• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Laplace transforms, heaviside step functions

  • Thread starter missavvy
  • Start date
82
0
Hey guys! I just have a question regarding some notes I've been looking at and trying to understand...

1. Homework Statement
eq0018MP.gif

Taking the transform:

eq0019MP.gif


Just wondering, can someone explain to me why the e-3s/s is positive, and not negative because it is originally -u3(t)...

Another quick one:

Write this in terms of Heaviside function:

f(t) =
{ 0, 0 <= t <= pi
{ -sin3t, t >= pi

1) So you would get f(t) = -(upit)sin3t

And somehow it ends up as f(t) = (upit)sin(3(t-pi))
Can someone explain this step as well? Just what happened to the negative in front of it, and why it became 3(t-pi)

Thanks!
 

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,414
1,102
Hey guys! I just have a question regarding some notes I've been looking at and trying to understand...

1. Homework Statement
eq0018MP.gif

Taking the transform:

eq0019MP.gif


Just wondering, can someone explain to me why the e-3s/s is positive, and not negative because it is originally -u3(t)...
It's probably just a typo in the original problem.

Another quick one:

Write this in terms of Heaviside function:

f(t) =
{ 0, 0 <= t <= pi
{ -sin3t, t >= pi

1) So you would get f(t) = -(upit)sin3t

And somehow it ends up as f(t) = (upit)sin(3(t-pi))
Can someone explain this step as well? Just what happened to the negative in front of it, and why it became 3(t-pi)
Try expanding sin[3(t-pi)] using the angle addition formula for sine. The reason it was written this way is probably because it's now in the form where the time-shift property of the Laplace transform can be applied.
 

Attachments

82
0
Thanks for the first one that makes sense.

For the second one, I attempted using the identity and get:

f(t) = -upi(t)sin(3t)
= -L{upi(t)sin3(t-pi+pi)}
= -L{upi(t)[sin3(t-pi)cos(3pi) + sin(3pi)cos3(t-pi)]}
taking the transform:
= -e-pi*s(3cos(3pi) + 3s*sin(3pi) / s2 + 9

Hmm I don't understand how it is suppose to be just upi(t)sin(3(t-pi)) = e-pi*s(3) / (s2 + 9)(s2 + 1)

What happened when you add the pi and subtract it, and what happened to the negative in front of it all?
 

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,414
1,102
What are cos(3π) and sin(3π) equal to? Also, I didn't mean that you needed to use the time-shift property to get that result. It's just basic trig.
 

Related Threads for: Laplace transforms, heaviside step functions

Replies
4
Views
5K
Replies
1
Views
3K
Replies
5
Views
3K
Replies
3
Views
2K
Replies
3
Views
2K
Replies
11
Views
712
Replies
7
Views
1K
Replies
7
Views
2K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top