A step response is calculated by taking the TF and multiplying it by the s-domain of a step ($$\frac{1}{s}$$). For instance the TF of a 2nd order LPF multiplied by $$\frac{1}{s}$$ is: -   $$\dfrac{1}{s}\cdot\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s + \omega_n^2}$$   For a 2nd order HPF this is: -   $$\dfrac{1}{s}\cdot\dfrac{s^2}{s^2+2\zeta\omega_n s + \omega_n^2} = \dfrac{s}{s^2+2\zeta\omega_n s + \omega_n^2}$$   For convenience the formulas are normalized with $$\omega_n$$ = 1
 LPF under-damped response   $$\dfrac{1}{s}\cdot\dfrac{1}{s^2+2\zeta s + 1}\Longrightarrow \dfrac{1}{s\left[(s+a)^2 + b^2\right]}$$   Where a = $$\zeta$$ and b = $$\sqrt{1-\zeta^2}$$   Using Laplace tables this converts to: -   $$1 - \dfrac{1}{\omega_d}\cdot e^{-\zeta t}\cdot \sin(\omega_d t + \phi)$$   Where $$\omega_d$$ (the normalized damped frequency)   = $$\sqrt{1-\zeta^2}$$      and $$\phi=\arccos(\zeta)$$ HPF under-damped response   $$\dfrac{s}{s^2+2\zeta s + 1}\Longrightarrow \dfrac{s}{(s+\zeta)^2 + (1-\zeta)}$$   $$\small =\dfrac{s+\zeta}{(s+\zeta)^2 + (1-\zeta)} - \dfrac{\zeta}{(s+\zeta)^2 + (1-\zeta)}\normalsize$$   Using Laplace tables this converts to: -   $$e^{-\zeta t}\left[ \cos(\omega_d t)-\frac{\zeta}{\omega_d}\cdot\sin(\omega_d t)\right]$$   Where $$\omega_d$$ (the normalized damped frequency)   = $$\sqrt{1-\zeta^2}$$ LPF critically-damped response   $$\dfrac{1}{s}\cdot\dfrac{1}{s^2+2\zeta s + 1} \Longrightarrow \dfrac{1}{s^3+2s^2 + s}$$  This reduces to: -   $$\dfrac{1}{s}-\dfrac{1}{s+1} - \dfrac{1}{(s+1)^2}$$ Using Laplace tables this converts to: - $$1- e^{-t}\cdot(1 + t)$$ HPF critically-damped response   $$\dfrac{s}{s^2+2\zeta s + 1} \Longrightarrow \dfrac{s}{s^2+2s + 1}$$   This reduces to: -   $$\dfrac{1}{s+1}-\dfrac{1}{(s+1)^2}$$   Using Laplace tables this converts to: -   $$e^{-t}\cdot(1 - t)$$ LPF over-damped response   $$\dfrac{1}{s}\cdot\dfrac{1}{s^2+2\zeta s + 1} \Longrightarrow \dfrac{1}{s}\cdot \dfrac{1}{(s+a)(s+b)}$$   where $$ab=1$$  thus, it follows that: -   $$a=\zeta-\sqrt{\zeta^2 - 1}$$ $$b=\zeta+\sqrt{\zeta^2 - 1}$$   And $$(b-a) = 2\sqrt{\zeta^2-1}$$   Using Laplace tables this converts to: -   $$1 - \dfrac{b}{b-a}\cdot e^{-at} + \dfrac{a}{b-a}\cdot e^{-bt}$$ HPF over-damped response   $$\dfrac{s}{s^2+2\zeta s + 1} \Longrightarrow \dfrac{s}{(s+a)(s+b)}$$   where $$ab=1$$  thus, it follows that: -   $$a=\zeta-\sqrt{\zeta^2 - 1}$$ $$b=\zeta+\sqrt{\zeta^2 - 1}$$   And $$(a-b) = -2\sqrt{\zeta^2-1}$$   Using Laplace tables this converts to: -   $$\dfrac{a}{a-b}\cdot e^{-at} - \dfrac{b}{a-b}\cdot e^{-bt}$$

Ω σ μ τ ω β δ η θ λ π ζ ∞ ° √ Δ ∂ ∫ Ʃ ± ≈ ≠ µ Ω ± ÷ Θ « » ≦ ≧ Σ Φ ε • θ λ ρ ω κ ² ³ ½ ¼ ¾ ° ∞ ∫≈ ≠ ≡ Ξ ∑ π Π © ®