I am reading a book and I couldn't understand an equality. Actually, I realized I don't really understand the property of convolution with Dirac's delta function. I hope you can help me. I know the property: $(\delta*f)(t)=f(t)$
First question, is it correct to write:
$\delta(t)*f(t)=f(t)$
Does this also work for a shifted $\delta$ function?
Here is what I am trying to understand from the book: $$\rho_p(x)=\sum_{k=-\infty}^{\infty}\rho(x-kp)=_{(\#)}\sum_{k=-\infty}^{\infty}\delta(x-kp)*\rho(x)=\left(\sum_{k=-\infty}^{\infty}\delta(x-kp)\right)*\rho(x)$$
I can't understand the equality I marked with a $\#$.
In case you need to know:
$\rho(x)$- is a function describing the density of electrons.
$p$- is the period of the the new function (they are periodizing the function $\rho$)
Thanks.
First you need to be aware of the following property,
$$\int_{-\infty}^\infty \delta(x) f(x) \ dx = f(0),$$
which implies that,
$$\int_{-\infty}^\infty \delta(x-a) f(x) \ dx = f(a).$$
Note that the $\delta$ function forces the integration variable $x$ to equal $a$ in the above example.
The definition of convolution is,
$$ (F(\tau)*G(\tau))(t) = \int_{-\infty}^{\infty} F(\tau) G(t-\tau) \ d\tau,$$
We will apply this definition to your expression. In this case $F(\tau) = \delta(\tau-kp)$ and $G(\tau)=f(\tau)$.
$$(F*G)(x) = \int_{-\infty}^{\infty} F(\tau) G(x-\tau) \ d\tau = \int_{-\infty}^{\infty} \delta(\tau-kp) f(x-\tau) \ d\tau = f(x-kp),$$
Where in the last equality we used the property of the delta function to collapse the integral and force the integration variable $\tau$ to equal $kp$.