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<H2><A NAME="SECTION001431000000000000000">
Sawtooth wave</A>
</H2>
First we apply this to the sawtooth wave <IMG
 WIDTH="29" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img1311.png"
 ALT="$s[n]$">.  For <IMG
 WIDTH="77" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
 SRC="img1312.png"
 ALT="$0 \le n &lt; N$"> we have:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
s[n] - s[n-1] = -{1 \over N} + 
    \left \{
    \begin{array}{ll}
        {1} & {n = 0} \\
        0 & \mbox{otherwise}
    \end{array}
    \right .
\end{displaymath}
 -->

<IMG
 WIDTH="279" HEIGHT="45" BORDER="0"
 SRC="img1313.png"
 ALT="\begin{displaymath}
s[n] - s[n-1] = -{1 \over N} +
\left \{
\begin{array}{ll}
{1} &amp; {n = 0} \\
0 &amp; \mbox{otherwise}
\end{array} \right .
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
Ignoring the constant offset of <IMG
 WIDTH="30" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img1314.png"
 ALT="$-{1 \over N}$">, this gives an
<A NAME="14349"></A><I>impulse</I>,
zero everywhere except one sample per cycle.  The summation in
the Fourier transform only has one term, and we get:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
{\cal FT}\{ s[n] - s[n-1] \} (k) = 1 , \; k \neq 0, \; -N < k < N\
\end{displaymath}
 -->

<IMG
 WIDTH="344" HEIGHT="28" BORDER="0"
 SRC="img1315.png"
 ALT="\begin{displaymath}
{\cal FT}\{ s[n] - s[n-1] \} (k) = 1 , \; k \neq 0, \; -N &lt; k &lt; N\
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
We then apply the difference formula backward to get:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
{\cal FT}\{ s[n] \} (k) \approx {1 \over {i \omega k}} 
    = {{-iN} \over {2 \pi k}}
\end{displaymath}
 -->

<IMG
 WIDTH="194" HEIGHT="39" BORDER="0"
 SRC="img1316.png"
 ALT="\begin{displaymath}
{\cal FT}\{ s[n] \} (k) \approx {1 \over {i \omega k}}
= {{-iN} \over {2 \pi k}}
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
valid for integer values of <IMG
 WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img58.png"
 ALT="$k$">, small compared to <IMG
 WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img3.png"
 ALT="$N$">, but with <IMG
 WIDTH="41" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
 SRC="img1317.png"
 ALT="$k \neq 0$"> .  
(To get the second
form of the expression we plugged in <!-- MATH
 $\omega = 2 \pi / N$
 -->
<IMG
 WIDTH="74" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img1047.png"
 ALT="$\omega=2\pi/N$"> and <IMG
 WIDTH="63" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img1318.png"
 ALT="$1/i = -i$">.)

<P>
This analysis doesn't give us the DC component <!-- MATH
 ${\cal FT}\{ s[n] \}(0)$
 -->
<IMG
 WIDTH="91" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img1319.png"
 ALT="${\cal FT}\{ s[n] \}(0)$">,
because we would have had to divide by <IMG
 WIDTH="41" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img1092.png"
 ALT="$k=0$">.  Instead, we can evaluate the DC
term directly as the sum of all the points of the waveform: it's approximately
zero by symmetry.

<P>
To get a Fourier series in terms of familiar real-valued sine and cosine functions,
we combine corresponding terms for negative and positive values of <IMG
 WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img58.png"
 ALT="$k$">.  The
first harmonic (<IMG
 WIDTH="53" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
 SRC="img1320.png"
 ALT="$k = \pm 1$">) is:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
{1 \over N}\left [
        {\cal FT}\{ s[n] \} (1) \cdot {U^n} + 
        {\cal FT}\{ s[n] \} (-1) \cdot {U^{-n}}
    \right ]
\end{displaymath}
 -->

<IMG
 WIDTH="313" HEIGHT="38" BORDER="0"
 SRC="img1321.png"
 ALT="\begin{displaymath}
{1 \over N}\left [
{\cal FT}\{ s[n] \} (1) \cdot {U^n} +
{\cal FT}\{ s[n] \} (-1) \cdot {U^{-n}}
\right ]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
\approx {{-i} \over {2 \pi}} \left [ U^n - U^{-n} \right ]
\end{displaymath}
 -->

<IMG
 WIDTH="122" HEIGHT="39" BORDER="0"
 SRC="img1322.png"
 ALT="\begin{displaymath}
\approx {{-i} \over {2 \pi}} \left [ U^n - U^{-n} \right ]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
= {{\sin ( \omega n)} \over {\pi}}
\end{displaymath}
 -->

<IMG
 WIDTH="69" HEIGHT="40" BORDER="0"
 SRC="img1323.png"
 ALT="\begin{displaymath}
= {{\sin ( \omega n)} \over {\pi}}
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
and similarly the <IMG
 WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img58.png"
 ALT="$k$">th harmonic is
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
{{\sin ( k \omega n)} \over {k \pi}}
\end{displaymath}
 -->

<IMG
 WIDTH="61" HEIGHT="40" BORDER="0"
 SRC="img1324.png"
 ALT="\begin{displaymath}
{{\sin ( k \omega n)} \over {k \pi}}
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
so the entire Fourier series is:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
s[n] \approx {1 \over \pi} \left [
        {\sin ( \omega n )}
        + {{\sin ( 2 \omega n)} \over 2}
        + {{\sin ( 3 \omega n)} \over 3}
        + \cdots
    \right ]
\end{displaymath}
 -->

<IMG
 WIDTH="332" HEIGHT="45" BORDER="0"
 SRC="img1325.png"
 ALT="\begin{displaymath}
s[n] \approx {1 \over \pi} \left [
{\sin ( \omega n )}
+ ...
...\over 2}
+ {{\sin ( 3 \omega n)} \over 3}
+ \cdots
\right ]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>

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<ADDRESS>
Miller Puckette
2006-12-30
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