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337 lines
8.6 KiB
337 lines
8.6 KiB
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original version by: Nikos Drakos, CBLU, University of Leeds
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HREF="node139.html">Designing filters</A>
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HREF="node132.html">Elementary filters</A>
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HREF="node137.html">Real outputs from complex</A>
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HREF="node4.html">Contents</A></B>
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<H2><A ID="SECTION001226000000000000000">
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Two recirculating filters for the price of one</A>
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</H2>
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<P>
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When pairing recirculating elementary filters, it is possible to avoid
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computing one of each pair, as long as the input is real-valued (and so,
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the output is as well.) Supposing the input is a real sinusoid 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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A{Z^n} + \overline{A}{Z^{-n}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="93" HEIGHT="25" BORDER="0"
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SRC="img900.png"
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ALT="\begin{displaymath}
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A{Z^n} + \overline{A}{Z^{-n}}
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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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and we apply a single recirculating filter with coefficient <IMG
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WIDTH="15" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img880.png"
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ALT="$P$">. Letting <IMG
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WIDTH="30" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img503.png"
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ALT="$a[n]$">
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denote the real part of the output, we have:
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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[n] =
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\mathrm{re} \left[ {
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{1 \over {1 - {P}{Z^{-1}}}} {A{Z^n} +
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{1 \over {1 - {P}{Z}}} \overline{A}{Z^{-n}}}
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} \right ]
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\end{displaymath}
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-->
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<IMG
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WIDTH="298" HEIGHT="45" BORDER="0"
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SRC="img901.png"
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ALT="\begin{displaymath}
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a[n] =
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\mathrm{re} \left[ {
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{1 \over {1 - {P}{Z^{-1}}}} {...
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...n} +
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{1 \over {1 - {P}{Z}}} \overline{A}{Z^{-n}}}
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} \right ]
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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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\mathrm{re} \left[ {
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{1 \over {1 - {P}{Z^{-1}}}} {A{Z^n}} +
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{1 \over {1 - \overline{P}{Z^{-1}}}} {A{Z^n}}
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} \right ]
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\end{displaymath}
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-->
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<IMG
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WIDTH="274" HEIGHT="45" BORDER="0"
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SRC="img902.png"
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ALT="\begin{displaymath}
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=
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\mathrm{re} \left[ {
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{1 \over {1 - {P}{Z^{-1}}}} {A{Z^n}} +
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{1 \over {1 - \overline{P}{Z^{-1}}}} {A{Z^n}}
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} \right ]
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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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\mathrm{re} \left[ {
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{{
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2 - 2 \, \mathrm{re} (P) {Z^{-1}}
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} \over {
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(1 - {P}{Z^{-1}}) (1 - {\overline{P}}{Z^{-1}})
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}}
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{A{Z^n}}
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} \right ]
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\end{displaymath}
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-->
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<IMG
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WIDTH="243" HEIGHT="46" BORDER="0"
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SRC="img903.png"
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ALT="\begin{displaymath}
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=
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\mathrm{re} \left[ {
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{{
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2 - 2 \, \mathrm{re} (P) {Z^{-...
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...^{-1}}) (1 - {\overline{P}}{Z^{-1}})
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}}
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{A{Z^n}}
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} \right ]
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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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\mathrm{re} \left[ {
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{{
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1 - \mathrm{re} (P) {Z^{-1}}
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} \over {
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(1 - {P}{Z^{-1}}) (1 - {\overline{P}}{Z^{-1}})
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}}
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{A{Z^n}}
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+
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{{
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1 - \mathrm{re} (P) {{\overline{Z}}^{-1}}
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} \over {
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(1 - {\overline{P}}{{\overline{Z}}^{-1}}) (1 - {P}{{\overline{Z}}^{-1}})
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}}
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{\overline{A}{{\overline{Z}}^{-n}}}
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} \right ]
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\end{displaymath}
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-->
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<IMG
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WIDTH="474" HEIGHT="54" BORDER="0"
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SRC="img904.png"
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ALT="\begin{displaymath}
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=
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\mathrm{re} \left[ {
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{{
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1 - \mathrm{re} (P) {Z^{-1}}
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...
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...}}^{-1}})
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}}
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{\overline{A}{{\overline{Z}}^{-n}}}
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} \right ]
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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 the second step we used the fact that you can conjugate all or part of an
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expression, without changing the result, if you're just going to take the real
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part anyway. The fourth step did the same thing backward.) Comparing the input
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to the output, we see that the effect of passing a real signal through a complex
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one-pole filter, then taking the real part, is equivalent to passing the signal
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through a two-pole, one-zero filter with transfer function 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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{H_{\mathrm{re}}}(Z) = {{
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1 - \mathrm{re} (P) {Z^{-1}}
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} \over {
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(1 - {P}{Z^{-1}}) (1 - {\overline{P}}{Z^{-1}})
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}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="235" HEIGHT="46" BORDER="0"
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SRC="img905.png"
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ALT="\begin{displaymath}
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{H_{\mathrm{re}}}(Z) = {{
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1 - \mathrm{re} (P) {Z^{-1}}
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} \over {
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(1 - {P}{Z^{-1}}) (1 - {\overline{P}}{Z^{-1}})
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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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A similar calculation shows that taking the imaginary part (considered as a real
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signal) is equivalent to filtering the input with the transfer function:
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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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{H_{\mathrm{im}}}(Z) = {{
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\mathrm{im} (P) {Z^{-1}}
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} \over {
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(1 - {P}{Z^{-1}}) (1 - {\overline{P}}{Z^{-1}})
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}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="238" HEIGHT="46" BORDER="0"
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SRC="img906.png"
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ALT="\begin{displaymath}
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{H_{\mathrm{im}}}(Z) = {{
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\mathrm{im} (P) {Z^{-1}}
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} \over {
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(1 - {P}{Z^{-1}}) (1 - {\overline{P}}{Z^{-1}})
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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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So taking either the real or imaginary part of a one-pole filter output gives
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filters with two conjugate poles. The two parts can be combined
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to synthesize filters with other possible numerators; in other words, with
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one complex recirculating filter we can synthesize a filter that acts on
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real signals with two (complex conjugate) poles and one (real) zero.
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<P>
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This technique, known as
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<I>partial fractions</I>, may be repeated for
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any number of stages in series as long as we compute the appropriate
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combination of real and imaginary parts of the output of each stage to
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form the (real) input of the next stage.
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No similar shortcut seems to exist for non-recirculating filters; for them
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it is necessary to compute each member of each complex-conjugate pair
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explicitly.
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<P>
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<BR>
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<B> Next:</B> <A ID="tex2html2636"
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HREF="node139.html">Designing filters</A>
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<B> Up:</B> <A ID="tex2html2630"
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HREF="node132.html">Elementary filters</A>
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<B> Previous:</B> <A ID="tex2html2626"
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HREF="node137.html">Real outputs from complex</A>
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<B> <A ID="tex2html2632"
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HREF="node4.html">Contents</A></B>
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<B> <A ID="tex2html2634"
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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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</ADDRESS>
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