What is the purpose of wrapping the negative times of a response function in discrete convolution?
It's about zero-phase filtering.
First, the continuous-time response function $r(t)$ is an even-symmetric, noncausal filter so that its Fourier transform will be a zero-phase function; i.e.,
$R(\omega)$ is a real-valued function of frequency $\omega$.
Note that there's almost a mistake in the figure 13.1.1; the output should begin before $t=0$, even for a zero-phase filter. Because this is a non-causal filter, it will anticipate future inputs before they happen, and put out the response before $t=0$.
I think, however, they wanted to justify this condition, by shifting the input $s(t)$ sufficiently to the right (or assuming that $s(t)=0$ until $t=t_1$) so that the output response before $t=0$ will be zero anyway, and thus eventually we are interested at the output of the filter beginning from $t=0$.
Now, sample the signals $s(t)$ and $r(t)$ (assuming they are bandlimited) and obtain a sampled version of the continuous-time output by performing a discrete-time linear convolution, or equivalently use a DFT based efficient implementation of linear convolution by circular convolution of proper length.
If you choose the first approach, and perform direct discrete-time linear convolution using the sequences $s[n]$ and $r[n]$, then you won't need to shift anything. But remember that a practical implementation will still need some shifts to represent the noncausal impulse response $r[n]$ in the proper index frame.
But if you choose the second approach; using DFT to implement a linear filtering, you have two choices for the filter response $r[n]$ to be registered; (because in the DFT domain, there's no negative time-index $n$, and you are restricted to the range $0 \leq n &lt; N$) first you can linearly shift $r[n]$ sufficiently right to make it a causal filter $r_c[n]$, and follow the procedure as usual; with this choice, the effective discrete-time filter loses its zero-phase (and zero group delay) property and your effective output begins at the group delay position index.
If, instead, you wrap the negative-index portion of $r[n]$ to its far right edge, in the DFT frame of $N$ samples, where $N$ is chosen properly to aviod aliasing in the time-domain to ensure that computed circular convolution will match the desired linear one, then you will maintain the zero-phase property of $r[n]$ while performing the DFT based convolution; your first output at $n=0$ will be what you would get at $t=0$ in the continuous-time case.