#maths
# lecture 1
fourier transforms let you have a function that is defined on the entire real line. this means that they dont need to be periodic
one of the disadvantages is that you always need to integrate with complex integrals.

fourier transforms are useful becuase $\mathcal{F}[\frac{df}{dt}] = \bf j \omega \mathcal{F}[f(t)]$. This means that taking the fourier transform of a differential equation will lead to an algebraic equation.

$\mathcal{F}[\frac{d^nf}{dt^n}] = (\bf j \omega)^n \mathcal{F}[f(t)]$
$\mathcal{F} [\alpha f(t) + \beta g(t)] = \alpha \mathcal{F}[f(t)] + \beta \mathcal{F}[g(t)]$

working with frequency space vs time space is useful because with the transfer function, you can see when you will get the most/least amount of signal through the system.
to find the maximum response, you just need the maximum of the transfer function

# lecture 2

a laplace transform is like a fourier transform but with $\int_0 ^\infty e^{-st}f(t) dt$. the $\omega$ in the laplace transform is like the $s$ in a laplace transform.
laplace transforms don't exist for all functions

a laplace transform always exists if a function grows slower than an exponential. if the function grows slower than an exponential (like $e^{x^2}$) the transform never exists

it is given that $\mathcal{L}[f(x)] = \tilde{f}(s)$
$\mathcal{L}[e^{-ax}f(x)] = \tilde{f}(s+a)$
this is known as the first shift theorem

$\mathcal{L}[\frac{df}{dx}] = s\tilde{f}(s) - f(0)$
$\mathcal{L}[\frac{d^2f}{dx^2}] =s^2\tilde f(s) - sf(0) - f'(0)$


# lecture 3
it is easier to have a bunch of laplace transforms to undo a transform than to use an inverse function
![[laplace transforms]]
using $\mathcal L [xf(x)]=-\frac{d\tilde f}{ds}$ gives the polynomial ones