# Homework Help: Laplace Transform problems

1. Jul 24, 2008

### alex73

Ok i have two questions, one i am unsure of and one i don't have a clue how to correctly find it.

1. The problem statement, all variables and given/known data
You have to work out the inverse function of:

4s/ s^2+4s+8

3. The attempt at a solution
I think the answer is:
4e^(-2t)*(cos2t+sin2t)
Is this correct?

1. The problem statement, all variables and given/known data
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)

3. 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

2. Jul 24, 2008

### Defennder

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.

You got that right. Just need to figure out what is f(0) here.

3. Jul 25, 2008

### alex73

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.

Last edited: Jul 25, 2008
4. Jul 25, 2008

### Defennder

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" [Broken] 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.

Yep that's right.

Last edited by a moderator: May 3, 2017
5. Jul 25, 2008

### alex73

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. Jul 25, 2008

### alex73

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. Jul 25, 2008

### Defennder

Yep I got that as well.
You're welcome. That's PF policy by the way.

8. May 24, 2011

### Mirnal Dev

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

Last edited: May 24, 2011
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