Question

Use the result of the previous problem to find the Laplace Transform of the given function:

Answer 1

\[\large f(t)=\left\{\begin{matrix} 1, & 0 \le t <1\\ 0, & 1 \le t <2 \end{matrix}\right.\]\[\large f(t+2)=f(t)\]


`PREVIOUS PROBLEM:`
Show that for f(t+T)=f(t):\[\large \mathcal{L}[f(t)]=\frac{\int\limits_{0}^{T} e^{-st}f(t)dt}{1-e^{-sT}}\]

Answer 2

@dan815 @amistre64 @Directrix

Answer 3

my bandwidth is so bad that i cannot see this properly, but i think you need the transform of the starting function before you go to the second bit.

guy to ask is @ganeshie8

Answer 4

Hey @IrishBoy123 !
Let me get a screenshot!

And yeah but he's not online at the moment! =/

Answer 5

Do you see why \(T=2\)?

Answer 6

With that in mind,
\[\mathcal{L}\{f(t)\}=\frac{\displaystyle\int_0^2e^{-st}f(t)\,\mathrm{d}t}{1-e^{-2s}}=\frac{\displaystyle\int_0^1e^{-st}\,\mathrm{d}t}{1-e^{-2s}}=\cdots\]

Answer 7

@SithsAndGiggles I'm not sure I understand what you did there =/

Answer 8

Oh I think I see. How did you get rid of f(t) though?

Answer 9

Just taking advantage of the definition of \(f(t)\). Over the interval \((0,1)\), you have \(f(t)=1\), while over \((1,2)\), \(f(t)=0\). You can thus split up the integral like so:
\[\int_0^2e^{-st}f(t)\,\mathrm{d}t=\int_0^1e^{-st}\times1\,\mathrm{d}t+\int_1^2e^{-st}\times0\,\mathrm{d}t=\int_0^1e^{-st}\,\mathrm{d}t\]

Answer 10

Ahhh yeah that makes sense. Thanks!!