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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.

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  1. Loser66
    • one year ago
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    @oldrin.bataku

  2. anonymous
    • one year ago
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    so it's a sawtooth wave varying between \(a\) and \(2a\) with a period of \(a\)?

  3. Loser66
    • one year ago
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    I think period is a, also

  4. anonymous
    • one year ago
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    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 \(f(x)=x+a\)

  5. Loser66
    • one year ago
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    I have to include the nodes on it. I am sorry |dw:1436467689944:dw|

  6. anonymous
    • one year ago
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    now, let's try to find way to make the function 'reset' with a period of \(a\)

  7. Loser66
    • one year ago
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    How about a piece wise function?

  8. anonymous
    • one year ago
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    consider the function \(\lfloor x\rfloor\), which looks like a stepped ladder: |dw:1436467877418:dw|

  9. anonymous
    • one year ago
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    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|

  10. anonymous
    • one year ago
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    |dw:1436467977849:dw|

  11. anonymous
    • one year ago
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    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: $$f(x)=a+a(x-\lfloor x\rfloor )$$

  12. anonymous
    • one year ago
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    and lastly, to change its period from \(1\) to \(a\), we'll substitute \(x\to x/a\), giving: $$f(x)=a+a(x/a-\lfloor x/a\rfloor)=a+x-a\lfloor x/a\rfloor $$

  13. anonymous
    • one year ago
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    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

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

  15. anonymous
    • one year ago
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    so here we have \(T=a\) and if we restrict \(f\) to a single period \([0,T)\) we have \(f(t)=a+t\), so $$\mathcal{L}\{f\}=\frac1{1-e^{-as}}\int_0^a e^{-st}(a+t)\ dt$$

  16. Loser66
    • one year ago
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    I don't get how you go from the second line to the third line, the second part

  17. anonymous
    • one year ago
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    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 \(\int_T^\infty \dots\ dt\to\int_0^\infty \dots\ du\)

  18. Loser66
    • one year ago
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    nevermind, I got it

  19. Loser66
    • one year ago
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    Hold on, you got f(t) = a+t from where?

  20. anonymous
    • one year ago
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    now we can integrate by parts: $$\begin{align*}\int_0^a e^{-st}(a+t)\ dt&=-\frac1s e^{-st}(a+t)\bigg|_{t=0}^a+\frac1s\int_0^ae^{-st}\ dt\\&=\left[-\frac1se^{-st} (a+t)-\frac1{s^2}e^{-st}\right]_{t=0}^a\\&=-\frac{2a}s e^{-as}-\frac1{s^2}e^{-as}+\frac{a}s+\frac1{s^2}\end{align*}$$

  21. anonymous
    • one year ago
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    \(f(t)=a+t \) is the function restricted to its first period; look at my first reply, it's a line passing through \((0,a)\) and \((a,2a)\)

  22. Loser66
    • one year ago
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    got it

  23. anonymous
    • one year ago
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    so our Laplace transform is: $$\mathcal{L}\{f\}=\frac1{1-e^{-as}}\cdot\frac{as+1-e^{-as}(2as+1)}{s^2}$$

  24. Loser66
    • one year ago
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    Woah!!! part 3, please

  25. anonymous
    • one year ago
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    now, part 3 concerns an inverse Laplace transform of \(\dfrac{k}{as+b}\):$$\mathcal{L}^{-1}\left\{\frac{k}{as+b}\right\}=\frac{k}a\mathcal{L}^{-1}\left\{\frac1{s+b/a}\right\}$$

  26. anonymous
    • one year ago
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    now, consider the Laplace transform of \(e^{kt}\) $$\mathcal{L}\{e^{kt}\}=\int_0^\infty e^{-st} e^{kt}\ dt=\int_0^\infty e^{-(s-k)t}\ dt=\left[-\frac1{s-k}e^{-(s-k)t}\right]_0^\infty=\frac1{s-k}$$

  27. anonymous
    • one year ago
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    so it follows that \(\mathcal{L}^{-1}\left\{\frac1{s-k}\right\}=e^{kt}\); for \(k=-b/a\) we get \(e^{-bt/a}\), so: $$\mathcal{L}\left\{\frac{k}{as+b}\right\}=\frac{k}ae^{-bt/a}$$

  28. Loser66
    • one year ago
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    Thank you sooo sooo sooo much. I do appreciate.

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