Question

\[\begin{array}{cc} \text{Find} \; \mathcal L \lbrace \phi (t) \rbrace \; \text{where} \; \phi (t) &= 1 \quad 0 < t < 2\\&=t \quad \quad \; \; t >2 \end{array}\]

Answer 1

uhh o.0 it looks hard
is this like parametrics ? :o

Answer 2

this is laplace transform haha lol

Answer 3

Try the same here..
As you did earlier..

Answer 4

\[\large \implies \int\limits_{0}^{2}(e^{-st})dt + \int\limits _{2}^{\infty} (t \cdot e^{-st}) dt\]

Answer 5

uhh wait...i posted the wrong question...

Answer 6

nevermind...i'll just solve this one

Answer 7

You will be having problem in second integral I guess..

If no then Very Good for you..

Answer 8

i got something like \[\huge \left[-\frac{e^{-st}}{s}\right|_0^2 + \left[-\frac{te^{-st}}{s} - \frac{e^{-st}}{s^2}\right|_2^\infty\]
is that right?

Answer 9

Sorry here it is t so you can solve it easily sorry..

But if it was t^{-1} then you will be having problems in evaluating this..

Answer 10

Let me check.

Answer 11

...i have no idea what you just said...

Answer 12

@waterineyes still here?

Answer 13

i got \[\huge \frac{e^{-2s} + 1}{s} + \frac{e^{-2s}}{s^2}\]
can anyone check if im right?

Answer 14

looks right;

;p
http://www.wolframalpha.com/input/?i=inverse+laplace+%28%28e%5E%28-2s%29%2B1%29%2Fs%2Be%5E%28-2s%29%2Fs%5E2%29

Answer 15

so wolfram can solve laplace o.O >.<

Answer 16

wolfram can do al lot of things
http://www.wolframalpha.com/input/?i=blue+%2B+red

Answer 17

but how did you know my answer is right?

Answer 18

i compared your piecewise function to the graph

Answer 19

oh. well thanks

Answer 20

My internet sucked that time...

So I was unable to connect...

Answer 21

maybe it was just OS...we never know