Inverse Laplace with given limits

In summary: GFDQDSKLIn summary, the function f(t) at t=0 and t=∞ is given by F(s)=(2s+3)/(s(s2+7s+10)), where s is the slope of the graph of f(t) at those points.
  • #1
caliboy
15
0

Homework Statement


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=∞?

Homework Equations


Inverse laplace transform

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 +0.1666e2t-0.465e5t

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?
 
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  • #2
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?
 
  • #3
caliboy said:
I solved F(t) down to F(t)= [itex].3/s[/itex]+[itex]0.166/s+2[/itex]-[itex]0.465/s+5[/itex]
That should be F(s). Use parentheses. What you wrote means
$$F(s) = \frac{0.3}{s}+\frac{0.166}{s} +2+\frac{0.465}{s}+5$$
 
  • #4
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??
 
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  • #5
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.
 
  • #6
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
 
  • #7
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.)
 
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  • #8
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.
 
  • #9
caliboy said:

Homework Statement


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=∞?


Homework Equations


Inverse laplace transform


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 +0.1666e2t-0.465e5t

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?

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
 

Related to Inverse Laplace with given limits

What is the inverse Laplace transform?

The inverse Laplace transform is a mathematical operation that takes a function in the frequency domain and converts it into a function in the time domain. It is the reverse of the Laplace transform, which converts a function in the time domain into the frequency domain.

How is the inverse Laplace transform calculated?

The inverse Laplace transform can be calculated using various techniques such as partial fraction decomposition, the method of residues, or by using a table of Laplace transforms. The specific method used depends on the complexity of the function being transformed.

What are the limits of the inverse Laplace transform?

The inverse Laplace transform is typically defined for functions that have a Laplace transform in the complex plane. This means that the function must be analytic in a region known as the region of convergence. If the function does not meet this condition, the inverse Laplace transform may not exist.

What are the applications of inverse Laplace transform?

The inverse Laplace transform has many applications in engineering, physics, and other fields. It is commonly used for solving differential equations, analyzing linear systems, and in signal processing. It can also be used to find the solutions to problems involving complex variables.

How can I check if my inverse Laplace transform is correct?

You can check your inverse Laplace transform by using properties and theorems of Laplace transforms, such as linearity, differentiation, and convolution. You can also use mathematical software programs or tables to verify the correctness of your result.

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