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Shah Function


\begin{displaymath}
\amalg\mkern-10.5mu\amalg (x)\equiv \sum_{n=-\infty}^\infty \delta(x-n)
\end{displaymath} (1)

where $\delta(x)$ is the Delta Function, so $\amalg\mkern-10.5mu\amalg (x)=0$ for $x\not\in \Bbb{Z}$ (i.e., $x$ not an Integer). The shah function obeys the identities
$\displaystyle \amalg\mkern-10.5mu\amalg (ax)$ $\textstyle =$ $\displaystyle {1\over a} \sum_{n=-\infty}^\infty \delta\left({x-{n\over a}}\right)$ (2)
$\displaystyle \amalg\mkern-10.5mu\amalg (-x)$ $\textstyle =$ $\displaystyle \amalg\mkern-10.5mu\amalg (x)$ (3)
$\displaystyle \amalg\mkern-10.5mu\amalg (x+n)$ $\textstyle =$ $\displaystyle \amalg\mkern-10.5mu\amalg (x),$ (4)

for $2n\in \Bbb{Z}$ (i.e., $n$ a half-integer).


It is normalized so that

\begin{displaymath}
\int_{n-1/2}^{n+1/2} \amalg\mkern-10.5mu\amalg (x)\,dx=1.
\end{displaymath} (5)

The ``sampling property'' is
\begin{displaymath}
\amalg\mkern-10.5mu\amalg (x)f(x)=\sum_{n=-\infty}^\infty f(n)\delta(x-n)
\end{displaymath} (6)

and the ``replicating property'' is
\begin{displaymath}
\amalg\mkern-10.5mu\amalg (x)*f(x)=\sum_{n=-\infty}^\infty f(x-n),
\end{displaymath} (7)

where $*$ denotes Convolution.

See also Convolution, Delta Function, Impulse Pair




© 1996-9 Eric W. Weisstein
1999-05-26