8.7 KiB
Laplace transform
From now on, LT is short for Laplace Transform #LT
What is LT? It's denoted as \mathcal{L} and it's an operator defined as the following:
{\underbrace{ \mathcal{L}\{f(t)\}(s) }_{ =F(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.
Btw, an operator is something that takes in a function and spits out a new function. For example, integration is a kind of operator.
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. (Don't worry, you will find these properties and many more on the table provided in the next lectures)
#end of lec 14 #start of lec 15
#ex #LT
compute the LT of this funny function:
f(t)=\begin{cases}1 &\text{if } 0\leq t\leq 1 \\ 2 &\text{if } 1<t\leq 2 \\0 & \text{if } 2<t \end{cases}
F(s)=\int _{0}^1 e^{-st}\, dt+2\int _{1}^2 e^{-st}\, dt+0
F(s)=-\frac{1}{s}e^{-st}|_{t=0}^{t=1}-\frac{2}{s}e^{-st}|_{t=1}^{t=2}
F(s)=-\frac{1}{s}(e^{-s}-1)-\frac{2}{s}(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.
So what is jump discontinuity look at the picture:
!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.
What if we calculate the LT of f' ?
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}+ \underbrace{s \int e^{-st}f(t) \, dt }_{ sF(s) }=sF(s)-f(0)
\mathcal{L}\{f''\}=s^2F(s)-sf(0)-f'(0)
in general, we can use proof by induction to show the following (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))'
F'(s)=\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
so that means:
-\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
#start of lec 16
He advises us to learn the table of common LT's, however a sheet will be provided for the exam.
#ex #LT
lets try to compute the LT of:
\mathcal{L}\{t\cos (t)e^t\}= ?
Look at the table, which one would be useful? (the bottom one?)
We know \mathcal{L}\{t^nf(t)\}=(-1)^n \frac{d^nF}{ds^n} this property was shown last lecture, can be proven by induction. It's good to learn this by heart.
\mathcal{L}\{t\cdot\underbrace{ \cos (t)e^t }_{ f(t) }\}=-\frac{d}{ds}\mathcal{L}\{e^t\cos t\}
how do we compute that \mathcal{L}? Well now we can use that last row in the table!
=-\frac{d}{ds}\left( \frac{{s-1}}{(s-1)^2+1^2} \right)
=\frac{{(s-1)^2-1}}{((s-1)^2+1)^2}Doing this normally would be hard, with the table its a piece of cake!
Inverse Laplace Transform
we know LT is a operator, we can also define an inverse transform! The inverse of LT. #inv_LT Here's the definition:
\mathcal{L}^{-1}\{F(s)\}=f(t)Recall the forward LT was: \mathcal{L}\{f(t)\}=F(s)
Next time hopefully, we will see how this will all be useful for solving DE!
We know the forward transform is linear, the inverse is also linear.
\mathcal{L}^{-1}\{\alpha F(s)+\beta G(s)\}=\alpha \mathcal{L}^{-1}\{F\}+\beta \mathcal{L}^{-1}\{G\}
This can be proven rather easily due to the linearity of the forward transform. (wasn't done in class unfortunately.)
#ex
\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \frac{{s-1}}{s^2+2s+10} \right\} notice that all these terms approach 0 as s approaches inf.
using linearity:
second term doesn't look like its in the table, 3rd term looks like second last row
\mathcal{L}^{-1}\left\{ \frac{1}{s^5} \right\}+\frac{3}{4}\mathcal{L}^{-1}\left\{ \frac{1}{\left( s+\frac{5}{2} \right)^2} \right\}+\frac{1}{2}\mathcal{L}^{-1}{\frac{2}{(s+2)^2+2^2}}+\mathcal{L}^{-1}\left\{ \frac{{s+1}}{(s+1)^2+3^2} \right\}
multiply the first term by 4!, divide outside by 4!
=\frac{1}{4!}t^4+\frac{3}{4}te^{-5t/2}+\frac{1}{2}\sin(2t)e^{-2t}+e^{-t}\cos(3t)
it's kinda fun, try to match each term with something in the table, like a puzzle.
We are done.
#ex #inv_LT
\mathcal{L}^{-1}\{\frac{1}{(s-3)(s^2+2s+2)}\} notice the numerator is at least one degree lower than the denominator. The limit of the overall term is zero as s->inf
it doesn't look like anything in the table, can we factor the denominator? not really, they have complex solutions. So maybe split the terms using partial fractions!
=\mathcal{L}^{-1}\left\{ \frac{A}{{s-3}}+\frac{{Bs+C}}{s^2+2s+2} \right\}
=\mathcal{L}^{-1}\left\{ \frac{{As^2+2As+2A+Bs^3+3Bs+Cs-3C}}{(s-3)(s^2+2s+2)} \right\}
we get a linear system of equations:
\begin{bmatrix}A+B&=0 \\2A-3B+C&=0 \\ 2A-3C&=1\end{bmatrix}
skipping some computations:
A=\frac{1}{17},\ B=-\frac{1}{17},\ C=-\frac{5}{17}
=\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s-3)} \right\}-\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{s}{(s+1)^2+1^2} \right\}-\frac{5}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s+1)^2+1^2} \right\}
=\frac{1}{17}e^{3t}-\frac{1}{17}
for second term we use an important identity: s=s+1-1:
-\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{{s+1-1}}{(s+1)^2+1^2} \right\}
final answer:
\frac{1}{17}e^{3t}-\frac{1}{17}e^{-t}\cos(t)-\frac{4}{17}\sin(t)e^{-t}partial fractions, hopefully you remember from math 101:
For each term (s+a)^k
we include \frac{A_{1}}{s+a}+\frac{A_{2}}{(s+a)^2}+\dots + \frac{A_{k}}{(s+a)^k}
for each term (s^2+as+b)^k
we include: \frac{B_{1}s+c_{1}}{(s^2+as+b)}+\frac{{B_{2}s+c_{2}}}{(s^2+as+b)^2}+\dots+\frac{{B_{k}s+c_{k}}}{(s^2+as+b)^k}
#ex #inv_LT
\mathcal{L}^{-1}\left\{ \frac{{3s^2+5s+3}}{s^4+s^3} \right\}
factor out s^3 (s+1)s^3
\mathcal{L}^{-1}\left\{ \frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{s+1} \right\}
=\mathcal{L}^{-1}\left\{ \frac{{As^2(s+1)+Bs(s+1)+C(s+1)+Ds^3}}{(s+1)s^3} \right\}
we get a linear system:
\begin{matrix}A+D=0 \\A+B=3 \\B+C=5 \\C=3\end{matrix}
solving the linear system yeilds:
A=1,\ B=2,\ C=3,\ D=-1
so:
\mathcal{L}^{-1}\left\{ \frac{1}{s}+\frac{2}{s^2}+\frac{3}{s^3}+\frac{-1}{s+1} \right\}
=1+2t+\frac{3}{2}t^2-e^{-t}is the final answer.
#ex #inv_LT
\mathcal{L}^{-1}\left\{ \ln \frac{{s^2+9}}{s^2+1} \right\}
how do we get rid of the ln? work with the derivative.
property: \mathcal{L}\{tf(t)\}=-\frac{dF}{ds}
\mathcal{L}^{-1}\left\{ \frac{d}{ds}\ln \frac{{s^2+9}}{s^2+1} \right\}=-tf(t)
partial fractions:
\mathcal{L}^{-1}\left\{ -\frac{16s}{(s^2+1)(s^2+9)} \right\}=2\mathcal{L}^{-1}\left\{ \frac{s}{{s^2+9}}-\frac{s}{{s^2+1}} \right\}=2(\cos 3t-\cos t)
f(t)=-\frac{2}{t}(\cos 3t-\cos t)
#end of lec 16