Thread: Laplace Transforms

Given the following:
f(t) = 1 for 0 <= t <= 1
f(t) = -1 for 1 < t <= 2
f(t) = 0 elsewhere

Find the Laplace Transform

Solution thus far:
The problem needs to be broken into two separate functions, one each for the time when f(t) = 1 and f(t) = -1.

For 0 <= t <= 1

L(s) = Integral (0 to 1) of e**-st f(t) dt

L(s) = (1-e**-s)/s

For 1 < t <= 2

L(s) = Integral (1 to 2) of e**-st f(t) dt

L(s) = (e**-2s - e**-s)/s

Right?

The solution in the back of the book (Advanced Engineering Mathematics, Kreyszig) says [(1 - e**-s) **2]/s.

I need help understanding how to go from my answer (or correcting my answer) to the final answer. I'm missing something. Thanks in advance.

Originally Posted by dsprice

Given the following:
f(t) = 1 for 0 <= t <= 1
f(t) = -1 for 1 < t <= 2
f(t) = 0 elsewhere

Find the Laplace Transform

Solution thus far:
The problem needs to be broken into two separate functions, one each for the time when f(t) = 1 and f(t) = -1.

For 0 <= t <= 1

L(s) = Integral (0 to 1) of e**-st f(t) dt

L(s) = (1-e**-s)/s

For 1 < t <= 2

L(s) = Integral (1 to 2) of e**-st f(t) dt

L(s) = (e**-2s - e**-s)/s

Right?

The solution in the back of the book (Advanced Engineering Mathematics, Kreyszig) says [(1 - e**-s) **2]/s.

I need help understanding how to go from my answer (or correcting my answer) to the final answer. I'm missing something. Thanks in advance.