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

Question

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.

Answer 1

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.