Unit step response

Coert Vonk

Derives the unit step response of RC low-pass filter. Part of a series about the properties of the RC low-pass filter.\(\)

Unit Step Response

The step response gives an impression of the system behavior when the input signal going from \(0\) to \(1\) volt at time \(t=0\). This input is called the Unit Step Function, here represented by \(u(t)=\gamma(t)\).

$$ \begin{align} \gamma(t)&=\begin{cases} 0 & t\lt 0 \\ 1 & t\geq0 \end{cases}\nonumber\\ \Rightarrow\ \Gamma(s)&=\mathcal{L}\left\{\gamma(t)\right\}=\frac{1}{s} \end{align} \label{eq:unitstep} $$

Combining the Laplace transform of \(\gamma(t)\), \(\Gamma(s)\), with the transfer function, gives the unit step response \(Y(s)\)

$$ \begin{align} Y(s)&=\Gamma(s)\cdot H(s)\nonumber \\ &= \frac{1}{s}\cdot K\frac{1}{(s-p)}\nonumber \\ &= K\frac{1}{s(s-p)} \end{align} \label{eq:multiplication} $$

Split up this complicated fraction into forms that are in the Laplace Transform table. According to Heaviside, this can be expressed as partial fractions. [swarthmore, MIT-cu]

$$ Y(s)=K\frac{1}{s(s+a)}\equiv\frac{c_0}{s}+\frac{c_1}{s+a} \label{eq:heaviside} $$

Substitute \(K=-p\) from the transfer function and find expressions for the constants \(c_{0,1}\), by multiplying with respectively \(s\) and \((s-p)\)

$$ \left\{ \begin{eqnarray} -p\frac{\cancel{s}}{\cancel{s}(s-p)} &\equiv& \frac{\cancel{s}c_0}{\cancel{s}}+\frac{sc_1}{s-p}\nonumber \\ -p\frac{\cancel{s-p}}{s\cancel{(s-p)}} &\equiv& \frac{c_0(s-p)}{s}+\frac{c_1\cancel{(s-p)}}{\cancel{s-p}}\nonumber \end{eqnarray} \right. $$

Given that these equations are true for any value of \(s\), choose two convenient values of \(s\) that help us find \(c_0\) and \(c_1\).

$$ \begin{eqnarray} c_0&=&\left.\frac{-p}{s-p}\right|_{s=0}=1 \\ c_1&=&\left.\frac{-p}{s}\right|_{s=p}=-1 \end{eqnarray} \label{eq:constants1} $$

The unit step response \(y(t)\) follows from the inverse Laplace transform of \(\eqref{eq:heaviside}\)

$$ \begin{align} y(t)&= \mathcal{L}^{-1}\left\{\frac{c_0}{s}\right\}+\mathcal{L}^{-1}\left\{\frac{c_1}{s-p}\right\}, & t\geq0\nonumber \\ &=c_0+c_1e^{pt}, & t\geq0 \end{align} $$

Substituting the constants gives the unit step response

$$ \shaded{ \begin{array}{ccr} y(t)=1-e^{pt}, & p=-\frac{1}{RC}, & t\geq0 \\ \end{array} } $$

As shown in the graph, the unit step response is a relatively slow decaying exponential curve (with \(p\lt 0\)).

own work
Unit step response

The GNU Octave code used to print this figure is listed in Appendix A.

Moving on from this unit step response of RC low-pass filter, continue reading about the Frequency Response.

Moving on from this unit step response of RC low-pass filter, continue reading about the Frequency Response.

Leave a Reply

Your email address will not be published. Required fields are marked *

 

This site uses Akismet to reduce spam. Learn how your comment data is processed.