In order to be able to work with functions representing the on/off behaviour of a switch, we define the following functions.

The **unit step function** or **Heaviside function**, is the real function #u# defined by

\[u(x) =\begin{cases} 0 &\text{if } x\lt 0\\ 1&\text{if }x\ge 0\end{cases}\]

This is a *piecewise continuous function*.

Let #c# be a real number number. The shifted function #u_c(x) = u(x-c)# is ifo called a unit step function or a Heaviside function; it satisfies

\[u_c(x) =\begin{cases} 0 &\text{if } x\lt c\\ 1&\text{if }x\ge c\end{cases}\]

We often write #t# for the argument of a function on the domain #t\ge0#. The unit step function #u_{-1}(t)# is thus the function which is constantly #1# on #\ivco{0}{\infty}# while #u_{-1}(x)# is the function with domain #\mathbb{R}# making a jump (in other words: having a discontinuity) at #-1#.

A negative unit step function may be described by \(1-u_c(t)\)

De functie \(u_{\pi}-u_{2\pi}\) fires up at #t=\pi# and goes down at #t=2\pi#.

The unit step function can be combined with another function #f# to make parts of that function disappear. For example, the function #g(t) = u(t-c)\cdot f(t-c)# can be described by

\[g(t ) =\begin{cases} 0 &\text{if } t\lt c\\ f(t-c)&\text{if }t\ge c\end{cases}\]

Conversely, unit step functions may be used for a compact way of writing the function rule of a piecewise continuous function.

The #u# in the notation of the Heaviside function is the first letter of the nowadays widely used name *unit step function*.

The notation #H_c# for #u_c# is also common and refers to Heaviside.

The Laplace transforms of piecewise continuous functions can be expressed in terms of exponential functions and the Laplace transforms of the continuous functions:

If #c\ge0# and #f# is a function of which the Laplace-transform exists, then \[ \laplace \left(u_c(t)\cdot f(t-c)\right) = \ee^{-c s}\cdot\laplace(f)\]

In particular, \[\laplace {\left(u_c\right) }(s)= \dfrac{\ee^{-c s}}{s}\]

\[\begin{array}{rcl}\laplace\left(u_c(t)\cdot f(t-c)\right)(s) &=& \int_0^{\infty}\ee^{-st}\cdot u_c(t)\cdot f(t-c)\dd t\\ &&\phantom{xx}\color{blue}{\text{definition of Laplace transform}}\\ &=& \int_0^c\ee^{-st}\cdot0\cdot f(t-c)\dd t+\int_c^{\infty}\ee^{-st}\cdot f(t-c)\dd t\\ &&\phantom{xx}\color{blue}{\text{definition used of }u_c}\\ &=&\int_c^{\infty}\ee^{-st}\cdot f(t-c)\dd t\\ &&\phantom{xx}\color{blue}{\text{first integrand is equal to }0}\\ &=&\int_0^{\infty}\ee^{-s\cdot (z+c)}\cdot f(z)\dd z\\ &&\phantom{xx}\color{blue}{\text{substitution }t=z+c}\\ &=&\ee^{-c\cdot s}\int_0^{\infty}\ee^{-s\cdot z}\cdot f(z)\dd z\\ &&\phantom{xx}\color{blue}{\text{constant term extracted from integral}}\\ &=&\ee^{-c\cdot s}\laplace( f)(s)\\ &&\phantom{xx}\color{blue}{\text{definition of Laplace transform}}\end{array}\]

The special case occurs when #f# is the constant function #1#.

The Laplace transform of #u_2 (t) \cdot t # cdot can thus be calculated as follows:

\[\begin{array}{rcl}\laplace\left(u_2(t)\cdot t\right)(s) &=&\ee^{-2s}\laplace( t+2)(s)\\ &&\phantom{xx}\color{blue}{\text{theorem}}\\ &=&\ee^{-2s}\left(\laplace( t)(s)+\laplace(2)(s)\right)\\ &&\phantom{xx}\color{blue}{\text{linearity of the Laplace transformation}}\\ &=&\ee^{-2s}\left(\dfrac{1}{s^2}+\dfrac{2}{s}\right)\\ &&\phantom{xx}\color{blue}{\text{known Laplace transforms}}\\ &=&\ee^{-2s}\left(\dfrac{1+2s}{s^2}\right)\\ &&\phantom{xx}\color{blue}{\text{fractions brought under common denominator}}\end{array}\]

Many processes in which an on/off state plays a role, can be described by linear second-order differential equations in which the inhomogeneous term is described by means of a Heaviside function. The above rule often makes it possible to solve the differential equation with the Laplace transform. We give some examples.

Compute the Laplace transform of the function \[f(t) = 5 u_{5}(t)- 4 u_{8}(t)\] on #\ivco{0}{\infty}#. Here, #u_c(t)# is the Heaviside function on #\ivco{0}{\infty}#.

Give your answer as a function of #s# for #s\gt 0#.

Solution #\laplace{(f)}(s) = # #{{5 \euler^ {- 5 s }-4 \euler^ {- 8 s }}\over{s}}#

This answer can be found by use of the

*linearity of Laplace transform* and the

*theorem on the Laplace transform of a function which involves a Heaviside function*:\[\begin{array}{rcl} \laplace{(f)}(s) &=&\displaystyle 5\laplace{(u_{5})}(s)- 4\laplace{( u_{8})}(s)\\

&&\phantom{xxxxxxxx1234}\color{blue}{\text{linearity of the Laplace transform}}\\

&=&\displaystyle 5\frac{\e^{-5 s}}{s}- 4\frac{\e^{-8 s}}{s}\\

&&\phantom{xxxxxxxx1234}\color{blue}{\laplace {\left(u_c\right) }(s)= \dfrac{\ee^{-c s}}{s}}\\

&=&\displaystyle {{5 \euler^ {- 5 s }-4 \euler^ {- 8 s }}\over{s}}

\end{array}\]