Lesson 25 - Transforms of Heaviside Functions

We can also take the Laplace transform of Heaviside functions. Let's look at this using the definition of Laplace transforms.

Recall the definition of the Laplace transform.

L [f (t)] = F (s) := \int_{0}^{\infty} e^{- s t} f (t) d t = lim_{b \to \infty} \int_{0}^{b} e^{- s t} f (t) d t

Let's use this definition to find $L [u (t - a)]$ .

L [u (t - a)] = \int_{0}^{\infty} (e^{- s t} u (t - a)) d t

From the last lesson, we know that when $t < a$ , $u (t - a) = 0$ and when $t > a,$ $u (t - a) = 1$ .

\int_{0}^{\infty} (e^{- s t} u (t - a)) d t = \int_{0}^{a} (e^{- s t} (0)) d t + \int_{a}^{\infty} (e^{- s t} (1)) d t

= \int_{a}^{\infty} (e^{- s t} (1)) d t

Since $\frac{1}{s}$ is a constant, let's just pull it out of the integral.

= lim_{b \to \infty} \int_{a}^{b} e^{- s t} d t = - \frac{1}{s} [e^{- s t}]_{a}^{b}

= \frac{1}{s} [e^{- s t}]_{a}^{b} = \frac{1}{s} [e^{- s b} - e^{- s a}]

As $b \to \infty$ , the term $e^{- s b}$ goes to zero. So we are left with

\frac{1}{s} e^{- s a}

L [u (t - a)] = \frac{e^{- a s}}{s}

Try For Yourself

Find $L [u (t - a) f (t - a)]$

Additional definitions:

$L [u (t - a)] = \frac{e^{- a s}}{s}$

$L [u (t - a) f (t - a)] = e^{- a s} F (s)$ , where $F (s) = L [f (t)]$

$L^{- 1} [e^{- a s} F (s)] = u (t - a) f (t - a)$

It's important to note that $u (t - a)$ is a time shift. This shift affects all $t$ variables.

Find $L^{- 1} [\frac{e^{- 2 s}}{s - 1}]$

L^{- 1} [\frac{e^{- 2 s}}{s - 1}]

I like to pull out the time shift.

L^{- 1} [e^{- 2 s}] L^{- 1} [\frac{1}{s - 1}]

This notation is not convention, but it helps me to understand what's going on when we take these inverse Laplace transforms.

We know that $L^{- 1} [\frac{1}{s - 2}] = e^{t}$ . Now how do we apply the shift?

L^{- 1} [e^{- 2 s}] = u (t - 2)

Now to apply the shift, combine the two transforms, and shift the $t$ . Since $a$ is $- 2$ , we will shift $t$ by $- 2$ .

L^{- 1} [\frac{e^{- 2 s}}{s - 1}] = e^{t - 2} u (t - 2)

Again, the shift applies to all variables of $t$ . It's also important to note that the term $e^{- 2 s}$ is not part of the Laplace transform function, it just acts as a time shift and when we take the inverse Laplace transform we convert the time shift into the form of a Heaviside function.

Try For Yourself

$L^{- 1} [\frac{e^{- s} (3 s^{2} - s + 2)}{(s - 1) (s + 1)}]$ (Hint, you will have to perform Partial Fraction Decomposition).
$L^{- 1} [\frac{s e^{- 3 s}}{s^{2} + 4 s + 5}]$

Next Lesson: Lesson 26 - Convolution

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