# Convolution of a sum of shifted delta functions

In the lecture notes for Fourier Transforms and it's Applications on page 212 by Bracewell he talks about representing a signal as a sum of distributions evenly spaced out by a distance p.

$$\rho_p(x) = \sum_{k=-\infty}^{\infty} \rho(x-kp)$$

He then goes on to say that you can write the periodized density as a convolution with a sum of shifted $\delta$'s:

$$\rho_p(x) = \sum_{k=-\infty}^{\infty} \rho(x-kp) = \sum_{k=-\infty}^{\infty} \delta(x - kp)*\rho(x)$$

This doesn't make sense to me. According to the sifting propery - integrating a shifted delta function by kp should give you $\rho(kp)$. However, for all values > p the distribution should be 0 so you end up just getting the first distribution $\rho(x)$ and it ends up not being periodized.

Can someone explain how those are equivalent?

I'm not sure about the filtering, but I think this is mostly a question of notation. To simplify, define $$\delta_y(x) := \delta(x-y),$$ where $\delta$ is such that $\int_{-\infty}^\infty f(x)\delta(x)dx = f(0)$. Then, $\int f\delta_y = f(y)$, so this is your statement that integrating against a shifted $\delta$ function should give you $\rho(kp)$. Regardless of what the distributions are doing for $p > 0$, you simply have the following identity: $$\rho(x-kp) = (\delta_{kp}*\rho)(x).$$ [Note that $\delta_{kp}*\rho$ is itself a function of $x$.] To see this 'formally' (this is not a proper argument) \begin{align*} (\delta_{y}*\rho)(x) &= \int_{-\infty}^\infty \rho(z)\delta_y(x-z)dz \\ &= \int_{-\infty}^\infty \rho(z)\delta(x-z-y)dz \\ &= \int_{-\infty}^\infty \rho(x-w-y)\delta(w)dw \\ &= \rho(x-y). \end{align*} [Remark: two minus signs cancelled out during the change of variables.] Take $y = kp$ for your result.

• Thank you that was very useful and clear – coderdave May 26 '14 at 15:09