Laplace Transforms 101 :explode: (Functions we know how to do laplace to…
Laplace Transforms 101 :explode:
w0t is dis
why is dis
Compare to a standard function (exp function)
we chose exp because it cant be attacked by the differential operator. exp is always there for us. :<3:
why does integral from inf to 0
tell us how similar two functions are??
I'm taking math2010 this sem and have a questions about how laplace transforms work. More specifically, if laplace transforms compare f(t) to a standard exponential function, then why does integral transform from inf to 0 tell us how similar f(t) is to function e^(-st)?
I'm aware you're not the math2010 lecturer, but I thought your background in algebraic methods might help contextualise laplace transforms.
Theorems and stuff
First shifting theorem
Allows you to scalar multiply and add Laplace transforms (of component functions)
how to do dis?
By definition (like a mathematician)
By looking up in a table (like an engineer)
Remember to check assumptions hold
t^n where n is an integer greater than 0
e^at: a is a constant, but not necessarily real. If complex, Re(s)>Re(a)
The integral transform only works if lim t-> infinity then e^-st -> 0
so real(s) > 0
first principle: t>=0 ; Re(s) > 0
Functions we know
how to do laplace to :
sine and cosine
When does it fail?
You can take the L of a rational function, but it won't be defined for the section of the domain where the demoninator part is zero
Functions cannot be Laplace transformed if it makes the improper integral in the defn does not converge I.E. if your f(t) grows faster than expotential growth
When is it defined?
The laplace of f(t) exists and is defined for all s>a
if f(t) is piecewise cts
and for t>= 0 there exists M, a >0 such that for all t>0, |f(t)| <= Me^(at)