info prev up next book cdrom email home

Bartlett Function

\begin{figure}\begin{center}\BoxedEPSF{Bartlett.epsf scaled 825}\end{center}\end{figure}

The Apodization Function

f(x)=1-{\vert x\vert\over a}
\end{displaymath} (1)

which is a generalization of the one-argument Triangle Function. Its Full Width at Half Maximum is $a$. It has Instrument Function

$\displaystyle I(x)$ $\textstyle =$ $\displaystyle \int_{-a}^a e^{-2\pi ikx}\left({1-{\vert x\vert\over a}}\right)\,dx$  
  $\textstyle =$ $\displaystyle \int_{-a}^0 e^{-2\pi ikx}\left({1+{x\over a}}\right)\,dx+\int_0^a e^{-2\pi ikx}\left({1-{x\over a}}\right)\,dx.$ (2)

Letting $x'\equiv -x$ in the first part therefore gives
$\displaystyle \int_{-a}^0 e^{-2\pi ikx}\left({1+{x\over a}}\right)\,dx$ $\textstyle =$ $\displaystyle \int_a^0 e^{2\pi ikx'}\left({1-{x'\over a}}\right)(-dx')$  
  $\textstyle =$ $\displaystyle \int_0^a e^{2\pi i kx}\left({1-{x\over a}}\right)\,dx.$ (3)

Rewriting (2) using (3) gives
$\displaystyle I(x)$ $\textstyle =$ $\displaystyle (e^{2\pi ikx}+e^{-2\pi ikx})\left({1-{x\over a}}\right)\,dx$  
  $\textstyle =$ $\displaystyle 2\int_0^a \cos(2\pi kx)\left({1-{x\over a}}\right)\,dx.$ (4)

Integrating the first part and using the integral
\int x\cos(bx)\,dx={1\over b^2}\cos(bx)+{x\over b}\sin(bx)
\end{displaymath} (5)

for the second part gives

$\displaystyle I(x)$ $\textstyle =$ $\displaystyle 2\left[{{\sin(2\pi kx)\over 2\pi k}-{1\over a}\left\{{{1\over 4\pi^2 k^2}\cos(2\pi kx)+{x\over 2\pi k}\sin(2\pi kx)}\right\}}\right]_0^a$  
  $\textstyle =$ $\displaystyle 2\left\{{\left[{{\sin(2\pi ka)\over 2\pi k}-0}\right]-{1\over a}\...
...{\cos(2\pi ka)-1\over 4\pi^2 k^2}+{a\sin(2\pi ka)\over 2\pi k}}\right]}\right\}$  
  $\textstyle =$ $\displaystyle {1\over 2\pi^2 ak^2} [\cos(2\pi ka)-1] = a{\sin^2(\pi ka)\over \pi^2 k^2a^2}$  
  $\textstyle =$ $\displaystyle a\mathop{\rm sinc}\nolimits ^2(\pi ka),$ (6)

where $\mathop{\rm sinc}\nolimits x$ is the Sinc Function. The peak (in units of $a$) is 1. The function $I(x)$ is always positive, so there are no Negative sidelobes. The extrema are given by letting $\beta\equiv\pi ka$ and solving
{d\over d\beta} \left({\sin\beta\over\beta}\right)^2 = 2{\sin\beta\over\beta}{\sin\beta-\beta\cos\beta\over\beta^2}=0
\end{displaymath} (7)

\end{displaymath} (8)

\end{displaymath} (9)

\end{displaymath} (10)

Solving this numerically gives $\beta=4.49341$ for the first maximum, and the peak Positive sidelobe is 0.047190. The full width at half maximum is given by setting $x\equiv \pi ka$ and solving
\mathop{\rm sinc}\nolimits ^2 x={\textstyle{1\over 2}}
\end{displaymath} (11)

for $x_{1/2}$, yielding
x_{1/2}=\pi k_{1/2} a=1.39156.
\end{displaymath} (12)

Therefore, with $L\equiv 2a$,
{\rm FWHM}=2k_{1/2} ={0.885895\over a} ={1.77179\over L}.
\end{displaymath} (13)

See also Apodization Function, Parzen Apodization Function, Triangle Function


Bartlett, M. S. ``Periodogram Analysis and Continuous Spectra.'' Biometrika 37, 1-16, 1950.

info prev up next book cdrom email home

© 1996-9 Eric W. Weisstein