We have: $$ s(t) = 4u(t)−u(t−1)−2u(t−2)−2u(t−3.5)+2u(t−4) $$ where $u$ is the Heaviside function.
I had to draw the graph and find the minimum sampling period so we wont have loss of signal. I have drawn the graph.
$t=kT$ where $T$ is the sampling period.So $T=t/k$ so how do I find T?
The smallest feature that you need to capture is the dip between 3.5 and 4. If you use a sampling period which is greater than 0.5, then you may end up not capturing this feature. The sampling period must therefore be $T\leq 0.5$ in order not to lose any features of the signal.
To prevent loss of signal you need to sample at twice the highest frequency presented in the signal. This is also known as Nyquist rate.
Assuming that time t is in seconds, the smallest time period is about 0.5s or the highest time frequency is 2Hz. So if sampled between 2Hz and 4Hz the processed signal will be subjected to what is known as aliasing. If the signal is sampled above 4Hz, there will no data loss. Below are few references that gives better detailed explanation.
References:
