when we turn a continuous signal into a time discret signal we use this function:
$\hat{s}(t)=s(t) \sum_{n=-\infty}^{\infty} \delta\left[t-n T_{a}\right]$, with $\delta\left[t-n T_{a}\right]=\left\{\begin{array}{ll}1 & t=n T_{a} \\ 0 & \text { else. }\end{array}\right.$
the $\sum_{n=-\infty}^{\infty} \delta\left[t-n T_{a}\right]$ part is what I dont get. When we want to get the specific value of s(t) for a general but fixed t- why do we multiply $s(t)$ with the infinite sume where $\delta\left[t-n T_{a}\right]$ will be $=1$ exactly one time only? (When our t is a multiple of $T_{a}$)
I understand that we have a sampling rate $T_{a}$ and every $n*T_{a}$ steps we take the value of $s(t)$ but why do we use the sum I mentioned above? The structure of this formula is not very clear to me yet and I would appreciate any explanation.
A dirac delta can be thought of as an infinitely-high "spike" whose integral is normalized (we can call it a "unit impulse", on account of the integral being normalized to 1). This procedure takes a continuous function, and replaces it with a countable sum of unit impulses across the time domain, spaced regularly at intervals of $T_a$, scaled by the value of the function at their location.
That is to say, if we have $s(nT_a) = k$, then $\hat{s}(nT_a)$ is an impulse with with mass $k$.
It may, at this point, be unclear why we replace the function with a sequence of scaled impulses rather than a sequence of finite values. This is because the Fourier transform maps an impulse to a pure sinusoid, and so this construction gives us a clean analog to the Fourier sum, with the dirac delta taking the place of the kronecker delta. Were we to simply discretize the function as a sequence of finite values, its Fourier transform would be zero (any function equal almost-everywhere to zero has zero integral - our dirac delta construction can avoid this because dirac deltas are not strictly "functions").