Express function given in the t-domain (deals with laplace)?

Question

Having trouble with a homework question.....The chapter is about laplace transforms. But i'm given a picture....i wrote it out below...and have to express it in the t-domain (also in the s....but am only worrying if i got the t-domain right for now)

Left column is going to be f(t)...right is going to the the range of t's.
f(t) t
1 1 t -1 2 1 3 6-t 5 0 6 < t

This is the part i'm confused if i did correctly...written as u(t)

F(t) = u(t-1)- (t-1) u(t-2) - u(t-3) - (6-t) u(t-5) - u(t-6)


if anyone could verify that....this is the example problem that was given...expected to learn this from the book.

0 t<0
1/h 0 0 t>h
..the answer is then
f(t) = 1/h [u(t) - u(t-h)]

thanks in advance.

the column didn't display the way i wanted it to

Left column is going to be f(t)...right is going to the the range of t's.
f(t)................ t
1 ..................1 t -1 ...............2 1 ...................3 6-t .................5 0 ...................6 < t


for the example problem

0..............t<0
1/h ........0 0 ...........t>h

Answer 1

Best idea is just to break it down into regions.

For t < 1, it's 0, so that's good.

For 1 < t < 2, it's the unit function, so that's good too.

For 2 < t < 3, it would be 1 - (t - 1) = 1 - t + 1 = 2 - t...but you were looking for t - 1, so that's a problem.

What you want to do instead is to say, "OK, I've got +1, and I want to end up with t - 1, so what do I need to do?" The answer is, "Subtract 2 and add t."

So far it should be:

f(t) = u(t - 1) + (t - 2) × u(t - 2)

Then from 3 < t < 5, you've got t - 1, and you want to have just 1, so you'd subtract t and add 2:

f(t) = u(t - 1) + (t - 2) × u(t - 2) + (2 - t) × u(t - 3)

Then from 5 < t < 6, you've got 1, and you want 6 - t, so you add 5 and subtract t:

f(t) = u(t - 1) + (t - 2) × u(t - 2) + (2 - t) × u(t - 3) + (5 - t) × u(t - 5)

And then, finally, for t > 6, you've got 6 - t and you want 0, so you subtract 6 and add t:

f(t) = u(t - 1) + (t - 2) × u(t - 2) + (2 - t) × u(t - 3) + (5 - t) × u(t - 5) + (t - 6) × u(t - 6)

Make sense?

Double check for t > 6, substitute 1 for all u(t) functions:

f(t) = 1 + (t - 2) + (2 - t) + (5 - t) + (t - 6) = (t - t - t + t) + (1 - 2 + 2 + 5 - 6) = 0, check.