# 6.8: Step Functions

- Page ID
- 395

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In this discussion, we will investigate piecewise defined functions and their Laplace Transforms. We start with the fundamental piecewise defined function, the *Heaviside function*.

Definition: The Heaviside Function

The Heaviside function, also called the unit step function, is defined by

$$ u_c(t) = \left\{\begin{aligned}

&1 && t <c \\

&0 && t \ge c

\end{aligned}

\right.$$

for \(c \ge 0\).

The Heaviside function \(y = u_c(t)\) and \(y = 1 - u_c(t)\) are graphed below.

Example \(\PageIndex{1}\)

We can write the function

$$ f(t) = \left\{\begin{aligned}

&3 && 0 \le x < 2 \\ &e^x && 2 \le x < 5 \\ &0 && 5 \le x

\end{aligned}

\right.$$

in terms of Heaviside functions.

**Solution**

We tackle the functions in parts. The function that is 1 from 0 to 2, and 0 otherwise is

\[1 - u_2(x) .\]

Multiplying by 3 gives

\[ 3\left(1 - u_2(x)\right) = 3 - 3u_2(x) .\]

To get the function that is 1 between 2 and 5 and 0 otherwise, we subtract

\[ u_2(x) - u_5(x) .\]

Now multiply by \(e^x\) to get

\[ e^x(u_2(x) - u_5(x)) = e^x u_2(x) - e^x u_5(x). \]

Adding these together gives

\[\begin{align} f(x) &= 3 - 3u_2(x) + e^x\, u_2(x) - e^x \,u_5(x) \\ &= 3 + (e^x - 3)\,u_2(x) - e^x\, u_5(x) . \end{align}\]

We can find the Laplace transform of \(u_c(t)\) by integrating

\[ \mathcal{L}\{u_c(t)\} = \int_0^{\infty} e^{st} u_c(t) \, dt \]

\[ = \int _c^{\infty} e^{st} dt = \dfrac{e^{-cs}}{s} \]

\[ L \{ u_c(t)\} = \dfrac{e^{-cs}}{s}. \]

In practice, we want to find the Laplace transform of a more general piecewise defined function such as

$$ f(t) = \left\{\begin{aligned}

&0 && x < \pi \\ &\sin x && x \ge \pi

\end{aligned}

\right.$$

This type of function occurs in electronics when a switch is suddenly turned on after one second and a forcing function is applied. We can write

\[ f(x) = u_p(x) \sin x.\]

We will be interested in the Laplace transform of a product of the Heaviside function with a continuous function. The result that we need is

\[ \mathcal{L}\{(u_c(t) f(t - c)\} = e^{-cs} \mathcal{L}\{f(t)\} .\]

By taking inverses we get that if \(F(s) = \mathcal{L}\{f(t)\}\), then

\[ L^{-1}\{e^{-cs}F(s)\} = u_c(t)f(t - c) .\]

**Proof**

We use the definition to get

\[ \mathcal{L}\{u_c(t)f(t-c)\} = \int _0^{\infty} e^{-st}u_c(t)f(t-c)\,dt\]

substitute \(\nu=t-c\)

\[ = \int _c^{\infty} e^{-st}f(t-c)\,dt = \int _0^{\infty} e^{-(\nu +c)s}f(\nu)\,d\nu \]

\[ e^{-cs}\int _0^{\infty} e^{-s\nu}f(\nu)\,d\nu = e^{-cs}F(s).\]

Example \(\PageIndex{2}\)

Find the Laplace transform of

$$ f(t) = \left\{\begin{aligned}

&0 && x < \pi \\ &\sin x && x \ge \pi.

\end{aligned}

\right.$$

**Solution**

We use that fact that

\[ f(x) = u_p(x) \sin x = -u_p(x) \sin[(x - p)] . \]

Now we can use the formula to get that

\[ \mathcal{L}\{f(x)\} = -\mathcal{L}\{u_p(x) \sin[(x - p)]\} = -e^{-cs} \mathcal{L}\{\sin x\}. \]

By the table of Laplace Transforms, we get

\[ = \dfrac{-e^{-cs}}{s^2 + 1} .\]

## Contributors

- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.