FIR filter reverse engineering from step response
The relation of the discrete-time step response, $s[n]$, and discrete-time impulse response, $h[n]$, is

$$ s[n] = \sum\limits^{n}_{m=-\infty} h[m] $$

and

$$ h[n] = s[n] - s[n-1] $$

from $h[n]$ you can compute the DFT (of a large finite segment of $h[n]$) and get the frequency response of this filter:

$$ H(e^{j \omega}) = \sum\limits_{n=-N/2}^{N/2-1} h[n] e^{-j \omega n} $$

remember, with a finite delay allowed for the filter (this is your 68 samples), you can model both your impulse response $h[n]$ and step response $s[n]$ as symmetric about $n=0$ and define the origin for $n$ there at the 68th sample.  doesn't matter if that makes the filter acausal.  don't sweat that.

also, keep in mind that the operation that turns the step response into the impulse response is a discrete-time differentiator, which means the information about DC is lost.  the DC component in your impulse response that is derived from your step response is the step size of the step response, $s[+68]-s[-68]$.  you can add to both any constant and that will not change the impulse response $h[n]$ at all.

oh, and expect 2×68 + 1 taps.