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<H1><A ID="SECTION001420000000000000000">
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Dissecting classical waveforms</A>
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</H1>
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<P>
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Among the several conclusions we can draw from the even/odd harmonic
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decomposition of the sawtooth wave (Figure <A HREF="node186.html#fig10.02">10.2</A>), one is that a
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square wave can be decomposed into a linear combination of two sawtooth
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waves. We can carry this idea further, and show how to compose any classical
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waveform having only jumps (discontinuities in value) but no corners
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(discontinuities in slope) as a sum of sawtooth waves of various phases and
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amplitudes. We then develop the idea further, showing how to generate waveforms
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with corners (either in addition to, or instead of, jumps) using
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another elementary waveform we'll call the
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<A ID="14273"></A><I>parabolic wave</I>.
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<P>
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Suppose first that a waveform of period <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="$N$"> has discontinuities at <IMG
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WIDTH="10" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img769.png"
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ALT="$j$">
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different points, <!-- MATH
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${L_1}, \ldots, {L_j}$
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-->
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<IMG
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WIDTH="74" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1278.png"
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ALT="${L_1}, \ldots, {L_j}$">, all lying on the cycle between 0 and
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<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="$N$">, at which the waveform jumps by values <!-- MATH
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${d_1}, \ldots, {d_j}$
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-->
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<IMG
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WIDTH="69" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1279.png"
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ALT="${d_1}, \ldots, {d_j}$">. A negative
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value of <IMG
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WIDTH="18" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1280.png"
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ALT="$d_1$">, for instance, would mean that the waveform jumps from a higher
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to a lower value at the point <IMG
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WIDTH="21" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1281.png"
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ALT="${L_1}$">, and a positive value of <IMG
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WIDTH="18" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1280.png"
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ALT="$d_1$"> would mean
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a jump from a lower to a higher value.
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<P>
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<DIV ALIGN="CENTER"><A ID="fig10.03"></A><A ID="14282"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 10.3:</STRONG>
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Dissecting a waveform: (a) the original waveform with two
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discontinuities; (b and c) the two component sawtooth waves.</CAPTION>
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<TR><TD><IMG
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WIDTH="479" HEIGHT="293" BORDER="0"
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SRC="img1282.png"
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ALT="\begin{figure}\psfig{file=figs/fig10.03.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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For instance, Figure <A HREF="#fig10.03">10.3</A> (part a) shows a classical waveform with
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two jumps: <!-- MATH
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$({L_1}, {d_1}) = (0.3N, -0.3)$
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-->
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<IMG
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WIDTH="163" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1283.png"
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ALT="$({L_1}, {d_1}) = (0.3N, -0.3)$"> and <!-- MATH
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$({L_2}, {d_2}) = (0.6N, 1.3)$
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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="img1284.png"
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ALT="$({L_2}, {d_2}) = (0.6N, 1.3)$">.
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Parts (b) and (c) show sawtooth waves, each with one of the two jumps. The
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sum of the two sawtooth waves reconstructs the waveform of part (a), except for
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a possible constant (DC) offset.
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<P>
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The sawtooth wave with a jump of one unit at the point zero 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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s[n] = n/N - 1/2
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\end{displaymath}
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-->
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<IMG
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WIDTH="121" HEIGHT="28" BORDER="0"
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SRC="img1285.png"
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ALT="\begin{displaymath}
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s[n] = n/N - 1/2
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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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over the period <!-- MATH
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$0 \le n \le N-1$
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-->
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<IMG
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WIDTH="105" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1286.png"
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ALT="$0 \le n \le N-1$">, and repeats for other values of
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<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="$n$">. A sawtooth wave with a jump <IMG
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WIDTH="42" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1287.png"
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ALT="$(L, d)$"> is given by <!-- MATH
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$s'[n] = d s[n-L]$
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-->
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<IMG
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WIDTH="119" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1288.png"
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ALT="$s'[n] = d s[n-L]$">.
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The sum of all the component sawtooth waves 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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x[n] = {d_1} s[n - {L_1}] + \cdots + {d_j} s[n - {L_j}]
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\end{displaymath}
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-->
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<IMG
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WIDTH="260" HEIGHT="29" BORDER="0"
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SRC="img1289.png"
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ALT="\begin{displaymath}
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x[n] = {d_1} s[n - {L_1}] + \cdots + {d_j} s[n - {L_j}]
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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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The slopes of the segments of the waveform of part (a) of the figure are all
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the same, equal to the sum of the slopes of the component sawtooth waves:
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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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-{{{d_1} + \cdots + {d_j}} \over {N}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="101" HEIGHT="39" BORDER="0"
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SRC="img1290.png"
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ALT="\begin{displaymath}
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-{{{d_1} + \cdots + {d_j}} \over {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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Square and rectangle waves have horizontal line segments (slope zero); for
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this to happen in general the jumps must add to zero:
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<!-- MATH
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${d_1} + \cdots + {d_j} = 0$
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-->
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<IMG
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WIDTH="120" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1291.png"
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ALT="${d_1} + \cdots + {d_j} = 0$">.
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<P>
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<DIV ALIGN="CENTER"><A ID="fig10.04"></A><A ID="14301"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 10.4:</STRONG>
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The parabolic wave.</CAPTION>
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<TR><TD><IMG
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WIDTH="480" HEIGHT="109" BORDER="0"
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SRC="img1292.png"
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ALT="\begin{figure}\psfig{file=figs/fig10.04.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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To decompose classical waveforms with corners we use the parabolic wave, which,
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over a single period from 0 to <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="$N$">, is 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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p[n] = {1 \over 2} {{({n\over N} - {1\over 2})}^2} - {1 \over {24}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="167" HEIGHT="42" BORDER="0"
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SRC="img1293.png"
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ALT="\begin{displaymath}
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p[n] = {1 \over 2} {{({n\over N} - {1\over 2})}^2} - {1 \over {24}}
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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 shown in Figure <A HREF="#fig10.04">10.4</A>. It is a second-degree (quadratic) polynomial
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in the variable <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="$n$">, arranged so that it reaches a maximum halfway through the
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cycle at <IMG
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WIDTH="63" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1246.png"
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ALT="$n=N/2$">, the DC component is zero (or in other words, the average
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value over one cycle of the waveform is zero), and so that the slope changes
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discontinuously by <IMG
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WIDTH="45" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1294.png"
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ALT="$-1/N$"> at the beginning of the cycle.
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<P>
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To construct a waveform with any desired number of corners (suppose they are
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at the points <!-- MATH
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${M_i}, \ldots, {M_l}$
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-->
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<IMG
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WIDTH="79" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1295.png"
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ALT="${M_i}, \ldots, {M_l}$">, with slope changes equal to
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<!-- MATH
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${c_1}, \ldots, {c_l}$
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-->
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<IMG
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WIDTH="64" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img1296.png"
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ALT="${c_1}, \ldots, {c_l}$">), we sum up the necessary parabolic waves:
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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 c_1} p[n - {M_1}] - \cdots - {N c_l} p[n - {M_l}]
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\end{displaymath}
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-->
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<IMG
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WIDTH="305" HEIGHT="28" BORDER="0"
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SRC="img1297.png"
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ALT="\begin{displaymath}
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x[n] = {-N c_1} p[n - {M_1}] - \cdots - {N c_l} p[n - {M_l}]
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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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An example is shown graphically in Figure <A HREF="#fig10.05">10.5</A>.
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<P>
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<DIV ALIGN="CENTER"><A ID="fig10.05"></A><A ID="14320"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 10.5:</STRONG>
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Decomposing a triangle wave (part a) into two parabolic waves
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(b and c).</CAPTION>
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<TR><TD><IMG
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WIDTH="479" HEIGHT="271" BORDER="0"
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SRC="img1298.png"
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ALT="\begin{figure}\psfig{file=figs/fig10.05.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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If the sum <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="$x[n]$"> is to contain line segments (not segments of curves), the
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<IMG
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WIDTH="20" HEIGHT="16" ALIGN="BOTTOM" BORDER="0"
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SRC="img1299.png"
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ALT="$n^2$"> terms in the sum must sum to zero. From the expansion of <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="$x[n]$"> above,
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this implies that <!-- MATH
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${c_1} + \cdots + {c_l} = 0$
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-->
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<IMG
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WIDTH="115" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
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SRC="img1300.png"
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ALT="${c_1} + \cdots + {c_l} = 0$">. Sums obtained from
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existing classical
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waveforms (as in the figure) will always satisfy this condition because the
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changes in slope, over a cycle, must all add to zero for the waveform to
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connect with itself.
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<P>
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Miller Puckette
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2006-12-30
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