Inverse Laplace : with Convolution

  • #1
212
2

Homework Statement


Use the convolution property to obtain the inverse Laplace transform of
F(s)= e-3s * ((3s+15)/s2+25)

Homework Equations




The Attempt at a Solution


= (3*(s/s2+52) + 15*(1/s2+52)) *e-3s

Using table of Laplace:
3*(s/s2+52) = 3*cos(5*t) = T7
15*(1/s2+52) = 15/5*sin(5*t) =T18
e-3s = s(t-a) = s(t-3) (T16)

=(3*cos(5*t)+15/5*sin(5*t))*s(t-3)

Convolution : int 0 to t (f(u)*g(t-u))du
integral 0 to t ((3*cos(5*u)+15/5*sin(5*u))*s((t-u)-3)) du

Something is wrong in my work here, and I cant find where. :([/SUP]
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770

Homework Statement


Use the convolution property to obtain the inverse Laplace transform of
F(s)= e-3s * ((3s+15)/s2+25)

Homework Equations




The Attempt at a Solution


= (3*(s/s2+52) + 15*(1/s2+52)) *e-3s

Using table of Laplace:
3*(s/s2+52) = 3*cos(5*t) = T7
15*(1/s2+52) = 15/5*sin(5*t) =T18
e-3s = s(t-a) = s(t-3) (T16)

None of the above equalities are true; you have functions of s "equal" to functions of t.

=(3*cos(5*t)+15/5*sin(5*t))*s(t-3)

Convolution : int 0 to t (f(u)*g(t-u))du
integral 0 to t ((3*cos(5*u)+15/5*sin(5*u))*s((t-u)-3)) du

Something is wrong in my work here, and I cant find where. :([/SUP]
And you are apparently using s for two different things. With 156 posts, it is past time for you to learn to use latex. It isn't worth the effort to decipher what you have written.
 
  • #3
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,169
1,766

Homework Statement


Use the convolution property to obtain the inverse Laplace transform of
F(s)= e-3s * ((3s+15)/s2+25)
Is the asterisk supposed to denote multiplication or convolution?

You need to use parentheses as appropriate. Assuming the asterisk is supposed to be multiplication, what you wrote means
$$F(s) = e^{-3s}\left(\frac 3s + \frac{15}{s^2} + 25\right),$$ which is quite different from what you intended.

Using table of Laplace:
3*(s/s2+52) = 3*cos(5*t) = T7
15*(1/s2+52) = 15/5*sin(5*t) =T18
e-3s = s(t-a) = s(t-3) (T16)

=(3*cos(5*t)+15/5*sin(5*t))*s(t-3)
Is there some reason you're not simplifying 15/5?

Convolution : int 0 to t (f(u)*g(t-u))du
integral 0 to t ((3*cos(5*u)+15/5*sin(5*u))*s((t-u)-3)) du

Something is wrong in my work here, and I can't find where. :([/SUP]
Why do you think something's wrong?

Besides the problems LCKurtz pointed out with what you've written, did you notice that when you wrote "=(3*cos(5*t)+15/5*sin(5*t))*s(t-3)," you never said what it was equal to and that you're using * to mean both multiplication and convolution? You should make an effort to be less sloppy, especially when you're trying to communicate with others.
 

Related Threads on Inverse Laplace : with Convolution

Replies
1
Views
896
  • Last Post
Replies
4
Views
657
  • Last Post
Replies
10
Views
282
  • Last Post
Replies
2
Views
2K
Replies
2
Views
2K
Replies
12
Views
3K
Replies
6
Views
790
  • Last Post
Replies
2
Views
798
  • Last Post
Replies
0
Views
4K
Replies
1
Views
2K
Top