I am reading the paper http://web.mit.edu/2.14/www/Handouts/Convolution.pdf because I want to learn about convolutions. With $$ \delta_T (t) = \begin{cases} 0 & t \leq 0 \\ 1/T & 0 < t \leq T \end{cases} $$ the following is said:
We now assume that the system response to $\delta_T (t)$ is a known function and is designated $h_T (t)$. Then if the system is linear and time-invariant, the response to a delayed unit pulse, occurring at time nT, is simply a delayed version of the pulse response: $y_n (t) = > h_T (t- nT)$
The linear system with the output function $y_n (t)$ is time-invariant. I think I get the definition of being time-invariant(not directly depend on time, possible indirectly from the input function). However, I still have no idea why $y_n (t)$ should be equal to $h_T (t - nT)$ -- like where does the $- nT$ part come from?
If you wanna take a look, it is formula 12 in the paper :))
The system is time-invariant, i.e., if the input is delayed by $\tau$, the output is also delayed by $\tau$, nothing else changes. So, if the response of the system to $\delta(t)$ is $h(t)$, then its response to $\delta(t-nT)$, which is $\delta(t)$ shifted to the right by $nT$, is simply $h(t-nT)$, which is $h(t)$ shifted to the right by $nT$.