# What is the sampling rate for the continuous unit step signal?

I would like to know how I can find the sampling rate for the continuous-time signal: u(t) - u(t-2014), where u(t) is the continuous unit step function can be uniquely recovered from its sample.

If you have prior knowledge, e.g., you know that it is a box function, then you just need three values: the times of the two edges, in your example $$t_1=0$$ and $$t_2=2014$$, and the amplitude of the signal.
Sampling a continuous signal into discrete samples can be cast as finding an appropriate function to convert $$x(t)$$ ($$t\in \mathbb{R}$$) into $$x[k]$$ ($$k\in \mathbb{Z}$$). Finding functions $$f: \mathbb{R}\mapsto \mathbb{Z}$$ is easy, but they are generally not inversible, as stated by Matt (one theoretical reason could be that there is not bijection between $$\mathbb{R}$$ and $$\mathbb{Z}$$).
However, if you can restrict the subspace or subset of functions in $$\mathbb{R}$$, (by prior knowledge) then there are inversible solutions, like the Nyquist-Shannon version. Essentially, band-limited functions can (theoretically) be recovered with discrete samples, with something like "two samples per period".
Other frameworks exist: for instance, FRI or finite-rate of innovation provides you with means to perfectly sample a signal made of piece-wise polynomials. They are non-bandlimited signals, hence cannot be retrieved by traditional Nyquist rate sampling. However, FRI method can recover $$K$$ delta diracs with a minimum sampling rate, or sampled at the rate of innovation, of $$2K$$. This method extends to integrals of the above, such as a piece-wise constant signal.