# Deconvolution of two delta functions (solving $y(t) = A x(t-a) + B x(t-b)$)

I would like to calculate $x(t)$, when only $y(t)$ with

$y(t) = A x(t-a) + B x(t-b)$

is known.

Since this is a linear shift invariant operation (convolution), the inverse relation must be of the form

$x(t) = (y * g)(t) = \int_{-\infty}^\infty y(t-\tau) g(\tau) \mathrm d \tau$

The question is: what is $g(\tau)$? I am sure this problem has been solved before, but I cannot find any suitable references.

Also I would be interested in solutions to the equation with three or more terms (the shifts can be equidistant though).

I know that there may be (some) situations where this equation has no solution, but I think not in the majority of cases.

You can use the theorem that convolution in the time domain is the multiplication of the Fourier transforms. So $X(\omega)=Y(\omega)G(\omega)$ where the capital letters represent the Fourier Transforms of the lower case letters.
We can also use the shift theorem. What is $Y(\omega)$? $Y(\omega)=Ae^{-i\omega a}X(\omega) + Be^{-i\omega b}X(\omega)$
Factoring out the $X(\omega)$ we can solve for $G(\omega)$ If you find a inverse transform of $G$ then you have found $g$.
By the way, are you sure there's an easily expression for $g$? Based on what you're taking the inverse transform of, I'm not sure there is.
• Thanks, but I would like to have an (as explicit as possible) expression for $g$. – Andreas H. Nov 14 '14 at 3:57
• @AndreasH. In general, this cannot be done. Suppose that $x$ is a triangle waveform, and that $a$ and $b$ shift $x$ by 1/2. Let $A=B=1$. Then $y=0$. So was $x$ a triangle waveform shifted by 1/2? Or was $x=0$, which would yield the same result? One can never know. – NicNic8 Jan 29 at 17:45