## 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.

1. Loser66

@oldrin.bataku

2. anonymous

so it's a sawtooth wave varying between $$a$$ and $$2a$$ with a period of $$a$$?

3. Loser66

I think period is a, also

4. 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 $$f(x)=x+a$$

5. Loser66

I have to include the nodes on it. I am sorry |dw:1436467689944:dw|

6. anonymous

now, let's try to find way to make the function 'reset' with a period of $$a$$

7. Loser66

How about a piece wise function?

8. anonymous

consider the function $$\lfloor x\rfloor$$, which looks like a stepped ladder: |dw:1436467877418:dw|

9. 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|

10. anonymous

|dw:1436467977849:dw|

11. 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: $$f(x)=a+a(x-\lfloor x\rfloor )$$

12. anonymous

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

13. 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

14. 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*} 15. 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$$16. Loser66 I don't get how you go from the second line to the third line, the second part 17. anonymous i replaced the variable of integration $$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$$ 18. Loser66 nevermind, I got it 19. Loser66 Hold on, you got f(t) = a+t from where? 20. anonymous now we can integrate by parts:$$\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*}$$21. anonymous $$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)$$ 22. Loser66 got it 23. anonymous so our Laplace transform is:$$\mathcal{L}\{f\}=\frac1{1-e^{-as}}\cdot\frac{as+1-e^{-as}(2as+1)}{s^2}$$24. Loser66 Woah!!! part 3, please 25. anonymous now, part 3 concerns an inverse Laplace transform of $$\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\}$$26. anonymous now, consider the Laplace transform of $$e^{kt}$$$$\mathcal{L}\{e^{kt}\}=\int_0^\infty e^{-st} e^{kt}\ dt=\int_0^\infty e^{-(s-k)t}\ dt=\left[-\frac1{s-k}e^{-(s-k)t}\right]_0^\infty=\frac1{s-k}$$27. 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}

28. Loser66

Thank you sooo sooo sooo much. I do appreciate.