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Uniform Apodization Function

\begin{figure}\begin{center}\BoxedEPSF{UniformApodization.epsf scaled 800}\end{center}\end{figure}

An Apodization Function

\begin{displaymath}
f(x)=1,
\end{displaymath} (1)

having Instrument Function
$\displaystyle I(x)$ $\textstyle =$ $\displaystyle \int_{-a}^a e^{-2\pi ikx}\,dx = -{1\over 2\pi ik}(e^{-2\pi ika}-e^{2\pi ikx})$  
  $\textstyle =$ $\displaystyle {\sin(2\pi ka)\over \pi k} = 2a\mathop{\rm sinc}\nolimits (2\pi ka).$ (2)

The peak (in units of $a$) is 2. The extrema are given by letting $\beta\equiv 2\pi ka$ and solving
\begin{displaymath}
{d\over d\beta} (\beta\sin\beta)={\sin\beta-\beta\cos\beta\over\beta^2}=0
\end{displaymath} (3)


\begin{displaymath}
\sin\beta-\beta\cos\beta=0
\end{displaymath} (4)


\begin{displaymath}
\tan\beta=\beta.
\end{displaymath} (5)

Solving this numerically gives $\beta_0=0$, $\beta_1=4.49341$, $\beta_2=7.72525$, ...for the first few solutions. The second of these is the peak Positive sidelobe, and the third is the peak Negative sidelobe. As a fraction of the peak, they are 0.128375 and $-0.217234$. The Full Width at Half Maximum is found by setting $I(x)=1$
\begin{displaymath}
\mathop{\rm sinc}\nolimits (x)={\textstyle{1\over 2}},
\end{displaymath} (6)

and solving for $x_{1/2}$, yielding
\begin{displaymath}
x_{1/2}=2\pi k_{1/2} a=1.89549.
\end{displaymath} (7)

Therefore, with $L\equiv 2a$,
\begin{displaymath}
{\rm FWHM}=2k_{1/2} = {0.603353\over a} ={1.20671\over L}.
\end{displaymath} (8)

See also Apodization Function




© 1996-9 Eric W. Weisstein
1999-05-26