The Convolution of Detla Functions

[EngWiPy]

Hi,

I have encountered with this:

\delta[y-a]*\delta[y-b]

where a and b are positive real numbers, and * denotes convolution. How to do this in both continuous and discrete cases? In Wikipedia, they say that:

\int_{-\infty}^{\infty}\delta(\zeta-x)\delta(x-\eta)\,dx=\delta(\zeta-\eta)

Can I use this result, so that:

\delta[y-a]*\delta[y-b]=\delta[y-b-a]?

Thanks in advance

 

[Goongyae]

>Can I use this result

Yes. The convolution delta(y-a)*delta(y-b) is an integral over s of delta(s-a) delta(y-s-b). This integrand is zero everywhere except where s = a, thus it yields delta(y-a-b).

It's the same whether it's a continuous delta function (an infinitely thin and high peak) undergoing an integral, or a discrete delta (of height 1 at an integer) undergoing a summation. When integrated, delta(zeta - x) picks out the value of the integrand where x=zeta, yielding delta(zeta - eta). It's the same if you consider the other delta function picking out the integrand at a particular x.

 

[EngWiPy]

>Can I use this result

Yes. The convolution delta(y-a)*delta(y-b) is an integral over s of delta(s-a) delta(y-s-b). This integrand is zero everywhere except where s = a, thus it yields delta(y-a-b).

It's the same whether it's a continuous delta function (an infinitely thin and high peak) undergoing an integral, or a discrete delta (of height 1 at an integer) undergoing a summation. When integrated, delta(zeta - x) picks out the value of the integrand where x=zeta, yielding delta(zeta - eta). It's the same if you consider the other delta function picking out the integrand at a particular x.

Ok, thank you very much.

Regards