The signal is binned into time bins with a width ##δt##

[arcTomato]

TL;DR

The signal are binned into time bins with a width $δt$

Hi all.I would like to know about "binning window".
This paper I'm reading says like this.
Why do "convolving the data with the $b(t)$ before the sampling" and "binning into time bins with a width $δt$" have the same meaning?

I know I'm addicted to post to PF
But this forum is so meaningful for me, so please help me if you can!

Thank you

 

[.Scott]

As an example, let's say that δt= 1 sec.
What it is saying is that the data is not being sampled at 1 second intervals.
Rather, each value is an proportional to the average data value across a 1-second interval.

The author is using a convolution to compute that average value across the bin period for each bit.

 

[arcTomato]

Thank you @.Scott !

That is the point I don't know why.
Why do convolution and computing the average value have same meaning??

 

[.Scott]

It's not just any convolution, it's a convolution with the specific bin function they are using - combined with the sampling.
That bin function is zero over most of the range (-$\infty$ to $\infty$) and and N/T within T/2N of the sampling time. The result is that, the function that results from the convolution (fc(t)) will the integral all all data outside the bin multiplied by zero and all data within the bin multiplied by N/T. So that fc(t) generates a running average of the samples that land within a bin centered at time t.

What the author is saying is you are not sampling the original function, but this convolution result (fc(t)).

 

[arcTomato]

Thank you @.Scott !
Ok I think I got it.
so, is this right?
$\bar{a}{(t)}=\frac{1}{\delta t} \int_{t-\frac{1}{2 \delta{t}}}^{t+\frac{1}{2\delta{t}}} a{(\tau)} d \tau=\frac{N}{T} \int_{t-\frac{N}{2 T}}^{t+\frac{N}{2 T}} a{(\tau)}d \tau=\int_{-\infty}^{\infty} a{(\tau)} b(t-\tau) d \tau=a{(t)}*b(t)$