Need Laplace transform help

  • Thread starter RafiS
  • Start date
1
0

Main Question or Discussion Point

iv got a problem i cant seem to understand. if anyone could help me out it would be great

f(t)= sin2tcos2t

im just not sure what to do when i have the product of 2 trig functions.

the correct answer is (2/(s^2 + 16))

thanks for any help
 
Last edited:

Answers and Replies

exk
119
0
Is there something you can do with f(t) to make it look different?
 
1,631
4
yeah, like exk suggested, try to use this trig identity somehow:

[tex] sin2x=2isnxcosx[/tex] can you figure it out how to transform your f(t) into a similar form?
 
Vid
401
0
Use f(x) = 1/4sin^2(2t)
f'(x) = sin(2t)cos(2t)
 
1,631
4
Use f(x) = 1/4sin^2(2t)
f'(x) = sin(2t)cos(2t)
I don't see how would this help!
If i have gotten the op right, he just needst to take the laplace transform of

[tex]f(t)=sin2tcos2t=\frac{1}{2}sin4t[/tex]
so

[tex]L{f(t)}=L{\frac{1}{2}sin4t}=\int_{0}^{\infty}\frac{1}{2}sin(4t)e^{-st}dt[/tex]
 
Vid
401
0
I was thinking that sin^2(t) was a common laplace transform, but I was mistaken.
 
exk
119
0
another strategy for this would be to look up in a transform table what your answer corresponds to and work backwards.

sutupidmath's solution is correct.
 

Related Threads for: Need Laplace transform help

  • Last Post
Replies
2
Views
2K
Replies
0
Views
1K
Replies
1
Views
1K
Replies
0
Views
862
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
5K
  • Last Post
Replies
1
Views
1K
Top