Laplace Transformations Step Functions

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 2K views
Mikesgto
Messages
18
Reaction score
0

Homework Statement



The attachment is the problem.

Homework Equations


The Attempt at a Solution



I understand how to go about solving the laplace transformations but I have no idea how to start with the Heaviside functions for the 5t and the 30. What I got was 5t+30U6(t) but it turned out to be wrong. Any help would be greatly appreciated.
 

Attachments

  • problem.png
    problem.png
    1.6 KB · Views: 472
Physics news on Phys.org
5t-5U6(t)(t-6)

I got this one after a ton of work, but I am failing to grasp the concept going behind this.
 
You can look at it this way. You only want the 5t term to contribute between t=0 and t=6, so after t=6, you have to subtract out 5t to cancel the other 5t term:

F(t) = 5t U0(t) + 30 U6(t) - 5t U6(t)

(Often, it's understood that the input is 0 for t<0, so you don't write the U0(t) in explicitly.) When t>6, the first and last terms cancel, and you're just left with F(t)=30.

You can also look at it like this, which is the way I prefer. If you rearrange the terms slightly, you can write

F(t) = 5t (U0(t) - U6(t)) + 30 U6(t)

The factor U0(t)-U6(t) is equal to 1 between t=0 and t=6 and is equal to 0 everywhere else, so it turns the 5t term on between t=0 and t=6 and disables it for the other times.
 
ok, so the next problem i have is attached and the solution i came up with is
3t-3tU9(t)

Am I on the right path there?
 

Attachments

  • problem11.png
    problem11.png
    1.4 KB · Views: 417
Thanks so much for all the help. The next part of my solutions are finding out F(s). Do I treat the -3t attached to the U9(t) as my f(t)?