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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>
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<LI><A NAME="tex2html3033"
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<LI><A NAME="tex2html3034"
  HREF="node166.html">Fourier transform as additive synthesis</A>
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