I'm trying to compute the Dirac delta of a function $d(t)$ in order to evaluate its Fourier transform, however I am not sure about the result. The function is: $$ \delta[d(t)] = \delta \left[ \left(\sum_{k \in \mathbb{Z}} \mathrm{tri}_4(t - 12k) \right) - \frac{1}{2} \right] $$
I know that, defining $t_k$ as the $k$-th simple root of $d(t)$ and for $d'(t_k) \neq 0 \ \forall t_k$, it is:
$$
\delta[ d(t) ] = \sum_{k \in \mathbb{Z}} \frac{1}{\left | d'(t_k) \right |} \delta(t - t_k)
$$
In order to find the roots of $d(t)$, I have plotted it
and I have also written its analytical expression:
$$
\mathrm{tri}_4(t - 12k) - \frac{1}{2} =
\begin{cases}
\frac{1}{2} - \frac{\left| t - 12k \right|}{4} \quad &|t-12k| < 4 \\
-\frac{1}{2} &\mathrm{otherwise}
\end{cases}
$$
The derivative of $d(t)$ is:
$$
d'(t) = \begin{cases}
- \frac{ t - 12k}{4\left| t - 12k \right|} \quad &|t-12k| < 4 \\
0 &\mathrm{otherwise}
\end{cases}
$$
The roots of $d(t)$ can be found solving the following equation: $$ \frac{1}{2} - \frac{|t - 12k|}{4} = 0 \iff t_k = \begin{cases} 2 + 12k \quad &t-12k > 0\\ -2 + 12k \quad &t-12k < 0 \end{cases} \iff t_k = \begin{cases} 2 + 12k \quad &0 < t-12k < 4\\ -2 + 12k \quad & -4 < t-12k < 0 \end{cases} $$
At this point, I don't know how to handle the two different expressions of $t_k$ properly in the sum which defines $\delta[d(t)]$. I mean, is it possible to define the function using two sums? In that case, I'd write: $$ \delta[d(t)] = \sum_{k \in \mathbb{Z}} 4\delta(t-2-12k) + \sum_{k \in \mathbb{Z}}4\delta(t+2-12k) $$ $\rule{8 cm}{0.44pt}$
The triangular function here, being $\Delta$ half the base of the triangle, is defined as: $$ \mathrm{tri}_\Delta(t) = \begin{cases} 1 - \frac{|t|}{\Delta} &|t| < \Delta \\ 0 &|t| > \Delta \end{cases} $$