Finding Inverse Functions Using Laplace Transform

In summary, the conversation is about finding the inverse function of a given expression and solving a problem involving the Laplace transform. The correct answer for the inverse function is 4e^(-2t) (cos(2t)+sin(2t)). The conversation also discusses the value of f(0) and how it affects the solution.
  • #1
alex73
4
0
Ok i have two questions, one i am unsure of and one i don't have a clue how to correctly find it.

Homework Statement


You have to work out the inverse function of:

4s/ s^2+4s+8


The Attempt at a Solution


I think the answer is:
4e^(-2t)*(cos2t+sin2t)
Is this correct?


Homework Statement


I have tryed to do this and keep getting it wrong, so could someone please show working of how to do this.
If f(t)=cos2t.u(t) then find the laplace transform for:
3(df/dt)-f(t)

The Attempt at a Solution


L{cos2t}=s/(s^2+4)
L{3(df/dt)-f(t)}=3(sF(s)-f(0))-F(s)
I think f(0)=1 but not sure
After that i get confused

Thanks
Alex
 
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  • #2
alex73 said:

Homework Statement


You have to work out the inverse function of:
4s/ s^2+4s+8

The Attempt at a Solution


I think the answer is:
4e^(-2t)*(cos2t+sin2t)
Is this correct?
No it's not. Try finding the Laplace transform of that answer. Can you get back the original expression?

One way of approaching this type of problem is first by trying to factorise the denominator and then separate the partial fractions. But in this case it doesn't work because the denominator can't be broken down further. So what you need to do is to try rewrite it in a way which allows you to use the Laplace transform of sin wt and cos wt. Hint: you nearly got it right.

Homework Statement


I have tryed to do this and keep getting it wrong, so could someone please show working of how to do this.
If f(t)=cos2t.u(t) then find the laplace transform for:
3(df/dt)-f(t)

The Attempt at a Solution


L{cos2t}=s/(s^2+4)
L{3(df/dt)-f(t)}=3(sF(s)-f(0))-F(s)
I think f(0)=1 but not sure
After that i get confused
You got that right. Just need to figure out what is f(0) here.
 
  • #3
thanks,
How do you work out what f(0) is?
i thought it was f(t) at 0. And 0 on the cos2x graph is 1, isn't it?
Or am i incorrect on how to work it out.

On the first question i get to here:
4(s/((s+2)^2 +4).
ok i just worked backwards and i think the answer is:
4e^(-2t) * (cos2t-sin2t) is this correct.
With this the sin2t takes away the 2 on top that you have extra.
 
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  • #4
alex73 said:
thanks,
How do you work out what f(0) is?
i thought it was f(t) at 0. And 0 on the cos2x graph is 1, isn't it?
Or am i incorrect on how to work it out.
I think this depends heavily on how you define u(t). My notes define it as u(t) = 1 for t>0 and 0 everywhere else, but http://mathworld.wolfram.com/HeavisideStepFunction.html" defines it as 1/2 when t=0. Wikipedia says it depends on convention:
http://en.wikipedia.org/wiki/Unit_step_function

So the best thing to do is to consult your notes for what is acceptable.

alex73 said:
On the first question i get to here:
4(s/((s+2)^2 +4).
ok i just worked backwards and i think the answer is:
4e^(-2t) * (cos2t-sin2t) is this correct.
With this the sin2t takes away the 2 on top that you have extra.
Yep that's right.
 
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  • #5
Normally they use 1.
But i can't get it correct if i use that.
Unless i am doing something wrong.
here is how i am doing it:
3(-1+sF(s))-F(s)
3-3sF(s)-F(s)
(-3s-1)*F(s)+3
((-3s-1)*s/(s^2 +4)) +3
(-3s^2 - s + 3s^2 +12)/(s^2 +4)
= (-s+12)/(s^2 +4)
 
  • #6
ok, i just noticed where i went wrong.
Using f(0)=1
I think its:
(-s-12)/(s^2 +4)
Is this correct?
Thanks.
I like that you don't just tell me the answer, it helps me improve and makes you feel better when you have done it yourself.
 
  • #7
alex73 said:
ok, i just noticed where i went wrong.
Using f(0)=1
I think its:
(-s-12)/(s^2 +4)
Is this correct?
Thanks.
Yep I got that as well.
alex73 said:
I like that you don't just tell me the answer, it helps me improve and makes you feel better when you have done it yourself.
You're welcome. That's PF policy by the way.
 
  • #8
work out the inverse function of:

4s/ s^2+4s+8

4(s+2)/[(s+2)^2 + 4] + 2/[(s+2)^2 + 4]

it seems to
4e^(-2t) (cos(2t)+sin(2t))... ANS

Mirnal
India
 
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FAQ: Finding Inverse Functions Using Laplace Transform

1. What is a Laplace Transform?

A Laplace Transform is a mathematical operation that transforms a function of time into a function of complex frequency, making it easier to solve differential equations and analyze systems in the frequency domain.

2. What types of problems can be solved using Laplace Transforms?

Laplace Transforms are commonly used to solve initial value problems and boundary value problems involving differential equations in engineering, physics, and other scientific fields.

3. How do I perform a Laplace Transform?

The Laplace Transform can be performed using integral calculus. The function is multiplied by the exponential function e^(-st) and integrated from 0 to infinity, where s is a complex variable.

4. What are some common applications of Laplace Transforms?

Laplace Transforms have many practical applications, including signal processing, control systems, electrical circuits, and heat transfer. They are also used in image and video compression algorithms.

5. Are there any limitations to using Laplace Transforms?

While Laplace Transforms are a powerful tool for solving differential equations, they may not always be applicable or feasible, especially for functions with discontinuities or those that do not have a finite limit as t approaches infinity.

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