Finding Inverse Functions Using Laplace Transform

[alex73]

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

 

[Defennder]

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.

 

[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.

 

[Defennder]

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.

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.

 

[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)

 

[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.

 

[Defennder]

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.

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.

 

[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