Laplace Transform Calculator

The Laplace transform converts a function of time $f(t)$ into a function of complex frequency $F(s)$:

$F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t)\,dt$

The transform is defined for $s$ in some right half-plane $\operatorname{Re}(s) > \sigma$ where the integral converges.

Why this is useful: Laplace converts differentiation into multiplication by $s$, turning linear ODEs with constant coefficients into algebraic equations in $s$. You solve the algebra, then take the inverse Laplace transform to get the answer in the time domain.

Laplace transforms also handle discontinuous and impulsive inputs (step functions, Dirac deltas) elegantly, which makes them indispensable in control theory, signal processing, and electrical engineering.

Basic Transform Pairs

Memorize the core table:

Key Properties

Linearity:

$\mathcal{L}\{a f(t) + b g(t)\} = a F(s) + b G(s)$

First Shifting (s-shift):

$\mathcal{L}\{e^{at} f(t)\} = F(s - a)$

This is how $e^{at}\sin(\omega t) \to \frac{\omega}{(s-a)^2 + \omega^2}$.

Differentiation in the $t$-domain:

$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$

$\mathcal{L}\{f''(t)\} = s^2 F(s) - s f(0) - f'(0)$

This is what converts ODEs into algebra: derivatives become polynomials in $s$ multiplied by $F(s)$, with initial conditions baked in.

Multiplication by $t$:

$\mathcal{L}\{t f(t)\} = -F'(s)$

Inverse Laplace Transform

Given $F(s)$, find $f(t)$ such that $\mathcal{L}\{f(t)\} = F(s)$. Standard techniques:

1. Partial fractions: decompose $F(s)$ into simple rational pieces matching the table.
2. Completing the square: for $\frac{1}{s^2 + bs + c}$ shapes, rewrite as $\frac{1}{(s - a)^2 + \omega^2}$ to match the shifted sine table entry.
3. Look up and combine using linearity.

Solving ODEs with Laplace

For $y'' + 3y' + 2y = e^{-t}$, $y(0) = 0, y'(0) = 1$:

1. Apply Laplace: $s^2 Y - s \cdot 0 - 1 + 3(sY - 0) + 2Y = \frac{1}{s+1}$
2. Solve for $Y$: $Y(s^2 + 3s + 2) = 1 + \frac{1}{s+1}$, so $Y = \frac{s + 2}{(s+1)(s^2+3s+2)} = \frac{1}{(s+1)^2}$ (after simplification).
3. Invert: $y(t) = t e^{-t}$.

Clean and mechanical — the same problem with variation of parameters takes twice the work.

• Forgetting initial conditions: $\mathcal{L}\{f'\} = sF(s) - f(0)$. Skipping $f(0)$ is the single most common error.
• Wrong sign in s-shift: $\mathcal{L}\{e^{at}f(t)\} = F(s - a)$, not $F(s + a)$. Sign matters.
• Mishandling discontinuities: For step inputs, use the unit-step function $u(t-a)$ and the time-shift theorem $\mathcal{L}\{u(t-a)f(t-a)\} = e^{-as}F(s)$.
• Inverse transform without partial fractions: $\frac{1}{(s-1)(s+2)}$ does not invert directly — decompose first.
• Confusing $F(s)$ with $\mathcal{L}^{-1}\{F\}$: $F(s)$ is the transform, $f(t)$ is the original. Always end ODE problems back in the time domain.