3.7 KiB
Laplace transform
From now on, LT is short for Laplace Transform
What is LT? It's denoted as \mathcal{L} and it's an operator defined as the following:
\underset{ =F(s) }{ \mathcal{L}\{f(t)\}(s) }:=\int _{0}^\infty e^{-st}f(t) dt=\lim_{ T \to \infty }\int _{0} ^T e^{-st}f(t)\, dtThis doesn't look like anything useful, but later on we will learn how it is.
Here are some trivial example computations:
\mathcal{L}\{0\}=0 Look at your bank account, integrate 0 you still get 0 :D
\mathcal{L}\{1\}=\int_{0}^\infty e^{-st} \, dt=-\frac{1}{s}e^{-st}|_{0}^\infty=\frac{1}{s} if s>0
\mathcal{L}\{e^{at}\}=\int_{0}^{\infty} e^{at}e^{-st}\, dt=\int_{0}^{\infty}e^{-(s-a)t}dt=\frac{1}{s-a} if s-a>0
\mathcal{L}\{\sin bt\}=\int _{0}^{\infty}e^{-st}\sin(bt) \, dt=\frac{b}{s^2+b^2} by integration by parts
similarly can be done for cos, but we have run out of time.
#end of lec 14 #start of lec 15
compute the LT of this funny function:
f(t)=\begin{pmatrix}1, & 0\leq t\leq 1 \\ 2, & 1<t\leq 2 \\0, & 2<t \end{pmatrix}
F(s)=\int _{0}^1 e^{-st}\, dt+2\int _{1}^2 e^{-st}\, dt+0
=-\frac{1}{s}e^{-st}|_{t=0}^{t=1}-\frac{2}{s}e^{-st}|_{t=1}^{t=2}=-\frac{1}{s}(e^{-s}-1)-\frac{2}{5}(e^{-2s}-e^{-s})
We have shown how to compute the LT of a choppy function.
\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f\}+\beta y\mathcal{L}\{g\}
the LT is a linear operator. as shown above. Proof:
\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \int_{0}^{\infty}f(t)e^{-st} \, dt+\beta \int_{0}^{\infty}g(t)e^{-st} \, dt \quad \Box
Def: f(t) is piecewise continuous on an interval I if f(t) is continuous on I, except possibly at a /finite/ number of points of /jump/ discontinuity
what is continuouity? the limit exists and equals the value at that point.
!Drawing 2023-10-11 13.17.32.excalidraw
Def: f(t) is of exponential order \alpha if there \exists\ T,\ M, T\geq0 such that \forall\ t>T : f(t)\leq Me^{\alpha t}
this is important so that f(t)e^{-st} doesn't go into infinity,
this can be proven, but not necessary
Theorem: If f(t) is piecewise continuous and of an exponential order \alpha on [0, \infty), then \mathcal{L}\{f\} exists s>\alpha
and we are guarenteed that s>\alpha
e^{\alpha t} is this of exponential order? Of course, M=1,\alpha=\alpha
what about \sin(t)? yes, sin is bounded b/w -1 and 1
What about e^{t^2}? No, t^2 always outgrows \alpha t eventually. This function does not have a LT.
!Drawing 2023-10-11 13.21.18.excalidraw
Theorem if F(s) is a LT then \lim_{ s \to \infty }F(s)=0
properties:
assume LT of f(t) exists:
first property: \mathcal{L}\{e^{\alpha t}f(t)\}=\int _{0}^\infty e^{\alpha t}e^{-st}\, dt=\int _{0}^\infty e^{-(s-a)t}f(t)\, dt=F(s-a)
\mathcal{L}\{e^{\alpha t}\cdot 1\}=\frac{1}{s-\alpha}
\mathcal{L}\{e^{\alpha t}\sin(bt)\}=\frac{b}{(s-a)^2+b^2} these properties are essential for the midterm.
using integration by parts:
\mathcal{L}\{f'(t)\}(s)=\int _{0}^\infty e^{-st}f(t)\, dt=e^{-st}f(t)|_{t=0}^{t\to \infty}+s \underbrace{ \int e^{-st}f(t) \, dt }_{ F(s) }=sF(s)-f(0)
\mathcal{L}{f''}=s^2F(s)-sf(0)-f'(0))
using proof by induction (try at home!):
\mathcal{L}\{f^{(m)}\}=s^mF(s)-s^{m-1}f(0)-\dots-f^{m-1}(0)
trig stuff:
\mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2}
\mathcal{L}\{\cos bt\}=\frac{s}{s^2+b^2}
\cos(bt)=\frac{1}{b}(\sin(bt))'
\frac{dF}{ds}(s)=\frac{d}{ds}\int _{0 } ^\infty e^{-st}f(t)\, dt=\int _{0} ^\infty f(t) \frac{d}{ds}(e^{-st}) \, dt
=-\int _{0} ^\infty e^{-st}tf(t)\, dt
-\frac{dF}{ds}=\mathcal{L}\{tf(t)\}; again we can use induction to prove:
\mathcal{L}\{t''f(t)\}=(-1)^{n} \frac{d^nF}{ds^n}
\mathcal{L}\{t 1\}=-(s^-1)^1=s^-2; \mathcal{L}\{t^n\}= \frac{n!}{s^{n+1}} (can be proven by induction)
today covers all midterm material. Yay!
#end of lec 15