List of Notes - Category: Laplace Transform

Definition of the Laplace Transform

The Laplace Transform of a function f(t), defined for all `t>=0`, is the function F(s), defined as follows:

`F(s)=L(f(t))=int_0^ooe^(-st)f(t)dt`, where s is a complex parameter.

Let's go through a couple of examples.

Table of the Laplace Transforms

This is not complete list of Laplace Transforms, but it contains all common transforms:

` f(t)=L^(-1)(F(s)) ` `F(s)=L(f(t))` 1 `1/s` `t^n`, n=0, 1, 2, 3... `(n!)/(s^(n+1))` `t^n`, n>-1 ` (Gamma(n+1))/s^(n+1) ` `e^(at)` `1/(s-a)` `t^(n-1/2)`, n=1,2,3.. `(1*3*5*...*(2n-1)*sqrt(pi))/(2^ns^(n+1/2))` `sqrt(t)` `sqrt(pi)/(2s^(3/2))` `sin(at)` `a/(s^2+a^2)` `cos(at)` `s/(s^2+a^2)` `sinh(at)` `a/(s^2-a^2)` `cosh(at)` `s/(s^2-a^2)` `tsin(at)` `(2as)/(s^2+a^2)^2` `tcos(at)` `(s^2-a^2)/(s^2+a^2)^2` `sin(at+b)` `(s*sin(b)+a*cos(b))/(s^2+a^2)` `cos(at+b)` `(s*cos(b)-a*sin(b))/(s^2+a^2)` `e^(at)sin(bt)` `b/((s-a)^2+b^2)` `e^(at)cos(bt)` `(s-a)/((s-a)^2+b^2)` `e^(at)sinh(bt)` `b/((s-a)^2-b^2)` `e^(at)cosh(bt)` `(s-a)/((s-a)^2-b^2)` `t^n e^(at)`, n=1,2,3... `(n!)/(s-a)^(n+1)` `f(ct)` `1/cF(s/c)` `u_c(t)=u(t-c)` `e^(-cs)/s` `u_c(t)f(t-c)` `e^(-cs)F(s)` `\delta(t-c)` `e^(-cs)` `\e^(ct)f(t)` `F(s-c)` `t^nf(t)`, n=1,2,3... `(-1)^nF^((n))(s)` `int_0^tf(\tau)d\tau` `(F(s))/s` `int_0^tf(t-tau)g(tau)dtau` `F(s)G(s)` `f'(t)` `sF(s)-f(0)` `f''(t)` `s^2F(s)-s f(0)-f'(0)` `f^((n))(t)` `s^nF(s)-sum_(k=0)^(n-1)(s^(n-1-k)f^((k))(0))`

Properties of the Laplace Transform

Laplace transform has some interesting properties.

Property 1. Linearity of Laplace transform: `L(af(t)+bg(t))=aL(f(t))+L(g(t))`, where a and b are some constants.

Proof is straightforward through definition:

Inverse Laplace Transform

As was seen previously Laplace transform is `F(s)=L(f(t))` , in other words we are given function `f(t)` and we need to find `F(s)` . Inverse Laplace transform is operation of finding `f(t)` given `F(s)` .

Convolution Integral

To find inverse Laplace transform partial fraction decomposition is very useful, but sometimes it can be very difficult to find partial fraction decomposition, so in some cases inverse Laplace transform can be found with the help of convolution integral.

Unit (Heaviside) Step Function

Heaviside function is defined as follows:

`u_c(t)=u(t-c)=H(t-c)={(1 if t>=c),(0 if t<c):}`

Unit step function is useful in sense that piecewise continuous functions can be written in terms of step functions.

Dirac Delta Function

Heaviside function represents switches from one value to another at some point, but what if we need instant change to very big value? This is what Dirac Delta function means.

Dirac Delta function is defined as follows: `delta(t-c)={(+oo if t=c),(0 if t!=c):}`

Solving IVP's with Laplace Transform

You probably asked yourself why Laplace transform is in Differential Equations section. Answer is simple. Because we can solve initial-value problems with the help of Laplace transform.

Let's see how it is done.