Laplace Transform

Make a flowchart for the cheat sheet to help decide Fourier vs Laplace transform!

The Laplace transform of a function $f(t)$ is defined by $\tilde{f}(s)={\int }_0^{\infty }f(t)e^{-st}dt$, used for purely real functions.

Laplace transforms can exist when ${\lim }_{t\to \infty }f(t)\ne 0$, when the Fourier transform as $t$ tends towards $\infty$ it does not exist.

Laplace transforms are often used when $t>0$, e.g. initial-value problems.

Laplace transforms are often denoted by $\mathcal{L}[f(t)]$.

Inverting Laplace transforms is often done so with table lookups rather than contour integrals, by breaking a function into known ones to hence find ${\mathcal{L}}^{-1}$ of everything to then consolidate.

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Example One

Calculate $\mathcal{L}(\cos \omega t)$::$\frac{s}{s^2+{\omega }^2}$.

Example Two

Calculate $\mathcal{L}(\sin \omega t)$::$\frac{\omega }{s^2+{\omega }^2}$.

Example Three

Using the Table of Laplace transforms, find $f(t)$ if $\tilde{f}(s)=\frac{s+3}{s(s+1)}$::$f(t)=3-2e^{-t}$.

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Properties of Laplace Transforms

Laplace transforms have four main properties: the shift theorem, scaling, exponential product, and convolution theorem.

The shift theorem of Laplace transforms is defined as::$\mathcal{L}[e^{-at}f(t)]=\tilde{f}(s+a)$, proven by using the laws of exponentials and definitions.

The scaling property of Laplace transforms is defined as::$\mathcal{L}[f(at)]=\frac{1}{a}\tilde{f}\left(\frac{s}{a}\right)$.

The exponential product property of Laplace transforms is defined as::$\mathcal{L}[t^{n}f(t)]=(-1{)}^n\frac{d^n}{d{s}^n}\tilde{f}(s)$, which relates products within transforms with differentiation.

This exponential product property of Laplace transforms can also defined the Laplace transform of a first derivative as $\mathcal{L}\left[\frac{df}{dt}\right]=-f(0)+s\tilde{f}(s)$. Similarly, the second derivative is defined as $\mathcal{L}\left[\frac{d^{2}f}{d{t}^2}\right]=s^2\tilde{f}(s)-sf(0)-f^{\prime }(0)$.

Using Laplace transforms to solve differential equations can be much easier when: 1. initial conditions are supplied and 2. when the right-hand side is not zero.

The convolution theorem for Laplace transforms is defined as::$\mathcal{L}\left[{\int }_0^{t}f(t-u)g(u)du\right]=\tilde{f}(s)\tilde{g}(s)$, which can reflect a sort of memory in systems.

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Example One

Find the solution of $y^{\prime \prime }+4y^{\prime }+4y=t^2{e}^{-2t}$::$y=\frac{t^4{e}^{-2t}}{12}$.