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<H1><A NAME="SECTION001310000000000000000">
Fourier analysis of periodic signals</A>
</H1>
<P>
Suppose <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> is a complex-valued signal that repeats every <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$"> samples. (We
are continuing to use complex-valued signals rather than real-valued ones
to simplify the mathematics.) Because of the period <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$">, the
values of <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> for <!-- MATH
$n=0,\ldots,N-1$
-->
<IMG
WIDTH="119" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img1042.png"
ALT="$n=0,\ldots,N-1$"> determine <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> for all integer values
of <IMG
WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img75.png"
ALT="$n$">.
<P>
Suppose further that <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> can be written as a sum of complex sinusoids of
frequency <IMG
WIDTH="11" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img179.png"
ALT="$0$">, <IMG
WIDTH="42" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img184.png"
ALT="$2\pi/N$">, <IMG
WIDTH="42" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1043.png"
ALT="$4\pi/N$">, <IMG
WIDTH="22" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img239.png"
ALT="$\ldots$">, <IMG
WIDTH="97" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1044.png"
ALT="$2(N-1)\pi/N$">. These are the
partials, starting with the zeroth, for a signal of period <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$">. We stop at
the <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$">th term because the next one would have frequency <IMG
WIDTH="21" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img16.png"
ALT="$2\pi $">, equivalent
to frequency <IMG
WIDTH="11" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img179.png"
ALT="$0$">, which is already on the list.
<P>
Given the values of <IMG
WIDTH="17" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img670.png"
ALT="$X$">, we wish to find the complex amplitudes of the
partials. Suppose we want the <IMG
WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img58.png"
ALT="$k$">th partial, where <IMG
WIDTH="77" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img1045.png"
ALT="$0 \leq k &lt; N$">. The
frequency of this partial is <IMG
WIDTH="51" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1046.png"
ALT="$2\pi k / N$">. We can find its complex amplitude
by modulating <IMG
WIDTH="17" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img670.png"
ALT="$X$"> downward <IMG
WIDTH="51" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1046.png"
ALT="$2\pi k / N$"> radians per sample in frequency, so
that the <IMG
WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img58.png"
ALT="$k$">th partial is modulated to frequency zero. Then we pass the signal
through a low-pass filter with such a low cutoff frequency that nothing but the
zero-frequency partial remains. We can do this in effect by averaging over a
huge number of samples; but since the signal repeats every <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$"> samples, this
huge average is the same as the average of the first <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$"> samples. In short, to
measure a sinusoidal component of a periodic signal, modulate it down to DC and
then average over one period.
<P>
Let <IMG
WIDTH="74" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1047.png"
ALT="$\omega=2\pi/N$"> be the fundamental frequency for the period <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$">, and
let <IMG
WIDTH="16" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img1048.png"
ALT="$U$"> be the unit-magnitude complex number with argument <IMG
WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img27.png"
ALT="$\omega $">:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
U = \cos(\omega) + i \sin(\omega)
\end{displaymath}
-->
<IMG
WIDTH="146" HEIGHT="28" BORDER="0"
SRC="img1049.png"
ALT="\begin{displaymath}
U = \cos(\omega) + i \sin(\omega)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
The <IMG
WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img58.png"
ALT="$k$">th partial of the signal <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> is of the form:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
{P_k}[n] = {A_k}{{\left [ {U^k} \right ]} ^ {n}}
\end{displaymath}
-->
<IMG
WIDTH="119" HEIGHT="30" BORDER="0"
SRC="img1050.png"
ALT="\begin{displaymath}
{P_k}[n] = {A_k}{{\left [ {U^k} \right ]} ^ {n}}
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where <IMG
WIDTH="23" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img1051.png"
ALT="${A_k}$"> is the complex amplitude of the partial, and the frequency
of the partial is:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
\angle({U^k}) = k \angle(U) = k\omega
\end{displaymath}
-->
<IMG
WIDTH="146" HEIGHT="28" BORDER="0"
SRC="img1052.png"
ALT="\begin{displaymath}
\angle({U^k}) = k \angle(U) = k\omega
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
We're assuming for the moment that the signal <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> can actually be written
as a sum of the <IMG
WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img75.png"
ALT="$n$"> partials, or in other words:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
X[n] =
{A_0}{{\left [ {U^0} \right ]} ^ {n}}
+ {A_1}{{\left [ {U^1} \right ]} ^ {n}}
+ \cdots
+ {A_{N-1}}{{\left [ {U^{N-1}} \right ]} ^ {n}}
\end{displaymath}
-->
<IMG
WIDTH="355" HEIGHT="30" BORDER="0"
SRC="img1053.png"
ALT="\begin{displaymath}
X[n] =
{A_0}{{\left [ {U^0} \right ]} ^ {n}}
+ {A_1}{{\l...
...n}}
+ \cdots
+ {A_{N-1}}{{\left [ {U^{N-1}} \right ]} ^ {n}}
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
By the heterodyne-filtering argument above, we expect to be able to measure
each <IMG
WIDTH="23" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img1054.png"
ALT="$A_k$"> by multiplying by the sinusoid of frequency <IMG
WIDTH="35" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img1055.png"
ALT="$-k\omega$"> and
averaging over a period:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
{A_k} = {1\over N} \left (
{{\left [ {U^{-k}} \right ]} ^ {0}} X[0] +
{{\left [ {U^{-k}} \right ]} ^ {1}} X[1] +
\cdots +
{{\left [ {U^{-k}} \right ]} ^ {N-1}} X[N-1]
\right )
\end{displaymath}
-->
<IMG
WIDTH="457" HEIGHT="38" BORDER="0"
SRC="img1056.png"
ALT="\begin{displaymath}
{A_k} = {1\over N} \left (
{{\left [ {U^{-k}} \right ]} ^ ...
...dots +
{{\left [ {U^{-k}} \right ]} ^ {N-1}} X[N-1]
\right )
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
This is such a useful formula that it gets its own notation. The
<A NAME="12325"></A><I>Fourier transform</I>
of a signal <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$">, over <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$"> samples, is defined as:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
{\cal FT}\left \{ X[n] \right \} (k) =
{V ^ {0}} X[0] +
{V ^ {1}} X[1] +
\cdots +
{V ^ {N-1}} X[N-1]
\end{displaymath}
-->
<IMG
WIDTH="406" HEIGHT="28" BORDER="0"
SRC="img1057.png"
ALT="\begin{displaymath}
{\cal FT}\left \{ X[n] \right \} (k) =
{V ^ {0}} X[0] +
{V ^ {1}} X[1] +
\cdots +
{V ^ {N-1}} X[N-1]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where <IMG
WIDTH="67" HEIGHT="17" ALIGN="BOTTOM" BORDER="0"
SRC="img1058.png"
ALT="$V = {U^{-k}}$">. The Fourier transform is a function of the variable <IMG
WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img58.png"
ALT="$k$">,
equal to <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$"> times the amplitude of the input's <IMG
WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img58.png"
ALT="$k$">th partial. So far <IMG
WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img58.png"
ALT="$k$">
has taken integer values but the formula makes sense for any value of <IMG
WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img58.png"
ALT="$k$"> if we
define <IMG
WIDTH="16" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img1059.png"
ALT="$V$"> more generally as:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
V = \cos(-k\omega) + i\sin(-k\omega)
\end{displaymath}
-->
<IMG
WIDTH="188" HEIGHT="28" BORDER="0"
SRC="img1060.png"
ALT="\begin{displaymath}
V = \cos(-k\omega) + i\sin(-k\omega)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where, as before, <IMG
WIDTH="74" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1047.png"
ALT="$\omega=2\pi/N$"> is the (angular) fundamental
frequency associated with the period <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$">.
<P>
<BR><HR>
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<LI><A NAME="tex2html3033"
HREF="node165.html">Periodicity of the Fourier transform</A>
<LI><A NAME="tex2html3034"
HREF="node166.html">Fourier transform as additive synthesis</A>
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<B> Next:</B> <A NAME="tex2html3032"
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<ADDRESS>
Miller Puckette
2006-12-30
</ADDRESS>
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