Sum of Dirac functions

Question

The Dirac function is commonly used to model the action potential of neurons, and in the text I am reading the "neural response function" is defined as:

$$\rho(t) = \sum_{i=1}^n \delta(t - t_i)$$

The author then follows to say that any well-behaved function h(t) can be expressed as such:

$$\sum_{i=1}^n h(t - t_i) = \int_0^T h(\tau)\rho(t-\tau)d\tau$$

I am confused on two points

1. Wouldn't the first function simply be 1 for all $t$, since the integral of the delta function is 1 by definition? If so, I do not understand the purpose of the function $\rho(t)$.

2. Related to my confusion on the first point I am simply not following this identity. I understand the left half of the equation to essentially be an integral of the shifted function $h(t)$, but then the right hand side is the integral of $h(\tau)$ multiplied by $\rho(t-\tau)$, the latter of which is one for every $\tau$. So as I follow it the right hand side is simply the integral of $h(\tau)$ and the $\rho(t-\tau)$ contributes nothing.

Does anyone understand this enough to explain why this identity is true or useful?

Thank you!

Answer 1

You could convince yourself the identity works as follows:

$$\int_0^T h(\tau) \rho (t - \tau) \mathrm{d} \tau = \int_0^T h(\tau) \sum_{i=1}^n \delta (t - t_i - \tau) \mathrm{d} \tau = \\ \int_0^T h(\tau) \delta (t -t_1 - \tau) \mathrm{d} \tau + \int_0^T h(\tau) \delta (t -t_2- \tau) \mathrm{d} \tau +\cdots +\int_0^T h(\tau) \delta (t -t_n- \tau) \mathrm{d} \tau = \\ h(t -t_1) + h(t-t_2) + \cdots h(t-t_n) = \sum_{i_1}^n h(t-t_i)$$

hopefully this hints at how the "Dirac comb", the sum of Dirac deltas, is sampling out values in the integral and yielding the sum with the function $h$

Answer 2

observe that :

$\delta(t)$ concentrates at $t=0$, so $\delta(t-t_i)$ (as a function of $t$) concentrates at $t=t_i$, so $\rho(t)$ concentrates at $t=t_1,t_2,\dots,t_n$, so $\rho(t-\tau)$ (as a function of $\tau$) concentrates at $\tau=t-t_1,t-t_2,\dots,t-t_n$.

assume $t_1,t_2,\dots,t_n\in [0,T]$, hence $\int_0^Th(\tau)\rho(t-\tau)d\tau=\sum_{i=1}^n h(t-t_i)$.