Math 555: Differential Equations

6.3 Step Functions

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6.3.1 Step Functions

As mentioned at the beginning of the chapter, one of the great advantages to using Laplace transforms is that they are well suited to handle discontinuous forcing functions and ones that include "impulses." This is especially common in applications for electric circuits, because the forcing function may be turned on or turned off at different times. In this section, we introduce the unit step function , also known as the Heaviside function , to give us a mathematical way to represent this behavior.

The unit step function (which we will refer to as just the step function for brevity) $u_c$ is defined as

$$ u_c(t) = \left\{\begin{matrix} 0, & t\lt c, \\ 1, & t\ge c. \end{matrix}\right. $$
Qualitatively, this means that the function is $0$ until it reaches the value $t=c$, then "steps up" to $1$. We only concern ourselves with $t\gt 0$, since the Laplace transform is only defined for values of $t\in [0,\infty)$. It is also possible to have a "step down" function, given by the formula $1 - u_c(t)$. Both are pictured here.

Typically, the value $1$ in this sense is associated with the forcing function being "on" and $0$ is associated with "off." In this way, a function with a switch that is turned on at time $c$ can be constructed using $u_c$, and a function with a switch that turns off at time $c$ is made using $1-u_c$.

Alternative Notation

The text uses $u_c(t)$ to represent a step up one that occurs at the value $t = c$. Another perfectly acceptable way to represent the unit step function is to define the Heaviside function as

$$ H(t) = \left\{\begin{matrix} 0, & t\lt 0 \\ 1, & t\ge 0 \end{matrix}\right. $$
and then represent the shift to the right using the typical translation notation of $H(t-c)$.

Example 6.3.1

Sketch the graph of

$$ y(t) = u_\pi (t) - u_{2\pi} (t),\quad t\ge 0. $$

Solution

The best way to think of this is to first represent it in piecewise form. It is $0$ until $t = \pi$, where it is $1$ until $t = 2\pi$, and function returns to $0$.

$$ y(t) = \left\{\begin{matrix} 0, & \ 0 \le t \lt\ \pi \\ 1, & \ \pi\le t\lt 2\pi \\ 0, & 2\pi\le t \lt\infty \end{matrix}\right. $$
The plot of this function is This is known as a rectangular pulse .

In this figure, the vertical lines are drawn, connecting the "jumps." Technically, this is incorrect. The best representation uses the open circle at the beginning of the jump and a closed circle at the end. It is drawn this way here for demonstration purposes, because many computer graphing utilities will not skip drawing the vertical lines and it is important that you recognize that they should not be there.

Example 6.3.2

Sketch the graph of

$$ f(t) = \left\{\begin{matrix} \ 3, & 0\le t\lt 3 \\ -1, & 3\le t\lt 5 \\ \ 5, & 5 \le t \lt 9 \\ \ 2, & t\ge 9,\end{matrix}\right. $$
and express $f(t)$ in terms of $u_c(t)$.

Solution

Since the function is already written in piecewise form, sketching the figure is simple:

Rewriting the piecewise $f(t)$ into step functions is simple, as long as scale the jumps correctly. Our function is "on" with a value of $3$ to begin with, so we start there. It then jumps down $4$ to $-1$ at $t = 3$. Next, it jumps up $6$ to $5$ at $t = 5$, and then finally down $3$ to $2$ at $t = 9$. Therefore,

$$ f(t) = 3 - 4u_3(t) + 6u_5(t) - 3u_9(t). $$

6.3.2 Laplace Transform of the Step Function

For $c\ge 0$, the Laplace transform of the step function is given by

$$ \begin{aligned} \mathcal{L}\left\{u_c(t)\right\} &= \int_0^\infty e^{-st} u_c(t)\, dt = \int_c^\infty e^{-st} \, dt \\ \\ &= \dfrac{e^{-cs}}{s},\quad s\gt 0. \end{aligned} $$
This reduces to Laplace transform for $1$ when $c = 0$, since

$$ \mathcal{L}\left\{u_0 (t)\right\} = \dfrac{e^0}{s} = \dfrac{1}{s} = \mathcal{L}\left\{1\right\}. $$

Horizontal Shift

A primary use of the step function is that it may be used to conveniently represent a horizontal shift in a function, where the function is taken to be $0$ until it is "turned on." For a given function $f(t)$, we would have its translation to the right by $c$ given by

$$ \begin{aligned} g(t) &= \left\{\begin{matrix} 0, & t\lt c \\ f(t-c), & t\ge c \end{matrix}\right. \\ \\ &= u_c(t) f(t-c). \end{aligned} $$

Exercise 1

Show that
$$\mathcal{L}\{u_c(t)f(t-c)\} = e^{-cs}F(s). $$

Follow Along

Beginning with the definition of the Laplace transform, we have
$$ \begin{aligned} \mathcal{L}\{u_c(t)f(t-c)\} &= \int_0^\infty e^{-st}u_c(t)f(t-c)\, dt \\ &= \int_c^\infty e^{-st}f(t-c)\, dt. \end{aligned} $$ Here, we may make a substitution to simplify in the integral. Choosing $r = t-c$, so that the limits of integration are now $0$ and $\infty$:
$$ \begin{aligned} \mathcal{L}\{u_c(t)f(t-c)\} &= \int_0^\infty e^{-s(c+r)}f(r)\, dr \\ &= \int_c^\infty e^{-sr}\left( e^{-cs}f(r)\right )\, dr \\[5pt] &= e^{-cs}F(s). \end{aligned} $$

Example 3

Suppose

$$ f(t) = \left\{\begin{matrix} 1 - e^{-t}, & 0\le t \lt \pi \\ 1 - e^{-t} + \sin(t-\pi), & \pi \le t \lt \infty\end{matrix}\right. $$
Sketch the graph of this function for $t\in [0,12]$ and determine $\mathcal{L}\left\{ f(t) \right\}$.

Solution

The $\sin t$ element of the function does not activate until $t\gt \pi$, so we may use the Heaviside function to represent this. I'm using the $H(t)$ notation mentioned at the beginning of the section for this example.

$$ f(t) = 1 - e^{-t} + H(t-\pi)\sin(t-\pi) $$

With $f(t)$ in terms of step functions, finding the Laplace transform is straightforward utilizing the Laplace Transform Table

$$ \begin{aligned} \mathcal{L}\{f(t)\} &= \mathcal{L}\left\{1 - e^{-t} + H(t-\pi)\sin(t-\pi)\right\} \\ \\ &= \mathcal{L}\left\{1\right\} - \mathcal{L}\left\{e^{-t}\right\} + \mathcal{L}\left\{H(t-\pi)\sin(t-\pi)\right\} \\ \\ &= \dfrac{1}{s} - \dfrac{1}{s+1} + \dfrac{e^{-\pi s}}{s^2 + 1} \end{aligned} $$

6.3.3 Computing Inverse Transforms

Practice referencing the Laplace transform table to make sure that the expression matches a form that appears there. If not an exact match, perform the necessary algebraic manipulations to make it a match.

Example 6.3.4

Find the inverse Laplace transform of the function

$$ F(s) = \dfrac{3!}{(s-2)^4}. $$

Solution

Referencing the Laplace Transform Table , we see that this is close in form to line 11 with $n = 3$ and $a = 2$, so

$$ \mathcal{L}^{-1}\left\{F(s)\right\} = f(t) = t^3e^{2t}. $$

Exercise 6.3.2

Find

$$\mathcal{L}^{-1}\left\{ \dfrac{e^{-2s}}{s^2 + s - 2} \right\}. $$

Check Your Work

$$ f(t) = \frac{1}{3}u_2(t)\left(e^{t-2} - e^{-2(t-2)}\right) $$

Video Solution

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