I have to find the laplace transform of $\delta(t^2-3t+2)$.
The answer is : $e^{-s}+e^{-2s}$
I tried using the definition of laplace transform :
$$\int_{-\infty}^{\infty}\delta(t^2-3t+2)e^{-st}\mathrm{d}t$$
Tried taking $m=t^2-3t+2, \mathrm{d}m=(2t-3)\mathrm{d}t$
But that makes the limit and the integral more complicated. I don't know if that's the correct way or not
Seeing the answer I tried factoring the the inner equation inside delta as: $\delta((t-1)(t-2))$, but again I am stuck here unless there are some properties of the delta function that makes $$\delta((t-1)(t-2))=\delta(t-1)+\delta(t-2)$$
Now that I think about it, it makes sense as delta has a value at $t=1$ and $t=2$ but I don't know if $\delta((t-1)(t-2))=\delta(t-1)+\delta(t-2)$ is actually correct.
Note: $\delta(.)$ is the dirac delta function.
Best Answer
Your intuition is true. In general, the delta function of a smooth function of $t$ is given by (see eq. (7) in the link): $$\delta(g(t))=\sum_{i}\frac{\delta(t-t_i)}{|g'(t_i)|}$$ Where the $t_i$'s are the roots of $g$. Multiple proofs are provided in the following Math.SE post: Dirac Delta Function of a Function
Here $g(t)=t^2-3t+2$, thus $g'(t)=2t-3$. The roots are given by $t_1=1$ and $t_2=2$, thus giving $|g'(t_i)|=1$ in both cases. Hence: $$\delta(t^2-3t+2)=\delta(t-1)+\delta(t-2)$$ Thus, the Laplace Transform is indeed $e^{-s}+e^{-2s}$.