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original version by: Nikos Drakos, CBLU, University of Leeds
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HREF="node193.html">Predicting and controlling foldover</A>
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HREF="node188.html">Fourier series of the</A>
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HREF="node191.html">Square and symmetric triangle</A>
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HREF="node4.html">Contents</A></B>
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<B> <A ID="tex2html3439"
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HREF="node201.html">Index</A></B>
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<H2><A ID="SECTION001434000000000000000"></A>
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<A ID="sect10.gentriangle"></A>
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<BR>
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General (non-symmetric) triangle wave
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</H2>
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<P>
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2022-04-12 23:32:40 -03:00
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<DIV ALIGN="CENTER"><A ID="fig10.07"></A><A ID="14434"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 10.7:</STRONG>
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Non-symmetric triangle wave, with vertices at <IMG
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WIDTH="47" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img69.png"
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ALT="$(M, 1)$"> and
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<IMG
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WIDTH="93" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img70.png"
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ALT="$(N-M, -1)$">.</CAPTION>
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<TR><TD><IMG
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WIDTH="474" HEIGHT="123" BORDER="0"
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SRC="img1342.png"
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ALT="\begin{figure}\psfig{file=figs/fig10.07.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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A general, non-symmetric triangle wave appears in Figure <A HREF="#fig10.07">10.7</A>. Here
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we have arranged the cycle so that, first, the DC component is zero (so that
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the two corners have equal and opposite heights), and second, so that the
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midpoint of the shorter segment goes through the point <IMG
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WIDTH="38" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1343.png"
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ALT="$(0,0)$">.
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<P>
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The two line segments have slopes equal to <IMG
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WIDTH="35" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1344.png"
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ALT="$1/M$"> and <IMG
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WIDTH="103" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1345.png"
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ALT="$-2/(N-2M)$">, so the
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decomposition into component parabolic waves is given by:
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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] = {{N^2} \over {MN - 2{M^2}}} (p[n-M] - p[n+M])
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\end{displaymath}
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-->
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<IMG
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WIDTH="291" HEIGHT="42" BORDER="0"
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SRC="img1346.png"
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ALT="\begin{displaymath}
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x[n] = {{N^2} \over {MN - 2{M^2}}} (p[n-M] - p[n+M])
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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 we're using the periodicity of <IMG
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WIDTH="29" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1347.png"
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ALT="$p[n]$"> to replace <IMG
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WIDTH="112" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1348.png"
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ALT="$p[n-(N-M)]$"> by
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<IMG
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WIDTH="66" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1349.png"
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ALT="$p[n+M]$">).)
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<P>
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The most general way of dealing with linear combinations of elementary
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(parabolic and/or sawtooth) waves is to go back to the complex Fourier
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series, as
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we did in finding the series for the elementary waves themselves. But in this
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particular case we can use a trigonometric identity to avoid the extra work of
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converting back and forth. First we plug in the real-valued Fourier series:
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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] = {{N^2} \over {2{\pi ^ 2} (MN - 2{M^2})}} \left [
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\parbox[t][0.12in]{0in}{\mbox{}}
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{\cos ( \omega (n-M))} - {\cos ( \omega (n+M))}
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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="407" HEIGHT="45" BORDER="0"
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SRC="img1350.png"
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ALT="\begin{displaymath}
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x[n] = {{N^2} \over {2{\pi ^ 2} (MN - 2{M^2})}} \left [
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\p...
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...{}}
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{\cos ( \omega (n-M))} - {\cos ( \omega (n+M))}
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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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\left .
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+ {{\cos ( 2 \omega (n-M)) - \cos ( 2 \omega (n+M))} \over 4}
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+ \cdots
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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="img1351.png"
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ALT="\begin{displaymath}
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\left .
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+ {{\cos ( 2 \omega (n-M)) - \cos ( 2 \omega (n+M))} \over 4}
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+ \cdots
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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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Now we use the identity,
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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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\cos(a) - \cos(b) = 2 \sin({{b-a}\over 2}) \sin({{a+b}\over 2})
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\end{displaymath}
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-->
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<IMG
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WIDTH="275" HEIGHT="39" BORDER="0"
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SRC="img1352.png"
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ALT="\begin{displaymath}
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\cos(a) - \cos(b) = 2 \sin({{b-a}\over 2}) \sin({{a+b}\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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so that, for example,
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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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{\cos ( \omega (n-M))} - {\cos ( \omega (n+M))} =
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2 \sin (2 \pi M / N) \sin ( \omega n)
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\end{displaymath}
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-->
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<IMG
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WIDTH="396" HEIGHT="28" BORDER="0"
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SRC="img1353.png"
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ALT="\begin{displaymath}
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{\cos ( \omega (n-M))} - {\cos ( \omega (n+M))} =
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2 \sin (2 \pi M / N) \sin ( \omega 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 again we used the definition of <!-- MATH
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$\omega = 2 \pi / N$
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-->
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<IMG
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WIDTH="74" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1047.png"
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ALT="$\omega=2\pi/N$">.) This is a
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simplification since the first sine term does not depend on <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="$n$">; it's
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just an amplitude term. Applying the identity to all the terms of the expansion
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for <IMG
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WIDTH="31" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img80.png"
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ALT="$x[n]$"> 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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x[n] = a[1] \sin(\omega n) + a[2] \sin(2 \omega n) + \cdots
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\end{displaymath}
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-->
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<IMG
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WIDTH="271" HEIGHT="28" BORDER="0"
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SRC="img1354.png"
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ALT="\begin{displaymath}
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x[n] = a[1] \sin(\omega n) + a[2] \sin(2 \omega n) + \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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where the amplitudes of the components are given by:
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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[k] = {1 \over {{\pi ^ 2} (M/N - 2{{(M/N)}^2})}}
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\cdot {{\sin (2 \pi k M / N) } \over {k^2}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="311" HEIGHT="47" BORDER="0"
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SRC="img1355.png"
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ALT="\begin{displaymath}
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a[k] = {1 \over {{\pi ^ 2} (M/N - 2{{(M/N)}^2})}}
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\cdot {{\sin (2 \pi k M / N) } \over {k^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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Notice that the result does not depend separately on the values of <IMG
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WIDTH="20" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img86.png"
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ALT="$M$"> and <IMG
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WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img3.png"
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ALT="$N$">,
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but only on their ratio, <IMG
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WIDTH="41" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1356.png"
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ALT="$M/N$"> (this is not surprising because the shape of
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the waveform depends on this ratio). If we look at small values of <IMG
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WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img58.png"
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ALT="$k$">:
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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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k < {{1} \over {4 M/N}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="77" HEIGHT="42" BORDER="0"
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SRC="img1357.png"
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ALT="\begin{displaymath}
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k < {{1} \over {4 M/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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the argument of the sine function is less than <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="$\pi /2$"> and using the
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approximation <!-- MATH
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$\sin(\theta) \approx \theta$
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-->
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<IMG
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WIDTH="72" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1358.png"
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ALT="$\sin(\theta) \approx \theta$"> we find that <IMG
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WIDTH="29" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1359.png"
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ALT="$a[k]$"> drops off
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as <IMG
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WIDTH="28" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img72.png"
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ALT="$1/k$">, just as the partials of a sawtooth wave. But for larger values of
|
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<IMG
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WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img58.png"
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ALT="$k$"> the sine term oscillates between 1 and -1, so that the amplitudes drop off
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|
irregularly as <IMG
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WIDTH="35" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
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SRC="img73.png"
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ALT="$1/{k^2}$">.
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<P>
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|
Figure <A HREF="node194.html#fig10.08">10.8</A> shows the partial strengths with <IMG
|
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|
WIDTH="41" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1356.png"
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ALT="$M/N$"> set to
|
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|
0.03; here, our prediction is that the <IMG
|
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WIDTH="28" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
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|
SRC="img72.png"
|
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|
ALT="$1/k$"> dependence should extend to
|
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|
<!-- MATH
|
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|
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|
$k \approx 1/(4\cdot 0.03) \approx 8.5$
|
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|
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|
-->
|
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|
<IMG
|
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|
WIDTH="151" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
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|
SRC="img1360.png"
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ALT="$k \approx 1/(4\cdot 0.03) \approx 8.5$">, in rough agreement with the figure.
|
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|
<P>
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|
Another way to see why the partials should behave as <IMG
|
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|
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|
WIDTH="28" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img72.png"
|
|
|
|
|
ALT="$1/k$"> for low values of
|
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|
|
<IMG
|
|
|
|
|
WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img58.png"
|
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|
ALT="$k$"> and <IMG
|
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|
|
|
WIDTH="35" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img73.png"
|
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|
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|
ALT="$1/{k^2}$"> thereafter, is to compare the period of a given partial with
|
|
|
|
|
the length of the short segment, <IMG
|
|
|
|
|
WIDTH="28" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img1361.png"
|
|
|
|
|
ALT="$2M$">. For partials numbering less than
|
|
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|
|
<IMG
|
|
|
|
|
WIDTH="49" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
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|
|
SRC="img1362.png"
|
|
|
|
|
ALT="$N/4M$">, the period is at least twice the length of the short segment, and at
|
|
|
|
|
that scale the waveform is nearly indistinguishable from a sawtooth wave.
|
|
|
|
|
For partials numbering in excess of <IMG
|
|
|
|
|
WIDTH="49" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img1363.png"
|
|
|
|
|
ALT="$N/2M$">, the two corners of the triangle
|
|
|
|
|
wave are at least one period apart, and at these higher frequencies the two
|
|
|
|
|
corners (each with <IMG
|
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|
|
|
WIDTH="35" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img73.png"
|
|
|
|
|
ALT="$1/{k^2}$"> frequency dependence) are resolved from each
|
|
|
|
|
other. In the figure, the notch at partial 17 occurs at the wavelength
|
|
|
|
|
<!-- MATH
|
|
|
|
|
$N/2M \approx 1/17$
|
|
|
|
|
-->
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="102" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img1364.png"
|
|
|
|
|
ALT="$N/2M \approx 1/17$">, at which wavelength the two corners are one cycle apart;
|
|
|
|
|
since the corners are opposite in sign they cancel each other.
|
|
|
|
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|
<P>
|
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|
<HR>
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