Lesson 24 - Introduction to Discontinuous Functions (Heaviside Functions)

Let's introduce a concept that will be useful in physics or engineering. The Heaviside function describes the behavior of a switch graph, in which a switch might be turned on or off at a specific point in time. It is often denoted

Example

In this function, $u(t)$ is "off" for $t<a$ and is "on" for $t>a$. We can visualize this with a graph.

\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
\begin{axis}[
    axis lines = middle,
    xlabel = {$t$},
    ylabel = {$u(t-a)$},
    grid = both,
    domain=-2:5,
    samples=100,
    legend style={at={(0.5,-0.15)}, anchor=north},
    title = {Step Function: $u(t-a)$},
    xtick=\empty,
    ytick=\empty,
]
% Step function with parameter a = 2
\addplot[blue, thick] {x < 2 ? 0 : 1}; 
\addlegendentry{$u(t-a)$}

\end{axis}
\end{tikzpicture}
\end{document}

Notice that the slope at $t<a$ is constant at 0, while the slope at $t>a$ is constant at $1$.

Let's look at another.

To find the piecewise values that define this function, we know that when $t<a$, then $u(t-a)=0,$ and when $t>a$, $u(t-a)=1$. So for $tu(t-a),$ when $t<a$, we plug in $0$ for $u(t-a)$ and when $t>a$, we plug in $1$ for $u(t-a)$. We should get the following:

We can describe this as a switch that is "off" for $t<a$ and "on" with a slope of $t$ for $t>a$. Here is the graph for this case.

\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
\begin{axis}[
    axis lines = middle,
    xlabel = {$t$},
    ylabel = {$tu(t-a)$},
    grid = both,
    domain=-2:5,
    samples=100,
    legend style={at={(0.5,-0.15)}, anchor=north},
    title = {Step Function: $tu(t-a)$},
    xtick=\empty,
    ytick=\empty,
]
% Step function: tu(t-a) = 0 for t < 2, and tu(t-a) = t for t >= 2
\addplot[blue, thick] {x < 2 ? 0 : x}; 
\addlegendentry{$tu(t-a)$}

\end{axis}
\end{tikzpicture}
\end{document}

Try For Yourself

Find the piecewise definition and graph each of the functions.

1. $1-u(t-a)$
2. $u(t-a)-u(t-b)$

We can also write a piecewise function in terms of a Heaviside function.

We can simply multiple the function $f(t)$ by the step function $u(t-a)$. So we will have:

$f(t)=3\to 3u(t-a)$

Now, we know that $f(t)$ is true for $t<2$, so we have to replace $a$ with $2$.

$f(t)=3\to 3u(t-2)$

In the next piece of the function, we will have a range of values for which $t$ is true.

$f(t)=1,2<t<5\to 1[u(t-2)-u(t-5)]$

Let's convert the remaining functions:

$f(t)=t,5<t<9\to t[u(t-5)-u(t-9)]$

$f(t)=t^4,t>9\to t^4[u(t-9)]$

Now we can combine these into one single equation.

$H(t)=3u(t-2)+u(t-2)-u(t-5)+t[u(t-5)-u(t-9)]+t^{4}u(t-9)$

Try for Yourself

Write in terms of Heaviside functions:

\begin{cases} e^{-t} & 0 \leq 2 \ 3 & 2 < t < 3 \ 1 & t > 3
\end

f(t) =
\begin{cases} 0 & 0 \leq t < 1 \ t & 1 < t < 3 \ 1 & t > 3
\end