Get the solution to "I'm having difficulties calculating a simple Laplace inverse..."

Haylee Bowen

Answered question

2022-04-07

I'm having difficulties calculating a simple Laplace inverse :

\frac{S - 4}{S^{2} - 2 S - 11}

enchantsyseq

Beginner2022-04-08Added 19 answers

When Laplace transforms are rational functions, the typical line of attack is to put it into the form of partial fractions. Note that

S^{2} - 2 S - 11 = {(S - 1)}^{2} - 12

has roots

1 \pm \sqrt{12} = 1 \pm 2 \sqrt{3}

, so:

\frac{S - 4}{S^{2} - 2 S - 11} = \frac{S - 4}{(S - 1 - 2 \sqrt{3}) (S - 1 + 2 \sqrt{3})}

So, we try to write it in the form

\frac{A}{S - 1 - 2 \sqrt{3}} + \frac{B}{S - 1 + 2 \sqrt{3}}

, with A,B constants. One way of solving for these constants is the cover-up method. We apply this here:

A = \frac{S - 4}{S - 1 + 2 \sqrt{3}} ∣_{S = 1 + 2 \sqrt{3}} = \frac{- 3 + 2 \sqrt{3}}{4 \sqrt{3}} = \frac{2 - \sqrt{3}}{4}

B = \frac{S - 4}{S - 1 - 2 \sqrt{3}} ∣_{S = 1 - 2 \sqrt{3}} = \frac{- 3 - 2 \sqrt{3}}{- 4 \sqrt{3}} = \frac{2 + \sqrt{3}}{4}

Thus, our function's Laplace transform can be written as

\frac{1}{4} (\frac{2 - \sqrt{3}}{S - 1 + 2 \sqrt{3}} + \frac{2 + \sqrt{3}}{S - 1 - 2 \sqrt{3}})

We now recall that

e^{a t}

has Laplace transform

\frac{1}{s - a}

, and this allows us to recover our function as:

f (t) = \frac{1}{4} ((2 - \sqrt{3}) e^{(1 - 2 \sqrt{3}) t} + (2 + \sqrt{3}) e^{(1 + 2 \sqrt{3}) t})

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