RLC Filter Circuit step response

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: -

\(\dfrac{a}{a-b}\cdot e^{-at} - \dfrac{b}{a-b}\cdot e^{-bt}\)