Lesson 22 - Inverse Laplace Transforms

Now let's learn about the inverse Laplace transform. This function will start in the $s$ domain and be transformed back to the $t$ domain.

Let $f (t)$ be a continuous function on $[0, \infty)$ and satisfies $L [f (t)] = F (s)$ then $f (t)$ is the inverse Laplace transform of $F (s)$ and we use the notation $f (t) = L^{- 1} [F (s)] .$ This operator also follows the linearity property.

Let's recall the table from before for reference.

\begin{array}{cc} L [f (t)] & F (s) \\ L [e^{a t}] & \frac{1}{s - a} \\ L [t^{n}] & \frac{n!}{s^{n + 1}} \\ L [\sin (b t)] & \frac{b}{s^{2} + b^{2}} \\ L [\cos (b t)] & \frac{s}{s^{2} + b^{2}} \\ L [e^{a t} f (t)] & F (s - a) \\ L [e^{a t} t^{n}] & \frac{n!}{(s - a)^{n + 1}} \\ L [e^{a t} \sin (b t)] & \frac{b}{(s - a)^{2} + b^{2}} \\ L [e^{a t} \cos (b t)] & \frac{s - a}{(s - a)^{2} + b^{2}} \end{array}

Examples

$L^{- 1} [\frac{3}{s^{3}} + \frac{1}{s^{2} + 9}]$

Impose the linearity property:

$3 L^{- 1} [\frac{1}{s^{3}}] + L^{- 1} [\frac{1}{s^{2} + 9}]$

Using the table above, starting on the right column, we find each of the forms and transform back to $t$ .

Starting with $3 L^{- 1} [\frac{1}{s^{3}}]$ , we notice that there isn't a form that this fits quite into. However, it is most closely related to the $F (s)$ function $\frac{n!}{s^{n + 1}}$ . Since the denominator contains $s^{3}$ , we can deduce that if $n + 1 = 3$ , then $n = 2$ .

However, the numerator does not contain $n = 2$ . We can implement a fix to this by multiplying the numerator by $2$ inside the Laplace transform.

$3 L^{- 1} [\frac{2}{s^{3}}]$

But we can't just alter the inside of the Laplace transform without doing the same to the outside. On the outside, we must multiple by the reciprocal on the outside.

$3 L^{- 1} [\frac{1}{s^{3}}] = \frac{3}{2} L^{- 1} [\frac{2}{s^{3}}]$

Now, the Laplace inverse transform is in the correct form.

$32 L^{- 1} [\frac{2}{s^{3}}] = \frac{3 t^{2}}{2}$

Let's look at the second function now. We can identify that the function looks closely related to the $F (s)$ function $\frac{b}{s^{2} + b^{2}}$ . To get in the proper form, let's substitute $9$ with $3^{2}$ .

$L^{- 1} [\frac{1}{s^{2} + 9}] = L^{- 1} [\frac{1}{s^{2} + 3^{2}}]$

We can implement a similar fix as we did before, but multiply by $3$ inside and by $\frac{1}{3}$ on the outside.

$L^{- 1} [\frac{1}{s^{2} + 3^{2}}] = \frac{1}{3} L^{- 1} [\frac{3}{s^{2} + 3^{2}}]$

We can now take the Laplace inverse transform.

$\frac{1}{3} L^{- 1} [\frac{3}{s^{2} + 3^{2}}] = \frac{\sin (3 t)}{3}$

So,

$L^{- 1} [\frac{3}{s^{3}} + \frac{1}{s^{2} + 9}] = \frac{3 t^{2}}{2} + \frac{\sin (3 t)}{3}$

$L^{- 1} [\frac{1}{s^{2} + 4 s + 8}]$

In this situation, we can complete the square in the denominator. $s^{2} + 4 s + 8 = (s^{2} + 4 s + 4) + 8 - 4 = (s^{2} + 4 s + 4) + 4$ .

This means we can factor the denominator now.

$(s^{2} + 4 s + 4) + 4 = (s + 2)^{2} + 4$

We now have:

$L^{- 1} [\frac{1}{(s + 2)^{2} + 4}]$

And we can now identify that this is the form $\frac{b}{(s - a)^{2} + b^{2}}$ .

Let's implement a fix of scale $2$ .

$\frac{1}{2} L^{- 1} [\frac{2}{(s + 2)^{2} + 4}]$

Now comparing to the table, we can find the proper Laplace inverse transform.

$\frac{1}{2} L^{- 1} [\frac{2}{(s + 2)^{2} + 4}] = \frac{1}{2} e^{- 2 t} \sin (2 t)$

Try for Yourself

$L^{- 1} [\frac{1}{(s + 3) (s + 5)}]$
$L^{- 1} [\frac{3 s + 2}{s^{2} + 2 s + 10}]$
$L^{- 1} [\frac{1}{(s^{2} + 1) (s + 3)}]$

Hint, some of these might require partial fractions in order to find the Laplace transform.

Next Lesson: Lesson 23 - Solving IVPs using Laplace Transforms

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