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279 lines
8.1 KiB
279 lines
8.1 KiB
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original version by: Nikos Drakos, CBLU, University of Leeds
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* revised and updated by: Marcus Hennecke, Ross Moore, Herb Swan
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<H2><A ID="SECTION001323000000000000000">
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Fourier transform of a sinusoid</A>
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</H2>
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<P>
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We can use the phase shift formula above to find the Fourier transform of
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any complex sinusoid <IMG
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WIDTH="24" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img1115.png"
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ALT="${Z^n}$"> with frequency <IMG
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WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img7.png"
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ALT="$\alpha $">, simply by setting
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<IMG
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WIDTH="65" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1077.png"
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ALT="$X[n]=1$"> in the formula and using the Fourier transform for DC:
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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} \left \{ {Z^n} \right \} (k) =
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{\cal FT} \left \{ 1 \right \}(k - {{\alpha } \over {\omega}})
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\end{displaymath}
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-->
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<IMG
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WIDTH="218" HEIGHT="35" BORDER="0"
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SRC="img1116.png"
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ALT="\begin{displaymath}
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{\cal FT} \left \{ {Z^n} \right \} (k) =
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{\cal FT} \left \{ 1 \right \}(k - {{\alpha } \over {\omega}})
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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 [ \cos(\Phi(k)) + i \sin(\Phi(k))\right ]
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{D_N}(k - {{\alpha } \over {\omega}})
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\end{displaymath}
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-->
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<IMG
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WIDTH="265" HEIGHT="35" BORDER="0"
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SRC="img1117.png"
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ALT="\begin{displaymath}
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= \left [ \cos(\Phi(k)) + i \sin(\Phi(k))\right ]
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{D_N}(k - {{\alpha } \over {\omega}})
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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 <IMG
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WIDTH="28" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1118.png"
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ALT="${D_N}$"> is the Dirichlet kernel and <IMG
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WIDTH="15" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img1119.png"
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ALT="$\Phi$"> is an ugly phase term:
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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\begin{displaymath}
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\Phi(k) = - \pi \cdot (k - {{\alpha } \over {\omega}}) \cdot (N-1)/N
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\end{displaymath}
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-->
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<IMG
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WIDTH="228" HEIGHT="35" BORDER="0"
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SRC="img1120.png"
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ALT="\begin{displaymath}
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\Phi(k) = - \pi \cdot (k - {{\alpha } \over {\omega}}) \cdot (N-1)/N
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\end{displaymath}">
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</DIV>
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<BR CLEAR="ALL">
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<P></P>
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<P>
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<DIV ALIGN="CENTER"><A ID="fig09.03"></A><A ID="12484"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 9.3:</STRONG>
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Fourier transforms of complex sinusoids, with <IMG
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WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img3.png"
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ALT="$N$"> = 100: (a)
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with frequency <IMG
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WIDTH="21" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img59.png"
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ALT="$2\omega $"> ; (b) with frequency <IMG
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WIDTH="34" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img60.png"
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ALT="$1.5\omega $">. (The effect of
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the phase winding term is not shown.)
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</CAPTION>
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<TR><TD><IMG
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WIDTH="423" HEIGHT="384" BORDER="0"
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SRC="img1121.png"
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ALT="\begin{figure}\psfig{file=figs/fig09.03.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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If the sinusoid's frequency <IMG
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WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img7.png"
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ALT="$\alpha $"> is an integer multiple of the fundamental
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frequency <IMG
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WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img27.png"
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ALT="$\omega $">, the Dirichlet kernel is shifted to the left or right by an
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integer. In this case the zero crossings of the Dirichlet kernel line up with
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integer values of <IMG
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WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img58.png"
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ALT="$k$">, so that only one partial is nonzero. This is pictured
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in Figure <A HREF="#fig09.03">9.3</A> (part a).
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<P>
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<DIV ALIGN="CENTER"><A ID="fig09.04"></A><A ID="12490"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 9.4:</STRONG>
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A complex sinusoid with frequency <!-- MATH
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$\alpha=1.5\omega=3\pi/N$
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-->
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<IMG
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WIDTH="126" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img61.png"
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ALT="$\alpha =1.5\omega =3\pi /N$">, forced to
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repeat every <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="$N$"> samples. (<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 arbitrarily set to 100; only the real
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part is shown.)
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</CAPTION>
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<TR><TD><IMG
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WIDTH="440" HEIGHT="136" BORDER="0"
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SRC="img1122.png"
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ALT="\begin{figure}\psfig{file=figs/fig09.04.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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Part (b) shows the result when the frequency <IMG
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WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img7.png"
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ALT="$\alpha $"> falls halfway between two
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integers. The partials have amplitudes falling off roughly 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="$1/k$"> in both
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directions, measured from the actual frequency <IMG
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WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img7.png"
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ALT="$\alpha $">. That the energy
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should be spread over many partials, when after all we started with a single
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sinusoid, might seem surprising at first. However, as shown in Figure
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<A HREF="#fig09.04">9.4</A>, the signal repeats at a period <IMG
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WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img3.png"
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ALT="$N$"> which disagrees with the
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frequency of the sinusoid. As a result there is a discontinuity at the
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beginning of each period, and energy is flung over a wide range
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of frequencies.
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<P>
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HREF="node167.html">Properties of Fourier transforms</A>
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Miller Puckette
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2006-12-30
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