Inverse Laplace Transform

The inverse Laplace transform

The inverse Laplace transform reverses the forward transform: $\mathcal{L}^{-1}\{F(s)\}(t) = f(t) \quad \iff \quad \mathcal{L}\{f(t)\}(s) = F(s).$

In practice, we rarely use the formal inversion integral (the Bromwich contour in the complex plane). Instead, we use a mix of:

1. A small table of known pairs $(f, F)$.
2. Linearity to decompose $F(s)$ into simpler pieces.
3. Partial fractions to break rational $F$ into sums of table entries.
4. Completing the square to expose $s$-shifts in the denominator.
5. Shifting theorems to recognise $e^{a t} f(t)$ patterns.

This problem exercises completing the square and the first shifting theorem in reverse.

The given transform

$F(s) = \frac{s + 1}{s^{2} + 2 s + 5}$

Denominator has no real roots (discriminant $= 4 - 20 = -16 < 0$), so the natural form is sine/cosine times an exponential — exactly what completing the square reveals.

Step-by-step inversion

Step 1 — Complete the square in the denominator.

$s^{2} + 2 s + 5 = (s^{2} + 2 s + 1) + 4 = (s + 1)^{2} + 4.$

So $F(s) = \frac{s + 1}{(s + 1)^{2} + 4} = \frac{s + 1}{(s + 1)^{2} + 2^{2}}.$

Step 2 — Recognise the shifted-cosine pattern.

Recall the table entry $\mathcal{L}\{\cos(b t)\} = \frac{s}{s^{2} + b^{2}}.$

With $b = 2$: $\mathcal{L}\{\cos(2 t)\} = \frac{s}{s^{2} + 4}.$

Step 3 — Apply the first shifting theorem in reverse.

First shifting theorem: $\mathcal{L}\{e^{a t} f(t)\} = F(s - a)$, where $F = \mathcal{L}\{f\}$. Equivalently, $\mathcal{L}^{-1}\{F(s - a)\} = e^{a t} \, f(t).$

In our case, the structure $\dfrac{s + 1}{(s + 1)^{2} + 4}$ is exactly the cosine transform $\dfrac{s}{s^{2} + 4}$ with $s \to s + 1$ — i.e. $s \to s - (-1)$, so $a = -1$: $\boxed{\, \mathcal{L}^{-1}\!\left\{ \frac{s + 1}{(s + 1)^{2} + 4} \right\} = e^{-t} \cos(2 t) \,}$

Verification

Take the forward Laplace of $f(t) = e^{-t} \cos(2 t)$ to check we hit $F(s)$ back: $\mathcal{L}\{e^{-t} \cos(2 t)\} = F(s)\big|_{s \to s + 1} \text{ where } F(s) = \mathcal{L}\{\cos(2t)\} = \frac{s}{s^{2} + 4}$ $= \frac{s + 1}{(s + 1)^{2} + 4} = \frac{s + 1}{s^{2} + 2 s + 5}. \; \checkmark$

Why completing the square

When the denominator of $F(s)$ is an irreducible quadratic $s^{2} + B s + C$ (i.e. complex-conjugate poles), completing the square is the cleanest way to spot the shifted sine/cosine form hiding inside. The general recipe:

$\frac{\text{Numerator}}{s^{2} + B s + C} \;\;\Longrightarrow\;\; \frac{\text{Numerator}}{(s + B/2)^{2} + (C - B^{2}/4)}$

and rewrite the numerator in terms of $(s + B/2)$ and the constant $1$ (or equivalently $\sqrt{C - B^{2}/4}$): $\frac{A (s + B/2) + D}{(s + B/2)^{2} + \omega^{2}}$ where $\omega = \sqrt{C - B^{2}/4}$. Now each piece matches the shifted cosine or sine directly:

• $\dfrac{A (s + B/2)}{(s + B/2)^{2} + \omega^{2}} \;\longleftrightarrow\; A \, e^{-B t / 2} \cos(\omega t)$
• $\dfrac{D}{(s + B/2)^{2} + \omega^{2}} \;\longleftrightarrow\; \dfrac{D}{\omega} e^{-B t / 2} \sin(\omega t)$

When partial fractions take over

If the denominator does factor over the reals, use partial fractions. Example: $\frac{1}{s^{2} - 4} = \frac{1}{(s - 2)(s + 2)} = \frac{1/4}{s - 2} - \frac{1/4}{s + 2}$ $\mathcal{L}^{-1}\left\{\frac{1}{s^{2} - 4}\right\} = \tfrac{1}{4} e^{2 t} - \tfrac{1}{4} e^{-2 t} = \tfrac{1}{2} \sinh(2 t).$

The decomposition mirrors the pole structure: each real pole contributes an $e^{r t}$.

Heaviside cover-up for fast coefficients

For simple poles, the partial-fraction coefficient at pole $s = r$ is $A_r = \lim_{s \to r} (s - r) \, F(s).$

This lets you read coefficients off by "covering up" the relevant factor — no need for simultaneous linear equations.

Quick inverse-Laplace reference (complementary direction of #189)

• $\dfrac{1}{s - a} \;\to\; e^{a t}$
• $\dfrac{1}{s^{2} + b^{2}} \;\to\; \dfrac{1}{b} \sin(b t)$
• $\dfrac{s}{s^{2} + b^{2}} \;\to\; \cos(b t)$
• $\dfrac{1}{(s - a)^{2} + b^{2}} \;\to\; \dfrac{1}{b} e^{a t} \sin(b t)$
• $\dfrac{s - a}{(s - a)^{2} + b^{2}} \;\to\; e^{a t} \cos(b t)$ (this problem)
• $\dfrac{n!}{(s - a)^{n + 1}} \;\to\; t^{n} e^{a t}$

Common mistakes

• Reading the shift direction wrong. $F(s - a) \;\leftrightarrow\; e^{a t} f(t)$ — a positive $a$ in $(s - a)$ corresponds to growing $e^{a t}$; a negative $a$ (as here, giving $(s + 1)$) corresponds to decaying $e^{-t}$.
• Missing the $1/b$ when inverting $\dfrac{1}{s^{2} + b^{2}}$. The inverse is $\dfrac{\sin(b t)}{b}$, not $\sin(b t)$.
• Forgetting to split the numerator. If the numerator isn't already lined up with the shifted-square form, algebraically split it: $s + 3 = (s + 1) + 2$ gives one cosine piece and one sine piece.
• Partial-fractioning irreducible quadratics. Don't try to factor $s^{2} + 2 s + 5$ over the reals; complete the square instead.

Try it in the visualization

Place a pole pair on the $s$-plane and read the time-domain waveform off the pole location: real part → decay rate, imaginary part → oscillation frequency. Move the pole pair horizontally (change damping) or vertically (change frequency) and watch the time-domain curve update continuously.