I disagree with the assertion in the answer by user7090 that the calculation in question is not a convolution; it is indeed a convolution. Let us consider the case $D=2$. Ignoring $Z$ for now,
the five terms in that sum
$\displaystyle \sum_{k = i-2}^{i+2} x(t_k)e^{-|i-k|/\Delta}$ on the left in the OP's question are
$$\begin{align}
x(t_{i-2})e^{-|i - (i-2))|/\Delta} &= x(t_{i-2})e^{-2/\Delta}\\
x(t_{i-1})e^{-|i - (i-1))|/\Delta} &= x(t_{i-1})e^{-1/\Delta}\\
x(t_{i-0})e^{-|i - (i-0))|/\Delta} &= x(t_i)e^{-0/\Delta} = x(t_i)\\
x(t_{i+1})e^{-|i - (i+1))|/\Delta} &= x(t_{i+1})e^{-1/\Delta}\\
x(t_{i+2})e^{-|i - (i-2))|/\Delta} &= x(t_{i+2})e^{-2/\Delta}
\end{align}$$
and so the sum is very much a convolution of the sequence $x$
whose $i$-th term is $x(t_i)$
with a length-$5$ sequence $h$ whose $n$-th term is $e^{-|n|/\Delta}$
for $|n| \leq 2$ in this case (and a length-$(2D+1)$ sequence
with $|n| \leq D$ more generally).
What you, the OP, need to understand is the following (and don't
get bogged down in words like convolution). You are doing a
smoothing of the data that you have collected. Each $x(t_i)$ of the
original data is being replaced by a $\tilde{x}(t_i)$ which is formed
as a weighted sum of $x(t_i)$ and its immediate neighbors. The
weights assigned to $x(t_i)$ is the largest, the weight assigned to
its nearest neighbors $x(t_{i+1})$ and $x(t_{i-1})$ are smaller (but equal),
the weights assigned to slightly more distant neighbors $x(t_{i+2})$ and $x(t_{i-2})$ is yet smaller, and so on, till the we get to $x(t_{i+D})$ and $x(t_{i-D})$ which get the smallest weights of all. Anything more
than $D$ away gets a weight of $0$. The weights in question are
normalized in the sense that the total weight is $1$, but let
us look at the raw weights which are alleged to be those $e^{-|i-k|/\Delta}$.
What you really have (before normalization of the weights) is
that sum (let us use $D = 3$ for simplicity) which is
$x(t_i) + \left.\left.e^{-1/\Delta}\right[x(t_{i+1}) + x(t_{i-1)}\right]
+ \left.\left.e^{-2/\Delta}\right[x(t_{i+2}) + x(t_{i-2})\right]
+ \left.\left.e^{-3/\Delta}\right[x(t_{i+3}) + x(t_{i-3})\right]$
$= e^{-3/\Delta}x(t_{i-3}) + e^{-2/\Delta}x(t_{i-2}) +
e^{-1/\Delta}x(t_{i-1}) + x(t_i) + e^{-1/\Delta}x(t_{i+1}) + e^{-2/\Delta}x(t_{i+2})
+ e^{-3/\Delta}x(t_{i+3})\quad (1)$
where on the right side we have the elements occurring in the
natural order, flanking the central element $x(t_i)$ on both sides.
Now, the sum of the raw weights exhibited in $(1)$ is
$$e^{-3/\Delta} + e^{-2/\Delta} +
e^{-1/\Delta} + 1 + e^{-1/\Delta} + e^{-2/\Delta}
+ e^{-3/\Delta} \tag{2}$$
which is just the element $Z$ in the question, and so the
normalized weights are just the ones exhibited in $(2)$
divided by $Z$. A better way of writing $Z$ might be
$$Z = \sum_{k=-D}^D e^{-|k|/\Delta} = 1 + 2\sum_{k=1}^De^{-k/\Delta}
\tag{3}$$
and perhaps a better way of expressing the first sum might be
$$\tilde{x}(t_i) = \frac{1}{Z}\sum_{k=-D}^D x(t_{i-k})e^{-|k|/\Delta}$$
which expresses more clearly that we are forming a weighted sum
of $x(t_i)$ and its nearest neighbors and that the weights assigned
to neighbors at distance $k$ decrease exponentially as a function
of $k$ (here $k \geq 0$ of course since we are not distinguishing
between the directions of separation).
