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<H2><A ID="SECTION001238000000000000000">
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Butterworth band-pass filter</A>
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</H2>
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<P>
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We can apply the transformation <IMG
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WIDTH="92" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
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SRC="img952.png"
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ALT="$R(Z) = -Z^2$"> to convert the Butterworth filter into a high-quality
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band-pass filter with center frequency <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="$\pi /2$">. A further transformation
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can then be applied to shift the center frequency to any desired value <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 $">
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between 0 and <IMG
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WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img41.png"
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ALT="$\pi $">. The transformation will be of the form,
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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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S(Z) =
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{{
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aZ + b
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} \over {
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bZ + a
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}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="106" HEIGHT="41" BORDER="0"
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SRC="img953.png"
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ALT="\begin{displaymath}
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S(Z) =
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{{
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aZ + b
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} \over {
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bZ + a
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}}
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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="img4.png"
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ALT="$a$"> and <IMG
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WIDTH="10" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img21.png"
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ALT="$b$"> are real numbers and not both are zero. This
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is a particular case of the general form given above for unit-circle-preserving
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rational functions. We have <IMG
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WIDTH="63" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img954.png"
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ALT="$S(1) = 1$"> and <IMG
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WIDTH="88" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img955.png"
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ALT="$S(-1) = -1$">,
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and the top and bottom halves of the unit circle are transformed
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symmetrically (if <IMG
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WIDTH="15" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img20.png"
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ALT="$Z$"> goes to <IMG
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WIDTH="20" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img31.png"
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ALT="$W$"> then <IMG
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WIDTH="15" HEIGHT="17" ALIGN="BOTTOM" BORDER="0"
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SRC="img956.png"
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ALT="$\overline{Z}$"> goes to <IMG
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WIDTH="20" HEIGHT="17" ALIGN="BOTTOM" BORDER="0"
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SRC="img957.png"
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ALT="$\overline{W}$">).
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The qualitative effect of the transformation <IMG
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WIDTH="14" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
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SRC="img958.png"
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ALT="$S$"> is to squash points of the
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unit circle toward <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 <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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In particular, given a desired center 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 $">,
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we wish to choose <IMG
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WIDTH="14" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
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SRC="img958.png"
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ALT="$S$"> so that:
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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\begin{displaymath}
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S(\cos(\omega) + i \sin(\omega)) = i
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\end{displaymath}
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-->
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<IMG
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WIDTH="163" HEIGHT="28" BORDER="0"
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SRC="img959.png"
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ALT="\begin{displaymath}
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S(\cos(\omega) + i \sin(\omega)) = i
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\end{displaymath}">
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</DIV>
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<BR CLEAR="ALL">
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<P></P>
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If we leave <IMG
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WIDTH="68" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
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SRC="img960.png"
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ALT="$R = - {Z^2}$"> as before, and let <IMG
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WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img25.png"
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ALT="$H$"> be the transfer
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function for a low-pass Butterworth filter, then the combined filter
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with transfer function <IMG
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WIDTH="90" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img961.png"
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ALT="$H(R(S(Z)))$">
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will be a band-pass filter with center frequency <IMG
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WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img27.png"
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ALT="$\omega $">. Solving for <IMG
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WIDTH="11" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img4.png"
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ALT="$a$">
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and <IMG
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WIDTH="10" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img21.png"
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ALT="$b$"> gives:
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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\begin{displaymath}
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a = \cos({{\pi}\over 4} - {{\omega} \over 2}) , \;
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b = \sin({{\pi}\over 4} - {{\omega} \over 2})
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\end{displaymath}
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-->
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<IMG
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WIDTH="227" HEIGHT="35" BORDER="0"
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SRC="img962.png"
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ALT="\begin{displaymath}
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a = \cos({{\pi}\over 4} - {{\omega} \over 2}) , \;
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b = \sin({{\pi}\over 4} - {{\omega} \over 2})
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\end{displaymath}">
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</DIV>
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<BR CLEAR="ALL">
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<P></P>
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The new transfer function, <IMG
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WIDTH="90" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img961.png"
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ALT="$H(R(S(Z)))$">, will have <IMG
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WIDTH="21" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img963.png"
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ALT="$2n$"> poles and <IMG
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WIDTH="21" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img963.png"
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ALT="$2n$">
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zeros (if <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="$n$"> is the degree of the Butterworth filter <IMG
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WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img25.png"
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ALT="$H$">).
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<P>
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Knowing the transfer function is good, but even better is knowing the locations
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of all the poles and zeros of the new filter, which we need to be able to
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compute it using elementary filters. If <IMG
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WIDTH="15" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img20.png"
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ALT="$Z$"> is a pole of the transfer
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function <!-- MATH
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$J(Z) = H(R(S(Z)))$
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-->
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<IMG
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WIDTH="146" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img964.png"
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ALT="$J(Z) = H(R(S(Z)))$">, that is, if <IMG
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WIDTH="75" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img965.png"
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ALT="$J(Z)=\infty$">, then <IMG
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WIDTH="63" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img966.png"
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ALT="$R(S(Z))$"> must
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be a pole of <IMG
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WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img25.png"
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ALT="$H$">. The same goes for zeros. To find a pole or zero of <IMG
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WIDTH="13" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img945.png"
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ALT="$J$">
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we set <IMG
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WIDTH="101" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img967.png"
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ALT="$R(S(Z)) = W$">, where <IMG
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WIDTH="20" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img31.png"
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ALT="$W$"> is a pole or zero of <IMG
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WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img25.png"
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ALT="$H$">, and solve for <IMG
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WIDTH="15" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img20.png"
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ALT="$Z$">.
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This gives:
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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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- {
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{ \left [ {
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{
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aZ + b
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} \over {
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bZ + a
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}
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} \right ] }
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^ 2
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} = W
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\end{displaymath}
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-->
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<IMG
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WIDTH="124" HEIGHT="48" BORDER="0"
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SRC="img968.png"
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ALT="\begin{displaymath}
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- {
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{ \left [ {
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{
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aZ + b
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} \over {
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bZ + a
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}
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} \right ] }
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^ 2
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} = W
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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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<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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{
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{
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aZ + b
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} \over {
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bZ + a
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}
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} = {
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\pm \sqrt {
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- W
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}
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}
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\end{displaymath}
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-->
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<IMG
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WIDTH="126" HEIGHT="41" BORDER="0"
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SRC="img969.png"
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ALT="\begin{displaymath}
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{
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{
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aZ + b
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} \over {
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bZ + a
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}
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} = {
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\pm \sqrt {
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- W
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}
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}
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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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<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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Z = { {
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\pm a \sqrt { - W }
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- b
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} \over {
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\mp b \sqrt { - W }
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+ a
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}
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}
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\end{displaymath}
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-->
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<IMG
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WIDTH="125" HEIGHT="46" BORDER="0"
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SRC="img970.png"
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ALT="\begin{displaymath}
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Z = { {
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\pm a \sqrt { - W }
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- b
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} \over {
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\mp b \sqrt { - W }
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+ a
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}
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}
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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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(Here <IMG
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WIDTH="11" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img4.png"
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ALT="$a$"> and <IMG
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WIDTH="10" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img21.png"
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ALT="$b$"> are as given above and we have used the fact that
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<IMG
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WIDTH="81" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
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SRC="img971.png"
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ALT="${a^2}+{b^2}=1$">). A sample pole-zero plot and frequency response of <IMG
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WIDTH="13" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img945.png"
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ALT="$J$">
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are shown in Figure <A HREF="#fig08.20">8.20</A>.
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<P>
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<DIV ALIGN="CENTER"><A ID="fig08.20"></A><A ID="10465"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 8.20:</STRONG>
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Butterworth band-pass filter: (a) pole-zero diagram; (b)
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frequency response. The center frequency is <IMG
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WIDTH="29" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img52.png"
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ALT="$\pi /4$">. The bandwidth depends
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both on center frequency and on the bandwidth of the original Butterworth
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low-pass filter used.</CAPTION>
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<TR><TD><IMG
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WIDTH="545" HEIGHT="244" BORDER="0"
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SRC="img972.png"
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ALT="\begin{figure}\psfig{file=figs/fig08.20.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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HREF="node148.html">Time-varying coefficients</A>
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<B> Up:</B> <A ID="tex2html2767"
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HREF="node139.html">Designing filters</A>
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<B> Previous:</B> <A ID="tex2html2761"
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HREF="node146.html">Stretching the unit circle</A>
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HREF="node4.html">Contents</A></B>
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HREF="node201.html">Index</A></B>
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<ADDRESS>
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Miller Puckette
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2006-12-30
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