Laplace Transform of the Step Function 📂Odinary Differential Equations

Laplace Transform of the Step Function

Definition¹

Let’s denote the unit step function unit step function shifted by $c$ as follows:

$u_{c}(t)=\begin{cases} 0 & t<c \\ 1 & t \ge c \end{cases}$

Formula

The Laplace transform of the step function $u_{c}(t)$ is as follows.

$\begin{equation} \mathcal{L} \left\{ u_{c}(t) \right\} = \dfrac{e^{-cs}}{s},\quad s>0 \label{eq1} \end{equation}$

Let’s assume that $c$ is an arbitrary constant, and when $s > a \ge 0$ , the Laplace transform of $f(t)$ , $F(s)$ exists, which means $F(s)=\mathcal{L} \left\{ f(t) \right\}$ . Then, the Laplace transform of the product of $f$ and $u_{c}$ is as follows.

$\begin{equation} \mathcal {L} \left\{ u_{c}(t) f(t-c) \right\} = e^{-cs} F(s) \label{eq2} \end{equation}$

Explanation

Since the variable $t$ we deal with is time, which means $t>0$ , let’s consider this to be obvious without special mention. Functions like $u_{c}(t)$ are useful in describing values that suddenly appear or disappear after a certain time when switching on or off in electrical circuits.

Derivation

(1)

$\begin{align*} \mathcal{L} \left\{ u_{c}(t) \right\} &= \int_{0}^\infty e^{-st}u_{c}(t)dt \\ &=\int_{0}^c e^{-st}\cdot 0 dt + \int_{c}^\infty e^{-st}\cdot 1 dt \\ &= \lim _{A \to \infty} -\dfrac{1}{s} \left[ e^{-st} \right]_{c}^A \\ &= \dfrac{e^{-cs}}{s} \end{align*}$

At this point, since $\lim \limits_{A \to \infty } e^{-sA}=0$ , it follows that $s>0$ .

■

(2)

$\begin{align*} \mathcal{L} \left\{ u_{c}(t) f(t-c) \right\} &= \int_{0} ^\infty e^{-st}u_{c}(t) f(t-c)dt \\ &= \int_{c}^\infty e^{-st} f(t-c) dt \end{align*}$

Let’s substitute $t-c=\tau$ here. Then, $0 \le \tau \le \infty$ and $dt=d\tau$ . Therefore,

$\begin{align*} \int _{0} ^\infty e^{-s(\tau+c)}f(\tau)d\tau &= e^{-cs}\int_{0}^\infty e^{-s\tau}f(\tau)d\tau \\ &= e^{-cs}F(s) \end{align*}$

■

Example

Let $f(t)=\sin t + u_{\pi /4} (t) \cos (t-\frac{\pi}{4})$ .

$\begin{align*} \mathcal{L} \left\{ f(t) \right\} &= \mathcal{L} \left\{ \sin t \right\} + \mathcal{L} \left\{ u_{\pi /4} \cos (t-\frac{\pi}{4}) \right\} \\ &=\mathcal{L} \left\{ \sin t \right\} + e^{-\frac{\pi }{ 4 }s} \mathcal{L} \left\{ \cos t \right\} \\ &=\dfrac{1}{s^2+1} + e^{ -\frac{\pi}{4}s}\dfrac{s}{s^2+1} \\ &=\dfrac{ 1+ se^{ -\frac{\pi}{4}s}}{s^2+1} \end{align*}$

■