Loser66
  • Loser66
Find the equation for the periodic function and its period (picture in comment) 2) find the laplace transform of the function in part 1 3) find \(L^{-1}(\dfrac{k}{as+b})\) where a, b,k are in Real/{0} and constants.
Mathematics
schrodinger
  • schrodinger
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Loser66
  • Loser66
@oldrin.bataku
anonymous
  • anonymous
so it's a sawtooth wave varying between \(a\) and \(2a\) with a period of \(a\)?
Loser66
  • Loser66
I think period is a, also

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anonymous
  • anonymous
well, first look at just the shape of the linear pieces: a line passing through \((0,a)\) and \((a,2a)\) has slope \(1\) and is given by \(f(x)=x+a\)
Loser66
  • Loser66
I have to include the nodes on it. I am sorry |dw:1436467689944:dw|
anonymous
  • anonymous
now, let's try to find way to make the function 'reset' with a period of \(a\)
Loser66
  • Loser66
How about a piece wise function?
anonymous
  • anonymous
consider the function \(\lfloor x\rfloor\), which looks like a stepped ladder: |dw:1436467877418:dw|
anonymous
  • anonymous
if we took the normal linear function \(x\) and subtracted its floor \(\lfloor x\rfloor\), we'd get a periodic saw tooth function: |dw:1436467952972:dw|
anonymous
  • anonymous
|dw:1436467977849:dw|
anonymous
  • anonymous
now, notice that this saw-tooth varies between \(0\) and \(1\) -- if we want it to vary between \(a\) and \(2a\) we'll need to scale it by \(a\) and then shift it up by \(a\) as well: $$f(x)=a+a(x-\lfloor x\rfloor )$$
anonymous
  • anonymous
and lastly, to change its period from \(1\) to \(a\), we'll substitute \(x\to x/a\), giving: $$f(x)=a+a(x/a-\lfloor x/a\rfloor)=a+x-a\lfloor x/a\rfloor $$
anonymous
  • anonymous
so this gives us a formula describing our sawtooth function, but it's not particularly 'nice' for the Laplace transform. the good news is that we don't even need such an explicit definition for us to be able to use the transform
anonymous
  • anonymous
recall that the Laplace transform has 'nice' properties for periodic functions \(f\) with period \(T\): $$\begin{align*}\mathcal{L}\{f\}&=\int_0^\infty e^{-st} f(t)\ dt\\&=\int_0^T e^{-st} f(t) \ dt+\int_T^{\infty} e^{-st} f(t)\ dt\\&=\int_0^T e^{-st} f(t)\ dt+\int_0^\infty e^{-s(t+T)} f(t+T)\ dt\\&=\int_0^T e^{-st} f(t)\ dt+e^{-sT}\int_0^\infty e^{-st}f(t)\ dt\\&=\int_0^Te^{-st}f(t)\ dt+e^{-sT}\mathcal{L}\{f\}\\(1-e^{-sT})\mathcal{L}\{f\}&=\int_0^T e^{-st} f(t)\ dt\\\mathcal{L}\{f\}&=\frac1{1-e^{-sT}}\int_0^Te^{-st} f(t)\ dt\end{align*}$$
anonymous
  • anonymous
so here we have \(T=a\) and if we restrict \(f\) to a single period \([0,T)\) we have \(f(t)=a+t\), so $$\mathcal{L}\{f\}=\frac1{1-e^{-as}}\int_0^a e^{-st}(a+t)\ dt$$
Loser66
  • Loser66
I don't get how you go from the second line to the third line, the second part
anonymous
  • anonymous
i replaced the variable of integration \(t\) with \(t+T\); if it's easier for you to see, I used a substitution \(u=t-T\) so \(\int_T^\infty \dots\ dt\to\int_0^\infty \dots\ du\)
Loser66
  • Loser66
nevermind, I got it
Loser66
  • Loser66
Hold on, you got f(t) = a+t from where?
anonymous
  • anonymous
now we can integrate by parts: $$\begin{align*}\int_0^a e^{-st}(a+t)\ dt&=-\frac1s e^{-st}(a+t)\bigg|_{t=0}^a+\frac1s\int_0^ae^{-st}\ dt\\&=\left[-\frac1se^{-st} (a+t)-\frac1{s^2}e^{-st}\right]_{t=0}^a\\&=-\frac{2a}s e^{-as}-\frac1{s^2}e^{-as}+\frac{a}s+\frac1{s^2}\end{align*}$$
anonymous
  • anonymous
\(f(t)=a+t \) is the function restricted to its first period; look at my first reply, it's a line passing through \((0,a)\) and \((a,2a)\)
Loser66
  • Loser66
got it
anonymous
  • anonymous
so our Laplace transform is: $$\mathcal{L}\{f\}=\frac1{1-e^{-as}}\cdot\frac{as+1-e^{-as}(2as+1)}{s^2}$$
Loser66
  • Loser66
Woah!!! part 3, please
anonymous
  • anonymous
now, part 3 concerns an inverse Laplace transform of \(\dfrac{k}{as+b}\):$$\mathcal{L}^{-1}\left\{\frac{k}{as+b}\right\}=\frac{k}a\mathcal{L}^{-1}\left\{\frac1{s+b/a}\right\}$$
anonymous
  • anonymous
now, consider the Laplace transform of \(e^{kt}\) $$\mathcal{L}\{e^{kt}\}=\int_0^\infty e^{-st} e^{kt}\ dt=\int_0^\infty e^{-(s-k)t}\ dt=\left[-\frac1{s-k}e^{-(s-k)t}\right]_0^\infty=\frac1{s-k}$$
anonymous
  • anonymous
so it follows that \(\mathcal{L}^{-1}\left\{\frac1{s-k}\right\}=e^{kt}\); for \(k=-b/a\) we get \(e^{-bt/a}\), so: $$\mathcal{L}\left\{\frac{k}{as+b}\right\}=\frac{k}ae^{-bt/a}$$
Loser66
  • Loser66
Thank you sooo sooo sooo much. I do appreciate.

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