Convolution of Gaussian with exponential decay?
Looks like the Laplace transform of a Gaussian function, which is well known, e.g. done here, and here. You will have to expand the square to make use of the identities, of course.
Related videos on Youtube
SuperCiocia
Updated on December 10, 2020Comments
-
SuperCiocia almost 3 years
I need to convolve an exponential decay (defined as the exponential $Ae^{-\lambda t}$ from $0$ to $+\infty$) with a Gaussian of known standard deviation $\sigma$, in other words I need to compute the following integral:
$$ g(\tau) = \int_{0}^\infty \exp(-\lambda t) \exp\left(-\frac{(t-\tau)^2}{2\sigma^2} \right)\mathrm d t $$
which is almost the same integral as in this question, but with a $0$ as the lower limit.
The answer to the Math exchange question above does not seem to apply here, or at least not for the whole range of the convolution: naively I would expect an exponential increase up to $\tau=0$, then a Gaussian-like peak and finally an exponential decay for large $\tau$.
Anyway know how to get the full mathematical expression? SOLVED
REAL QUESTION: And by the way - am I doing this right? I have a lot of data for particle momenta, creation and end vertices: from these I'm calculating the rest lifetime for each particle event and plotting it in a histogram. The tail does look like an exponential decay but close to $t =0$ there's a Gaussian-like behaviour. If we assume that the detector resolution will smear the data with a Gaussian, the final plot of lifetimes should look like the convolution of an exponential decay with a Gaussian, right?
-
SuperCiocia almost 9 yearsThanks, so that's solved. What about the physical interpretation question?
-
lionelbrits almost 9 yearsI don't really have a good feel for what you are measuring, but probably because I haven't had time to give it enough thought. Are you sure you don't need a Poissonnian distribution?