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345 lines
8.4 KiB
345 lines
8.4 KiB
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original version by: Nikos Drakos, CBLU, University of Leeds
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<H2><A ID="SECTION001431000000000000000">
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Sawtooth wave</A>
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</H2>
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First we apply this to the sawtooth wave <IMG
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WIDTH="29" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1311.png"
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ALT="$s[n]$">. For <IMG
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WIDTH="77" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1312.png"
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ALT="$0 \le n < N$"> we have:
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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] - s[n-1] = -{1 \over N} +
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\left \{
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\begin{array}{ll}
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{1} & {n = 0} \\
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0 & \mbox{otherwise}
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\end{array}
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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="279" HEIGHT="45" BORDER="0"
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SRC="img1313.png"
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ALT="\begin{displaymath}
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s[n] - s[n-1] = -{1 \over N} +
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\left \{
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\begin{array}{ll}
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{1} & {n = 0} \\
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0 & \mbox{otherwise}
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\end{array} \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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Ignoring the constant offset of <IMG
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WIDTH="30" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
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SRC="img1314.png"
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ALT="$-{1 \over N}$">, this gives an
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<A ID="14349"></A><I>impulse</I>,
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zero everywhere except one sample per cycle. The summation in
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the Fourier transform only has one term, and we get:
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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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{\cal FT}\{ s[n] - s[n-1] \} (k) = 1 , \; k \neq 0, \; -N < k < N\
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\end{displaymath}
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-->
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<IMG
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WIDTH="344" HEIGHT="28" BORDER="0"
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SRC="img1315.png"
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ALT="\begin{displaymath}
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{\cal FT}\{ s[n] - s[n-1] \} (k) = 1 , \; k \neq 0, \; -N < k < 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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We then apply the difference formula backward to get:
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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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{\cal FT}\{ s[n] \} (k) \approx {1 \over {i \omega k}}
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= {{-iN} \over {2 \pi k}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="194" HEIGHT="39" BORDER="0"
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SRC="img1316.png"
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ALT="\begin{displaymath}
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{\cal FT}\{ s[n] \} (k) \approx {1 \over {i \omega k}}
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= {{-iN} \over {2 \pi k}}
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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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valid for integer 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="$k$">, small compared 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="$N$">, but with <IMG
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WIDTH="41" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1317.png"
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ALT="$k \neq 0$"> .
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(To get the second
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form of the expression we plugged in <!-- 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="$\omega=2\pi/N$"> and <IMG
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WIDTH="63" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1318.png"
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ALT="$1/i = -i$">.)
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<P>
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This analysis doesn't give us the DC component <!-- MATH
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${\cal FT}\{ s[n] \}(0)$
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-->
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<IMG
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WIDTH="91" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1319.png"
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ALT="${\cal FT}\{ s[n] \}(0)$">,
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because we would have had to divide by <IMG
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WIDTH="41" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img1092.png"
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ALT="$k=0$">. Instead, we can evaluate the DC
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term directly as the sum of all the points of the waveform: it's approximately
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zero by symmetry.
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<P>
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To get a Fourier series in terms of familiar real-valued sine and cosine functions,
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we combine corresponding terms for negative and positive 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="$k$">. The
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first harmonic (<IMG
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WIDTH="53" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1320.png"
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ALT="$k = \pm 1$">) 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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{1 \over N}\left [
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{\cal FT}\{ s[n] \} (1) \cdot {U^n} +
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{\cal FT}\{ s[n] \} (-1) \cdot {U^{-n}}
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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="313" HEIGHT="38" BORDER="0"
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SRC="img1321.png"
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ALT="\begin{displaymath}
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{1 \over N}\left [
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{\cal FT}\{ s[n] \} (1) \cdot {U^n} +
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{\cal FT}\{ s[n] \} (-1) \cdot {U^{-n}}
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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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\approx {{-i} \over {2 \pi}} \left [ U^n - U^{-n} \right ]
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\end{displaymath}
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-->
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<IMG
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WIDTH="122" HEIGHT="39" BORDER="0"
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SRC="img1322.png"
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ALT="\begin{displaymath}
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\approx {{-i} \over {2 \pi}} \left [ U^n - U^{-n} \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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= {{\sin ( \omega n)} \over {\pi}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="69" HEIGHT="40" BORDER="0"
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SRC="img1323.png"
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ALT="\begin{displaymath}
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= {{\sin ( \omega n)} \over {\pi}}
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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 similarly the <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="$k$">th harmonic 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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{{\sin ( k \omega n)} \over {k \pi}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="61" HEIGHT="40" BORDER="0"
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SRC="img1324.png"
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ALT="\begin{displaymath}
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{{\sin ( k \omega n)} \over {k \pi}}
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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 entire Fourier series 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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s[n] \approx {1 \over \pi} \left [
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{\sin ( \omega n )}
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+ {{\sin ( 2 \omega n)} \over 2}
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+ {{\sin ( 3 \omega n)} \over 3}
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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="332" HEIGHT="45" BORDER="0"
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SRC="img1325.png"
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ALT="\begin{displaymath}
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s[n] \approx {1 \over \pi} \left [
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{\sin ( \omega n )}
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+ ...
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...\over 2}
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+ {{\sin ( 3 \omega n)} \over 3}
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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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Miller Puckette
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2006-12-30
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