Laplace Transform Properties
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## 1 Properties and theorems

Given the functions f(t) and g(t), and their respective Laplace transforms F(s) and G(s):

$LaTeX: f(t) = \mathcal{L}^{-1} \{ F(s) \}$
$LaTeX: g(t) = \mathcal{L}^{-1} \{ G(s) \}$

the following table is a list of properties of unilateral Laplace transform:

Properties of the unilateral Laplace transform
Time domain Frequency domain Comment
Linearity $LaTeX: a f(t) + b g(t) \$ $LaTeX: a F(s) + b G(s) \$ Can be proved using basic rules of integration.
Frequency differentiation $LaTeX: t f(t) \$ $LaTeX: -F'(s) \$
Frequency differentiation $LaTeX: t^{n} f(t) \$ $LaTeX: (-1)^{n} F^{(n)}(s) \$ More general form
Differentiation $LaTeX: f'(t) \$ $LaTeX: s F(s) - f(0^-) \$ Write the exact integral form of the given function, and add another integral to complement the former to deduce the sum to indefinite integration of a differential. Next few steps are simple.
Second Differentiation $LaTeX: f''(t) \$ $LaTeX: s^2 F(s) - s f(0^-) - f'(0^-) \$ Apply the Differentiation property to $LaTeX: f'(t)$.
General Differentiation $LaTeX: f^{(n)}(t) \$ $LaTeX: s^n F(s) - s^{n - 1} f(0^-) - \cdots - f^{(n - 1)}(0^-) \$ Follow the process briefed for the Second Differentiation.
Frequency integration $LaTeX: \frac{f(t)}{t} \$ $LaTeX: \int_s^\infty F(\sigma)\, d\sigma \$
Integration $LaTeX: \int_0^t f(\tau)\, d\tau = u(t) * f(t)$ $LaTeX: {1 \over s} F(s)$ $LaTeX: u(t)$ is the Heaviside step function.
Scaling $LaTeX: f(at) \$ $LaTeX: {1 \over |a|} F \left ( {s \over a} \right )$
Frequency shifting $LaTeX: e^{at} f(t) \$ $LaTeX: F(s - a) \$
Time shifting $LaTeX: f(t - a) u(t - a) \$ $LaTeX: e^{-as} F(s) \$ $LaTeX: u(t)$ is the Heaviside step function
Convolution $LaTeX: (f * g)(t) \$ $LaTeX: F(s) \cdot G(s) \$
Periodic Function $LaTeX: f(t) \$ $LaTeX: {1 \over 1 - e^{-Ts}} \int_0^T e^{-st} f(t)\,dt$ $LaTeX: f(t)$ is a periodic function of period $LaTeX: T$ so that $LaTeX: f(t) = f(t + T), \; \forall t$
• Initial value theorem:
$LaTeX: f(0^+)=\lim_{s\to \infty}{sF(s)}$
• Final value theorem:
$LaTeX: f(\infty)=\lim_{s\to 0}{sF(s)}$, all poles in left-hand plane.
The final value theorem is useful because it gives the long-term behaviour without having to perform partial fraction decompositions or other difficult algebra. If a function's poles are in the right hand plane (e.g. $LaTeX: e^t$ or $LaTeX: \sin(t)$) the behaviour of this formula is undefined.

### 1.1 Proof of the Laplace transform of a function's derivative

It is often convenient to use the differentiation property of the Laplace transform to find the transform of a function's derivative. This can be derived from the basic expression for a Laplace transform as follows:

$LaTeX: \mathcal{L} \left\{f(t)\right\} =\int_{0^-}^{+\infty} e^{-st} f(t)\,dt$
$LaTeX: ~~ = \left[ \frac{f(t)e^{-st}}{-s} \right]_{0^-}^{+\infty} -

\int_{0^-}^{+\infty} \frac{e^{-st}}{-s} f'(t)dt$

(by parts)

$LaTeX: ~~ = \left[-\frac{f(0)}{-s}\right] +

\frac{1}{s}\mathcal{L}\left\{f'(t)\right\},$

yielding

$LaTeX: \mathcal{L}\left\{\frac{df}{dt}\right\} = s\cdot\mathcal{L} \left\{ f(t) \right\}-f(0),$

and in the bilateral case, we have

$LaTeX: \mathcal{L}\left\{ { df \over dt } \right\}
 = s \int_{-\infty}^{+\infty} e^{-st} f(t)\,dt = s \cdot \mathcal{L} \{ f(t) \}.$