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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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HREF="node108.html">Delay networks</A>
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HREF="node4.html">Contents</A></B>
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<B> <A ID="tex2html2184"
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<H1><A ID="SECTION001140000000000000000"></A>
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<A ID="sect7.recirculatingcomb"></A>
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<BR>
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Recirculating delay networks
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</H1>
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<P>
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It is sometimes desirable to connect the outputs of one or more delays in a
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network back into their own or each others' inputs. Instead of getting one
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or several echos of the original sound as in the example above, we
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can potentially get an infinite number of echos, each one feeding back into
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the network to engender yet others.
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<P>
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The simplest example of a recirculating network is the
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<A ID="7935"></A>
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<I>recirculating comb filter</I>
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whose block diagram is shown in Figure <A HREF="#fig07.07">7.7</A>. As with the
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earlier, simple comb filter, the input signal is sent down a delay line whose
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length is <IMG
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WIDTH="11" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img28.png"
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ALT="$d$"> samples. But now the delay line's
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output is also fed back to its input; the delay's input is the sum of
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the original input and the delay's output. The output is
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multiplied by a number <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img29.png"
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ALT="$g$"> before feeding it back into its input.
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<P>
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<DIV ALIGN="CENTER"><A ID="fig07.07"></A><A ID="7940"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 7.7:</STRONG>
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Block diagram for a recirculating comb filter. Here <IMG
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WIDTH="11" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img28.png"
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ALT="$d$"> is the
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delay time in samples and <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img29.png"
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ALT="$g$"> is the feedback coefficient.</CAPTION>
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<TR><TD><IMG
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WIDTH="125" HEIGHT="221" BORDER="0"
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SRC="img708.png"
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ALT="\begin{figure}\psfig{file=figs/fig07.07.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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The time domain behavior of the recirculating comb filter is shown in Figure
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<A HREF="#fig07.08">7.8</A>. Here we consider the effect of sending an impulse into the
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network. We get back the original impulse, plus a series of echos, each
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in turn <IMG
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WIDTH="11" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img28.png"
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ALT="$d$"> samples after the previous one, and multiplied each time by the
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gain <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img29.png"
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ALT="$g$">. In general, a delay network's output given an impulse as input is
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called the network's
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<A ID="7944"></A><I>impulse response</I>.
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<P>
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<DIV ALIGN="CENTER"><A ID="fig07.08"></A><A ID="7948"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 7.8:</STRONG>
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Time-domain analysis of the recirculating comb filter, using
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an impulse as input.</CAPTION>
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<TR><TD><IMG
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WIDTH="440" HEIGHT="257" BORDER="0"
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SRC="img709.png"
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ALT="\begin{figure}\psfig{file=figs/fig07.08.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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Note that we have chosen a gain <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img29.png"
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ALT="$g$"> that is less than one in absolute value.
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If we chose a gain greater than one (or less than -1), each echo would have
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a larger magnitude than the previous one. Instead of falling exponentially
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as they do in the figure, they would grow exponentially. A recirculating
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network whose output eventually falls toward zero after its input terminates
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is called
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<A ID="7951"></A><I>stable</I>;
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one whose output grows without bound is called <I>unstable</I>.
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<P>
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We can also analyse the recirculating comb filter in the frequency domain.
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The situation is now quite
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hard to analyze using real sinusoids, and so we get the first big payoff
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for having introduced complex numbers, which greatly simplify the analysis.
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<P>
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If, as before, we feed the input with the signal,
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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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X[n] = Z^n
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\end{displaymath}
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-->
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<IMG
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WIDTH="74" HEIGHT="28" BORDER="0"
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SRC="img710.png"
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ALT="\begin{displaymath}
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X[n] = 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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with <IMG
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WIDTH="53" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img22.png"
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ALT="$\vert Z\vert=1$">, we can write the output as
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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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Y[n] = (1 + g{Z^{-d}} + {g^2}{Z^{-2d}} + \cdots) X[n]
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\end{displaymath}
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-->
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<IMG
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WIDTH="270" HEIGHT="28" BORDER="0"
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SRC="img711.png"
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ALT="\begin{displaymath}
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Y[n] = (1 + g{Z^{-d}} + {g^2}{Z^{-2d}} + \cdots) X[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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Here the terms in the sum come from the series of
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discrete echos.
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It follows that the amplitude of the output is:
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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 = 1 + g{Z^{-d}} + {(g{Z^{-d}})} ^ 2 + \cdots
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\end{displaymath}
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-->
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<IMG
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WIDTH="214" HEIGHT="28" BORDER="0"
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SRC="img712.png"
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ALT="\begin{displaymath}
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H = 1 + g{Z^{-d}} + {(g{Z^{-d}})} ^ 2 + \cdots
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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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This is a geometric series; we can sum it using the standard technique. First
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multiply both sides by <IMG
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WIDTH="41" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
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SRC="img713.png"
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ALT="$g{Z^{-d}}$"> to give:
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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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g{Z^{-d}} H = g{Z^{-d}} + {(g{Z^{-d}})} ^ 2 + {(g{Z^{-d}})} ^ 3 + \cdots
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\end{displaymath}
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-->
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<IMG
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WIDTH="300" HEIGHT="28" BORDER="0"
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SRC="img714.png"
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ALT="\begin{displaymath}
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g{Z^{-d}} H = g{Z^{-d}} + {(g{Z^{-d}})} ^ 2 + {(g{Z^{-d}})} ^ 3 + \cdots
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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 subtract from the original equation to give:
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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 - g{Z^{-d}} H = 1
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\end{displaymath}
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-->
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<IMG
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WIDTH="115" HEIGHT="27" BORDER="0"
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SRC="img715.png"
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ALT="\begin{displaymath}
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H - g{Z^{-d}} H = 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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Then solve for <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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<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 = {1 \over {1 - g{Z^{-d}}}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="102" HEIGHT="41" BORDER="0"
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SRC="img716.png"
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ALT="\begin{displaymath}
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H = {1 \over {1 - g{Z^{-d}}}}
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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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A faster (but slightly less intuitive) method to get the same result is to
|
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examine the recirculating network itself to yield an equation for <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="$H$">, as
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follows. We named the input <IMG
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WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img669.png"
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ALT="$X[n]$"> and the output <IMG
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WIDTH="34" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img717.png"
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ALT="$Y[n]$">. The signal going
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into the delay line is the output <IMG
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WIDTH="34" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img717.png"
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ALT="$Y[n]$">, and passing this through the delay
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line and multiplier 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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|
Y[n] \cdot g{Z^{-d}}
|
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\end{displaymath}
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-->
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<IMG
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WIDTH="79" HEIGHT="28" BORDER="0"
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SRC="img718.png"
|
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ALT="\begin{displaymath}
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|
Y[n] \cdot g{Z^{-d}}
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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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This plus the input is just the output signal again, so:
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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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|
Y[n] = X[n]+Y[n] \cdot g{Z^{-d}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="185" HEIGHT="28" BORDER="0"
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SRC="img719.png"
|
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|
|
ALT="\begin{displaymath}
|
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|
|
Y[n] = X[n]+Y[n] \cdot g{Z^{-d}}
|
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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 dividing by <IMG
|
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|
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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|
SRC="img669.png"
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ALT="$X[n]$"> and using <IMG
|
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|
WIDTH="111" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
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|
|
SRC="img720.png"
|
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|
|
ALT="$H=Y[n]/X[n]$"> 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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|
|
H = 1 + Hg{Z^{-d}}
|
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|
|
\end{displaymath}
|
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|
|
|
-->
|
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|
|
<IMG
|
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|
|
WIDTH="115" HEIGHT="27" BORDER="0"
|
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|
|
SRC="img721.png"
|
|
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|
|
ALT="\begin{displaymath}
|
|
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|
|
H = 1 + Hg{Z^{-d}}
|
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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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|
|
This is equivalent to the earlier equation for <IMG
|
|
|
|
|
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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|
Now we would like to make a graph of the frequency response (the gain as a
|
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|
|
function of frequency) as
|
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|
|
we did for non-recirculating comb filters in
|
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|
|
Figure <A HREF="node108.html#fig07.06">7.6</A>. This
|
|
|
|
|
again requires that we make a preliminary picture in the complex plane. We
|
|
|
|
|
would like to estimate the magnitude of <IMG
|
|
|
|
|
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img25.png"
|
|
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|
|
ALT="$H$"> equal to:
|
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|
|
<BR><P></P>
|
|
|
|
|
<DIV ALIGN="CENTER">
|
|
|
|
|
<!-- MATH
|
|
|
|
|
\begin{displaymath}
|
|
|
|
|
|H| = {1 \over {|1 - g{Z^{-d}}|}}
|
|
|
|
|
\end{displaymath}
|
|
|
|
|
-->
|
|
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|
|
|
|
<IMG
|
|
|
|
|
WIDTH="118" HEIGHT="42" BORDER="0"
|
|
|
|
|
SRC="img722.png"
|
|
|
|
|
ALT="\begin{displaymath}
|
|
|
|
|
\vert H\vert = {1 \over {\vert 1 - g{Z^{-d}}\vert}}
|
|
|
|
|
\end{displaymath}">
|
|
|
|
|
</DIV>
|
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|
|
<BR CLEAR="ALL">
|
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|
|
|
<P></P>
|
|
|
|
|
where we used the multiplicative property of magnitudes to conclude that the
|
|
|
|
|
magnitude of a (complex) reciprocal is the reciprocal of a (real) magnitude.
|
|
|
|
|
Figure <A HREF="#fig07.09">7.9</A> shows the situation graphically. The gain <IMG
|
|
|
|
|
WIDTH="26" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img30.png"
|
|
|
|
|
ALT="$\vert H\vert$"> is
|
|
|
|
|
the reciprocal of the length of the segment reaching from the point 1 to the
|
|
|
|
|
point <IMG
|
|
|
|
|
WIDTH="41" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img713.png"
|
|
|
|
|
ALT="$g{Z^{-d}}$">. Figure
|
|
|
|
|
<A HREF="#fig07.10">7.10</A> shows a graph of the
|
|
|
|
|
frequency response <IMG
|
|
|
|
|
WIDTH="26" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img30.png"
|
|
|
|
|
ALT="$\vert H\vert$"> as a function of the angular frequency
|
|
|
|
|
<!-- MATH
|
|
|
|
|
$\omega = \angle(Z)$
|
|
|
|
|
-->
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="69" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img678.png"
|
|
|
|
|
ALT="$\omega = \angle(Z)$">.
|
|
|
|
|
|
|
|
|
|
<P>
|
|
|
|
|
|
2022-04-12 23:32:40 -03:00
|
|
|
<DIV ALIGN="CENTER"><A ID="fig07.09"></A><A ID="7976"></A>
|
2022-04-12 21:54:18 -03:00
|
|
|
<TABLE>
|
|
|
|
|
<CAPTION ALIGN="BOTTOM"><STRONG>Figure 7.9:</STRONG>
|
|
|
|
|
Diagram in the complex plane for approximating the output gain <IMG
|
|
|
|
|
WIDTH="26" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img30.png"
|
|
|
|
|
ALT="$\vert H\vert$">
|
|
|
|
|
of the
|
|
|
|
|
recirculating comb filters at three different frequencies: 0, and the
|
|
|
|
|
arguments of two unit complex numbers <IMG
|
|
|
|
|
WIDTH="20" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img31.png"
|
|
|
|
|
ALT="$W$"> and <IMG
|
|
|
|
|
WIDTH="15" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img20.png"
|
|
|
|
|
ALT="$Z$">; <IMG
|
|
|
|
|
WIDTH="20" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img31.png"
|
|
|
|
|
ALT="$W$">
|
|
|
|
|
is chosen to give a gain about 3 dB below the peak.</CAPTION>
|
|
|
|
|
<TR><TD><IMG
|
|
|
|
|
WIDTH="339" HEIGHT="379" BORDER="0"
|
|
|
|
|
SRC="img723.png"
|
|
|
|
|
ALT="\begin{figure}\psfig{file=figs/fig07.09.ps}\end{figure}"></TD></TR>
|
|
|
|
|
</TABLE>
|
|
|
|
|
</DIV>
|
|
|
|
|
|
|
|
|
|
<P>
|
|
|
|
|
|
2022-04-12 23:32:40 -03:00
|
|
|
<DIV ALIGN="CENTER"><A ID="fig07.10"></A><A ID="7981"></A>
|
2022-04-12 21:54:18 -03:00
|
|
|
<TABLE>
|
|
|
|
|
<CAPTION ALIGN="BOTTOM"><STRONG>Figure 7.10:</STRONG>
|
|
|
|
|
Frequency response of the recirculating comb filter with <IMG
|
|
|
|
|
WIDTH="53" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img32.png"
|
|
|
|
|
ALT="$g = 0.8$">.
|
|
|
|
|
The peak
|
|
|
|
|
gain is <IMG
|
|
|
|
|
WIDTH="67" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img33.png"
|
|
|
|
|
ALT="$1/(1-g)$"> = 5. Peaks are much narrower than for the non-recirculating
|
|
|
|
|
comb filter.</CAPTION>
|
|
|
|
|
<TR><TD><IMG
|
|
|
|
|
WIDTH="312" HEIGHT="148" BORDER="0"
|
|
|
|
|
SRC="img724.png"
|
|
|
|
|
ALT="\begin{figure}\psfig{file=figs/fig07.10.ps}\end{figure}"></TD></TR>
|
|
|
|
|
</TABLE>
|
|
|
|
|
</DIV>
|
|
|
|
|
|
|
|
|
|
<P>
|
|
|
|
|
Figure <A HREF="#fig07.09">7.9</A> can be used to analyze how the frequency response
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="49" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img725.png"
|
|
|
|
|
ALT="$\vert H(\omega)\vert$"> should
|
|
|
|
|
behave
|
|
|
|
|
qualitatively as a function of <IMG
|
|
|
|
|
WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img29.png"
|
|
|
|
|
ALT="$g$">. The height and bandwidth of the peaks
|
|
|
|
|
both depend on <IMG
|
|
|
|
|
WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img29.png"
|
|
|
|
|
ALT="$g$">. The maximum value that <IMG
|
|
|
|
|
WIDTH="26" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img30.png"
|
|
|
|
|
ALT="$\vert H\vert$"> can attain is
|
|
|
|
|
when
|
|
|
|
|
<BR><P></P>
|
|
|
|
|
<DIV ALIGN="CENTER">
|
|
|
|
|
<!-- MATH
|
|
|
|
|
\begin{displaymath}
|
|
|
|
|
{Z^{-d}} = 1
|
|
|
|
|
\end{displaymath}
|
|
|
|
|
-->
|
|
|
|
|
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="56" HEIGHT="24" BORDER="0"
|
|
|
|
|
SRC="img726.png"
|
|
|
|
|
ALT="\begin{displaymath}
|
|
|
|
|
{Z^{-d}} = 1
|
|
|
|
|
\end{displaymath}">
|
|
|
|
|
</DIV>
|
|
|
|
|
<BR CLEAR="ALL">
|
|
|
|
|
<P></P>
|
|
|
|
|
This occurs at the frequencies <!-- MATH
|
|
|
|
|
$\omega = 0, 2\pi/d , 4\pi/d , \ldots$
|
|
|
|
|
-->
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="150" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img727.png"
|
|
|
|
|
ALT="$\omega = 0, 2\pi/d , 4\pi/d , \ldots$"> as in the
|
|
|
|
|
simple comb filter above. At these frequencies the gain reaches
|
|
|
|
|
<BR><P></P>
|
|
|
|
|
<DIV ALIGN="CENTER">
|
|
|
|
|
<!-- MATH
|
|
|
|
|
\begin{displaymath}
|
|
|
|
|
|H| = {1 \over {1 - g}}
|
|
|
|
|
\end{displaymath}
|
|
|
|
|
-->
|
|
|
|
|
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="80" HEIGHT="41" BORDER="0"
|
|
|
|
|
SRC="img728.png"
|
|
|
|
|
ALT="\begin{displaymath}
|
|
|
|
|
\vert H\vert = {1 \over {1 - g}}
|
|
|
|
|
\end{displaymath}">
|
|
|
|
|
</DIV>
|
|
|
|
|
<BR CLEAR="ALL">
|
|
|
|
|
<P></P>
|
|
|
|
|
|
|
|
|
|
<P>
|
|
|
|
|
The next important question is the bandwidth of the peaks in the frequency
|
|
|
|
|
response. So we would like to find sinusoids <IMG
|
|
|
|
|
WIDTH="29" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img729.png"
|
|
|
|
|
ALT="$W^n$">, with frequency
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="43" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img730.png"
|
|
|
|
|
ALT="$\angle(W)$">, giving rise to a value of <IMG
|
|
|
|
|
WIDTH="26" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img30.png"
|
|
|
|
|
ALT="$\vert H\vert$"> that is, say, 3 decibels below the
|
|
|
|
|
maximum. To do this, we return to Figure <A HREF="#fig07.09">7.9</A>, and try to place <IMG
|
|
|
|
|
WIDTH="20" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img31.png"
|
|
|
|
|
ALT="$W$">
|
|
|
|
|
so that the distance from the point 1 to the point <IMG
|
|
|
|
|
WIDTH="46" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img731.png"
|
|
|
|
|
ALT="$gW^{-d}$"> is about
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="24" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img732.png"
|
|
|
|
|
ALT="$\sqrt{2}$"> times the distance from 1 to <IMG
|
|
|
|
|
WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img29.png"
|
|
|
|
|
ALT="$g$"> (since <IMG
|
|
|
|
|
WIDTH="24" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img732.png"
|
|
|
|
|
ALT="$\sqrt{2}$">:1 is a ratio of
|
|
|
|
|
approximately 3 decibels).
|
|
|
|
|
|
|
|
|
|
<P>
|
|
|
|
|
We do this by arranging for the imaginary part of
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="46" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img731.png"
|
|
|
|
|
ALT="$gW^{-d}$"> to be roughly <IMG
|
|
|
|
|
WIDTH="39" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img733.png"
|
|
|
|
|
ALT="$1-g$"> or its negative, making a nearly isosceles right
|
|
|
|
|
triangle between the points 1, <IMG
|
|
|
|
|
WIDTH="39" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img733.png"
|
|
|
|
|
ALT="$1-g$">, and <IMG
|
|
|
|
|
WIDTH="46" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img731.png"
|
|
|
|
|
ALT="$gW^{-d}$">. (Here we're supposing that
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img29.png"
|
|
|
|
|
ALT="$g$"> is at least 2/3 or so; otherwise this approximation isn't very good). The
|
|
|
|
|
hypotenuse of a right isosceles triangle is always <IMG
|
|
|
|
|
WIDTH="24" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"
|
|
|
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|
SRC="img732.png"
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ALT="$\sqrt{2}$"> times the leg,
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and so the gain drops by that factor compared to its maximum.
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<P>
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We now make another approximation, that the imaginary part of <IMG
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WIDTH="46" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
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SRC="img731.png"
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ALT="$gW^{-d}$">
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is approximately the angle in radians it cuts from the real axis:
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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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\pm(1-g) \approx \mathrm{im}({gW^{-d}}) \approx \angle({W^{-d}})
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\end{displaymath}
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-->
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<IMG
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WIDTH="231" HEIGHT="28" BORDER="0"
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SRC="img734.png"
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ALT="\begin{displaymath}
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\pm(1-g) \approx \mathrm{im}({gW^{-d}}) \approx \angle({W^{-d}})
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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 the region of each peak reaching within 3 decibels of the maximum value
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is about
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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-g)/d
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\end{displaymath}
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-->
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<IMG
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WIDTH="63" HEIGHT="28" BORDER="0"
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SRC="img735.png"
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ALT="\begin{displaymath}
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(1-g)/d
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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 radians) to either side of the peak. The bandwidth narrows (and the filter
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peaks become sharper) as <IMG
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WIDTH="11" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img29.png"
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ALT="$g$"> approaches its maximum value of <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="$1$">.
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<P>
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As with the non-recirculating comb filter of Section <A HREF="node108.html#sect7.network">7.3</A>, the
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teeth of the comb are closer together for larger values of the delay <IMG
|
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WIDTH="11" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img28.png"
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ALT="$d$">. On
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the other hand, a delay of <IMG
|
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WIDTH="40" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
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SRC="img736.png"
|
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|
ALT="$d=1$"> (the shortest possible) gets only one tooth
|
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|
(at zero frequency) below 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 $"> (the next tooth, at
|
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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 $">, corresponds again to a frequency of zero by foldover).
|
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|
So the recirculating comb filter with <IMG
|
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|
WIDTH="40" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
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|
SRC="img736.png"
|
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|
ALT="$d=1$"> is just a low-pass filter.
|
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|
Delay networks
|
|
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|
|
with one-sample delays will be the basis for designing many other kinds of
|
|
|
|
|
digital
|
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|
|
|
filters in Chapter <A HREF="node127.html#chapter-filter">8</A>.
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<P>
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<HR>
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