Unit step response

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 $$p\lt 0$$).

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.

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