Inverse Laplace with given limits

caliboy

1. The problem statement, all variables and given/known data
Problem reads: find inverse Laplace transform of f(t) of F(s)=(2s+3)/(s(s²+7s+10) What is the value of the function f(t) at t=0 and t=∞?


2. Relevant equations
Inverse laplace transform


3. The attempt at a solution
I solved F(t) down to F(t)= [itex].3/s[/itex]+[itex]0.166/s+2[/itex]-[itex]0.465/s+5[/itex]

thus lead me to the inverse of f(t)=0.3 +0.1666e^2t-0.465e^5t

Correct me on the math if i am wrong or whether i am on the wrong track up to this point. So, I’m guessing that I put 1 in for t which would be f(1)=-67.486 and for f(∞) I’m getting it as "undef" can someone shed light as to whether I’m on the right track?

algebrat

I think you should expect decaying exponentials, so check the signs on those puppies. Also, you mentioned both t=0 and t=1, did you do what you meant to do?

vela

That should be F(s). Use parentheses. What you wrote means
$$F(s) = \frac{0.3}{s}+\frac{0.166}{s} +2+\frac{0.465}{s}+5$$

caliboy

I apologize I have been using the "latex reference" but am still getting use to it. WhatI meant to say was that F(s)=[itex]0.3/\left(s\right)[/itex]+[itex]0.166/\left(s+2\right)[/itex] -[itex]0.465/\left(s+5\right)[/itex]. Now looking at this am I correct in assuming that when the problem asks for f(t) at t=0 and t=∞. So, @t=0 f(0)=-66.1498 and f(∞)=undefined??

vela

Did you check your work like algebrat suggested? You should get a finite answer for t=∞. Your answer for f(0) doesn't look correct either.

caliboy

Yeah I have been working Laplace transforms for a while now and am still making stupid mistakes. I think I caught it: f(t)=0.3+0.166e^-2t-0.465e^-5t; thus f(0)=0.01 and f(∞)=0.3

vela

It might have been intended for you to use the Laplace transform limit theorems. They say that
\begin{align*}
\lim_{t \to 0^+} f(t) &= \lim_{s \to \infty} sF(s) \\
\lim_{t \to \infty} f(t) &= \lim_{s \to 0} sF(s)
\end{align*} Try those and see if you get the same answers.

(Or maybe not, now that I've reread the original post.)

caliboy

Im would not think I would have to use the limit theorem, mainly because it has not been in the course work I have been studying. I will look into it though and thank you very much for your help.

Ray Vickson

You should get out of the bad habit of extreme roundoff, which you have performed above. Your F(s) is NOT
[tex] \frac{0.3}{s} +\frac{0.166}{s+2} - \frac{0.465}{s+5}.[/tex] It is
[tex] F(s) = \frac{3}{10 s} + \frac{1}{6(s+2)} - \frac{7}{15(s+5)}
\doteq \frac{0.3}{s} + \frac{0.1666667}{s+2} - \frac{0.4666667}{s+5}.[/tex] In fact, there is no reason at all to convert to decimal numbers; using rationals is just as easy.

RGV