Question

Hi--I am having a problem on laplace transforms. I have a stepfunction graph (attached), but I don't know how translate it into a formula that I can solve for.

Answer 1

Sorry guys for the wait.

Answer 2

I know one eqn: u(t) = u(t-c), but I don't know how to use it properly for this situation.

Answer 3

I'm sorry. Don't think I know. :(

Answer 4

Thanks okay. Thanks for looking at my question though :)

Answer 5

f only goes to 2 between x=2 and x=5, and for all other values the function is zero
to turn the "switch" on at 2 we have\[2u_2(t)\]to turn it off we have\[-2u_5(t)\]so\[2u_2(t)-2u_5(t);0\le t\le8\]seems to be your graph
you want the laplace of that...

Answer 6

When you say "switch" what do you mean? Between open and closed?

Answer 7

I mean that step functions act like switches:
when x reaches a certain value, addig a step function can turn a function on\[u_c(t)f(t)\]has the effect of "turning on" the function f(t) at x=c, because that is when the value of \(u(t)\) changes from 0 to 1

Answer 8

when t reaches a certain....*

Answer 9

oh, okay. I think that you have the u_5 (t) as another equation? I know that it represents something, but I don't know how to calculated it so that it can be evaluated from 0 to 8

Answer 10

right now I have 2u_2/s^2 and 2u_5/s^2, which I was going to evaluate from 0 to 8, but that doesn't make sense

Answer 11

we need to turn the "switch" off at t=5, right?
so we can do that by "turning on" the inverse operation (subtraction here) at t=5 with another step function\[u_3(t)f(x)-u_5(t)f(x)\]think about what happens here at t=3:
f(x) turns on
think about what happens then at t=5:
-f(x) turns on, which effectively cancels f(x)
i.e. it turns f(x) off

Answer 12

Where does the t = 3 come from?

Answer 13

so we can even write this as\[f(x)[u_3(t)-u_5(t)]\]inside the brackets we have 1 between t=3 and t=5
for all other values we have zero
hence, f(x) works only on that range

now about ending at t=8....

Answer 14

I got disconnected! I still don't know where t comes from, I am sorry

Answer 15

I have t =2...

Answer 16

Sorry, Open study froze for 5 mins for some reason on my end

Answer 17

mine too, they're having problems

Answer 18

I am doing this now---in my integral, is the lower bound 2 b/c u_2_(t)?

Answer 19

what integral?

Answer 20

I though you were doing laplace?

Answer 21

Your welcome! Good to know its not just my computer! :)

Answer 22

\[ \int\limits_{2}^{infinity}\]

Answer 23

Right---I am doing laplace, but I have to the integral of e^(-st)*u_2_(t) dt right?

Answer 24

but that is the hard way
do you \(have\) to do it that way?

Answer 25

and then subtract that from whatever the one were u_5_(t) dt?

Answer 26

...using the definition of the Laplace transform?

Answer 27

I don't have too, but I don't know how to do this at all, so I was trying to follow my book but its not clear

Answer 28

I don't know the derivation for the laplace of a step function
but I do have a formula!

Answer 29

I think the formula you are talking about is e^(-as)/s ?

Answer 30

so it would be e^(-2s)/s ? something like this?

Answer 31

yes, but with the 2 still there

Answer 32

oh, so the 2 stays out front--and it's two because that's when the switch happens on the u(t) axis or something?

Answer 33

no that's a coincidence
the function in this case f(x)=2 in the formula I gave above

Answer 34

\[(2*e^(-2s))/s - (2*e^(-5s))/s\]

Answer 35

\[\frac2s(e^{-2x}-e^{-5s})\]yeah

Answer 36

...is that x supposed to be a s? :)

Answer 37

yep

Answer 38

Okay. And I understand your explanation about the 2 now, looking back at this graph

Answer 39

and now I just take the laplace of that. Thanks for helping me!

Answer 40

wait... I can't because it is in s, not t....

Answer 41

so that eqn IS the laplace transform?

Answer 42

yes :)

Answer 43

Thanks! You have been a big help. I really need to study this chapter for Laplace. Oh, and the website is reloading soon, so hopefully we won't have any more freezes!

Answer 44

here's a ref:
http://tutorial.math.lamar.edu/Classes/DE/LaplaceIntro.aspx
good luck !

Answer 45

Thanks!