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190 lines
5.9 KiB
190 lines
5.9 KiB
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<H2><A ID="SECTION001232000000000000000"></A>
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<A ID="sect08.highpass"></A>
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<BR>
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One-pole, one-zero high-pass filter
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</H2>
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<P>
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Sometimes an audio signal carries an unwanted constant offset, or in other
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words, a zero-frequency component. For example, the waveshaping spectra of
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Section <A HREF="node78.html#sect5.waveshaping">5.3</A> almost always contain a constant component.
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This is inaudible, but, since it specifies electrical power that is sent
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to your speakers, its presence reduces the level of loudness you can
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reach without distortion. Another name for a constant signal component is
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<A ID="10330"></A>"DC", meaning "direct current".
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<P>
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An easy and practical way to remove the zero-frequency component from an audio
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signal is to use a one-pole low-pass filter to extract it, and then subtract the
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result from the signal. The resulting transfer function is one minus the
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transfer function of the low-pass filter:
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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(Z) = 1 - {{{1-p} \over {1 - p{Z^{-1}}}}}
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= p{{{1-{Z^{-1}}} \over {1 - p{Z^{-1}}}}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="252" HEIGHT="44" BORDER="0"
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SRC="img912.png"
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ALT="\begin{displaymath}
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H(Z) = 1 - {{{1-p} \over {1 - p{Z^{-1}}}}}
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= p{{{1-{Z^{-1}}} \over {1 - p{Z^{-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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The factor of <IMG
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WIDTH="38" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img910.png"
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ALT="$1-p$"> in the numerator of the low-pass transfer function is the
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normalization factor needed so that the gain is one at zero frequency.
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<P>
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By examining the right-hand side of the equation (comparing it to the general
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formula for compound filters), we
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see that there is still a pole at the real number <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 there is now also
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a 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="$1$">. The pole-zero plot is shown in Figure
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<A HREF="#fig08.13">8.13</A>
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(part a), and the frequency response in part (b). (Henceforth, we
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will only plot frequency responses to the Nyquist frequency <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 $">;
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in the previous example we plotted it all the way up to the sample rate,
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<IMG
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WIDTH="21" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img16.png"
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ALT="$2\pi $">.)
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<P>
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<DIV ALIGN="CENTER"><A ID="fig08.13"></A><A ID="10338"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 8.13:</STRONG>
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One-pole, one-zero high-pass filter: (a) pole-zero diagram; (b)
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frequency response (from zero to Nyquist frequency).</CAPTION>
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<TR><TD><IMG
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WIDTH="542" HEIGHT="245" BORDER="0"
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SRC="img913.png"
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ALT="\begin{figure}\psfig{file=figs/fig08.13.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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HREF="node142.html">Shelving filter</A>
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<B> Up:</B> <A ID="tex2html2683"
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HREF="node139.html">Designing filters</A>
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HREF="node140.html">One-pole low-pass filter</A>
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<ADDRESS>
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Miller Puckette
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2006-12-30
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