Laplace Transform

Converts a time-domain function $f(t)$ into a frequency-domain function $F(s)$.

$$F(s) = \mathcal{L}\{ f(t) \} = \int_{0}^{\infty} e^{-st} f(t) \,\text{d}t$$

Mainly used to solve linear ODEs with initial conditions.

For all below-mentioned formulas, the Laplace transform of $f(t)$, is denoted as $F(s)$.

Important formulas

For $n\ge 0$ and $s \gt 0$.

$$\mathcal{L} \{ t^n \} = \frac{n!}{s^{n+1}}$$ $$\mathcal{L} \{ e^{at} \} = \frac{1}{s-a}$$ $$\mathcal{L} \{ \sin (at) \} = \frac{a}{s^2 + a^2}$$ $$\mathcal{L} \{ \cos (at) \} = \frac{s}{s^2 + a^2}$$ $$\mathcal{L} \{ \sinh (at) \} = \frac{a}{s^2 - a^2}$$ $$\mathcal{L} \{ \cosh (at) \} = \frac{s}{s^2 - a^2}$$ $$\mathcal{L}\{u(t-a)\}= \frac{e^{-as}}{s}$$

Properties

Linearity

$$\mathcal{L} \{ cf(t) + g(t) \} = c\mathcal{L}\{ f(t) \} + \mathcal{L} \{ g(t) \}$$

Shiting

Suppose $F(s)$ is the Laplace transform of $f$, where $s \gt \alpha$.

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

For $s \gt \alpha + a$. $$

Time scaling

$$\mathcal{L} \{ f(\alpha t) \} = \frac{1}{\alpha} F\left(\frac{s}{\alpha}\right)$$

Transform with t

$$\mathcal{L} \{ t^n f(t) \} = (-1)^n \frac{\text{d}^n}{\text{d}s^n} {F(s)}$$ $$\mathcal{L} \left\{ \frac{1}{t} f(t) \right\} = \int_s^\infty F(x)\,\text{d}x$$

Time shifting

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

Here $u$ is the unit step function.

Derivatives and Integrals

Derivative

$$\mathcal{L} \{ f'(t) \} = s F(s) - f(0)$$ $$\mathcal{L} \{ f''(t) \} = s^2 F(s) - sf(0) - f'(0)$$ $$\mathcal{L} \{ f^{(n)}(t) \} = s^n F(s) - \sum_{k=0}^{n-1} s^k f^{(n - 1 - k)}(0)$$

Integral

$$\mathcal{L} \left\{ \int_0^t f(x) \,\text{d}x \right\} = \frac{1}{s} F(s)$$

Provided that $s \neq 0$. $$

For periodic functions

Suppose $f(t)$ is a periodic function with period $T$.

If the Laplace transform of $f(t)$ exists then:

$$\mathcal{L}\{ f(t) \} = \frac{1}{1 - e^{-sT}} \int_0^T e^{-st} f(t) \,\text{d}t$$