Laplace transform of piecewise function - making it to become heaviside unitstep function

Question

The piecewise function of $f(t)$ is as follows-

$f(t) = t , 0\le t < 1 $

$f(t) = 2-t , 1\le t \le 2$

$f(t) = 0 , t>2$

I made a quick sketch -

now, I need help on making this into a formula f(t) = something.

I am not too sure on this shape of the graph. The function is ‘ON’ from 0 to 2. If I am not wrong, it is called the heaviside unitstep function

I need to get a function of f(t) before I can apply the laplace transform of second shifting to get the answer for Laplace transform of that function.. thanks for the help!!

Answer 1

With Heaviside 's function..

$$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))+0H(t-2)$$ $$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))$$ $$f(t)=t+(2-2t)H(t-1)-(2-t)H(t-2))$$ $$f(t)=t+2(1-t)H(t-1)-(2-t)H(t-2))$$ $$f(t)=t-2(t-1)H(t-1)+(t-2)H(t-2))$$

Then take the Lpalace transform

Note that : $$\mathcal{L}(f(t-c)H(t-c))=e^{-cs}F(s)$$ Therefore we have : $$\boxed {F(s)=\frac 1{s^2}-2\frac {e^{-s}}{s^2} +\frac {e^{-2s}}{s^2}}$$

Answer 2

I think it's easier to calculate the laplace transform from the definition: $$F(s)=\int_{0}^{\infty}f(t)e^{-st}\mathrm{d}t$$ $$F(s)=\int_{0}^{1}te^{-st}\mathrm{d}t+\int_{1}^{2}(2-t)e^{-st}\mathrm{d}t$$

Answer 3

Don't worry about Heaviside, just break it into two integrals and evaluate it.

Botond has it right. $$ F(s)=\int_{0}^{1}te^{-st}\mathrm{d}t+\int_{1}^{2}(2-t)e^{-st}\mathrm{d}t$$

Now you have to do integration by parts which I hope you still remember from your calculus class.

Answer 4

If you want to use the step function then you take a case which applies on $(a,b)$ and multiply it by $H(t-a)-H(t-b)$ (note the order, and use the convention $H(t-\infty)=0$) and then sum over cases. This is useful because it matches the form used for Laplace transform tables.

Answer 5

Note that $$ t(H(0)-H(1)) + (2-t)(H(1)-H(2)) =tH(0)-2(t-1)H(1)+(t-2)H(2)$$

You should be able to finish it.