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<H2><A NAME="SECTION001323000000000000000">
Fourier transform of a sinusoid</A>
</H2>
<P>
We can use the phase shift formula above to find the Fourier transform of
any complex sinusoid <IMG
WIDTH="24" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img1115.png"
ALT="${Z^n}$"> with frequency <IMG
WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img7.png"
ALT="$\alpha $">, simply by setting
<IMG
WIDTH="65" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1077.png"
ALT="$X[n]=1$"> in the formula and using the Fourier transform for DC:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
{\cal FT} \left \{ {Z^n} \right \} (k) =
{\cal FT} \left \{ 1 \right \}(k - {{\alpha } \over {\omega}})
\end{displaymath}
-->
<IMG
WIDTH="218" HEIGHT="35" BORDER="0"
SRC="img1116.png"
ALT="\begin{displaymath}
{\cal FT} \left \{ {Z^n} \right \} (k) =
{\cal FT} \left \{ 1 \right \}(k - {{\alpha } \over {\omega}})
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
= \left [ \cos(\Phi(k)) + i \sin(\Phi(k))\right ]
{D_N}(k - {{\alpha } \over {\omega}})
\end{displaymath}
-->
<IMG
WIDTH="265" HEIGHT="35" BORDER="0"
SRC="img1117.png"
ALT="\begin{displaymath}
= \left [ \cos(\Phi(k)) + i \sin(\Phi(k))\right ]
{D_N}(k - {{\alpha } \over {\omega}})
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where <IMG
WIDTH="28" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img1118.png"
ALT="${D_N}$"> is the Dirichlet kernel and <IMG
WIDTH="15" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img1119.png"
ALT="$\Phi$"> is an ugly phase term:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
\Phi(k) = - \pi \cdot (k - {{\alpha } \over {\omega}}) \cdot (N-1)/N
\end{displaymath}
-->
<IMG
WIDTH="228" HEIGHT="35" BORDER="0"
SRC="img1120.png"
ALT="\begin{displaymath}
\Phi(k) = - \pi \cdot (k - {{\alpha } \over {\omega}}) \cdot (N-1)/N
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<P>
<DIV ALIGN="CENTER"><A NAME="fig09.03"></A><A NAME="12484"></A>
<TABLE>
<CAPTION ALIGN="BOTTOM"><STRONG>Figure 9.3:</STRONG>
Fourier transforms of complex sinusoids, with <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$"> = 100: (a)
with frequency <IMG
WIDTH="21" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img59.png"
ALT="$2\omega $"> ; (b) with frequency <IMG
WIDTH="34" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img60.png"
ALT="$1.5\omega $">. (The effect of
the phase winding term is not shown.)
</CAPTION>
<TR><TD><IMG
WIDTH="423" HEIGHT="384" BORDER="0"
SRC="img1121.png"
ALT="\begin{figure}\psfig{file=figs/fig09.03.ps}\end{figure}"></TD></TR>
</TABLE>
</DIV>
<P>
If the sinusoid's frequency <IMG
WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img7.png"
ALT="$\alpha $"> is an integer multiple of the fundamental
frequency <IMG
WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img27.png"
ALT="$\omega $">, the Dirichlet kernel is shifted to the left or right by an
integer. In this case the zero crossings of the Dirichlet kernel line up with
integer values of <IMG
WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img58.png"
ALT="$k$">, so that only one partial is nonzero. This is pictured
in Figure <A HREF="#fig09.03">9.3</A> (part a).
<P>
<DIV ALIGN="CENTER"><A NAME="fig09.04"></A><A NAME="12490"></A>
<TABLE>
<CAPTION ALIGN="BOTTOM"><STRONG>Figure 9.4:</STRONG>
A complex sinusoid with frequency <!-- MATH
$\alpha=1.5\omega=3\pi/N$
-->
<IMG
WIDTH="126" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img61.png"
ALT="$\alpha =1.5\omega =3\pi /N$">, forced to
repeat every <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$"> samples. (<IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$"> is arbitrarily set to 100; only the real
part is shown.)
</CAPTION>
<TR><TD><IMG
WIDTH="440" HEIGHT="136" BORDER="0"
SRC="img1122.png"
ALT="\begin{figure}\psfig{file=figs/fig09.04.ps}\end{figure}"></TD></TR>
</TABLE>
</DIV>
<P>
Part (b) shows the result when the frequency <IMG
WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img7.png"
ALT="$\alpha $"> falls halfway between two
integers. The partials have amplitudes falling off roughly as <IMG
WIDTH="28" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img72.png"
ALT="$1/k$"> in both
directions, measured from the actual frequency <IMG
WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img7.png"
ALT="$\alpha $">. That the energy
should be spread over many partials, when after all we started with a single
sinusoid, might seem surprising at first. However, as shown in Figure
<A HREF="#fig09.04">9.4</A>, the signal repeats at a period <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$"> which disagrees with the
frequency of the sinusoid. As a result there is a discontinuity at the
beginning of each period, and energy is flung over a wide range
of frequencies.
<P>
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<ADDRESS>
Miller Puckette
2006-12-30
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