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Loser66
 one year ago
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.
Loser66
 one year ago
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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it's a sawtooth wave varying between \(a\) and \(2a\) with a period of \(a\)?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I think period is a, also

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, 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
 one year ago
Best ResponseYou've already chosen the best response.0I have to include the nodes on it. I am sorry dw:1436467689944:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now, let's try to find way to make the function 'reset' with a period of \(a\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0How about a piece wise function?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider the function \(\lfloor x\rfloor\), which looks like a stepped ladder: dw:1436467877418:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436467977849:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now, notice that this sawtooth 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
 one year ago
Best ResponseYou've already chosen the best response.0and lastly, to change its period from \(1\) to \(a\), we'll substitute \(x\to x/a\), giving: $$f(x)=a+a(x/a\lfloor x/a\rfloor)=a+xa\lfloor x/a\rfloor $$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0recall that the Laplace transform has 'nice' properties for periodic functions \(f\) with period \(T\): $$\begin{align*}\mathcal{L}\{f\}&=\int_0^\infty e^{st} f(t)\ dt\\&=\int_0^T e^{st} f(t) \ dt+\int_T^{\infty} e^{st} f(t)\ dt\\&=\int_0^T e^{st} f(t)\ dt+\int_0^\infty e^{s(t+T)} f(t+T)\ dt\\&=\int_0^T e^{st} f(t)\ dt+e^{sT}\int_0^\infty e^{st}f(t)\ dt\\&=\int_0^Te^{st}f(t)\ dt+e^{sT}\mathcal{L}\{f\}\\(1e^{sT})\mathcal{L}\{f\}&=\int_0^T e^{st} f(t)\ dt\\\mathcal{L}\{f\}&=\frac1{1e^{sT}}\int_0^Te^{st} f(t)\ dt\end{align*}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so here we have \(T=a\) and if we restrict \(f\) to a single period \([0,T)\) we have \(f(t)=a+t\), so $$\mathcal{L}\{f\}=\frac1{1e^{as}}\int_0^a e^{st}(a+t)\ dt$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I don't get how you go from the second line to the third line, the second part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i replaced the variable of integration \(t\) with \(t+T\); if it's easier for you to see, I used a substitution \(u=tT\) so \(\int_T^\infty \dots\ dt\to\int_0^\infty \dots\ du\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Hold on, you got f(t) = a+t from where?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now 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
 one year ago
Best ResponseYou've already chosen the best response.0\(f(t)=a+t \) is the function restricted to its first period; look at my first reply, it's a line passing through \((0,a)\) and \((a,2a)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so our Laplace transform is: $$\mathcal{L}\{f\}=\frac1{1e^{as}}\cdot\frac{as+1e^{as}(2as+1)}{s^2}$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Woah!!! part 3, please

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now, 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
 one year ago
Best ResponseYou've already chosen the best response.0now, consider the Laplace transform of \(e^{kt}\) $$\mathcal{L}\{e^{kt}\}=\int_0^\infty e^{st} e^{kt}\ dt=\int_0^\infty e^{(sk)t}\ dt=\left[\frac1{sk}e^{(sk)t}\right]_0^\infty=\frac1{sk}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it follows that \(\mathcal{L}^{1}\left\{\frac1{sk}\right\}=e^{kt}\); for \(k=b/a\) we get \(e^{bt/a}\), so: $$\mathcal{L}\left\{\frac{k}{as+b}\right\}=\frac{k}ae^{bt/a}$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thank you sooo sooo sooo much. I do appreciate.
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