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Parabolic wave

The same analysis, with some differences in sign and normalization, works for parabolic waves. First we compute the difference:

\begin{displaymath}
p[n] - p[n-1] = {
{
{{({n\over N} - {1\over 2})}^2} -
{{({{n-1}\over N} - {1\over 2})}^2}
} \over {
2
}}
\end{displaymath}


\begin{displaymath}
= {
{
{{({n\over N} - {N\over {2N}})}^2} -
{{({{n}\over N} - {{N - 2}\over {2N}})}^2}
} \over {
2
}}
\end{displaymath}


\begin{displaymath}
= {
{
{{{2n}\over {N^2}} - {1\over {N}}} + {1\over {N^2}}
} \over {
2
}}
\end{displaymath}


\begin{displaymath}
\approx - s[n] / N .
\end{displaymath}

So (again for $k \neq 0$, small compared to $N$) we get:

\begin{displaymath}
{\cal FT}\{ p[n] \} (k) \approx
{{-1} \over N} \cdot {{-iN} \over {2 \pi k}} \cdot {\cal FT}\{ s[n] \} (k)
\end{displaymath}


\begin{displaymath}
\approx {{-1} \over N} \cdot {{-iN} \over {2 \pi k}}
\cdot {{-iN} \over {2 \pi k}}
\end{displaymath}


\begin{displaymath}
= {N \over {4 {\pi ^2} {k^2}}}
\end{displaymath}

and as before we get the Fourier series:

\begin{displaymath}
p[n] \approx {1 \over {2 {\pi^2}}} \left [
{\cos ( \omega ...
...\over 4}
+ {{\cos ( 3 \omega n)} \over 9}
+ \cdots
\right ]
\end{displaymath}



Miller Puckette 2006-09-24