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<H2><A NAME="SECTION001321000000000000000">
Fourier transform of DC</A>
</H2>

<P>
Let <IMG
 WIDTH="65" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img1077.png"
 ALT="$X[n]=1$"> for all <IMG
 WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
 SRC="img75.png"
 ALT="$n$"> (this repeats with any desired integer period
<IMG
 WIDTH="47" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
 SRC="img1078.png"
 ALT="$N&gt;1$">).  From the preceding discussion, we expect to find that
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) = 
    \left \{
    \begin{array}{ll}
        N & {k=0} \\
        0 & {k=1, \ldots, N-1}
    \end{array}
    \right .
\end{displaymath}
 -->

<IMG
 WIDTH="287" HEIGHT="45" BORDER="0"
 SRC="img1079.png"
 ALT="\begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) =
\left \{
\begin{...
...}
N &amp; {k=0} \\
0 &amp; {k=1, \ldots, N-1}
\end{array} \right .
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
We will often need to know the answer for non-integer values of <IMG
 WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img58.png"
 ALT="$k$"> however,
and for this there is nothing better to do than to calculate the value
directly:
<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="16" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img1059.png"
 ALT="$V$"> is, as before, the unit magnitude complex number with argument
<IMG
 WIDTH="35" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
 SRC="img1055.png"
 ALT="$-k\omega$">.  This is a geometric series; as long as <IMG
 WIDTH="45" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
 SRC="img1080.png"
 ALT="$V \not= 1$"> we get:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) = 
    {{
        {V^N} - 1
    } \over {
        V - 1
    }}
\end{displaymath}
 -->

<IMG
 WIDTH="177" HEIGHT="43" BORDER="0"
 SRC="img1081.png"
 ALT="\begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) =
{{
{V^N} - 1
} \over {
V - 1
}}
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
We now symmetrize the top and bottom in the same way as we earlier did in
Section <A HREF="node108.html#sect7.network">7.3</A>.  To do this let:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
\xi = \cos(\pi k / N) - i \sin(\pi k / N)
\end{displaymath}
 -->

<IMG
 WIDTH="201" HEIGHT="28" BORDER="0"
 SRC="img1082.png"
 ALT="\begin{displaymath}
\xi = \cos(\pi k / N) - i \sin(\pi k / N)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
so that <IMG
 WIDTH="52" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
 SRC="img1083.png"
 ALT="${\xi^2} = V$">.  Then factoring appropriate powers of <IMG
 WIDTH="11" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
 SRC="img1084.png"
 ALT="$\xi$"> out of the
numerator and denominator gives:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) = 
        {\xi^{N-1}}
    {{
        {\xi^N} - {\xi^{-N}}
    } \over {
        \xi - {\xi^{-1}}
    }}
\end{displaymath}
 -->

<IMG
 WIDTH="230" HEIGHT="45" BORDER="0"
 SRC="img1085.png"
 ALT="\begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) =
{\xi^{N-1}}
{{
{\xi^N} - {\xi^{-N}}
} \over {
\xi - {\xi^{-1}}
}}
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
It's easy now to simplify the numerator:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
{\xi^N} - {\xi^{-N}} =
        \left (\cos(\pi k) - i \sin(\pi k) \right ) - 
        \left (\cos(\pi k) + i \sin(\pi k) \right )
        = - 2 i \sin(\pi k)
\end{displaymath}
 -->

<IMG
 WIDTH="493" HEIGHT="28" BORDER="0"
 SRC="img1086.png"
 ALT="\begin{displaymath}
{\xi^N} - {\xi^{-N}} =
\left (\cos(\pi k) - i \sin(\pi k) ...
...eft (\cos(\pi k) + i \sin(\pi k) \right )
= - 2 i \sin(\pi k)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
and similarly for the denominator, giving:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) = 
    \left ( {
        \parbox[t][0.1in]{0in}{\mbox{}}
        \cos(\pi k (N-1)/N) - i \sin(\pi k (N-1)/N)
    } \right ) 
    {{
        \sin(\pi k)
    } \over {
        \sin(\pi k / N)
    }}
\end{displaymath}
 -->

<IMG
 WIDTH="502" HEIGHT="44" BORDER="0"
 SRC="img1087.png"
 ALT="\begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) =
\left ( {
\parbo...
...
} \right )
{{
\sin(\pi k)
} \over {
\sin(\pi k / N)
}}
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
Whether <IMG
 WIDTH="45" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img1088.png"
 ALT="$V=1$"> or not, we have
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) = 
    \left ( {
        \parbox[t][0.1in]{0in}{\mbox{}}
        \cos(\pi k (N-1)/N) - i \sin(\pi k (N-1)/N) 
    } \right )
        {D_N}(k)
\end{displaymath}
 -->

<IMG
 WIDTH="473" HEIGHT="35" BORDER="0"
 SRC="img1089.png"
 ALT="\begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) =
\left ( {
\parbo...
...(\pi k (N-1)/N) - i \sin(\pi k (N-1)/N)
} \right )
{D_N}(k)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where <IMG
 WIDTH="49" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img1090.png"
 ALT="${D_N}(k)$">, known as the
<A NAME="12394"></A><I>Dirichlet kernel</I>,
is defined as
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
{D_N}(k) = 
    \left \{
    \begin{array}{ll}
        N & {k= 0} \\
            {{
                \sin(\pi k)
            } \over {
                \sin(\pi k / N)
            }}
        & {k\not=0,\; -N < k < N}
    \end{array}
    \right .
\end{displaymath}
 -->

<IMG
 WIDTH="307" HEIGHT="54" BORDER="0"
 SRC="img1091.png"
 ALT="\begin{displaymath}
{D_N}(k) =
\left \{
\begin{array}{ll}
N &amp; {k= 0} \\
{...
...pi k / N)
}}
&amp; {k\not=0,\; -N &lt; k &lt; N}
\end{array} \right .
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>

<P>
Figure <A HREF="#fig09.01">9.1</A> shows the Fourier transform of <IMG
 WIDTH="65" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img1077.png"
 ALT="$X[n]=1$">, with <IMG
 WIDTH="63" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img63.png"
 ALT="$N=100$">.  The
transform repeats every 100 samples, with a peak at <IMG
 WIDTH="41" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img1092.png"
 ALT="$k=0$">, another at
<IMG
 WIDTH="57" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img1093.png"
 ALT="$k=100$">, and so on.  The figure endeavors to show both the magnitude and phase
behavior using a 3-dimensional graph projected onto the page.  The phase
term
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{displaymath}
\cos(\pi k (N-1)/N) - i \sin(\pi k (N-1)/N)
\end{displaymath}
 -->

<IMG
 WIDTH="280" HEIGHT="28" BORDER="0"
 SRC="img1094.png"
 ALT="\begin{displaymath}
\cos(\pi k (N-1)/N) - i \sin(\pi k (N-1)/N)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
acts to twist the values of <!-- MATH
 ${\cal FT} \left \{ X[n] \right \} (k)$
 -->
<IMG
 WIDTH="104" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img1095.png"
 ALT="${\cal FT} \left \{ X[n] \right \} (k)$"> around
the <IMG
 WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img58.png"
 ALT="$k$"> axis with a period of approximately two.  The Dirichlet kernel
<IMG
 WIDTH="49" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img1090.png"
 ALT="${D_N}(k)$">, shown in Figure <A HREF="#fig09.02">9.2</A>, controls the magnitude of
<!-- MATH
 ${\cal FT} \left \{ X[n] \right \} (k)$
 -->
<IMG
 WIDTH="104" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img1095.png"
 ALT="${\cal FT} \left \{ X[n] \right \} (k)$">.  It has a peak, two units wide, around
<IMG
 WIDTH="41" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img1092.png"
 ALT="$k=0$">.  This is surrounded by one-unit-wide 
<A NAME="12409"></A><I>sidelobes</I>,
alternating in sign and gradually decreasing in magnitude as <IMG
 WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img58.png"
 ALT="$k$"> increases or
decreases away from zero.  The phase term rotates by almost <IMG
 WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
 SRC="img41.png"
 ALT="$\pi $"> radians
each time the Dirichlet kernel changes sign, so that the product of the
two stays roughly in the same complex half-plane for <IMG
 WIDTH="41" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
 SRC="img1096.png"
 ALT="$k&gt;1$"> (and in the
opposite half-plane for <IMG
 WIDTH="53" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
 SRC="img1097.png"
 ALT="$k &lt; -1$">).  The phase rotates by almost <IMG
 WIDTH="21" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
 SRC="img16.png"
 ALT="$2\pi $">
radians over the peak from <IMG
 WIDTH="53" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
 SRC="img1098.png"
 ALT="$k=-1$"> to <IMG
 WIDTH="41" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img259.png"
 ALT="$k=1$">.

<P>

<DIV ALIGN="CENTER"><A NAME="fig09.01"></A><A NAME="12413"></A>
<TABLE>
<CAPTION ALIGN="BOTTOM"><STRONG>Figure 9.1:</STRONG>
The Fourier transform of a signal consisting of all ones.  Here
N=100, and values are shown for <IMG
 WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img58.png"
 ALT="$k$"> ranging from -5 to 10.  The result
is complex-valued and shown as a projection, with the real axis pointing up the
page and the imaginary axis pointing away from it.</CAPTION>
<TR><TD><IMG
 WIDTH="470" HEIGHT="265" BORDER="0"
 SRC="img1099.png"
 ALT="\begin{figure}\psfig{file=figs/fig09.01.ps}\end{figure}"></TD></TR>
</TABLE>
</DIV>

<P>

<DIV ALIGN="CENTER"><A NAME="fig09.02"></A><A NAME="12418"></A>
<TABLE>
<CAPTION ALIGN="BOTTOM"><STRONG>Figure 9.2:</STRONG>
The Dirichlet kernel, for <IMG
 WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
 SRC="img3.png"
 ALT="$N$"> = 100.</CAPTION>
<TR><TD><IMG
 WIDTH="448" HEIGHT="175" BORDER="0"
 SRC="img1100.png"
 ALT="\begin{figure}\psfig{file=figs/fig09.02.ps}\end{figure}"></TD></TR>
</TABLE>
</DIV>

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