150 lines
4.1 KiB
HTML
150 lines
4.1 KiB
HTML
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<H2><A NAME="SECTION001235000000000000000"></A>
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<A NAME="sect8.peaking"></A>
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<BR>
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Peaking and stop-band filter
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</H2>
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<P>
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In the same way, a peaking filter is obtained from a shelving filter by rotating the
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pole and the zero, and by providing a conjugate pole and zero, as shown in
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Figure <A HREF="#fig08.16">8.16</A>. If the desired center frequency is <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 $">, and
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the radii of the pole and zero (as for the shelving filter) are
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<IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img57.png"
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ALT="$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$">, then we place the the upper pole and zero at
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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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{P_1} = p \cdot (\cos \omega + i \sin \omega)
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\end{displaymath}
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-->
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ALT="\begin{displaymath}
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{P_1} = p \cdot (\cos \omega + i \sin \omega)
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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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{Q_1} = q \cdot (\cos \omega + i \sin \omega)
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\end{displaymath}
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-->
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<IMG
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WIDTH="166" HEIGHT="28" BORDER="0"
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SRC="img926.png"
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ALT="\begin{displaymath}
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{Q_1} = q \cdot (\cos \omega + i \sin \omega)
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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 a special case, placing the zero on the unit circle gives a stop-band filter;
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in this case the gain at the center frequency is zero. This is analogous
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to the one-pole, one-zero high-pass filter above.
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<P>
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<DIV ALIGN="CENTER"><A NAME="fig08.16"></A><A NAME="10371"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 8.16:</STRONG>
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A peaking filter: (a) pole-zero diagram; (b)
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frequency response. Here the filter is set to attenuate by 6 decibels at
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the center frequency.</CAPTION>
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<TR><TD><IMG
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WIDTH="536" HEIGHT="245" BORDER="0"
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SRC="img927.png"
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ALT="\begin{figure}\psfig{file=figs/fig08.16.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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<BR><HR>
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<ADDRESS>
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Miller Puckette
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2006-12-30
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</ADDRESS>
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</HTML>
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