How to calculate poles of overdamped, 2nd order system if only given experimental data?
Write the TF as \$ \small G(s)=\large\frac{ab}{(s+a)(s+b)}\$, where \$a\$ and \$b\$ are real and distinct.

This factorises to: \$\small G(s)=\large\frac{ab}{(b-a)} \large(\frac{1}{s+a}-\frac{1}{s+b}\large)\$

From which the inverse LT, i.e. the unit impulse response, is:

\$g(t)= \large\frac{ab}{(b-a)} \normalsize (e^{-at}-e^{-bt})\$

Differentiating and setting to zero determines the time to peak value, \$t_P\$:

\$t_P =\large\frac{ln(\frac{b}{a})}{(b-a)}\$, with corresponding peak value: \$g(t_P)=be^{-b t_P}\$

Hence, measuring \$t_P\$ and \$g(t_P)\$ will yield the value of \$b\$; and substituting back into the expression for \$t_P\$ will yield \$a\$

Note that the unit impulse response can be obtained by differentiating the unit step response, but the differentiator must be carefully designed to avoid introducing excessive noise.

Typical overdamped step and impulse responses are shown below.