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<H2><A NAME="SECTION001312000000000000000"></A>
<A NAME="sect9-IFT"></A>
<BR>
Fourier transform as additive synthesis
</H2>
<P>
Now consider an arbitrary signal <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> that repeats every <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$">
samples. (Previously we
had assumed that <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> could be obtained as a sum of sinusoids, and we haven't
yet found out whether every periodic <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> can be obtained that way.) Let
<IMG
WIDTH="33" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1062.png"
ALT="$Y[k]$"> denote the Fourier transform of <IMG
WIDTH="17" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img670.png"
ALT="$X$"> for <!-- MATH
$k = 0, ..., N-1$
-->
<IMG
WIDTH="110" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img1063.png"
ALT="$k = 0, ..., N-1$">:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
Y[k] = {\cal FT}\left \{ X[n] \right \} (k)
\end{displaymath}
-->
<IMG
WIDTH="151" HEIGHT="28" BORDER="0"
SRC="img1064.png"
ALT="\begin{displaymath}
Y[k] = {\cal FT}\left \{ X[n] \right \} (k)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
= {{\left [ {U^{-k}} \right ]} ^ {0}} X[0] +
{{\left [ {U^{-k}} \right ]} ^ {1}} X[1] +
\cdots +
{{\left [ {U^{-k}} \right ]} ^ {N-1}} X[N-1]
\end{displaymath}
-->
<IMG
WIDTH="393" HEIGHT="30" BORDER="0"
SRC="img1065.png"
ALT="\begin{displaymath}
= {{\left [ {U^{-k}} \right ]} ^ {0}} X[0] +
{{\left [ {U^...
...X[1] +
\cdots +
{{\left [ {U^{-k}} \right ]} ^ {N-1}} X[N-1]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
= {{\left [ {U^{0}} \right ]} ^ {k}} X[0] +
{{\left [ {U^{-1}} \right ]} ^ {k}} X[1] +
\cdots +
{{\left [ {U^{-(N-1)}} \right ]} ^ {k}} X[N-1]
\end{displaymath}
-->
<IMG
WIDTH="395" HEIGHT="39" BORDER="0"
SRC="img1066.png"
ALT="\begin{displaymath}
= {{\left [ {U^{0}} \right ]} ^ {k}} X[0] +
{{\left [ {U^{...
...1] +
\cdots +
{{\left [ {U^{-(N-1)}} \right ]} ^ {k}} X[N-1]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
In the second version we rearranged the exponents to show that <IMG
WIDTH="33" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1062.png"
ALT="$Y[k]$"> is a sum
of complex sinusoids, with complex amplitudes <IMG
WIDTH="40" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1067.png"
ALT="$X[m]$"> and frequencies <IMG
WIDTH="40" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
SRC="img1068.png"
ALT="$-m\omega$">
for <!-- MATH
$m = 0, \ldots, N-1$
-->
<IMG
WIDTH="123" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img1069.png"
ALT="$m = 0, \ldots, N-1$">. In other words, <IMG
WIDTH="33" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1062.png"
ALT="$Y[k]$"> can be considered as a
Fourier series in its own right, whose <IMG
WIDTH="17" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img111.png"
ALT="$m$">th component has strength <IMG
WIDTH="53" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1070.png"
ALT="$X[-m]$">.
(The expression <IMG
WIDTH="53" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1070.png"
ALT="$X[-m]$"> makes sense because <IMG
WIDTH="17" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img670.png"
ALT="$X$"> is a periodic signal).
We can also express the amplitude of the partials of <IMG
WIDTH="33" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1062.png"
ALT="$Y[k]$"> in terms of its own
Fourier transform. Equating the two gives:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
{1 \over N} {\cal FT} \left \{ Y[k] \right \} (m) = X[-m]
\end{displaymath}
-->
<IMG
WIDTH="190" HEIGHT="38" BORDER="0"
SRC="img1071.png"
ALT="\begin{displaymath}
{1 \over N} {\cal FT} \left \{ Y[k] \right \} (m) = X[-m]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
This means in turn that <IMG
WIDTH="53" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1070.png"
ALT="$X[-m]$"> can be obtained by summing sinusoids with
amplitudes <IMG
WIDTH="55" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1072.png"
ALT="$Y[k]/N$">. Setting <IMG
WIDTH="60" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
SRC="img1073.png"
ALT="$n = -m$"> gives:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
X[n] = {1 \over N} {\cal FT} \left \{ Y[k] \right \} (-n)
\end{displaymath}
-->
<IMG
WIDTH="183" HEIGHT="38" BORDER="0"
SRC="img1074.png"
ALT="\begin{displaymath}
X[n] = {1 \over N} {\cal FT} \left \{ Y[k] \right \} (-n)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
= {{\left [ {U^{0}} \right ]} ^ {n}} Y[0] +
{{\left [ {U^{1}} \right ]} ^ {n}} Y[1] +
\cdots +
{{\left [ {U^{N-1}} \right ]} ^ {n}} Y[N-1]
\end{displaymath}
-->
<IMG
WIDTH="360" HEIGHT="30" BORDER="0"
SRC="img1075.png"
ALT="\begin{displaymath}
= {{\left [ {U^{0}} \right ]} ^ {n}} Y[0] +
{{\left [ {U^{...
...Y[1] +
\cdots +
{{\left [ {U^{N-1}} \right ]} ^ {n}} Y[N-1]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
This shows that any periodic <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> can indeed be obtained as a sum of
sinusoids. Further, the formula explicitly shows how to reconstruct <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$">
from its Fourier transform <IMG
WIDTH="33" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1062.png"
ALT="$Y[k]$">, if we know its value for the integers <!-- MATH
$k=0,
\ldots, N-1$
-->
<IMG
WIDTH="118" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img1076.png"
ALT="$k=0,
\ldots, N-1$">.
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<B> Next:</B> <A NAME="tex2html3060"
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<B> Up:</B> <A NAME="tex2html3054"
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<B> Previous:</B> <A NAME="tex2html3050"
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Miller Puckette
2006-12-30
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