19" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img305.png"
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ALT="$\infty$">. If <IMG
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WIDTH="40" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img929.png"
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ALT="$r=0$"> this is the point
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<IMG
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WIDTH="11" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img262.png"
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ALT="$1$">, and if <IMG
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WIDTH="48" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img930.png"
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ALT="$r=\infty$"> it's <IMG
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WIDTH="23" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img401.png"
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ALT="$-1$">.
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<P>
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Then, for reasons which will remain mysterious, we replace the point (whether
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pole or zero) by <IMG
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WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img75.png"
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ALT="$n$"> points given by:
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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 - {r^2}) - (2 r \sin(\alpha)) i }
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\over
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{ 1 + {r^2} + 2 r \cos(\alpha))}
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\end{displaymath}
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-->
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<IMG
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WIDTH="152" HEIGHT="45" BORDER="0"
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SRC="img931.png"
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ALT="\begin{displaymath}
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{ (1 - {r^2}) - (2 r \sin(\alpha)) i }
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\over
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{ 1 + {r^2} + 2 r \cos(\alpha))}
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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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where <IMG
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WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img7.png"
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ALT="$\alpha $"> ranges over the values:
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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\begin{displaymath}
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{\pi \over 2} ({1 \over n} - 1) , \;
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{\pi \over 2} ({3 \over n} - 1) , \; \ldots , \;
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{\pi \over 2} ({{2n-1} \over n} - 1)
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\end{displaymath}
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-->
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<IMG
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WIDTH="288" HEIGHT="38" BORDER="0"
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SRC="img932.png"
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ALT="\begin{displaymath}
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{\pi \over 2} ({1 \over n} - 1) , \;
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{\pi \over 2} ({3 \ov...
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...} - 1) , \; \ldots , \;
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{\pi \over 2} ({{2n-1} \over n} - 1)
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\end{displaymath}">
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</DIV>
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<BR CLEAR="ALL">
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<P></P>
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In other words, <IMG
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WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img7.png"
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ALT="$\alpha $"> takes on <IMG
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WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img75.png"
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ALT="$n$"> equally spaced angles between
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<IMG
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WIDTH="41" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img283.png"
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ALT="$-\pi/2$"> and <IMG
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WIDTH="29" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img5.png"
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ALT="$\pi /2$">. The points are arranged in the complex plane as shown in
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Figure <A HREF="#fig08.17">8.17</A>. They lie on a circle through the original real-valued
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point, which cuts the unit circle at right angles.
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<P>
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A good estimate for the cutoff or transition frequency defined by
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these circular collections of poles or zeros is simply the spot where the
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circle intersects the unit circle, corresponding to <!-- MATH
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$\alpha = \pi/2$
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-->
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<IMG
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WIDTH="60" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img933.png"
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ALT="$\alpha = \pi/2$">. This gives
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the point
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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 - {r^2}) - 2 r i }
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\over
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{ 1 + {r^2} }
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\end{displaymath}
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-->
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<IMG
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WIDTH="95" HEIGHT="42" BORDER="0"
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SRC="img934.png"
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ALT="\begin{displaymath}
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{ (1 - {r^2}) - 2 r i }
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\over
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{ 1 + {r^2} }
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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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which, after some algebra, gives an angular frequency equal to
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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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\beta = 2 \arctan (r)
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\end{displaymath}
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-->
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<IMG
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WIDTH="105" HEIGHT="28" BORDER="0"
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SRC="img935.png"
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ALT="\begin{displaymath}
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\beta = 2 \arctan (r)
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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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<P>
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<DIV ALIGN="CENTER"><A NAME="fig08.17"></A><A NAME="10393"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 8.17:</STRONG>
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Replacing a real-valued pole or zero (shown as a solid dot) with an
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array of four
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of them (circles) as for
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a Butterworth filter. In this example we get four new poles or zeros as
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shown, lying along the circle where <IMG
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WIDTH="52" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img45.png"
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ALT="$r=0.5$">.</CAPTION>
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<TR><TD><IMG
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WIDTH="406" HEIGHT="291" BORDER="0"
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SRC="img936.png"
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ALT="\begin{figure}\psfig{file=figs/fig08.17.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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Figure <A HREF="#fig08.18">8.18</A> (part a) shows a pole-zero diagram and frequency
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response for a Butterworth low-pass filter with three poles and three zeros.
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Part (b) shows the frequency response of the low-pass filter and three other
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filters obtained by choosing different values of <IMG
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WIDTH="13" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img8.png"
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ALT="$\beta $"> (and hence <IMG
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WIDTH="11" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img467.png"
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ALT="$r$">) for
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the zeros, while leaving the poles stationary. As the zeros progress from
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<IMG
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WIDTH="44" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img47.png"
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ALT="$\beta =\pi $"> to <IMG
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WIDTH="42" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img50.png"
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ALT="$\beta =0$">, the filter, which starts as a low-pass filter,
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becomes a shelving filter and then a high-pass one.
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<P>
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<DIV ALIGN="CENTER"><A NAME="fig08.18"></A><A NAME="10399"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 8.18:</STRONG>
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Butterworth low-pass filter with three poles
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and three zeros: (a) pole-zero plot. The poles are chosen for a cutoff frequency
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<IMG
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WIDTH="60" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img46.png"
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ALT="$\beta = \pi /4$">;
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(b) frequency responses for four filters with the same
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pole configuration, with different placements of zeros (but leaving the poles
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fixed). The low-pass filter
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results from setting <IMG
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WIDTH="44" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img47.png"
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ALT="$\beta =\pi $"> for the zeros; the two shelving filters
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correspond to <IMG
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WIDTH="76" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img48.png"
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ALT="$\beta =3\pi /10$"> and <IMG
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WIDTH="76" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img49.png"
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ALT="$\beta =2\pi /10$">, and finally the high-pass
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filter is obtained setting <IMG
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WIDTH="42" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img50.png"
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ALT="$\beta =0$">. The high-pass filter is normalized for
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unit gain at the Nyquist frequency, and the others for unit gain at DC. </CAPTION>
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<TR><TD><IMG
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WIDTH="588" HEIGHT="252" BORDER="0"
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SRC="img937.png"
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ALT="\begin{figure}\psfig{file=figs/fig08.18.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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<HR>
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Miller Puckette
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2006-12-30
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