1. ## 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.

2. 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.
$\mathcal{L}(f(t))=\int_{0}^{\infty}e^{-st}f(t)dt=$
$\int_{0}^{1}e^{-st}dt-\int_{1}^2 e^{-st}dt$

$-\frac{1}{s}(e^{-s}-1)+\frac{1}{s}(e^{-2s}-e^{-s})=\frac{1}{s}(e^{-2s}-2e^{-2}+1)=\frac{1}{s}(e^{-s}-1)^2$