Shah Function

$$\amalg\mkern-10.5mu\amalg (x)\equiv \sum_{n=-\infty}^\infty \delta(x-n)$$ (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

$$\amalg\mkern-10.5mu\amalg (ax) = {1\over a} \sum_{n=-\infty}^\infty \delta\left({x-{n\over a}}\right)$$ (2)

$$\amalg\mkern-10.5mu\amalg (-x) = \amalg\mkern-10.5mu\amalg (x)$$ (3)

$$\amalg\mkern-10.5mu\amalg (x+n) = \amalg\mkern-10.5mu\amalg (x),$$ (4)

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

It is normalized so that

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

The ``sampling property'' is

$$\amalg\mkern-10.5mu\amalg (x)f(x)=\sum_{n=-\infty}^\infty f(n)\delta(x-n)$$ (6)

and the ``replicating property'' is

$$\amalg\mkern-10.5mu\amalg (x)*f(x)=\sum_{n=-\infty}^\infty f(x-n),$$ (7)

where $*$ denotes Convolution.

See also Convolution, Delta Function, Impulse Pair