Laplace question help?

DJW11

Hi,

Can anyone do this?

Find the inverse laplace of (tan^-1(s+2))

I get -sint/t x e^-2t, not sure of im right.

Much appreciated,

Michael Collins

I'd say you're looking for the inverse Laplace transform of

[latex] \displaystyle F(s)= \tan^{-1}\left(\frac{1}{s+2}\right)[/latex]

yeh? (Note the 1/(s+2) instead of s+2.) In which case you're right if you leave out the minus sign:

[latex] \displaystyle f(t)= \frac{\sin(t)}{t}e^{-2t}[/latex]

bassey

Wondering if anyone can help me, I'm trying to find the inverse laplace of

9 / s^2 + 4s + 32

I've gotten it down to

3^2 / (s + 2) + 28

And can't figure out what to do now, I think I may have gone wrong at some point as well, I've attached what I've got done so far

Michael Collins

bassey wrote: »

Wondering if anyone can help me, I'm trying to find the inverse laplace of

9 / s^2 + 4s + 32

I've gotten it down to

3^2 / (s + 2) + 28

And can't figure out what to do now, I think I may have gone wrong at some point as well, I've attached what I've got done so far

There's two ways to tackle a problem like this: using complex numbers, or completion of the square. You seem to have tried both in the attachment -- I think the completion of the square method is easier, so I'll outline that here.

You have a function F(s) can perhaps be manipulated to look like the Laplace transform of the sin(wt) function:

[latex] \displaystyle \sin(\omega t) \to \frac{\omega}{s^2+\omega^2} [/latex]

Completing the square of the denominator

[latex] \displaystyle \frac{9}{s^2+2s+32} = \frac{9}{(s+2)^2+28}[/latex]

don't forget to square the (s+2) bit -- otherwise it doesn't make sense!

It's looking a bit more like the sin(wt) transform, now but we need an [latex] \omega [/latex] on the top and an [latex] \omega^2 [/latex] on the bottom:

[latex] \displaystyle \frac{9}{(s+2)^2+28}=\frac{9}{(s+2)^2+(\sqrt{28})^2} = \frac{9}{\sqrt{28}}\frac{\sqrt{28}}{(s+2)^2+(\sqrt{28})^2} [/latex]

where I divided the numerator by 9 and multiplied it by [latex]\sqrt{28} [/latex] to get it like the sin(wt) numberator, and then since I did this I had to do the opposite in front of the fraction i.e. multiply by 9 and divide by [latex]\sqrt{28} [/latex].

Now you try the last bit yourself, it involves using the shifting property of the Laplace transform, namely

[latex] \displaystyle e^{-at}f(t) \to F(s+a) [/latex],

this means, you can treat that (s+2) above as simply 's' (in which case it matches the sin transform above exactly) provided you multiply the result by [latex] e^{-2t} [/latex]

bassey

Thanks for that, I ended up with

f(t) = 9/√28 (e^(-2t).sin√28.t)

I copied that in from MS word but don't know how to make it come out as an equation

Michael Collins

bassey wrote: »

Thanks for that, I ended up with

f(t) = 9/√28 (e^(-2t).sin√28.t)

I copied that in from MS word but don't know how to make it come out as an equation

Perfect!

One small point: the Laplace transform usually assumes the time domain signal is zero before t=0, so you should either multiply all the above by the unit step u(t), or just put "for t>0" beside it, and "0 otherwise" below.

The unit step u(t)=0 for t<0, and u(t)=1 for t>0, hence it has the same effect.