Step Response of 2nd Order HPF
The double zero at \$ s\small =0\$ gives rise to the undershoot.

To make the analysis easier, it's best to normalise the TF to \$\omega_n=1\$, thus divide \$\omega_n\$ by \$\small 1000\$ to give: $$G(s)=\dfrac{s^2}{s^2+3s+1}$$

Multiply by \$\dfrac{1}{s}\$ to obtain the step response: \$R(s)=s\left(\dfrac{1}{s^2+3s+1}\right)\$

Now find the inverse Laplace transform of the bracketed term and then differentiate (multiply by \$s\$ = differentiation) to determine \$\small r(t)\$, thus:

$$\dfrac{1}{s^2+3s+1}=\dfrac{0.45}{s+0.38}-\dfrac{0.45}{s+2.62}\rightarrow 0.45e^{-0.38t}-0.45e^{-2.62t}$$
differentiate to give the step response:

$$r(t)=\dfrac{d}{dt}\left( 0.45e^{-0.38t}-0.45e^{-2.62t}\right)=-0.17e^{-0.38t}+1.17e^{-2.62t}$$

This starts at \$\small r(0)=1.0\$, then falls to a minimum (undershoot), \$\small r(1.72)=-0.0755\$, and settles at \$\small r(\infty)=0\$

Finally, scale the time axis by the normalising factor, \$\small 1/1000\$, giving the undershoot of \$\small -0.0755\$ at \$\small t=1.72ms\$